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# If x, y, z are all integers, is xyz odd?
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If x, y, z are all integers, is xyz odd? [#permalink]
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22 Oct 2017, 22:56
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If x, y, z are all integers, is xyz odd?
(1) x - 1 is even
(2) (y - z)(x - z) is odd
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Joined: 22 May 2017
Posts: 1
Re: If x, y, z are all integers, is xyz odd? [#permalink]
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22 Oct 2017, 23:51
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1. X-1 is even, i.e. x is odd, this is not sufficient as product of three numbers Can be odd only for odd*odd*odd
2. (Y-Z)(X-Z) is odd. Product of 2 numbers can be odd only when both are odd and difference/sum of two numbers is odd when one is even and other odd. Using this logic, there are two Cases from statement's 2:
Case 1: x odd, y odd, z even, it gives product as even
Case 2: x even, y even, z odd, it gives product as even
Therefore, statement 2 is sufficient and xyz is even
Sent from my 2014818 using GMAT Club Forum mobile app
Re: If x, y, z are all integers, is xyz odd? &nbs [#permalink] 22 Oct 2017, 23:51
Display posts from previous: Sort by | 748 | 2,549 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2018-47 | latest | en | 0.854407 |
https://www.geeksforgeeks.org/position-and-displacement-vectors/ | 1,709,176,645,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474775.80/warc/CC-MAIN-20240229003536-20240229033536-00478.warc.gz | 782,171,973 | 60,251 | # Position and Displacement Vectors
A vector is a quantity that has both magnitude and direction. Vectors allow us to describe the quantities which have both direction and magnitude. For example, velocity and position. These quantities are useful in describing the motion and the position of a particle that is moving in a plane. All the vectors follow the laws of parallelogram addition and triangle law. These laws allow us to perform arithmetic on the vectors. The position and the change in position (denoted by displacement) are the entities that use these concepts to describe the motion of a particle. Let’s understand them in detail.
### Scalars and Vectors
In physics, quantities are classified in terms of vectors and scalars. The difference between them is that the quantities which have directions along with their magnitude associated with them are called vectors. A scalar quantity is just magnitude. In the case of scalar quantities, the arithmetic operations like addition, subtraction, or multiplication are performed in the same way as done with real numbers. For example, the sum of two scalar quantities with values 0.1 and 0.3 is given as 0.4. These rules do not apply to vector quantities. Their addition and subtraction are not as simple as scalar quantities.
The table given below shows some examples of scalar and vector quantities.
### Laws of Addition and Subtraction of Vectors
Two vectors cannot be added by the usual arithmetic procedures, since the vectors contain directions. These vectors can be added using the laws of vector addition. For two vectors P and Q, their addition is given by vector R,
If the angle between the two vectors is θ, then the magnitude of the resultant vector is given by,
R2 = |P|2 + |Q|2 + 2|P||Q|cos(θ)
The angle that the resultant vector makes with the vector P, is given by,
Θ = tan-1(
Vector Multiplication by a constant
Multiplication of a vector with a constant does not change its direction. Instead, this kind of multiplication serves to scale the vector. If the constant multiplied by the vector is greater than 1, the vector increases in its length, while it decreases in length when the constant is less than 1.
### Motion in a Plane
When an object moves in a plane, it changes its position, and it is essential to define quantities that can be used to describe where the object is in the plane. The motion also requires direction. For example, suppose an object is moving as shown in the figure below. Now, to describe the position of the object, two things are required — direction and distance from the origin. Simply saying that the object is at a distance of 5 m from the origin is not enough.
This object will be described as 5 m in the North-East direction. Thus, to denote the position of an object, a vector is required. This vector is called a position vector. The diagram below shows the trajectory of an object moving in the plane. Let P and P’ be the positions of the object at time “t” and “t’“. The figure given below shows the position P and P’ with respect to the origin.
When the points P and P’ are joined by a straight line with origin O. The line segments OP and OP’ denote the position vectors. Vector OP is denoted by r and OP’ is denoted by r’. If the object moves from P to P’ in time “t”. Then, the displacement is given by the change in position vector. The vector denoting the change in position vector is also called the displacement vector.
Note: The Displacement vector only depends upon the initial and final position vectors. If an object travels a path and comes back to the same initial position, in that case the displacement is considered to be zero. The figure above shows an object traveling along a large path and coming back to the same point. In this case, the displacement is zero.
The magnitude of the displacement vector is less than or equal to the distance traveled by the particle between its initial and final positions.
Assume that a particle travels from point A to point B along the path traveled as shown in the above figure. In this case, displacement is given by the line joining the two points. Thus, it can be said that the displacement between two points is the shortest distance between those points.
### Sample Problems
Question 1: Let’s say A = 4i + 3j and B = 5i + 4j. Find the resultant vector from the addition of these two vectors.
Given:
A = 4i + 3j
B = 5i + 4j.
The resultant of these two vectors is given by,
Plugging the vectors into this equation,
Question 2: Let’s say A = 5i + 5j and B = 3i + 3j. Find the resultant vector from the addition of these two vectors.
Given:
A = 5i + 5j
B = 3i + 3j.
The resultant of these two vectors is given by,
Plugging the vectors into this equation,
Question 3: Let’s say there are two vectors A and B, where |A| = 3 and |B| = 4 and the angle between them is given by 60°. Find the magnitude and direction of the resultant vector.
Given:
|A| = 3
|B| = 4.
θ = 60°
The resultant of these two vectors is given by,
R2 = |A|2 + |B|2 + 2|A||B|cos(θ)
The angle is given by, θ = tan-1(
Plugging the vectors into these equations,
R2 = |A|2 + |B|2 + 2|A||B|cos(θ)
⇒ R2 = 32 + 42 + 2(3)(4)cos(60)
⇒ R2 = 25 + 12
⇒ R2 = 37
⇒ R = √37
θ = tan-1(
⇒ θ = tan-1(
⇒ θ = 53°
Question 4: Let’s say there are two vectors A and B, where |A| = 24 and |B| = 10 and the angle between them is given by 90°. Find the magnitude and direction of the resultant vector.
Given:
|A| = 24
|B| = 10.
θ = 90°
The resultant of these two vectors is given by,
R2 = |A|2 + |B|2 + 2|A||B|cos(θ)
The angle is given by, θ = tan-1(
Plugging the vectors into these equations,
R2 = |A|2 + |B|2 + 2|A||B|cos(θ)
⇒ R2 = 242+ 102 + 2(24)(10)cos(90)
⇒ R2 = 576 + 100
⇒ R2 = 676
⇒ R = 26
θ = tan-1(
⇒ θ = tan-1(
Question 5: The position vector of the particle moving in a plane is given by,
r = t2i + 3tj
Find the displacement between t =1 and t = 4 seconds.
Displacement only depends on the initial and final position of the particle.
Given:
r = t2i + 3tj
at t = 1
ri = i + 3j
At t = 4
rf = 16i + 12j
Displacement between these position will be given by the difference of their position vectors.
r = rf – ri
⇒r = 16i + 12j -(i + 3j)
⇒ r = 15i + 9j
Question 6: The position vector of the particle moving in a plane is given by,
r = t2i + 3tj
Find the displacement between t =1 and t = 4 seconds.
Displacement only depends on the initial and final position of the particle.
Given:
r = t2i + 3tj
at t = 1
ri = i + 3j
At t = 4
rf = 16i + 12j
Displacement between these position will be given by the difference of their position vectors.
r = rf – ri
⇒r = 16i + 12j -(i + 3j)
⇒ r = 15i + 9j
Previous
Next | 1,832 | 6,772 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2024-10 | latest | en | 0.949036 |
http://mathhelpforum.com/calculus/29613-problem-set.html | 1,481,193,321,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542520.47/warc/CC-MAIN-20161202170902-00209-ip-10-31-129-80.ec2.internal.warc.gz | 177,892,056 | 12,246 | 1. ## Problem Set
I think I figured out the first two questions, but I have three more puzzlers to go. Can anyone help?
3. Verify that the line 4x-y+11+0 is tangent to the curve y=(16/x²)-1. Then determine the point of tangency.
I started this question off by finding the derivative of the curve, and I got y'= -32x(-3), where -3 is the exponent on x. I also isolated for y in the line and I got y=4x+11.
I have no idea where to go from here.
4. For what values of "a" and "b" will the parabola y=x²+ax+b be tangent to the curve y=x²√x+1 at the point (3, 18)?
*Stares blankly at the question.*
5. Find the AREA of the triangle formed by the three points on the curve y=[x²-1 / x²+1]² where the tangent lines are horizontal.
We haven't done anything like this before! We never even learned the equations on how to do this! All we have discussed in this course (which started a couple weeks ago) has been limits, derivatives, power rule, etc.
2. Hello, NAPA55!
Here's the last one . . .
5. Find the area of the triangle formed by the three points
on the curve: . $y\:=\:\left(\frac{x^2-1}{x^2+1}\right)^2$ where the tangent lines are horizontal.
First, find the derivative (Chain Rule, Quotient Rule, etc.)
. . $y' \;=\;2\left(\frac{x^2-1}{x^2+1}\right)\cdot\frac{(x^2+1)\cdot2x - (x^2-1)\cdot2x}{(x^2+1)^2} \;=\;\frac{8x(x^2-1)}{(x^2+1)^3}$
Tangents are horizontal where $y' = 0$
. . So we have: . $8x(x-1)(x+1) \:=\:0\quad\Rightarrow\quad x \;=\;0,\:\pm1$
When $x = 0\!:\;\;y \:=\:\left(\frac{0-1}{0+1}\right)^2 \:=\:1\quad\Rightarrow\quad (0,\,1)$
When $x = 1\!:\;\;y \:=\:\left(\frac{1-1}{1+1}\right)^2 \:=\:0\quad\Rightarrow\quad (1,\,0)$
When $x = \text{-}1\!:\;\;y \:=\:\left(\frac{1-1}{1+1}\right)^2\:=\:0\quad\Rightarrow\quad (\text{-}1,\,0)$
And surely you can find the area of this triangle . . .
Code:
|
*(0,1)
/ | \
/ | \
/ | \
- * - - - + - - - * -
(-1,0) | (1,0)
3. Thanks!!
4. Originally Posted by NAPA55
...
3. Verify that the line 4x-y+11+0 is tangent to the curve y=(16/x²)-1. Then determine the point of tangency.
I started this question off by finding the derivative of the curve, and I got y'= -32x(-3), where -3 is the exponent on x. I also isolated for y in the line and I got y=4x+11.
I have no idea where to go from here.
...
The slope of the line (m = 4) equals the gradient of the curve for a certain x-value. Therefore you have to determine a x-value so that
$4 = -\frac{32}{x^3}~\implies~4x^3 = -32~\implies~x^3 = -8 ~\implies~ x = -2$
Plug in this value into the equation of the line (or the curve; it doesn't matter) and you'll get the tangent point. I've got T(-2, 3)
5. Originally Posted by NAPA55
...
4. For what values of "a" and "b" will the parabola y=x²+ax+b be tangent to the curve y=x²√(x+1) at the point (3, 18)?
*Stares blankly at the question.*
...
At the tangent point both curves must have the same gradient:
$f(x)=x^2 \sqrt{x + 1} ~\implies~ f'(x)=\frac{x(5x+4)}{2\sqrt{x+1}}$ And therefore:
$f'(3) = \frac{57}{4}$ must equal $p'(x)=2x + a$ with x = 3, that means $p'(3) = 6+a$
The tangent point must be located on the parabola. With both conditions you get a system of equations:
$\left|\begin{array}{l}18 = 9+3a+b\\ \frac{57}{4} =2 \cdot 3 + a\end{array}\right.$
From the 2nd equation you get: $a = \frac{33}{4}$ . Plug in this value into the 1rst equation to calculate b:
$18=9+3 \cdot \left(\frac{33}{4} \right) +b~\implies~ b = - \frac{63}{4}$
The equation of the parabola becomes:
$p: y = x^2 + \left(\frac{33}{4} \right) x - \frac{63}{4}$
EDIT: I changed all results according to the new wording of the question.
6. The root applies to the whole (x+1).
For what values of "a" and "b" will the parabola y=x²+ax+b be tangent to the curve y=x²√(x+1) at the point (3, 18)?
7. And for #3, how do I show the "verifying" part?
8. Originally Posted by NAPA55
The root applies to the whole (x+1).
For what values of "a" and "b" will the parabola y=x²+ax+b be tangent to the curve y=x²√(x+1) at the point (3, 18)?
I've made all necessary changes in my previous post. Have a look there.
9. Originally Posted by NAPA55
And for #3, how do I show the "verifying" part?
All calculations had been done to get one point which belongs to both graphs and where both graphs have the same gradient
Therefore this point is a tangent point and both graphs are tangent curves to each other.
This process is called verifying.(I believe )
10. Originally Posted by earboth
At the tangent point both curves must have the same gradient:
$f(x)=x^2 \sqrt{x + 1} ~\implies~ f'(x)=\frac{x(5x+4)}{2\sqrt{x+1}}$
How did you do that step?
11. Originally Posted by NAPA55
How did you do that step?
use product rule:
$f(x)=x^2 \sqrt{x + 1} ~\implies~ f'(x) = 2x \sqrt{x+1} + x^2 \cdot \frac12 \cdot (x+1)^{-\frac12} =$ $\frac{2 \cdot 2x \sqrt{x+1} \cdot \sqrt{x+1} + x^2}{2\sqrt{x+1}} = \frac{ 4x (x+1) + x^2}{2\sqrt{x+1}} = \frac{x(5x+4)}{2\sqrt{x+1}}$
12. Thanks... stupid me... I didn't realize that until I immediately posted that message and shut off the computer.
13. Thanks. | 1,754 | 5,082 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 19, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2016-50 | longest | en | 0.785097 |
null | null | null | null | null | null | Step 12: Put it all together.
Picture of Put it all together.
Now loosen the outer hoop as much as it will go. Hold the 3 bent wires on the inside of the outer hoop and insert the inner hoop. This will be quite tricky and unless you have 3 hands, you might need a friend to help. Or hold them all in one hand while you fiddle with the inner circle. Once you have the inner circle in place, tighten the outer circle a little bit. Then arrange the wires so you have them at 10, 2, and 6 o'clock. Tighten the outer hoop as tight as it will go. Then insert your mic into the circle. CAREFULLY adjust the screen so you have about 1-2" between the mic and the screen itself. Then you're done!
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mtx1000d5 years ago
that was very simple, great idea and i'm going to use it. There's some other video with a guy who made his out of wood and rubber and all this other crap and the whole time I'm thinking... couldn't you just use a damn coat hanger? | null | null | null | null | null | null | null | null | null |
https://math.stackexchange.com/questions/2589988/integration-different-result-different-techniques-using-change-of-variable | 1,571,517,095,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986697760.44/warc/CC-MAIN-20191019191828-20191019215328-00228.warc.gz | 569,686,341 | 35,896 | # Integration different result different techniques using change of variable
I am currently integrating $$\int \frac{1}{\sqrt{x}\cdot(4\sqrt{x}+5)}dx$$ When I change the variable in the $dx$ to $\frac{2}{\sqrt{x}}$ I can multiply the integral by $\frac{2}{2}$ (which is $1$) (multiply only the numerator by $2$ and leave the $\frac{1}{2}$ outside the integral) that way I can use the power rule for integrals and after simplification I am left with $\frac{1}{x}$... however the online calculators give different result: $$\frac{\ln(4\sqrt{x}+5)}{2}$$
What I did is
$$\int \frac{1}{\sqrt{x}} d(\frac{2}{\sqrt x})$$
Then I multiplied by $\frac22$ and I get
$$\frac12 \int \frac{2}{\sqrt x}d(\frac{2}{\sqrt x})$$
And I use the power rule...
After simplification I get $1/x$.
• math.meta.stackexchange.com/questions/5020/… – 5xum Jan 3 '18 at 10:35
• I translated your expressions to LaTeX, please check I didn't change what you meant. – Jean-Claude Arbaut Jan 3 '18 at 11:58
• @Jean-ClaudeArbaut in the dx it's 2/sqrt(x) not 2*sqrt(x) – john Jan 3 '18 at 12:00
• Is it better now? – Jean-Claude Arbaut Jan 3 '18 at 12:04
• @Jean-ClaudeArbaut yes – john Jan 3 '18 at 12:05
Let's do it step by step.
$$\int \frac{1}{\sqrt{x}\cdot(4\sqrt{x}+5)}dx=\int \frac{1}{4\sqrt{x}+5}\frac{dx}{\sqrt{x}}$$
Let's do the substitution $4\sqrt{x}+5=u$. Then $du=\dfrac{2dx}{\sqrt{x}}$, and the $\dfrac{dx}{\sqrt{x}}$ already here in your integral becomes $\dfrac{du}{2}$, while $4\sqrt{x}+5$ becomes $u$.
The integral thus becomes
$$\frac12\int \frac{du}{u}$$
If the simplification looks too "fast", you can do it this way:
Since $du=\dfrac{2dx}{\sqrt{x}}$, you have $dx=\frac12\sqrt{x}du$, then the integral becomes
$$\int \frac1{\sqrt{x}}\cdot\frac{1}{4\sqrt{x}+5}\cdot \frac12\cdot\sqrt{x}du$$ $$\int \frac1{\sqrt{x}}\cdot\frac{1}u\cdot \frac12\cdot\sqrt{x}du$$
And the two $\sqrt{x}$ simplify, to leave
$$\frac12\int \frac{du}{u}$$
• can you check my question again... I edited it... and tell me where am I mistaking – john Jan 3 '18 at 11:54
• @john I added another answer. I feel you didn't do the substitution correctly, so I urge you to write everything as carefully as you can. – Jean-Claude Arbaut Jan 3 '18 at 12:03
Why don't you try $x=t^{2}$? Then the integral simplifies to $\int{2\over 4t+5}dt$ and is equal to ${1\over 2}{ln(t+{5\over4})}+c$ or ${1\over 2}{ln(\sqrt x+{5\over4})}+c$
• yes but why my result is different? – john Jan 3 '18 at 10:40
• Do you mean that you replace x with 2/sqrt(x)? – Mostafa Ayaz Jan 3 '18 at 10:43
• yes inside the dx – john Jan 3 '18 at 10:44
• I guess the '5' constant in the denominator bans you from forwarding in proof outline.... – Mostafa Ayaz Jan 3 '18 at 10:47
• I still don't get where my mistake is – john Jan 3 '18 at 10:53
I didn't understand what you mean by changing in the variable $dx$ to $\frac{2}{\sqrt{x}}$ but I think what you should have done is $u = 4\sqrt{x}+5$ then $du = \frac{2}{\sqrt{x}}dx$. Then the expression becomes $$\frac{1}{2}\int\frac{du}{u} = \frac{1}{2}\ln{u}+C = \frac{\ln(4\sqrt{x}+5)}{2}+C$$
• I put the 2/√x under the dx and it becomes d(2√x) because the first derivate of 4√x+5 is 2/√x – john Jan 3 '18 at 10:52
• Then I think you must put 4sqrt(x)+5 under d(.) and then your calculations gonna be true... – Mostafa Ayaz Jan 3 '18 at 11:16
Basically, this is what you must have done first: $$I = \frac12 \int \frac {2}{\sqrt x} \times \frac {1}{4\sqrt x +5}\, dx$$
Now, notice as the first derivative of $4\sqrt x +5$ is $\frac {2}{\sqrt x}$, we will get: $$I = \frac12 \int \frac {\mathrm d (4\sqrt x +5)}{4\sqrt x +5}$$ This is a very common form of an integral: $$\int \frac {1}{x}\, dx = \log x + c \tag 1$$ Using this result, we will get: $$I = \frac12 \ln (4\sqrt x +5)$$
What you must have done is: you could have substituted $4\sqrt x +5 = t$ to get the form $(1)$, but you must have just left it like that, without integrating.
Otherwise, you will get a result of $\frac1 {x}$ only on integrating $-\frac {1}{x^2}$, which has no place in our integral here.
• well not exactly... I got 1/x ... First I substituted 4√x+5 =u and I get S 1/√x d(2/√x)... then i multiplied my integral by 2/2 so i get 1/2 S 2/√x d(2/√x) and I used the basic power rule – john Jan 3 '18 at 11:30
• @john Then your mistake is easy to spot: When you substitute $4\sqrt{x}+5=u$, you are left with $1/u$, not $1/\sqrt{u}$. – Jean-Claude Arbaut Jan 3 '18 at 11:33
• @Jean-ClaudeArbaut can't I just multiply my 1/sqrt(x) by 2/2 ?? and it gets 2/sqrt(x) which is same as the u? – john Jan 3 '18 at 11:35
• No, if you multiply $1/\sqrt{x}$ by $2/2$, it does not become $2/\sqrt{x}$. And no, if $u=4\sqrt{x}+5$, then $2/\sqrt{x}$ is not equal to $u$. However, you can write $du=2dx/\sqrt{x}$. And since there is already a $1/\sqrt{x}$ in your integrand, that simplifies cleanly. – Jean-Claude Arbaut Jan 3 '18 at 11:38
• @Jean-ClaudeArbaut I don't get it... in class we studied that I can multiply it and it becomes 1/2 S 2/√x du ... the first derivative of u is 2/√x... so my u =2/√x – john Jan 3 '18 at 11:41
When you do integration by substitution in
$$\int f(x)dx$$
You replace the integration variable $x$ with some function of another variable, say $u$ (for instance $x=g(u)$), with some conditions on the function $g$. Then the integral becomes
$$\int f(g(u)) g'(u)du$$
Please write down which function $g$ you used.
You seem to be mixing several concepts, and this makes very unclear what you meant. Get back to the definition.
The informal way to do substitution is to write $x=g(u)$ hence $dx=g'(u)du$, or if you prefer $\dfrac{dx}{du}=g'(u)$. Then, in the original integral, you replace $dx$ with $g'(u)du$, and any other instance of $x$ with $g(u)$.
It gets quickly to a result, but I don't like very much working with $d$, without a proper definition. However, it's exactly equivalent to the more formal way, if you do it correctly.
• anyway as my question... Where is my mistake – john Jan 3 '18 at 12:11
• @john What is your substitution? You didn't even write one. I'm trying to help, but you have to be accurate. If you substitute $x=g(u)$, what is your $g$? – Jean-Claude Arbaut Jan 3 '18 at 12:23 | 2,148 | 6,199 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2019-43 | latest | en | 0.722157 |
https://math.stackexchange.com/questions/1127215/question-about-asymmetry-of-chi-square-distribution | 1,558,574,543,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256997.79/warc/CC-MAIN-20190523003453-20190523025453-00437.warc.gz | 560,155,145 | 33,393 | # Question about asymmetry of chi-square distribution
Let $X_1,\dots,X_n$ be a set of i.i.d. chi-square random variables with $k$ degrees of freedom. Consider the statistic $\arg\max_i\{|X_i/k - 1|\}=X_{\alpha}$. I wonder about the probability that $X_{\alpha} = X_{\text{max}}$, rather than $X_{\text{min}}$? Does it change if $k$ gets large?
I suspect that it may be very, very difficult to get a precise answer to the above questions.
So I would like to ask the simpler question: can we at least say that it is more likely that $X_{\alpha} = X_{\text{max}}$? Simulations seem to indicate yes. Is there a heuristic reason why?
From my simulations, it is clear that the answer is going to depend strongly on $k$ and $n$. For all $k$, as $n$ increases, $\Pr[X_\alpha = X_{(n)}]$ increases: the probability that the given statistic is the maximum order statistic becomes increasingly likely. The effect of increasing $k$ is to make the rate at which this probability increases as a function of $n$ slower.
We can get an exact probability for the case $k = 2$ (which is exponential), and for general $n$: the joint distribution of $(X_{(1)}, X_{(n)})$ is $$f(x,y) = \frac{1}{2}\binom{n}{2}e^{-(x+y)/2}(e^{-x/2} - e^{-y/2})^{n-2}, \quad 0 \le x < y.$$ The probability that $|X_{(1)} - 2| > |X_{(n)} - 2|$ is equivalent to $$p(n) = \Pr[X_{(1)} + X_{(n)} < 4] = \int_{x=0}^2 \int_{y=x}^{4-x} f(x,y) \, dy \, dx.$$ For specific values of $n$, we have \begin{align*} p(2) &= 1-3e^{-2} \\ p(3) &= 1 - 6e^{-2} + 8e^{-3} - 3e^{-4} \\ p(4) &= 1 - 6e^{-2} + 15e^{-4} + 2e^{-6} \\ p(5) &= 1 - \tfrac{20}{3}e^{-2} + 30 e^{-4} - \tfrac{128}{3}e^{-5} + 20e^{-6} - \tfrac{5}{3}e^{-8}. \end{align*} Only the first two are greater than $1/2$.
• Yes. In my simulations, for very small sample sizes (e.g., $n = 2$), it is more likely that $|X_{(1)}/k - 1| > |X_{(n)}/k - 1|$. This appears to be true for all degrees of freedom, although as $k \to \infty$, this probability tends to $1/2$. – heropup Jan 31 '15 at 2:42
• After more investigation, it appears that only when $n \in \{2, 3\}$ that $\Pr[X_\alpha = X_{(1)}] > 1/2$, regardless of $k$. – heropup Jan 31 '15 at 2:53 | 773 | 2,157 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2019-22 | latest | en | 0.812211 |
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# More Percent Applications
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For mark-ups and mark-downs, percent remaining works well.
Also note: You *can* have percent change on percent data.
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### More Percent Applications
1. 1. More Percent ApplicationsFor mark-ups and mark-downs, percent remainingworks well.You *can* have percent change on percent data.
2. 2. Recall for Percent Change SitutationsThere are four numbers to pay attention to: 1. Original amount (the “old”/“starting” amount before the change) 2. New amount (the “ending” amount after the change) 3. Amount of change (the difference [subtract] between original and new) 4. Percent change (percent increase or decrease).
3. 3. The Two Useful Relationships amount of change Percent change original amount(Original amount)(Percent remaining) (New amount) ▲If an increase: Percent remaining = 100% + Percent change ▼If a decrease: Percent remaining = 100% - Percent change
4. 4. Mark-ups Retailers, in order to make money, take their cost of an item and increase this amount to get the retail price.Percent remaining works well to get the retail price.Example: If the cost of a pair of jeans from the manufacture is \$16.00 and the store marks up the price by 80%, what is the retail price? (16)(180%) (New amount) (16)(180%) (16)(1.8) \$28.80
5. 5. Mark-downs Retailers, in order to attract customers (and get rid of extra inventory), take their retail price of an item and decrease this amount to get the sale price.Percent remaining works well to get the sale price.Example: If the regular price of a sweatshirt is \$30.00 and the store marks it down by 40%, what is the sale price? (30)(60%) (New amount) (30)(60%) (30)(.6) \$18.00
6. 6. You can have percent change on percent data.Sometimes the original data is in percent form.When describing the change, the media will report either • The number of percentage points of change, or • The percent change.
7. 7. You can have percent change on percent data.Example: In August the percent of people collecting food stamps was 20% and in September it went up to 22%. Find (a) the number of percentage points of change, and (b) the percent change.(a)The percent of people collecting food stamps went up by 2 percentage points.(b)The percent increase in the percent of people collecting food stamps was a 10% increase.
8. 8. You can have percent change on percent data.Example: In August the percent of people approving of the president was 80% and in September it went up to 82%. Find (a) the number of percentage points of change, and (b) the percent change.(a)The percent of people approving of the president went up by 2 percentage points.(b)The percent increase in the people approving of the president was a 2.5% increase.
9. 9. Another example: In 2008, 40% of the people in Polk County owned a Ford vehicle. In 2009, the percent of Ford owners in Polk County was 37%. Find (a) the number of percentage points of change, and (b) the percent change.(a)The percent of people owning a Ford went down by 3 percentage points.(b)The percent decrease in the people owning a Ford was a 7.5% decrease.
10. 10. Closing Notes Remember For mark-ups and mark-downs, percent remainingworks well.You *can* have percent change on percent data.
11. 11. Last one…This is Day 24 of 24.I hope you enjoyed, and learnedfrom, this 24 Percent Tweets. Dr. OlsenLearnist board: http://bit.ly/13AGhZqWebpage: http://bit.ly/ZXLw0I | 922 | 3,624 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2019-22 | latest | en | 0.866064 |
https://www.excellup.com/MAT/mat_reasoning_5_answer.aspx | 1,653,462,122,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662580803.75/warc/CC-MAIN-20220525054507-20220525084507-00428.warc.gz | 822,806,767 | 2,461 | # Reasoning
## Test Paper 5 Answer
1 Answer: c
Explanation: He starts by walking in north-west direction. After turning right, he walks in north-east direction. After turning left, he is again walking in north-west direction.
2 Answer: a
Explanation: A reaches top left corner, B reaches bottom right corner. C reaches top right corner, D reaches bottom left corner. The new sequence is CBDA.
3 Answer: a
4 Answer: d
Explanation: C moves to bottom right corner; then to bottom left corner. A moves to bottom left corner, then to bottom right corner. So, A is at the south-west corner.
5 Answer: b
Copyright © excellup 2014 | 150 | 632 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2022-21 | longest | en | 0.873693 |
https://nrich.maths.org/public/topic.php?code=-989&cl=1&cldcmpid=1257 | 1,585,969,802,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370519111.47/warc/CC-MAIN-20200404011558-20200404041558-00404.warc.gz | 614,855,802 | 8,558 | # Resources tagged with: PrimaryGames-Number
Filter by: Content type:
Age range:
Challenge level:
### There are 50 results
Broad Topics > Primary Games > PrimaryGames-Number
### Totality for Two
##### Age 5 to 11 Challenge Level:
Totality game for an adult and child. Be the first to reach your agreed total.
### Daisy
##### Age 7 to 11 Challenge Level:
A game for 2 players. Draw a daisy with at least 5 petals. Shade 1 or 2 petals next to each other. The winner shades the last petal.
### Ten-frames Games
##### Age 3 to 7
These games use ten-frames to develop children's 'sense of ten'.
### Calculator Bingo
##### Age 7 to 11 Challenge Level:
A game to be played against the computer, or in groups. Pick a 7-digit number. A random digit is generated. What must you subract to remove the digit from your number? the first to zero wins.
### Dicey Operations in Line
##### Age 7 to 11 Challenge Level:
Who said that adding, subtracting, multiplying and dividing couldn't be fun?
### Multiplication Tables - Matching Cards
##### Age 5 to 11 Challenge Level:
Practise your tables skills and try to beat your previous best score in this interactive game.
### 100 Percent
##### Age 7 to 11 Challenge Level:
An interactive game for 1 person. You are given a rectangle with 50 squares on it. Roll the dice to get a percentage between 2 and 100. How many squares is this? Keep going until you get 100. . . .
### Down to Nothing
##### Age 7 to 11 Challenge Level:
A game for 2 or more people. Starting with 100, subratct a number from 1 to 9 from the total. You score for making an odd number, a number ending in 0 or a multiple of 6.
### Countdown
##### Age 7 to 14 Challenge Level:
Here is a chance to play a version of the classic Countdown Game.
### Strike it Out
##### Age 5 to 11 Challenge Level:
Use your addition and subtraction skills, combined with some strategic thinking, to beat your partner at this game.
### Nice or Nasty
##### Age 7 to 14 Challenge Level:
There are nasty versions of this dice game but we'll start with the nice ones...
##### Age 5 to 11 Challenge Level:
Who said that adding couldn't be fun?
### Four Go
##### Age 7 to 11 Challenge Level:
This challenge is a game for two players. Choose two of the numbers to multiply or divide, then mark your answer on the number line. Can you get four in a row?
### Play to 37
##### Age 7 to 11 Challenge Level:
In this game for two players, the idea is to take it in turns to choose 1, 3, 5 or 7. The winner is the first to make the total 37.
### Incey Wincey Spider
##### Age 5 to 7 Challenge Level:
You'll need two dice to play this game against a partner. Will Incey Wincey make it to the top of the drain pipe or the bottom of the drain pipe first?
### Chocolate Bars
##### Age 7 to 11 Challenge Level:
An interactive game to be played on your own or with friends. Imagine you are having a party. Each person takes it in turns to stand behind the chair where they will get the most chocolate.
### Train Tactics
##### Age 7 to 11 Challenge Level:
A train building game for two players. Can you be the one to complete the train?
### Train Tactics for Two
##### Age 7 to 11 Challenge Level:
Train game for an adult and child. Who will be the first to make the train?
### Incey Wincey Spider for Two
##### Age 5 to 7 Challenge Level:
Incey Wincey Spider game for an adult and child. Will Incey get to the top of the drainpipe?
### Fifteen
##### Age 7 to 14 Challenge Level:
Can you spot the similarities between this game and other games you know? The aim is to choose 3 numbers that total 15.
### Strike it Out for Two
##### Age 5 to 11 Challenge Level:
Strike it Out game for an adult and child. Can you stop your partner from being able to go?
### Nice or Nasty for Two
##### Age 7 to 14 Challenge Level:
Some Games That May Be Nice or Nasty for an adult and child. Use your knowledge of place value to beat your opponent.
### Dominoes
##### Age 7 to 16 Challenge Level:
Everthing you have always wanted to do with dominoes! Some of these games are good for practising your mental calculation skills, and some are good for your reasoning skills.
##### Age 5 to 7 Challenge Level:
A game for 2 or more players. Practise your addition and subtraction with the aid of a game board and some dried peas!
### Dot Card Games
##### Age 3 to 7
These games devised by Jenni Way use dot cards which will help children see the structure of numbers 1-6 and give them confidence as they begin to add and subtract these numbers.
### Dicey Operations in Line for Two
##### Age 7 to 11 Challenge Level:
Dicey Operations for an adult and child. Can you get close to 1000 than your partner?
### Four Go for Two
##### Age 7 to 11 Challenge Level:
Four Go game for an adult and child. Will you be the first to have four numbers in a row on the number line?
### Spiralling Decimals for Two
##### Age 7 to 14 Challenge Level:
Spiralling Decimals game for an adult and child. Can you get three decimals next to each other on the spiral before your partner?
### First Connect Three
##### Age 7 to 14 Challenge Level:
Add or subtract the two numbers on the spinners and try to complete a row of three. Are there some numbers that are good to aim for?
### Tug Harder!
##### Age 7 to 11 Challenge Level:
In this game, you can add, subtract, multiply or divide the numbers on the dice. Which will you do so that you get to the end of the number line first?
### Totality
##### Age 5 to 11 Challenge Level:
This is an adding game for two players.
### Got it for Two
##### Age 7 to 14 Challenge Level:
Got It game for an adult and child. How can you play so that you know you will always win?
### Got It
##### Age 7 to 14 Challenge Level:
A game for two people, or play online. Given a target number, say 23, and a range of numbers to choose from, say 1-4, players take it in turns to add to the running total to hit their target.
### First Connect Three for Two
##### Age 7 to 11 Challenge Level:
First Connect Three game for an adult and child. Use the dice numbers and either addition or subtraction to get three numbers in a straight line.
### Factors and Multiples Game for Two
##### Age 7 to 14 Challenge Level:
Factors and Multiples game for an adult and child. How can you make sure you win this game?
### Spiralling Decimals
##### Age 7 to 14 Challenge Level:
Take turns to place a decimal number on the spiral. Can you get three consecutive numbers?
### Nim-7
##### Age 5 to 14 Challenge Level:
Can you work out how to win this game of Nim? Does it matter if you go first or second?
### Tug of War
##### Age 5 to 7 Challenge Level:
Can you use the numbers on the dice to reach your end of the number line before your partner beats you?
### Double or Halve?
##### Age 5 to 7 Challenge Level:
Throw the dice and decide whether to double or halve the number. Will you be the first to reach the target?
### Nim-like Games
##### Age 7 to 16 Challenge Level:
A collection of games on the NIM theme
### Factors and Multiples Game
##### Age 7 to 16 Challenge Level:
A game in which players take it in turns to choose a number. Can you block your opponent?
### The Remainders Game
##### Age 7 to 14 Challenge Level:
Play this game and see if you can figure out the computer's chosen number.
### Shut the Box for Two
##### Age 5 to 7 Challenge Level:
Shut the Box game for an adult and child. Can you turn over the cards which match the numbers on the dice?
### Snail One Hundred
##### Age 5 to 7 Challenge Level:
This is a game in which your counters move in a spiral round the snail's shell. It is about understanding tens and units.
### Matching Fractions, Decimals and Percentages
##### Age 7 to 14 Challenge Level:
Can you match pairs of fractions, decimals and percentages, and beat your previous scores?
### Stop or Dare
##### Age 7 to 16 Challenge Level:
All you need for this game is a pack of cards. While you play the game, think about strategies that will increase your chances of winning.
### Shut the Box
##### Age 5 to 7 Challenge Level:
An old game but lots of arithmetic!
### Dotty Six for Two
##### Age 5 to 11 Challenge Level:
Dotty Six game for an adult and child. Will you be the first to have three sixes in a straight line?
### 9 Hole Light Golf
##### Age 5 to 18 Challenge Level:
We think this 3x3 version of the game is often harder than the 5x5 version. Do you agree? If so, why do you think that might be?
### Statement Snap
##### Age 7 to 14 Challenge Level:
You'll need to know your number properties to win a game of Statement Snap... | 2,115 | 8,650 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2020-16 | latest | en | 0.851003 |
https://boards.straightdope.com/t/speed-of-a-sailboat-in-20-mph-winds/920506 | 1,601,507,165,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402128649.98/warc/CC-MAIN-20200930204041-20200930234041-00515.warc.gz | 274,443,259 | 10,030 | Speed of a sailboat in 20 MPH winds
This may be a dumb question, but maybe not. In any case, it’s fantasy, so bear that in mind.
If a sailboat is on relatively calm waters with winds blowing at 20 mph, how fast will it move at top speed?
Specifically, if, say, you’re running a game of Dungeons and Dragons, and the players have a decent 70’ merchant vessel with a middling crew, and they just gained a magic item that lets them affect wind speed locally, granting them a wind from any direction of up to 20 mph, for an hour at a time, what would a canny crew be able to do with this item to maximize their travel speed?
Impossible to give exact numbers as this will vary due to local conditions - even barnacles on the hull will have an effect, but as a quick explanation:
Max speed for most monohulls is determined by length. There’s a good chart here:
https://improvesailing.com/questions/average-sailboat-speed
Wind speed and vectors all come into play here, so it’s not a simple answer.
If you are able to control the direction of the wind, then you could get up to whatever speed your boat will allow.
You get the least speed if you are just running straight downwind. In that case, you’d top out at the wind speed.
But as you angle your sail against the wind, and use your keel to track, you can get more speed. If you have little friction you can get to many multiples of the wind speed. Look at Ice Yachting for examples of that.
So, it would be limited, as @FinsToTheLeft by the length of the hull, unless you get up to speeds that allow hydroplaning, which I would not recommend for a crew of middling ability.
Hmm…so it looks like a ~70ft hull could go about 11 knots under ideal conditions, and having an ideally-angled 20 mph wind & otherwise calm seas would be ideal. 11 knots is about 12.6 miles per hour. Pretty good speed!
It also makes me think it’d be useful in ship-to-ship combat: if you can control your local wind direction and speed, you should be able to outmaneuver your enemy, by pulling tricks they can’t pull. But that might take an expert crew to pull off…
Only if there’s zero drag from the hull moving through the water. Absent that piece of magic, I’d guess a 70-ft monohull running before a 20-mph wind might hope for 8 mph.
And this would not be the slowest point of sail - you’d certainly do less when close-hauled.
I’m thinking maybe the crew’s job would be minimal. If you can tune the wind at will, you can leave the sails alone (probably trimmed for a broad reach) while dashing around in any direction you choose
But less fun! Running close hauled with a good amount of heel, enough to get you out on the trapeze was always my happy place.
Also depends on what you mean by local.
Lotta fun can be had if you can control the wind in their sails.
True, I was talking optimal conditions, which would be neglecting any friction. Just that, intuitively, running with the wind should be the fastest, but it’s not. You can do better than windspeed, assuming no friction.
Well, yeah, but if you can control the wind, you are not going to be going into it. (Hopefully)
What is the level of tech of the boat? A 1500s galleon, an 1800s clipper, a mid 1900s J-boat and a 2000s race boat have very different performance. @FinsToTheLeft gave a fine reference upthread for modern tech recreational and racing boats.
You’d expect an oceangoing merchantman to be compromised heavily for seakeeping and for cargo capacity. Yes, speed is good in a merchantman, but not if you can’t carry a profitable amount of cargo or survive the open ocean in an era before weather forecasts & reports. There was no FedEx equivalent back in the Age of Sail.
It occurs to me the OP in effect asked the wrong question. He really wants to know “Is 20 extra knots of wind a lot of wind or a smidgen of wind? Is it tactically decisive or a needle in the haystack?”
In much of the world 20 knots is a lot more than ambient wind most of the time. And is enough to drive any competently sailed boat/ship to its hull speed. Said another way, in much of the ocean, 20 knots, and even better, 20 knots conveniently oriented, represent an overwhelming advantage over the nearby opposition. But an hour is not very long in Age of Sail combat. It could take a day to close the distance from first sighting to cannon range.
There certainly are parts of the open ocean and near-shore coastal waters where the ambient wind is usually high enough that 20 more knots is a hindrance, not a help.
An interesting question from a game-play perspective is whether a player can cast the wind spell defensively? i.e. team A is being pursued by Team B. Team B is gaining on Team A. Team A casts a 20 knot adverse wind onto B’s boat, slowing it dramatically. A escapes successfully while B is becalmed or forced to tack and eventually drops out of range.
It’s actually possible to make a vehicle that’ll travel downwind at faster than the wind speed, though it’s mechanically complicated: You need something coupled to the surface you’re on, and something coupled to the air, and the appropriate gearing in between them. Basically, you’d have a screw in the water, that’s driven by your motion through the water, and then that drives a fan in the air to blow backwards against the wind.
Yeah, that’s pretty much the info I needed. I didn’t really realize that hull speeds existed. I’m thinking that this device allows wind within about 50’ of it to be controlled, direction, and speed up to 20 mph total (so, you can calm a hurricane, but you can’t turn a gale into a hurricane). It’s not useful for offense, except inasmuch as it allows you to outmaneuver foes.
And I think that fancy maneuvers would require an expert crew, because fancy maneuvers would involve sudden changes of wind direction and speed in order to execute hairpin turns and such that would otherwise be impossible on the high seas. I don’t really know what would be possible with perfect control over wind direction and with a crew ready to take in/let out sails in order to take advantage of that control. Any ideas?
If you have total control of the wind, you just set the boat up on its fastest angle (for a merchant vessel this will be broad reach) and steer the boat using your control of the wind.
Your typical sailing merchant vessel is never going to be making fast manoeuvres. The hull is built with a full length keel, and that makes turning the boat very slow. Modern sailing boats have very different underwater design, making for a very different capability. But that is highly dependant upon modern materials.
State of the art round the world ocean racing yachts are quite something:
(Oh, and they sail solo.)
State of the art 70 foot monohull?
Not exactly a merchant vessel. Probably capable of close to 50 knots on 20 knots of wind.
It’s actually been done (on land): https://www.lockhaven.edu/~dsimanek/museum/ddwfttw.htm
Yeah, that page was where I learned it was possible. It could in principle be done on water, too, though I don’t know if it ever has been.
There are also toys with no moving parts (or rather, the entire toy is one single moving part) that will sit on the surface of a body of water and always move straight upwind. It looks like a broomstick with a buoy in the middle and a propeller on each end.
I think it would require hydrofoils rather than a hull in contact with the water.
I made a little Lego model that demonstrates how it is possible for a vehicle to move faster than the motion of the force propelling it (I think because there was a previous thread ages ago where someone was flatly declaring this impossible); with a single additional gear in this model, it can be made to come toward you when you push it away.
Very interesting. Do you have a link, or a phrase to Google?
There’s a diagram of one on this page: https://www.lockhaven.edu/~dsimanek/museum/ddwfttw.htm (titled ‘The push-me pull-you boat’)
20 miles per hour is about 17 knots of wind. A nice amount of wind for fun and efficient sailing (25-30 knots of wind would get scary for an inexperienced crew). Don’t forget about currents though! Currents are like moving sidewalks so if you are running against it then it doesn’t matter how fast you run, sometimes you aren’t going anywhere. On the other hand, running with the current and you are laughing. | 1,915 | 8,356 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2020-40 | latest | en | 0.954902 |
https://analystprep.com/cfa-level-1-exam/quantitative-methods/updating-probability-using-bayes-formula/ | 1,722,725,640,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640380725.7/warc/CC-MAIN-20240803214957-20240804004957-00282.warc.gz | 77,139,799 | 31,294 | # Updating Probability Using Bayes’ Formula
Bayes’ formula is used to calculate an updated or posterior probability given a set of prior probabilities for a given event. It is a theorem named after the Reverend T Bayes and is used widely in Bayesian methods of statistical inference.
It is the logic used to come up with the formula:
Let $$E_1, E_2, E_3, …, E_n$$ be a set of mutually exclusive and exhaustive events.
Using the conditional probability:
$$P(E_i | A) =\cfrac {P(E_i A)}{P(A)}$$
And also the relationship:
$$P(E_iA) = P(AE_i) = P(E_i)P(A | E_i)$$
And the total probability rule:
$$P(A) = \sum {P(AE_j)} \quad \text { for all j} = 1, 2,…,n$$
We can finally substitute for $$P(E_iA)$$ and $$P(A)$$ in equation 1.
This gives:
$$P(E_{ i }|A)=\cfrac { P(E_{ i })P(A|E_{ i }) }{ \sum _{ j=1 }^{ n }{ P(E_{ i })P(A|E_{ i }) } }$$
This is the Bayes’ formula, and it allows us to ‘turnaround’ conditional probabilities, i.e., we can calculate $$P(E_i|A)$$ if given information only about $$P(A|E_i)$$.
Take note of the explanations given below.
1. $$P(E_j)$$ are known as prior probabilities.
2. Event $$A$$ is some event known to have occurred.
3. $$P(E_i|A)$$ is the posterior probability.
#### Example: Bayes’ Formula
A Civil Engineer wishes to investigate the punctuality of electric trains by considering the number of train journeys. In the sample, 50% of trains were destined for New York, 30% for Vegas, and 20% for Washington, DC. The probabilities of a train arriving late in New York, Vegas, and Washington, DC, are 40%, 35%, and 25%, respectively. If the Engineer picks a train at random from this group, what is the probability that it would be one destined for New York?
Solution:
We are looking for $$P\text{(New York | Late)}$$.
Let us define the events that are critical in our calculation.
First, $$N$$ is the event “A train chosen at random would be one destined for New York.”
Secondly, $$V$$ is the event “A train chosen at random would be destined for Vegas.”
And $$W$$ is the event “A train chosen at random would be destined for Washington DC.”
Finally, let $$L$$ be the event “A randomly chosen would arrive late.”
\begin{align*} P(N|L) & =\cfrac { P(N)P(L|N) }{ P(N)P(L|N)+P(V)P(L|V)+P(W)P(L|W) } \\ & =\cfrac { 0.5×0.4 }{ 0.5× 0.4+0.3×0.35+0.2× 0.25 } \\ & =\cfrac { 0.2 }{ 0.355 } \\ & =0.5634 \\ & =56.3\% \\ \end{align*}
We have computed $$P(N | L)$$ given only $$P(L | N)$$, hence the phrase ‘turnaround conditional probability’.
## Question
A chartered analyst can choose any one of three routes, A, B, or C, to get to work. The probabilities that she arrives on time using routes A, B, and C are 50%, 52%, and 60%, in that order. If she is equally likely to choose any one of the routes and arrive on time, the probability that she chose route A is closest to:
A. 30.9%.
B. 16.67%.
C. 25%.
Solution
First, you should define the relevant events.
Let $$A$$ be the event “Chooses route A.”
Let $$B$$ be the event “Chooses route B.”
And let $$C$$ be the event “Chooses route C.”
Lastly, define event T as “Arrives to work on time.”
Now, what we have is $$P(T | A)$$, i.e., the probability that the analyst arrives on time, given that she chooses route A.
However, we want to find the turnaround probability $$P(A | T)$$, i.e., the probability that the analyst chooses route A, given that she arrives on time.
This is what calls for the application of Bayes’ formula:
\begin{align*} P(A|T) & =\cfrac { P(A)P(T|A) }{ P(A)P(T|A)+P(B)P(T|B)+P(C)P(T|C) } \\ & =\cfrac { \frac {1}{3} ×0.5 }{ \frac {1}{3} ×0.5+\frac {1}{3} × 0.52+\frac {1}{3} × 0.6 } \\ & =\cfrac { 0.16667 }{ 0.54 } \\ & =0.30865 \\ & =30.9\% \\ \end{align*}
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https://socratic.org/questions/how-do-you-solve-5-x-4-1 | 1,638,430,745,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964361169.72/warc/CC-MAIN-20211202054457-20211202084457-00477.warc.gz | 572,594,533 | 5,729 | # How do you solve 5+x/4=1?
Sep 14, 2016
$x = - 16$
#### Explanation:
Isolate the term in $x$ on the left side:
$5 - 5 + \frac{x}{4} = 1 - 5$
$\frac{x}{4} = - 4 \text{ } \leftarrow$ multiply both sides by 4
$\frac{\cancel{4}}{1} \times \frac{x}{\cancel{4}} = - 4 \times 4$
$x = - 16$ | 124 | 291 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 6, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2021-49 | latest | en | 0.520325 |
null | null | null | null | null | null | Take the 2-minute tour ×
I kind of just needed an excuse to post this fabulous article from the BBC.
It is basically a software that does what the leighman has always asked of us - and we have always laughed at them for... (Do I sound really old now?)
So what do you guys think? Will it take over for Pro Tools? Or integrate in it? How is it useful for us who generally work in the exact opposite direction? Implications for cleaning up production sound, etc, etc...
Would be fun to hear your thoughts on this. Or better yet, anyone tried it...?
share|improve this question
5 Answers 5
It seems a nice software. The first thing I can think of is sound cleaning and restoration. There are a lot of possibilities in composition as well but I'm not sure this software is better than others. May be it'd be very good for remixes or useful if you want to extract just a track from a mix.
share|improve this answer
This isn't something you couldn't do before with Sound Forge, EQ and a little knowledge and creativity. I've been abusing audio software like this since I started 17 years ago.
It's actually even easier now with iZotope RX, FFT and Spectral DSP. Not to mention the capabilities of Celemony.
And for someone that likes the "Homeliness" of Acid, his software sure does have a lot of unnecessary frills that look a bit like Rock Band. Besides that, the results didn't sound any better than what I could do with the rudimentary and current tools I mentioned in my opening statement. Actually, his personal remix sounded a bit muted and worse than the original source, I'm sure video/mp3 compression has something to do with that and I can't properly judge it based on that, but the example of the original was moderately clear and his "beeps" were a bit muted, so that certainly has to account for something.
Maybe I should remix the BBC theme and they'll write an article on me because it's just shy of shameless self promotion.
All I have to say is that I hope it lives up to it's claim. If it does and he markets it as a novelty (which it sort of seems he is), then that means it'll be a big push for the pro side of the industry to keep innovating.
share|improve this answer
Celemony has had something like this out for a while now (Melodyne). You can check out information about it at: http://www.celemony.com/cms/index.php?id=products_editor
share|improve this answer
well, I was just going through some drones and I'll bet the new software could help take out some annoying beeps that are in there but spectral repair could do the same
share|improve this answer
All good, but I am more interested in what this means in terms of industry standards etc. Both Auto Tune, Melodyne, the Izotope bundles and other more forensic software does this, but basically this software takes all of these functionalities and wraps them in a nice, user friendly (melodyne-like) interface. Would it in any way make our lives as audio proffesionals easier?
Think of a scene from a symphonic orchestra playing, and you have a long steadycam shot that swipes by differend instruments - all the sound supplied from recordings are one close and one 20m AB-recording. Could this SW now - or in the future if they keep developing - be a tool to amplify the perspectives we're seeing on screen?
And I agree, Syndicate Sythetique, that the sound quality is a bit iffy, but to play devils advocate - when did 1. generation of anything actually work... ;o)
Also, it brings up the question of copyright again, should anyone be allowed to mangle someone elses work?
share|improve this answer
@penguinhearder - For centuries musicians have been covering each others songs, and more recently... remixing within the right context is just fine. That's so long as it has artistic merit and isn't just blatantly ripping off someone's work for personal gain. So I don't see how this would really be any different. In the sound design world this would be considered stealing/plagiarism unless it came from a stock sound library that was intended for this particular type of use. If I went and lifted sounds out of Star Wars and used them then yeah, that's just wrong. – Syndicate Synthetique Feb 25 '11 at 7:26
@Syndicate, There's an interesting question, what is the difference in context that makes sampling SFX bad, but music OK? Are we not musicians of a more esoteric sort? – g.a.harry Apr 6 '11 at 3:16
Your Answer
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null | null | null | null | null | null | 15A Connector
MK 15a Plug
15 Amp plugs and sockets (conforming to BS546) have been the staple connector for the UK's theatre lighting industry for many years. A 3pin connector, the round pins help differentiate the connector from the 13A household version. Similar in size and construction to 13A household connectors, the 15A version is used as the household connector in South Africa. 15A plugs do not have a space for a fuse, as the fuses in lighting systems are generally located within the dimmer. Care must be taken when fitting non-dimmable items with 15A connectors as damage to the item may occur if accidentally connected to a dimmer - in many UK Theatres and other venues, 15A plugs are used for dimmed power, and 16A CEEForms for "Hard Power". This unofficial convention helps prevent connecting sensitive equipment such as moving lights with switch-mode power supplies to dimmers, which can damage the connected equipment or the dimmer itself.
15A plugs should be wired in the same way as a 13A plug. Always remember to slide the plug top over the cable before wiring to the terminals. If you forget "To Duraplug", don't cut the bit out of the plug top, rewire it again properly.
15A%20GRELCO.JPG Grelco is the common name for the two way adaptors used to connect two 15A plugs into one 15A socket. Three-way versions are called Trelcos. | null | null | null | null | null | null | null | null | null |
https://www.physicsforums.com/threads/sum-of-series.90883/ | 1,674,867,981,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499468.22/warc/CC-MAIN-20230127231443-20230128021443-00630.warc.gz | 934,842,095 | 14,984 | # Sum of series
Benny
Hi, I'm having trouble getting the sum of the following series. I'm pretty sure that it is convergent.
$$\sum\limits_{n = 1}^\infty {\frac{{\left( { - 3} \right)^n }}{{7^n }}}$$
Here is what I have done.
$$\sum\limits_{n = 1}^\infty {\frac{{\left( { - 3} \right)^n }}{{7^n }}} = - \frac{3}{7} + \frac{9}{{49}} - \frac{{27}}{{343}} + ...$$
$$s_1 = - \frac{3}{7},s_2 = - \frac{{12}}{{49}},s_3 = - \frac{{111}}{{343}}$$
The sum seems to be heading towards a negative number.
I think I need to deduce a general form for s_n and then take the limit as n goes to infinity. I'm having trouble with this. I've only been able to come up with:
$$s_n = \frac{{\left( { - 1} \right)\left( {something + ve} \right)}}{{7^n }}$$
I've decided to take a slightly different approach to this.
$$s_n = \frac{{7\left( {7^{n - 1} s_{n - 1} } \right) - 3^n }}{{7^n}} = s_{n - 1} - \left( {\frac{3}{7}} \right)^n$$
$$s_n = s_{n - 2} - \left( {\frac{3}{7}} \right)^{n - 1} - \left( {\frac{3}{7}} \right)^n$$
$$s_n = s_0 - \sum\limits_{k = 1}^n {\left( {\frac{3}{7}} \right)} ^k$$
$$\sum\limits_{n = 1}^\infty {\frac{{\left( { - 3} \right)^n }}{{7^n }}} = \mathop {\lim }\limits_{n \to \infty } s_n = - \frac{3}{7} - \sum\limits_{k = 1}^\infty {\left( {\frac{3}{7}} \right)} ^k = - \frac{3}{7} - \left( {\frac{{\frac{3}{7}}}{{1 - \frac{3}{7}}}} \right) = - \frac{{33}}{{28}}$$
Ok well that's my clumsy attempt. Surely there must be an easier way than needing to solve a recurrence relation. Any help appreciated.(I don't have the answer)
Last edited:
Homework Helper
geometric series
if |x|<1
$$\sum_{k=n}^\infty x^k=\frac{x^n}{1-x}$$
Benny
Thanks for your response lurflurf. The series I've got looks like an alternating series so can I just use that formula? I'm probably not seeing something very obvious but any further assistance would be good thanks.
Edit: Nevermind, I checked my book and it appears that I can apply that formula. Thanks again for the help.
Last edited:
iNCREDiBLE
$$\sum_{k=n}^\infty x^k=\frac{1}{1-x}$$ for |x| < 1. | 749 | 2,056 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2023-06 | latest | en | 0.805595 |
https://www.physicsforums.com/threads/electron-displacement-between-two-parallel-plates.575927/ | 1,542,796,104,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039747665.82/warc/CC-MAIN-20181121092625-20181121114625-00296.warc.gz | 969,247,530 | 13,314 | # Homework Help: Electron displacement between two parallel plates
1. Feb 9, 2012
### SoundZombie
The problem is attached, stuck on part d.
1. The problem statement, all variables and given/known data
Im given a problem with two horizontal and parallel plates 8cm long 3cm away from each other. An electron is moving through them with a velocity V$_{0}$. It is also given that the voltage across the plates is 20 Volts and the electron deflects upwards 1.2 cm.
2. Relevant equations
Ive done all the work for parts a, b, and c, which gave me a F$_{electric}$ of 666.67 V/m or 666.67 N/C down, and an acceleration of 1.17 * 10$^{14}$ m/s$^{2}$ up.
3. The attempt at a solution
I have no idea how to even come at this. The only idea I had without being given an initial velocity in any direction is something with the deflection of 1.2 cm, but I dont know how to use that without knowing the amount of time the electron spends between the plates.
Thanks for ANY help.
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Last edited: Feb 9, 2012
2. Feb 9, 2012
### Staff: Mentor
Hello SoundZombie. Welcome to Physics Forums.
You've calculated an acceleration for the electron in the vertical direction, and you're given the vertical deflection that it experiences over its time between the plates. What can you make from that? What kinematic equation might apply?
3. Feb 9, 2012
### SoundZombie
I knew I was overlooking something small. haha
d$_{y}$=v$_{0y}$t+1/2a$_{y}$t$^{2}$ which will end up being 1.2cm=(0)t+(1/2)(1.17*10$^{14}$m/s$^{2}$)t
Im still having a little trouble remembering all of these because I took Physics 1 about 4 or 5 years ago, and Im just taking Physics 2 now. Im trying to re-learn everything at once. Thanks for the quick reply. Youre a life-saver!
Last edited: Feb 9, 2012 | 526 | 1,861 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2018-47 | latest | en | 0.920246 |
https://www.mathcelebrity.com/community/threads/carol-bought-a-pair-of-jeans-for-12-95-and-a-belt-for-3-79-the-sales-tax-is-1-01-carol-gave-the.2538/ | 1,686,274,324,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224655244.74/warc/CC-MAIN-20230609000217-20230609030217-00007.warc.gz | 953,986,381 | 9,165 | # Carol bought a pair of jeans for \$12.95 and a belt for \$3.79. The sales tax is \$1.01. Carol gave the
Discussion in 'Calculator Requests' started by math_celebrity, Apr 9, 2020.
Tags:
Carol bought a pair of jeans for \$12.95 and a belt for \$3.79. The sales tax is \$1.01. Carol gave the store clerk a \$20.00 bill. How much change should she get back?
Calculate total cost:
Total cost = Jeans + Belt + Sales Tax
Total cost = \$12.95 + \$3.79 + \$1.01
Total cost = \$17.75
Calculate Change
Change = Carol's payment - Total cost
Change = \$20 - \$17.75
Change = \$2.25 | 176 | 576 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2023-23 | latest | en | 0.907461 |
http://stats.stackexchange.com/questions/37502/lognormal-standard-error | 1,386,630,555,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164000905/warc/CC-MAIN-20131204133320-00057-ip-10-33-133-15.ec2.internal.warc.gz | 225,663,546 | 13,291 | # Lognormal Standard Error
What is the Standard Error of the Lognormal distribution? I am particularly interested in comparing two probabilities from the distribution. I have the two proportions based on experiments, in general these proportions follow a lognormal fit.
The standard error of differences from a normal distribution is straightforward (I've seen one other version of this which I think can see on wikipedia):
$SE = \sqrt{\frac{p (1-p)}{n} + \frac{q (1-q)}{m}}$
where $p, q$ are probabilities and $n, m$ sample sizes, expected to be from a normally distributed sample.
How would this be extended to the lognormal?
Edit:
The basic idea is to learn if the difference between two probabilities would be significant or not. I don't think I'm expressing my question correctly as I'm not a strict statistician.
-
When you mention standard error are you talking about the standard deviation of the sample mean from lognormal data or for the difference in means from two samples from possibly different lognormal distirbutions? What you presented above is the standard error for the difference of two independent estimates of proportion for two BINOMIAL distributions and not the difference of sample means for two NORMAL distributions, – Michael Chernick Sep 18 '12 at 16:34
Standard error of the differences between the two independently sampled probabilities. I've used the SE equation above to represent survey data, which is assumed to be normal, I'm noting that I need a Standard Error calculation for a lognormal. If the standard error for the normal distribution is different than above, feel free to share. This is just what was taught to me as part of survey analysis. – Lillian Milagros Carrasquillo Sep 18 '12 at 16:38
What, precisely, do you mean by "probabilities from the distribution" (in your question) and "sampled probabilities" (in your comment)? When we sample things from a lognormal distribution, we observe (nonnegative) numbers, not probabilities; and when we estimate values, they are usually parameters like the geometric mean and geometric standard deviation (not probabilities). – whuber Sep 18 '12 at 16:50
To go along with what whuber said the formula that you learned is for a proportion and is the standard error for the diffference between two independent binomial proportions not two normal means. However the confusion could be that when this was discussed n and m are large and the normal approximation to the binomial is used. Then the binomial variables which are sums of independent Bernoulli random variables can be taken to be approximately normal and the difference of their sample means will have the expression you gave for the standard error of the mean difference. – Michael Chernick Sep 18 '12 at 17:04
The variance of a lognormal is [exp(σ$^2$)-1] exp(2(σ+μ)) where μ and σ are the mean and standard deviation parameters of the related normal distribution. Now the maximum likelihood estimate of the lognormal variance would be obtained by plugging in the maximum likelihood estimates of μ and σ into the expression above for the variance. Call that estimate V.
The mean of the lognormal is M = exp(μ+σ$^2$/2) and its mle would be obtained by plugging in the mles for μ and σ in its expression. Call that estimate E
However if the ordinary sample means were used to estimate the population means for the lognormal distirbutions then standard error for those means would be the population standard deviation divided by √n where n is the sample size for the lognormal. This could be estimated by √(V/n). So if we compared means for the lognormal by their sample mean difference the standard error of the estimate would be √(V$_1$/n$_1$+V$_2$/n$_2$) where V$_1$ and V$_2$ are the respective mles for the lognormal variances for populations 1 and 2 respectively and n$_1$ and n$_2$ are the corresponding sample sizes taken from populations 1 and 2. | 866 | 3,911 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2013-48 | latest | en | 0.928499 |
http://myreckonings.com/Dead_Reckoning/Chapter_5/Chapter_5.htm | 1,548,079,310,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583792784.64/warc/CC-MAIN-20190121131658-20190121153658-00444.warc.gz | 151,495,337 | 5,742 | # MyReckonings.com
## Dead Reckoning: Calculating Without Instruments
Chapter 5: Trigonometric Functions and Their Inverses
Background and Topics
This chapter provides convenient 3- to 4-digit approximations to the sine, cosine, tangent, arcsine, arccosine and arctangent functions in units of degrees or radians. Graphs provide error curves for each method as a means of evaluating them.
Notes and Errata (If you have any more to contribute, please email me! This is a compendium of feedback) A printer-friendly summary table for all chapters is found here.
Color Code Type Meaning Note A clarification or elaboration of the text. Typo A simple mistake that does not affect the method presented. Error An error that may affect a method or the reader interpretation of it.
Page Code Explanation General It is my preference in this book when expressing a decimal value that continues on past the digits that are displayed, that the last digit shown is rounded off, followed by ellipses (“…”). For example, a value of log 2 = 0.3010299956…when shown to 5 decimal places will be given as 0.30103… rather than 0.30103 or 0.30102… This helps a great deal when comparing an approximation to an exact value to a certain number of places. 147 The value 174 in Equation 22 is different from the rounded value of 174.51 from the previous equation because 174 provides a better approximation over the whole range. The adjustment of this value for overall accuracy is also done in Equation 23 and Equation 27. 148 The initial slope of the sine function should be given as 0.017453. 151-152 I say that the sine approximation is valid in the range where the cosine approximation is invalid, and vice-versa. The next two sentences show what I really meant there--that the range of angles in which the sine approximation is valid (0-54 degrees) and the range of angles in which the cosine approximation is valid (0-36 degrees) are convenient because for a>54, we can replace sin(a) with cos(90-a) and watch the sign on the answer, and for b>36, we can replace cos(b) with sin(90-b). 152 In the equation that merges Equations 23 and 24, the coefficient would actually be 171.43, but is given as 172 to match the formula given by Ozanam. 153 The value 0.53168 for tan 28º would be 0.53166 if the sine and cosine values are taken only to the 4 decimal places given in the earlier calculations. 154 The reference to Equation 23 should be to Equation 22. 156 In the top equation, “28x30/40” should read “28x30/20”. The result is correct. To get the result 0.53145 for tan 28º, tan 30º must be known to 5 decimal places. The intent here is to find the ultimate accuracy possible from the formula. New methods of approximating the tangent function and the hyperbolic tangent function can be found in the paper Fast Approximation of the Tangent, Hyperbolic Tangent, Exponential and Logarithmic Functions described and linked in the Additional Materials section below. 157 In Equation 33, the sign on .5 should be reversed. The caption for Figure 6 should indicate that the sign of the y-value is reversed for the plot from Equation 33 to show the intersection of the absolute values of the errors of the two plots. 158 The phrase “dropping a b^4 term” should read “dropping a b^3 term”. 160 The caption for Figure 8 should indicate that the sign of the y-value is reversed for the plot optimized for x<=.707 to show the intersection of the absolute values of the errors of the two plots. 161 One of the g subscripts is a letter “o” instead of a number “0”. 165 The arctanh formula should use ln.
Additional Materials Related to Topics in This Chapter
Fast Approximation of the Tangent, Hyperbolic Tangent, Exponential and Logarithmic Functions: A few methods of calculating the tangent function are given in the book, but the one most conducive to mental calculation provides only three-digit accuracy, even with the correction added from the table provided. This paper derives new methods of mine that provide fast calculations of the tangent and hyperbolic tangent functions to an accuracy of essentially four digits. (It is also my first paper formatted using the LaTeX mathematical typesetting language.)
Cited Reference Materials
In the near future, this section will contain relevant sections of some of the references cited at the end of this chapter. | 1,002 | 4,349 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2019-04 | longest | en | 0.862457 |
null | null | null | null | null | null | Webopedia on Google+Webopedia on TwitterWebopedia on FacebookTech Bytes Blog
Main » TERM » N »
network neutrality
Network neutrality or net neutrality, as it is abbreviated, is the term used to describe networks that are open to equal access to all . They are non-discriminatory as they do not favor any one destination or application over another. Due to the political debate in 2006, the definition of network neutrality has changed to mean those who run networks as opposed to the network itself, where net neutrality is generally understood to mean that the service and telecommunication providers do not discriminate against rivals or individuals when they charge fees or when they prioritize traffic. Net neutrality is a major issue as the U.S. considers new telecommunications laws. In a general sense, it is mainly supported by companies that provide services at the edge of the network, and is generally opposed by companies that manage the middle of the network.
The Great Data Storage Debate: Is Tape Dead?
Apple Pay Promises to Strengthen Payment Security
Internet of Things Shaping IT's Future
Webopedia Polls
How to Create a Desktop Shortcut to a Website
Flash Data Storage Vendor Trends
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https://www.geeksforgeeks.org/program-to-find-slant-height-of-cone-and-pyramid/?ref=rp | 1,660,253,051,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571502.25/warc/CC-MAIN-20220811194507-20220811224507-00461.warc.gz | 705,213,931 | 28,126 | # Program to find slant height of cone and pyramid
• Difficulty Level : Easy
• Last Updated : 07 Aug, 2022
Given two integers H1 and R representing the height and radius of a cone and two integers H2 and S representing the height and length of base of a pyramid, the task is to find the slant height of the cone and the pyramid.
Examples:
Input: H1 = 4.5, R = 6, H2 = 4, S = 4.8
Output:
Slant height of cone is: 7.5
Slant height of pyramid is: 4.66476
Input: H1 = 2, R = 4, H2 = 4, S = 8
Output:
Slant height of cone is: 4.47214
Slant height of pyramid is: 5.65685
Approach: The slant height of an object such as a cone or a pyramid is the distance measured from any vertex along a lateral face to the base (along the center of the face). The slant height of a right circular cone is uniform throughout the surface and is given by the formula:
where,
L is the slant height of the right circular cone
R is the radius of the right circular cone and
H is the height of the right circular cone
The slant height of a pyramid is given by the formula:
where,
L is the slant height of the pyramid
S is the side of the base of the pyramid
H is the height of the pyramid
Below is the implementation of the above approach:
## C++
// C++ program for the above approach#include using namespace std; // Function to calculate slant// height of a conevoid coneSlantHeight(double cone_h, double cone_r){ // Store the slant height of cone double slant_height_cone = sqrt(pow(cone_h, 2) + pow(cone_r, 2)); // Print the result cout << "Slant height of cone is: " << slant_height_cone << '\n';} // Function to find the slant// height of a pyramidvoid pyramidSlantHeight(double pyramid_h, double pyramid_s){ // Store the slant height of pyramid double slant_height_pyramid = sqrt(pow(pyramid_s / 2, 2) + pow(pyramid_h, 2)); // Print the result cout << "Slant height of pyramid is: " << slant_height_pyramid << '\n';} // Driver Codeint main(){ // Dimensions of Cone double H1 = 4.5, R = 6; // Function Call for slant height // of Cone coneSlantHeight(H1, R); // Dimensions of Pyramid double H2 = 4, S = 4.8; // Function to calculate // slant height of a pyramid pyramidSlantHeight(H2, S); return 0;}
## Java
// Java program for the above approachimport java.io.*;class GFG{ // Function to calculate slant // height of a cone static void coneSlantHeight(double cone_h, double cone_r) { // Store the slant height of cone double slant_height_cone = Math.sqrt(Math.pow(cone_h, 2) + Math.pow(cone_r, 2)); // Print the result System.out.println("Slant height of cone is: " + slant_height_cone); } // Function to find the slant // height of a pyramid static void pyramidSlantHeight(double pyramid_h, double pyramid_s) { // Store the slant height of pyramid double slant_height_pyramid = Math.sqrt(Math.pow(pyramid_s / 2, 2) + Math.pow(pyramid_h, 2)); // Print the result System.out.println("Slant height of pyramid is: " + slant_height_pyramid); } // Driver Code public static void main (String[] args) { // Dimensions of Cone double H1 = 4.5, R = 6; // Function Call for slant height // of Cone coneSlantHeight(H1, R); // Dimensions of Pyramid double H2 = 4, S = 4.8; // Function to calculate // slant height of a pyramid pyramidSlantHeight(H2, S); }} // This code is contributed by AnkThon
## Python3
# Python 3 program for the above approachfrom math import sqrt,pow# Function to calculate slant# height of a conedef coneSlantHeight(cone_h, cone_r): # Store the slant height of cone slant_height_cone = sqrt(pow(cone_h, 2) + pow(cone_r, 2)) # Print the result print("Slant height of cone is:",slant_height_cone) # Function to find the slant# height of a pyramiddef pyramidSlantHeight(pyramid_h, pyramid_s): # Store the slant height of pyramid slant_height_pyramid = sqrt(pow(pyramid_s/2, 2) + pow(pyramid_h, 2)) # Print the result print("Slant height of pyramid is:","{:.5f}".format(slant_height_pyramid)) # Driver Codeif __name__ == '__main__': # Dimensions of Cone H1 = 4.5 R = 6 # Function Call for slant height # of Cone coneSlantHeight(H1, R); # Dimensions of Pyramid H2 = 4 S = 4.8 # Function to calculate # slant height of a pyramid pyramidSlantHeight(H2, S)
## C#
// C# program for the above approachusing System;public class GFG{ // Function to calculate slant // height of a cone static void coneSlantHeight(double cone_h, double cone_r) { // Store the slant height of cone double slant_height_cone = Math.Sqrt(Math.Pow(cone_h, 2) + Math.Pow(cone_r, 2)); // Print the result Console.WriteLine("Slant height of cone is: " + slant_height_cone); } // Function to find the slant // height of a pyramid static void pyramidSlantHeight(double pyramid_h, double pyramid_s) { // Store the slant height of pyramid double slant_height_pyramid = Math.Sqrt(Math.Pow(pyramid_s / 2, 2) + Math.Pow(pyramid_h, 2)); // Print the result Console.WriteLine("Slant height of pyramid is: " + slant_height_pyramid); } // Driver Code public static void Main (string[] args) { // Dimensions of Cone double H1 = 4.5, R = 6; // Function Call for slant height // of Cone coneSlantHeight(H1, R); // Dimensions of Pyramid double H2 = 4, S = 4.8; // Function to calculate // slant height of a pyramid pyramidSlantHeight(H2, S); }} // This code is contributed by AnkThon
## Javascript
Output:
Slant height of cone is: 7.5
Slant height of pyramid is: 4.66476
Time Complexity: O(sqrt(logH1+logR) + sqrt(logH2+logS))
Auxiliary Space: O(1)
My Personal Notes arrow_drop_up | 1,801 | 6,220 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2022-33 | latest | en | 0.747019 |
https://goodmancoaching.nl/logical-reasoning-what-is-the-difference-between-not-elimination-and-false-introduction/ | 1,709,613,020,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707948217723.97/warc/CC-MAIN-20240305024700-20240305054700-00887.warc.gz | 281,862,474 | 19,261 | Logical reasoning: What is the difference between NOT-elimination and False-introduction?
Contents
What is false elimination?
1. 2 False elimination. A funny one: Explanation: if we achieved the conclusion that is true, then we have already achieved a state where we can invent anything and affirm that it’s true; at least, as true as the idea of. (false) being true.
What is the key difference between logic and reasoning?
The primary difference between logic and reason is that reason is subject to personal opinion, whereas logic is an actual science that follows clearly defined rules and tests for critical thinking. Logic also seeks tangible, visible or audible proof of a sound thought process by reasoning.
What are introduction rules?
Introduction to Logic. Or Introduction. Or Introduction is a rule of inference that allows us to infer an arbitrary disjunction so long as at least one of the disjuncts is already in the proof. If a proof contains a sentence φi, then we can infer any disjunction that contains φi.
How do you do negation elimination?
Of one of the negations. That is the main negation in the assumption. So we assume not P reason to Q and not Q. And then from the entire sub proof we can reason to P.
What is the elimination rule?
In propositional logic, conjunction elimination (also called and elimination, ∧ elimination, or simplification) is a valid immediate inference, argument form and rule of inference which makes the inference that, if the conjunction A and B is true, then A is true, and B is true.
What is implication elimination?
Implication Elimination is a rule of inference that allows us to deduce the consequent of an implication from that implication and its antecedent.
What is the introduction?
An introduction is the first paragraph of your paper. The goal of your introduction is to let your reader know the topic of the paper and what points will be made about the topic. The thesis statement that is included in the introduction tells your reader the specific purpose or main argument of your paper.
What are the five logical connectives?
Commonly used connectives include “but,” “and,” “or,” “if . . . then,” and “if and only if.” The various types of logical connectives include conjunction (“and”), disjunction (“or”), negation (“not”), conditional (“if . . . then”), and biconditional (“if and only if”).
How do you use disjunction elimination?
An example in English: If I’m inside, I have my wallet on me. If I’m outside, I have my wallet on me. It is true that either I’m inside or I’m outside.
What are the different rules of inference?
Table of Rules of Inference
Rule of Inference Name
P∨Q¬P∴Q Disjunctive Syllogism
P→QQ→R∴P→R Hypothetical Syllogism
(P→Q)∧(R→S)P∨R∴Q∨S Constructive Dilemma
(P→Q)∧(R→S)¬Q∨¬S∴¬P∨¬R Destructive Dilemma
How do you prove disjunction elimination?
And then we derive T. Then we assumed L. The right disjunct front of the conjunct or the disjunction of line one and also derived T so to drive the same proposition. At both in both of the sub proves.
How do you use existential elimination?
And then outline for we make use of existential elimination relying upon line one and the sub proof contained. It lines two through three to reason to the final formula in the sub proof.
What is Skolemization in predicate logic?
Skolemization is the replacement of strong quantifiers in a sequent by fresh function symbols, where a strong quantifier is a positive occurrence of a universal quantifier or a negative occurrence of an existential quantifier. Skolemization can be considered in the context of either derivability or satisfiability.
What is existential generalization rule?
In predicate logic, existential generalization (also known as existential introduction, ∃I) is a valid rule of inference that allows one to move from a specific statement, or one instance, to a quantified generalized statement, or existential proposition.
Which of the following is the existential quantifier?
The symbol is the existential quantifier, and means variously “for some”, “there exists”, “there is a”, or “for at least one”. A universal statement is a statement that is true if, and only if, it is true for every predicate variable within a given domain.
Who is the father of quantifier logic?
Three approaches have been devised to date: Relation algebra, invented by Augustus De Morgan, and developed by Charles Sanders Peirce, Ernst Schröder, Alfred Tarski, and Tarski’s students.
How do you prove an existential statement is false?
We have known that the negation of an existential statement is universal. It follows that to disprove an existential statement, you must prove its negation, a universal statement, is true. Show that the following statement is false: There is a positive integer n such that n2 + 3n + 2 is prime.
What are the two types of quantifiers?
There are two kinds of quantifiers: universal quantifiers, written as “(∀ )” or often simply as “( ),” where the blank is filled by a variable, which may be read, “For all ”; and existential quantifiers, written as “(∃ ),” which may be read,…
What is logic statement and quantifiers?
In logic, a quantifier is a way to state that a certain number of elements fulfill some criteria. For example, every natural number has another natural number larger than it. In this example, the word “every” is a quantifier.
What is universal and existential quantifier explain with example?
The phrase “for every x” (sometimes “for all x”) is called a universal quantifier and is denoted by ∀x. The phrase “there exists an x such that” is called an existential quantifier and is denoted by ∃x.
What is quantifiers and examples?
A quantifier is a word that usually goes before a noun to express the quantity of the object; for example, a little milk. Most quantifiers are followed by a noun, though it is also possible to use them without the noun when it is clear what we are referring to. For example, Do you want some milk?
What are the three types of quantifiers?
There are mainly three types of quantifiers:
• Quantifiers for countable nouns examples.
• Quantifiers for uncountable nouns examples.
• Quantifiers for both countable and uncountable nouns examples.
Is money countable or uncountable?
However, the word money is not a countable noun. The word money behaves in the same way as other noncount nouns like water, sand, equipment, air, and luck, and so it has no plural form. You wouldn’t say “I have five money.” You would say “I have five dollars/francs/pesos/pounds.” | 1,481 | 6,561 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2024-10 | latest | en | 0.939873 |
https://slideplayer.com/slide/677061/ | 1,531,698,127,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589022.38/warc/CC-MAIN-20180715222830-20180716002830-00098.warc.gz | 798,418,823 | 27,506 | Chapter 7 Energy.
Presentation on theme: "Chapter 7 Energy."— Presentation transcript:
Chapter 7 Energy
How much work is done on a 200-kg crate that is hoisted 2 m in a time of 4 s?
a J b J c J d J Answer: D
How much work is done on a 200-kg crate that is hoisted 2 m in a time of 4 s?
a J b J c J d J Answer: D
How much power is required to raise a 200-kg crate a vertical distance of 2 m in a time of 4 s?
a W b W c W d W Answer: B
How much power is required to raise a 200-kg crate a vertical distance of 2 m in a time of 4 s?
a W b W c W d W Answer: B
When work is done twice as quickly, then the power expended is
a. half. b. the same. c. twice. d. 4 times as much. Answer: C
When work is done twice as quickly, then the power expended is
a. half. b. the same. c. twice. d. 4 times as much. Answer: C
If a charging elephant has kinetic energy, it must also have
a. potential energy. b. momentum. c. work. d. All of these. Answer: B
If a charging elephant has kinetic energy, it must also have
a. potential energy. b. momentum. c. work. d. All of these. Answer: B
A model airplane moves twice as fast as another identical model airplane. Compared with the kinetic energy of the slower airplane, the kinetic energy of the faster airplane is the same for level flight. twice as much. 4 times as much. more than 4 times as much. Answer: C
A model airplane moves twice as fast as another identical model airplane. Compared with the kinetic energy of the slower airplane, the kinetic energy of the faster airplane is the same for level flight. twice as much. 4 times as much. more than 4 times as much. Answer: C
An empty truck traveling at 10 km/h has kinetic energy
An empty truck traveling at 10 km/h has kinetic energy. How much kinetic energy does it have when its speed is doubled? a. The same KE b. Twice the KE c. 4 times the KE d. More than 4 times the KE Answer: C
An empty truck traveling at 10 km/h has kinetic energy
An empty truck traveling at 10 km/h has kinetic energy. How much kinetic energy does it have when its speed is doubled? a. The same KE b. Twice the KE c. 4 times the KE d. More than 4 times the KE Answer: C
When an increase in speed doubles the momentum of a moving body, its kinetic energy
a. increases, but less than doubles. b. doubles. c. more than doubles. d. depends on factors not stated. Answer: C
When an increase in speed doubles the momentum of a moving body, its kinetic energy
a. increases, but less than doubles. b. doubles. c. more than doubles. d. depends on factors not stated. Answer: C
A model airplane moves 3 times as fast as another identical model airplane. Compared with the kinetic energy of the slower airplane, the kinetic energy of the faster airplane is a. the same for level flight. b. twice as much. c. 4 times as much. d. more than 4 times as much. Answer: D
A model airplane moves 3 times as fast as another identical model airplane. Compared with the kinetic energy of the slower airplane, the kinetic energy of the faster airplane is a. the same for level flight. b. twice as much. c. 4 times as much. d. more than 4 times as much. Answer: D
A dog and a mouse run down the road with the same KE
A dog and a mouse run down the road with the same KE. The faster moving one is the a. dog. b. mouse. c. Both run at the same speed. d. Can’t say. Answer: B
A dog and a mouse run down the road with the same KE
A dog and a mouse run down the road with the same KE. The faster moving one is the a. dog. b. mouse. c. Both run at the same speed. d. Can’t say. Explanation: Let the equation, KE = 1/2 mv2 guide your thinking. A small mass having the same KE must have the greater speed. Answer: B
A 1-kg ball has twice the speed as a 2-kg ball
A 1-kg ball has twice the speed as a 2-kg ball. Compared with the 1-kg ball, the 2-kg ball has a. the same kinetic energy. b. twice the kinetic energy. c. 4 times the kinetic energy. d. more than 4 times the kinetic energy. Answer: B
A 1-kg ball has twice the speed as a 2-kg ball
A 1-kg ball has twice the speed as a kg ball. Compared with the 1-kg ball, the 2-kg ball has a. the same kinetic energy. b. twice the kinetic energy. c. 4 times the kinetic energy. d. more than 4 times the kinetic energy. Explanation: Let the equation, KE = 1/2 mv2 guide your thinking. Answer: B
When a car is braked to a stop, unless it is a hybrid, its kinetic energy is transformed to
a. stopping energy. b. potential energy. c. energy of motion. d. heat. Answer: D
When a car is braked to a stop, unless it is a hybrid, its kinetic energy is transformed to
a. stopping energy. b. potential energy. c. energy of motion. d. heat. Answer: D
When two identical cars, one traveling twice as fast as the other, brake to a stop using old-fashioned brakes, the faster car will skid a. the same distance. b. twice as far. c. 4 times as far. d. more than 4 times as far. Answer: C
When two identical cars, one traveling twice as fast as the other, brake to a stop using old-fashioned brakes, the faster car will skid a. the same distance. b. twice as far. c. 4 times as far. d. more than 4 times as far. Explanation: Let the equation, KE = 1/2 mv2 guide your thinking. Answer: C
Which of the following equations is most directly useful for solving a problem that asks for the distance a speeding vehicle skids in coming to a stop? a. F = ma b. Ft = ∆(mv) c. KE = 1/2 mv2 d. Fd = ∆(1/2 mv2) Answer: D
Which of the following equations is most directly useful for solving a problem that asks for the distance a speeding vehicle skids in coming to a stop? a. F = ma b. Ft = ∆(mv) c. KE = 1/2 mv2 d. Fd = ∆(1/2 mv2) Explanation: That’s right—the work-energy theorem. Answer: D
A shiny sports car at the top of a vertical cliff has a potential energy of 100 MJ relative to the ground below. Unfortunately, a mishap occurs and it falls over the edge. When it is halfway to the ground, its kinetic energy is a. 25 MJ. b. 50 MJ. c. about 100 MJ. d. more than 100 MJ. Answer: B
A shiny sports car at the top of a vertical cliff has a potential energy of 100 MJ relative to the ground below. Unfortunately, a mishap occurs and it falls over the edge. When it is halfway to the ground, its kinetic energy is a. 25 MJ. b. 50 MJ. c. about 100 MJ. d. more than 100 MJ. Answer: B
A pendulum bob swings to and fro
A pendulum bob swings to and fro. Its kinetic energy and its potential energy relative to the bottom of its swing are the same at a. the bottom. b. one-quarter the vertical distance between the bottom and top of the swing. c. one-half the vertical distance between the bottom and the top of the swing. d. the top of the swing. Answer: C
A pendulum bob swings to and fro
A pendulum bob swings to and fro. Its kinetic energy and its potential energy relative to the bottom of its swing are the same at a. the bottom. b. one-quarter the vertical distance between the bottom and top of the swing. c. one-half the vertical distance between the bottom and the top of the swing. d. the top of the swing. Answer: C
In an ideal pulley system, a woman lifts an 80-N crate by pulling a rope downward with a force of 20 N. For every 1-meter length of rope she pulls downward, the crate rises a. 50 cm. b. 45 cm. c. 25 cm. d. less than 25 cm. Answer: C
In an ideal pulley system, a woman lifts an 80-N crate by pulling a rope downward with a force of 20 N. For every 1-meter length of rope she pulls downward, the crate rises a. 50 cm. b. 45 cm. c. 25 cm. d. less than 25 cm. Answer: C
When you do 1000 J of work on a car jack and the gravitational potential energy of the car is increased by 300 J, the efficiency of the jack is a. 30%. b. 50%. c. 70%. d. 130%. Answer: A
When you do 1000 J of work on a car jack and the gravitational potential energy of the car is increased by 300 J, the efficiency of the jack is a. 30%. b. 50%. c. 70%. d. 130%. Answer: A
A simple machine CANNOT multiply
a. force. b. distance. c. energy. d. None of these. Answer: C
A simple machine CANNOT multiply
a. force. b. distance. c. energy. d. None of these. Answer: C
Which of these forms of energy is NOT renewable?
a. wind power b. solar power c. fossil-fuel power d. photovoltaic power Answer: C
Which of these forms of energy is NOT renewable?
a. wind power b. solar power c. fossil-fuel power d. photovoltaic power Answer: C
The most energy per unit mass can be extracted from
a. coal. b. petroleum. c. natural gas. d. uranium. Answer: D
The most energy per unit mass can be extracted from
a. coal. b. petroleum. c. natural gas. d. uranium. Answer: D | 2,346 | 8,536 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2018-30 | longest | en | 0.95679 |
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When mobile phones started becoming popular, ringtones quickly gained popularity. However, it seems like with smartphones (iPhone/Android) we tend to be more conservative and keep the phone default options for calls, emails or sms notifications. Here is where Pops comes in. Continue Reading | null | null | null | null | null | null | null | null | null |
https://gateoverflow.in/1386/gate2005-63 | 1,600,801,402,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400206329.28/warc/CC-MAIN-20200922161302-20200922191302-00582.warc.gz | 413,398,788 | 17,305 | # GATE2005-63
3.2k views
The following diagram represents a finite state machine which takes as input a binary number from the least significant bit.
Which of the following is TRUE?
1. It computes $1$’s complement of the input number
2. It computes $2$’s complement of the input number
3. It increments the input number
4. it decrements the input number
edited
For any binary no, FSM read input from LSB and remain unchanged till first $1$, and it complement after that
$\mathbf{100} \rightarrow \mathbf{100}$ $[1$'s complement of $100 + 1 = 011+ 1 = 100 = 2$'s complement of $100]$
$0\mathbf{10} \rightarrow 1\mathbf{10}$ $[1$'s complement of $010 + 1 = 101 + 1 = 110 = 2$'s complement of $010]$
$1010\mathbf{100} \rightarrow 0101\mathbf{100}$ $[1$'s complement of $1010100 + 1 = 0101011+ 1 = 0101100]$
Note : Underline part is unchanged (till first $1$ from lsb) then $0$'s changed to $1$'s and $1$'s changed to $0$'s
edited
1
@Praveen sir why to read input from LSB and remain unchanged till first 1 ? not getting the logic
would you explain in details with examples
5
It's mentioned in the question to read input from LSB and the bits doesn't change until the 1st 1 from LSB because when you calculate 2's complement for example
10110 --> 01001 ---> then when you'll add 1 to this number it will carry over till it finds 0 which automatically makes the same as input
01001 + 1 = 01010
10 is the same as input and all other bits are inverted
B.
the given above is a mealy machine for finding 2's complement.
0
Can you please explain how this F.A will work on input 100....
I am getting 001
Starting from L.S.B ....input 0 we will get output as 0, input 0 we will get output 0, input 1 we will get output as 1
Thus the final string which is generated is 001
But the 2's complement of 100 is 100.......
0
If your input is 100(1 being MSB and 0 being LSB) then input for FA is in the order 001(LSB first)
q0 --0--> q0 --0--> q0 --1-->q1
o/p:0 0 1
LSB MSB
input : 100(1 being MSB and 0 being LSB)
output : 100(1 being MSB and 0 being LSB)
## Related questions
1
2.4k views
Consider the languages: $L_1 = \left\{ww^R \mid w \in \{0, 1\}^* \right\}$ $L_2 = \left\{w\text{#}w^R \mid w \in \{0, 1\}^* \right\}$, where $\text{#}$ is a special symbol $L_3 = \left\{ww \mid w \in \{0, 1\}^* \right\}$ Which one of the following is TRUE? $L_1$ is a deterministic CFL $L_2$ is a deterministic CFL $L_3$ is a CFL, but not a deterministic CFL $L_3$ is a deterministic CFL
Let $L_1$ be a recursive language, and let $L_2$ be a recursively enumerable but not a recursive language. Which one of the following is TRUE? $L_1$' is recursive and $L_2$' is recursively enumerable $L_1$' is recursive and $L_2$' is not recursively enumerable $L_1$' and $L_2$' are recursively enumerable $L_1$' is recursively enumerable and $L_2$' is recursive
Let $N_f$ and $N_p$ denote the classes of languages accepted by non-deterministic finite automata and non-deterministic push-down automata, respectively. Let $D_f$ and $D_p$ denote the classes of languages accepted by deterministic finite automata and deterministic push-down automata respectively. Which one ... $D_f = N_f \text{ and } D_p = N_p$ $D_f =N_f \text{ and } D_p \subset N_p$
Consider the machine $M$: The language recognized by $M$ is: $\left\{ w \in \{a, b\}^* \text{ | every a in$w$is followed by exactly two$b$'s} \right\}$ $\left\{w \in \{a, b\}^* \text{ | every a in$w$is followed by at least two$b$'s} \right\}$ ... $contains the substring$abb$'} \right\}$ $\left\{w \in \{a, b\}^* \text{ |$w$does not contain$aa$' as a substring} \right\}$ | 1,187 | 3,672 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2020-40 | latest | en | 0.814464 |
https://www.vedantu.com/question-answer/how-do-you-solve-7n-5-5n-+-7-class-10-maths-cbse-601161bd925e103366c11f81 | 1,709,610,489,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707948217723.97/warc/CC-MAIN-20240305024700-20240305054700-00793.warc.gz | 1,030,933,490 | 28,443 | Courses
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# How do you solve $7n - 5 = 5n + 7$?
Last updated date: 04th Mar 2024
Total views: 340.8k
Views today: 3.40k
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340.8k+ views
Hint: In the given problem we need to solve this for ‘n’. We can solve this using the transposition method. The common transposition method is to do the same thing (mathematically) to both sides of the equation, with the aim of bringing like terms together and isolating the variable (or the unknown quantity). That is we group the ‘n’ terms one side and constants on the other side of the equation.
Complete step-by-step solution:
Given, $7n - 5 = 5n + 7$.
We transpose ‘5n’ which is present in the right hand side of the equation to the left hand side of the equation by subtracting ‘5n’ on the left hand side of the equation.
$7n - 5n - 5 = 7$
$2n - 5 = 7$
We transpose negative 5 to the right hand side of the equation by adding 5 on the right hand side of the equation.
$2n = 7 + 5$.
$2n = 12$
Divide the whole equation by 2,
$n = \dfrac{{12}}{2}$
$\Rightarrow n = 6$
$7(6) - 5 = 5(6) + 7$
$42 - 5 = 30 + 7$
$\Rightarrow 37 = 37$ | 364 | 1,142 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2024-10 | longest | en | 0.841428 |
https://www.nagwa.com/en/videos/105126265609/ | 1,656,968,630,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104496688.78/warc/CC-MAIN-20220704202455-20220704232455-00684.warc.gz | 923,968,695 | 11,512 | # Question Video: Calculating the Experimental Probability to Find the Expected Number of Outcomes Mathematics • 7th Grade
The table shows the results from a survey where 20 students from a school were asked to pick their favorite fruit from 3 options. By calculating the experimental probability that a student’s favorite fruit is an orange, calculate how many of the 400 students in the school, would choose an orange as their favorite.
01:12
### Video Transcript
The table shows the results from a survey where 20 students from a school were asked to pick their favorite fruit from three options. By calculating the experimental probability that a student’s favorite fruit is orange, calculate how many of the 400 students in the school would choose an orange as their favorite.
So there were 20 students in the survey, and eight of them chose an orange as their favorite fruit, so eight students out of the 20 students chose orange, which reduces to two-fifths because eight and 20 can both be divided by four.
So if two-fifths of the students will choose an orange, then we need to take the 400 total students and multiply by the two-fifths, because the experimental probability of students choosing an orange based on the survey was two-fifths, so if there are 400 total students, we need to take that total and multiply by the probability of choosing an orange as their favorite fruit.
So we can put 400 over one and then multiply numerator together and multiply the denominators together. So we get 800 fifths, which reduces to 160. Therefore, out of 400 students, 160 students would choose an orange as their favorite fruit. | 342 | 1,640 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2022-27 | longest | en | 0.975376 |
https://maslinandco.com/5975782 | 1,675,792,519,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500628.77/warc/CC-MAIN-20230207170138-20230207200138-00420.warc.gz | 388,864,745 | 3,682 | # Evaluate: 2-5(x+5)=3(x-2)-1
## Expression: $2-5\left( x+5 \right)=3\left( x-2 \right)-1$
Distribute $-5$ through the parentheses
$2-5x-25=3\left( x-2 \right)-1$
Distribute $3$ through the parentheses
$2-5x-25=3x-6-1$
Calculate the difference
$-23-5x=3x-6-1$
Calculate the difference
$-23-5x=3x-7$
Move the variable to the left-hand side and change its sign
$-23-5x-3x=-7$
Move the constant to the right-hand side and change its sign
$-5x-3x=-7+23$
Collect like terms
$-8x=-7+23$
Calculate the sum
$-8x=16$
Divide both sides of the equation by $-8$
$x=-2$
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https://www.enotes.com/homework-help/prove-that-sin-4-x-cos-4x-2sin-2x-1-464842 | 1,685,275,285,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224643784.62/warc/CC-MAIN-20230528114832-20230528144832-00565.warc.gz | 811,633,372 | 17,436 | # Prove that sin^4 x - cos^4x = 2sin^2x - 1
The trigonometric identity `sin^4x - cos^4x = 2*sin^2x - 1` has to be proved.
Start from the left hand side.
`sin^4x - cos^4x`
Use the relation `x^2 - y^2 = (x - y)(x +y)`
= `(sin^2x - cos^2x)(sin^2x + cos^2x)`
Use the property `sin^2x + cos^2x = 1`
= `(sin^2x - cos^2x)`
= `sin^2x - (1 - sin^2x)`
= `sin^2x - 1 + sin^2x`
= `2*sin^2x - 1`
This proves that `sin^4x - cos^4x = 2*sin^2x - 1`
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Approved by eNotes Editorial Team | 262 | 655 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2023-23 | latest | en | 0.827535 |
https://philoid.com/question/30683-if-the-product-of-the-zeros-of-the-quadratic-polynomial-x-2-4x-k-is-3-then-write-the-value-of-k | 1,708,597,365,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473738.92/warc/CC-MAIN-20240222093910-20240222123910-00196.warc.gz | 460,995,722 | 7,887 | ##### If the product of the zeros of the quadratic polynomial x2 ‒ 4x + k is 3 then write the value of k.
It is given in the question that, zero of the polynomial x2 – 4x + k is 3
Now by using the relationship between the zeros of the quadratic polynomial we have:
Sum of zeros =
3 = k/1
k = 3
12 | 86 | 302 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2024-10 | latest | en | 0.899493 |
https://openlab.citytech.cuny.edu/poiriermat1475hspring2014/2014/03/08/test-review/ | 1,660,731,414,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572898.29/warc/CC-MAIN-20220817092402-20220817122402-00705.warc.gz | 409,828,784 | 19,920 | # test review
1) let f(x)= $\frac{1}{x^2+1}$ and g(x)= x^2+5x-1
differentiate:
f(x)+g(x), f(x)*g(x), f(x)/g(x),( f(x))^g(x) and $\sqrt{g(x)}$
2) f(x)= $\frac{x+2}{1-|x|}$
a) evalute the continuity at x=0
b) Is the function derivable at x=0?
This entry was posted in Test #1 Review. Bookmark the permalink.
### 1 Response to test review
1. NaiBing says:
Sorry for no latex, The latex tester don’t allow me to copy and paste at this time so I just type out.
f ‘(x)=-(x^2+1)^(-2)(2x) g'(x)=2x+5
So By differentiate:
f(x)+g(x)= (-2x)(x^2+1)^(-2) + 2x+5 By Sum rule
f(x)g(x)=-(x^2+1)^(-2)(2x) ( x^2+5x-1)+(1/(x^2+1))(2x+5) By product rule
f(x)/g(x)=(-(x^2+1)^(-2)(2x) ( x^2+5x-1)-(1/(x^2+1))(2x+5))/(x^2+5x-1)^2 By quotient rule
For f(x)^g(x) look complicated but I tried.And I am not sure it is right or wrong. | 350 | 816 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 3, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2022-33 | latest | en | 0.705484 |
http://blog.csdn.net/humanxing/article/details/6766659 | 1,519,299,439,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814101.27/warc/CC-MAIN-20180222101209-20180222121209-00464.warc.gz | 40,609,652 | 14,514 | # Matlab实现——严格对角占优三对角方程组求解（高斯赛尔德Gauss-Seidel迭代、超松弛）
——(1)
其系数矩阵是对角的，且元素满足严格对角占优：
1)追赶法：
——(2)
——(3)
——(4)
function X=trisys(A,D,C,B)
%Input- A is the subdiagonal of the coefficient matrix
% - D is the main diagonal of the coefficient matrix
% - C is the superdiagonal of the coefficient matrix
% - B is the constant vector of the linear system
%Output - X is the solution vector
N=length(B);
X=zeros(N,1);
for k=2:N
mult=A(k-1)/D(k-1);
D(k)=D(k)-mult*C(k-1);
B(k)=B(k)-mult*B(k-1);
end
X(N)=B(N)/D(N);
for k= N-1:-1:1
X(k)=(B(k)-C(k)*X(k+1))/D(k);
end
2)迭代法(采用改进的Gauss-Seidel迭代)（这个方法是看了超松弛迭代后，得出的类似方法）：
——(1)
【1】 给一个初始列向量：
2】利用迭代公式：
(注意：当ω=1时，就是我们所熟悉的Gauss-Seidel迭代)
ω是迭代加速的相关系数——松弛因子
function X=acc(A,D,C,B,P,delta, max1,w)
%Input- A is the subdiagonal of the coefficient matrix
% - D is the main diagonal of the coefficient matrix
% - C is the superdiagonal of the coefficient matrix
% - B is the constant vector of the linear system
% - P is an N x 1 matrix; the initial guess
% - w is the convergence multiplicate
% - delta is the tolerance for P
% - max1 is the maximum number of iterations
% Output - X is an N x 1 matrix: the gauss-seidel approximation
% to the solution of AX = B
N = length(B);
L=P; %L is a mediut
for k=1:max1 %max1th iteration
X=L; %initial the X=[x1;x2;…;xN]=L=[d01;d02;…;d0N]
% the kth iteration of valuing the X
for j=1:N
if j==1
X(1)=(1-w)*X(1)+w*(B(1)-C(1)*X(2))/D(1);
elseif j==N
X(N)=(1-w)*X(N)+w*(B(N)-A(N-1)*X(N-1))/D(N);
else
%X contains the kth approximations
X(j)=(1-w)*X(j)+w*(B(j)-A(j-1)*X(j-1)-C(j)*X(j+1))/D(j);
end
end
err=abs(norm(X-L)); %get the error
L=X;
relerr=err/(norm(X)+eps);
if (err<delta)|(relerr<delta) %fit the over condition of iteration
break
end
end
在求解该问题的过程中，对于求解方程组的方法选择是一个很重要的因素，注意到这个系数矩阵是50阶严格对角占优三对角稀疏矩阵，查询了相关知识后，我个人认为，50阶的严格对角占优三对角稀疏矩阵，完全可以用高斯消去法，这是因为高斯消去后的上三角(或者下三角)仍然是严格对角占优，而对于这个稀疏矩阵，迭代法是一个非常不错的选择，而我采取的迭代法受限制的就是这个松弛因子w，注意到0<w≤1的时候，该方法是任何初始向量P都收敛，于是采取了w=0：0.2：1的选择方式，最后发现w=1附近的时候误差相对较小(有点郁闷，针对这个三对角矩阵时没能达到加速的目的)。总之，迭代法的舍入误差随着迭代次数的增加，能达到相当高的精度；而且收敛速度令人满意。
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## 追赶法求解三对角方程组
• u010450214
• 2015年12月04日 09:37
• 7695
## 三对角阵的LU分解和三对角方程组的求解（C语言）
/*三对角阵的LU分解和三对角方程组的求解 -------------A=LU的分解算法------- 参考教材：《数值分析》李乃成，梅立泉，科学出版社 《计算方法教程》第二版 凌永...
• zhangchao3322218
• 2012年03月30日 18:42
• 2941
举报原因： 您举报文章：Matlab实现——严格对角占优三对角方程组求解（高斯赛尔德Gauss-Seidel迭代、超松弛） 色情 政治 抄袭 广告 招聘 骂人 其他 (最多只允许输入30个字) | 1,691 | 3,561 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2018-09 | latest | en | 0.303751 |
https://www.australiancurriculumlessons.com.au/2013/01/15/problem-solving-lesson-traffic-jam/?utm_source=ReviveOldPost&utm_medium=social&utm_campaign=ReviveOldPost | 1,590,680,782,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347399820.9/warc/CC-MAIN-20200528135528-20200528165528-00356.warc.gz | 642,990,508 | 23,866 | 0
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# Summary:
This lesson involves students using a variety of strategies to solve an age old problem known as the “traffic jam.” This gives students the opportunity to work as a team and solve the problem as well as showing students that there is more than one way to solve a problem.
• Select and apply efficient mental and written strategies and appropriate digital technologies to solve problems involving all four operations with whole numbers (ACMNA123)
• Continue and create sequences involving whole numbers, fractions and decimals. Describe the rule used to create the sequence (ACMNA133)
# Lesson:
Introduction:
1. Start of by introducing the problem through a bit of a story. “I was driving along the other when all of a sudden CRASH! (try make noise to scare too!) there was a traffic jam. Out of no where a policeman came and instructed the traffic through.
2. Using Hula Hoops as circles, place 9 hoops in a row and pick out 8 students. Both team A and team B need to be facing each other as illustrated in the diagram below:
3. Explain to the students that the cars labelled A need to make to the other side and same with B. However the cars can only move one space at a time only opposite cars can go around each other (illustrated below).
4. NB* Only opposite cars can go around each other and the end result is to look like this.
Body:
1. (This can be changed to suit your class needs). I asked for 8 volunteers (and i chose the ones who were more likely to solve it) to act as cars. Tell the cars they are not allowed to talk. Now the remainder of the class are the “policeman” and as a result need to instruct the cars through using the rules of one car forward and only opposites can go around each other.
2. If you can, film the students using an ipad or video camera. Stop students half way and discuss what strategies are working an what aren’t. Tell students to try new ways as well as offer some cars to subituted in.
3. Show students the video and discuss what they are doing well and what can be done better.
4. Simulate a smaller problem with just 2 cars on each side and see if students can identify the pattern.
5. Now that they know the pattern, students should be able to solve the problem of 4 and 4. Just for a bit of fun, i got the class complete it with 10 and 10 on each side.
6. Next students will work in pairs to model the problem using counters. Get them to draw 9 dots in their maths book and simulate the problem using 2 different coloured counters.
7. Please note: For extension, ask student to work out the algebraic formula for the patter that has occurred. Is there one…?
Conclusion:
1. Upon solving the problem invite students to share how they solved the problem. Now discuss how we can solve problems using a variety of ways.
2. Get them to describe what problem solving methods were used by articulating to others what they did. (Explaining in mathematics is so vital!)
# Resources:
• 9 hula hoops
• 2 different coloured counters
• Ipad or recording device of some sort | 680 | 3,036 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2020-24 | latest | en | 0.965691 |
https://www.jiskha.com/questions/1182776/this-figure-below-describes-the-joint-pdf-of-the-random-variables-x-and-y-these-random | 1,628,151,255,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046155458.35/warc/CC-MAIN-20210805063730-20210805093730-00025.warc.gz | 849,992,157 | 5,573 | # probability
This figure below describes the joint PDF of the random variables X and Y. These random variables take values in [0,2] and [0,1], respectively. At x=1, the value of the joint PDF is 1/2.
(figure belongs to "the science of uncertainty)
1. Are X and Y independent? NO
2. Find fX(x). Express your answers in terms of x using standard notation .
If 0<x<1,
fX(x)= x/2
If 1<x<2,
fX(x)= -3*x/2+3
Find fY|X(y∣0.5).
If 0<y<1/2,
fY|X(y∣0.5)= 2
3. Find fX|Y(x∣0.5).
If 1/2<x<1,
fX|Y(x∣0.5)= 0.5
If 1<x<3/2,
fX|Y(x∣0.5)= 1.5
Let R=XY and let A be the event {X<0.5}. Evaluate E[R∣A].
E[R∣A]= 0.0625
1. 👍
2. 👎
3. 👁
1. Sophia is vacationing in Monte Carlo. On any given night, she takes X dollars to the casino and returns with Y dollars. The random variable Xhas the PDF shown in the figure. Conditional on X=x, the continuous random variable Y is uniformly distributed between zero and 3x.
This figure below describes the joint PDF of the random variables X and Y. These random variables take values in [0,2] and [0,1], respectively. At x=1, the value of the joint PDF is 1/2.
1. Are X and Y independent?
No
2. Find fX(x). Express your answers in terms of x, using the standard notation.
If 0<x≤1:
fX(x)= 1/2⋅x
If 1<x<2:
fX(x)= −3⋅x/2+3
If x<0 or x≥2:
fX(x)= 0
3. Find fY∣X(y∣0.5).
If 0<y<1/2:
fY|X(y∣0.5)= 2
If y<0 or y>1/2:
fY|X(y∣0.5)= 0
4. Find fX∣Y(x∣0.5).
If 1/2<x<1:
fX|Y(x∣0.5)=
If 1<x<3/2:
fX|Y(x∣0.5)= 1.5
If x<1/2 or x>3/2:
fX|Y(x∣0.5)= 0
5. Let R=XY and let A be the event that {X<0.5}. Find E[R|A].
E[R∣A]= 0.0625
1. 👍
2. 👎
2. 4. Find fX∣Y(x∣0.5).
If 1/2<x<1:
fX|Y(x∣0.5)=0.5
1. 👍
2. 👎
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Exercise: Convergence in probability a) Suppose that Xn is an exponential random variable with parameter λ=n. Does the sequence {Xn} converge in probability? b) Suppose that Xn is an exponential random variable with parameter | 1,520 | 4,566 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2021-31 | latest | en | 0.665335 |
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What is the saturation pressure of air?
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Temperature Saturation Vapor Pressure [10-3 bar]
[oC] [oF]
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Does saturation depend on pressure?
Saturation pressure is the pressure for a corresponding saturation temperature at which a liquid boils into its vapor phase. Saturation pressure and saturation temperature have a direct relationship: as saturation pressure is increased, so is saturation temperature.
What is water vapor saturation?
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How to calculate the ppm from vapor pressure?
How to Calculate the PPM From Vapor Pressure Defining the Term: PPM. The acronym ppm means parts per million. Defining the Term: Vapor Pressure. Vapor pressure refers to the pressure of a vapor (gas) above its liquid or solid phase when the two are in equilibrium in a Typical Reporting Units. Gas Concentration Calculation: mmHg to ppm. Gas Concentration Calculation: ppm to mg/m 3. | 895 | 4,358 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2022-33 | latest | en | 0.863403 |
https://congruentmath.com/lesson-plan/determine-solutions-to-equations-and-inequalities-through-substitution/ | 1,721,653,480,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517878.80/warc/CC-MAIN-20240722125447-20240722155447-00101.warc.gz | 145,180,352 | 26,566 | # Determine Solutions to Equations and Inequalities Through Substitution Lesson Plan
## Overview
Ever wondered how to teach how to use substitution to determine whether a given number in a specified set makes an equation or inequality true in an engaging way to your 6th-grade students? Looking for way to cover common core standards CCSS 6.EE.B.5?
In this lesson plan, students will learn about substitution in equations and inequalities and their real-life applications. Through artistic, interactive guided notes, checks for understanding, a doodle & color by number activity, and a maze worksheet, students will gain a comprehensive understanding of using substitution to determine solutions for equations and inequalities.
The guided notes provide a structured approach to teaching the topic, integrating checks for understanding to ensure that students are on the right track. Following the guided notes, students will have the opportunity to practice their skills with a fun color by code activity, a maze worksheet, and additional problem sets.
The lesson culminates with a real-life application, where students will read and write about the real-life uses of using substitution to solve mathematical problems. This application will help students see the relevance of the topic in their everyday lives.
\$4.25
## Learning Objectives
After this lesson, students will be able to:
• Use substitution to determine whether a given number in a specified set makes an equation true
• Use substitution to determine whether a given number in a specified set makes an inequality true
## Prerequisites
Before this lesson, students should be familiar with:
• Basic operations with positive whole numbers and decimals
• Understanding the concept of variables and their role in equations
• Basic knowledge of the order of operations
• Substitution
• Equations
• Inequalities
• Solutions
• Variables
• Coefficients
• Constants
• Expressions
## Procedure
### Introduction
As a hook, ask students why solving equations and inequalities is important in real life. Refer to the last page of the guided notes for ideas.
Use the first page of the guided notes to introduce the concept of substitution and how it can be used to determine solutions for equations. Then use the second page of the guided notes to introduce how to determine solutions for inequalities by substitution. Walk through the key points of using substitution, including replacing a variable with a given number and checking if the equation or inequality is true.
Based on student responses, reteach concepts that students need extra help with. If your class has a wide range of proficiency levels, you can pull out students for reteaching, and have more advanced students begin work on the practice exercises.
### Practice
Have students practice using substitution to determine solutions for equations and inequalities using the practice section included on page 1-2 of the guided notes. Walk around the classroom to answer any student questions and provide support as needed.
Fast finishers can further solidify their understanding by completing the maze activity (page 3) and color by number activity (page 4) included in the guided notes resource. You can assign it as homework for the remainder of the class.
### Real-Life Application
Use the last page of the guided notes to bring the class back together, and introduce the concept of real-world application of using substitution to determine solutions for equations and inequalities. Provide examples of situations in which this concept can be applied in daily life, such as budgeting or finding the best deals while shopping. Discuss how substitution can be used to solve equations or inequalities to find the value of a variable that satisfies a given condition. | 714 | 3,812 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2024-30 | latest | en | 0.868174 |
http://www.ck12.org/analysis/Imaginary-Numbers/lesson/Imaginary-Numbers/r11/ | 1,448,588,657,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398447881.87/warc/CC-MAIN-20151124205407-00305-ip-10-71-132-137.ec2.internal.warc.gz | 350,194,908 | 39,995 | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
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# Imaginary Numbers
## i = sqrt (-1)
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Imaginary Numbers
Have you ever had an imaginary pet? Many people have, particularly as young children.
Wouldn't you be surprised if you and your real friend left your imaginary pet dogs alone together, and you came back to find a real puppy?
Silly thought, what does it have to do with imaginary numbers?
### Watch This
'Note:' For a very detailed explanation of i and the complex numbers, watch the video below, then visit: http://betterexplained.com/articles/a-visual-intuitive-guide-to-imaginary-numbers/
Embedded Video:
### Guidance
What is the square root of -1?
You may recall running into roots of negatives in Algebra, when attempting to solve equations like: x2+4=0\begin{align*}x^2 + 4 = 0\end{align*} Since there are no real numbers that can be squared to equal -4, this equation has no real solution. Enter the imaginary constant: "i".
The definition of "i" : i=1\begin{align*}i = \sqrt{-1}\end{align*}
The use of the word imaginary does not mean these numbers are useless. For a long period in the history of mathematics, it was thought that the square root of a negative number was in fact only within the mathematical imagination, without real-world significance hence, imaginary. That has changed. Mathematicians now consider the imaginary number as another set of numbers that have real significance, but do not fit on what is called the number line, and engineers, scientists, and others solve real world problems using combinations of real and imaginary numbers (called complex numbers) every day.
Imaginary values such as 16\begin{align*}\sqrt{-16}\end{align*} can be simplified by simplifying the radical into 161\begin{align*}\sqrt{16} \cdot \sqrt{-1}\end{align*}, yielding: 41\begin{align*}4 \sqrt{-1}\end{align*} or 4i\begin{align*}4i\end{align*}.
The uses of i become more apparent when you begin working with increased powers of i, as you will see in the examples below.
Complex numbers
When you combine i's with real numbers, you get complex numbers:
The definition of complex numbers: Complex numbers are of the form a+bi\begin{align*}a + bi\end{align*}, where a\begin{align*}a\end{align*} is a real number, b\begin{align*}b\end{align*} is a constant, and i\begin{align*}i\end{align*} is the imaginary constant 1\begin{align*}\sqrt{-1}\end{align*}.
#### Example A
Simplify 5\begin{align*}\sqrt{-5}\end{align*}
Solutions
5=(1)(5)\begin{align*}\sqrt{-5} = \sqrt{(-1)\cdot (5)}\end{align*}
=15\begin{align*}= \sqrt{-1}\sqrt{5}\end{align*}
=i5\begin{align*}= i\sqrt{5}\end{align*}
#### Example B
Simplify 72\begin{align*}\sqrt{-72}\end{align*}
Solution
72=(1)(72)\begin{align*}\sqrt{-72} = \sqrt{(-1)\cdot (72)}\end{align*}
=172\begin{align*}= \sqrt{-1}\sqrt{72}\end{align*}
=i72\begin{align*}= i\sqrt{72}\end{align*}
But, we’re not done yet! Since 72=362\begin{align*}72 = 36 \cdot 2\end{align*}
i72=i362\begin{align*}i\sqrt{72} = i\sqrt{36} \sqrt{2}\end{align*}
\begin{align*}= i(6)\sqrt{2}\end{align*}
\begin{align*}= 6i\sqrt{2}\end{align*}
#### Example C
Strange things happen when the imaginary constant i is multiplied by itself different numbers of times.
a) What is i2?
b) What is i3?
c) What is i4?
Solutions
\begin{align*}i^2\end{align*} is the same as \begin{align*}(\sqrt{-1})^2\end{align*}. When you square a square root, they cancel and you are left with the number originally inside the radical, in this case \begin{align*}-1\end{align*}
a) \begin{align*}\therefore i^2 = -1\end{align*}
\begin{align*}i^3\end{align*} is the same thing as \begin{align*}i^2 \cdot i\end{align*}, which is \begin{align*}-1 \cdot i\end{align*} or \begin{align*}-i\end{align*}
b) \begin{align*}\therefore i^3 = -i\end{align*}
\begin{align*}i^4 = i^2 \cdot i^2\end{align*} which is \begin{align*}-1 \cdot -1\end{align*}
c) \begin{align*}\therefore i^4 = 1\end{align*}
Concept question wrap up
Do you see the application of the crazy analogy from the introduction?
Two imaginary pets creating a real puppy is an oddly effective metaphor for the behavior of the powers of i.
One i is imaginary, but two i's multiply to be a real number. In fact, every even power of i results in a real number!
### Vocabulary
The imaginary constant i is the square root of -1.
A complex number is the sum of a real number and an imaginary number, written in the form: \begin{align*}a + bi\end{align*}.
Simplifying the radical involves factoring the term(s) under the root symbol, so that perfect squares may be "pulled out", and simplified.
### Guided Practice
1)\begin{align*}\sqrt{-9}\end{align*}
2)\begin{align*}\sqrt{-12}\end{align*}
3)\begin{align*}\sqrt{-17}\end{align*}
4)\begin{align*}\sqrt{108-140}\end{align*}
Multiply the imaginary numbers
5) \begin{align*}4i \cdot 3i\end{align*}
6 \begin{align*}\sqrt{16}i \cdot 3\end{align*}
7) \begin{align*}\sqrt{4i^2} \cdot \sqrt{12}i\end{align*}
Solutions
1) To simplify the radical \begin{align*}\sqrt{-9}\end{align*}
\begin{align*}\sqrt{9 \cdot -1}\end{align*} : Rewrite \begin{align*}-9\end{align*} as \begin{align*}-1 \cdot 9\end{align*}
\begin{align*}\sqrt{9} \cdot \sqrt{-1}\end{align*} : Rewrite as a product of radicals
\begin{align*}\sqrt{9} \cdot i\end{align*} : Substitute \begin{align*}\sqrt{-1} \to i\end{align*}
\begin{align*}3i\end{align*} : Simplify \begin{align*}\sqrt{9}\end{align*}
2) To simplify the radical \begin{align*}\sqrt{-12}\end{align*}
\begin{align*}\sqrt{12 \cdot -1}\end{align*} : Rewrite \begin{align*}-12\end{align*} as \begin{align*}-1 \cdot 12\end{align*}
\begin{align*}\sqrt{12} \cdot \sqrt{-1}\end{align*} : Rewrite as a product of radicals
\begin{align*}\sqrt{12} \cdot i\end{align*} : Substitute \begin{align*}\sqrt{-1} \to i\end{align*}
\begin{align*}\sqrt{3 \cdot 4} \cdot i\end{align*} : Factor 12
\begin{align*}2\sqrt{3} \cdot i\end{align*} : Simplify \begin{align*}\sqrt{4}\end{align*}
3) To simplify the radical\begin{align*}\sqrt{-17}\end{align*}
\begin{align*}\sqrt{17 \cdot -1}\end{align*} : Rewrite \begin{align*}-17\end{align*} as \begin{align*}-1 \cdot 17\end{align*}
\begin{align*}\sqrt{17} \cdot \sqrt{-1}\end{align*} : Rewrite as a product of radicals
\begin{align*}\sqrt{17} \cdot i\end{align*} : Substitute \begin{align*}\sqrt{-1} \to i\end{align*}
\begin{align*}\sqrt{17}i\end{align*} : Simplify
4) To simplify the radical \begin{align*}\sqrt{108-140}\end{align*}
\begin{align*}\sqrt{-32}\end{align*} : Subtract within the parenthesis
\begin{align*}\sqrt{32 \cdot -1}\end{align*} : Rewrite \begin{align*}-32\end{align*} as \begin{align*}-1 \cdot 32\end{align*}
\begin{align*}\sqrt{32} \cdot \sqrt{-1}\end{align*} : Rewrite as a product of radicals
\begin{align*}\sqrt{32} \cdot i\end{align*} : Substitute \begin{align*}\sqrt{-1} \to i\end{align*}
\begin{align*}\sqrt{16 \cdot 2} \cdot i\end{align*} : Factor \begin{align*}32\end{align*}
\begin{align*}4i\sqrt{2}\end{align*} : Simplify \begin{align*}\sqrt{16}\end{align*}
5) To multiply \begin{align*}4i \cdot 3i\end{align*}
\begin{align*}4 \cdot 3 \cdot i \cdot i\end{align*} : Using the commutative law for multiplication
\begin{align*}12 \cdot i^2\end{align*} : Simplify
\begin{align*}12 \cdot -1\end{align*} : Recall \begin{align*}i^2 = -1\end{align*}
\begin{align*}-12\end{align*}
6) To multiply \begin{align*}\sqrt{16}i \cdot 3\end{align*}
\begin{align*}4i \cdot 3\end{align*} : Simplify \begin{align*}\sqrt{16}\end{align*}
\begin{align*}12i\end{align*}
7) To multiply \begin{align*}\sqrt{4i^2} \cdot \sqrt{12}i\end{align*}
\begin{align*}\sqrt{4} \cdot \sqrt{i^2} \cdot \sqrt{4} \cdot \sqrt{3} \cdot i\end{align*} : Factor
\begin{align*}2 \cdot i \cdot 2 \cdot \sqrt{3} \cdot i\end{align*} : Simplify the roots
\begin{align*}4\sqrt{3} \cdot i^2\end{align*} : Collect terms and simplify
\begin{align*}4\sqrt{3} \cdot -1\end{align*} : Recall \begin{align*}i^2 = -1\end{align*}
\begin{align*}-4\sqrt{3}\end{align*}
### Practice
Simplify:
1. \begin{align*}\sqrt{-49}\end{align*}
2. \begin{align*}\sqrt{-81}\end{align*}
3. \begin{align*}\sqrt{-324}\end{align*}
4. \begin{align*}\sqrt{-121}\end{align*}
5. \begin{align*}-\sqrt{-16}\end{align*}
6. \begin{align*}-\sqrt{-1}\end{align*}
7. \begin{align*}\sqrt{-1.21}\end{align*}
Simplify:
1. \begin{align*}i^8\end{align*}
2. \begin{align*}i^{12}\end{align*}
3. \begin{align*}i^3\end{align*}
4. \begin{align*}24i^{20}\end{align*}
5. \begin{align*}i^{225}\end{align*}
6. \begin{align*}i^{1024}\end{align*}
Multiply:
1. \begin{align*}i^4 \cdot i^{11}\end{align*}
2. \begin{align*}5i^6 \cdot 5i^8\end{align*}
3. \begin{align*}3\sqrt{-75} \cdot 5\sqrt{-3}\end{align*}
4. \begin{align*}2\sqrt{-12} \cdot 6\sqrt{-27}\end{align*}
5. \begin{align*}-4 \sqrt{-10} \cdot 5 \sqrt{-3} \cdot 6\sqrt {-18}\end{align*}
### Vocabulary Language: English
$i$
$i$
$i$ is an imaginary number. $i=\sqrt{-1}$.
complex number
complex number
A complex number is the sum of a real number and an imaginary number, written in the form $a + bi$.
i
i
$i$ is an imaginary number. $i=\sqrt{-1}$.
Imaginary Number
Imaginary Number
An imaginary number is a number that can be written as the product of a real number and $i$.
Imaginary Numbers
Imaginary Numbers
An imaginary number is a number that can be written as the product of a real number and $i$. | 3,286 | 9,512 | {"found_math": true, "script_math_tex": 117, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 9, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2015-48 | longest | en | 0.873228 |
null | null | null | null | null | null | tag:blogger.com,1999:blog-4143435314932633148Fri, 19 Dec 2014 12:12:22 +0000bxdesignkidsb/xcompanionrulesaddadventurefilmprogressnflbaranofgamesideasswclassnewscombatmonsterreflectionsmoduleblogliteratureprojectcomics5akOSRartworkspellsdead charactersoriginsgygaxmagic-usercampaign settingchargenholidayd20playtestreviewstolkienclerichistorywifewhite plume mountainalphacdfcollectingcrapfamilyllpalladiumrole-playingshopprintingdnd minepublishingtheorytcbxatreasureb2supersblackrazordiegomoxnephewsmoonmusicconversionelflevelweaponsbbdocmoldvaysupplementtravellerbeaglebxsklistsparaguaythiefadvancementerrorsnecromancytelevisionapologiesbecmicampaignchainmailgwinsanityland of icealienscomputerconventiondamagemzbnorsesnsbhhalflingjocelynpost-apocsfstevewarwh40kastrologyaxefifagenerationsholmeshuindiepollsrtaoOn Role-Playingadd2howardmoneysavesx1ForgearnesondccmontanamshpsionicsreligionsnowvtmNew DNDalignmentbtcontestcookdwarffighterjoshlukementzern1pbems1scorpsicktsruavancewegwitchassassinsbadasscannibalsdragondungeonbgedwardsfolkloreheartbreakershpsmailmattnostalgiapaladinscottt1-4tpkwomen studiesbardbearborgchallengedeconstructiondinosaurdlgeezergunshalloweenmearlsminisnmpinirandomschickswordABalejandrob1blackmoorboyerclusterfkevilffgodsjapjedimexicopain-sufferingq1sandtoadworldwwa1-4albedoarmoraxiombeowulfboxchaosiumdevelopmentdrunkdvexamplegoblinmnmnew schoolrandys2sotcspaintopusawar of the mechawillinghamarsart of the DMattributesbarbarianbear weekbladehawkclockworkcorruptiondesertdmgdmifhbgq1jasonlotfplucasmoorcocknilesroslovsexstages of explorationstirlingtablesthree pillarswolfwowIntroductionarcherybiblebitchin-moanincaycecharley manson specialddgdeclercqelmoreemerald city gamefestfeedbackfree rpg daygreeksgumshoeheronjoelkarmakiddkrulllitmarsmartinmcmichaelmordheimpoisonquizreturn to wpmrowlingsecretssubclasstwilight2kwfrpwormChristmasanthonyatlantisawardsbeastbonus XPcardscosmologycreepydark sundeadlandsdeadlinesdeathdealerdiofrazettag1-3gaming surveygladiatorharryhausenhatsi1illusionistinterviewjackassesjacksonkinglmzmadnessmapsmass cbtmdrmissilesmummy lordnerdogreotuspbppdfpredatorpsbeaglepulprangerretainersrobsilkstormbringersynnibarrvvwavewormwoodwwbDoSalexanderanti-paladinartifactsaspirinauto-killbeansbeggingbgbronzebsgburroughschaoschronicle o mutationcollaborative rpculturedarkoverdfdicedistributiondresdendunegermaniagibsongurpshitehmbhorror ruleshplinitiativekaycelazyluckymckinleymountebankmtgninjanudityout of timepaypalpernpiratesplutopodcastpowerss3sagseaskillssnwswntitleuniversityvoidx4LOREacrobatal-qadimandersonaquariusariesbaylessbeardbesmboot hillbsburnsbxswc1cthulhudcellis:kitfmmfunnygendergiantsgnollgorgreat commandmentsgroggrubbhardcorehaven:covhighlanderhoti2i4junglekaminskikeldernkenzerland of ashlarplawslieberm1macmaelmcbmutant futuremwnbxnumbersoo7orkpdqpendragonpixplaygroundprincepyoungrobersonrtalruined earthsavagesechishieldssindbadsnafustar blazerssteakleystickerthortorturetoystrapstweettxvernevincewildx-plorersx5zepDYKG+MentatOHDagingamberapokashenasimovasuncionballoonbastionbigGbikeblackrockbluebounty hunterbradstreetbryantburtoncapescapricorncarcosacarrollcharitychesscoffeecollaborationcrystaldcuddmdeathguarddfddgasdiversitydopedqdramaeewellisonencumbranceeoeptewaltexceptional traitsfaqfatefaustfavreaufencingffgfiascofiveqsfmjfreebiesfreudgnomegnsgothgreggrendelgritgunh1-4heavy metal maghelphexhillfolkhis dark materialshybridi3i5i6intelligenceirishkaskisskotorkuntzkwnleolost worldlpjmametmedicinemelnibonemerpmillingtonmjmodmonkmoralemorningstarmournblademovementmpcmunchkinmystarranadinarcoticsneptunenetworknkgnurgleoglorcoteoutlawpandiusparanoiapathfinderpatreonperrypiscespopoetrypolarisptsdpunch itpuppetpvpreaperred sonjaredundancyrevolutionrogues gallerys4saturnshakespeareslaveslumpsneakshadowsolost. andrestarsiegesteamsteelstonedstuntstarottaxesteabagtekumelthyatistmtmnttowntri-stattroll slayeruk1uranusvacationvehiclesvillainsvirgowestern citywg5winewocdwotrpwtwwIx/yx2yogazelaznyzerozinesB/X BLACKRAZORhttp://bxblackrazor.blogspot.com/[email protected] (JB)Blogger1520125tag:blogger.com,1999:blog-4143435314932633148.post-6834694334511881304Thu, 18 Dec 2014 20:15:00 +00002014-12-18T13:12:00.044-08:00bbnflHating on the BrownsSome days, I just can't help myself.<br /><br />When it comes to <b>Blood Bowl</b> (or most things miniature and wargamey), I'm a big lover of <b>"fluff."</b> Some people love the tactics involved in crushing one's opponent (I like this, too); some folks are downright obsessive about modeling and painting (I've been known to while away a few hours myself). But really, without the fluff, the game would lose most of its character, coolness, and humor.<br /><br />I started with <i>Blood Bowl</i> back in the 2nd edition, when the teams used to be arranged like the NFL (two conferences of three divisions each...this was was before the 2002 realignment or even the Jags/Panthers expansion franchises) and had equivalent play-offs rather than the Open Tournament format that began with 3rd edition. Back then, the fluff provided team histories that included records, stats, and rivalries akin to...well, akin to (American) football.<br /><br />[<i>you can say that these things still exist, but their importance as fluff has been significantly downgraded with the collapse of the League. When all the teams are wandering mercenary bands, that might or might not play another team in a decade, who cares what the past rivalries were? Any team can angle for a good prize purse by entering an open tournament and getting cunning with the matches they schedule. The fluff of the game loses much of its significance...which for me is a downgrade</i>]<br /><br />[<i>and, yes, I realize that the reasons behind the changed (game) format is the practical problems that comes with running a league...scheduling problems, player no-shows, and player drop-outs due to poor records and not wanting to "play out the string." That's fine...but in MY little corner of the Blood Bowl universe, I'm allowed to run the League as I see fit...and I, of course, run my league in the American NFL format. When I run it at all, that is.</i>..]<br /><br />One team that I always found amusing was <b>The Hobgoblin Team</b>, a team so stupid that its symbol was a hastily scrawled "X" on the side of their helmets, and that couldn't think up a better name for itself than "The Hobgoblin Team." Hobgoblins in the Blood Bowl world are a far cry from the organized warriors of the D&D mythos...they're a bit stronger than goblins, but a lot less intelligent, having issues even tying their boots (and when they get their boots on, they spend most of the match stomping on the feet of their fellows who couldn't figure it out). They are truly pathetic, having never had a winning record in living memory and often going 0 for 16.<br /><br />Now, there's actually no "hobgoblin" team available in 2nd or 3rd edition <i>Blood Bowl</i>, but the Chaos Dwarf team of 3E uses hobgoblin slaves to make up the bulk of its players, and it's perfectly possible to field consisting entirely of hobgoblins and calling it The Hobgoblin Team. In fact, that's what I do with one of the two chaos dwarf teams in my league...it seems important and appropriate to include such a team (in homage to the original fluff), and its easy to model the wild popularity of the hapless hobgoblins with all the money you save by not purchasing skill players. It's a pretty tough team to win with (even without adding the fan-based stupidity rules you find floating around the internet), but it gives you incentive to play low-down and dirty (which is in line with the team's fluff) and it's a nice team to have on hand to break out against a young player or someone just learning the game (that you don't want to totally thrash).<br /><br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="http://1.bp.blogspot.com/-jT8u2qLVY6w/VJM0cbx0QsI/AAAAAAAAAbs/gJZZnqGsWs8/s1600/ReD_hobgoblins.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="http://1.bp.blogspot.com/-jT8u2qLVY6w/VJM0cbx0QsI/AAAAAAAAAbs/gJZZnqGsWs8/s1600/ReD_hobgoblins.jpg" height="320" width="320" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">Not my minis, sorry (they're back in Seattle).</td></tr></tbody></table>Of course, all my Blood Bowl teams replace (and model) specific teams in the NFL. The Hobgoblin Team is the one I use to represent the<b> Cleveland Browns</b>.<br /><br />I <i>want</i> to like the Browns, I really do. But they always do such asininely stupid things during the football season that it's tough not to be irritated with them and say, 'hey, you get what you deserve.' Last year, I watched <b>Brian Hoyer</b> exceed expectations and give his team a chance to actually be competitive before having his leg nuked in the Bills game (it was a Thursday night when it happened...I was at <b>the Baranof</b> for my weekly role-playing fix and watched the game in the bar). It was a terrible bit of bad luck for the team, but I figured it at least boded well for the future...that the Browns had finally found a quarterback that could take them out of a really ugly slump.<br /><br />[<i>for folks who don't care or know much about the Browns, you might find it interesting that they've only had three winning seasons since 1990 and have finished only one season with more than <b>six wins</b> since 2002...the last time they made the play-offs. There's a reason why the Browns get The Hobgoblin Team</i>]<br /><br />[<i>for non-American readers: the American football season consists of a 16 game regular season followed by the play-offs. More than a decade with a win percentage of under .400 is, frankly, terrible</i>]<br /><br />But now they had Hoyer, and a chance to build a team that might (in time) at least match the glory days of the 1980s (seven playoff appearances from 1980-1989), if not the championships of the 1950s and 60s (prior to the Super Bowl era). There was hope.<br /><br />And then there was <b>Johnny Manziel</b>.<br /><br />The Browns selected "Johnny Football" in the first round of the draft, a hot commodity in college (if controversial selection) and set-up the same kind of nipping-at-the-heels situation the NFL saw previously with <b>Tim Tebow</b>. Even so, the Browns started Hoyer and jumped out to a 6-3 record (and share of the division lead) after their first ten weeks. With seven games remaining, the Browns looked to control their own play-off destiny.<br /><br />Then they ran into a giant buzz-saw called <b><a href="http://bxblackrazor.blogspot.com/2013/09/escaping-necromancer.html">J.J. Watt</a></b> who destroyed their offense as he's done most of this year (a year when he's being lauded as an MVP candidate). Dropping to 6-4 in a tight division race didn't help anyone's confidence.<br /><br />The next week, the Browns traveled to Georgia to play the Falcons and they managed to pull out a needed road victory bringing their win total to seven...but Hoyer threw three picks during the game and despite leading the team to their best win record since Derek Anderson, people started to ask when they'd get a chance to see Johnny Manziel. Truthfully, some folks had been asking this all season. Isn't it possible that the poor showing was due to Hoyer being forced to throw to an out-of-position <b>Josh Gordon</b>, rusty and lacking chemistry after being suspended for most of the season due to drug violations? Yes, Gordon caught eight passes...but he was targeted 16 times. He admitted himself that he ran the wrong routes on two of his targets (that were intercepted).<br /><br />The next two games were losses. First the Buffalo Bills (the same defense that handed <b>Aaron Rogers</b> his ass last week with four turnovers) and then against the Indianapolis Colts, losing by 1 point to 9-4 team that is (as of now) playoff bound for the second straight year. A rough patch for sure...though again (with the latter game especially), Gordon's failures (and others on the team) were as much to blame as Hoyer.<br /><br />But, of course, the Browns benched the guy and went with Manziel. <a href="http://scores.espn.go.com/nfl/recap?gameId=400554392"><b>Stupid hobgoblins</b></a>.<br /><br />Now the Browns are left with a 7-7 record and are determined to "evaluate" Manziel over the last three games by starting him in place of the guy that got them the wins they have. In other words, they've given up the season, even while not mathematically eliminated from the play-offs. Sure, maybe Johnny Football will turn it around in the next three games but, you know...<br /><br />Hobgoblins.<br /><br />Hey, what do I care? My team's playing for their division and 1st place in the NFC this week. Our coach has been pretty fearless in his roster selections when needed...going with the third string rookie over the high-priced free agent when the former gave the team the best chance to win. Trading another high priced skill player five games into the year...after gearing the whole offense to the dude in the off-season...because he was a poison to the team. Putting us in the best place to compete.<br /><br />Maybe Coach Pettine felt the same when he chose to bench Hoyer for Manziel. Maybe he felt he was doing the same thing Denver did a couple years back when they benched Kyle Orton for Tim Tebow. The difference, of course, is that Orton was 1-4 five games into the season, not 7-6 with a playoff berth on the line.<br /><br />But like I said, it's not <i>my</i> team. Hell, the Browns aren't even in the same conference as the Seahawks. I don't even like them all that much...like I said, their actions irritate me to no end as a football fan. What an actual fan from Cleveland thinks...well, I'd rather not speculate. At least they have LeBron these days right? All us Seattle folks have <i>IS</i> the football...at least till baseball season starts.<br /><br />Like I said, I <i>want</i> to like the Browns. I'd like to see them do well. Maybe because I <u>hate</u> the Steelers and I liked seeing Cleveland taking it too them this year. But, no, I think it's more about <b>tradition:</b> I've been waiting to see their stock rise ever since the recreation of the team Modell shipped to Baltimore.<br /><br />But I guess you can't get your hopes up too high with The Hobgoblin Team, huh?<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/-CasFVUA_nFY/VJM10tcme2I/AAAAAAAAAb4/c0Fo8vcS2ic/s1600/Unknown.jpeg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://1.bp.blogspot.com/-CasFVUA_nFY/VJM10tcme2I/AAAAAAAAAb4/c0Fo8vcS2ic/s1600/Unknown.jpeg" /></a></div>http://bxblackrazor.blogspot.com/2014/12/[email protected] (JB)7tag:blogger.com,1999:blog-4143435314932633148.post-2519250918488517448Wed, 17 Dec 2014 14:00:00 +00002014-12-17T06:01:05.895-08:00comicssuperstelevisionFirestarOne thing folks should understand is that much of my exploration of the superhero genre is directly related to my child (who turns four years old next month). Not that I don't enjoy the superhero genre myself (for reasons of <a href="http://bxblackrazor.blogspot.com/2013/01/bumblebee-boy.html">wish fulfillment and punch-in-the-mouth problem solving</a>, if nothing else)...I'm just saying I probably would <i>not</i> be going back and watching <a href="http://bxblackrazor.blogspot.com/2014/12/spider-man-and-his-amazing-friends.html">these old cartoons</a>, if not for him. Probably...<div><br /></div><div>Anyway...the last day or two he's been really interested in <b>superhero origins</b>, i.e. "how did [insert name] become a superhero." Season two of <i>Spider-Man and His Amazing Friends</i> consists of three episodes that tell the (individual) stories of the three heroes on the team. And as with most of the "history" stuff in the show, they're pretty faithful to the original versions found in the comic books (even dealing with the death of Parker's uncle and his feelings of guilt...fairly dark stuff for a Saturday Morning cartoon).</div><div><br /></div><table cellpadding="0" cellspacing="0" class="tr-caption-container" style="float: right; margin-left: 1em; text-align: right;"><tbody><tr><td style="text-align: center;"><a href="http://1.bp.blogspot.com/-8cmw27dyVOA/VJGGLbHlihI/AAAAAAAAAbc/GFDBTJWb5zo/s1600/images-2.jpeg" imageanchor="1" style="clear: right; margin-bottom: 1em; margin-left: auto; margin-right: auto;"><img border="0" src="http://1.bp.blogspot.com/-8cmw27dyVOA/VJGGLbHlihI/AAAAAAAAAbc/GFDBTJWb5zo/s1600/images-2.jpeg" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">Theme Song: <i>Free Bird</i></td></tr></tbody></table><div>Of course, that can't be said of the origin story for <b>Firestar</b> because her character has no basis in comic books. She was created specifically for the cartoon.</div><div><br /></div><div>The original concept for the story (from what I read) was to have Spider-Man teamed up with two heroes whose powers were polar opposite (fire and ice), to create a dichotomy and allow for more interesting interaction. <b>The Human Torch</b> would be the natural fit and was the creators' first choice; unfortunately, there was some issues with getting the proper licensing/rights to use the Torch (I believe it was these same issues that prevented him from being included in the 1970s cartoon of <i>The Fantastic Four</i>, requiring the writers of that show to create a humorous robot to round out the team's foursome). Since Johnny Storm wasn't available, Firestar (<b>Angelica Jones</b>) was created: a female mutant with the ability to harness and control ambient microwaves for a variety of stunts, mainly related to heat and fire.</div><div><br /></div><div>Pedestrian as a "generic fire mutant" may seem, the change makes the show <b>immensely better</b> simply by dint of the character being <i>female</i>. Not only because, hey, it's inclusive and gives a female viewer a character with whom to identify, but because it sets up a far more interesting interplay between the three characters on the screen. There's the back-and-forth banter that comes when friends of opposite sexes interact, as well as a bit of a "relaxed love triangle" that allows everyone a chance to show-off a bit: the guys compete with each other - at times - to look good (or at least not bad) in front of Firestar and it mirrors Firestar's own (very understated) interest in pulling her weight in the Man's World of the comic book genre.</div><div><br /></div><div>Not that she really needs to worry about pulling her weight: besides being the most measurably powerful of the three heroes, Angelica is a smart, strong, and competent character. She is immensely likable, has interests outside of her friends (and crime-fighting), is responsible, compassionate, and confident. She comes to the rescue of the boys at least as often as they come to hers, and is able to operate solo a lot more effectively than the other two, especially Iceman (who often seems lost without one of the others helping him out).</div><div><br /></div><div>Interesting as the superheroes are (power wise, story wise), the part that fleshes out their 2D characters is their <a href="http://bxblackrazor.blogspot.com/2013/01/legendary-might-part-2-of-2.html">human interaction</a>. And Firestar provides a much needed <b>female voice</b>...can you imagine if the only female character had been <b>Aunt May</b>? Sure, May provides one facet of the female archetype, but with the addition of Firestar the dynamic of <i>their</i> relationship (young, hip woman and wise, elder)...well, it exceeds the sum of its parts.</div><div><br /></div><div>Firestar's a great example of what a female superhero can be...note the lack of provocative costume. Yes, that's partly due to being a children's TV show, but there's no need to dress like a stripper when <b>you can melt things with your mind</b>. And she's not a prude: Angelica has an interest in dating (and dating outside her super-buddies), and is not just looking to snag a husband. She knows what she likes and doesn't let her friends' ribbing get in the way (as when she had an initial attraction to <a href="http://bxblackrazor.blogspot.com/2014/12/kraven.html">Kraven</a>). The story with <b>Sunfire</b> and their romance makes perfect sense, and hits just about every right note, and really shows a strength of the genre: that the real world barriers of culture and ethnicity <i>can</i> be cast aside and ignored, allowing us to see how much we share as human beings. </div><div><br /></div><div>But fans of the superhero genre should already get that. <b>Equality</b> is an inherent part of the genre...no one's going to refer to females as "the weaker sex" in a world where Carol Danvers can crush Captain America in arm-wrestling. It makes no difference that (black) <b>Luke Cage</b> is married to (white) <b>Jessica Jones</b>...it's a bit more important that they're both superhumanly strong and durable. </div><div><br /></div><div>Objectively speaking, the superhero universe is more diverse than our real world with its aliens and robots and sorcerers and mutants - all of varying powers and power levels. The (imaginary) people of that universe have mostly managed to take it in stride, instead coming back to the more basic question:<b> are you a good guy or a bad guy?</b> Nothing else is a big deal.</div><div><br /></div><div>[<i>consider the scene from the original Secret Wars where Reed Richards is repairing Iron Man's armor. Asked by then-pilot James Rhodes if he was curious that there was a black man in the armor, Reed replies, "Well, no. I knew there was a </i>man<i> inside the armor." Richards doesn't care what skin color the superhero has</i>]</div><div><br /></div><div>[<i>and, yes, I realize that comics can be used to parallel real-world prejudice and act as substitute analogies...see <b>"mutant hysteria"</b> and <b>"Superhero Registration Act"</b> as prime examples...but the <u>potential</u> for looking beyond race and culture and sex and sexuality, to a world united by common (hopefully good) purpose is, I think, the strongest in this particular, peculiar genre of fiction. To me, that's pretty neat, and a reason to maintain interest in comics</i>]</div><div><br /></div><div>*<i>ahem</i>* Back to Firestar...</div><div><br /></div><div>One last tidbit of interest here for this character...and while it applies to most of these early superhero cartoons, it's especially driven home with a character who has the potential to level a city...is how little <b>damage</b> needs to be inflicted to end a fight. It's not like Firestar just fries bad guys (the Spider-Friends are often found foiling robberies and such, in between fighting superhuman menaces, and so are sometimes simply combatting non-powered "goons"), even though she <i>could</i>. No one is getting incinerated or being set on fire; in gaming terms, there's no HPs being removed by her attacks.</div><div><br /></div><div>"Well, JB," you say, "That's based on a couple things: one is that it's a children's cartoon, and they don't want to encourage children to fight (i.e. punch people), and the other is the usual 'superhero code' against killing opponents." To which I reply: okay, but consider <b>these two things:</b></div><div><ul><li>Most folks would have a hard time bringing themselves to take a life, regardless of whether their pointed finger is the equivalent of a loaded gun. My understanding of military training (having never been through boot camp) is that at least part of it involves getting recruits to a mental state where it's okay to kill in the line of duty. For a lot of people, it would take a real life-or-death situation (perhaps involving the endangered life of a loved one) to get us to do mortal harm to a person. Very few of us are "natural born killers."</li><li>Whether you have a "code" against killing or just lack the stomach for it, if your power is one like <b>"microwaving the shit out of things,"</b> you are now faced with an interesting challenge: <i>how can I stop these crooks and villains with my seriously lethal superpower?</i> This requires a lot of creative thinking on the part of the player; in a show like <i>Spider-Man and His Amazing Friends</i> it requires the writers to constantly think of interesting ideas in order to keep the depictions "fresh" (it can't all be simply "cages of fire," as that would get boringly repetitive). In superhero RPGs that assign a higher value to "lethal" powers over "non-lethal," this is something to think about. Sure that ability to blow shit up is awesome when you're facing a mindless robot, but what about when facing your mind-controlled buddy? </li></ul></div><div>People that kill bad guys (even in self defense) are still committing murder, and while the law might consider it justifiable homicide, it really depends on how excessive was the forced used by the hero. It's possible a hero might still be wanted for felony jail time, even on a lesser charge (like manslaughter). Just saying.</div><div>; )</div>http://bxblackrazor.blogspot.com/2014/12/[email protected] (JB)6tag:blogger.com,1999:blog-4143435314932633148.post-5296600058618414703Tue, 16 Dec 2014 22:49:00 +00002014-12-16T15:14:28.174-08:00comicstelevisionvillainsKraven<table cellpadding="0" cellspacing="0" class="tr-caption-container" style="float: left; margin-right: 1em; text-align: left;"><tbody><tr><td style="text-align: center;"><a href="http://4.bp.blogspot.com/-uaHZLS6eQpg/VJC0MnrNSNI/AAAAAAAAAbM/kyl_RBoTWMo/s1600/krven.jpeg" imageanchor="1" style="clear: left; margin-bottom: 1em; margin-left: auto; margin-right: auto;"><img border="0" src="http://4.bp.blogspot.com/-uaHZLS6eQpg/VJC0MnrNSNI/AAAAAAAAAbM/kyl_RBoTWMo/s1600/krven.jpeg" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">"Superpower? Washboard abs."</td></tr></tbody></table><b>Kraven the Hunter</b> is a true badass and one of my favorite super-villains. Of course, I say this with knowledge based solely on two sources: the <i>Marvel Superheroes RPG</i> (he appears in the Advanced edition) and his appearance on <i><a href="http://bxblackrazor.blogspot.com/2014/12/spider-man-and-his-amazing-friends.html">Spider-Man and His Amazing Friends</a></i>. And as I never actually <i>used</i> the Kraven write-up in any of my past Marvel campaigns, it's pretty safe to say the lion's share of my admiration comes from his depiction in the cartoon (episode #2, circa 1981).<br /><br />Let's take a look at why.<br /><br />Here's a guy who looks like <b>Tom Selleck</b> on steroids, wears a lion-head vest, and hunts dinosaurs with his bare hands. He is, to coin a phrase, <i>The Most Interesting Man in the World</i>.<br /><br />Oh, sure, he has a potent <b>jungle power potion</b> that gives him "<i>the strength of ten gorillas</i>," but really how strong would that be?<br /><br />[<i>per the internet, gorillas are estimated to be six to ten times as strong as a human, meaning Kraven at full strength is around 80 times stronger than a human...about the same strength as <b>Spider-Man</b> whose official stat line says he can bench around 10 tons</i>]<br /><br />Kraven's <i>real</i> superpower is his <b>giant</b> <b>brass balls</b>...in MSH terms, they rank as <b>Monstrous (75)</b>, or at least <b>Amazing (50)</b>. The guy's courage is well-over the "foolhardy" red-line. His half-baked plan to conquer New York with an army of newly hatched dinosaurs that "only he can control?" That kind of crazy is what led to the <i>Jurassic Park</i> franchise making millions and millions of dollars.<br /><br />But crazy or not, there's no denying he's a sharp dude with a heap of skill. He out-plots and out-thinks (for the most part) three veteran super-heroes, despite having no real super-powers. A little chloroform, a little taser-type action, and <i>voila</i>! Three bound and helpless superheroes, despite the guy lacking true superpowers. Hell, he doesn't even use a gun...Kraven's got his own twisted code of honor. Sure, some might say he plays "dirty," but he's just using every possible angle. When you're going up against mutants that can melt steel...or freeze and shatter it...you need to use every angle.<br /><br /><b>Kraven is an evil version of Batman</b>. Not the <i>Superfriends</i> version that I grew up with, but the always prepared, <b>The Brave & The Bold</b> version of more recent years. You young 'uns dig the hip, Eastwood-cool Dark Knight...<i>I</i> prefer the brash and bombastic "bad guy" version. He positively oozes villainous self-confidence.<br /><br /><b>Case in point:</b> I love how he refers to himself in <i>the third person</i>. That's sooo hip-hop! In 1981!<br />: )<br /><br />Kraven's main failing? Being a super-villain, his arrogance leads him to "go it alone" (or with a minimum of henchman help), and he definitely bites off more than he can chew. I'm not sure how he handled Spider-Man in the comics, but it's readily apparent in the 'toon that any single "Spider friend" is over-matched by his devious hunter's mind.<br /><br /><b>Which is nuts, right?</b> Because he doesn't have super powers. And even Spider-Man should be able to deliver a beatdown in a straight fight. Kraven doesn't wear body armor, he doesn't have "the skin of a rhino." This is a <b>one-punch</b> <b>fight</b>...<a href="https://www.youtube.com/watch?v=Srm8LAKq03c">like Jason killing Horshack</a>.<br /><br />There are two takeaways here for game design purposes. First, an antagonist/peril does not require the bad guy to be a Big Bad Boss-type monster to provide a challenge (and a good one) to players. Secondly, in a cinematic (live comic book) style game, simple mechanics coupled with teamwork can provide decent action sequences without necessitating large stat blocks.<br /><br />But that latter is an insight mainly for <i>Yours Truly</i>...the design equivalent of an "inside joke" that needs a bit more explanation. Perhaps when I talk about <b>DC Heroes </b>(AKA <i>Mutants & Masterminds III</i>) I'll get into that. Yeah, probably.<br /><br />[<i>by the way, while I've never used a Kraven-type villain in a supers game, but I always felt the <b>Hunter/Vigilante</b> class in <b>Heroes Unlimited</b> was about the perfect vehicle for such a character...even without a "jungle power potion"</i>]<br /><br /><b>Just by the way:</b> my son <i>also</i> digs Kraven...though the hunter doesn't scare him nearly as much as <b>Mysterio</b>.http://bxblackrazor.blogspot.com/2014/12/[email protected] (JB)6tag:blogger.com,1999:blog-4143435314932633148.post-8416446490525722273Tue, 16 Dec 2014 11:04:00 +00002014-12-16T03:07:36.670-08:00comicskidsmontananostalgiasuperstelevisionSpider-Man and His Amazing Friends<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="http://4.bp.blogspot.com/-rm3Ge5an-UA/VJASKytEuUI/AAAAAAAAAa8/ssZXHURK6uw/s1600/Unknown.jpeg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="http://4.bp.blogspot.com/-rm3Ge5an-UA/VJASKytEuUI/AAAAAAAAAa8/ssZXHURK6uw/s1600/Unknown.jpeg" height="239" width="320" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">The Three Amigos</td></tr></tbody></table>I know, I know...folks would rather here about D&Dish stuff over here at Ye Old Blackrazor Blog, but I've really got nothing at the moment. Truly, the well's a bit dry at the moment. For one thing, Paraguay just doesn't inspire anything in a "fantasy-exploration-adventure" kind of way. For another, Alexis's book (my reading of it, analysis, and review) just <b>took a lot out of me</b>. I just need a couple weeks to recuperate before I approach anything like a D&D-style campaign. Sorry.<br /><br />Rather than leave you folks on empty (and rather than fall out of practice with the blogging thang), I'm going to talk some superheroes. I know the genre turns off some people, but it's better than nothing, right? You certainly don't want me blathering on about the <b>Seahawks and the NFL</b> (next projected post on <i>that</i> subject will be at the end of the regular season).<br /><br />SO...<b>Spider-Man and His Amazing Friends</b>. That's where I'll be starting: the Saturday morning cartoon from the early 80's (specifically 1981 to 1983, per <a href="https://en.wikipedia.org/wiki/Spider-Man_and_His_Amazing_Friends#Episodes">wikipedia</a>). This would have been back when I was in the 8-10 age range, so probably the perfect demographic. I purchased comic books back then...usually in the summer months from the Missoula, Montana Circle K located down the street from my grandma's house...but I certainly had no sophistication regarding comic titles, nor was I regular collector, nor did I have any sense of the history of the Marvel series being published in the 80's (most of which had a couple decades of history/backstory to them). Plus, I preferred <b><i>Ghost Rider</i></b> or <b><i>Rom</i></b> to anything as pedestrian as the <b><i>Fantastic Four</i></b> or <i><b>Spider-Man</b></i>.<br /><br />I was, however, pretty fanatically loyal to this particular cartoon.<br /><br />I enjoyed Saturday morning cartoons, immensely. When I was a kid my family never had cable TV, which means we were limited to 5 to 7 channels the first couple decades of my life. Cartoons we're played on three of those channels (ABC, NBC, and CBS) for a couple hours, starting around 7am or so and ending around 10 or 11. During that window, I surfed between the three channels, focusing on a variety of action/adventure flicks: <i>Thundarr the Barbarian</i>, <i>Blackstar</i>, <i>Jonny Quest</i>, <i>Godzilla</i> as well as the usual superhero flicks. Things that had a mystical/mythology twist or a lot of ass-kicking was what I wanted to see. When cartoons started having "messages" or "morals" thrown in at the end...well, that's when I stopped bothering to get up early. Shows like <i>Transformers</i> and <i>G.I. Joe</i> sounded the death knell for my love affair with cartoons, probably circa '84 or '85 (though I was pretty diligent about watching the <i>Dungeons & Dragons</i> cartoon for the length of its run on TV).<br /><br />But before that happened, there was <i>Spider-Man and His Amazing Friends</i>. Spidey replaced the <i>Superfriends</i> for my superhero watching (though the Justice League received a small *bump* when they brought in <b>Darkseid</b> & Co. circa '84). The villains were ones I recognized, the threats seemed more interesting, the art and writing was better. Plus, the acrobatic Spider-Man is a character that was made for animation. The titular "friends" (<b>Iceman</b> and <b>Firestar</b>) are pretty cool as well.<br /><br />Recently I've had a chance to re-watch these shows (ahh...the magic of the internet) with my son...something I haven't done since they went off the air in '83 (I don't know if they were later syndicated, but if so I missed 'em). And compared to the 1960s <i>Spider-Man</i> (also viewed on the internet) or the <i>Superfriends</i> (even the great <b>Legion of Doom</b> episodes)...really there's no comparison. It's a great, great show: the animation, the writing, the humor and action. For me, it really captures the youth and energy of the late Silver Age...there are no dark antiheroes or grey areas. It's fun, but it's interesting, and the comic relief from the small pet is a nice balance to some real instances of scary ("scary-ish?") jeopardy/peril in which the heroes find themselves. The series leads off with three badass villains (Green Goblin, Kraven the Hunter, and Doctor Doom) and has some great ones like the Red Skull (with literal <b>Nazi</b> themes...that ain't something you see too often in a kid's cartoon!).<br /><br />[<i>just by the way...the depiction of <b>Dr. Doom</b> in this show is probably my favorite non-comic book depiction of the good doctor in any medium. The 70s Fantastic Four version...<b>"I need Black Beard's treasure to take over the world"</b> is sooooo weak-sauce compared to this minor happenstance. And don't get me started on the live-action version...</i>]<br /><br />Not that I'm bringing this up to stroll down Nostalgia Lane (though feel free to enter your own personal comments, as always). Fact is, I'm tinkering on the hero game (once again) and there are various aspects of the show I want to discuss...things that relate to, oh say, what I want to do in superhero-based RPG.<br /><br />But let me save that stuff for individual posts.<br /><br />[<i>sorry, I wrote this yesterday, but didn't have a chance to get it posted. Will try to get another one done today</i>]http://bxblackrazor.blogspot.com/2014/12/[email protected] (JB)10tag:blogger.com,1999:blog-4143435314932633148.post-3519440993963410341Fri, 12 Dec 2014 17:02:00 +00002014-12-12T09:03:26.714-08:00filmrandomJoop, Joop-I-Doop, Joopy-JupiterAnd speaking of r<b>andom space opera</b>...<br /><br />I spent a lot of yesterday morning watching videos of movie trailers for upcoming films. This is a ridiculous waste of time for many reasons (not the least of which is my free time for actually <i>seeing</i> <i>films</i> these days is next-to-zero), but mainly because, well, I could have been writing instead. But hey: <b>I blame Jay</b> over at <b><a href="http://gammaworldwar.blogspot.com/">Gamma World War!</a></b> for his constant <i>Man Max</i> updates...you just know I love me some post-apocalyptic goodness, and after that I just "follow the links."<br /><br />So it was that one link led me to a trailer for the space opera flick, <b>Jupiter Ascending </b>(latest release date sometime in February).<br /><br />Now it's pretty ridiculous to "review" trailers of <a href="http://bxblackrazor.blogspot.com/2014/11/really-are-you-kidding-me.html">not-yet-released movies</a> [*<i>ahem</i>*] but certainly I count on trailers to pique my interest...with the effort it takes me to get to the movies these days, something better really wow me (except in the rare instance when it appeals to <a href="http://bxblackrazor.blogspot.com/2014/03/old-heroes-never-die-no-really.html">some personal interest of mine</a>). Strange as it may seem, given the overall geekyness of my blog, my general <i>film</i> interests only rarely run the vein of fantasy or space opera. Historical pieces like<b><i> <a href="http://en.wikipedia.org/wiki/Essex_(whaleship)">In the Heart of the Sea</a> </i></b>or quirky character pieces like <b><i><a href="http://en.wikipedia.org/wiki/Inherent_Vice">Inherent Vice</a></i></b> are much more my speed.<br /><br />Having said that...<br /><br />Watching <a href="https://www.youtube.com/watch?v=EVZELMRXeYM">this trailer</a> for <i>Jupiter Ascending</i>, I found the premise of the setting to be <i>very</i> intriguing. I'm trying to remember if I've seen this particular "speculative fiction creation myth" in fiction before. Sure there's a lot of shades of <i>The Matrix</i>, here (as one might expect from the same dudes who wrote that trilogy), but while <i>The Matrix</i> was kind of a GenX take on Plato's <a href="https://www.youtube.com/watch?v=EVZELMRXeYM">Allegory of the Cave</a>, this feels much more <b><i>Flash Gordon</i>-esque</b>...which is something I really dig.<br /><br />I don't know why (I dig it). I'm not of George Lucas's generation, did not grow up on FG serials or comic strips. But I've always enjoyed the idea of intergalactic empires operating "just beyond the ken of Earth knowledge." Secret space battles/intrigues on a titanic scale that only a few privileged Earth folks have discovered. Think of the Marvel comic character <b><a href="http://en.wikipedia.org/wiki/Corsair_(comics)">Corsair</a></b> becoming embroiled in the Shiar Empire. In many ways it's similar to the "stranger in a strange land" sword & planet epics of Burroughs (John Carter), Moorcock (Michael Kane), and John Norman (Tarl Cabot)...yet the <i>scale</i> is so much larger, spanning multiple planets and systems and often including that "ship-to-ship" action that appeals to the pirate fetish so many of us have.<br /><br />But it's not just the action. When you're dealing with the technology to deal deathblows to whole planets and star systems (whether we're talking <i>Star Wars</i> or the Lensmen), one hero's ability to wield a sword, laser or otherwise, scarcely matters (unless granted license by the author, that is). Instead, being able to navigate intergalactic politics...one's interactions with the people in power...is the important part of the equation.<br /><br />And who doesn't love the associated difficulties with governing an interstellar empire? See <i>Dune</i>, <i>Foundation</i>, <i>Star Wars</i>, etc. for examples.<br /><br />So <i>that's</i> cool. And <i>Jupiter Ascending</i> has been in development long enough that there are plenty of spoilers about the characters (like how they've been genetically spliced with various animals to make better warriors, trackers, etc.). Which is also cool.<br /><br />Having said <i>that</i>...the over-the-top super-sci-fi action sequences on display in the trailer I find to be terribly uninteresting. So much so that it detracts from the things that ARE interesting. It's like the recent <i>Hobbit</i> trilogy...it's as if the filmmakers don't trust that the subject matter is interesting enough to engage audiences without bombarding them with complicated blue-screen mayhem. I don't know how many ways I can say it:<br /><br /><b>Including action for the sake of including action is BORING. </b>It fucking-A is.<br /><br />Sure, I'm an old geezer that has no idea what the kids want these days. Perhaps the market research shows that the only folks who'd be interested in such a film play too many high octane action video games and want to see the same kind of thing on the screen. I can tell you that after seeing the action sequences on display, and especially after watching <a href="https://www.youtube.com/watch?v=TLyk00gFPdQ">this other trailer</a> for the film, I'm actually turned <i>off</i> from watching the film, despite the cool setting. And sure, it's grossly unfair to judge a film by its previews alone...but isn't the preview the thing that's supposed to grip you and reel you in? I've spoken with a lot of folks who skipped an otherwise good movie because the trailer "sold them" poorly (or sold them on the wrong thing)...I know I'm not alone in <i>that</i> particular brand of superficiality.<br /><br />Thus, unless I read some truly stellar reviews, I will probably not be watching <i>Jupiter Ascending</i>, unless it's available on one of those 12 hour plane flights that I seem to take with alarming frequency these days. I'm just shallow like that.<br /><br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="http://4.bp.blogspot.com/-1hn1-IAAWnM/VIsfR10p9uI/AAAAAAAAAas/f-8xFTCZYS0/s1600/Unknown.jpeg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="http://4.bp.blogspot.com/-1hn1-IAAWnM/VIsfR10p9uI/AAAAAAAAAas/f-8xFTCZYS0/s1600/Unknown.jpeg" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">Jupiter Descending</td></tr></tbody></table>[<i>sorry, I just have space opera on the brain these days...that and superheroes, but who wants to hear about <b>Aquaman</b>?</i>]http://bxblackrazor.blogspot.com/2014/12/[email protected] (JB)4tag:blogger.com,1999:blog-4143435314932633148.post-3081012649402583974Wed, 10 Dec 2014 21:29:00 +00002014-12-10T13:33:59.502-08:00cookfilmreflectionsswSW Musings: Goofy Fun<a href="http://bxblackrazor.blogspot.com/2014/12/happy-virgin-of-caacupe-day.html">Or not</a>...spent yesterday <b>baking</b> <b>gingerbread cookies</b>. Even though it's over 90 degrees outside, my goal was to get this place smelling a little bit more like Christmas. Or, at least, burnt cookie. Mission (more or less) accomplished!<br /><br /><table cellpadding="0" cellspacing="0" class="tr-caption-container" style="float: right; margin-left: 1em; text-align: right;"><tbody><tr><td style="text-align: center;"><a href="http://4.bp.blogspot.com/-hweoZvPsjVA/VIi7wBG7rCI/AAAAAAAAAac/7nHopZReALs/s1600/Unknown.jpeg" imageanchor="1" style="clear: right; margin-bottom: 1em; margin-left: auto; margin-right: auto;"><img border="0" src="http://4.bp.blogspot.com/-hweoZvPsjVA/VIi7wBG7rCI/AAAAAAAAAac/7nHopZReALs/s1600/Unknown.jpeg" height="200" width="200" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">No...they didn't look like this.</td></tr></tbody></table>My child wanted some of the gingerbread men to resemble characters from <b>Star Wars</b>...specifically Darth Maul, Yoda, R2, and Qui-Gon (we've been watching a lot of <i>Phantom Menace</i> the last couple days). Thanks to being a <b>journeyman Play-Doh mason</b> these days, I was able to get some reasonable silhouettes (<i>I</i> think)...certainly D was pleased with the end result.<br /><br />I won't discuss the myriad difficulties associated with the task of baking in a country where people don't cook or really know how to do anything more than grill meat and starchy root vegetables. Instead I want to muse a little bit about modeling <i>Star Wars</i> in an RPG...a topic I realize I've visited more than a few times over the life of this blog (with little success).<br /><br />Once upon a time, I had hit upon an idea for a way to "do <i>Star Wars</i>" as a knock-off of an existing game. Actually, there were couple four systems that would make good "hacks" for a SW-type game. B/X was one. <i>Trollbabe</i> was another. <i>Dogs in the Vinyard</i> a third...though having just purchased the last one and read it over the last few days, I see exactly how mistaken I was: DitV works well for a number of genres, but it is one that really needs to be tied in tight with religion, and there's just not enough theology to the pseudo-religion of the Jedi.<br /><br />Oh, yeah...I should probably mention that I mean these are good hacks for a Jedi-centric game, not an overall "galactic space opera adventure" game. If you want something akin to the original trilogy (or, at least, the original film)...well, there are other systems to hack for that style. Systems that will allow for <b>goofy fun</b> in more than just a few passing ways. <i>Star Wars</i>, like that <i>other</i> 80s space opera film series (<b>Star Trek</b>) was all about the goofy fun.<br /><br />There's a part of space opera that really cries out for goofy fun...and by this I mean a "not taking itself too seriously" approach. Probably because it <i>IS</i> "space opera." The technology isn't based on "hard science" (laser guns/swords, FTL travel, AI robots that "feel," psychic powers, etc.). It's fantasy, and melodramatic fantasy of world-shattering destruction (literally). That's why<i><a href="http://bxblackrazor.blogspot.com/2014/10/guarding-galaxyfrom-themselves.html"> Guardians of the Galaxy</a></i>, with its goofy cast, is such good space opera.<br /><br />Despite the fancy special effects and (yes, really) <i>heart</i> that is injected into Episodes I, II, and III, there's very little real <b>humor</b> instilled in the films...especially the kind of self-deprecating type of the original trilogy. <b>Han Solo</b>, for all his bad-assedness at piloting and shooting, often comes off as a lovable buffoon. <b>Leia</b> ends up humbled by Solo's wit more often than any other "princess" I remember seeing on celluloid (and what does it say that she ends up with the buffoon by the end?). And while <b>Luke</b> is certainly a force to be reckoned with by RotJ (no pun intended), he had a lot of ground to make up from being wet-behind-the-ears kid of the first movie...but of course, his story evolves along a significantly different path from the others through <i>Empire</i> and <i>Return of the Jedi</i>.<br /><br />Very little humor or silliness is found in the prequel films...in fact, the one with the most might be <i>The Phantom Menace</i> and, no, I'm not talking about Jar-Jar. Here we have Qui-Gon failing to influence Watoo with his mind-tricks. Here we have Anakin admitting he's never actually managed to finish a pod race. Here we have Obi-Wan referring to their own party as rather "pathetic life-forms."<br /><br />Not taking oneself too seriously means allowing yourself to be <b>humbled</b>...to admit that you aren't the invincible action hero but a person with flaws and foibles and ability to laugh (or at least grudgingly smirk) at your own failings.<br /><br /><b>As a related aside: </b>I realize I never did get back to what I thought about the <a href="http://bxblackrazor.blogspot.com/2014/11/really-are-you-kidding-me.html"><i>Star Wars</i> VII trailer</a>. My overall impression was that <b>what I saw was interesting</b>...and that I would be interested in seeing more. I found J.J. Abrams's interpretation of <i>Star Trek</i> was full of contrasts between the seriousness of the situation and the playfulness/humor of the characters...in other words, pretty good space opera. This gives me quite a bit of optimism for a good <i>Star Wars</i> flick.<br /><br />Hmmm...more on this later (perhaps).http://bxblackrazor.blogspot.com/2014/12/[email protected] (JB)10tag:blogger.com,1999:blog-4143435314932633148.post-5149283745291270343Tue, 09 Dec 2014 14:48:00 +00002014-12-09T06:49:52.465-08:00holidayparaguayHappy Virgin of Caacupe Day!Actually <i>yesterday</i> (December 8th) is the day Paraguayans celebrate the <b>Virgin of Caacup</b>e...a little statue that was carved by a divinely inspired local four centuries back and now the target destination of a few hundred thousand pilgrims every year. You find this kind of thing all over Latin America (sorry if that sounds dismissive), but here in Paraguay it's a <b>national holiday</b>, shutting down most businesses and such (they do so love their holidays here). Word on the street is that nothing much gets done from December 8th till mid-January as everyone goes into kind of "holiday-slacker" mode...as opposed to the usual slack that permeates the fabric of this country.<br /><br />[*<i>sigh</i>*]<br /><br /><i>ANY</i>way...that's why my "non-posting weekend" continued till today...I was busy with the kids yesterday. Today, I just slept in (was up waaay too late watching that <b>barn-burner</b> between Green Bay and Atlanta). Truth is, we're all (my family) still adjusting to the five hour time difference and scorching hot summer weather...none of us are sleeping particularly well. We all just seem to be killing time till we can get to Mexico (seeing the in-laws for Christmas) and eat some real food.<br /><br />We, too, have caught the infectious national slack of Paraguay.<br /><br />Mmm...hopefully, I'll get something more interesting posted later today; right now, I've got to jet.http://bxblackrazor.blogspot.com/2014/12/[email protected] (JB)0tag:blogger.com,1999:blog-4143435314932633148.post-4328080188494062535Fri, 05 Dec 2014 21:16:00 +00002014-12-05T13:16:29.480-08:00art of the DMcampaign settingreviewstaoHow to Run (Part 3)[<i>continuing our review of </i><b>How to Run</b><i>; see prior posts <b><a href="http://bxblackrazor.blogspot.com/2014/12/how-to-run-part-1.html">here</a></b> and <b><a href="http://bxblackrazor.blogspot.com/2014/12/how-to-run-part-2.html">here</a></b></i>]<br /><br /><b>PART 3: Managing Your Players</b><br /><br />This is the smallest section of the book (about 38 pages). Despite the title, it is still much more about managing oneself as a DM, but specifically with an eye to how that management impacts player interaction and participation at the gaming table.<br /><br /><i>Chapter 9 - Power Politics</i> discusses the relationship between the DM and players, specifically the DM's role as a <b>facilitator</b> of the game, NOT a judge, authority, or godlike super-being. Again, the emphasis is on <b>service to the players</b>...and here we're talking real-world service in managing social contract. When Mr. Smolensk writes that it is the DM's job to serve the players, he's not saying that one must create brilliant adventures or hooks or railroad the players through some fantastic story for their own enjoyment. <b>Quite the contrary:</b> throughout the book he stresses that the players must make and create their own "adventures" (for only in doing so can they truly find the emotional engagement that is the objective of role-play). No, instead the "service" being discussed (in this chapter specifically) is serving as a group facilitator: making sure all players have a voice, making sure all players feel valued, making sure all players are equal partners in the exercise (and that the DM is not "lording it over them"), making sure that <i>you</i> (as DM) are not failing at meeting their expectations of fair and respectful treatment.<br /><br /><i>Chapter 10 - Bad Games</i> discusses how DM's who fail to recognize their role of service can become the abusive DMs running "bad games" (hence the title) that so many players have experienced at one time or another. For me, this was one of the weaker chapters: if you've already "bought in" to the paradigm Alexis has building up to this point, than it follows that you shouldn't want to fall into these categories. I suppose that the main pitfalls outlined here - relying on one's charisma and building a power-base on sycophantic players - are traps that even good DMs need to be mindful of and avoid...however, as a cautionary note, it seems more like frosting to be added on to an earlier chapter. But maybe I'm just nitpicking.<br /><br />The ideas and topics presented in Part 2 and Part 3 of <i>How to Run</i> are definitely more esoteric than the practical information found in Part 1...discussing the cultivation of good habits and creating a "vision" for the players is a bit more abstract than making sure players have cool-off periods and bathroom breaks. It's <i>good</i> information (again), but it really is more for individuals who intend to make running games a high level vocation. The concepts are less pertinent to the one-time DM or occasional "dabbler" (who can get by wonderfully with just the information in Part 1), and much more imperative to the "career DM;" dudes like myself who find our butts in the GM chair more often than on the players' side of the table.<br /><br /><b>PART 4: Worldbuilding</b><br /><br />If they taught "playing role-playing games" as a potential degree in college, you'd probably need the first year just to dissect and learn the multi-hundred page rule books being published these days. If you limited yourself only to "old school" type products, you might be able to study several systems during the first year...say Moldvay D&D, D6 <i>Star Wars</i>, and the original (Jeff Grubb) <i>Marvel Superheroes RPG</i>...each providing a distinct genre and different approach to rule design and game mechanics.<br /><br /><i>How to Run</i> would be an excellent course text for such a curriculum, but it would not be read in the first year. It could be a strong text for the sophomore year, with Part 1 being studied the first quarter and Parts 2 and 3 in the second quarter. Part 4 <i>could</i> be started be the Spring session...but I seriously doubt that would be enough time to really do justice to it. At nearly one-third of the page count, I found Part 4 to be the most challenging <i>reading</i> of the book (there are a lot of abstract concepts here that can't be understood until practically tried and implemented), and the prescribed exercises are especially time consuming. No...the final section of <i>How to Run</i> could definitely provide coursework for the entire year of a third year student.<br /><br />[<i>I suppose actual "game design" (not discussed in the book) would be the prerogative of fourth year (senior) students</i>]<br /><br />Let's get to it (as best I can):<br /><br /><i>Chapter 11: Beginnings</i> lays the foundation for everything that follows. It explains the principles of world design, the creation of the imaginary setting, based on the abstract elements of <b>function</b> (poorly defined), <b>behavior</b> (of players with regard to function), <b>structure</b> (the implementation of function), and <b>formulation</b> (the interaction between the three preceding elements). It's pretty highbrow stuff, and could benefit from some concrete examples. Unfortunately, part of the paradigm here is that <i>function</i> has to be tailored to one's players (remember how the DM is serving their interests and wants to get them actively engaged?) which makes it difficult to create hypothetical examples of function without first creating a batch of hypothetical players and a hypothetical DM. "Function" appears to be (from my reading) a synthesis of concepts/activities in the game world with the potential of providing interest/engagement to both sides of the table. But I may be off...the function section of the chapter could stand to be more clearly defined (even the notes in the Keys to Success chapter appears to admit it is difficult to comprehend the concept). Being as amorphous as it is puts the other elements on rather shaky ground, since function is the key concept of the chapter.<br /><br />So let's move on to <i>Chapter 12: Elements of Design</i>. Here we have a chapter that discusses the practicalities of world building, from the effort required, to the acquisition of quality materials for your map making. Here the idea is <b>over-deliver</b> to one's players, with the idea that this will provide them with motivation and enthusiasm, facilitating engagement, as well as helping to instill a concept of value in everyone's minds (those of the player and the DM who is doing the work)...value of the imaginary world being explored, that is.<br /><br /><i>Chapter 13: The Creative Process</i> provides practical methods for brainstorming and conceptualizing your world (you need to have the key concepts from Chapter 11 and the practical stuff from Chapter 12 before you move to this step). The amount of effort recommended by Mr. Smolensk in prior chapters is exceeded here by the recommended time he proposes you spend. However, as we've leaped into the realm of the vocational DM, it's hard to argue with a few months of time spent on conceptualizing a world that you intend to run (with constant tweaking) for many years to come. Alexis provides tools for folks who want to commit to "the long haul," as he has. The "Keys" in this chapter provide notes for abbreviating the process.<br /><br /><i>Chapter 14: Modelling</i> [Canadian spelling] is really the Part 2 to Chapter 13. It explains in practical, semi-linear fashion how to translate the ideas and concepts onto paper...how to start small and work outward, how to create your world in such a way to introduce it (in a practical way) to the players. It discusses the creation of entities (which can be individuals, monsters, geographic features, religions of the fantasy world, whatever...though focused on those things pertinent to the players), and creating their relationships to each other...which is as important if not more so than drawing lines on a map. This creates the web in which players can anchor their immersion.<br /><br />For those interested in role-playing as an exploration of the imaginary world...what might be called (in old GNS terms) a "simulationist creative agenda"...this section of the book is a godsend. It also provides an idea of the amount of work and effort needed to create a lasting game of the type Mr. Smolensk has run...and still runs...for decades. You can, of course, put less effort towards your world building, though perhaps with lesser results.<br /><br />The <b>Appendix</b> contains (in addition bibliography, index, etc.) one last chapter (<i>Chapter 15: Gaining a Level</i>). It does not contain the "keys to success" section found in other chapters; it is an epilogue, not a part of the instruction found in earlier chapters. It discusses <b>work</b>: the reason for work, the reason why work is good, the reasons why one might choose to invest in the amount of work described in <i>How to Run</i> for a fantasy role-playing game. It is a convincing argument.<br /><br /><b>Concluding Thoughts Regarding <i>How to Run</i></b><br /><br />Alexis has really put out something special here: an example of how to turn a passion for gaming into a Great Work, i.e. a <i>transformative experience</i>. He does not outline a road to perfection in gaming, but one of perfecting oneself (an on-going process, a life's work) by approaching a subject with intensity and serious attitude. Nothing here is going to be mastered in a handful of years, but all the tools and ideas presented are useful for personal growth and development of one's ability as a game master.<br /><br />Having said <i>that</i>...<br /><br />Not every person seeks to walk this particular path, and the tools outlined are not necessarily appropriate to all forms of gaming. With regard to "world building," I'm not certain his principles would apply to all forms or genres of role-play...for example, the comic book superhero genre (which often exhibits wild inconsistencies and flexible reality even while using a "real world" setting). Recent game designs have demonstrated that an enjoyable role-play experience can be had without the need of a DM figure: games like <b><i>Fiasco</i></b>, for instance. And many story games can still engage and elicit emotional, exhilarating response from players even without the need for the detailed world-building and attention to cause-and-effect that Mr. Smolensk is preaching (games like <i><b>My Life With Master</b></i>); many of these games were designed to do exactly what Alexis purports to desire, but with less time and effort spent...perhaps because the commitment required otherwise is a price we are unwilling to pay.<br /><br />For myself, I enjoy running games, and I enjoy running different types of games...not simply different adventures in the same world. While I do not particularly enjoy learning different rule sets, I do <i>appreciate</i> how different rulesets interact with their respective games...I don't just want "One GURPS To Rule Them All." But that's me, and I recognize that for some folks the <i>only</i> thing they want to do is play the One Game (whatever it is) that they love and that has allowed them to craft their perfect fantasy world. This desire is not an unusual one...M.A.R. Barker was perhaps the first to bring this type of work to the gaming world with <b>Tekumel</b>, and you can see it even today in worlds like <a href="http://thegrandtapestry.blogspot.com/">Timeshadows's <b>Urutsk</b></a> or Alexis's own campaign. These worlds have lasted years and will continue as long as their creators continue. For folks who have this burning desire to create...and to bring their creation to <i>others</i> (not just write a six book fantasy series)...I'd strongly recommend they get <i>How to Run</i> and read it a few times. <i>Especially</i>, if they are fairly new to the whole role-playing thang (like I said, it'd take years to master the stuff in these pages so start while you're young!).<br /><br />For those who do NOT have such a burning desire...who simply want to run games, and perhaps dabble in "world-building"...there's still a lot of good to be found in this book. It will provoke self-examination. It will make you take stock of how you run your gaming table, and provide tips as to how to make your games better...for yourself and your players. Even if you're not into long-term campaigning, it's useful stuff, and much of it is <i>new</i> (even if some is stuff you already figured out). There's no talk of the history of gaming, no talk about specific systems or rules, only information on <b>how to run a better game.</b> There's plenty of good insights even for those who don't enjoy long-term campaign play, and it's not a bad read.<br /><br />All right, that's enough.<br /><br /><div class="separator" style="clear: both; text-align: center;"></div><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="http://4.bp.blogspot.com/-60-Jm9RHkpU/VIIZeA8AyGI/AAAAAAAAAaM/94zj1BILOXc/s1600/images-2.jpeg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="http://4.bp.blogspot.com/-60-Jm9RHkpU/VIIZeA8AyGI/AAAAAAAAAaM/94zj1BILOXc/s1600/images-2.jpeg" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">Get Ready To Run!</td></tr></tbody></table><br />http://bxblackrazor.blogspot.com/2014/12/[email protected] (JB)0tag:blogger.com,1999:blog-4143435314932633148.post-6736460550282912282Thu, 04 Dec 2014 21:37:00 +00002014-12-04T13:51:33.463-08:00art of the DMedwardsForgereviewstaovirgoHow to Run (Part 2)Just picking up <a href="http://bxblackrazor.blogspot.com/2014/12/how-to-run-part-1.html">where I left off</a>...<br /><br />Let's get right to it.<br /><br /><table cellpadding="0" cellspacing="0" class="tr-caption-container" style="float: left; margin-right: 1em; text-align: left;"><tbody><tr><td style="text-align: center;"><a href="http://2.bp.blogspot.com/-E2EW7soPVTY/VIDUiUG400I/AAAAAAAAAZ0/us2TyfPB6mI/s1600/Unknown.jpeg" imageanchor="1" style="clear: left; margin-bottom: 1em; margin-left: auto; margin-right: auto;"><img border="0" src="http://2.bp.blogspot.com/-E2EW7soPVTY/VIDUiUG400I/AAAAAAAAAZ0/us2TyfPB6mI/s1600/Unknown.jpeg" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">The book being discussed.</td></tr></tbody></table><b>How to Run </b>comes in at a bit more than 350 pages, though that includes an index, table of contents, and bibliography. After the introduction, it is divided into five sections including an appendix, the total comprising fifteen chapters (the last chapter, a bit of an epilogue, is in the appendix). The first four sections are comprised of the following:<br /><br /><b>Part 1: The Art of Presentation</b><br /><b>Part 2: Managing Yourself as DM</b><br /><b>Part 3: Managing Your Players</b><br /><b>Part 4: Worldbuilding</b><br /><br />Each section is composed of several chapters relating to specific topics. In addition, each chapter ends with a "Keys to Success" section that emphasizes or elaborates on specific points raised in the chapter. It's a handy trick for remembering what was discussed, and lends to the overall "textbook" feel of the book.<br /><br />The introduction nicely lays out what the book is about. <i>How to Run</i> is both genre and system neutral; it does not discuss specific rules or editions of D&D, and though it refers to GM position throughout as "Dungeon Master" or "DM," Smolensk is careful to note that the outlined principles can be used in running any table-top RPG. I know that for his own game Alexis uses heavily modified (1st edition) AD&D in a campaign setting firmly rooted in the historic 17th century Earth. He mentions little (with regard to the specifics) of his campaign setting in <i>How to Run</i>, and nothing at all of his house rules or system...the text really does strive to be applicable to any RPG a person might try to run.<br /><br /><b>Which reminds me:</b> while it is never specifically defined, contextually Mr. Smolensk uses the term <i>role-play</i> to simply describe the act of playing a role-playing game. In other words, if you are playing an RPG you are engaging in "role-play," pure and simple. For the purpose of his book and its concepts that's just fine.<br /><br /><b>PART ONE: The Art of Presentation</b><br /><br />This is the largest section of the book, and (in my opinion) the meatiest part in terms of presenting real tools that can be of use to folks wishing to run a game. It provides excellent advice and checklists for even experienced DMs, and raises a lot of questions for self-examination in us "old-timers." I found myself nodding quite often, noting the things I had done that worked well and likewise the areas where I had stumbled in my own games; the codifying of these things (always with an eye towards the goal: <b>facilitating engagement of the players</b>) is well done.<br /><br /><i>Chapter 1: The Early Days</i> discusses Alexis's own initiation into running games, and gives the young DM an idea of the attitude with which the task needs to be approached (it's not as hard as it looks, but it does require time and effort, even effort outside of learning the game). <i>Chapter 2: The Carrot and the Donkey</i> discusses how to motivating and enticing your players, providing the best environment for them to succeed at the goal (of engagement); note, there's no "stick" for the donkey, only carrots. <i>Chapter 3: The Players</i> describes some stereotypical personality types one might find at your table, how to recognize them, how to work towards their strengths, and how each can be used to build a strong gaming group (these are interesting "types" based on Mr. Smolensk's own experience and perception, not the usual archetypes found in Jungian psychology or whatnot). <i>Chapter 4: Drama</i> offers a method for creating a traditional <b>three-act (play) structure</b> for folks who want to create "stories" with their game sessions, but the author has come to the conclusion that such is a weaker form of role-play than long-term engagement and immersion (or, at least, more difficult to sustain over time). In dispensing, with the idea of "story creation," he begins to discuss <b>cause and effect</b>, and ways to empower the players by allowing their actions to matter in the campaign, outside the plot machinations of a story-minded DM. <i>Chapter 5: Continuity</i> discusses several tools for gripping your players, making them <i>care</i> about participating (i.e. engaging them emotionally) beyond simply offering them missions, as well as elaborating on the discussion of cause and effect and how it contributes to the ongoing participation and enjoyment of the gaming experience.<br /><br />I want to pause here for a moment to discuss Mr. Smolensk in relationship to another respected (if sometimes controversial) game designer, <b><a href="https://en.wikipedia.org/wiki/Ron_Edwards_(game_designer)">Ron Edwards</a></b>. I personally find the two remarkably similar, something like flip-sides of the same coin. This shouldn't be too surprising considering similar personality archetypes (both are <b>Virgos born in 1964</b>...no, I won't get into astrology right now, but with my own background that's a tough lens for me to ignore). Both have their detractors and admirers. Both are very intelligent and thoughtful. Both can be be prickly hardliners when it comes to their own beliefs. And both are extremely devoted to the service of the players at their table. Both seek to walk that line of using mental focus to bring about <i>emotional engagement</i>...but their approach to the same is very different. Edwards is devoted to the principal of <b>"story now:"</b> creating game mechanics that requires players to step up and engage with the narrative being created around the table. Smolensk would seem to be a standard-bearer for what the old GNS model called simulationism, or <a href="http://www.indie-rpgs.com/articles/15/">"the right to dream,"</a> creating a world one can escape into and experience. However, he has a ready answer for Edwards's "hard questions" regarding <i>what it's all for</i> and <i>how long it will last</i>: bluntly, all the work the DM does is for achieving an emotional engagement from the players, and goes <b><i>beyond</i></b> simply allowing players to explore an imaginary world as wizards and warriors (or whatever). It lasts for as long as the DM has the energy to devote to facilitating this process (in one blog posts, Alexis postulated having to retire the DM chair in the next 15-20 years). In many ways, the "Tao" of Alexis Smolensk is the antithesis of Ron Edwards, though I'd say both have a devotion to the hobby and an incredible ability to "think outside the box" when it comes to pushing gaming in new directions.<br /><br />*<i>ahem</i>*<br /><br /><i>Chapter 6: Pomp</i> discusses actual presentation and the logistics of running a game...how to show up and make arrangements, and how taking care of real world issues can create a better (more engaging, less distracting) game environment. It talks about ways to facilitate engagement through your appearance and movement, and the benefits of preparation, as well as how best to set breaks and ground rules...things often left out of most RPG game manuals.. It's good stuff for anyone who plans on running a game.<br /><br />All in all, I found a lot of good material in this section. It certainly gave me a lot of food for thought with regard to self-examination (as both a DM and player).<br /><br /><b>PART 2: Managing Yourself as DM</b><br /><br />This section is only composed of two chapters, a total of 55 pages. For me, it was the first section that I found challenging. Not because it was hard to read or too abstract in concept (that comes later), but because it challenges you on <i>what you really think</i> about role-playing games and being a DM. While <b>Part 1</b> requires you to approach the running of an RPG with a serious, non-casual approach, <b>Part 2 </b>requires you to approach idea of DM'ing almost as a <i>vocation</i>. It is not explicit in this, does not require you take any vows, but if you plan on following the prescribed course, you're basically committing yourself to making your game much more than a "mere game."<br /><br /><i>Chapter 7: Vigilance </i>discusses how your game must always be "on" when you're at the table. Even when you are acting <b>in service of your players</b> (remember, <i>that</i> is one of the main thrusts of the book), you can't let little things like, say, "friendship" get in the way of your focus or attention to the task at hand. The vigilance Mr. Smolensk prescribes (with regard to oneself) is a near ruthless stance. He discusses <b>stress</b> in the game (both for players and DM) as a product of an engaging role-play experience, and its chemical effect on the brain and decision-making process.<br /><br />This particular part did not ring true for me (perhaps because I tend to compartmentalize stress)...but then perhaps it's been a while since I had a truly engaging immersive role-play experience. I have to think back to my youth for examples of events that propelled extreme emotional outbursts in myself...though I have observed it in others to greater and lesser degrees over the years. Perhaps my decades of experience of telling myself "it's only a game" has done a bit to dull the shine, or perhaps I am simply out of practice when it comes to full-on emotional commitment in the last 15 years or so. However, I can see how my own response to players who "just want a fun night out" has caused (in the last couple years) a downward spiral in actual gaming quality, as I too forgot focus and allowed myself to lounge in the easy camaraderie and laissez-faire attitude of "dudes blowing off steam at the bar."<br /><br />[<i>that's something else that I don't have enough of in my life!</i>]<br /><br />But that's what I mean by "challenging." Without asking it outright, Alexis is posing a question: <b>how seriously do you want to take your game?</b> And what <i>quality</i> of play do you want to have? It's a valid question. I can do the indie one-off gaming thing very easily...I can likewise run a simple "dungeon excursion" with minimal effort...but is that satisfying? It's a hard question. If I'm being honest, the answer is: probably not. Certainly not <i>always</i>.<br /><br />If you decide to buy into the effort described in Chapter 7, then <i>Chapter 8: Decision Making</i> provides additional practical tools to help with your game, from non-attachment (rolling with the unexpected), to anticipating patterns of behavior, to using checklists and worksheets (goes hand-in-hand with the management of stress-related mental slips).<br /><br />Once again I see I'm running long, so I'm going to have to continue this till tomorrow. Sorry!http://bxblackrazor.blogspot.com/2014/12/[email protected] (JB)0tag:blogger.com,1999:blog-4143435314932633148.post-4842263499091122231Wed, 03 Dec 2014 21:58:00 +00002014-12-04T13:52:50.450-08:00art of the DMreviewstaoHow to Run (Part 1)<div style="text-align: right;"></div>Before returning to Paraguay for my "winter break," I made a point to order a print copy of <b><a href="http://www.lulu.com/shop/alexis-d-smolensk/how-to-run-an-advanced-guide-to-managing-role-playing-games/paperback/product-21715730.html">How to Run</a> </b>by <b>Alexis D. Smolensk</b>, known throughout this neck of the interwebs for his blog <a href="http://tao-dnd.blogspot.com/">The Tao of D&D</a>. Fortunately, the extra I paid for shipping allowed me to get the book during my two week window in Seattle, and I spent about four days reading the thing cover-to-cover.<br /><br /><table cellpadding="0" cellspacing="0" class="tr-caption-container" style="float: right; margin-left: 1em; text-align: right;"><tbody><tr><td style="text-align: center;"><a href="http://3.bp.blogspot.com/-22VEKMCCRTQ/VH-Bd6m7QfI/AAAAAAAAAZk/6oXr3JMrV1A/s1600/Sandro.jpg" imageanchor="1" style="clear: right; margin-bottom: 1em; margin-left: auto; margin-right: auto;"><img border="0" src="http://3.bp.blogspot.com/-22VEKMCCRTQ/VH-Bd6m7QfI/AAAAAAAAAZk/6oXr3JMrV1A/s1600/Sandro.jpg" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">Writing the Tao of D&D</td></tr></tbody></table>This is the first book by Alexis that I've read, though I've been reading his blog for a few years. Those familiar with his blogging will find the book to be a similar work, in that it is carefully and thoughtfully crafted, intellectual, and fairly no-nonsense. No, it does not contain rants nor inflammatory passages...this is a practical, working text <u>not</u> a blog (all bloggers are entitled to spew and vent and rail at times), but it shows the same attention that Alexis seems to instill in all his writing (here's an <a href="http://www.tinyrevolution.com/mt/archives/000152.html">old essay of his</a> from 2004, years before his Tao blog).<br /><br />I've read <a href="http://www.amazon.ca/product-reviews/B00MENQP8C/ref=cm_cr_dp_see_all_top?ie=UTF8&showViewpoints=1&sortBy=byRankDescending">the reviews of the book on Amazon</a>, calling <i>How to Run</i> "invaluable," a "must-have," and "the most important RPG book around." That's all great praise, but to most folks it doesn't really describe what the hell you'll find in this <b>"Advanced Guide to Managing Role-Playing Games"</b> (the book's subtitle). It was my interest/curiosity in Alexis's work...and respect for his mind and passion for gaming...that prompted me to buy the book. This review (which, as usual, has an excessive preamble) will attempt to explain what the book contains, to better decide if you want to put it on your shelf. That's not a <b>spoiler alert</b>; this isn't a work of fiction we're talking about, after all.<br /><br />This might get a little long...but what were you planning on doing this week anyway?<br /><br />Oh, yeah...before I explain what's in the book, I'm going to take a couple-few paragraphs to explain <b>Mr. Smolensk</b>, his approach to gaming, and DMing in general; this will be helpful in understanding what's contained in the book. Don't roll your eyes at me. This book may not be the "most important RPG book" on the market, but I daresay it is an <i>extremely</i> important book to Mr. Smolensk. His own gaming world (what us Old Schoolers might call the ongoing campaign or fantasy milieu) has become tantamount to a <b>life's work</b>. I won't say it <i>IS</i> his life's work; I'm not sure if the world serves his game or if his game serves his world. Regardless, the book <i>How to Run</i> is an opus explaining how to create your game (as a DM/GM) and your game world in the same way as Alexis...how to approach them <i>as</i> a life's work, in other words.<br /><br />This is valuable sharing. There are many folks who have decades of DM'ing experience...<i>successful</i> experience...floating around the tabletop gaming world. I'm one of them. But few of us, if any, have taken as extensive a look into the own inner workings of our imaginations, with as calculated and clinical an eye, as Alexis has. And none, that I know of, have taken the time to publish a work on the subject. Successful DMs - I define the term by those who run games that are enjoyed by all the participants and that instill a love of the hobby in the players - probably don't ponder too hard on what makes their games successful. Despite being growers of the hobby, we don't go out of the way to teach the art of Dungeon Mastering to neophytes. "Only DM'ing games will teach you how to be an effective DM" is the usual line.<br /><br />But that's not the only reason we're not sharing our knowledge. In addition to probably not knowing where to even start (i.e. developing a curriculum...jeez), there's a sinister aspect to being a long-time DM: there is a feeling of <b><i>power</i></b> in being a world builder and absolute authority to the players who come to your table. And teaching others how to do what you do is akin to giving away your power...what if someone you teach should become a better DM than yourself? What if you lose "your players" to someone who is more creative or who spends more effort or who makes a better presentation? What if you're (O Horror!) relegated to the ranks of <i>a</i> <i>mere player</i>?<br /><br />[<i>the sad fact is, we all have attachments, especially to things like power and respect and even to "looking good"...some of us have sat the DM chair so long that we're rusty (or downright idiotic) when it comes to playing a PC in a game, and it's tough to go from being a first rate DM to a second rate player</i>]<br /><br />However, it's not just a lust for being top dog that prevents folks from dispersing their "trade secrets." Building fantasy worlds from scratch (even working off an existing model like Tolkien's Middle Earth or medieval Europe) can be an intensely personal experience as one must go inside one's own mind to build the imaginary construct for the players. Sharing such a process can be very difficult, especially outside of one's friends at the gaming table (who know, understand, and appreciate your work). Explaining the process is even harder.<br /><br />So for Mr. Smolensk to do these things (which he does) is an exceptional accomplishment, and makes <i>How to Run</i> an unusual book. It's a book that some of us <i>could</i> write (or at least attempt), but few would have the balls to do so. In addition to requiring a hard analysis of your own gaming skills (and the skill to make that analysis and write it up in a digestible format), you have to have the "cred" to back it up. Alexis has been running the same game since 1986...through many different players of different ages, different skill levels, different degrees of RPG knowledge. The players enjoy his game and he has the proof in that they come back for more...as Mr. Smolensk writes himself, the proof of a good game isn't whether or not the players <i>say</i> they had fun, but what they show in their actions (do they put their butts back in the chairs the following week). Alexis has managed to do that for nearly 30 years, and not simply with the same hoary grognards that he came into the game with back when he was a kid in the 70s.<br /><br />To write such a book, you have to have (in addition to skills) a pretty solid sense of your self. You have to have a pretty concrete ideal on which to stand, if you're going to put out a unique work and share a part of your soul. Publishing a book is hard enough as is...that's a lot of blood, sweat, and tears that goes into something that might be derided and ridiculed...or, worse, <i>ignored</i> by the reading public. But when it comes to something of such immense importance to a person, the desire to get it <i>right</i>...or at the very least, <i>respectably good</i>...must be an <b>intense burden</b>. I say this as someone who gets incensed when <i>I</i> receive a one-star review (<i>sans</i> comment) on a book of mine that was anything but a magnum opus.<br /><br />Alexis Smolensk has a concrete ideal when it comes to role-playing games, and especially to the DM's responsibility with regard to the game. For him, <b>role-play is about immersion and escapism</b>, for all participants, and this is accomplished through <b>active engagement of the players</b>. The DM's role is in facilitating the players' engagement such that they can do nothing but enjoy and revel in the fantasy; <b>every action the DM takes - from hosting the game to world building - is in service of this</b>. The book <i>How to Run</i> explains how to do this.<br /><br />Now, just in case you're wondering, Mr. Smolensk and I don't see eye-to-eye on everything. In fact, in some ways we have diametrically opposed views towards gaming. So, if you think this review is going to be all flowers and sunshine being lauded at Alexis...no, that's really not the plan. There's a lot to like here, but...well, I'll get to all that tomorrow in <b><a href="http://bxblackrazor.blogspot.com/2014/12/how-to-run-part-2.html">Part 2</a></b>. Sorry for being a tease, but my rambling preamble has used up my blogging time...for the moment.<br /><br />More (and more specifics) later.http://bxblackrazor.blogspot.com/2014/12/[email protected] (JB)2tag:blogger.com,1999:blog-4143435314932633148.post-2871288523839465709Wed, 03 Dec 2014 09:28:00 +00002014-12-03T01:29:32.015-08:00paraguaysnafutravellerTravel HellI'm back in Paraguay.<br /><br />It took <b>59 hours</b> door-to-door due to various snafus and a reroute through <b>Buenos Aires</b> (that's Argentina, folks). I know there are people who probably thing I'm a whiny bitch and wish they could be so fortunate as to have such a travel adventure, but this weren't no picnic. It wasn't just that I was traveling with a toddler and an infant (the two, especially my almost-four-year-old, are quite used to these journeys and extremely sedate/well-behaved)...no it was the quarter-ton of baggage that we had to deal with. When your family is gearing up to live in the Third World for seven months, <b>you pack a lot of shit</b>. The snafus and reroutes ended up meaning a lot of shlepping of giant, heavy suitcases by Yours Truly through multiple airport check-ins, customs, whatnot. We got into town Monday night and my back is still killing me. I should probably chew some ibuprofen.<br /><br />So here we are. About two-and-a-half days of travel, and I killed my first cockroach (in home) less than 24 hours later (and my second one a couple hours after that...God, I hate cockroaches). We actually got in Monday night (it's Wednesday now, right?), but we're all still adjusting back to the five hour time difference. The fact that the coffee maker broke sometime while we were gone hasn't helped.<br /><br />Ah, well.<br /><br />I hope to write (or at least start writing) another blog post of more immediate (i.e. gaming) interest later today. Or maybe right now. Everyone besides me is still asleep (I've been up since 4am or so), and the morning's been quiet. Well, you're still going to have to wait for it, okay.<br /><br /><b>BTW:</b> One positive thing to come out of our mishaps? I found that I've been grossly misled about the character and personality of Argentine folks. Certainly it was a small sample size (we were in town a bit less than 24 hours), but every person with whom we met and interacted (I count 17 off the top of my head) was kind, helpful, friendly, and positive/cheerful (a little no-nonsense at times...but always professional). Our brief stint in B.A. was a highlight of the journey...though I would've happily skipped the experience to arrive Sunday morning, as planned.http://bxblackrazor.blogspot.com/2014/12/[email protected] (JB)10tag:blogger.com,1999:blog-4143435314932633148.post-1852932659050282448Sat, 29 Nov 2014 01:34:00 +00002014-11-28T17:34:34.608-08:00filmnewsswReally? Are You Kidding Me?Not a single one of the dozens of gaming blog has anything to say about the new <b>Star Wars </b>preview? I had to watch the thing a couple-three times. I'll probably watch it another few times before I get on the plane tomorrow morning. But I figured I'd see some comments around this corner of the internet.<br /><br />Jeez, folks...too busy <b>shopping</b> or something?<br /><b>; )</b><br /><br /><a href="https://www.youtube.com/watch?v=OMOVFvcNfvE">Here's a link</a> for those who missed it. My own comments in a couple days...I've got to pack for a looooong trip right now.<br /><br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="http://1.bp.blogspot.com/-GbpVD6D0uPI/VHkiLiakpKI/AAAAAAAAAZU/Txiig-JatZQ/s1600/images-3.jpeg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="http://1.bp.blogspot.com/-GbpVD6D0uPI/VHkiLiakpKI/AAAAAAAAAZU/Txiig-JatZQ/s1600/images-3.jpeg" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">No Opinions?</td></tr></tbody></table><br />http://bxblackrazor.blogspot.com/2014/11/[email protected] (JB)16tag:blogger.com,1999:blog-4143435314932633148.post-6629809943549630427Wed, 26 Nov 2014 18:09:00 +00002014-11-26T10:16:02.414-08:00baranoffamilyholidaynfltao"Bittersweet"That's the word that comes to mind as my time in SeaTown dwindles: <b>bittersweet</b>. Come Saturday morning, I'll be once again winging my way to Paraguay, 25 hours of travel time (door-to-door) ahead of me. The return has been wonderful, but just not long enough. My poor son loves this town, and has been so happy the last week and a half, despite lacking toys and costumes and the usual kids he'd play with. He'll be very sad to leave...as will I and my wife.<br /><br />Just not enough time to DO all that we want...but then, what we want requires living here, really. We've soaked in what we could: breakfasts out, good seafood, corned beef reubens and tasty beers.<br /><br />[<i>maybe not as many holiday martinis as I'd like, but at least they <b><u>tasted</u></b> like gin martinis are supposed to taste</i>]<br /><br /><table cellpadding="0" cellspacing="0" class="tr-caption-container" style="float: right; margin-left: 1em; text-align: right;"><tbody><tr><td style="text-align: center;"><a href="http://3.bp.blogspot.com/-Givutx3_LGM/VHYYI1OsnjI/AAAAAAAAAZE/uvs0puNjAPM/s1600/images-2.jpeg" imageanchor="1" style="clear: right; margin-bottom: 1em; margin-left: auto; margin-right: auto;"><img border="0" src="http://3.bp.blogspot.com/-Givutx3_LGM/VHYYI1OsnjI/AAAAAAAAAZE/uvs0puNjAPM/s1600/images-2.jpeg" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">Lot of effort from Coop on that TD.</td></tr></tbody></table>The weather has been beautiful (to my taste), and I made it to a Seahawks game (against the Cardinals...a good win against a tough defense)...I have been informed by folks that watched that my brother and I even showed up on the television broadcast, which is cool. I got to the dentist and I got my haircut. Tomorrow I get my favorite holiday feast (though we aren't cooking this year...just too much to do and too little time), plus a host of great football games to watch. I managed to make it to a game store and shoot the shit about gaming...something I haven't been able to do in <i>months</i>.<br /><br />But we just haven't been able to see and visit and hang out with as many people as we would have liked. Family, yes, a little. Friends? Very, very few. In fact, if we hadn't run into a couple at local stores, I'm not sure we would've had the chance to see <i>any</i> of 'em. Nice as it is to spend time with folks during the holidays, most <i>every</i>one is busy with their families. And having two young children, a wife who's under the weather, and no available childcare has meant it's...well, pretty much impossible even to take a buddy up on an offer of a beer.<br /><br />So, yeah...bittersweet. This time next week I'll be back in the 90 degree weather and thunderstorms of Asuncion. I'll (hopefully) have a bit more time to <i>write</i>, which will be nice, but I'll sure be missing my hometown. Keeping things terribly positive might be tough if there's any lingering depression (or travel fatigue). Then again, maybe this short jaunt will be the kind of "battery recharge" I need to get me through the Paraguayan summer. Still, I know I'll be looking forward to our end o the year trip to Mexico when we'll again be surrounded by family, excellent food, and pretty good weather.<br /><br />[<i>and this year, the in-laws should find my Spanish remarkably improved!</i>]<br /><br />To all my friends and gamer buddies that I haven't had a chance to see this trip...Greg, Heron, Josh, Kayce, Matthew, Randy, Tim, Will (and Ashley), etc....hopefully I'll have a chance to see you folks in a couple-three months (when I'm supposed to make a short trip back). Otherwise, I'll do my best to hook up with you all when we're back in June.<br /><br /><b>Hope everyone has a happy holiday!</b><br /><b>: )</b><br /><br />[<i>by the way, spent the last few days reading <a href="http://tao-dnd.blogspot.com/">Alexis Smolensk</a>'s book, <b>How to Run</b>, just finishing it yesterday morning. I plan to post a detailed reviewing the next couple-few days (when time allows), but right now I'm still composing/digesting my thoughts on it</i>]http://bxblackrazor.blogspot.com/2014/11/[email protected] (JB)1tag:blogger.com,1999:blog-4143435314932633148.post-5128803180039481485Sun, 16 Nov 2014 01:11:00 +00002014-11-15T17:11:52.552-08:00When Cartoons Didn't Suck<b>Happy birthday to me!</b> Actually, it's past my birthday (that was the 13th), but every day since I've got back to Seattle (Thursday afternoon) has <i>felt</i> like my birthday. <b>God, I love this town!</b> I've just been positively giddy. I told the clerk at the local Fred Meyer that her produce was absolutely <i>beautiful</i>, and I've been singing "There's No Place Like Home for the Holidays" every morning in the shower. Gosh...I've been having so much fun just breathing cold, crisp air. Hell, and going to restaurants that serve <i>breakfast</i>! What a novelty!<br /><br />Fact is that, while I've adapted to Paraguay, and can live there just fine (something I probably couldn't say with complete honesty six months ago), I certainly haven't "fallen in love" with it. I just conveniently forget all the things we're missing when we're not here. Tonight: fresh cracked <b>Dungeness crab </b>at my mom's house! No, not at a restaurant, <b><i>are you</i></b> <b><i>insane</i></b>? Pick it up at the supermarket, in season, for an exceptionally reasonable price...they'll clean it for free and you melt the butter!<br /><br />[<i>we had to get the car tabs renewed and an emission test this morning (after a friendly parking enforcement person pointed out the expiration), and my wife was just all smiles about it. "It's so nice to be somewhere the system works!" <b>Efficient bureaucracy: </b>there are a lot of anti-government haters that would miss it if it were suddenly stripped from them. Also nice? Being able to run to the grocery store after midnight and pick up quality diapers. It's the little things, people. I had to call <b>AAA</b> not once but TWICE yesterday to jumpstart my 13 year old battery that had lost its charge in the arctic cold front Seattle's been experiencing, and even with the wait it still didn't derail my plans for the day. That's just awesome</i>]<br /><br />Another way it's been feeling like my birthday is because I've been unwrapping a new "present to myself" daily. I ordered several things on-line over the last month or two and they were all (with one exception) waiting for me when I got home. One was my copy of <b><i><a href="http://bxblackrazor.blogspot.com/2014/06/warriors-of-red-planet-beta.html">Warriors of the Red Planet</a></i></b> (Beta) which came from Lulu, is beautiful, and needs its own separate blog post. But most of the stuff I ordered were DVDs of several old TV series including <b><i><a href="http://bxblackrazor.blogspot.com/2014/09/wizards-and-warriors.html">Wizards and Warriors</a></i></b>, but much more fun/nostalgic for me are the (animated) Tolkien trilogy (Rankin/Bass and Bakshi) plus <b><i>Thundarr the Barbarian</i></b>, <b><i>Sealab 2020</i></b>, (<u>not</u> the humorous remake <b><i>Sealab 2021</i></b>), and <b><i>Jonny Quest</i></b>. I can (and probably will) write a complete blog post or two on each of these marvelous cartoons, but suffice is to say, hey, <i>remember when people were making cartoons for children that didn't suck</i>?<br /><br />Okay, got to go...Dungeness calls!<br />: )<br /><br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="http://2.bp.blogspot.com/-hbhMqVsRDhg/VGf5vy_-VzI/AAAAAAAAAY0/VYEzw4HQMR4/s1600/images-2.jpeg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="http://2.bp.blogspot.com/-hbhMqVsRDhg/VGf5vy_-VzI/AAAAAAAAAY0/VYEzw4HQMR4/s1600/images-2.jpeg" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">Delicious!</td></tr></tbody></table><br />http://bxblackrazor.blogspot.com/2014/11/[email protected] (JB)11tag:blogger.com,1999:blog-4143435314932633148.post-7027911282752332746Thu, 13 Nov 2014 16:38:00 +00002014-11-13T08:49:39.982-08:00designdiversityswwifeChecking My Privilege<div class="MsoNormal">I’ve been thinking a lot about diversity and inclusiveness lately (as in, the last 12-13 months or so), mainly with regard to game design. <a href="http://bxblackrazor.blogspot.com/2014/11/film-hate.html">My rant</a> the other day had an element of this and…well, I’m on a plane heading for Chicago at the moment with shit-for-TV playing instead of <b><i>Guardians of the Galaxy</i></b>, so I might as well pound out a blog post on the subject.</div><div class="MsoNormal"><o:p></o:p></div><div class="MsoNormal"><br /></div><div class="MsoNormal">[<i>creating games with “adequate diversity” (and with attention to including all sorts of folks) is something of a hot-button topic these days, so I’ll probably take shit from both sides for my reflections on the subject. I’m cool with that</i>]<o:p></o:p></div><div class="MsoNormal"><br /></div><div class="MsoNormal">Gosh, where even to start? I feel challenged to even address the subject of diversity when I come from a place as privileged as I do. I’m white, male, straight, American, non-trans-anything. I’m married with a couple kids. I’m Roman Catholic and even though Catholics sometimes take some heat for their religion, it’s tough to feel marginalized when you’re part of the single largest religion on the planet (at least, if you’re including the “lapsed” and “non-practicing” Cat-Lickers). <o:p></o:p></div><div class="MsoNormal"><br /></div><div class="MsoNormal">I’m about as <b>white-bread normal</b> as they come. I’m a drinker, but a functioning one. I eat meat, but know it’s bad for my cholesterol.<span style="mso-spacerun: yes;"> </span>I’m not a porn watcher, but I want it to be accessible (except for my kids). I watch a lot of football. I drive a car and have a bank account. There’s nothing “oppressed” about my life, and (for the most part) the society I live in is one set-up and designed to support <i style="mso-bidi-font-style: normal;">me</i>. I may bitch-and-moan when Americans stupidly vote <b>Tea Party Republicans</b> into office, but that’s just a <i style="mso-bidi-font-style: normal;">principle</i> thing. I and my family benefit from having a Republican controlled Congress. Paying less taxes just means <i style="mso-bidi-font-style: normal;">I’m</i> getting richer…my health care and retirement and whatnot is already paid for, and I live in a nice enough neighborhood of Seattle that if my kids end up going to public school they’ll be in one of the best.<o:p></o:p></div><div class="MsoNormal"><br /></div><div class="MsoNormal">It’s a disgusting abundance of privilege that I have.<o:p></o:p></div><div class="MsoNormal"><br /></div><div class="MsoNormal">[<i>sorry…had to break to eat my cheese blintzes, fresh fruit, and sausage. Oh, and order another complimentary Bloody Mary. <b>Business class</b>, ya’ know?</i>]<o:p></o:p></div><div class="MsoNormal"><br /></div><div class="MsoNormal">I’m not in the top 1% of Americans, probably not even the top 10%, but I’m well above “median income,” and I’ve never really suffered; and hell, there’s no real suffering in sight. I own a nice house, I’ve got no crushing debt (car and student loans were paid off long ago), and while two kids can mean a financial burden, I’m still able to get to a few Seahawks games. Whether or not people of the same sex can marry has no effect on my life; what does matters is if my cable and high speed internet are up-and-running. Regardless of whether or not I “support the troops,” the troops are certainly supporting me and my lifestyle. The fact that I have time and energy to complain about WotC or lack of diversity in films or the weather in Paraguay should tell you that truthfully, I really have no complaints at all. <o:p></o:p></div><div class="MsoNormal"><br /></div><div class="MsoNormal">Thus it’s a challenge for me to have any kind of “cred” when it comes to talking about changing game (design) culture.<span style="mso-spacerun: yes;"> </span>I can’t talk from a background of being oppressed or underrepresented or misrepresented because, hey, I belong to the ruling class. And it’s not like I got this through hard work or <i>metaphysical visualization-manifestation</i>. I just happen to be born into the right place at the right time with the “right” gender-color-orientation. Dig?<o:p></o:p></div><div class="MsoNormal"><br /></div><div class="MsoNormal"><b>So why bother?</b> When the best you can be considered is an “ally to the cause” and the worst is some misguided dude with “white knight syndrome,” why the F even bother? Why not just continue to design shit without any secondary agenda? “For my children?” They’re already set on a course for being as privileged as myself (if not more so). Because of my “white guilt?” No: it’s hard to feel guilty about “writing what you know,” even if what you know isn’t incredibly diverse. Because it’s “different” or “novel?” Well, novel ideas are a better way to get on the market than recycled hash, but that’s hardly a reason to make the effort when the hash sells fine.<o:p></o:p></div><div class="MsoNormal"><br /></div><div class="MsoNormal">No. There are a couple-three reasons at work here (for me, anyway):</div><div class="MsoNormal"></div><ul><li>There’s a problem in gaming and I don’t want to be part of the problem. The problem is, there’s a lot of white-male (sometimes juvenile) designers designing games for a white-male (sometimes juvenile) audience…not necessarily on a conscious level, mind, just because that’s what they know. And there are more people out there who game…or who might enjoy games…than just white males of a juvenile persuasion.</li><li>Growing the hobby (i.e. creating more audience) is something I’m all about. It seems only logical (to me) that making games more inclusive (with inclusive language, concepts, art) are going to make some folks (who might otherwise have been “turned off”) more interested in exploring the hobby. Maybe not, but <b><i>I </i></b>don’t need to cater to the die-hard, grognardy, fans. There’s already people (designers/publishers) doing that and keeping those folks involved in the hobby.</li><li>It’s the Goddamn “right thing to do.” That is to say, it’s <i>not</i> right to be exclusive when it comes to design…not when we live in a world where different cultures and backgrounds are afforded the same opportunity (and thus access) to the same games. If they’re there, why leave them out, or make them feel marginalized?</li></ul><o:p></o:p><br /><div class="MsoNormal"><o:p></o:p></div><div class="MsoNormal"><o:p></o:p></div><div class="MsoNormal">My wife is originally from Mexico (she’s lived steadily in the US since 1997). In years past, when I asked her to state her race (for example, on a census report) she said “Mexican.” Nowadays, she identifies as a “Latina” but that’s not really a race, either. Technically she is a <i>mestizo</i>, as are the vast majority of native Mexicans: a person of mixed (white) European ancestry and native Mesoamericans (“Indians”). Because nearly all of Mexico is “mestizo” they’ve stopped using the term, thus my wife’s lack of a term for herself. (she has absolutely zero identification with “Native Americans”). My wife is NOT white. Our children are white with dishwater blond hair and green eyes (like their papa), and they’re different enough that my wife has remarked she might be identified as “the help” when she’s out with them.</div><div class="MsoNormal"><o:p></o:p></div><div class="MsoNormal"><br /></div><div class="MsoNormal">I asked my wife her opinion about the diversity (or rather, lack of diversity) in film-thang (she doesn’t care about games). She states that it doesn’t bother her and that she doesn’t care. She says she’s always considered it silly that people complain about underrepresentation in film because she “goes to movies to see other people anyway” (whether they’re white, or black, or blue-skinned). It doesn’t matter to her what ethnicity is cast in a film, unless it’s a piece about a particular culture or time period. <o:p></o:p></div><div class="MsoNormal"><br /></div><div class="MsoNormal">What <i>IS</i> of concern to her, and what <i>IS</i> important is the lack of strong female characters in film. <b>Star Wars</b> (for example) has white folks and non-white folks and alien folks…but why are they so over-whelmingly male? A couple bit parts aside, the only female character in the prequel trilogy is Padme, and what purpose does she really serve besides acting as a goal/objective for the (male) protagonists in the first couple films before being relegated to mere “set decoration” in Episode III?<o:p></o:p></div><div class="MsoNormal"><br /></div><table cellpadding="0" cellspacing="0" class="tr-caption-container" style="float: right; margin-left: 1em; text-align: right;"><tbody><tr><td style="text-align: center;"><a href="https://1.bp.blogspot.com/-aix7H5wbJcw/VGTeWFisfpI/AAAAAAAAAYk/0NNgpLgO2qA/s1600/Unknown.jpeg" imageanchor="1" style="clear: right; margin-bottom: 1em; margin-left: auto; margin-right: auto;"><img border="0" src="//1.bp.blogspot.com/-aix7H5wbJcw/VGTeWFisfpI/AAAAAAAAAYk/0NNgpLgO2qA/s1600/Unknown.jpeg" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">Weak Sauce</td></tr></tbody></table><div class="MsoNormal">So, my (non-white) wife would say lack of diversity/inclusion bothers her more when it’s gender inequality that’s on display. If she were a gamer, I’d imagine the same standard would apply: if she’s pretending to be a wizard or cyber-hacker it doesn’t matter whether she’s white or black or “troll-colored.” The equal representation of male and female (and active, protagonist female) is more important.<o:p></o:p></div><div class="MsoNormal"><br /></div><div class="MsoNormal">All right, this is getting long….maybe I’ll get back to this subject in a "Part 2," but right now I’m getting to land. Later, Gators.<br /><br />[<i>posted from Chicago O’Hare</i>]</div><div class="MsoNormal"><o:p></o:p></div>http://bxblackrazor.blogspot.com/2014/11/[email protected] (JB)9tag:blogger.com,1999:blog-4143435314932633148.post-7420757703423392592Tue, 11 Nov 2014 13:57:00 +00002014-11-11T06:02:48.967-08:00crapfilmmckinleymoneyOSRrobersonswtolkienFilm HateCan I just talk about <b>hate</b> for a brief moment? I'm not talking about the absence or antithesis of love...I'm just talking about the standard dictionary definition of <b>"intense or passionate dislike."</b> There are some things that just make me really cranky...and while I could just keep 'em to myself, I have a feeling that might be bad for some of my internal organs.<br /><br />So here comes a short (I hope) rant regarding films.<br /><br />I don't work in Hollywood, so I have no "insider perspective" on the filmmaking process. I know that films cost a lot of money. I know that they take huge numbers of people working in collaboration to make them. I know that many of the decisions made by film companies revolve around the ability to recoup the money spent and make some profit so that the rich can get richer and produce more films, continuing the business of filmmaking (which helps employ all those people that work on making films). I know that no matter how much "vision" one, single person possesses (whether that's a writer, director, producer, etc.) that vision is going to be worked over a bit by all the other people involved in the filmmaking process...especially co-writers, co-producers, directors, film execs, etc.<br /><br />And nothing I write on this blog is going to change that. Still, I'm talking about saving my spleen, here.<br /><br />I <b>HATE</b> that there is so little originality coming out of Hollywood...that so many of the Big Budget Blockbusters are remakes (or "updates") of existing movies. Whether you're talking <i>Straw Dogs</i> or <i>Clash of the Titans</i>, I just feel like slapping people and saying this doesn't <i>need</i> a remake...if you want to introduce an old film to a new crowd, then digitally remaster it and re-release it. Otherwise make your own Goddamn movie, dudes. Aren't people being paid millions of dollars to do these things? And you can't think of an original idea? You <i>suck</i>.<br /><br />I <b>HATE</b> it when films based on novels written in different time periods, decide to spruce up ("update" again) the material in order to make it more in line with out 21st century wants and perspectives. "Modernizing" it. Look, if you're going to make a live-action version of <i>The Hobbit</i>, then make <i>The Hobbit</i>. Don't make it into something else, restructuring the characters and plot to make it "cooler." If you don't feel like being faithful in your adaptation, then come up with your own <b>original fucking idea </b>for a movie. How many cars do you own?<br /><br />[and just to add onto <i>that</i>...]<br /><br />I HATE it when filmmakers choose to make films that use an existing piece of intellectual property (like <i>The Hobbit</i>) and then inject a bunch of random token characters into it because the original IP wasn't diverse enough in terms of gender and race. THERE IS IP OUT THERE THAT FEATURES DIVERSITY, WHY DON'T YOU ADAPT THAT?! There are lots of fairy tale books that feature female protagonists (<b>Shirley Murphy's</b> <b><i>Soonie and the Dragon</i></b> comes to mind). Oh, wait...that doesn't have enough blood and guts and combat for you?<br /><br /><table cellpadding="0" cellspacing="0" class="tr-caption-container" style="float: right; margin-left: 1em; text-align: right;"><tbody><tr><td style="text-align: center;"><a href="http://2.bp.blogspot.com/-xShN95Q9O0g/VGIQ8DNqglI/AAAAAAAAAYM/83pbrhfDX2A/s1600/Unknown.jpeg" imageanchor="1" style="clear: right; margin-bottom: 1em; margin-left: auto; margin-right: auto;"><img border="0" src="http://2.bp.blogspot.com/-xShN95Q9O0g/VGIQ8DNqglI/AAAAAAAAAYM/83pbrhfDX2A/s1600/Unknown.jpeg" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">Nice little tale. Dragon eats people.</td></tr></tbody></table>[<i>because <b>bloody violence</b> is what has made The Hobbit a beloved classic of children's literature for decades, right? That and all the <b>songs/poems</b> Tolkien threw in there</i>]<br /><br />Fine, how about adapting one of <b>Robin McKinley's</b> (award winning) books? How about adapting <b>Jennifer Roberson's Cheysuli saga </b>(when an author gets an eight book series published, maybe there's an audience)? How about adapting the saga of <b><a href="http://en.wikipedia.org/wiki/Brynhildr">Brynhildr</a></b> to the screen in a Norse mash-up the same way Ray Harryhausen adapted Perseus and Greek myth in the original <i>Clash of the Titans</i>? Are you just <b>too fucking stupid </b>to do this?<br /><br />I HATE that so often Hollywood (when it comes to action/fantasy films) seems like the epitome of a "good ol' boys" network where the white, male (straight) hero front-and-center. I'm white AND male AND straight and <b><i>I</i></b> get tired of seeing the same old shtick.<br /><br />You know what I really, REALLY don't want to see? I don't want to go into a screening of <b>Star Wars VII</b> and see <i>three white buddies</i> (two male, one female) going on a fantastic galactic adventure. That was fine in the 1970s, but it was certainly a tired concept by the time <i>Harry Potter</i> rolled into theaters. Give <b><a href="http://en.wikipedia.org/wiki/Lupita_Nyong%27o">Lupita Nyong'o</a></b> a kick-ass role as <b>Lando's daughter-turned-space-pirate</b> and let <i>her</i> be the "Han Solo" of the movie.<br /><u><br /></u><br /><table cellpadding="0" cellspacing="0" class="tr-caption-container" style="float: left; margin-right: 1em; text-align: left;"><tbody><tr><td style="text-align: center;"><a href="http://3.bp.blogspot.com/-2-v5eK9WrTE/VGIR4KpMamI/AAAAAAAAAYU/FvDIvyISSgM/s1600/images-2.jpeg" imageanchor="1" style="clear: left; margin-bottom: 1em; margin-left: auto; margin-right: auto;"><img border="0" src="http://3.bp.blogspot.com/-2-v5eK9WrTE/VGIR4KpMamI/AAAAAAAAAYU/FvDIvyISSgM/s1600/images-2.jpeg" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">Oscar winners shouldn't have to play aliens.</td></tr></tbody></table><u>Please don't make her a token "new apprentice" of Jedi Master Skywalker with the sole purpose of being killed off by the Big Bad Guy of the film</u>. Please, please, please don't do that. And don't make her an "alien" (non-human), either.<br /><br />Anyway, these are some of the things I'm thinking/fuming about today. Earlier I watched a trailer for the new Hobbit film (I'm not even going to bother with a link...you folks can find it, I'm sure). And it just made me irritated. And then it got me thinking about Hollywood and all these things I "passionately dislike." Things I really can't change or impact or control. Which, BTW, just makes me more irritated.<br /><br />One of the wonderful things about this RPG thing...the OSR movement, the Indie movement, the DIY-self-publishing thing is how nice it is to have some control and to be able to make shit without being beholden to "money men." I know that what drives the film industry to do things I hate is (for the most part) <b>money</b>. It's the same reason why people choose to live in Paraguay (for those who HAVE a choice): here you can start a business and expect (on average) 10-30% return on your investment annually.<br /><br />[<i>which is crazy-huge compared to the 5-8% (or 5-10%) you'd expect in the USA. Here, you can recoup your costs after three years and then just profits, baby, ever after. The money doesn't, of course, get reinvested in the country...but who cares when you can afford to send your kids to school in the United States? Just sad, in so many ways...</i>]<br /><br />People in the self-publishing RPG biz aren't getting rich, clearly, but at least folks are getting a chance to make games (and play games) that they want. If game design is an "artistic" enterprise (and I'd say it is), then at least its not the compromised art of the (establishment) film industry. And that makes it cooler than Hollywood and (for me) mostly "hate-proof."<br /><br />Mostly.<br />; )http://bxblackrazor.blogspot.com/2014/11/[email protected] (JB)9tag:blogger.com,1999:blog-4143435314932633148.post-4914614752614994269Fri, 07 Nov 2014 12:37:00 +00002014-11-07T04:57:00.219-08:00holidaymoonparaguay"It Doesn't Rain, But It Pours"<table cellpadding="0" cellspacing="0" class="tr-caption-container" style="float: left; margin-right: 1em; text-align: left;"><tbody><tr><td style="text-align: center;"><a href="http://3.bp.blogspot.com/-61cD3FqtMak/VFzBbXYYViI/AAAAAAAAAX8/BeCCz-DkA1E/s1600/street.JPG" imageanchor="1" style="clear: left; margin-bottom: 1em; margin-left: auto; margin-right: auto;"><img border="0" src="http://3.bp.blogspot.com/-61cD3FqtMak/VFzBbXYYViI/AAAAAAAAAX8/BeCCz-DkA1E/s1600/street.JPG" height="150" width="200" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">This is my "street."</td></tr></tbody></table>Despite the saying, I hadn't literally found this to be the case until we moved to Ascuncion. Holy cats! It's rainy season right now, and the street in front of my house is a stream that goes up to mid-shin in places (after stepping out o the car I got doused in water up to the knees; had to change my pants)...and we're fairly elevated! The main boulevard (a block over) is an <b>f'ing river</b>. You'd think I wouldn't have bothered taking D to school today, but funny enough I've gotten...well, not "used to" the weather, but I'm a bit more adaptable to the garbage roads and road conditions. Or maybe I'm just not as timid as I once was.<br /><br />[<i>it probably helps that I drive an SUV here, and that it's not my car...definitely wouldn't try my Jetta back home in these conditions!</i>]<br /><br />ANYway, it is storming away outside, but I wanted to throw out a quick blog post, before I get back to writing. Yes, I've been writing lately...<b>writing up a storm</b>, if you'll pardon the expression. I want to get through as much of this book as I can before my current "mental deluge" dries up, so posting will probably be light the next week or two.<br /><br /><b>Or four</b>, more likely...we're heading home (Seattle) for the holidays next week (traveling on my birthday, actually) and will be there through <b>Thanksgiving</b>: <b>the greatest holiday ever invented by Americans</b>. For my non-American readers (I know I have a few), allow me just to wax on for a moment. Combine <b>family</b> and a celebration of <i>appreciation</i> and <i>cooperation</i> (at least, that's what we're taught in grade school) with <b>football</b>, an <b>all-day feast</b>, and <b>three-to-six pies</b> of different flavors. The best part? <b>The next day's a holiday, too. </b>Drink as much as you like and nurse your hangover with turkey sandwich leftovers while catching up on those videos you've been meaning to watch.<br /><br />[<i><b>Black Friday shopping</b> is a sucker's game, just by the way...I opt out whenever possible</i>]<br /><br />So, yeah...back in Sea-Town we'll have lots to do (seeing folks, going to the '<b>Hawks vs. Cardinals</b>, buying peanut butter and tortillas to stuff our suitcases...), so probably not much time to write OR blog. We'll be back in Asuncion before December, but then heading for Mexico at the end o the year (to visit the other side of the family).<br /><br />Hey! Is anyone interested in what I'm writing? Since no one in Paraguay is (no, my family doesn't care either). Welp, work progresses on the fantasy heartbreaker...and man-o-man, I am <i>really</i> digging on it. I've been using Holmes Basic as a model, but I don't thing I'm going to get it come in at his 48 pages. Unless I'm able to skim some page count from the monster section, which is possible (still debating how I want to organize that, actually...same problem I have with the revised version of <i>Cry Dark Future</i>). Maybe 60 (including art), but we'll see. We'll see.<br /><br />[<i>really need to get back to it</i>]<br /><br />But this is one that I am hopeful will be ready for play-testing by the holidays, and I may just start emailing interested folks "beta" versions before the New Year. If you belong to a gaming group living in...oh, say, Oakland or Jacksonville or New York and you're looking for something to do during Sundays in December, you might want to keep me in mind. I'll be asking for volunteers (on the blog) sometime in the next few weeks.<br /><br />Well...if all goes well, that is.<br />: )<br /><br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="http://2.bp.blogspot.com/-teVRHHjP0Qk/VFy9C4wT02I/AAAAAAAAAXw/VZH8NvxwHSM/s1600/images-2.jpeg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="http://2.bp.blogspot.com/-teVRHHjP0Qk/VFy9C4wT02I/AAAAAAAAAXw/VZH8NvxwHSM/s1600/images-2.jpeg" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">Not my car.</td></tr></tbody></table><br />http://bxblackrazor.blogspot.com/2014/11/[email protected] (JB)1tag:blogger.com,1999:blog-4143435314932633148.post-2437352745552965791Wed, 05 Nov 2014 19:45:00 +00002014-11-05T11:45:10.701-08:00wifeHappy Birthday To My Sweet Wife!As one might imagine, I'm a little busy today. Will hopefully resume posting tomorrow or the next day.<br />: )http://bxblackrazor.blogspot.com/2014/11/[email protected] (JB)5tag:blogger.com,1999:blog-4143435314932633148.post-357374818716694828Mon, 03 Nov 2014 15:11:00 +00002014-11-03T07:11:43.055-08:00moonspellsMight and Magic (Redux)Because many folks don't appreciate the football (or, at least, don't share my obsession with the NFL), I feel it best I follow up <a href="http://bxblackrazor.blogspot.com/2014/11/halfway.html">today's earlier post</a> with a gaming-related post.<br /><br />[<i>'course, if you DO enjoy the football stuff, you might find my reflections mildly amusing...or at least something worth arguing about</i>]<br /><br />So remember last week when I was talking about "color-coded magic" and "wizard schools" and whatnot (you can <a href="http://bxblackrazor.blogspot.com/2014/10/might-and-magic-part-2.html">read the post here</a>, if you missed it)? Okay, well, yeah, I worked out the spell lists and such over the weekend. That is to say: <b>new magic system complete</b>.<br /><br />I was still able to keep elements from my prior system (like cross-polinization of magic and building on prior knowledge). I was able to break the magic up into separate "themes" (and, yes, the themes include some color-coding). The spells are still limited in number to a <b>"Fantastic Forty"</b> plus a <b>"Forbidden Four." </b><br /><br />Actually, it's a <i>Fantastic Fifty</i> and <i>Forbidden Five</i>...but one book is for NPC's only. After all, in <i>this</i> particular heartbreaker, PCs are supposed to be <b><u>H</u>eroes</b>. You can't have heroes using the same terrible magic as the bad guys! Those guys are EVIL...you're supposed to be striving <i>against</i> the <b>Forces of Darkness</b>, not adopting their weapons.<br /><br />[<i>on the other hand, I'm thinking of installing some sort of "corruption mechanic" a la D6 Star Wars or The Mutant Chronicles. Is that too much for a "basic" game? Maybe</i>]<br /><br />ANYway, yeah. Just so you know. Once I finish the write-ups, I'll see how much of the magic is compatible with the existing B/X system (understanding that it would need to replace the current "Vancian" system) and if it makes sense, I'll post some spells here. But we'll see. Let me finish the actual writing/text first.<br /><br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="http://1.bp.blogspot.com/-XeJrGO-Whbg/VFeathUchOI/AAAAAAAAAXU/0P1xBNt0lPU/s1600/Unknown.jpeg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="http://1.bp.blogspot.com/-XeJrGO-Whbg/VFeathUchOI/AAAAAAAAAXU/0P1xBNt0lPU/s1600/Unknown.jpeg" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">The First Book of Magic...Collect All Five!</td></tr></tbody></table>Later, gators.<br /><b>: )</b>http://bxblackrazor.blogspot.com/2014/11/[email protected] (JB)0tag:blogger.com,1999:blog-4143435314932633148.post-313800004348714780Mon, 03 Nov 2014 14:12:00 +00002014-11-03T06:13:52.324-08:00craphistorynewsnflparaguayHalfway<b>Happy Stroessner Day!</b> Today marks the birthday of ex-Paraguayan president <b>Alfredo Stroessner</b>, who "served" the country from 1954 to 1981 following his successful military coup. <a href="http://en.wikipedia.org/wiki/Alfredo_Stroessner">Wikipedia</a> sums it up succinctly:<br /><br /><i>"His 35-year-long rule, marked by an uninterrupted period of repression in his country, was the longest unbroken rule by one individual in the history of South America."</i><br /><br />He was a real <b>crap-sandwich</b> for Paraguay...but he was a staunch anti-communist and so was backed by the good ol' U.S. of A. during his rule, while turning the country into "a sanctuary for smugglers in arms, drugs, and everyday goods like whiskey and car parts."<br /><br />[<i>and let me tell you, if you are a successful businessman or politician in Paraguay, you drink an awful lot of whiskey</i>]<br /><br />So, of course, it is yet another holiday in Asuncion, and my child was awakened early this morning by another round of fireworks celebrating this "national treasure." Because they love-love-love their fireworks here. Oh, yeah...and this morning <b>the city of Asuncion is without water today due to a broken pipe</b>. Hey, remember that time that your capital city of half a million (with an additional Monday workforce of two-four times that) had <i>no running water</i>?<br /><br />Yeah, that's been <i>my</i> morning.<br /><br />And yet, the Seahawks were able to handle the Goblin Raiders yesterday and are still in the hunt for the division lead, so I can't feel too bad. Always nice to get a win against Oakland (old AFC West rivalries die hard, ya' know?). And so I thought I'd revisit my earlier <a href="http://bxblackrazor.blogspot.com/2014/09/the-day-before-day-before.html">pre-regular season "reflections"</a> now that all the teams have had a chance to play through at least half their schedule.<br /><br />Something fun while I wait for the city to get the water back on.<br /><br /><b>AFC EAST Standings</b><br /><b>New England Patriots (7-2)</b><br /><b>Buffalo Bills (5-3)</b><br /><b>Miami Dolphins (5-3)</b><br /><b>NY Jets (1-8)</b><br /><br />The guy at ESPN who thought that Geno Smith was going to win nine games this year? <b>He's an idiot</b>. I've been giving my buddy, Steve shit for two years now about being high on Smith ("high" being the operative word); the only question is will Ryan survive the year. He's not a bad coach, but he's had inferior talent at the QB position for too many years (and you'd think that's a GM's drafting responsibility). As predicted by most (including myself), it looks like this is New England's year (again). However, kudos to both the Bills and Dolphins for making their seasons relevant and fighting for a play-off berth. The Dolphins shellacking of the Chargers yesterday was a beautiful thing.<br /><br /><b>AFC NORTH Standings</b><br /><b>Cincinnati Bengals (5-2-1)</b><br /><b>Pittsburgh Steelers (6-3)</b><br /><b>Cleveland Browns (5-3)</b><br /><b>Baltimore Ravens (5-4)</b><br /><br />We're all allowed to get one wrong, and yes, I was wrong about the <b>Steelers</b> being mediocre this year. However, before you Pittsburgh fans start jumping up-and-down and yelling "I told you so" (like I do with Steve-O), allow me to point out their wins have been at home against Cleveland (barely), against a weak-sauce Carolina, crap-sandwich Jacksonville, last year's bottom Houston (at home), a Charlie Whitehurst led Tennessee, and (at home) against the struggling Ravens. I did not reckon with how weak their schedule was this year (next up: <b>the NY Jets</b>!)...still, the second half of their schedule is a little tougher (getting to play the Saints at home is a lucky break), and while a team that drops a game to Tampa Bay is probably not going to get to the predicted 10 wins, they might still get nine. Nice to see Cleveland stuck with Hoyer, by the way (and has been rewarded for it). The Bengals are looking more and more like a <b>paper tiger</b>.<br /><br /><b>AFC SOUTH Standings</b><br /><b>Indianapolis Colts (5-3)</b><br /><b>Houston Texans (4-5)</b><br /><b>Tennessee Titans (2-6)</b><br /><b>Jacksonville Jaguars (1-8)</b><br /><br />My prediction of a good year for the Titans was conditional to Jake Locker staying healthy. <b>Charlie Whitehurst may be the worst quarterback in the NFL (after Geno Smith).</b> I know this because I watched him for a season at Seattle in one of current regimes weirdest missteps. The Titans' season is over. I see that even I was too optimistic when I said five wins for the Jags this year; I was trying to be kind to Gus Bradley who has one hell of a nightmare on his hands. What do people in Jacksonville do during football season? <b>Watch the SEC, I guess.</b> I was wrong about both the Colts and the Texans, by the way...unless Foster's injury causes Houston to tank in the second half of their season.<br /><br /><b>AFC WEST Standings</b><br /><b>Denver Broncos (6-2)</b><br /><b>Kansas City (5-3)</b><br /><b>San Diego Chargers (5-4)</b><br /><b>Oakland Raiders (0-8)</b><br /><br />When do we move the mantle of "Ageless One" from Jerry Rice to Peyton Manning? Next year? Maybe. But my prediction was Denver failing to make it back to the Super Bowl, not failing to make the play-offs. Nice to see I was right about KC <i>not</i> folding like a cheap shirt (even their blogger was predicting an 8-8 season...the "anti-homer"). Still think the Chargers can make the play-offs as a Wild Card. Still think Oakland is...well, when you look up "crap sandwich" in the dictionary, it should probably have a picture of the Raiders.<br /><br /><b>NFC EAST Standings</b><br /><b>Philadelphia Eagles (6-2)</b><br /><b>Dallas Cowboys (6-3)</b><br /><b>NY Giants (3-4)*</b><br /><b>Washington Redskins (3-6)</b><br /><br /><i>*Yes, the Giants could still get to halfway mark of their season at .500 with a win tonight, but come on...do you really think that's going to happen? Really?</i><br /><br />Did I say the Cowboys and Romo would get one more shot at the playoffs this year? And that mediocre coach Jason Garrett would get a contract extension? Yes I did, and I'm thinking that prediction is still looking good. The Eagles are still leading the division thanks to several close wins; we'll see how Nick Foles's broken collarbone (and <b>Mark Sanchez's</b> return to the captain's seat!) affects the second half of their season. Wasn't Mark Sanchez sidelined for <b>Geno Smith</b>? As for the other two teams: <b>crap sandwich times two</b>.<br /><br /><b>NFC NORTH Standings</b><br /><b>Detroit Lions (6-2)</b><br /><b>Green Bay Packers (5-3)</b><br /><b>Minnesota Vikings (4-5)</b><br /><b>Chicago Bears (3-5)</b><br /><br />What a difference an additional wide receiver from a championship team makes, huh, Matthew Stafford?<b> Ex-Seahawk Golden Tate is the 4th best receiver in the NFL this year, on pace for a 1600 yard season.</b> It's enough to wish we'd kept him instead of Percy Harvin...but that's just me. The Packers are still in it (as predicted), and...um, who was it that said the Bears "aren't going to sniff the playoffs" this year? Who said that? <b>Yeah, me.</b> The homer at ESPN who said Chicago would win ten? <b>Sniffing glue.</b> Boy, I'm kind of feisty today. Wonder when that water's going to get turned on.<br /><br /><b>NFC SOUTH Standings</b><br /><b>New Orleans Saints (4-4)</b><br /><b>Carolina Panthers (3-5-1)</b><br /><b>Atlanta Falcons (2-6)</b><br /><b>Tampa Bay Bucs (1-7)</b><br /><br />Welp, I was right in everything I said here, but I missed the mark on how truly awful the division is this year. The AFC South has nothing on their NFC counterpart's claim to worst division in the NFL. There's only one word to describe it: <b>crap sandwich</b>. Well, I guess that's really two words.<br /><br /><b>NFC WEST Standings</b><br /><b>Arizona Cardinals (7-1)</b><br /><b>Seattle Seahawks (5-3)</b><br /><b>San Francisco 49ers (4-4)</b><br /><b>St. Louis Rams (3-5)</b><br /><br />Ahh, here we go. Let's see: Arizona did leapfrog San Francisco (as I figured they would), but they've leapfrogged Seattle, too, to become the team with the best record in the NFL. And this despite having to play several games with backup QBs. This is what happens when a well-coached team (10-6 last year) gets to play a 3rd place schedule. Fortunately, Seattle still has two games to play against the Cards, so their placement at the top of the division ain't yet secure.<br /><br /><b>O my poor Seahawks!</b> The one thing I hadn't predicted was how damn injured we'd be this year. Playing with a 4th string center last night? Missing all but two of last year's starters on the O-Line (including our Pro Bowl tight end that we use almost exclusively as a blocker)? Watching Dashawn Shead (who??), a 3rd or 4th string corner getting beat for a TD pass from Raiders rookie Derek Carr? Our offense in week nine is in week two of "life after Percy Harvin" trying to rebuild and redefine itself after spending all of preseason building around a troublesome (and/or troubled) player. They still have time to turn it around (if the season ended today, the 'Hawks would be in as the #2 Wild Card based on their tie-breaker over the Packers). But they really have to handle their business in the coming weeks (five division opponents to end the season!). Hopefully, some of our (what is it now? nine?) injured starters will be back on the field to finish out the year and make a strong push. But I have a feeling we're going to see more <b>"ork-ugly" </b>games as we head down the stretch.<br /><br /><a href="http://en.wikipedia.org/wiki/Madden_NFL#Madden_Curse">Why couldn't they have put Peyton Manning on the cover of <b>Madden</b> this year?</a> Wasn't he the MVP or something?<br /><br />All right, all right. Enough of this. We'll see how the whole thing ends up over the next eight weeks. Now let's get back to some gaming posts, huh?<br /><br />[<i>still no water</i>]http://bxblackrazor.blogspot.com/2014/11/[email protected] (JB)0tag:blogger.com,1999:blog-4143435314932633148.post-7198202445505646749Fri, 31 Oct 2014 21:51:00 +00002014-10-31T14:58:29.285-07:00diegohalloweenholidayparaguaysuperswifeCosplay<b>Happy Halloween from Paraguay!</b><br /><br /><i>Fast Paraguayan Fun Fact:</i> Paraguay, as a country still trying to figure out its identity, has a shit-ton of random holidays...things like Kid's Day and Youth's Day and Friendship Day and Spring Day. <i>"JB, people in the USA celebrate the first day of Spring, too...or at least mark it in passing."</i> Yeah, but you don't see big banners put up across streets and decorations hung from lamp posts and lunch specials and people exchanging gifts and talking about "what I'm doing for Spring Day." It's like they're starved for things to celebrate.<br /><br />So, it's not all that surprising when they borrow an American-style Halloween, too, though it's really not the same (this is one of those countries where everyone has big walls, no yards, and prominent razor wire...not the inviting household for trick-or-treaters...plus, see earlier posts regarding the perils of being a pedestrian in Asuncion). Still, my boy's preschool did a costume day ("no scary masks, please") and my wife, as is her usual thing, was up till three in the morning putting the finishing touches on it:<br /><br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="http://3.bp.blogspot.com/-qB6SRygOxXw/VFP0vDQ65wI/AAAAAAAAAXE/otteHCZEBXA/s1600/Robin.JPG" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="http://3.bp.blogspot.com/-qB6SRygOxXw/VFP0vDQ65wI/AAAAAAAAAXE/otteHCZEBXA/s1600/Robin.JPG" height="240" width="320" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">Robin assumes control of the Justice League.</td></tr></tbody></table>That's my boy...the only one in the room who didn't have a store bought costume. You might not be able to tell, but the variety of suits on the rack was a little lacking. Superman, Iron Man, Hulk, Spider Man, and Buzz Lightyear for boys; Princess, Minnie Mouse, or Witch for girls. No female superheroes, no Star Wars, no monsters... No <i>monsters</i>?! Paraguayans don't do spicy food, and they don't do monster masks.<br /><br />[<i>but boy o boy do they <u><b>LOVE</b></u> candy!</i>]<br /><br /><b>Diego</b> wanted to be Robin after his Papa was dressed as Robin for a costume party a couple weeks ago. Why was Papa dressed as Robin? Because D loves to dress up as (and pretend to be/play) <b><i>Batman</i></b>, and he insists Papa play Robin. A 40-year old Boy Wonder with paunch and thinning hair. I'm <i>always</i> Robin...others in our house might be Superman or Supergirl or Wonder Woman or Batgirl (this was Mama's costume) or Green Lantern or Flash or whoever. JB never rates higher than "sidekick."<br /><br />[<i>when D was on his <b>Iron Man</b> kick for awhile, he DID let me play <b>Rhodey</b> to his Tony Stark. I didn't bother explaining I'm the <b>wrong color</b>...I was just happy to get a break from Robin</i>]<br /><br />Anyway, ever since since D saw me in my outfit (again, crafted by my wife), he's moved his obsession to Robin. Thus the new costume.<br /><br />'Course it's only ONE of his costumes. In addition to dressing up as Batman or Aquaman (his two favorite superheroes) or Superman, D also enjoys dressing as a police officer, firefighter, pirate, or (most recently) <b>a <i>caballero</i></b> (knight). In fact, we kit-bashed a knight costume for him from a bunch of plastic play-gear and this is what he plans on wearing when he goes <b>trick-or-treating</b> on Sunday (yes, that's <b>November 2nd</b>. It's even in a different neighborhood...well, really a gated <i>cul-de-sac</i>...that we have to drive to). My son loves to dress up and pretend and my wife and I support him in this, probably because we enjoy doing the same. We've often gone to absurd lengths to make our own Halloween costumes (<i>especially</i> my wife).<br /><br />And yet, we only do this <b>one time a year</b>. We don't do conventions or ren fairs or cosplay. Heck, my wife doesn't even <i>game</i>, and I've never been one to LARP. It's kind of crazy...like we won't admit (or buy into) our deeper impulse to <b>dress up and pretend </b>except at socially acceptable times (like Halloween). I'm sure my love of "pretending" is what led me to pursue a degree in performing arts (acting). 'Course the last show I was in was...what, 2007? 2005? Certainly it's been a while since I got to put on a wig and costume for an "extended engagement."<br /><br />Well, it is what it is. Perhaps the wife and I will start doing more costume events when we get back to Seattle, now that it's as much fun (if not more so) for our child. M is already in the process of making a "Wonder Woman" outfit for our six month old, though I'm seriously doubtful it'll be ready by Sunday. The question arises, "what <i>IS</i> such a costume for?" I honestly have no idea...it's my wife's thing. Far be it from me to rain on her creative expression.<br />; )<br /><br />Got to go...hope everyone's has a happy one!http://bxblackrazor.blogspot.com/2014/10/[email protected] (JB)4tag:blogger.com,1999:blog-4143435314932633148.post-1618918851548938503Thu, 30 Oct 2014 21:46:00 +00002014-10-30T14:48:07.488-07:00dragonfolklorehalloweenhistoryliteraturemoonwitchConsidering WitchesDidn't have much time yesterday (don't have much time today, either). I'm still considering what would be the "set spell lists" or even the choice of "schools" should I go that route (<a href="http://bxblackrazor.blogspot.com/2014/10/might-and-magic-part-2.html">as discussed a couple days ago</a>). In other words, haven't made any progress on dismantling the magic system already written for the new fantasy heartbreaker.<br /><div><br /></div><div>Maybe it's because (subconsciously?) I think it's a bad idea? Maybe.</div><div><br /></div><div>But <i><b>flavor</b></i>...I like flavor. Flavor is important. It makes a game <i>tasty</i>. Without flavor, you might have a robust game system, chock-full of nutritious, caloric-value (just to carry an analogy too far), but I want <i>more</i> than that. I'm not trying to create <i>Hero System: Fantasy</i> or <i>GURPS: Wizards</i> or something. Too bland for my taste. The idea of different schools of magic is <i>flavorful</i>.</div><div><br /></div><div>Anyhoo...part of the reason I didn't do any work/writing yesterday is that I was again perusing old <b>Dragon magazines</b>. In this case, I was reading every article I could find on <b>witches </b>and witchcraft (for those who're curious that includes issues <b>#5</b>, <b>#20</b>, <b>#43</b>, and <b>#114</b>). I didn't have access to these issues back when I wrote up a <a href="http://bxblackrazor.blogspot.com/2011/05/bx-witch.html">"B/X Witch"</a> for <i>The Complete B/X Adventurer</i>...but even if I <i>had</i>, I'm not sure I would've used much of the stuff here. Certainly not the <b>gemstone level titles</b> (a little too Amway-esque)...not sure where <i>that</i> idea came from. Maybe some of the more interesting NPC spells from issue <b>#5</b>; some of those are pretty cool.</div><div><br /></div><div>[<i>strange there's no author attached to that article. Wonder if anyone ever figured out the writer</i>]</div><div><br /></div><div>The point is, maybe because it's so close to <b>Halloween</b>, I've got witches on the mind. I dig the concept of witch mythology (the <i>fantasy</i> witch if you will) - both good and bad - and wouldn't mind seeing something witch-like in the new heartbreaker. The problem is, how to do it without being <b>offensive</b> to folks. </div><div><br /></div><div><table cellpadding="0" cellspacing="0" class="tr-caption-container" style="float: right; margin-left: 1em; text-align: right;"><tbody><tr><td style="text-align: center;"><a href="http://2.bp.blogspot.com/-j6oM9HzhFmw/VFKpRYgFtOI/AAAAAAAAAWc/EdmhmRUjm_U/s1600/wgg.png" imageanchor="1" style="clear: right; margin-bottom: 1em; margin-left: auto; margin-right: auto;"><img border="0" src="http://2.bp.blogspot.com/-j6oM9HzhFmw/VFKpRYgFtOI/AAAAAAAAAWc/EdmhmRUjm_U/s1600/wgg.png" height="320" width="262" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">I remember Long's book with much fondness.</td></tr></tbody></table>Modern witches, for those who don't know, are very different from the critters you find in classic fantasy literature... whether you're talking <i>The Wizard of Oz</i> or Narnia or those old school Halloween masterpieces. They're very different from the witches portrayed on television and 21st century film, too...but that's <i>not</i> the kind of witch I'm interested in (the witches of <i>Charmed</i> or whatnot are meant for a different RPG than D&D and its ilk). Nor am I talking about the Satanic, Black Mass coven-types of B-horror films, either.</div><div><br /></div><div><div style="text-align: right;"></div>For me, "old school" witches are more fun than frightening...even if the bad ones do (on occasion) <b>eat children.</b> From Baba Yaga and The Old Sea Hag to the beautiful Circe or Morgan Le Fey...the <i>solitary</i> witch is what I'm talking about. That chick in the first <i>Conan</i> movie or Glenda of Oz. In many ways, they are the female equivalent of the <b>solitary sorcerer</b>: someone who has removed herself from society (generally, by her own choosing) in order to practice her craft. Perhaps out of the (real medieval) fear of being burned at the stake by one's neighbors.</div><div><br /></div><div>When these Halloween-y witches get together at all, it's only once a year or every seven years or every century (depending on the story) to celebrate in a big brouhaha (<i>bruja</i>-ha?), otherwise staying out of each other's way unless engaged in some petty rivalry or magical dispute. Apart from these occasional gatherings of celebrated solidarity, these "fantasy witches" are private individuals, opting out of any sort of politics, mundane or magical. Any "Queen of Witches" title is more honorary (or a straight recognition of power) than an actual office to which other witches owe "fealty." I daresay the term might be one designed to poke fun at Earthly feudal titles...the witches are, after all, <i>opting out</i> of standard patriarchal society.</div><div><br /></div><div><table cellpadding="0" cellspacing="0" class="tr-caption-container" style="float: left; margin-right: 1em; text-align: left;"><tbody><tr><td style="text-align: center;"><a href="http://2.bp.blogspot.com/-lavW6ZancLk/VFKp3ZkIoyI/AAAAAAAAAWk/hO2_6CKieH4/s1600/Unknown.jpeg" imageanchor="1" style="clear: left; margin-bottom: 1em; margin-left: auto; margin-right: auto;"><img border="0" src="http://2.bp.blogspot.com/-lavW6ZancLk/VFKp3ZkIoyI/AAAAAAAAAWk/hO2_6CKieH4/s1600/Unknown.jpeg" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">Ah, vinyl. In rotation every Halloween.</td></tr></tbody></table><b>Does that all make sense?</b> I'm not <i>trying</i> to be offensive here, I'm talking about a tradition of folklore and fiction. I'm not trying to "perpetuate stereotypes" of witches, I'm talking about enjoying some of those stereotypes in a fun fashion...and a little Grimm-dark fantasy to a fantasy adventure game.<br /><br />Still, maybe that doesn't fly with some folks. Certainly, I've put my "pulp B/X adventure" game on-hold indefinitely because, no matter how one slices it, any game that includes "savages" (or even "natives") is going to tick someone off. It's borrowing from fiction that was created at a time when Colonialism and white privilege was "okay" (and being packaged and sold to folks of a white privilege persuasion). The pagan persecutions and witch-burnings of earlier centuries was also deemed "okay" at the time, and that is where the majority of our folklore on the subject (with its "wicked witches") comes from. If I do a "for fun" version of witches that buys into that folklore, I may be perpetuating harmful perspectives that some people will apply to real world witches and pagans (both present day and historical).</div><div><br /></div><div>Ugh. Ugh. Ugh.</div><div><br /></div><div>I don't want to offend folks. I don't want to contribute to ignorance. And I don't want to include "disclaimers" in my writing...it wouldn't be a big enough section of the game to warrant such singular treatment (in my opinion), anyway.</div><div><br /></div><div>Am I making too much out of this? People don't worry how elves or wizards are portrayed in RPGs because we consider these to be <i>fictional</i> creations...<i>magic</i> is considered fictional in general and real life hermetic magicians are considered delusional by most of the population (similarly, no one worries about offending people of the <a href="http://en.wikipedia.org/wiki/Jediism">"Jedi Religion"</a>). I don't <i>think</i> dwarves are offensive to <b>little people</b>, as they are based on a fairy race of Norse mythology. But witches...well, a lot of people really did get <i>tortured and murdered</i> back in the day for their non-Christian beliefs. Real people. And there are plenty of real people today that consider themselves witches, though they don't sport pointy hats and green skin. Making light of the history is a bit like making a game where your intrepid explorers (*<i>ahem</i>*) shoot "savages" (pick a continent). And running with folklore that demonized a particular group of individuals is kind of "making light," no?<br /><br />Maybe <i>I'M</i> just overly sensitive. But, well, that's what I'm thinking about today. More later, I'm sure.<br /><br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="http://1.bp.blogspot.com/-yCFOhPFRAPg/VFKu0jDbPSI/AAAAAAAAAW0/oY5DdttVtlg/s1600/images-2.jpeg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="http://1.bp.blogspot.com/-yCFOhPFRAPg/VFKu0jDbPSI/AAAAAAAAAW0/oY5DdttVtlg/s1600/images-2.jpeg" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">It's not easy being green.</td></tr></tbody></table><br /></div>http://bxblackrazor.blogspot.com/2014/10/[email protected] (JB)14tag:blogger.com,1999:blog-4143435314932633148.post-8583238066038342810Tue, 28 Oct 2014 16:19:00 +00002014-10-28T09:23:30.098-07:00eptskillsspellsQuick Addendum to Might and MagicAnd I mean <i>really</i> quick.<br /><br />In <a href="http://bxblackrazor.blogspot.com/2014/10/might-and-magic-part-2.html">my last (very early morning) post</a>, I mentioned (briefly) <b>M.A.R. Barker's <i>Empire of the Petal Throne</i></b> and its skill lists. For those who don't have EPT for reference, here's how it works:<br /><br />In EPT you receive a handful of skills during character creation that help round out your character. These come in a couple different varieties. First there are <b>background skills</b>, similar to AD&D's "secondary skills" in that they are professional skills with no real game mechanics attached. These include things like butcher, carpenter, and wheelwright, as well as physician, poet, scholar, and slaver. <br /><br />Actually, <i>some</i> have some mechanical benefits: the <i>assassin-spy-tracker</i> (that's one skill) can hide in shadows, and the <i>alchemist</i> can brew elixirs and poisons, for example. There are a few, but most don't have more effect then, "Oh, <i>fisherman</i>? You know how to fish." New characters receive a random number of these skills...as few as one, or as many as ten.<br /><br />In addition, characters receive from two to five <b>professional skills</b>. These come from a list based on the character's class, of which there are only three: warrior, priest, and magic-user. Warrior skills are pretty much weapon proficiencies (spear, axe, crossbow, etc.) but priests and magic-users have a selection that ranges from additional languages to spells. While the starting number of pro skills is random, characters receive an additional skill with every level, so eventually a character can claim all on the list (each list has a dozen or so). The interesting thing is that they must be chosen in order...a warrior can't learn bowman until he's learned crossbow, for example. The most advanced skills cannot be claimed until all the lesser skills have been learned.<br /><br /><table cellpadding="0" cellspacing="0" class="tr-caption-container" style="float: right; margin-left: 1em; text-align: right;"><tbody><tr><td style="text-align: center;"><a href="http://3.bp.blogspot.com/-1os7qO4LWCo/VE_CcEctqSI/AAAAAAAAAWM/iWyn7BmIBQU/s1600/Unknown-1.jpeg" imageanchor="1" style="clear: right; margin-bottom: 1em; margin-left: auto; margin-right: auto;"><img border="0" src="http://3.bp.blogspot.com/-1os7qO4LWCo/VE_CcEctqSI/AAAAAAAAAWM/iWyn7BmIBQU/s1600/Unknown-1.jpeg" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">Summoning <i>The Vapor of Death</i>. Ooo!</td></tr></tbody></table>In <i>addition</i> to this, priests and magic-users have a random chance per level of learning <b>bonus spells</b> from one of three separate lists ("groupings"). The spells in these groups are very similar to the ones in OD&D, and they are grouped by power (so Group I spells are the "weakest," though many are still plenty potent). Group III spells cannot be learned until 4th level, but with enough experience, EPT spell-casters can become quite powerful.<br /><br />Over-all, a very interesting system and very "magical" in feeling. I very much like the "knowledge needs to be built on knowledge" attitude of the professional skills. The random bonus spells are appropriate for a campaign setting that features psychic abilities (and the, perhaps, spontaneous development of such abilities), but probably doesn't make much sense for my current project.<br /><br />Ok...got to go. More later (I hope!).http://bxblackrazor.blogspot.com/2014/10/[email protected] (JB)1tag:blogger.com,1999:blog-4143435314932633148.post-1060993716528956426Tue, 28 Oct 2014 07:01:00 +00002014-10-28T00:02:03.228-07:00fmmillusionistLOREmagic-usermoonspellstcbxaMight and Magic (Part 2)[<i>continued from <b><a href="http://bxblackrazor.blogspot.com/2014/10/might-and-magic-part-1.html">here</a></b></i>]<br /><br />People may notice that the list of blogs listed in the sidebar include a random sampling of non-active blogs that probably need to be deleted...and perhaps someday they will, if I ever get around to <a href="http://bxblackrazor.blogspot.com/2014/10/patreon-fever-dreams.html">getting active on Patreon</a> (and want to make blog listing a "reward" for some level of support). However, I still hold out hope that these might "fire up" again, at some point in the future. And sometimes, I still reference them for their old posts.<br /><br />Such is the case with long dead <b><a href="http://Limiting them, perhaps, by theme.">Grognardia</a></b>. As I wrote in Part 1, I was up till the wee hours combing through old (digital) copies of <i>The Dragon</i>, and I was using some of James's old posts to add a bit of additional perspective. Just in random passing, I came across <a href="http://grognardia.blogspot.com/2009/09/lost-gygaxian-classes.html">this old post of his</a> and (especially in view of my recent thoughts, dissatisfaction with magic-users as is/was) it reminded me of something. I <b><i>hate</i></b> the lumping of spell-casters into one or (at most) two types "magic paradigm."<br /><br />See <a href="http://bxblackrazor.blogspot.com/2010/09/spell-casters.html">my post on this back in 2010</a>. At the time, I was working on <b><i>The Complete B/X Adventurer</i></b> (man, <i>THAT</i> has been selling like hotcakes the last two months, just by the way) and in an effort to add more content to a skinny book, decided to throw in some different types of spell-casting classes. I ended up with five total (<b>gnomes</b>, <b>mystics</b>, <b>summoners</b>, <b>tattoo mages</b>, and <b>witches</b>), each with their own <i>variant form of magic</i>...not just "magic-users with different spell lists," but completely different approaches to the form and function of magic. The bee in my bonnet (at the time) was this idea that <b>isn't it Goddamn boring</b> to have everything simply be <i>arcane</i> or <i>divine</i>?<br /><br />Let me answer that: <b>Yes</b>. Yes it is.<br /><br />This is the reason you don't find <b>illusionists</b> in D&D after 1st edition (at least, not as a core class). First, ya' fold all their spells under a heading called "arcane," then you say:<br /><br /><b>"Hey, if you want to specialize in illusion magic, pick illusion spells."</b><br /><br />Much as I want a certain cosmology in my game world, I don't want a <i>unified field theory</i> of magic.<br /><br />[<i>hmm...I say this after already creating a brand-new variant magic system for the current project consisting of a single list of spells: the <b>Forty Magnificent Marvels</b>. Sigh...back to the drawing board...again</i>]<br /><br />I like the idea of <b>different magical schools</b>, each dedicated to a different brand of enchantment. Fire mages, necromancers, druids, etc. It's not a terribly original concept, I realize: I believe I first saw this kind of paradigm circa 1983 with <b>DragonQuest</b> (I created a stone giant who was a member of the Earth Magic college...sadly, we never had the chance to do more than chargen that day...). <b>Ars Magica</b> does a little of this, too, and I've used the concept a couple times in past FHBs I was developing ("LORE," <a href="http://bxblackrazor.blogspot.com/2014/10/eowyn.html">which I briefly mentioned before</a>, had some of this). The original WHFRP had demonologists, necromancers, battle mages, etc. each with their own separate spell list, skills, and (in some cases) horrifying drawbacks.<br /><br />I dig this...it has a very old school (please, PLEASE forgive the use of that term!) pulp fantasy vibe. Like the rival wizard guilds in a Leiber story trying to show off why their magic is supreme (shades of 70's Hong Kong flicks with feuding martial art schools). Heck, it's the kind of thing that could work well with the concept of <a href="http://bxblackrazor.blogspot.com/2014/10/counter-spells-for-real-this-time.html">"wizard duels."</a> Forget counter-spells: an illusionist doesn't know the first thing about countering a fire mage's spell. But create a mechanic to simulate <i>dueling</i>, and you can still have two mages of different backgrounds duking it out.<br /><br />Now that I think of it, this is a big part of why I dug <b>Magic Cards</b>, waaaay back: the idea that you were a Red Mage or a Blue Mage or whatever, and the deck represented your spell book. I always liked working with a "theme;" but then, I've long been one of those people that prefer the <b>fluff</b> of a game over practical application (often to my detriment). Hmm...now that reminds me of the Rankin-Bass film <b>Flight of Dragons</b> with its different colored "wizard brothers."<br /><br />Of course, <i>that</i> just reminds me of Tolkien (again) with its grey, white, brown, and blue wizards...and weren't Frank Baum's witches color-coded as well? Differing magic by color has a long and distinguished tradition, I suppose...<br /><br />*<i>ahem</i>*<br /><br />And to tie this back to <a href="http://bxblackrazor.blogspot.com/2014/10/might-and-magic-part-1.html">the last post</a>... One thing I was considering (just an idea, mind you), is providing more <b>static</b> spell lists for magicians. <i>Limiting them</i> (I suppose you'd say) rather than throwing this huge list of spells at players, only to have them (mostly) choose the same spells over and over. There would be <i>some</i> variation, of course (just as fighters get to choose what weapons they want to carry), and perhaps different lists depending on <b>theme</b> (a druid style list versus the illusionist, for example). Magicians would still acquire effectiveness with experience, though perhaps not so much a greater repertoire of spells. And the spells that <i>would</i> be gained (with increase in level) would be equally limited...something akin to Barker's EPT (1st edition) skill trees, building on knowledge already known.<br /><br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="http://4.bp.blogspot.com/-CIMEItkqPgg/VE8_FOza14I/AAAAAAAAAV8/1ttN5sBrTkY/s1600/Unknown.jpeg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="http://4.bp.blogspot.com/-CIMEItkqPgg/VE8_FOza14I/AAAAAAAAAV8/1ttN5sBrTkY/s1600/Unknown.jpeg" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">No necromancers, though...otherwise everyone wants to be one.</td></tr></tbody></table><br />The trade-off here...or, rather, what would be <i>gained</i>...would be an increased effectiveness from the get-go. Your magician (or pyromancer or whatever), would have a number of spells at his or her disposal right from 1st level...perhaps the equivalent (in D&D terms) of seven or eight spells ranging in magnitude of 1st through 3rd level. Something fairly equivalent to <b>Gandalf</b>, in other words. None of these would be over-powering, "game changing" sort (unless applied in creative fashion)...most would be of the "utility" type (said utility being determined by the school or "theme" of magic).<br /><br />Anyway, just an idea I have...I'll see if I can work up some sample lists in the next couple days and post 'em to the blog. I have a strong suspicion that long-time players of the MU class in D&D might hate-hate-hate this concept for a number of reasons: it undermines the work they've put into mastering spell lists, it reduces the choices/options they have, it penalizes the creative strategies they've developed over years of play, it doesn't have the same feel as Vancian D&D, etc. And if folks DO raise these objections, I say: <b>FINE</b>. Go play D&D.<br /><br />Take your 1st level <i style="font-weight: bold;">sleep </i>bomber or <b><i>charm personer</i></b> with your bandolier of throwing knives and go play D&D. Pick your edition...it's all the same (except that recent ones let you shoot lasers...um, "cantrips"...just like Harry Potter). Go skulk behind the fighters and clerics for umpteen game sessions until you've acquired enough points that you can be effective. Go do it...I'm not going to stop you! And when you've reached the lofty level where you outclass the non-spell-casters and there's the potential threat of grumbling, you can always incorporate feats and maneuvers and weapon specialization and be a merry band spending an entire game session on a single battle that will be a challenge for your immensely talented party.<br /><br /><b>That's one way to play</b> (and a time honored one, to be sure). I'm just trying to work on a <i>different</i> way: one that appeals to me. I'm a guy who doesn't play magic-users...like never ever ever. And not because I don't like magic or something. I love stories of wizards and sorcerers and magicians and witches (well, most such stories...sorry, Ms. Rowling). Gandalf is a personal favorite...in fact, I lied! I did...<i>once</i>...play a wizard in 3rd Edition, and I modeled the character on Gandalf, right down to taking <i>Martial Weapon Proficiency: Sword</i>.<br /><br />[<i>I believe I related this story in the past? We couldn't get past the first obstacle in the adventure because I, as the party wizard, had not taken the "correct" spells. 'What do you mean you don't have <b>fly</b>? You're seventh level!' The DM, folded the adventure in disgust and our session ended. It was the last time I've ever played a straight wizard PC</i>]<br /><br />I don't like magic-users in D&D. I don't like the way they're conceptualized, I don't like their mechanics, I don't particularly like their steep power curve (and I'm a person that likes power!). So, I want to make a magician class that I'd like to play. That's all this is, folks.<br /><b>: )</b>http://bxblackrazor.blogspot.com/2014/10/[email protected] (JB)2tag:blogger.com,1999:blog-4143435314932633148.post-6036075113613504189Mon, 27 Oct 2014 22:33:00 +00002014-10-28T00:02:43.855-07:00dragonmagic-useroriginsreflectionstolkienMight and Magic (Part 1)I have so many different ideas for blog posts I want to get up, I feel a little stymied in where to start. But I guess I better do <i>some</i>thing...here goes:<br /><br />I was up till...oh...2:30 or something in the morning reading the first twelve issues of <b>Dragon magazine</b> (at the time called simply, <i>The Dragon</i>). I went looking for a particular article, which then led me to another, thence to another, and so on until I finally just said, 'whatever...I'll just read the first twelve issues and see where it takes me.'<br /><br />Minus the fiction, of course...boy, there was a LOT of fiction back in those days. Much more than what I remember back in the mid-late 80s (when I first started reading <i>Dragon</i>). Anyway, the main article of interest for purposes of this post is Bill Seligman's March '77 essay, <b><i>"Gandalf was only a fifth level magic-user,"</i></b> found in <i>The Dragon</i> #5.<br /><br />I love this article...perhaps not the way it is written (though the point here is not to critique Mr. Seligman), but the <i>concept</i>. That is, the <i>idea</i> the man is trying to express. The point (for those who don't have access to this particular issue) is that the abilities displayed by Gandalf throughout both <i>The Hobbit</i> and <i>The Lord of the Rings</i> trilogy can be modeled in OD&D (the only edition at the time) with the stats of a 5th level magic-user. That is to say, the majority of magic Gandalf displays in the text can be mimicked using existing 1st and 2nd level spells (or slight variations) with only the occasional use of a 3rd level <b><i>lightning bolt</i></b> spell...though the author points out the latter could possibly be explained by Gandalf's use of <b><a href="http://en.wikipedia.org/wiki/Three_Rings">Narya the Great</a></b>.<br /><br />Seligman also states that Sauron's displays of magic can be modeled with a 7th or 8th level caster's ability (or 12th level "if you're going to be nasty" and allow that he has the<b><i> control weather</i></b> spell). Personally, I'd say that <b>Sauron is at least 11th level</b> in D&D terms, given his ability to manufacture magic items (like, <i>ahem</i>, <b>magic rings</b>).<br /><br />As I said, <b>I really like this.</b> For one thing, if you look at the Tolkien books as adventure guides, you can see just how much is possible with a 5th level Magic-User carrying a bunch of utility spells and one (maybe two) "blasting" type dweomers. Gandalf is no slouch as an adventurer, being quite clever and not reliant on his magical abilities. It helps, of course, that he carries a sword like <b>Glamdring</b> (presumably the sword-equivalent of something like the <b><i>dagger +1, +2 vs. goblins</i></b> or similar)...but I've seen plenty of low-level (and not so low-level) magic-users that would be skulking around the back of the party, even with such a blade. Clever and yet bold: this is the kind of character I'd like to see in my own games...but herein lies the problem.<br /><br /><table cellpadding="0" cellspacing="0" class="tr-caption-container" style="float: left; margin-right: 1em; text-align: left;"><tbody><tr><td style="text-align: center;"><a href="http://4.bp.blogspot.com/-37dKa4qbeZc/VE7Gw0V1TJI/AAAAAAAAAVs/kyyInX6xw_E/s1600/Unknown.jpeg" imageanchor="1" style="clear: left; margin-bottom: 1em; margin-left: auto; margin-right: auto;"><img border="0" src="http://4.bp.blogspot.com/-37dKa4qbeZc/VE7Gw0V1TJI/AAAAAAAAAVs/kyyInX6xw_E/s1600/Unknown.jpeg" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">Being 5th level means needing an umbrella.</td></tr></tbody></table>Gandalf is a very cool character<i> in literature</i>, but how does one get to him in the game? If he starts at 1st level (with only the capability of casting a single spell), he's certainly not going to resemble "Gandalf." More like a very raw apprentice...and one who tires quickly (blows his wad with a single spell-casting). Of course, if he survives to level up (probably by ducking and skulking), you'll get there eventually...and then you'll pass that "Gandalf level" rather quickly and soar into the stratosphere of magical power: polymorphing Nazgul into rabbits, conjuring walls of fire and stone to shore up Minas Tirith, and teleporting back to the Shire as necessary.<br /><br />Just not quite right.<br /><br />[<i>"JB! D&D isn't Tolkien!" <b>Got it</b>...just bear with me, ok?</i>]<br /><br />The problem (or, more accurately, <i>MY</i> problem) is that D&D is too slow to get to (what I consider to be) a competent level of magic, is too quick to ascend to lofty superhero levels of power, and too focused on combat, in general...the latter due to the nature of the game.<br /><br />[<i><a href="http://bxblackrazor.blogspot.com/2013/09/cacodemon.html">plus, not enough geezers</a> (though I recognize that may not be everyone's style)</i>]<br /><br />That all counts as ONE problem, by the way...it's a problem of granularity that doesn't really exist in the other classes. Characters increase in effectiveness doesn't jump in the same leaps and bounds as other classes...the differences between a 1st level fighter and 5th level fighter are very minimal compared to the difference between a 1st level (raw apprentice) MU and a 5th level "Gandalf." The gulf between the 5th level Gandalf and the 12th level Sauron is <b><i>gigantic</i></b>, which is a <i>good</i> thing....until you consider that it's not so terribly hard to advance from 5th to 12th level. Certainly it doesn't take thousands of years (considering the age, experience, and power of Tolkien's "Big Bad Guy") of game time...depending on the amount of playtime, the generosity of the DM, and the skill of the party, and the particular edition being played, a player could reasonably expect to reach 12th level within one to three years of play. I know 3rd edition shot for about one level gained per month (assuming weekly sessions). That is a fast, fast road to power.<br /><br />How to rectify that?<br /><br />Much as I liked the essay about Gandalf being "only" 5th level, there IS a part of me that says "how weak sauce!" when you know that it doesn't take that much effort to get to (and beyond) 5th level. <i>Hey, Old Man:</i> he who falls behind gets left behind, ya' know?<br /><br />Okay, that's one thing I want to talk about...the article made me consider that Vancian magic isn't that terrible as put forward in the original LBBs. But there's tweaking that needs to happen with the advancement dynamic of the wizardly class to get to what I want to see. That <i>and</i> I think I'd like to <i>restrict</i> the variety of spells available to the mage...even more than I've already planned for my "basic" game.<br /><br />But that has to do with a different issue that <a href="http://bxblackrazor.blogspot.com/2014/10/might-and-magic-part-2.html">I'll be discussing in</a><b><a href="http://bxblackrazor.blogspot.com/2014/10/might-and-magic-part-2.html"> Part 2</a>.</b><br /><b>; )</b>http://bxblackrazor.blogspot.com/2014/10/[email protected] (JB)6 | null | null | null | null | null | null | null | null | null |
https://yanjiyu.com/dev/warwick-910/ | 1,718,737,352,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861773.80/warc/CC-MAIN-20240618171806-20240618201806-00387.warc.gz | 964,712,015 | 7,859 | # Data analysis module outline
For Warwick University module 910, Data analysis.
### 2. math back ground
• why frequency/rank for heavy tailed?
The frequency/rank plot because the CCDF is a monotonous function. Moreover, unlike the PDF function, the CCDF displays a significantly lower variability in the tail.
• independence: event and random variables
• HH and TT
• not independent because the mere occurrence of one would tell us for sure that the other can’t happen.
• Conditional Probability
• Heavy-Tailed Distributions roughly non-negligible probabilities at large values far to the right.
• CDF+CCDF=1
• Real Histograms
• log-log plot is also a great trick to visualize data spanning a wide range of data.
• Pareto Distribution
• Good (Enough) Method: Frequency/Rank Plot the sharp drop in variance in the tail is the fact hat the CCDF is a monotonous function
• the Q-Q curve is non-decreasing
### 3. statistics
• quantile-quantile plot
• Allows comparison of attributes of different cardinality
• Plot points corresponding to f-quantile of each attribute
• If close to line y=x, evidence for same distribution
• PMCC = (E(XY)-EXEY)/($\sigma x \sigma y$) = cov(xy)/($\sigma x \sigma y$)
• pmcc between -1,1
• Measures linear dependence in terms of simple quantities
> table, correlated. < table, no correlation.
#### 3.1 handle problems
1. numeric data, ranges vastly different
• range scaling, to 0,1.
• statistical scaling, (x-u)/a
1. ordered data: replace each point with its position in the ordering
2. mixed data: encode each dimension into the range 0,1. use Lp distance
3. handle missing values
• drop the whole record
• fill in missing values manually
• as unknown, ensure can handle
• no ideal solution
1. Noise, hard to tell. statistical tests
2. Outliers: extreme values
• finding in numerica data
• sanity check, mean ,max, std dev
• visual, plot , far from rest
• rule based, set limits
• data based, declare if > 6
• categoric data
• visual, look at frequency statistics, low frequency
dealing outliers: delete, clip(change to max/min), treat as missing
1. discretization: features have too many values
• binning, place data into bands
• use existing hierarchies in data
1. random sampleing, stratifies, take samples from subsets
2. feature selection
• principal components analysis
• greedy attribute selection
### 4.regression
• regression 4
• Assuming unbiased independent noise with bounded variance
• PMCC , Product-Moment Correlation Coefficient determines the quality of the linear regression between X and Y
• R2 = Cov2(X,Y) /(Var(X)Var(Y)) = PMCC^2
• measure the coefficient of determination, how much the model explains the variance in Y
• 1, good fit, 0 ,weak fit to the model
• Two perfectly correlated attributes can’t be used in regression
• Violates requirement of no linearly dependent columns
• Regularization can fix this: ensures matrix is non-singular
• dealing with categoric attributes
• numerically encode categoric explanatory variables
• simple case: 0/1, include this new variable in the regression
• general categoric attributes, create a binary variable for each possibility
• Regularization in regression: To handle the danger of overfitting the model of the data, include the parameters in the optimization
### 6. classification
• measures
• precision : tp/(tp+fp)
• recall: tp/(tp+fn)
• F1 2precision*recall/(precision+recall)
• ROC: recall-y, fp-x(fp/(fp+tn))
• k-fold cross validation
• Divide data into k equal pieces (folds)
• use k-1 for training, evaluate accuracy on 1 fold, repeat
• Entropy H= -plog(p),
• information gain=H-Hsplit
• gain ratio = (Hx-Hsplit)/Hsplit
• kappa
• higher is better
• compares accuracy to random guessing
• Po as the observed agreement between two labelers
• Pe is the agreement if both are labeling at random
• K = (Po-Pe)/(1-Pe)
• pruning
• Postpruning: build full tree, then decide which branches to prune
• SVM
• assume data is linearly separable
• seeks the plane that maximizes the margin between the classes
### 7. clustering
• k-center Furthest point for minimize the max distance between pairs in same cluster
• 2-approximation
• asuume the furthest point to all center>2opt
• then the distances between all centers are also>2opt
• we have k+1 points with distances>2opt between every pair
• since each point has a center with distance < opt
• there exists a pair of points with same center, the distance is at most 2opt
• k-medians minimize the average distance from each point to its closest centre
• k means Lloyd minimize the sum of squared distance
• assign each point to its cloest center
• compute new centroid of each cluster, until no change
• Hierarchical agglomerative HAC
• merge closest pair of clusters
• single: d(C1,C2) = min d(c1 in C1, c2 in C2)
• complete: d(C1,C2) = max d(c1 in C1, c2 in C2) max nodes between each cluster, then find the min
• avg
• Dc-Tc >= Db - Tb
• DBSCAN based on local density of data points
• Epsilon, the radius to search for neighbours
• MinPts, min start points to make one cluster
• density reachable .->.
• density connected .<-.->.
### 8. recommendation
• recommendation 8
• neighbourhood-based collaborative filtering
• item-based collaborative filtering
• matrix factoraization-singular value decomposition
• solving the optimization
• least squares
### 9. social network analysis
• social networks 9
• concept
• in-degree, out degree.
• distance
• diameter
• simplest: common neighbours will link
• weighting common neighbours
• finding important nodes
• graph centrality
• eccentrality
• betweenness centrality
• pageRank
• eigenvector formulation
• r=Mr
• power iteration method
• random walk interpretation
• classification in social networks
• clustering coefficient
• measures the fraction of triangles
• (number of triangles)/(number of pairs of common neighbours)
#### page-rank
here (1,2)(1,3)
• key shortcoming of simplified version?
The problem is the periodic behavior, meaning that PageRank cannot identify the most important web-page. One solution is to create a self-loop (e.g., (1, 1)), in which case the iterative procedure from (a) would be guaranteed to converge.
• main objective? how achieved?
PageRank computes the importance of web-pages. This is achieved by leveraging the transition matrix. | 1,557 | 6,303 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2024-26 | latest | en | 0.713938 |
http://bcid.ca/gcvw0k/bspxk5c.php?5f7812=method-of-sections-truss | 1,631,816,858,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780053717.37/warc/CC-MAIN-20210916174455-20210916204455-00240.warc.gz | 8,824,769 | 7,583 | About the sense of forces, you can always choose to draw an unknown force as tension. This is because the other three members at that cut (BD, EG, and GH) are all coincident through point G. This means that if we take a moment equilibrium around point G, then none of these member forces will contribute to the equilibrium, leaving only the force in AB, which can then be found easily. The method ofsections can also be used to “cut” or section the members of the entire truss. All copyrights are reserved. Example problem using method of sections for truss analysis - statics and structural analysis. This rafter truss calculator, has a range of applications including being used as a wood truss calculator, roof truss calculator, roof rafter calculator, scissor truss … The moment arm, as described previously in Section 1.2, is the perpendicular distance of force from the centre of rotation (in this case point F). The primary benefit of the method of sections is that, for a determinate truss, you can find the force in any individual member quickly without having to solve through the entire truss one joint at a time (which you must do when using the method of joints). This means that only two main methods are to applied for solving the unknown force on truss. Trusses: Method of Sections Frame 19-1 *Introduction In the preceding unit you learned some general facts about trusses as well as a method of solution called the "Method of Joints." Either the right or the left section could be used to solve for the three member forces of interest. For this problem, the moment arm for $F_{AD}$ is equal to $(6\mathrm{\,m})(\cos \theta)$. The next step is to draw a free body of one part or the other indicating all known and unknown forces. Hint: To apply the method of sections… The section must cut completely through the truss and should cut through no more than three members. Step 1: Identify the section 1-1 which passes through the members whose forces are required and note that the section is not passing through more than three members. Make a cut to divide the truss into section, passing the cut through members where the force is needed. When we go back to section cut a-a, we will look at the section to the right of the cut as well for the same reason. Figure 3-2(c) The free body diagram for the cut section to the right of section b-b is shown in Figure 3.11. This test is Rated positive by 93% students preparing for Civil Engineering (CE).This MCQ test is related to Civil Engineering (CE) syllabus, … This section was chosen deliberately because the other three forces ($F_{BG}$, $F_{EG}$, and $F_{GH}$) all point direction through point G. So, if we evaluate the moment equilibrium about point G, we can solve directly for $F_{AB}$: which is negative, meaning that the member is actually in compression. This will give the needed fourth equation. The Method of Sections involves analytically cutting the truss into sections and solving for static equilibrium for each section. What are trusses? It is expalined in this example. Since truss members are subjected to only tensile or compressive forces along their length, the internal forces at the cut member will also be either tensile or compressive with the same magnitude. This joint has an external vertical force of 300N which must be countered by the members attached to the joint. 2 Common Types of Trusses gusset plate Ł Roof Trusses top cord roof purlins knee brace bottom cord gusset plate span, 18 - 30 m, typical bay, 5-6 m typical. This will give the needed fourth equation. This is based on: a) where you need to determine forces, and, b) where the total number of unknowns does not exceed three (in general). The angle $\theta$ may be found using trigonometry: Since we $F_{DF}$ and $F_{FG}$ both point directly through point F, we can use a moment equilibrium around point F to find the third unknown force $F_{AD}$: The moment arm for $F_{AD}$ in the moment equilibrium above was found using the geometry shown on the right side of Figure 3.11. In these notes, we will see the method of sections (MOS), which is very different from the MOJ. The sections are obtained by cutting through some of the members of the truss to expose the force inside the members. For the truss shown in Fig. The method of Sections is used to solve larger truss structures in a fast, simple manner. Method of section cuts the whole structure of trusses into section and then uses the cut out portion for the calculations of the unknown forces. But the left part involves eight forces (VA, HA,F1, F2, F3, PFH, PFI,PGI). Method of Sections In this method, we will cut the truss into two sections by passing a cutting plane through the members whose internal forces we wish to determine. both the parts is drawn. View 04 Truss- Method of Joints and Sections.ppt from CIVIL ENGI 501 at U.E.T Taxila. Study of various instruments used in chain surveying and their uses, Determination of Moisture Content by Oven Drying Method, Determination of Moisture Content By Means of a Calcium Carbide Gas Pressure Moisture Tester, Determination of the normal consistency of hydraulic cement. In the method of joints, we are dealing with static equilibrium at a point. It starts by labeling the spaces between the forces and members with an example shown above; reaction Ra and applied force, P labeled as space 1 and continue moving clockwise around the truss. The method of joints is one of the simplest methods for determining the force acting on the individual members of a truss because it only involves two force equilibrium equations. If we looked at the equilibrium around point B, the force $F_{AB}$ would still push towards the joint (to the right). The remaining unknowns may be found using vertical and horizontal equilibrium: The information on this website is provided without warantee or guarantee of the accuracy of the contents. As mentioned previously, the point of this cut is only to find the force in member AB ($F_{AB}$). Problem 429 - Cantilever Truss by Method of Sections Problem 430 - Parker Truss by Method of Sections Problem 431 - Members in the Third Panel of a Parker Truss At each cut through a member, a force is shown Their direction helps us find the forces. So, in our example here would be our slice: Focussing on the left side only, you are left with the following structure: Now think of this structure as single standing structure. This result is based on the equilibrium principle and Newton’s third law. Method of Sections: If only a few of the member forces are of interest, and those members happen to be somewhere in the middle of the truss, it would be very inefficient to use the method of joints to solve for them. Method of Sections " The second method of truss analysis that we will consider is called the method of sections. " If positive values are obtained, the members are in tension. Nov 30,2020 - Test: Method Of Joints And Sections | 17 Questions MCQ Test has questions of Civil Engineering (CE) preparation. This free online truss and roof calculator generates the axial forces and reactions of completely customisable 2D truss structures. To perform a 2D determinate truss analysis using the method of sections, follow these steps: The truss shown in Figure 3.9 has external forces and boundary conditions provided. We will be focused here with the method of joints with the help of this post and further we will see method of sections in our next post. The desired member forces are then determined by considering the equilibrium of one of the two portions of the truss. Method of joints We can determine the forces in all the members of the truss by using the method of joints. The free body diagram of We're asked to determine the force in member GH here of the truss shown below. Step 2: Using this section 1-1, separate the truss into two parts. Select an appropriate section that cuts through the member that you want to find the axial force for. In this method, we will cut the truss into two sections by passing a cutting plane through the members whose internal forces we wish to determine. The method of joints consists of satisfying the equilibrium conditions for the forces exerted “on the pin” at each joint of the truss we can solve for a maximum of three forces. In the Method of Joints, we are dealing with static equilibrium at a point. The method ofsections can also be used to “cut” or section the members of the entire truss. Therefore, for truss members, it is often more convenient to think of the forces in terms of tension or compression instead of in terms of a specific direction. Analysis Of Trusses By Method Of Joints. This allows solving for up to three unknown forces at a time. This test is Rated positive by 93% students preparing for Civil Engineering (CE).This MCQ test is related to Civil Engineering (CE) syllabus, … Trusses: Method of Joints Frame 18-1 *Introduction A truss is a structure composed of several members joined at their ends so as to form a rigid body. Decide how you need to “cut” the truss. may be used as free body. P-424, determine the force in BF by the method of joints and then check this result using the method of sections. Since there are only three global equilibrium equations, we can only solve for three unknown member axial forces at a time using the method of sections. The method of sections is another method to determine forces in members of a truss structure. Here comes the most important part of solving a truss using the method of Sections. Tension or compression whose forces are minimum from the MOJ $) here relative! When you 're done reading this section shares member AB with the other method is given! Section 1-1, separate the truss shown below section, must satisfy the equilibrium principle and Newton ’ s law... Collinear than the task of determining the unknown forces solve a truss is in equilibrium the cutting.., as we did when using the method of sections is used to solve a! Section to find force in member becomes negative, the members are in,!, only two unknowns can be solved if the entire truss is in equilibrium, then whichever section of selected... Is tension or compression procedure as the method of sections involves cutting the truss sections. Would be needed if the force in BF by the method of joints and:, is.... Any member by analyzing the left or the right or the right section of the members of a structure whether! Members AB, AD, DF and FG has an external vertical force of 300N which must be by. A method of sections truss, a force is tension or compression Their direction helps find. The equilibrium principle and Newton ’ s third law 29, 2020 now more than members... That cuts through the truss: three equations but four unknown so another equation is needed tension forces always towards... Of 300N which must be countered by the members, compression forces always pull away method of sections truss and... The reactions ( VA, HA, F1, F2, F3,,. Intersected members have been cut may be used as free body where the member is in equilibrium for unknown... Save my name, email, and Therefore the truss obtained after the intersected members have been cut may used... Equilibrium then any part of the truss into section, it is the whole or section the are. Procedure as the forces acting in a specific member of the truss into two distinct sections or parts expose force... Simple truss that is simply supported ( with pin at one end a... 130 Engineering Mechanics - statics Lecture # 13 method of sections example problem method..., but today we 're going to go through and use the of... > > when you 're done reading this section, passing the cut section to find the internal at. Apparatus, procedure, Results forces in members FH and GI consider the equilibrium principle other! Section a-a from four to three unknown forces done by making a cut '' along selected! Member becomes negative, the truss by using the method of sections involves cutting the truss with at most unknown. Quiz at the other ) is produced and managed by Prof. Jeffrey,. Duration: 13:27. structurefree 43,398 views through the members whose forces are then determined by considering equilibrium! Solving method of sections truss gives, Standard Proctor Test ASTM: D698-91 Apparatus, procedure, Results for solving unknown... ’ s third law for a simpler problem, only two unknowns can be used to the! Erochko, PhD, P.Eng., Carleton University, Ottawa, Canada 2020... The unknown forces acting in a specific member of a truss consists of a structure whether! This tutorial we will determine here the force in member and is found means can! A body is also no internal instability, and GH, and Therefore the truss being considered must also in! Well as the forces in the drawing as a reference point for the unknown at... And state whether the force in member GH here of the cut truss will be in. Note that in most cases it is the whole or section, must the... The whole or section, it is compression truss, cutting it into sections and each. Always pull away from joints and members view Lecture 13_Method of sections.pdf from ENGG 100 University... Cut completely through the truss with at most three unknown forces at that section we did when the. Eight forces ( VA, HA and VL ) becomes negative, method! Please note that the right section of the truss to solve directly any member by analyzing the section. From ENGG 100 at University of Alberta four member cuts may be made ( will! Way of solving for static equilibrium for each section determine forces in all the members whose are! Two portions of the selected cut section to find the internal axial of. Members are in tension, as we did when using the method of sections consists of passing an imaginary must. Body is in tension, as we did when using the method of sections is used,... Be solved without further analysis ; however, joint B can be solved if the force is tension compression...: D698-91 Apparatus, procedure, Results see the method of joints 3-2! Ha and VL ) moment equations to solve for a simpler problem, only one cut would be needed the! Third law ( VA, HA and VL ) 501 at U.E.T Taxila step 2: using section... Find ) an method of sections truss to the right arrow ($ \rightarrow \$ ) here is revealing the inner axial in... To indicate the unknown force as tension and GI consider the section must be in equilibrium )... Desired member forces of members by dividing the truss, cutting it into sections support reactions ) as as. Section 2.5: Therefore, the member that you want to talk about by. 'Ve learned the method of joints and Sections.ppt from CIVIL ENGI 501 at Taxila... Common practice but not the eleventh commandment Prof. Jeffrey Erochko, PhD, P.Eng., Carleton,... Ce using the method of sections consists of passing an imaginary section must be countered by the method of involves... Test ASTM: D698-91 Apparatus, procedure, Results you need to “ cut ” the truss into.. We 've learned the method of sections 1 Previously: Trusses a truss structure must isolated! ( as will be easier to work with ( minimize the number unknown! Tutorial we will consider is called the method of sections. Ottawa,,... Cases, method of sections involves analytically cutting the truss is determinate > when you done! Duration: 13:27. structurefree 43,398 views at that section may be made ( as will be shown in 3.10. Entire truss is determinate F2, F3, PFH, PFI, PGI.! Side of the page produced and managed by Prof. Jeffrey Erochko, PhD, P.Eng., Carleton University Ottawa. Method is just given for the portion of truss to expose the force inside the members a! About this, visitez notre What is a simple truss that is simply supported ( with pin one... Gh, and CE using the method of sections involves analytically cutting the truss when using the method breaking... Section has finite size and this means that only two equations are involved, one.
## method of sections truss
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Aplastic Anemia
Aplastic anemia is a rare (5-10 cases per million), serious condition where the bone marrow fails to produce white cells, red cells and the platelets. Aplastic anemia is observed in children as well as in adults.
The term Aplastic comes from the word aplasia that means failure of generation or formation. Anemia means deficiency in the oxygen-carrying component of the blood.
The spongy material inside the bones is called as bone marrow and is like a factory that produces blood cells (red cells, white cells and platelets). In case of Aplastic anemia bone marrow fails to form or develop blood cells.
The red cells in the blood carry oxygen from the lungs to all areas of the body. In Aplastic anemia with less number of red cells, body parts do not get sufficient oxygen that is absolutely essential for their functioning.
White cells fight infection by attacking and destroying germs and their less number in Aplastic anemia leads to poor defense mechanism making the suffering individual prone to various and at times lethal infections.
Platelets are blood cells that control bleeding by forming blood clots in areas of injury. Reduced platelets lead to blood clotting disorder, whereby blood does not clot naturally, leading to uncontrolled bleeding.
Suggestion about ayurvedic treatment: Role of ayurveda in Aplastic anemia
1. ayurveda medicines attempt to stimulate the healthy portion of bone marrow to improve cell production. This may help to reduce the number of blood transfusions.
2. ayurveda medicines improve general vitality and well being of a patient to help him fight infections.
3. ayurveda medicines can be useful to control bleeding disorder associated with Aplastic anemia.
4. ayurveda medicines are effective in curtailing further course and pace of disease.
5. The diagnosis of Aplastic anemia can have terrible emotional impact on the patient, which can have detrimental effect on immunity, thus adding fuel to the fire. ayurveda medicines having positive influence on the psyche of patient can take care of this mind-body link.
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7. ayurveda medicines have very effective long term beneficial effect or preventive effect in terms of countering genetic tendencies and balancing disturbed immunity that are root causes of this serious malady. The chances of relapse significantly diminish with ayurveda treatment. The important point that should be emphasized is that when dealing with serious condition like Aplastic anemia, the beneficial effects of multiple therapeutic systems should be used in concordance with each other. There cannot be a single or exclusive line of treatment. ayurveda medicines, as they do not adversely interfere with conventional medication, are absolutely safe to have synergetic effect of treatment. Moreover, ayurveda medicines are absolutely safe and do not have any side effects what so ever. However, considering the pace and grievous nature of disease, ayurveda alone may have limitations in managing Aplastic anemia. One cannot expect miraculous cure for Aplastic anemia with ayurveda alone. Nonetheless, ayurveda treatment along with conventional treatment can add a lot to quality management of patient with Aplastic anemia. It should be emphasized that ayurvedic treatment is not a substitute to any mode of modern medicine such as blood transfusion or bone marrow transplant, etc. The role of ayurveda is more complementary than alternative, when it comes to managing Aplastic anemia.
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# Search results
Found 566 matches
Law of sines (related to circumdiameter)
The law of sines, sine law, sine formula, or sine rule relates the sine of an angle to the opposite side of an arbitrary triangle and the diameter of the ... more
Right triangle or right-angled triangle is a triangle in which one angle is a right angle (that is, a 90-degree angle). The incircle or inscribed circle of ... more
One of the legs of a right triangle related to the inradius and the other leg.
Right triangle or right-angled triangle is a triangle in which one angle is a right angle (that is, a 90-degree angle). The incircle or inscribed circle of ... more
Tangent function
The trigonometric functions (also called the circular functions) are functions of an angle. They relate the angles of a triangle to the lengths of its ... more
Cotangent function
The trigonometric functions (also called the circular functions) are functions of an angle. They relate the angles of a triangle to the lengths of its ... more
Worksheet 334
In a video game design, a map shows the location of other characters relative to the player, who is situated at the origin, and the direction they are facing. A character currently shows on the map at coordinates (-3, 5). If the player rotates counterclockwise by 20 degrees, then the objects in the map will correspondingly rotate 20 degrees clockwise. Find the new coordinates of the character.
To rotate the position of the character, we can imagine it as a point on a circle, and we will change the angle of the point by 20 degrees. To do so, we first need to find the radius of this circle and the original angle.
Drawing a right triangle inside the circle, we can find the radius using the Pythagorean Theorem:
Pythagorean theorem (right triangle)
To find the angle, we need to decide first if we are going to find the acute angle of the triangle, the reference angle, or if we are going to find the angle measured in standard position. While either approach will work, in this case we will do the latter. By applying the cosine function and using our given information we get
Cosine function
Subtraction
While there are two angles that have this cosine value, the angle of 120.964 degrees is in the second quadrant as desired, so it is the angle we were looking for.
Rotating the point clockwise by 20 degrees, the angle of the point will decrease to 100.964 degrees. We can then evaluate the coordinates of the rotated point
For x axis:
Cosine function
For y axis:
Sine function
The coordinates of the character on the rotated map will be (-1.109, 5.725)
Reference : PreCalculus: An Investigation of Functions,Edition 1.4 © 2014 David Lippman and Melonie Rasmussen
http://www.opentextbookstore.com/precalc/
Cosine function
The trigonometric functions (also called the circular functions) are functions of an angle. They relate the angles of a triangle to the lengths of its ... more
A quadrilateral is a polygon with four sides (or edges) and four vertices or corners. The interior angles of a simple (and planar) quadrilateral add up to ... more
Perimeter of a Triangle
A perimeter is a path that surrounds a two-dimensional shape. The word comes from the Greek peri (around) and meter (measure). The term may be used either ... more
Law of cotangents (in term of tangents)
In trigonometry, the law of cotangents is a relationship among the lengths of the sides of a triangle and the cotangents of the halves of the three angles. ... more
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# exam3retakeSP072810_make_up_test - BUS 2810 Exam 3 Name...
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Person Before After Ho: ud >= 0 1 320 340 H1: ud < 0 2 390 380 3 420 470 t-Test: Paired Two Sample for Means 4 510 510 5 610 630 Before After 6 410 400 Mean 472.5 490 7 630 640 Variance 9784.09 11090.91 8 560 560 Observations 12 12 9 360 370 Pearson Correlation 0.97 10 430 440 Hypothesized Mean Difference 0 11 510 540 df 11 12 520 600 t Stat -2.33 P(T<=t) one-tail 0.02 >.01 t Critical one-tail 2.72 P(T<=t) two-tail 0.04 t Critical two-tail 3.11 conclusion Do not Reject Null Hypothesis: There is evedence test scores increase if you take the GMAT prep course. BUS 2810 Exam 3 Name Brian Ayers RETEST SP07 100 points Section 9 am class Show ALL Null and Alternative Hypotheses, crtitical values, p values, calucated tests, and conclusions. ... Problem 1: 15 points A GMAT prep course claim that taking their course will result in improved scores. A random sample of 12 people took the GMAT before the course and than again after the GMAT prep course. At the .01 significance level, can we conclude GMAT results improved? Use an appropriate hypothesis test at the .01 significance level. What is the p-value?
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Critical 2.82 Sample Weight Test statistic -4.32 1 45.6 p-value 0 > .01 2 47.7 population mean 90 3 47.6 Ho: u <= 50 sample std dev 1.63 4 46.3 sample size 10 5 46.2 H1: u> 50 6 47.4 Weight 7 49.2 8 50.8 Mean 47.78 9 47.5 Standard Error 0.51 10 49.5 Median 47.55 Mode #N/A Standard Deviation 1.63 Sample Variance 2.64 Kurtosis -0.29 Skewness 0.58 Range 5.2 Minimum 45.6 Maximum 50.8 Sum 477.8 Conclusion Count 10 Do not reject Null Hypothesis: There is evidence that the mean weight of the bags is less than 50 pounds
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## This note was uploaded on 04/08/2008 for the course BUS 2801 taught by Professor Wilson during the Spring '07 term at E. Illinois.
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exam3retakeSP072810_make_up_test - BUS 2810 Exam 3 Name...
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# What does Mad stand for in Math? When we use a number, it can either stand for the entire number or it can also stand for the negative of the entire number.
But the meaning of both of those is distinct.
Without a doubt, the explanation why this type of mathematics is named as this can be since it features a massive and big effect on all our lives. There are many online Masters in Mathematics that make this subject the main subject taught in higher schools.
With this, it has been realized that that is indeed a very fantastic thought to possess a web based Master in Mathematics due to the fact this may enable you to operate from anywhere you want. essay writer Needless to say, you can not anticipate to obtain a job without performing some study. The truth is, you will discover a lot of other sources like professors and teachers which are out there by way of these internet sites that assist you fulfill your dreams.
These will be the motives why these Masters in Mathematics ought to be among the list of major priorities for you. It will likely be your very first step towards a superior future. So, if you have the ambition and determination to perform in a superior profession, then you need to be passionate about math. In case you know only basics, then you definitely will in no way get your self ready for a difficult job.
Math is amongst the most magical mathematical secrets that allow us to complete incredible factors like multiplication, division, quadratic equations, differential equations, and so on. This can be mainly because, math lets us use unique combinations of numbers in equations to resolve challenges also is definitely an important technique to test the correctness of any offered equation. essay help So, let us begin with an introduction for the point known as mathematics.
In mathematics, the basic point is its collection of rules. These rules would be the basis of all of the other calculations that comply with and depend on the fact that these guidelines are invariant.
Let us commence together with the ideal angle triangle. The angle involving two straight lines is 90 degrees. Working with trigonometry, we are able to obtain the following formula.
Angle = 60 * sin ((side)) / cos ((side)) Result: A straight line is often divided into two shorter lines by the formula shown above. Angle R=cos (side) -sin (side) = tan (side). Thus, when we apply the formula, we get the following outcome. Angle is definitely the angle between the second two sides.
Now, let us visit the negative quantity. write a term paper The adverse number equals for the positive quantity. If we take the middle point of your two constructive numbers, the result will be the adverse number.
If we divide the quantity by the number, we’ll get the number. So, in the left side with the equation above, we can find the unfavorable quantity. | 577 | 2,881 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2020-10 | latest | en | 0.953445 |
https://www.gradesaver.com/textbooks/math/algebra/intermediate-algebra-12th-edition/chapter-r-section-r-3-exponents-roots-and-order-of-operations-r-3-exercises-page-29/20 | 1,539,755,927,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583510998.39/warc/CC-MAIN-20181017044446-20181017065946-00177.warc.gz | 957,534,199 | 12,684 | ## Intermediate Algebra (12th Edition)
$a^{5}$
In algebra, exponents can be used to write products of repeated factors. Exponents take the form $a^{b}$, where $a$ is the repeated factor and $b$ is the amount of times that the factor is being multiplied by itself. In this case, a is being multiplied by itself 5 times. Therefore, $a\times a\times a\times a\times a=a^{5}$. | 101 | 373 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2018-43 | longest | en | 0.917707 |
http://betterlesson.com/lesson/reflection/23121/ask-don-t-answer | 1,487,648,890,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501170624.14/warc/CC-MAIN-20170219104610-00149-ip-10-171-10-108.ec2.internal.warc.gz | 26,009,425 | 21,955 | ## Reflection: Developing a Conceptual Understanding Scaling up, scaling down, scaling all around - Section 3: Exploration
One of the big shifts I have made that is really helpful in engaging my students in sense making is to stop answering their questions. Instead I find it most effective to help them find a way to answer the question themselves. This can be really hard to do 'in the moment'. Sometimes I turn the question to the table group, sometimes I paraphrase back their question, but in this lesson I referred to a familiar context. When one of my students asked about the procedure for calculating equivalent fractions I really wanted him to understand why the procedure works and not just be able to carry it out on command. By connecting the math to the familiar context of tile designs he was able to make sense of what was happening when he was multiplying numerators and denominators. I think when my students can understand the math as a model for the real world they don't need to memorize procedures.
# Scaling up, scaling down, scaling all around
Unit 6: Proportionality on a graph
Lesson 9 of 10
## Big Idea: Ratios can be compared with common denominators.
Print Lesson
Standards:
Subject(s):
Math, Number Sense and Operations, common denominator, multiple methods, peer instruction, comparing ratios, ratios, pattern
54 minutes
### Erica Burnison
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Environment: Suburban | 462 | 2,164 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2017-09 | latest | en | 0.916424 |
https://atcoder.jp/contests/abc271/editorial/4944 | 1,726,565,295,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651750.27/warc/CC-MAIN-20240917072424-20240917102424-00067.warc.gz | 92,022,196 | 5,655 | Contest Duration: - (local time) (100 minutes) Back to Home
Official
## G - Access Counter Editorial by en_translator
### Finding the probability that the $$1$$-st access is made at $$0$$, $$1$$, $$\ldots$$, and $$23$$ o’clock
If $$N=1$$, it is sufficient to find the probability that the $$1$$-st access is made at $$0$$, $$1$$, $$\ldots$$, and $$23$$ o’clock and find the sum over those corresponding to Aoki’s, but some of you may feel this is already difficult. That is, the desired value is $$\sum_{i=1}^{\infty } p_i$$ where $$p_i$$ is the probability that the $$1$$-st access is made at say $$0$$ o’clock on the $$i$$-th day, but we cannot find the sum of the infinite number of values with a for statement. Instead, we need a mathematical transformation.
Let $$q$$ be the probability that there is no access at $$0$$, $$1$$, $$\ldots$$, and $$23$$ o’clock. Then, $$p_i = q^ip_1$$, so $$\sum_{i=1}^{\infty } q^i p_1=(\sum_{i=1}^{\infty } q^i )\times p_1$$. By the constraints, we have $$q<1$$, so we have $$\sum_{i=1}^{\infty } q^i =\frac{1}{1-q}$$ by the formula of the sum of infinite geometric series, and thus we can express the probability that the $$1$$-st access is made at $$0$$ o’clock as $$\frac{p_1}{1-q}$$. Same applies for the other time.
By the way, we can check exhaustively that $$q \neq 1$$ modulo $$998244353$$ under the constraints.
### $$\mathrm{O}(N)$$ solution
Let $$\mathrm{dp}_{i,j}$$ be the probability that the $$i$$-th access is made at $$j$$ o’clock on the $$i$$-th day. Since the discussion above is equivalent to the probability that the next access is at $$0$$, $$1$$, $$\ldots$$, and $$23$$ o’clock when the last access is made at $$23$$ o’clock. We can find similarly when the last access is made at $$0$$, $$1$$, $$\ldots$$, and $$22$$ o’clock. How can we optimize it?
### $$\mathrm{O}(\log N)$$ solution
Let $$\mathrm{dp}_{i,j,k}$$ be the probability that the $$2^i$$-th access is made at $$k$$ o’clock when the last access is made at $$j$$ o’clock. $$i=0$$ can be solved by the discussion above, and $$i\geq 1$$ can be found as $$\mathrm{dp}_{i,j,k} = \sum_{l=0}^{23} \mathrm{dp}_{i-1,j,l} \times \mathrm{dp}_{i-1,l,k}$$. Therefore, we can use fast exponentiation trick to find the answer for $$N$$, for a complexity of $$\mathrm{O}(\log N)$$.
Note that this is equivalent to finding the $$N$$-th power of a matrix where the columns and rows correspond to $$j$$ and $$k$$, respectively. Since the optimization of $$\mathrm{dp}$$ using matrices are often seen, we recommend you to learn it.
Exercises from the past problems: https://atcoder.jp/contests/dp/tasks/dp_r | 824 | 2,615 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2024-38 | latest | en | 0.867233 |
https://www.physicsforums.com/threads/finding-spring-constant-energy-w-doubt-in-exercise.1001068/ | 1,716,047,648,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057440.7/warc/CC-MAIN-20240518152242-20240518182242-00398.warc.gz | 859,095,680 | 17,450 | # Finding Spring Constant & Energy w/ Doubt in Exercise
• Dunkodx
In summary: Is this what you meant by your "position where it is ##E_{\text{k}}=U##" question?Yes, I meant the position where ##E_{\text{k}}=U##.
Dunkodx
Thread moved from the technical forums, so no Homework Template is shown
Summary:: Doubt in a spring exercise
Text of the exercise "a mass of ##m = 0.4 \ \text{kg} ## is attached to a spring and it oscillates horizontally with period ##T = 1.57 \text{s}##; the amplitude of the oscillation is ##d = 0.4 \text{m}##. Determine the spring constant, the total energy of the system and the position where the kinetic energy is equal to the potential energy."
I have found the spring constant with the relation ##T = \frac{2\pi}{\omega}## and I've used the conservation of energy to say that ##E=\frac{1}{2} k \left(\frac{d}{2}\right)^2=\frac{1}{2}mv^2=\frac{1}{2} m\omega^2 \left(\frac{d}{2}\right)^2##; I have a doubt for the last request, that is the position where it is ##E_{\text{k}}=U##; my reasoning is that in a generic position it is, again for the conservation of energy, that ##\frac{1}{2}kx^2+\frac{1}{2}mv^2=\frac{1}{2}k \left(\frac{d}{2}\right)^2##, but since we want the position where ##E_{\text{k}}=U## and it is ##E=E_{\text{k}}+U## substituting ##E_{\text{k}}=U## leads to ##E_m=2U \implies E_{\text{k}}+U= 2\cdot \frac{1}{2} k \left(\frac{d}{2}\right)^2\implies \frac{1}{2}kx^2+\frac{1}{2}mv^2=k \left(\frac{d}{2}\right)^2##. Again, since ##\frac{1}{2}kx^2=\frac{1}{2}mv^2## because I am interested when kinetical and potential energy are the same, I get ##kx^2=k\left(\frac{d}{2}\right)^2 \implies x=\frac{d}{2}##.
However the solution says that ##x=\frac{d}{2\sqrt{2}}##, where is my mistake? Thanks.
What is ##E_m## in your equation soup above?
@hutchphd: It was a typo, that ##E_m## is the same as all the other ##E## in the post. Sorry for the confusion.
O.K. Why do you say ##E=\frac1 2 k (\frac d 2 )^2##
Because there are only conservative forces, so when the mass reaches the maximum amplitude (that is, when it is at position ##\frac{d}{2}##) the kinetical energy is ##0## because the mass has no velocity and there is only potential energy ##\frac{1}{2} k \left(\frac{d}{2}\right)^2##; that means that ##E=\frac{1}{2}k \left(\frac{d}{2}\right)^2##.
The amplitude is the size of the excursion from equilibrium in my vernacular. Check this.
Steve4Physics
The question explicitly states that the amplitude is d. And there is no "maximum amplitude". Amplitude is the maximum displacement from equilibrium. Unless otherwise stated (like in peak-to-peak amplitude).
Steve4Physics
Dunkodx said:
Because there are only conservative forces, so when the mass reaches the maximum amplitude (that is, when it is at position ##\frac{d}{2}##) the kinetical energy is ##0## because the mass has no velocity and there is only potential energy ##\frac{1}{2} k \left(\frac{d}{2}\right)^2##; that means that ##E=\frac{1}{2}k \left(\frac{d}{2}\right)^2##.
If the equilibrium position is x=0, then displacement (x) is in the range ## -d≤x≤d##.
Extremum-to-extremum distance is 2d (twice the amplitude).
That means maximum potential energy is ##\frac 1 2 kd^2##, not ##\frac 1 2 k(\frac d 2)^2##.
nasu
## 1. How do you find the spring constant in an exercise?
The spring constant, also known as the force constant, can be found by dividing the force applied to the spring by the displacement of the spring. This can be represented by the equation k = F/x, where k is the spring constant, F is the applied force, and x is the displacement of the spring.
## 2. What is the significance of finding the spring constant in an exercise?
The spring constant is a measure of the stiffness of a spring and is an important factor in understanding the behavior of the spring. It is used in various equations to calculate the potential energy, kinetic energy, and work done by the spring.
## 3. How can doubt affect the accuracy of finding the spring constant and energy in an exercise?
Doubt can affect the accuracy of the results when finding the spring constant and energy in an exercise by introducing errors in the measurements or calculations. It is important to minimize doubt by using precise and accurate equipment and techniques.
## 4. Can the spring constant and energy be determined without doubt in an exercise?
While it is impossible to completely eliminate doubt, it is possible to minimize it by using precise and accurate measurements and techniques. This will result in more accurate and reliable results for the spring constant and energy in an exercise.
## 5. What are some examples of exercises where finding the spring constant and energy is important?
Finding the spring constant and energy is important in various exercises, such as determining the force required to stretch a spring to a certain length, calculating the energy stored in a spring, and understanding the behavior of springs in machines and structures.
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826 | 1,523 | 5,552 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2024-22 | latest | en | 0.76899 |
null | null | null | null | null | null | Migrants Now Facing Greater Risk Of Death
June 05, 2013
TUCSON, Ariz. -- Immigrants are facing a greater risk of dying as they try to cross into the United States illegally, researchers said Wednesday. Their data shows deaths in the desert have remained at a constant even as arrests have dropped significantly
Researchers at the University of Arizona compared how many bodies were found in the desert every year since 1990, for a 13-year total of 2,238. Then they compared that number to the number of arrests made by the U.S. Border Patrol in each year. What they found was that the same number of dead bodies were discovered even as apprehensions along the border dropped.
That means migrants crossing the border into Arizona in 2012, for example, were twice as likely to die as they were in 2009.
Bodies are also left out in the desert for longer periods now. More bodies were found in advanced stages of decomposition. That makes it impossible to pinpoint a cause of death.
"Basically the cause of death change from "exposure" to "undetermined" probably has more to do with the remoteness they bodies were found in and the longer they’d been in the desert and less to do with the actual cause of death," said Greg Hess, the Pima County Chief Medical Examiner.
Every year, about 200 people are found dead in the desert.
"Certainly this summer is not going to be any different," said Raquel Rubio Goldsmith, one of the authors of Wednesday’s report.
So far this fiscal year, 102 people have been found dead in southern Arizona. But as health officials noted, migrant deaths are not merely an Arizona issue. This year for the first time, the Rio Grande Valley in Texas may surpass Arizona in migrant deaths. | null | null | null | null | null | null | null | null | null |
https://www.physicsforums.com/threads/eigenspaces-eigenvalues-and-eigenbasis.196238/ | 1,477,314,026,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719566.74/warc/CC-MAIN-20161020183839-00359-ip-10-171-6-4.ec2.internal.warc.gz | 975,764,808 | 18,422 | # Eigenspaces, eigenvalues and eigenbasis
1. Nov 5, 2007
### FunkyDwarf
Hey guys,
I was wondering what the difference between a generalized eigenspace for an eigenvalue and just an eigenspace is. I know that you can get a vector space using an eigenbasis ie using the eigenvectors to span the space but apart from that im kinda stumped.
Also with regard to this i was trying to answer the question: Show that if U is the generalised eigenspace for an eigenvalue a and V is the generalised eigenspace for an eigenvalue b then if a doesn't equal b, U intersects V only in the zero vector. Now i understand the basic premise, if you have two different eigenvalues you need to show that their eigenvectors are linearly independent and thus can be used to span two non overlapping spaces (i know overlapping is more for venn diagrams but thats how i think about many of these problems). What I dont understand is this: if we have an operator A on Rn the whole space is the direct sum of the generalised eigenspaces. I guess my question here is more about direct sums actually. If we 'add' two spaces together, we're not actually adding them are we? Instead we're constructing a new basis which is the union of the two basis sets of the two different spaces and building a new space from that. The reason i ask is if we have two LI vectors in R2, the union of those spaces would just be two lines rather than the whole space (yes i understand the concept of spanning spaces and stuff) so im assuming when we say direct sum we mean, effectively, the space spanned by those two vectors.
Does this sort of make sense?
Thanks
-Graeme
2. Nov 6, 2007
### HallsofIvy
Staff Emeritus
Yes, that makes sense and, yes, you are right that the union of two subspaces is not, in general, a subspace. The direct sum of two subspaces is the span of vectors in the two subspaces.
3. Nov 6, 2007
### llarsen
A generalized eigenvector is not an eigenvector, but returns an eigenvector or another generalized eigenvector. I would give a simple example matrix, but I don't know LaTeX well enough. However if $$q_1$$ is an eigenvector of the matrix A (i.e. $$A q_1 = \lambda_1 q_1$$, a vector $$q_2$$ that satisfies $$A q_2 = \lambda_2 q_1$$ is a generalized eigenvector associated with $$q_1$$. You might have another vector $$q_3$$ that satisfies $$A q_3 = \lambda_3 q_2$$ which also implies that $$A A q_3 = A^2 q_3 = \lambda_2 \lambda_3 q_1$$ so this is another generalized eigenvector associated with $$q_1$$. The vectors $$q_2$$ and $$q_3$$ are not eigenvectors themselves since they do not satisfy the eigenvector equation, but successive multiplications will result in an eigenvector, so they are called generalized eigenvectors. An eigenvector in combination with its associated generalized eigenvectors is a generalized eigenspace.
The basis for Rn is the generalized eigenspaces plus the basis of the Null Space (the space associated with the zero eigenvalues).
Last edited: Nov 6, 2007
4. Nov 6, 2007
### matt grime
v is an eigen vector with eigenvalue t of A if (A-t)v=0. It is generalized if some power of (A-t) sends it to zero. That is the difference. If you're still stuck just consider
[1 1]
[0 1]
5. Nov 7, 2007
### FunkyDwarf
Ok i understand its mathematical construction (sort of) what i dont understand is a graphical analog. Usually i think of eigenvalues as a 'stretching' factor along an eigenvector (really an eigenline). Where would a generalised eigenvector fit into this picture?
6. Nov 7, 2007
### matt grime
It is easier I suspect to think about A-t where t is an e-value of A: just look at the Jordan block description. In the case above
[1 1] =A
[0 1]
with respect to the standard basis {e,f}.
e is an e-vector: (A-1)e=0. And f is a generalized e-vector: (A-1)f=e.
I like to think of generalized e-vectors as being the preimage under A-t of an e-vector, then a preimage of that and so on. Thus they come along in sequences e_1,e_2,..e_r and (A-t)e_{i+1}=e_i and (A-t)e_1=0
7. Nov 7, 2007
### FunkyDwarf
Would it be fair to call ker(A-sI)^k as the area of affect of A with factor s on Rn? I still cant really see the difference between generalised eigenspaces and just eigenspaces im sorry, im sure its really stupid and obvious and i appriciate the help but i dont get it =(
I mean in R3 if i have s repeated twice and the other value t then we have two distinct eigenvectors (A-sI)u = 0 and (A-tI)v = 0 but there kernel of (A-sI)^2 would be a plane which means there must be another eigenvector (A-sI)a = 0 right with a and u linearly independent? So what i get from this circuitous route is that an eigenspace for an eigenvalue is a line of vectors for which the usual equation holds, but if you have repeated eigenvalues you have two linearly independant directions on which s is acting and so the generalised eigenspace is the plane defined by those...right?
8. Nov 7, 2007
### llarsen
I don't think that the matrix given above:
[1 1]
[0 1]
is a correct example. Try the matrix:
[1 1 0]
[0 0 0]
[0 0 1]
Try the vectors:
[1 1 0] [1] [1]
[0 0 0] [0] = [0]
[0 0 1] [0] [0]
Above is the first eigenvector:
[1 1 0] [0] [1]
[0 0 0] [1] = [0]
[0 0 1] [0] [0]
This is another vector that returns the first eigenvector. It is a generalized eigenvector associated with the first eigenvector. The generalized eigenspace is made of the two vector above.
[1 1 0] [0] [0]
[0 0 0] [0] = [0]
[0 0 1] [1] [1]
This is a second eigenvector. Note that three independent eigenvectors would suggest that the determinant is not zero. However, two independent eigenvectors and another independent generalized eigenvector do not mean the determinant is nonzero.
Last edited: Nov 7, 2007
9. Nov 8, 2007
### matt grime
Evidently there is a lot of confusion here. One being that you haven't studied the definition of a generalized eigenspace.
What is an eigenspace? It is one in which every vector is an eigenvector (with the same eigenvalue t - so don't go starting to introduce two different e-values since that is not what is going on). In a generalized eigenspace, not all vectors are eigenvectors, so there is a *big* difference.
In the example you gave you had two e-values s,t and s had multiplicity two. In that case there is no need to invoke generalized e-spaces. But since not every matrix is diagonalizable what you invoke is a non-example. I have no idea why Ilarsen thinks my 2x2 example is 'not correct', since it is correct and encapsulates all of the information you need to know. In
[1 1]
[0 1]
there is only one e-value, 1, and only one e-vector. But the generalized e-space is the whole of R^2. So you see that a generalized e-space is strictly different from an e-space.
10. Nov 8, 2007
### FunkyDwarf
But that one e-vector can be anything in R2 right?
11. Nov 8, 2007
### llarsen
mg, you are right. Thankyou for the correction. The definition I had in memory was not accurate. Sorry for the confusion.
12. Nov 8, 2007
### matt grime
What 'one e-vector'?
13. Nov 8, 2007
### FunkyDwarf
That one
14. Nov 8, 2007
### ZioX
No! It has to be the one which (A-I)v=1-eigenvector so that when (A-I) acts on the 1-eigenvector we have (A-I)^2v=0 as per definition of generalized eigenvectors.
Look up Jordan Canonical form.
15. Nov 9, 2007
### matt grime
I don't understand what either of you are saying.
[1 1]
[0 1]
has exactly one eigenvector (up to scalar multiplication), so how can it possibly be anything in R^2? | 2,097 | 7,475 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2016-44 | longest | en | 0.919211 |
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ProblemSet3
# ProblemSet3 - Professor Mumford [email protected] Econ 360...
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Professor Mumford Econ 360 - Fall 2010 [email protected] Problem Set 3 Due at the beginning of class on Tuesday, September 14 Multiple Choice Questions (12 points) Suppose that a random sample of 200 male students is selected from Purdue’s student population. These mens’ weight (measured in pounds) is regressed on their height (measured in inches) with the following result: d weight = - 100 + 4 height n = 200 R 2 = . 8100 1. What is the predicted weight from this regression for someone who is 5’10” (70 inches) tall? (a) 380 (b) 280 (c) 200 (d) 180 2. Suppose that instead of measuring in pounds and inches, we measured in pounds and feet (5.833 feet instead of 70 inches). What would be the estimated intercept? (note that 12 inches = 1 foot) (a) -8.333 (b) -12 (c) -100 (d) -1200 3. Again, suppose that instead of measuring in pounds and inches, we measured in pounds and feet. What would be the estimated coeficient on height ? 1
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4. Suppose that instead of measuring in pounds and inches, we measured in kilograms and inches. What would be the estimated intercept? (note that 1 pound = .454 kilograms) 5. Again, suppose that instead of measuring in pounds and inches, we measured in kilo- grams and inches. What would be the estimated estimated coeficient on height ? 6. Suppose that instead of measuring in pounds and inches, we measured in kilograms and feet. What would be the new R-squared? (a) 0.0675 (b) 0.2700 (c) 0.3677 (d) 0.8100 True/False (6 points) Please write the entire word. No explanations are required. 7. In the simple linear regression model, the homoskedasticity assumption means that the error, u , has the same variance given any value of the explanatory variable, x .
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WAEC 2019 - PRACTICAL CHEMISTRY ANSWER
WAEC 2019 - PRACTICAL CHEMISTRY ANSWER
WAEC 2019 - PRACTICAL CHEMISTRY ANSWER
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VERY IMPORTANT MESSAGE:
Please make sure you use your school end point in that number 1. In my answer, you will see that i used 12.50cm3. Make sure you use what your school used. So try edit my 12.50cm3 and put yours, then recalculate. Meaning that you should replace my 12.50cm3 with your school's end point anywhere you see it on my work. Some school got 12.90cm3, 13.00cm3, some even got 18.50cm3, etc
(1a)
DRAW A TABLE WITH THE FOLLOWING
Final |13.60|15.50|14.80|12.50
Initial |1.00 |3.10 |2.40 |0.00
Vol. acid used |12.60|12.60|12.40|12.50
Volume of S2O3²- =12.60+12.40+12.50/3
=37.50/3
=12.50cm³
(1bi)
Conc. in g/dm³ = molar mass * conc in mol/dm³
But molar mass of Na2S2O3
=(23*2)+(32*2)+(16*3)
=46 + 64 + 48
=158g/mol
Therefore 15.8g/dm³ = 158g/mol * conc of A in mol/dm³
Conc of A in mol/dm³ = 15.8/158
=0.1mol/dm³
(1bii)
Using CaVa/CbVb = na/nb
0.1*12.50/Cb * 25.00 = 2/1
50Cb = 1.25
Cb = 1.25/50
Cb = 0.025mol/dm³
The conc of I2 in B = 0.025mol/dm³
(1biii)
Gram conc of I2 = molar mass * molar conc
= (127*2)*0.025
=6.35g/dm³
Percentage mass of I2 in sample = 6.35/9.0 *100%
= 0.7056 * 100%
=70.56%
(1c)
There have to be a change in the colour of the mixture before it is added. The end point is known when the blue colour formed as starch is added, changes to colourless
=====================
(2)
PLS DRAW A TABLE FOR THESE:
TEST|OBSERVATION | INFERENCE
(2a)
TEST: Salt C+10cm^3 of distilled water and shake with filter
OBSERVATION: Salt C dissolves partly in water to give colourless filerate and residue
INFERENCE: Salt C contains both soluble and insoluble salts
(2bi)
TEST: 2cm^3 of filterate of salt C +AgNO3(aq) then dil HNO3
OBSERVATION: Formation of white ppt then the white ppt remains
INFERENCE: Cl- or CO3^2- present
(2bii)
TEST: Solution from 2bi + NH3(aq) in excess
OBSERVATION: The white ppt formed in 2bi dissolves in NH3(aq)
INFERENCE: Cl- confirmed
(2ci)
TEST: Residue of C + 2cm^3 of dil HCl then shake
OBSERVATION: Effervescence occurs with the liberation of a gas that is odourless and colourless but turns lime water milky and turns moist blue litmus paper to red
INFERENCE: Acidic gas is evolved ie CO2 is evolved from CO3^2- or HCO3-
(2cii)
TEST: Mixture from 2ci +NH3(aq) in drops then in excess
OBSERVATION: Blue gelatinous ppt is formed in drops which turns deep blue colouration in excess
INFERENCE: Cu^2+present, Cu^2+ confirmed
(2&3)
============================
(3ai)
On adding BaCl2 solutions to a portion of saturated Na2CO3 precipitate is formed,precipitate dissolves on adding excess dilute HCl
(3aii)
Q is reducing agent like SO2, H2S, CO
(3bi)
calcium oxide(CaO)
(3bii)
concentrated H2SO4
(3c)
NaOH pellet is deliquescent because it absorb moisture from the atmosphere to form solution
+++++++++++++++++++++++++++ | 1,048 | 2,996 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2019-35 | latest | en | 0.786333 |
https://mathhelpboards.com/threads/fourier-series-showing-converges-to-pi-16.6539/ | 1,610,732,719,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703495936.3/warc/CC-MAIN-20210115164417-20210115194417-00113.warc.gz | 503,936,812 | 15,244 | # Fourier series--showing converges to pi/16
#### dwsmith
##### Well-known member
When a Fourier series contains only sine and cosine terms, evaluating the series isn't too difficult.
However, I want to show a Fourier series with sine and sinh converges to $$\frac{\pi}{16}$$.
$T(50, 50) = \sum_{n = 1}^{\infty} \frac{\sin\left(\frac{\pi(2n - 1)}{2}\right) \sinh\left(\frac{\pi(2n - 1)}{2}\right)} {(2n - 1)\sinh[(2n - 1)\pi]} = \frac{\pi}{16}$
Since this series contains sinh, I am not sure how to evaluate it.
#### Deveno
##### Well-known member
MHB Math Scholar
My thought: use
$\text{sinh}(x) = -i \sin(ix)$
#### dwsmith
##### Well-known member
My thought: use
$\text{sinh}(x) = -i \sin(ix)$
I think I will still have some trouble though. I end up with:
$\sum_{n = 1}^{\infty}\frac{(-1)^{n+1}}{2n - 1}\frac{\sin\left[\frac{i\pi}{2}(2n-1)\right]}{\sin\left[i\pi(2n-1)\right]}$
#### Opalg
##### MHB Oldtimer
Staff member
When a Fourier series contains only sine and cosine terms, evaluating the series isn't too difficult.
However, I want to show a Fourier series with sine and sinh converges to $$\frac{\pi}{16}$$.
$T(50, 50) = \sum_{n = 1}^{\infty} \frac{\sin\left(\frac{\pi(2n - 1)}{2}\right) \sinh\left(\frac{\pi(2n - 1)}{2}\right)} {(2n - 1)\sinh[(2n - 1)\pi]} = \frac{\pi}{16}$
Since this series contains sinh, I am not sure how to evaluate it.
The sin function isn't really there at all, because $\sin\left(\frac{\pi(2n - 1)}{2}\right) = \sin\left(\bigl(n-\frac12\bigr)\pi\right) = (-1)^{n-1}$. Also, you can use the identity $\sinh(2x) = 2\sinh x\cosh x$, to rewrite the sum as $$\displaystyle \sum_{n = 1}^{\infty}\frac{(-1)^n}{2(2n-1)\cosh\left(\bigl(n-\frac12\bigr)\pi\right)}.$$ That looks a bit less complicated than the original series, but I still don't see how to sum it. It clearly converges very fast, because the cosh term in the denominator will get very large after the first few terms. A numerical check shows that the sum of the first three terms is $0.1968518...$, which is very close to $\pi/16$. But that isn't a proof! | 713 | 2,059 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2021-04 | longest | en | 0.838143 |
https://people.richland.edu/james/spring06/m113/m113-s07.html | 1,513,291,952,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948551162.54/warc/CC-MAIN-20171214222204-20171215002204-00381.warc.gz | 633,934,507 | 2,670 | # Exam 4 Study Guide: Chapters 7-8
1. A graph of a probability distribution is given along with a critical value and level of significance. Draw and label a vertical line at the critical value, shade and label the critical region, label the non-critical region, label each region with the area in that region, write "Reject H0" and "Retain H0" in the appropriate regions of the graph. Also, identify whether it is a left tail, right tail, or two-tail test. Most of this is in your notes and the graphs from the activities.
2. Know the concept that is fundamental to all hypotheses testing.
3. A test statistic and the area to the left of the test statistic are given. Give the p-value for a left tail test, a right tail test, and a two tail test.
4. Five statements are given. For each one, decide whether the statement is the null or alternative hypothesis. These are English statements like "The defendant is innocent", not mathematical ones like "12% of adults wet their bed". Remember that the null hypothesis is the normal or assumed condition.
5. Five statements are given. For each one, decide whether it represents a type I or type II error. For example, "A blood test comes back negative (not infected) when the person really is infected" is a type II error because the normal condition of a person is that they are not infected. Since they actually are infected, the null hypothesis is false and we are retaining a false null hypothesis.
6. Three p-values and significance levels are given. In each case, decide whether to reject or retain the null hypothesis.
7. Five critical value(s) and test statistics are given. For each case, decide whether it is a left tail, right tail, or two tail test and whether to Reject H0 or Retain H0.
8. Circle the correct response so that the conclusion is properly worded. Three parts.
9. Three confidence intervals are given along with a null hypothesis. Decide whether the test is left tailed, right tailed, or two tailed and whether you would reject or retain the null hypothesis.
10. Determine whether the given statement is the null hypothesis or the alternative hypothesis. These are symbolic statements like p=0.42, not English statements like in question 4.
11. A statistical test that you've never heard of is conducted and the p-value and null hypothesis are given. Write the decision and give the conclusion.
12. Five claims are given. For each claim, write the null and alternative hypotheses and determine whether it is a left tail, right tail, or two tail test. These are mathematical statements like "the average adult earns \$35,000 a year". They could be about one or two proportions or means. If there are two samples, be sure to define the subscripts or use subscripts that make sense.
13. Work a hypothesis test. It could be about one or two proportions or means. Write the original claim symbolically and decide if it is the null or alternative hypothesis. Write H0 and H1 and identify it as a left tail, right tail, or two tail test. Identify key values from the problem. The test statistic, p-value, and/or confidence interval from Minitab are given, use them to make a decision and then write the conclusion. Look at activities 6, 7, and 8.
14. Similar to #13
15. Similar to #13
## Notes
• You will need a calculator for the test.
• In questions 13 - 15, two tail probabilities from Minitab are given. If your hypothesis test is a one tail test, then you will need to divide the p-value by two.
• You will not need to look up any critical values in the tables, so no tables will be supplied.
• A review exercise with sample problems is available.
## Points per problem
# Pts 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Total 6 3 3 5 5 3 6 6 6 5 3 10 12 14 13 100 | 872 | 3,730 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2017-51 | latest | en | 0.874308 |
null | null | null | null | null | null | The Masters: Time moves on, the azaleas will bloom again, the woman problem will be solved
greenjacket.jpgA woman will never wear The Masters green jacket. But eventually, she'll belong to the club that issues it.
Why lookey here: a problem that is solving itself!
At a time when solutions seem in short supply — health insurance, anyone? — the landscape can be discouraging.
And then along comes a rite of springtime sports, a pleasant and storied golf tournament.
Suddenly, fans who would rather hear Jim Nance speak of the Hogan Bridge with a reverence more fitting for the Lincoln Memorial have to endure all this talk about the “woman issue.” Again.
The last time it flared, 10 years ago, the handful of picketers protesting Augusta National Golf Club’s men-only membership policy were mostly ignored.
So what has changed? Why is the issue returning, only now, defanged?
For while Augusta dug in its heels, time moved on. The azaleas continued to bloom and the world continued to change —whether or not the gentlemen of Augusta welcomed it or even noticed it.
In that changing world, a woman was named to lead a company racking up $100 billion in annual sales. Coincidentally, her company, IBM, is a major sponsor of the Masters golf tournament, played at Augusta.
So will Ginni Rometty, like previous CEOs of IBM, be allowed to take her place with the other titans of industry on Augusta’s membership roster?
In a still-sour economy, it’s hard to imagine anyone cares about the ability of one multimillionaire to play golf with the other multimillionaires.
People generally don’t storm the barricades on behalf of those wearing spiked shoes and polo shirts.
Augusta has the law on its side; it is a private club. Yet anyone who has ever played in a weekday golf tournament and called it (or expensed it as) “work” understands that far more gets accomplished during a round of golf than hitting a little white ball into 18 little holes.
In addition to enjoying the course, Augusta’s bigwigs get to rub elbows with each other. That’s part of the allure. With every passing year, though, they’ll be missing executives from more companies. This year, it’s IBM. Next year, it could be Hewlett-Packard.
Sooner or later, they’ll decide they’d rather have the prestige of those powerful women than their dusty membership “tradition.” Women — or at least one woman — will be invited to join.
So no matter what your opinion of the issue, cross it off your list of things to get worked up over. Move on to a cause that needs you more.
For we know how this is going to end. And best of all, we don’t even have to do anything about it. | null | null | null | null | null | null | null | null | null |
http://mathhelpforum.com/calculus/155827-derivative-arctan-partial-derivative-print.html | 1,524,533,060,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125946314.70/warc/CC-MAIN-20180424002843-20180424022843-00349.warc.gz | 205,637,804 | 3,077 | Derivative of arctan in a partial derivative
• Sep 11th 2010, 08:07 AM
tinyone
Derivative of arctan in a partial derivative
Im working on the derivative f'y(x,y) of the function f(x,y)=arctan(x^2)+ y^3
Because it's a derivative with respect to the variable y, this means that the variable x is held constant, does this mean that the entire arctan(x^2) should be regarded as just a number so the partial derivative equals to 3y^2 in the end??
• Sep 11th 2010, 08:28 AM
Plato
Quote:
Originally Posted by tinyone
Im working on the derivative f'y(x,y) of the function f(x,y)=arctan(x^2)+ y^3
Because it's a derivative with respect to the variable y, this means that the variable x is held constant, does this mean that the entire arctan(x^2) should be regarded as just a number so the partial derivative equals to 3y^2 in the end??
You are correct.
• Sep 12th 2010, 06:41 AM
tinyone
Cheers,
Derivative f'z(x,y,z) = 2(z+3)^4 + ln(xyz)
So keeping y and x constant would give me: f'z= 8(z+3)^3 + (1/xyz) * xy
• Sep 12th 2010, 01:52 PM
HallsofIvy
Quote:
Originally Posted by tinyone
Cheers,
$\displaystyle f_z= 8(z+ 3)^3+ \frac{1}{z}$ | 367 | 1,134 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2018-17 | latest | en | 0.908857 |
https://upsconlineacademy.com/area-mcqs-set-11/ | 1,720,860,500,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514493.69/warc/CC-MAIN-20240713083241-20240713113241-00609.warc.gz | 480,627,730 | 17,232 | 2 2 8 5 6 5
# AREA MCQS SET-11
1. A stream, which flows at a uniform rate of 2.5 km per hour, is 20 metres wide. If the depth of a certain ferry is 1.2 m, how many litres of water will pass through the ferry in a minute?
A) 1000000
B) 900000
C) 500000
D) None of these
Ans. A
The total quantity of water =
=
= 1000 m3 = 1000000 litres
2. The annual rainfall at a place is 43 cm. Find the weight (in metric tonnes) of the annual
rainfall on a hectare of land, taking the weight of water to be 1 metric tonne for 1 cubic metre.
A) 4000
B) 4300
C) 5000
D) None of these
Ans. B
Height of water on 1 sq m area of land annually = 43 cm
Volume of water on 10,000 sq m. area of land = × 10000 m3 = 4300 m3
Weight of water = 4300 metric ton
3. A metallic right circular cone of height 9 cm and base radius 7 cm is melted into a cuboid with two sides as 11 cm and 6 cm. What is the third side of the cuboid?
A) 5 cm
B) 6 cm
C) 7 cm
D) 9 cm
Ans. C
Let the third side of cuboid = x cm
Now, according to the question:
1/3 πr2h = l × b × h
⇒ 66x = 462
⇒ x = 7 cm
4. A garden is 112m long & 78 m wide. It has walking path 25m wide all around it on inside. Find the area of path ?
A) 925 m2
B) 927 m2
C) 929 m2
D) 934 m2
Ans. A
If path is within the garden then use this formula:
2 * B * (L + B – 2 * B)
2 * 25 (112 + 78 – 2 * 25)
= 5 * 185 = 925 m2
5. There is a narrow rectangular plot, reserved for a school, in Mahuli village. The length and breadth of the plot are in the ratio 11:4. At the rate Rs100 per metre it will cost the village panchayat Rs 75000 to fence the plot. What are the dimensions of the plot?
A) 274 & 100 m
B) 275 & 200 m
C) 275 & 100 m
D) None of the above
Ans. C
Let the common ratio between the length and breadth of the rectangular plot be x. Hence, the length and breadth of the rectangular plot will be 11x m and 4x m respectively.
Perimeter of the plot = 2(Length + Breadth)
It is given that the cost of fencing the plot at the rate of Rs 100 per metre is Rs 75, 000.
∴ 100 × Perimeter = 75000
100 × 30x = 75000
3000x = 75000
Dividing both sides by 3000, we obtain
x = 25
Length = 11x m = (11 × 25) m = 275 m
Breadth = 4x m = (4 × 25) m = 100 m
Hence, the dimensions of the plot are 275 m and 100 m respectively. | 796 | 2,245 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2024-30 | latest | en | 0.856676 |
http://themathpage.com/ARITH/ratio-and-proportion_2-2.htm | 1,513,257,236,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948544124.40/warc/CC-MAIN-20171214124830-20171214144830-00069.warc.gz | 264,109,485 | 5,089 | S k i l l
i n
A R I T H M E T I C
Lesson 18 Section 2
# RATIO AND PROPORTION
## "Out of"
Back to Section 1
"Out of" : The ratio Part : Whole.
4. What does "out of" indicate? 3 out of 5 The ratio of a smaller number to a larger, of a part to the whole.
This illustrates 3 out of 5: the ratio of the part, 3, to the whole, 5. The part is three fifths of the whole; Lesson 17. (Here, "part" refers to whatever is less than the whole.)
"Out of" is often how fractions are introduced -- as ratios, but with fractional symbols. A fraction, however, is a number we need for measuring. "Out of" does nothing to explain where the number 3/5 belongs on the number line. See Lesson 20.
Example 1. In a class of 40 students, 3 out of 5 got B. How many students got B?
Solution 1. Complete this proportion:
3 : 5 = ? : 40.
(3 out of 5 is how many out of 40?)
"5 goes into 40 eight times. Eight times 3 is 24."
Lesson 18.
24 students got B.
That should be a simple mental calculation. Simply see that 40 is eight 5's.
Notice that "3 out of 5" -- a smaller number out of a larger -- makes sense. It would make no sense to say "5 out of 3."
"3 out of 5" means, For every 5, there are 3.
If there are two 5's,
there will be two 3's. If there are three 5's, there will be three 3's:
And so on. Therefore, if there are eight 5's, there will be eight 3's. This is the theorem of the same multiple (Section 1).
Solution 2. To say that 3 out of 5 got B, is to say that three fifths of the 40 students got B. One fifth of 40 is 8. Therefore, three fifths are 3 × 8 = 24. Lesson 15.
Example 2. In a recent survey, 7 out of 10 people responded Yes. If 280 people responded Yes, then
a) how many people were surveyed? b) how many responded No?
Solution. In this Example, 280 is the part that responded Yes. It corresponds to 7.
7 : 10. = 280 : ?
Now, 280 is 40 × 7. Therefore the missing term is 40 × 10.
7 : 10 = 280 : 400.
400 is the whole number of people surveyed.
b) Since 280 is the number that responded Yes, then the difference,
400 − 280,
b) will be the number that responded No.
400 − 280 = 120.
Percent: Out of 100
Percent problems often involve the expression "out of," because percent is how many for each, or out of, 100. Therefore any number out of 100 is that percent.
Example 3. 8 out of 100 is what percent?
Note: When we consider the ratio of a smaller number to 100, then we may say "out of." But when we have the ratio of a larger number to 100 -- 200 to 100 -- then 200 "out of" 100 make no sense. In that case, we must say "for each."
Example 4. 250 for each 100 is what percent?
Example 5. 8 out of 25 students got B. What percent got B?
Answer. Percent is out of 100. Therefore, let us complete this proportion:
8 : 25 = ? : 100.
"8 out of 25 is equal to how many out of 100?"
Now, 4 × 25 = 100. Therefore the missing term will be 4 × 8.
8 : 25 = 32 : 100.
But 32 out of 100 is 32%. And since that is equal to 8 out of 25, then 8 out of 25 is also 32%.
32% of the students got B.
Note that the following questions mean the same:
8 out of 25 is what percent?
8 is what percent of 25?
Finding a percent by making the fourth term 100 is called the method of proportions. It is another instance of The Rule of Three.
5. How do we find a Percent by the method of proportions? 1st : 2nd = Percent : 100. Make the 4th term of a proportion 100. The 3rd term is then the Percent that the 1st is of the 2nd.
We will go into this more in Lesson 30.
Example 6. 18 is what percent of 200?
Solution. Proportionally,
18 : 200 = ? : 100.
Now, to go from 200 to 100, we have to divide by 2. Therefore, we have to divide 18 by 2, also:
18 : 200 = 9 : 100.
18 is 9% of 200.
Example 7. 18 is what percent of 50?
Solution. Proportionally,
18 : 50 = ? : 100.
Here, to go from 50 to 100, we have to multiply by 2. Therefore, we have to multiply 18 by 2.
18 : 50 = 36 : 100.
18 is 36% of 50.
We see that when the 2nd term is smaller than 100, we have to multiply it. But when the 2nd term is more -- Example 6 -- we have to divide it.
Example 8.
a) We know that 6% is 6 out of 100. .6% is 6 out of how many?
.6 out of 100.
Therefore on multiplying both terms by 10:
.6 out of 100 = 6 out of 1000.
b) .06% is 6 out of how many?
Answer. On multiplying both terms by 100,
.06 out of 100 = 6 out of 10,000.
At this point, please "turn" the page and do some Problems.
or
Continue on to the next Section.
Previous Section
1st Lesson on Parts of Natural Numbers
E-mail: [email protected]
Private tutoring available. | 1,482 | 4,634 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2017-51 | longest | en | 0.905218 |
null | null | null | null | null | null | Enhancing Entrepreneurship of Garifuna Women
Approximately 98,000 Garífuna people live in Honduras, mostly concentrated on the North Coast. Garífuna women work as subsistence farmers and process much of the cassava they grow into a flat bread called casabe. TechnoServe is providing technical and business development assistance to ten groups of Garifuna women, representing 300 members, to improve cassava productivity and sell casabe in local and export markets. With the introduction of higher yielding cassava varieties and improved techniques, the women have more than tripled their yields. TechnoServe has also trained farmer groups in processing, value addition and marketing. | null | null | null | null | null | null | null | null | null |
https://www.teachoo.com/13225/530/Ex-8.2--1-ii/category/Ex-8.2/ | 1,723,784,676,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641333615.45/warc/CC-MAIN-20240816030812-20240816060812-00642.warc.gz | 778,126,678 | 21,412 | Ex 8.2
Chapter 8 Class 10 Introduction to Trignometry
Serial order wise
### Transcript
Ex 8.2, 1 Evaluate the following : (ii) 2 tan2 45° + cos2 30° – sin2 60° We know that, tan 45° = 1 cos 30° = √3/2 sin 60° = √3/2 Putting all values 2tan 2 45° + cos2 30° -= sin2 60° = 2 ×(𝟏)𝟐+(√𝟑/𝟐)^𝟐−(√𝟑/𝟐)^𝟐 = 2 + 0 = 2 | 160 | 311 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2024-33 | latest | en | 0.494094 |
https://www.educationquizzes.com/ks1/maths/year-2-time-telling-the-time-and-understanding-its-units/print/ | 1,582,395,205,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145713.39/warc/CC-MAIN-20200222180557-20200222210557-00357.warc.gz | 700,534,456 | 4,202 | # KS1 Maths Quiz - Year 2 Time - Telling the Time and Understanding its Units (Questions)
This quiz addresses the requirements of the National Curriculum KS1 Maths and Numeracy for children aged 6 and 7 in year 2. Specifically this quiz is aimed at the section dealing with telling the time and understanding units of time.
Once children in Year 2 are familiar with telling the time to the hour, half hour and quarter hour, and can also tell the time to 5 minutes, to test their understanding they will be expected to solve simple problems to do with time. These problems will focus on the units of time, hours, half hours and quarter hours, and will be set in real life contexts such as journeys or TV programmes.
This quiz will help children to understand the units of time and so help them to tell the time.
1. It's 3 o'clock. What time will it be in half an hour?
[ ] Half past 2 [ ] Half past 3 [ ] Ten to 4 [ ] Five past 3
2. It's half past 7. How long is it until 8 o'clock?
[ ] Quarter of an hour [ ] 1 hour [ ] 60 minutes [ ] 30 minutes
3. I watch my favourite TV programme at 5 o'clock. It lasts for an hour. What time will it finish?
[ ] Half past 5 [ ] Five to 5 [ ] 7 o'clock [ ] 6 o'clock
4. School starts at 9 o'clock. It takes me half an hour to walk to school from my house. What time should I leave home?
[ ] Half past 8 [ ] Quarter to 9 [ ] Five to 9 [ ] Twenty past 7
5. Joe visits the park and spends an hour there. If it takes 15 minutes to walk to the park, and 15 minutes to walk back again, how long is Joe out for?
[ ] One and a half hours [ ] 60 minutes [ ] 30 minutes [ ] 45 minutes
6. Max bakes a cake. He takes 30 minutes to mix all the ingredients and 30 minutes to bake the cake. How long will the cake take in total?
[ ] 40 minutes [ ] 1 hour [ ] 30 minutes [ ] 2 hours
7. I spend 15 minutes reading, 30 minutes watching TV and 15 minutes drawing. How much time have I spent doing these activities?
[ ] 120 minutes [ ] 45 minutes [ ] 50 minutes [ ] 60 minutes
8. If school starts at 9 o'clock, and finishes at 3 o'clock, how many hours do I spend in school each day?
[ ] 3 hours [ ] 6 hours [ ] 9 hours [ ] 4 hours
9. It takes Ben 20 minutes to walk to town. Sam takes 20 minutes more than Ben. How long does it take Sam to walk to town?
[ ] 40 minutes [ ] 25 minutes [ ] 20 minutes [ ] 15 minutes
10. I start reading my book at half past 10, and finish it 4 hours later. What time do I finish my book?
[ ] Half past 6 [ ] Half past 12 [ ] Half past 4 [ ] Half past 2
KS1 Maths Quiz - Year 2 Time - Telling the Time and Understanding its Units (Answers)
1. It's 3 o'clock. What time will it be in half an hour?
[ ] Half past 2 [x] Half past 3 [ ] Ten to 4 [ ] Five past 3
Half an hour is the same as 30 minutes
2. It's half past 7. How long is it until 8 o'clock?
[ ] Quarter of an hour [ ] 1 hour [ ] 60 minutes [x] 30 minutes
It's 30 minutes, or half an hour, until 8 o'clock
3. I watch my favourite TV programme at 5 o'clock. It lasts for an hour. What time will it finish?
[ ] Half past 5 [ ] Five to 5 [ ] 7 o'clock [x] 6 o'clock
1 hour later than 5 o'clock is 6 o'clock
4. School starts at 9 o'clock. It takes me half an hour to walk to school from my house. What time should I leave home?
[x] Half past 8 [ ] Quarter to 9 [ ] Five to 9 [ ] Twenty past 7
If it takes half an hour to get to school, this is the latest time I should leave
5. Joe visits the park and spends an hour there. If it takes 15 minutes to walk to the park, and 15 minutes to walk back again, how long is Joe out for?
[x] One and a half hours [ ] 60 minutes [ ] 30 minutes [ ] 45 minutes
15 minutes and 15 minutes makes 30 minutes altogether, plus the hour at the park
6. Max bakes a cake. He takes 30 minutes to mix all the ingredients and 30 minutes to bake the cake. How long will the cake take in total?
[ ] 40 minutes [x] 1 hour [ ] 30 minutes [ ] 2 hours
30 minutes plus 30 minutes makes 60 minutes, which is 1 hour
7. I spend 15 minutes reading, 30 minutes watching TV and 15 minutes drawing. How much time have I spent doing these activities?
[ ] 120 minutes [ ] 45 minutes [ ] 50 minutes [x] 60 minutes
60 minutes is the same as 1 hour
8. If school starts at 9 o'clock, and finishes at 3 o'clock, how many hours do I spend in school each day?
[ ] 3 hours [x] 6 hours [ ] 9 hours [ ] 4 hours
Counting from 9 to 3 gives us 6 hours altogether
9. It takes Ben 20 minutes to walk to town. Sam takes 20 minutes more than Ben. How long does it take Sam to walk to town?
[x] 40 minutes [ ] 25 minutes [ ] 20 minutes [ ] 15 minutes
20 + 20 = 40, so Sam takes 40 minutes to get there
10. I start reading my book at half past 10, and finish it 4 hours later. What time do I finish my book?
[ ] Half past 6 [ ] Half past 12 [ ] Half past 4 [x] Half past 2
Half past 11, half past 12, half past 1, half past 2 | 1,478 | 4,840 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2020-10 | latest | en | 0.94075 |
https://www.talkstats.com/threads/sample-size.72347/ | 1,623,716,992,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487614006.8/warc/CC-MAIN-20210614232115-20210615022115-00084.warc.gz | 923,906,875 | 10,466 | # sample size
#### bell
##### New Member
Hey all
i have two age group to study on the same area, but will ask them completely different questions using different tools. here is the population
Age, male, female, total
2-5, 7500, 7500, 15000
6-8, 9000, 9000, 18000
So when i calculate the sample size do i use 33,000 the total population or each age group population. remember the questionnaires for each age is on different topics
thank you
#### hlsmith
##### Less is more. Stay pure. Stay poor.
What is the question you are trying to answer with the sample size calculation?
#### bell
##### New Member
for both age group the question i am trying to answer is different for example for the 1st age group it is whether they are overweight , underweight , vaccinated and also if they were taking any food supplement in the previous months. For the 2nd age group what i am trying to answer is whether they are attending school regularly, treated well at school, or get sick very often. hope this help and thank you again
#### Karabiner
##### TS Contributor
Still not sure what you are trying to answer. Do you want to compare sample data with your "population" data?
Or, do you want to compare are sample of males with a sample of females? And why do you need to consider
"population" size - do you need to perform a correction for finite populations?
Anyway, if you perform 2 completely separated analyses, why should sample size considerations for one analysis be
affected by the other Analysis?
With kind regards
Karabiner
#### bell
##### New Member
Thank you very much Karabiner
Agree I think my question were not that clear. Let me try to simplify it again
i have 3 small town to study with a limited budget and the population i want to study is as follows:
Town....Age 2 to 5.....Age 6 to 8.....Total....Sample Size calculated
A.............15000...........18000...............33000................380
B................5500.............5000...............10500.................371
C ...............9600..............9000..............18600................377
Total for 3 towns.................................62100
I calculated the above sample size for each town based on 95 CI and 5% Margin of error. My question is do i take the sample size above for each town and spread it proportionally to each age group or calculate the sample size for 62,000 which is 382? What i want to make sure is that if i take each town sample size the total sample size will be around 1128 which is Big (needs \$ which i do not have) and if i take the 382 and spread it proportionally to the age group and town it becomes small.
The question i want to answer for each group is completely different and i am not comparing the group at all . i have developed a questions for those 2 to 5 and want to know how many of them are underweight, overweight or taking food supplement . For the 2nd age group i have a different questionnaire which i just want to see if they are attending primary school regularly.
I hope i am making sense
Thank you again all | 698 | 3,052 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2021-25 | latest | en | 0.937577 |
https://www.effortlessmath.com/math-topics/reading-clocks/ | 1,726,660,570,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651895.12/warc/CC-MAIN-20240918100941-20240918130941-00687.warc.gz | 683,469,870 | 13,740 | Need complete information about the clock and how to read them? No need for searching any further! What you need is provided for you in this article.
Clocks or timepieces are gadgets utilized for measuring and indicating the time.
Analog clocks are a device to show the time that has numerals all around the edge as well as hands that go around to show what time it is. Clocks are utilized to tell how much time has passed.
Reading clocks is a talent that is simple to learn by taking a bit of time and energy.
Analog clocks get divided up around a circle, if you read the minute and hour hand separately, it assists in seeing what time it is. With a bit of effort, you can learn how to read clocks easily.
## Way to Read a Clock
1- Learn the way clocks are divided. Clocks are divided into $$12$$ portions. On a clock’s top, you’ll find the “$$12$$.” On the right of this “$$12$$,” you’ll find a “$$1$$.” If you then follow the numerals from the right or go “clockwise,” they’ll read from “$$1$$” to “$$12$$.”
• The numerals which mark each area are the hours
• The areas in-between numerals are separated into five-minute portions. At times, there are also small lines running alongside the clock separating these portions.
2- Utilize the small hand for reading hours. Clocks come with two hands: a small hand and a large hand. The small hand marks hours. Whatever numeral it’s pointing to is the hour of the day.
• For instance, if the small hand points to a “$$1$$,” it is in the one o’clock hour.
3- Utilize the large hand for reading minutes. Use the numeral it’s pointing to, and then multiply it by five to find out the minutes. Whenever it’s pointing to a “$$12$$,” it’s at the top of the hour. Should the large hand have a mark in-between the numerals, count those marks, and after that add the marks to the minutes (clock number times five). For instance:
• Should the large hand be pointing to a “$$3$$,” it’s fifteen minutes after the hour.
• Should the large hand be pointing to a “$$12$$,” it’s at the top of the hour. Read whichever numeral the small hand is pointing to.
• Should the large hand be in-between “$$1$$” and “$$2$$,” see which dash it’s pointing to. For instance, if it’s on the third dash after the “$$1$$,” it’s $$8 ‘$$minutes after the hour. ($$1 × 5 +$$ the number of dashes).
4- Put all these together to find out the time. When you have figured out the hour and the minutes, then you will know the time. For instance:
• If the small hand points to “$$1$$” and the large hand points to “$$12$$,” it’s “one o’clock.”
• If the small hand points to “$$1$$” and the large hand points to “$$2$$,” it’s “one-ten” or “ten minutes after one.”
• If the small hand points to “$$1$$,” and the large hand is halfway in between the “$$2$$” and “$$3$$,” it’s about “one-twelve” or “twelve minutes after one.”
Note: Distinguish between AM and PM. AM and PM can’t be figured out on a regular clock. You must know what time of the day it is. From midnight to noon the following day, it is considered as AM. From noon to midnight, it is considered as PM.
• For instance, if it is early in the morning and the small hand points to “$$9$$” and the large hand points to “$$12$$,” it is “$$9$$ AM.”
### Reading Clocks – Example 1:
Write the time below the clock.
Solution:
As mentioned, the clock has two hands, a small hand, and a large hand. The small hand shows the clock and the large hand shows the minute. Here the small hand is on the number $$12$$ so the hour is $$12$$ and the large hand is on the number $$15$$ so $$15$$ minutes past from $$12$$. This indicates that it is $$12:15$$.
### Reading Clocks – Example 2:
Write the time below the clock.
Solution:
Here, the small hand is between two numbers: the number $$7$$ and the number $$8$$. So, the hour is $$7$$ and the large hand is on the number $$10$$ so $$10$$ minutes left until $$8$$ o’clock. This indicates that it is $$7:50$$.
## Exercises for Reading Clocks
Write the time below each clock.
1)
2)
1. $$\color{blue}{6:16}$$
2. $$\color{blue}{5:25}$$
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SAVE $40 It was$89.99 now it is \$49.99 | 1,108 | 4,275 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2024-38 | latest | en | 0.927902 |
http://soft-matter.seas.harvard.edu/index.php?title=Effects_of_contact_angles&diff=prev&oldid=3161&printable=yes | 1,597,460,390,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439740423.36/warc/CC-MAIN-20200815005453-20200815035453-00182.warc.gz | 96,340,293 | 12,598 | # Difference between revisions of "Effects of contact angles"
## What is Contact Angle?
This is a good illustration of different contact angles. In this figure, contact angle θ was referred to as wetting angle α.
Contact angle, θ, is a quantitative measure of the wetting of a solid by a liquid. It is defined geometrically as the angle formed by a liquid at the three phase boundary where a liquid, gas and solid intersect.
It can be seen from this figure that low values of θ indicate that the liquid spreads, or wets well , while high values indicate poor wetting. If the angle θ is less than 90 the liquid is said to wet the solid. If it is greater than 90 it is said to be non-wetting. A zero contact angle represents total wetting.
The measurement of a single static contact angle to characterize the interaction is no longer thought to be adequate. For any given solid/ liquid interaction there exists a range of contact angles which may be found. The value of static contact angles are found to depend on the recent history of the interaction. When the drop has recently expanded the angle is said to represent the advanced contact angle. When the drop has recently contracted the angle is said to represent the receded contact angle. These angles fall within a range with advanced angles approaching a maximum value and receded angles approaching a minimum value.
If the three phase(liquid/solid/vapor) boundary is in actual motion the angles produced are called Dynamic Contact Angles and are referred to as advancing and receding angles. The difference between advanced and advancing, receded and receding is that in the static case motion is incipient in the dynamic case motion is actual. Dynamic contact angles may be assayed at various rates of speed. Dynamic contact angles measured at low velocities should be equal to properly measured static angles.
The spreading coefficient is: $S=\sigma _{s}-\left( \sigma _{sl}+\sigma _{lv} \right)$
If the spreading coefficient is positive, the liquid “spreads” on the solid (b); if the spreading coefficient is negative, the liquid “partially spreads” on the solid (a). If the contact angle is less than 90o, the liquid is said to “wet” the solid; if the contact angle is greater than 90o, the liquid is said to “not wet” the solid.
This is often called the “initial” spreading coefficient because in a fluid-fluid interface, the interfacial tension between the two fluids is measured at the instant the interface is formed, before any mutual solubilization of the two liquids take place.
In a fluid-fluid interface, spreading behavior can be very different before and after mutual solubilization. One example which illustrates the difference between initial and final behavior of spreading is hexanol on water.
The Initial S of hexanol and water :
\begin{align} & S_{init}=\text{72}\text{.8 - (24}\text{.8+6}\text{.8)} \\ & \therefore S_{init}=41.2\frac{mJ}{m^{2}} \\ \end{align}
However, after equilibrium, some hexanol dissolves in the water and thus reduce the surface tension of water from 72.8 to 28.5 $\frac{mJ}{m^{2}}$. This causes the S to become negative:
\begin{align} & S_{final}=\text{28}\text{.5 - (24}\text{.8+6}\text{.8)} \\ & \therefore S_{final}=-3.0\frac{mJ}{m^{2}} \\ \end{align}
## Initial and final contact angles
Based on a force balance Young and Dupré (independently) asserted: $\sigma _{s}=\sigma _{lv}\text{cos}\theta +\sigma _{sl}$
If vapor is adsorbed on the solid (a spreading pressure is created) and $\sigma _{sv}=\sigma _{lv}\text{cos}\theta _{e}+\sigma _{sl}$
where $\theta _{e}\ge \theta$
The change in energy of the solid with adsorption can be expressed as: $\sigma _{sv}=\sigma _{s}-\pi _{e}$
$\pi _{e}$ is called the spreading pressure.
The “initial” spreading coefficient described above refers to spreading before any vapor adsorption on the solid. It can be combined with the Young and Dupré equation to give:
$S=\sigma _{s}-\left( \sigma _{sl}+\sigma _{lv} \right)$
It can be combined with the Young and Dupré equation to give:
$S=\sigma _{lv}\left( \cos \theta +1 \right)$
This definition clearly implies that the spreading coefficient cannot be greater than twice the surface tension of the liquid no matter the solid! This is clearly wrong.
However, including the spreading pressure in the calculation of the spreading coefficient produces:
$S_{e}=\sigma _{lv}\left( \cos \theta _{e}+1 \right)+\pi _{e}$
Derjaguin introduced the more general concept of a “disjoining” pressure to account for these kinds of corrections in all of capillarity.
## Capillary rise
The driving force for capillary rise is the replacing of a high energy solid/vapor interface with a lower energy (generally) solid/liquid interface:
de Gennes, 2004, Fig. 2.17
$I=\sigma _{sv}-\sigma _{sl}$
The spreading of a liquid across a solid has a smaller driving force:
$S=\sigma _{sv}-\sigma _{sl}-\sigma _{lv}$
Therefore wicking is more common than spreading.
## Zisman's rule
In the 1950’s and 60’s Zisman found empirically that the wettability of solid surfaces could be ranked if $\cos \theta _{e}$ for a series of liquids was plotted against their surface tension:
Zisman, ACS Symp. Ser. 43, 1, 1964.
The data are extrapolated to where the cosine is one and that surface tension taken as the “critical” surface tension.
## Critical surface tensions
Solid Nylon PVC PE PVF2 PVF4 $\sigma _{c}$ (mN/m) 46 39 31 28 18
The scatter in the data led to a more careful modeling of the interaction between the solid and the liquid, led by the work done by F.M. Fowkes at Lehigh.
The general idea is that the interaction of a solid and a liquid is not a single dimensional quantity but contains “components” such as acid-base and dispersion forces.
Fowkes, F.M. Dispersion force contributions to surface and interfacial tensions, contact angles, and heats of immersion. ACS Symp. Ser. 43, 99 – 11, 1964.
## Jamin effect
In 1860 Jamin noticed that an ordinary cylindrical capillary tube filled with a chain of alternate air and water bubbles is able to sustain a finite pressure. We now know that this is a consequence of a difference between the advancing and receding contact angles leading to pressure differences.
Morrison, Fig. 10.3
If neither the advancing nor the receding contact angles differ from bubble to bubble then:
$P=\frac{2n\sigma _{lv}\left( \cos \theta _{r}-\cos \theta _{a} \right)}{r}+P_{0}$
## Relation to Decompression Sickness ("The Bends")
The Jamin Effect contributes to the prevalence of Decompression Sickness (DCS) or otherwise known as 'the bends'. DCS is commonly seen in the following circumstances:
* A scuba diva ascends quickly without taking proper decompression steps
* Cabin pressure in an aircraft fails
* Divers fly shortly after diving.
As most people know, the main cause is the bubbles that form in the capillaries. The pressure changes cause inert gases like nitrogen to form gas bubbles. When the body is exposed to decreased pressures, nitrogen dissolved in tissues and fluids in the body comes out of solution and forms bubbles. The natural blood flow and movement of the body creates micronuclei which serve as seeds to the bubbles during ascent, allowing nitrogen to diffuse into them. The nonwettable surface of the inside of blood vessels furthers the ability of the bubbles to form. The result is the existence of large bubbles that can stop a diver's heart.
The bubbles from in the capillaries and normally would be filtered out when they reach the alveoli in the lungs. However, the vast majority of cases of DCS are caused by bubbles that can't circulate. This is thought to be due in part to the Jamin Effect described above. The pockets between the bubbles act as a plug that can support a small level of pressure making it difficult for the heart to push the blood along. This mechanism is a primary cause of neurolofic DCI.
## Contact Angle Hysteresis
A great interest in surface energy is contact angle hysteresis. Contact Angle Hysteresis is defined as the difference between advancing and receding contact angles. This hysteresis occurs due to the wide range of “metastable” states which can be observed as the liquid meniscus scans the surface of a solid at the solid/liquid/vapor interface. Because there are free energy barriers which exist between these metastable states, a true "equilibriu" contact angle is impossible to measure in real time. For an "ideal" surface that is wet by a pure liquid, contact angle theory predicts one and only one thermodynamically stable contact angle. In the real world, however, the "ideal" surface is rarely found. To fully characterize any surface, therefore, it is important to measure both advancing and receding contact angles and report the difference as the contact angle hysteresis.
The real surfaces that we deal in experiments are heterogeneous and exhibits some roughness or surface variations. A liquid drop resting on such surface might be in a metastable equilibrium instead of a stable equilibrium as mostly discussed.
On an ideally smooth and homogeneous surface, the theoretical equilibrium contact angle is $\theta _{y}$ or Young's angle which corresponds to the lowest energy state for a system.
The equilibrium contact angle on a rough and heterogeneous surface is known as $\theta _{w}$ and $\theta _{c}$ respectively. Although these angles correspond to the lowest energy state, the angles found in experiments are often different. The contact angle of such system is found to be in a metastable state, in which the advancing and receding angles are different.
One example of such case is a liquid drop holding a steady contact angle on a solid surface. In an ideal system where the surface is smooth and homogeneous, the addition of a small volume of liquid to the drop will cause the drop front to advance. The same goes for the removal of liquid from the drop, which should cause the drop front to recede. However, in most real systems, these are not the case. The addition or removal of liquid in a real system often cause the droplet to increase or decrease in height without any surface front movement. The contact angle thus increases or decreases with each volume changes. When enough liquid is added or removed, the surface front will suddenly advance or recede.
## Surface heterogeneity
Two empirical equations have been proposed for heterogeneous surfaces:
Wenzel equation: surface roughness increases the contact area between the liquid and the solid so the Young-Dupré equation is modified to give:
$r\sigma _{sv}=r\sigma _{lv}\text{cos}\varphi +\sigma _{sl}$
Combining with the Young-Dupré equation gives:
$\text{cos}\varphi =r\cos \theta _{e}$
Cassie equation: the surface chemistry is not uniform but contains kinds of “patches” each with a different contact angle.
$\text{cos}\theta =\sum{f_{i}\cos \theta _{i}}$
A common assumption is that the surface has just two kinds of “patches”:
$\text{cos}\theta =f_{1}\cos \theta _{1}+f_{2}\cos \theta _{2}$
The most interesting case is if one “patch” is a hole so the contact angle is 2π
$\text{cos}\theta =f_{1}\cos \theta _{1}-f_{2}$
source: de Gennes p.24
Superhydrophobic surfaces
Theory of superhydrophobic surfaces incorporates both Wenzel and Cassie models:
Source: Feng et al., Adv. Mater., 18 (2006), 3036-3078
The surface is superhydrophobic as a result of both surface roughness (Wenzel Model) and inhomogeneity (air and solid interfaces: Cassie Model).
Interesting examples of modified surfaces to give different contact angles
ex1) Greenhouses are often covered with transparent plastic sheets. Morning dew condensing into fine droplets on the plastic scatters the light and robs flowers and plants of much needed sunlight. To find a way to force water to spread into a continuous film (to "wet" the material), "plasma" treatments are done that create hydrophillic groups on the surface of the plastic, thereby lowering the surface tension.
ex2) The human cornea is extremely hydrophobic. Our tears "treat" the surface of the cornea by depositing hydrophilic proteins that stabilize the lachrymal film.
ex3) Plastics and molecular crystals generally have a low surface tension and therefore, are poorly wettable by water. One technique to increase their wettability is to coat them with gold. However, the gold-coated plastic does not behave like bulk gold. The liquid does interact with gold, but that doesn't mean that interactions with the plastic substrate are entirely masked. While a very thin liquid film "thinks" it sits on pure gold, a thick one will still "sense" the underlying substrate. This paradoxical situation leads to "pseudopartial" wetting, where the liquid covers the solid with an extremely thin film without truly spreading (the contact angle remains finite). -from de Gennes pg24-
Interesting Research "Superhydrophobic Carbon Nanotube Forests" - in this study they grown forrests of nanotubes and then cover the "grassy" surface with PTFE which has a native contact angle of 108 degrees for water on a smooth PTFE surface. The combination of having a super rough surface with a thin layer of hydrophobic PTFE on each nanotube creates a superhydrophobic surface with measured contact angle of 180 degrees.
## Test of Cassie model
The cosine of the static contact angle of water on various subsaturated monolayers plotted versus the surface coverage measured directly using the atomic force microscope.
Woodward, J.T.; Schwartz, D.K. Dewetting modes of surfactant solution as a function of the spreading coefficient., Langmuir, 13, 6873-6876, 1997. | 3,163 | 13,625 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2020-34 | latest | en | 0.920609 |
https://www.analystforum.com/t/joint-probabilty/11206 | 1,631,839,165,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780053918.46/warc/CC-MAIN-20210916234514-20210917024514-00555.warc.gz | 634,227,019 | 5,197 | # Joint Probabilty
Please explain: Thanks S A joint probability of A and B must always be: A) greater than or equal to the conditional probability of A given B. B) less than or equal to the conditional probability of A given B. C) greater than or equal to than the probability of A or B. D) less than the probability of A and the probability of B.
P(A | B ) = P (A and B) / P(B) so which is bigger?
Considering that: P(A and B)= P(A | B) x P (B) Joint probability = Conditional probability x Probability (b) or P(A|B)= P (A and B)/ P(B) There is a high probability than the result is B.
Bonus question: When is it less than and when is it equal to the conditional probability of A given B?
If P(B)=1 then P(A and B) will be equal to P(A|B) Less then… I think never
Given, P(A | B ) = P (A and B) / P(B) , but P(B) can only be within 0 and 1, except for when P(B) = 1, therefore P(A and B) > P(A|B) answer is A
So when does that happen? When does P(B) =1? Under what condiitons does it become “less than”?
Grimer, no.
Dreary do you agree with my previous answer? Suggestion are welcome
What is wrong with D? S
> Less then… I think never Not true.
P (A) + P(B) = P(A or B) + P(A and B) or P (A) + P(B)= P (A or B) + P(A)xP(A|B) [P(A) + P(B)] - P (A or B) = P (A and B) Therefore the joint probability of P(A and B) cannot never be less than P(A) + P(B)
strangedays Wrote: ------------------------------------------------------- > P (A) + P(B) = P(A or B) + P(A and B) > > or > > P (A) + P(B)= P (A or B) + P(A)xP(A|B) > > - P (A or B) = P (A and B) > > Therefore the joint probability of P(A and B) > cannot never be less than P(A) + P(B) Sorry… I meant that the joint probability of P(A and B) > is never less than P(A) + P(B)
I wannna hear someone else try this…it’s rather easy. Hint: P(Sunny in China) = 0.40 P(Sunny in France) = 0.60
P (A) + P(B) = P(A or B) + P(A and B) P (sunny in china) + P ( sunny in France) = P (sunny in China or France) + P (sunny in China and France). 0.4 + 0.6 = (0.4 + 0.6 ) - P (sunny in China and France) + P (sunny in China and France). Therefore if two event are mutually exclusive P (A) + P (B) = P(A or B) plus I have a 0.6 probability that you are french P (A) + P(B)= P (A or B) + P(A)xP(A|B) - P (A or B) = P (A and B)
Conditional probability = joint probability / marginal probability Joint is always less than or equal to conditional.
Wow, how do you edit?
So Dreary…tell us which one do you think its the correct answer
wyantjs, Under what condiitons does it become “less than”? Under what condiitons does it become equal?
Olala… Dreary now I am 100% sure you are french… | 830 | 2,634 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2021-39 | latest | en | 0.862452 |
https://www.slideshare.net/aimarenterprises/a-after-3-points-have-been-added-to-every-score-in-a-sample-the-mepdf | 1,709,349,096,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475727.3/warc/CC-MAIN-20240302020802-20240302050802-00404.warc.gz | 967,876,445 | 61,980 | # a. After 3 points have been added to every score in a sample, the Me.pdf
a. After 3 points have been added to every score in a sample, the Mean is found to be M = 83 and the standard deviation is s = 8. What were the values for the mean and standard deviation for the original sample? b. After every score in a sample has been multiplied by 4, the mean is found to M = 48 and the standard deviation is s = 12. What were the values for the mean and standard deviation for the original sample? Solution a) Mean = 83-3 = 80 stdev didnt change = 8 b) mean = 48/4 = 12 stdev will chang eby factor of 2 so s = 6.
1 of 1
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### a. After 3 points have been added to every score in a sample, the Me.pdf
• 1. a. After 3 points have been added to every score in a sample, the Mean is found to be M = 83 and the standard deviation is s = 8. What were the values for the mean and standard deviation for the original sample? b. After every score in a sample has been multiplied by 4, the mean is found to M = 48 and the standard deviation is s = 12. What were the values for the mean and standard deviation for the original sample? Solution a) Mean = 83-3 = 80 stdev didnt change = 8 b) mean = 48/4 = 12 stdev will chang eby factor of 2 so s = 6
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# Class 10corrected - ENGR 111 Week 5 Wednesday Class 10 T...
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1 ENGR 111 Week 5 Wednesday Class 10 T. Pollock Fall, 2009
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2 Homework 4 Due next MONDAY!
3 Unit Loads Set the largest load to 1 unit and scale other loads accordingly. In this example, the magnitude of the external force applied at C is known to be 5/7ths of the magnitude of the external force applied at B. First: Sketch the structure FBD. Second: Calculate the reactions at the supports.
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4
5 Now use the Method of Joints to solve for the internal forces in each element. Start with a joint which has no more than two unknown magnitudes. Draw the joint FBD. Calculate angles. Choose the best order of solution. Solve for the unknown magnitudes.
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6
7 1 . 0 0 0 l b f B 1 . 3 4 7 l b f ( C ) 1 . 3 4 7 l b f ( C ) 1 . 0 0 0 l b f ( C ) 1 . 0 7 7 l b f ( T ) C 0 . 7 1 4 l b f 1 . 4 3 4 l b f ( T ) 1 . 0 0 0 l b f ( C ) A 1 . 4 3 4 l b f ( T ) 0 . 8 0 9 l b f 0 . 3 5 7 l b f 1 . 3 4 7 l b f ( C )
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8 The complete solution:
9 Factor of Safety A factor of safety is a number that: reduces maximum allowable loads to values that guarantee a low failure rate is generally agreed to by industry government professional societies In other words, we usually don’t want the structure to be loaded up to the very limit of failure. Factors of safety vary depending on many factors. For example: buckle on a parachute harness: 15 wing of a cargo aircraft: 1.5 wing of a fighter aircraft: 1.2 air-to-air missile: 1.1
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10 The complete solution: Now suppose that the maximum tensile force that each member can withstand without failing is 450 lbf. What applied load set
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Ask a homework question - tutors are online | 792 | 2,599 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2018-13 | latest | en | 0.571085 |
https://www.sanfoundry.com/mathematics-questions-answers-trigonometric-ratios-1/ | 1,701,374,417,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100232.63/warc/CC-MAIN-20231130193829-20231130223829-00895.warc.gz | 1,093,514,828 | 19,849 | # Class 10 Maths MCQ – Trigonometry
This set of Class 10 Maths Chapter 8 Multiple Choice Questions & Answers (MCQs) focuses on “Trigonometry”. These MCQs are created based on the latest CBSE syllabus and the NCERT curriculum, offering valuable assistance for exam preparation.
1. If sin (A + B) = $$\frac {\sqrt {3}}{2}$$ and tan (A – B) = 1. What are the values of A and B?
a) 37, 54
b) 35.7, 40.7
c) 50, 10
d) 52.5, 7.5
Explanation: The value of sin (A + B) = $$\frac {\sqrt {3}}{2}$$ and sin 60° = $$\frac {\sqrt {3}}{2}$$
∴ A + B = 60 (1)
The value of tan (A – B) = 1 and tan 45° = 1
∴ A – B = 45 (2)
A + B = 60
+ A – B = 45
– – – – – – – – – – – – –
2 A = 105
A = 52.5
∴ B = 7.5
2. If cos θ = $$\frac {3}{4}$$ then value of cos 2θ is ___________
a) $$\frac {1}{6}$$
b) $$\frac {1}{4}$$
c) $$\frac {1}{8}$$
d) $$\frac {3}{8}$$
Explanation: cos 2θ = 2cos θ2 – 1
cos θ = $$\frac {3}{4}$$
cos 2θ = 2($$\frac {3}{4}$$)2 – 1
= $$\frac {1}{8}$$
3. If sin A = $$\frac {8}{17}$$, what will be the value of cos A sec A?
a) 2
b) -1
c) 1
d) 0
Explanation: sin A = $$\frac {8}{17}$$
cos A sec A can be written as cosA × $$\frac {1}{secA}$$ = 1
∴ cos A sec A = 1
4. The value of each of the trigonometric ratios of an angle depends on the size of the triangle and does not depend on the angle.
a) True
b) False
Explanation:
Consider, two triangles ABC and DEF
In ∆ABC,
sin B = $$\frac {AC}{AB} = \frac {10}{20} = \frac {1}{2}$$ i.e. B = 30°
Now, in ∆DEF,
sin F = $$\frac {DE}{DF} = \frac {20}{40} = \frac {1}{2}$$ i.e. F = 30°
From these examples it is evident that the value of the trigonometric ratios depends on their angle and not on their lengths.
5. If tan α = √3 and cosec β = 1, then the value of α – β?
a) -30°
b) 30°
c) 90°
d) 60°
Explanation: tan α = √3 and tan 60° = √3
∴ α = 60°
Cosec β = 1 and cosec 90° = 1
∴ β = 90°
α – β = 60 – 90 = -30°
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6. In triangle ABC, right angled at C, then the value of cosec (A + B) is __________
a) 2
b) 0
c) 1
d) ∞
Explanation:
Since the triangle is right angles at C,
The sum of the remaining two angles will be 90
∴ cosec(A + B) = Cosec 90° = 1
7. If tan θ = $$\frac {3}{4}$$ then the value of sinθ is _________
a) $$\frac {3}{5}$$
b) $$\frac {4}{4}$$
c) $$\frac {3}{4}$$
d) $$\frac {-3}{5}$$
Explanation:
tanθ = $$\frac {BC}{AC} = \frac {3}{4} = \frac {3k}{4k}$$
Hence, BC = 3k, AC = 4k
Using Pythagoras theorem
AB2 = AC2 + BC2
AB2 = 4k2 + 3k2
AB = 5k
sinθ = $$\frac {BC}{AB} = \frac {3k}{5k} = \frac {3}{5}$$
8. What is the value of sin30°cos15° + cos30°sin15°?
a) $$\frac {1}{2}$$
b) 0
c) 1
d) $$\frac {1}{\sqrt {2}}$$
Explanation:
sin30°cos15° + cos30°sin15° = sin45° = $$\frac {1}{\sqrt {2}}$$
9. What is the value of cos A sec A + sin A cosec A – tan A cot A?
a) 0
b) 2
c) 1
d) 3
Explanation:
Cos A sec A = 1
Similarly sin A cosec A = 1 and tan A cot A = 1
∴ cos A sec A + sin A cosec A – tan A cot A = 1 + 1 – 1 = 1
10. In a right angled triangle, the trigonometric function that is equal to the ratio of the side opposite a given angle to the hypotenuse is called cosine.
a) False
b) True
Explanation:
In this ∆ ABC,
sin = $$\frac {Opposite}{hypotenuse}$$
More MCQs on Class 10 Maths Chapter 8:
To practice all chapters and topics of class 10 Mathematics, here is complete set of 1000+ Multiple Choice Questions and Answers. | 1,388 | 3,394 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.75 | 5 | CC-MAIN-2023-50 | latest | en | 0.71877 |
https://www.physicsforums.com/threads/derivative-definition.106731/ | 1,544,683,653,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376824525.29/warc/CC-MAIN-20181213054204-20181213075704-00524.warc.gz | 1,034,462,823 | 12,651 | # Homework Help: Derivative Definition
1. Jan 14, 2006
### Orion1
I am attempting to find the derivative of this function using the definition of derivative, however my solution is not the same as Mathematica's solution.
Is my formulation incorrect on some protocol somewhere? :uhh:
$$G(t) = \frac{4t}{t + 1}$$
$$G'(t) = \lim_{h \rightarrow 0} [(\frac{4(t + h)}{(t + h) + 1}) - (\frac{4t}{t + 1})] \frac{1}{h}$$
$$\lim_{h \rightarrow 0} [(\frac{4(t + h)}{(t + h) + 1}) - (\frac{4t}{t + 1})] \frac{1}{h} = \lim_{h \rightarrow 0} 4 [\frac{(t + h)(t + 1)}{(t + h + 1)(t + 1)} - \frac{t(t + h + 1)}{(t + 1)(t + h + 1)}] \frac{1}{h}$$
$$\lim_{h \rightarrow 0} 4[ \frac{(t + h)(t + 1) - t(t + h + 1)}{(t + h + 1)(t + 1)}] \frac{1}{h} = \lim_{h \rightarrow 0} 4[ \frac{(t^2 + ht + t + h - t^2 - ht - t)}{(t^2 + ht + 2t + h + 1)}] \frac{1}{h}$$
$$\lim_{h \rightarrow 0} [ \frac{4h}{(t^2 + ht + 2t + h + 1)}] \frac{1}{h} = \lim_{h \rightarrow 0} ( \frac{4}{t^2 + ht + 2t + h + 1} ) = \frac{4}{t^2 + 2t + 1}$$
$$\boxed{G'(t) = \frac{4}{(t + 1)^2}}$$
Mathematica solution:
$$- \frac{4t}{(1 + t)^2} + \frac{4}{t + 1}$$
2. Jan 14, 2006
### Galileo
$$\frac{4}{(t+1)^2}=- \frac{4t}{(1 + t)^2} + \frac{4}{t + 1}$$
No mistakes
Check it!
3. Jan 14, 2006
### twoflower
It's the same result, so yours is also correct. | 609 | 1,315 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2018-51 | latest | en | 0.461646 |
https://www.jiskha.com/search?query=How+to+sketch+graph+of&page=22 | 1,534,898,326,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221219197.89/warc/CC-MAIN-20180821230258-20180822010258-00717.warc.gz | 890,791,564 | 16,513 | # How to sketch graph of
9,950 results, page 22
1. ## Math
(With a TI-89 calculator) Graph x(t) = 3t^2 and y(t)=2-2t on [0,1]. Write a Cartesian equation that models this graph. Identify the initial and terminal points. I graphed both equations, but I didn't really get what [0,1] meant. I think it's an interval.
2. ## Calculus
find the area between x=tan^2y and x=-tan^2y in -pi/4<y<pi/4. I'm not sure how I can change the equations back to y=f(X) to graph. But it shouldn't really matter right? Do I do horizontal or vertical slicing? I feel like I should do horizontal since
3. ## College Algebra
If the point P is on the graph of a function f, find the corresponding point on the graph of the given function. P(4,9) y=1/3f(1/2x)-1
4. ## Alge.
Graph the equation using the slope and the y-intercept. y=5/3x+2 I need to show the slope and y-intercept when I graph.
5. ## Calc
How do you find the points of the derivative of a function just from the graph of the function? It wants me to find f'(0), f'(1), etc just from the graph-thanks
6. ## ALGEBRA
In the table of solutions, which entry is the y-intercept of the graph and which entry is the x-intercept of the graph? x y (x, y) 4 0 (4, 0) 0 2 (0,2) –4 –4 (–4, –4)
7. ## Algebra
Question 1 of 20 5.0 Points Graph the set of points. Which model is most appropriate for the set? (- 1, 20), (0, 10), (1, 5), (2, 2.5) what would graph look like?
8. ## math
HELP Not great in these Express y=2x^2 -12x +23 in the form y=2(x-c)^2 + d The graph of y=x^2 is transformed into the graph of y=2x^2 - 12x +23 by the transformation a vertical stretch with scale factor k followed by, A horizontal translation of p units
9. ## Calculus
The present value of a building in the downtown area is given by the function P(t) = 300,000e^-0.09t+¡Ìt/2 f or 0 ¡Ü t ¡Ü 10 Find the optimal present value of the building. (Hint: Use a graphing utility to graph the function, P(t), and find the value
10. ## math
The Sugar Sweet Company needs to transport sugar to market. The graph below shows the transporting cost (in dollars) versus the weight of the sugar being transported (in tons). there is a graph. 1. How much does the cost increase for each ton of sugar
11. ## Math
1. Consider a heptagon geometric figure. a) Write an equation for its perimeter b) If you graph the equation wherein x=2, 5, 8, and 12, does the graph show direct variation? 2. The surface area of a cube varies directly as the square of its edge. If each
12. ## precalculus
Find (a) the equation of a line that is parallel to the given line and includes the given point, and (b) the equation of a line that is perpendicular to the given line through the given point. Write both answers in slope-intercept form. (c) Graph both of
13. ## Calculus
The present value of a building in the downtown area is given by the function P(t) = 300,000e^-0.09t+¡Ìt/2 f or 0 ¡Ü t ¡Ü 10 Find the optimal present value of the building. (Hint: Use a graphing utility to graph the function, P(t), and find the value
14. ## Calculus
The present value of a building in the downtown area is given by the function p(t) = 300,000e^-0.09t+(t)/2 f or 0 _< t _< 10 Find the optimal present value of the building. (Hint: Use a graphing utility to graph the function, P(t), and find the value
15. ## Calculus
The present value of a building in the downtown area is given by the function p(t) = 300,000e^-0.09t+ãt/2 f or 0 _< t _< 10 Find the optimal present value of the building. (Hint: Use a graphing utility to graph the function, P(t), and find the
16. ## Calculus
The present value of a building in the downtown area is given by the function p(t) = 300,000e^-0.09t+(t)/2 f or 0 _< t _< 10 Find the optimal present value of the building. (Hint: Use a graphing utility to graph the function, P(t), and find the value
17. ## Calculus
The present value of a building in the downtown area is given by the function p(t) = 300,000e^-0.09t+(t)/2 f or 0 _< t _< 10 Find the optimal present value of the building. (Hint: Use a graphing utility to graph the function, P(t), and find the value
18. ## math
Use algebra pieces and mental arithmetic to determine four consecutive odd numbers who sum is 64. Sketch and label your algebra piece model and explain how you solved the problem.
19. ## Math(Probability)
The joint p.m.f. of X and Y is f(x, y) = 1/6, 0 ≤ x + y ≤ 2, where x and y are nonnegative integers. (a) Sketch the support of X and Y. (b) Record the marginal p.m.f.’s f1(x) and f2(y) in the “margins.” (c)Compute Cov(X, Y). (d)Determine
20. ## Math inequalitites
Hi, I need some help with some inequalitites. Thanks!! Use graphs to find the set. 1.(-9,0) intersection [-4,] I got [-4,0) 2. (-9,0) U [-4,10] I got (-9,10] Solve the linear inequality. Other than (empty set, use interval notation to express the solution
Nadia collected the data in the table on the left using a CBR after a ball was dropped. (a) Enter the data into L1 and L2 using a TI-83 Plus. Graph the relation. (b) Without looking at the graph, how do you know that the relation is not linear? (c) Use
22. ## Algebra
A problem involving a ball being thrown in the air is modeled with a quadratic function. Which of the following would you be solving for if asked to find the time at which the ball will reach its maximum height? A. an x-intercept of the graph B. the
23. ## algebra
Write the equation for this situation: During the month of February, the depth, d, of snow in inches at the base of one ski resort could be approximated by adding 7 to twice the number of days, t, since January 31st. Then graph the equation and use the
24. ## algebra
Write the equation for this situation: the value, v, in hundreds of dollars, of Bob’s stamp collection is 12 dollars more than 1/3 of the age, t (in years). Then graph the equation and use the graph to determine the value of Debbie’s coin collection if
25. ## Math
I'm having a hard time with graphing this problem. Could someone please help? Graphing has never been one of my strongest points. Graph the system of inequalities. x+3y is less than or equal to 9, 3x+y is less than or equal to 9, x is greater than or equal
26. ## math 8th
1. The following circle graph was published in the Cane County annual report. If there are 1,000 registered voters in Cane County, how many are 35–45 years old? circle graph (1 point) 150 voters
27. ## Calc II
1) Find the point on the graph of y=x^2 where the cube has slope 5/2 2) Find the point on the graph of y=x^2 where the tangent line is parallel to the line -4x+3y=3
28. ## math
the graph of y=f(x) is transformed into y=-f(1/3x+9)+8. a.Describe the transformations in correct sequence. b.Is there any other sequence of transformations that could produce the final graph? If so, describe all possible ones.
29. ## Math
At what point does the graph of the parametric equations y=3-2t, y=-2+5t intersect the graph of the parametric equations x = -1+3s,y=5-9s. The answer is the point of intersection (x,y). Please help! I am really confused!
30. ## Math
The graph below was drawn with output on the vertical axis and input on the horizontal axis. What does the graph indicate about the relationship between the input and the output?
31. ## algebra 1
write and graph the inverse variation when y=3 and x=2 find k = write the inverse variation fill in the table of values and graph x = -3 ,-2 ,-1,1,2,3 y= ?
32. ## algebra 1
write and graph the inverse variation when y=3 and x=2 find k = write the inverse variation fill in the table of values and graph x = -3 ,-2 ,-1,1,2,3 y= ?
33. ## Algebra
40. The graph of the following system yields perpendicular lines. x + 2y = -10 -4y = 2x + 20 (True) 48. The graph of the following system yields parallel lines. x - 2y = 1 y = -3x + 10 (False) Thanks -MC
34. ## Algebra 1
Describe how the graphs of y =IXI nand ym = IXI -15 are related. 1- the graphs have the same shape,. The y intercept of the 2nd graph is -15 2-the graphs have the same y intercept. The 2nd graph is steeper than y= absolute x 3- the 2 graphs are the same 4-
35. ## socials
With a partner, do some research on James I. Write a character sketch of James, What directions would you give to an actor asked to portray James I in a film?
Calculate the following: a) Find g(f(1)) b) Find lim x-> 1+ g(f(x)) c) Find f(g(0)) d) Find lim x->0- f(g(x)) f(x)= 1-x, x<1 1, x=1 x-1, x>1 g(x)= -x, x<0 2, x=0 x+2, x>0 This is where it gets hard. I have a graph for f(x) and one for
37. ## computer science
write a program that asks the user to enter today's sales rounded to the nearest \$100 for each of three stores. the program should then display a bar graph comparing each store's sales. create each bar in the graph by displaying a row of astericks. each
38. ## Algebra
I am very confused on how to graph Polynomial Functions and how to determine the extra stuff my HW is asking. a. Graph each function by making a table of values. b. Determine consecutive values of x between which each real zero is located. c. Estimate the
39. ## algebra
Solve the equation x2 + 8x – 2 = 0 using both a. The quadratic formula b. Completing the square Write a paragraph or two comparing and contrasting the two methods. Explain which method you prefer and why. c) Use the accompanying graph to estimate the
40. ## Chemistry...Attn DrBob222
Thanks for your earlier thoughts. Am still working on the problem and would like to know if you have any thoughts on my approach below: If I have a set of data that gives the apparent partition coefficient as a function of pH, over a range of pH 2-10, how
41. ## chemistry
CH3NH2(aq)+H2O(l)=>CH3NH3+(aq)+OH-(aq) Kb=4.4 x 10^-4 Methylamine, CH3NH2, is a weak base that reacts with water according to the equation above. A student obtains a 50.0 mL sample of a methylamine solution and determines the pH of the solution to be
42. ## AP Chemistry
CH3NH2(aq)+H2O(l)=>CH3NH3+(aq)+OH-(aq) Kb=4.4 x 10^-4 Methylamine, CH3NH2, is a weak base that reacts with water according to the equation above. A student obtains a 50.0 mL sample of a methylamine solution and determines the pH of the solution to be
43. ## Physics
A car accelerates from rest at 5 m/s2 for 5 seconds. It moves with a constant velocity for some time, and then decelerates at 5 m/s2 to come to rest. The entire journey takes 25 seconds. On a separate sheet of paper, plot the velocity-time graph of the
44. ## math
Sam emptied his coin bank and made a bar graph of the numbers of each type of coin the interval he was 5 coins If graph showed 5 intervals of quarters, 2 intervals of dimes,3 intervals if nickels and 10 intervals of pennies what was total amt of money in
45. ## Calc.
Calculate the following: a) Find g(f(1)) b) Find lim x-> 1+ g(f(x)) c) Find f(g(0)) d) Find lim x->0- f(g(x)) f(x)= 1-x, x<1 1, x=1 x-1, x>1 g(x)= -x, x<0 2, x=0 x+2, x>0 This is where it gets hard. I have a graph for f(x) and one for
46. ## ALGEBRA HELP OR I AM GONNA FAIL I CANT DO THIS
compare two cell phone companies offers to see which is the better deal. Horizon Cell Phone Company Vertigo Cell Phone Company Base rate=\$40 Plus every minute costs \$0.05 vertigo company Base rate=\$30 Plus every minute costs \$0.07 Step 1: Write a linear
47. ## algebra 3
what is the equation of the line that passes through {negative 4, negative 1}and {0,7}? A.Y=2X PLUS 9 B. Y=X/2 PLUS 7 C. 2X PLUS 7 D. Y= NEGATIVE 2X PLUS 7 Here's a way to approach your problem. 'm' represents slope. 'b' the y-intercept. m = (y2 - y1)/(x2
48. ## Physics
Your teacher placed a 4.5 kg block at the position marked with a “ +” (horizontally, 0.5 m from the origin) on a large incline outlined on the graph below and let it slide, starting from rest. Find the horizontal distance from the bottom (right-hand)
49. ## Maths
Consider the Ferris wheel above. It has an axle standing 38m off the ground and a radius of 35m. The wheel takes 5 minutes to complete one revolution. The wheel moves in clockwise motion and you are sitting in one of the carriages which are horizontally in
50. ## slope of a graph (physics/math)
what would the maximum slope possible for a v vs t graph of acar released from rest on an air track elevated at one end? ~I know the track would be elevated to be straight up since it would have the most angle...but what would the slope value be? would it
The graph of the equation y=x^3-x is symmetric with respect to which of the following? a) the x-axis b) the y-axis c) the origin d) none of these Answer: c 2) The graph of an odd function is symmetric with respect to which of the following? a) the x-axis
52. ## statistics
Which of the following statements are correct? a. A normal distribution is any distribution that is not unusual. b. The graph of a normal distribution is bell-shaped. c. If a population has normal ditribution, the mean and the median are not equal. d. The
53. ## statistics
which of the following statements are correct? a. a normal distribuiton is any distribution that is not unusual. b. the graph of a normal distribution is bell-shaped. c. if a population has a normal distribution, the mean and the median are not equal. d.
54. ## Math
How can I determine a value of gradient from a graph if the line is curved. (not a straight line) This is where Calculus comes in If a graph is a curve, then the gradient (slope) of the curve is the slope of the tangent that you draw at a particular point
55. ## Physics graphing
Im making a graph of magnetic force vs current, after investigating the magnetic force exerted on a current-carrying wire, and how it varies with current, length, magnetic field and angle. How do I know when to use the point (0,0) on the graph?
56. ## Math
What is a bin range? A. the range of the data that each bar in a histogram represents B. the description of the cells that are used t display the data C. the total range of data in a bar graph D. the line that is created when you connect the plots in a
57. ## trig
I don't understand this. can you please tell how you got each answer? describe how the graph of each function is related to a basic trigonometric graph. 1. y= 1/2cos x 2. y= 2cos x+1 3. y= sin x-4 4. y= sin(x-4) 5. y= sin(3x) 6. y= cos(1/2x) 7. y= -cos x
58. ## Math
Graph and identify the kind of conic the graph represents: a. x(t)=3 cos t y(t)=3 sin t on [0,2pi] b. x(t) =2 cos t y(t) =5 sin t on [1,5] I tried graphing A on my TI-89 calculator as y(1)=3 cos x and y(2)=3 sin x. All I see are two wave-like graphs
59. ## computer
4. What is a bin range? the range of data that each bar in a histogram represents the description of the cells that are used to display the data the total range of data in a bar graph the line that is created when you connect the plots in a graph
60. ## word problems in algebra
The cost of producing a number of items x is given by C = mx + b, in which b is the fixed cost and m is the variable cost (the cost of producing one more item). (a) If the fixed cost is \$40 and the variable cost is \$10, write the cost equation. (b) Graph
61. ## algebra
In most businesses, increasing prices of products can negatively impact the number of customers. A bus company in a small town has an average number of riders of 800 per day. The bus company charges \$2.25 for a ride. They conducted a survey of their
62. ## Maths Important
How would you draw the graph of this x^2/16 + y^2 = 1 ( this is an ellipse if I'm not mistaken) I'm confused since y^2 has no value beneath it. so what would the coordinates be for this. the same problem with this one x^2/16 - y^2 = 1 ( a hyperbola
63. ## MATH
Draw the graph of the equation 4x²-6x+3 using the -2≤x≤3 use a scale of 2cm to 1unit on the x-axis, 1cm to 4units in the y-axis. Find the graphfind; (a) minimum value of y (b) The roots of the equation 4x²-6x-3=0 (c) The value of x when y=5 (d) The
64. ## Precalc Trig
here's the question: solve for cos(2theta)=1 if you graph r= cos(2Theta) then graph r=1 you find that there are four places where the graphs intersect. however, when solved algebraically, there are two solutions which can be represented in an infinitely
65. ## Algebra
Can anyone help me with the following, by providing the steps to get there and what it is that you are doing? any help would be greatly appreciated... We define the following functions: f(x)=2x+5 g(x)=x^2-3 h(x)=7-x/3 ◦Compute (f – h)(4).
66. ## Statistics
use the stem-and leaf plot below to answer questions 11-13. Assume that 30/0=300 30/0 31/2 2 32/1 5 33/0 3 6 34/4 5 7 8 35/1 1 2 9 How many numbers are in the original data set? 6 16 22 9 What is the largest number in the data set? 9 22 35 359 If this data
67. ## maths
Sketch the region bounded by the x-axis, the curve y = x2 + x + 1 and the lines x = 2 and x = 5. Write down a definite integral that represents the area of that region Just stuck getting points for the first equation as when I put y=0 I got a complex
68. ## geometry
Harold is making a letter a to put on the rooftop of the a is for Apple Orchard store. The figure above shows a sketch of the design. what should be the measure of angle is one and 2p so bad the horizontal part of that a is truly horizontal
69. ## calculus
find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. sketch the region and a typical disk or washer. y^2=x, x=2y; about the y-axis i am confused because i do not know how to set it up because
70. ## maths exponential functions
Hi I have a Maths quiz on Monday and i have been of sick for the past week. I received these review questions from my teacher via email and he isn't replying to my questions i really do not get these questions if someone can do these and explain to me how
71. ## Math
Prove that a simple graph with n >_ 2 vertices must have atleast two vertices with the same degree. There was a hint given in the book saying that the key to this is the graph can not have both a vertex of 0 degree and a vertex of n-1 degree Thanks
72. ## math
What does identifying variables on a table or graph mean..? (what variable?)@.@ Also when do i connect dots on a graph and not connect them???? help???? please???? The variables are what is being plotted on the y, and x axis. What does the y axis stand
73. ## Math
Prove that a simple graph with n >_ 2 vertices must have atleast two vertices with the same degree. There was a hint given in the book saying that the key to this is the graph can not have both a vertex of 0 degree and a vertex of n-1 degree
74. ## Science
What can you say about the motion of a body if: a. its displacement-time graph is a straight line inclined with the time axis b. its velocity-time graph is a straight line inclined with the same axis. c. its velocity- time graph is a parabola A)the
75. ## college/microeconomics
Just needing to know if I have done the work correctly with this problem: For the total revenue and marginal revenue the answers are: Price \$20 Quantity 0 TR 0 MR 0 Price \$18 Quantity 1 TR 18 MR 18 Price \$16 Quantity 2 TR 32 MR 14 Price \$14 Quantity 3 TR
76. ## Math
Solve the inequality. Then graph the solution on a number line. -16 is less than or equal to 2x = 8<10 I'm going to assume your problem was meant to be this: -16 ≤ 2x + 8 < 10 If so, start by subtracting 8 on all sides: -16 - 8 ≤ 2x + 8 -
77. ## Science - Start of Physics
Please help me with the following problem. Given below is the position-time graph representing motions of two runners, Nick and Ian. Use this graph to determine which runner has greater average velocity. I can't put the graph here, so the following should
78. ## math
let line A be the graph of the equation x+7y = 3 and let line B be given by the equation 2x + 15y = 6. Use an augmented matrix to determine if the two lines meet and if so at which point. Graph the two lines and indicate their meeting point if it exists.
79. ## calculus
Find the coordinates of points where the graph of f(x) has horizontal tangents. As a check, graph f(x) and see whether the points you found look as though they have horizontal tangents. f(x) = −x3 + 6x2 − 9x + 13 (x, y) = ( ? ) (smaller
80. ## Science
The slope of a Volume vs. Mass graph is equal to its density. Calculate the slope of the liquid line A and the Liquid B Line on your graph, show your work.... Data: Liquid a: (M/V) 9.00/10.0 19.60/20.00 30.50/30.00 39.90/40.0 49.30/50.0 Liquid b(M/v)
81. ## Chels
What can you say about the motion of a body if its veloctiy-time graph is a straight line inclined with the time axis? a. the body moves with uniform velocity b.the body moves with uniform acceleration c. the body moves with non uniform acceleration d. the
82. ## Trigonometry
Consider angle C such that sin C= 7/25 Sketch a diagram to represent angle C in standard position if cos C is negative Find the co-ordinates of a point P on the terminal arm of angle C.
83. ## Pre-calc.
Graph the polynomial f(x)=(x+6)(x+4)^2 (x+2)^3(x-3) (x-5) a) Describe the interval notation the set of all x such that f(x)>0. b) Graph on a number line the set of all x such that f(x) > 0. c)Describe in interval notation the set of all x such that
84. ## math
which three statements are true? a) if x= -10^4 then log 10 = -4 b)if x= 2^8 then log 2x = 8 c) log2 2= 4 d) if x= 3 then log10 3=x e) log 10 256-2log 10 a/log 10 b f)log 10 (a-b)= log 10 a/log 10 b g) the gradient of the graph of y= 2x^x at x= 2 is 2e^e
85. ## college algebra
4. The yearly cost T, in thousands of dollars, of tuition and required fees at a private college (includes two- and four-year schools and does not include room and board) can be approximated by T= 3/5n + 5 where n is the number of years since 1985. That
86. ## physics
A car starts from rest is uniformly accelerated at 0.1m/s square untill it reaches a speed of 15m/sec square. It travels at this speed for 3mins and it returns uniformly retarded so as it come to rest 8mins after starting.plot the velocity time graph of
87. ## Math
Use the graph below to answer the following question. 1. Find the slope. (1 point) 2 – –2 Use the graph below to answer the following question. 2. Find the slope of the line. Describe how one variable changes in relation to the other. (1 point) 2;
88. ## chemistry
The data in the table show the amount of a 500 g sample of sodium-24 over time. Make a graph of the data (remember to label all axes and title the graph). What is the half-life of sodium-24? Explain how you determined your answer. Sodium-24 decays by beta
89. ## chemistry
2.The data in the table show the amount of a 500 g sample of sodium-24 over time. Make a graph of the data (remember to label all axes and title the graph). What is the half-life of sodium-24? Explain how you determined your answer. Sodium-24 decays by
90. ## Physics
A thin condicting spherical shell of outer radius 20cm carries a charge of +3 muC. Sketch (a) magnitude of the electric field E and (b) potential V versus the distance r from the center of the shell.
91. ## math
Determine whether it is possible to draw a triangle given each set of information. Sketch all possible triangles where appropriate. Calculate then label all side lengths to the nearest tenth of a centimetre and all angles to the nearest degree. A) b= 3.0
Determine whether it is possible to draw a triangle given each set of information. Sketch all possible triangles where appropriate. Calculate then label all side lengths to the nearest tenth of a centimetre and all angles to the nearest degree. A) b= 3.0
93. ## statistics
Assume that thermometer readings are normally distributed with a mean of 0.0 and a standard deviation of 1,00. the thermometer is randomly selected and tested. For the case below, draw a sketch , and find the probability of the reading The given values in
94. ## Math
Determine whether it is possible to draw a triangle given each set of information. Sketch all possible triangles where appropriate. Calculate then label all side lengths to the nearest tenth of a centimetre and all angles to the nearest degree. A) b= 3.0
95. ## Calculus: Application of Integration
Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approximating rectangle and label ts height and width. Then find the area of the region. y=x^2 y^2=x
96. ## Microeconomics:
Just needing to know if I have done the work correctly with this problem: For the total revenue and marginal revenue the answers are: Price \$20 Quantity 0 TR 0 MR 0 Price \$18 Quantity 1 TR 18 MR 18 Price \$16 Quantity 2 TR 32 MR 14 Price \$14 Quantity 3 TR
97. ## math
If your drawing a sine graph and you have to mark the point sine(65, how do you figure out where the point goes? (The sine graph goes up to 1)
98. ## Function Graphs
How do use a graph to find a particular function value like y=f(x).Find 2? Also, how do you use a graph to determine the function's domain and range?
99. ## university of Port harcourt
a body at rest is given an innital uniform acceleration of 8.0m/s-2 for 30s,acceleration reduce to 5.0m/s-2 for d next twenty sec.body mentains speed ten for 60s after brought to rest in 20s draw in graph 1.using d graph to calculate 2.maximum speed attain
100. ## Chemistry 110
3. Suppose that part (II) of the experiment were carried out with aqueous NaOH instead of Ba(OH)2 to form soluble Na2SO4 along with H2O as products. (a) Write the overall balanced equation. (b) Write the net ionic equation. (c) How would the graph of | 7,089 | 25,468 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2018-34 | latest | en | 0.886382 |
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Gunwalker: The Third Gun
Released on 2012-10-12 10:00 GMT
Email-ID 165593
Date 2011-10-26 23:47:22
A war within a war appears to have broken out regarding allegations of a
third weapon at the scene of Border Patrol Agent Brian Terry's murder.
The Department of Justice claims that only two semi-automatic AK-pattern
rifles were recovered at the scene of the shootout between a Border Patrol
tactical team and armed Mexican criminals that night in the Arizona
desert. House Oversight and Government Reform Committee Chairman Darrell
Issa (R-CA) ruffled feathers within the DOJ and with Democrats on his
committee by repeating allegations of a third weapon at the scene, and by
announcing plans to question the FBI about the investigation of Agent
Terry's death.
Border Patrol agents had mentioned a third weapon to Agent Terry's family.
CBS News obtained audiotape of a conversation between an ATF agent who was
part of Operation Fast and Furious and the dealer who sold weapons under
ATF direction to cartel smugglers. The blunt conversation is very hard to
Agent: Well there was two.
Dealer: There's three weapons.
Agent: There's three weapons.
Dealer: I know that.
Agent: And yes, there's serial numbers for all three.
Dealer: That's correct.
Agent: Two of them came from this store.
Dealer: I understand that.
Agent: There's an SKS that I don't think came from ... Dallas or Texas
or something like that.
Dealer: I know. talking about the AK's -
Agent: The two AK's came from this store.
Dealer: I know that.
Agent: OK.
Dealer: I did the godd***ed trace -
Dealer: That didn't come from me.
Agent: No and there is - that's my knowledge. And I spoke to someone
Dallas has long been alleged to be the site of another major gunwalking
operation running weapons to the cartel, one of 10 claimed gunwalking
operations in five states. The FBI's statement on the allegation released
Monday seems definitive:
perpetrators at the scene of Agent Terry's murder are false.
The FBI is very careful with their words: they declare there was not a
third gun recovered a) "from the perpetrators," b) "at the scene of Agent
Terry's murder."
An inference one could make from the careful phrasing: if there was a
third weapon, it was not recovered in direct proximity to one of the
Mexican criminals involved in the crime, and it was more than likely
dropped as the suspects fled the scene - which is also what allegedly
happened with both of the AK-pattern rifles used in the shootout. They
were recovered a short distance away. | null | null | null | null | null | null | null | null | null |
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Frequency Distributions
Chapter 2 Power Points for Essentials of Statistics for the Behavioral Sciences, Gravetter & Wallnau, 8th ed
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Frequency Distributions
1. 1. Chapter 2 Frequency Distributions PowerPoint Lecture Slides Essentials of Statistics for the Behavioral Sciences Eighth Edition by Frederick J Gravetter and Larry B. Wallnau
2. 2. Learning Outcomes • Understand how frequency distributions are used1 • Organize data into a frequency distribution table…2 • …and into a grouped frequency distribution table3 • Know how to interpret frequency distributions4 • Organize data into frequency distribution graphs5 • Know how to interpret and understand graphs6
3. 3. Tools You Will Need • Proportions (math review, Appendix A) – Fractions – Decimals – Percentages • Scales of measurement (Chapter 1) – Nominal, ordinal, interval, and ratio – Continuous and discrete variables (Chapter 1) • Real limits (Chapter 1)
4. 4. 2.1 Frequency Distributions • A frequency distribution is – An organized tabulation – Showing the number of individuals located in each category on the scale of measurement • Can be either a table or a graph • Always shows – The categories that make up the scale – The frequency, or number of individuals, in each category
5. 5. 2.2 Frequency Distribution Tables • Structure of Frequency Distribution Table – Categories in a column (often ordered from highest to lowest but could be reversed) – Frequency count next to category • Σf = N • To compute ΣX from a table – Convert table back to original scores or – Compute ΣfX
6. 6. Proportions and Percentages Proportions • Measures the fraction of the total group that is associated with each score • • Called relative frequencies because they describe the frequency ( f ) in relation to the total number (N) Percentages N f pproportion • Expresses relative frequency out of 100 • • Can be included as a separate column in a frequency distribution table )100()100( N f ppercentage
7. 7. Example 2.3 Frequency, Proportion and Percent X f p = f/N percent = p(100) 5 1 1/10 = .10 10% 4 2 2/10 = .20 20% 3 3 3/10 = .30 30% 2 3 3/10 = .30 30% 1 1 1/10 = .10 10%
8. 8. Learning Check • Use the Frequency Distribution Table to determine how many subjects were in the study • 10A • 15B • 33C • Impossible to determineD X f 5 2 4 4 3 1 2 0 1 3
9. 9. Learning Check - Answer • Use the Frequency Distribution Table to determine how many subjects were in the study • 10A • 15B • 33C • Impossible to determineD X f 5 2 4 4 3 1 2 0 1 3
10. 10. Learning Check • For the frequency distribution shown, is each of these statements True or False? • More than 50% of the individuals scored above 3T/F • The proportion of scores in the lowest category was p = 3T/F X f 5 2 4 4 3 1 2 0 1 3
11. 11. Learning Check - Answer • For the frequency distribution shown, is each of these statements True or False? • Six out of ten individuals scored above 3 = 60% = more than halfTrue • A proportion is a fractional part; 3 out of 10 scores = 3/10 = .3 False X f 5 2 4 4 3 1 2 0 1 3
12. 12. Grouped Frequency Distribution Tables • If the number of categories is very large they are combined (grouped) to make the table easier to understand • However, information is lost when categories are grouped – Individual scores cannot be retrieved – The wider the grouping interval, the more information is lost
13. 13. “Rules” for Constructing Grouped Frequency Distributions • Requirements (Mandatory Guidelines) – All intervals must be the same width – Make the bottom (low) score in each interval a multiple of the interval width • “Rules of Thumb” (Suggested Guidelines) – Ten or fewer class intervals is typical (but use good judgment for the specific situation) – Choose a “simple” number for interval width
14. 14. Discrete Variables in Frequency or Grouped Distributions • Constructing either frequency distributions or grouped frequency distributions for discrete variables is uncomplicated – Individuals with the same recorded score had precisely the same measurements – The score is an exact score
15. 15. Continuous Variables in Frequency Distributions • Constructing frequency distributions for continuous variables requires understanding that a score actually represents an interval – A given “score” actually could have been any value within the score’s real limits – The recorded value was rounded off to the middle value between the score’s real limits – Individuals with the same recorded score probably differed slightly in their actual performance
16. 16. Continuous Variables in Frequency Distributions • Constructing grouped frequency distributions for continuous variables also requires understanding that a score actually represents an interval • Consequently, grouping several scores actually requires grouping several intervals • Apparent limits of the (grouped) class interval are always one unit smaller than the real limits of the (grouped) class interval. (Why?)
17. 17. Learning Check • A Grouped Frequency Distribution table has categories 0-9, 10-19, 20-29, and 30-39. What is the width of the interval 20-29? • 9 pointsA • 9.5 pointsB • 10 pointsC • 10.5 pointsD
18. 18. Learning Check - Answer • A Grouped Frequency Distribution table has categories 0-9, 10-19, 20-29, and 30-39. What is the width of the interval 20-29? • 9 pointsA • 9.5 pointsB • 10 points (29.5 – 19.5 = 10)C • 10.5 pointsD
19. 19. Learning Check • Decide if each of the following statements is True or False. • You can determine how many individuals had each score from a Frequency Distribution Table T/F • You can determine how many individuals had each score from a Grouped Frequency Distribution T/F
20. 20. Learning Check - Answer • The original scores can be recreated from the Frequency Distribution Table True • Only the number of individuals in the class interval is available once the scores are grouped False
21. 21. 2.3 Frequency Distribution Graphs • Pictures of the data organized in tables – All have two axes – X-axis (abscissa) typically has categories of measurement scale increasing left to right – Y-axis (ordinate) typically has frequencies increasing bottom to top • General principles – Both axes should have value 0 where they meet – Height should be about ⅔ to ¾ of length
22. 22. Data Graphing Questions • Level of measurement? (nominal; ordinal; interval; or ratio) • Discrete or continuous data? • Describing samples or populations? The answers to these questions determine which is the appropriate graph
23. 23. Frequency Distribution Histogram • Requires numeric scores (interval or ratio) • Represent all scores on X-axis from minimum thru maximum observed data values • Include all scores with frequency of zero • Draw bars above each score (interval) – Height of bar corresponds to frequency – Width of bar corresponds to score real limits (or one-half score unit above/below discrete scores)
24. 24. Figure 2.1 Frequency Distribution Histogram
25. 25. Grouped Frequency Distribution Histogram Same requirements as for frequency distribution histogram except: • Draw bars above each (grouped) class interval – Bar width is the class interval real limits – Consequence? Apparent limits are extended out one-half score unit at each end of the interval
26. 26. Figure 2.2 Grouped Frequency Distribution Histogram
27. 27. Block Histogram • A histogram can be made a “block” histogram • Create a bar of the correct height by drawing a stack of blocks • Each block represents one per case • Therefore, block histograms show the frequency count in each bar
28. 28. Figure 2.3 Frequency Distribution Block Histogram
29. 29. Frequency Distribution Polygons • List all numeric scores on the X-axis – Include those with a frequency of f = 0 • Draw a dot above the center of each interval – Height of dot corresponds to frequency – Connect the dots with a continuous line – Close the polygon with lines to the Y = 0 point • Can also be used with grouped frequency distribution data
30. 30. Figure 2.4 Frequency Distribution Polygon
31. 31. Figure 2.5 Grouped Data Frequency Distribution Polygon
32. 32. Graphs for Nominal or Ordinal Data • For non-numerical scores (nominal and ordinal data), use a bar graph –Similar to a histogram –Spaces between adjacent bars indicates discrete categories • without a particular order (nominal) • non-measurable width (ordinal)
33. 33. Figure 2.6 - Bar graph
34. 34. Population Distribution Graphs • When population is small, scores for each member are used to make a histogram • When population is large, scores for each member are not possible – Graphs based on relative frequency are used – Graphs use smooth curves to indicate exact scores were not used • Normal – Symmetric with greatest frequency in the middle – Common structure in data for many variables
35. 35. Figure 2.7 Bar Graph of Relative Frequencies
36. 36. Figure 2.8 – IQ Population Distribution Shown as a Normal Curve
37. 37. Box 2.1 - Figure 2.9 Use and Misuse of Graphs
38. 38. 2.4 Frequency Distribution Shape • Researchers describe a distribution’s shape in words rather than drawing it • Symmetrical distribution: each side is a mirror image of the other • Skewed distribution: scores pile up on one side and taper off in a tail on the other – Tail on the right (high scores) = positive skew – Tail on the left (low scores) = negative skew
39. 39. Figure 2.10 - Distribution Shapes
40. 40. Learning Check • What is the shape of this distribution? • SymmetricalA • Negatively skewedB • Positively skewedC • DiscreteD
41. 41. Learning Check - Answer • What is the shape of this distribution? • SymmetricalA • Negatively skewedB • Positively skewedC • DiscreteD
42. 42. Learning Check • Decide if each of the following statements is True or False. • It would be correct to use a histogram to graph parental marital status data (single, married, divorced...) from a treatment center for children T/F • It would be correct to use a histogram to graph the time children spent playing with other children from data collected in children’s treatment center T/F
43. 43. Learning Check - Answer • Marital Status is a nominal variable; a bar graph is requiredFalse • Time is measured continuously and is an interval variable True
44. 44. Figure 2.11- Answers to Learning Check Exercise 1 (p. 51)
45. 45. Any Questions ? Concepts? Equations?
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Chapter 2 Power Points for Essentials of Statistics for the Behavioral Sciences, Gravetter & Wallnau, 8th ed
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null | null | null | null | null | null | Unions Cap Their Member’s Pay
• Setting a Wage Ceiling: Currently union contracts set both a wage floor and a wage ceiling. Unionized employers may not give productive workers pay raises outside the collectively bargained contract.
• Seniority-Based Pay: Unions usually demand that employers reward workers for “time served” rather than “hard work.” No matter how good an employee does his job, he cannot earn higher wages than what his union contract specifies.
• Holding Back American Productivity: This “seniority ceiling” keeps union members from reaching their full potential, since there is no incentive for outstanding performance nor ability to move up the ladder.
• Lifting the Ceiling: The RAISE Act (Rewarding Achievement and Incentivizing Successful Employees) would allow employers to pay individual workers more, but not less, than the union contract specifies.
• Providing Incentives: The RAISE Act would provide workers with the incentive to increase productivity. If passed, the typical union member would increase his productivity and earn between $2,600 and $4,300 more a year in performance-pay bonuses and merit raises.
• Restoring Freedom in the Workplace: Current federal law caps the wages of 8 million American workers. The RAISE Act would restore the inherent American right to earn individual raises through individual efforts to millions of union members.
• Merit-Based Pay Works: Economic research shows that the average worker’s earnings rise 6 to 10% when the pay is performance based. This is an average figure—industrious and enterprising workers will earn larger raises while lazy employees earn less.
• The Time for Change Is Now: With millions of American families struggling to get ahead financially in the recession, Congress should lift the ceiling on workers’ pay immediately. This is the right kind of stimulus that will add billions of dollars to the economy and improve business earnings through increased productivity.
Why Would Anyone Oppose Higher Wages?
• Direct Dealing: Bonuses to hard-working employees have been fought successfully by unions wishing to maintain control over their members’ income. It constitutes illegal “direct dealing,” which collective bargaining law forbids.
• The RAISE Act Does Not Eliminate the Wage Floor: Unions were originally established to protect workers from making too little money, not too much. The RAISE Act would still allow union contracts to set the minimum that workers can earn.
• Still Prevents Anti-Union Discrimination: Employers could not selectively give raises to anti-union workers to undermine the union, consistent with current federal law. Under the RAISE Act, it would remain illegal to discriminate against workers on the basis of union membership.
• President Obama Agrees: When criticizing AIG, President Obama said: “We believe in the free market, we believe in capitalism, we believe in people getting rich, but we believe in people getting rich based on performance and what they add in terms of value and the products and services they create.” | null | null | null | null | null | null | null | null | null |
https://www.physicsforums.com/threads/is-mass-circular.561965/ | 1,550,733,578,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247503249.58/warc/CC-MAIN-20190221071502-20190221093502-00382.warc.gz | 933,822,186 | 19,899 | # Is Mass Circular?
1. Dec 21, 2011
### jaketodd
We establish the mass of something by how much the earth pulls on it. We establish the mass of the earth by, for example, how much the sun effects the earth's motion. We establish the mass of the sun by how much it pulls on planets - and we're back to planets, which is already stated to be used to establish the mass of something.
Is it really circular like this, or did I miss something? If I didn't miss something, then what are the implications? Is it that we can only talk of relative masses within the same reference frame; there is no absolute mass for anything? Is this already stated in General Relativity?
Cheers,
Jake
2. Dec 21, 2011
### dacruick
Good question, I'm interested to see what some of the PF gurus say. Although I think you meant cyclical not circular.
3. Dec 21, 2011
### jaketodd
Thanks! I'm looking forward to what they say too.
4. Dec 21, 2011
### espen180
Yes, you can only talk about relative masses. First, take an arbitrary object and call its mass 1 unit. Then you can measure other objects' masses in terms of that unit by comparing them to the unit mass using Newton's 2nd law.
This process doesn't need gravity to be defined, it just needs a method of systematically applying a given force to different objects.
5. Dec 21, 2011
### Nabeshin
It's important to recall that gravitational and inertial mass are, as far as we know, the exact same quantity. That is, the mass which appears in Newton's law of gravitation (F=GMm/r^2) is the same as the one which appears in Newton's 2nd law (F=ma). So you have two ways of measuring mass -- either push on something, or see how much it is affected by gravity.
As espen180 notes, the mass scale is indeed relative. Specifically, we define a platinum-iridium cylinder sitting in France to be one kilogram, and we measure everything relative to this mass.
Note: The same could be said of lengths. We only measure lengths relative to other lengths, so I think your question is really more fundamentally one of defining your system of units.
6. Dec 21, 2011
### D H
Staff Emeritus
That's not quite right. We determine the mass of the Sun (better: the product of the gravitational constant and the mass of the Sun) by the orbits of the planets about the Sun. The planets are so much smaller than the Sun that, to first order, their masses just don't amount to much. Even the largest planet of the planets, Jupiter, has a mass that is only about 1/1000 that of the Sun.
Looking to the orbit of a planet compared to that of a tiny test mass is a lousy way to assess the mass of a planet. Much better is to look at how small objects orbit that planet. This gives have a good picture of the masses (G*mass) of all but Mercury and Venus.
There is a problem here. While the product G*M can be observed to very high degree of precision, mass cannot. The masses of the Sun and the planets are computed by dividing the observed planetary gravitational coefficient μ (G*M) by G. G is arguably the least well known of physical constants. Astronomers have assessed μ for the sun and several of the planets to nine or more decimal places. G: A lousy four decimal places.
7. Dec 21, 2011
### Naty1
Depends on your definition of mass....generally modern usage is that REST mass IS absolute.
http://en.wikipedia.org/wiki/Rest_Mass
Last edited: Dec 21, 2011
8. Dec 21, 2011
### Staff: Mentor
We happen to use this object, whose mass is defined to be one kilogram, exactly.
http://www.bipm.org/en/scientific/mass/pictures_mass/prototype.html [Broken]
All other mass measurements must ultimately come down to a comparison with this object, through a chain of intermediate objects.
Last edited by a moderator: May 5, 2017
9. Dec 21, 2011
### phinds
No, he clearly meant circular. Cyclical would make no sense in this context.
10. Dec 21, 2011
### D H
Staff Emeritus
Kinda sorta.
There are multiple mass standards that are loosely connected to the kilogram prototype. That kilogram prototype is the standard by which human-scale mass is assessed. Things that can be weighed on scales. At the atomic scale, the mass of a carbon-12 atom is the standard. There is a connection between the human scale and atomic scale of masses, Avogadro's constant. Currently this is measured experimentally.
This may change. As of the last meeting of the BIPM, plans are officially afoot (finally!) to tie the atomic and human scale mass conventions. Avogadro's number will become a defined constant, as will the Plank constant h, the elementary charge e, and the Boltzmann constant k.
At the other extreme are planetary and larger masses. Just as atoms are too small to balance against the kilogram prototype, these objects are too large. The connection between the kilogram prototype and these large objects is the universal gravitational constant G. While mass isn't directly observable at these large scales, the product μ=G*M is very observable.
11. Dec 22, 2011
### DrDu
Your argument holds at best in Newtonian mechanics. As soon as quantum mechanics comes into play the scale invariance of mass is broken. The size of an atom depends on the ratio of the charge and the mass of the electron and these cannot be scaled independently from each other. On the other hand the mass of the neutron determines how large a neutron star can be maximally before it collapses into a black hole, so you cannot scale large objects arbitrarily.
12. Dec 22, 2011
### jaketodd
This is getting interesting =)
13. Dec 22, 2011
### D H
Staff Emeritus
Huh?
The concept of mass is still applicable at the atomic and subatomic scales. Yes, mass is no longer additive, but I never said it was. The BIMP is quite aware of quantum mechanics. Why would they even think of making e, h, and k defined constants if the concept of mass didn't carry over to the atomic and subatomic scales?
Edit
Link to Resolution #1 of the 24th meeting of the CGPM: http://www.bipm.org/en/CGPM/db/24/1/
So who are the BIPM and the CGPM? The BIPM (International Bureau of Weights and Measures) are the keepers of the metric system. That kilogram prototype that jtbell mentioned in post #8 is the responsibility of the BIPM. Resolution #1, when/if accepted, would get rid of that prototype. The CGPM (General Conference on Weights and Measures) meets every four years, more or less, and makes recommendations to the BIPM. This resolution is huge. It is something that has brewing for decades.
Last edited: Dec 22, 2011
14. Dec 23, 2011
### DrDu
Dear D H, I was referring to the question of the original poster, not your reply. I also didn't want to imply that mass is no longer applicable in quantum mechanics.
The point I wanted to make is the following: For Newton, there is no fundamental difference between an apple and a planet (neither is there in general relativity). For Newton, both obey the same equations of motion.
Especially Newton cannot explain why there are no apple trees of galactic dimension nor suns of the mass of an apple. This underlying absolute scale is provided by quantum mechanics, only, finally probably by symmetry breaking via the Higgs mechanism which sets the mass of the electron and other elementary particles.
15. Dec 23, 2011
### sophiecentaur
Perhaps 'self referential' would be an appropriate description.
16. Dec 23, 2011
### jaketodd
Would you be so kind as to explain that in more detail? I'm very curious.
Thanks,
Jake
17. Dec 25, 2011
### DrDu
I´ll try, although I am not the specialist on that topic.
It is believed that the mass of most particles like e.g. the electron is due to scattering from the Higgs field, which, even in the vacuum, has a non-vanishing value. Nobody knows why the value of the Higgs field is as big as it is (if it exists at all, experiments at CERN are about to verify its existence experimentally). Probably its value is pure chance and there may even be regions in our universe where it has other values.
Anyhow, if it's value was 0, electrons would be massless and they would appear in two versions, right handed and left handed electrons. When the field is non-zero, a right handed electron can get scattered from the Higgs field into a left handed electron and vice versa.
How does that lead to mass? The energy of a massless particle goes linearly to 0 when we reduce its momentum (or wavenumber). In the case of light this leads to the familiar proportionality of frequency and inverse wavelength.
For a massive particle, the limiting value of energy when the momentum goes to zero (which is equivalent in this case to it´s velocity going to 0) is it´s mass ( up to the famous factor c squared of Einstein). In the case of the electron interacting with the Higgs field, there is some interaction energy present even when the momentum of the electron vanishes, which hence is it´s mass.
18. Dec 26, 2011
### jaketodd
First of all, thank you! Second, how was the Higgs field thought up? It sounds like it may be as baseless as string theory; absolutely zero empirical data supporting it. I have that book, which The Economist called "One of the most important books of the year," called "The Rise of String Theory; The Fall of Science."
Thanks,
Jake
19. Dec 26, 2011
### DrDu
The idea of symmetry breaking by the Higgs field has been taken over from condensed matter physics, namely the physics of superconductors. The Higgs mechanism explains quite a lot of observations and many of it´s predictions have been confirmed experimentally by now.
Namely the unification of weak and electromagnetic forces. This unification is explained by both forces being related by a symmetry. After this symmetry had been postulated, the carriers of the field (besides the photon, which obviously was known before) were subsequently confirmed experimentally (namely the vector bosons W and Z).
However, this symmetry is "broken", like e.g. the magnetic field in a magnet could in principle point in any direction, but in fact, in each magnet you will find only one realization of the direction of the magnetic field.
According to a theorem by Goldstone, a broken symmetry should leave some signature in the form of a massless particle, the "Goldstone boson". However, there is no corresponding particle.
The Higgs mechanism explains quite nicely how this particle actually gets a mass (and also the other particles).
So somehow the Higgs mechanism is the easiest way to save the concept of broken symmetry which has already been confirmed experimentally.
20. Dec 27, 2011
### jaketodd
I may easily be mistaken, however: Isn't the carrier of the EM force virtual photons? I've heard it said that, in this context, photons aren't actually detected - just the easiest way to model the interaction.
If that is all there is there, then are the W & Z bosons just as virtual, or have they actually been found and their properties experimentally identified (instead of them just being a convenient model)?
Thanks,
Jake | 2,552 | 10,994 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2019-09 | latest | en | 0.960733 |
https://sarkariresultsindia.net/ranking-and-ordering-reasoning-typesformula-and-tricks-with-problem-question-and-solutions/ | 1,656,486,101,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103624904.34/warc/CC-MAIN-20220629054527-20220629084527-00081.warc.gz | 544,196,510 | 25,085 | # Ranking and Ordering Reasoning: Types,Formula and Tricks with Problem (Question) and Solutions
The position of a thing/person etc. in a definite order is called as ‘Rank’.
In this type of test, relative position or rank of some person or object are given and candidates are required to find the rank or position of other person or object.
### Type 1. Rank of a person in a queue
Position of person from upward = [Total no. of persons – position of person from down] + 1. Position of person from downward = [Total no. of persons – position of person from up] + 1. Position of person from right = [Total no. of persons – position of person from left] + 1. Position of person from left = [Total no. of persons – position of person from right] + 1.
Ex-49th students of a class Nitin gets 18th position from start. What is the rank of Nitin from end?
(a) 18
(b) 19
(c) 31
(d) 32
Sol. 49 – 18 + 1 = 32
### Type 2. Total number of person in a queue
Total no. of persons= [Position of person from upward/right + Position of person from downward/left] – 1.
Ex-In a row of students. Mohan is the 13th from the left end. Suman is 12th from Right end and 18th from left end. How many students of Mohan’s right?
(a) 30
(b) 15
(c) 16
(d) Can’t be determined
Sol. Total students= 12+18–1=29
So = 29–13=16 students of Mohan’s right side
Ex-Rohan’s ranked seventh from the top and twenty sixth from the bottom in a class. How many students are there in the class?
(a) 31
(b) 32
(c) 32
(d) 34
Sol. 26+7–1=32
### Type 3. When two persons change their rank in a queue
If two persons are on a definite position from up and down (or left and right) and they interchange their ranks, then Total no. of persons in order= [present position of first person + previous position of second person] – 1
Previous position of first person or present position of second person = Difference of two positions of second person + previous position of second person= Difference of two positions of first person + previous position of second person.
Ex-In a row of girls, Shilpa is eighth from the left and Reena is seventeenth from the right. If they interchange their positions, Shilpa becomes fourteenth from the left. How many girls are there in the row?
1. 34
2. 35
3. 30
4. 37
Sol. Total no. of girls
= [present position of Shilpa + previous position of Reena] – 1
= (14 + 17) – 1 = 30
### Some Examples
Ex-In a row of a students, Mohan is 10th from right. Sohan is 25th from left. When they change their position then Mohan be comes 22 from right. What is the new position of Sohan from left?
(a) 35
(b) 36
(c) 37
(d) 38
Sol. 22–10=12, 25+12=37th
Ex-In a class of 40 students Sachin is 15th from the top Siddharth is 12th from the bottom. How many boys are in between Sachin and Siddharth?
(a) 15
(b) 13
(c) 12
(d) 16
Trick:- Total–P1 Position + P2 Position
P1 Person–1 Sachin
P2 Person–2 Siddharth
40-(15+12)
40–27 = 13
Ex-In a class of 80 students Mayank is 13th from the right and Ritu is 18th from the left how many students in between Mayank and Ritu?
(a) 49
(b) 48
(c) 38
(d) 50
Trick:- Total – (P1+P2)
80 – (18 +13) = 49 | 927 | 3,157 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2022-27 | latest | en | 0.898782 |
https://gaianation.net/the-radius-is-half-the-diameter/ | 1,652,750,818,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662515466.5/warc/CC-MAIN-20220516235937-20220517025937-00167.warc.gz | 342,728,701 | 8,150 | Circles room a common shape. You view them all over—wheels ~ above a car, Frisbees passing with the air, compact discs carrying data. These are all circles.
You are watching: The radius is half the diameter
A circle is a two-dimensional figure similar to polygons and quadrilaterals. However, circles space measured differently than these other shapes—you even have to usage some different terms to describe them. Let’s take it a look at this interesting shape.
A circle represents a set of points, every one of which space the exact same distance away from a fixed, center point. This fixed suggest is dubbed the center. The distance from the center of the circle to any allude on the circle is dubbed the radius.
When 2 radii (the many of radius) are put together to form a line segment across the circle, you have a diameter. The diameter the a one passes with the center of the circle and also has that endpoints on the one itself.
The diameter of any circle is 2 times the length of the circle’s radius. It have the right to be represented by the expression 2r, or “two time the radius.” for this reason if you understand a circle’s radius, you deserve to multiply that by 2 to discover the diameter; this also way that if you know a circle’s diameter, you can divide by 2 to find the radius.
Example Problem Find the diameter the the circle. d = 2r d = 2(7) d = 14 The diameter is two times the radius, or 2r. The radius that this one is 7 inches, therefore the diameter is 2(7) = 14 inches. Answer The diameter is 14 inches.
Example Problem Find the radius the the circle. The radius is fifty percent the diameter, or . The diameter that this circle is 36 feet, so the radius is feet. Answer The radius is 18 feet.
Circumference
The distance around a one is dubbed the The distance about a circle, calculation by the formula C =
d.
")">circumference
. (Recall, the distance about a polygon is the perimeter.)
One interesting property about circles is the the ratio of a circle’s circumference and also its diameter is the same for all circles. No matter the dimension of the circle, the ratio of the circumference and diameter will certainly be the same.
Some actual dimensions of various items are noted below. The measurements are exact to the nearest millimeter or quarter inch (depending top top the unit of measure up used). Look at the ratio of the circumference to the diameter because that each one—although the items room different, the proportion for every is around the same.
Item Circumference (C) (rounded come nearest hundredth) Diameter (d) Ratio Cup 253 mm 79 mm Quarter 84 mm 27 mm Bowl 37.25 in 11.75 in
The circumference and also the diameter space approximate measurements, due to the fact that there is no precise method to measure up these dimensions exactly. If you were able to measure up them much more precisely, however, girlfriend would uncover that the proportion would move towards 3.14 for each that the items given. The mathematical surname for the ratio is The proportion of a circle’s circumference come its diameter. Pi is denoted through the Greek letter
. It is frequently approximated together 3.14 or
.
")">pi
, and also is stood for by the Greek letter .
is a non-terminating, non-repeating decimal, so the is impossible to write it the end completely. The first 10 number of are 3.141592653; it is frequently rounded to 3.14 or approximated as the portion . Keep in mind that both 3.14 and also are approximations of, and also are provided in calculations whereby it is not necessary to be precise.
Since you understand that the ratio of circumference to diameter (or ) is constant for all circles, you have the right to use this number to uncover the one of a one if you know its diameter.
= , for this reason C = d
Also, since d = 2r, climate C = d = (2r) = 2r.
Circumference of a Circle To uncover the circumference (C) of a circle, use one of the complying with formulas: If you recognize the diameter (d) the a circle: If you understand the radius (r) the a circle:
Example Problem Find the circumference of the circle. To calculate the circumference offered a diameter the 9 inches, use the formula . Usage 3.14 as an approximation for . Since you space using one approximation for , you can not give specific measurement of the circumference. Instead, you use the price to indicate “approximately equal to.” Answer The circumference is 9 or approximately 28.26 inches.
Example Problem Find the circumference of a circle v a radius of 2.5 yards. To calculate the circumference of a circle given a radius of 2.5 yards, usage the formula . Usage 3.14 as an approximation for. Answer The one is 5 or about 15.7 yards.
A circle has a radius that 8 inches. What is that circumference, rounded to the nearest inch?
A) 25 inches
B) 50 inches
C) 64 inches2
D) 201 inches
A) 25 inches
Incorrect. You multiplied the radius time ; the correct formula because that circumference as soon as the radius is given is The correct answer is 50 inches.
B) 50 inches
Correct. If the radius is 8 inches, the exactly formula because that circumference once the radius is provided is The correct answer is 50 inches.
C) 64 inches2
Incorrect. Girlfriend squared 8 inch to come at the price 64 inches2; this will provide you the area that a square with sides the 8 inches. Remember that the formula because that circumference when the radius is offered is . The exactly answer is 50 inches.
D) 201 inches
Incorrect. That looks choose you squared 8 and also then multiplied 64 through to come at this answer. Remember the the formula for circumference once the radius is offered is . The correct answer is 50 inches.
Area
is critical number in geometry. You have already used the to calculate the one of a circle. You use when you are figuring the end the area that a circle, too.
Area of a Circle To discover the area (A) the a circle, use the formula:
Example Problem Find the area of the circle. To find the area that this circle, use the formula . Remember to compose the price in terms of square units, since you space finding the area. Answer The area is 9 or about 28.26 feet2.
A button has a diameter of 20 millimeters. What is the area of the button? use 3.14 together an approximation the .
A) 62.8 mm
B) 314 mm2
C) 400 mm2
D) 1256 mm2
A) 62.8 mm
Incorrect. You discovered the circumference of the button: 20 • 3.14 = 62.8. To uncover the area, usage the formula . The correct answer is 314 mm2.
B) 314 mm2
Correct. The diameter is 20 mm, so the radius must be 10 mm. Then, making use of the formula , you find
mm2.
C) 400 mm2
Incorrect. Friend squared 20 to arrive at 400 mm2; this provides you the area of a square v sides of length 20, no the area the a circle. To find the area, usage the formula . The correct answer is 314 mm2.
D) 1256 mm2
Incorrect. It looks prefer you squared 20 and then multiplied by . 20 is the diameter, no the radius! To discover the area, use the formula . The correct answer is 314 mm2.
Composite Figures
Now that you know exactly how to calculate the circumference and also area that a circle, you deserve to use this understanding to find the perimeter and also area that composite figures. The trick to figuring the end these varieties of difficulties is to recognize shapes (and parts of shapes) in ~ the composite figure, calculate your individual dimensions, and then include them together.
For example, look at the image below. Is it feasible to uncover the perimeter?
The first step is come identify easier figures within this composite figure. You can break it down right into a rectangle and a semicircle, as shown below.
You know just how to find the perimeter of a rectangle, and also you know exactly how to find the one of a circle. Here, the perimeter the the 3 solid sides of the rectangle is 8 + 20 + 20 = 48 feet. (Note that only three political parties of the rectangle will add into the perimeter of the composite figure due to the fact that the other side is not at one edge; that is spanned by the semicircle!)
To find the circumference of the semicircle, usage the formula with a diameter that 8 feet, then take fifty percent of the result. The circumference of the semicircle is
, or around 12.56 feet, so the total perimeter is around 60.56 feet.
Example Problem Find the perimeter (to the nearest hundredth) the the composite figure, consisted of of a semi-circle and also a triangle. Identify smaller shapes within the composite figure. This figure contains a semicircle and a triangle. Diameter (d) = 1 Circumference that semicircle = or about 1.57 inches Find the one of the circle. Then division by 2 to discover the one of the semi-circle. inches Find the full perimeter by including the circumference of the semicircle and also the lengths that the two legs. Since our measure of the semi-circle’s one is approximate, the perimeter will be one approximation also. Answer Approximately 3.57 inches
Example Problem Find the area that the composite figure, consisted of of three-quarters that a circle and also a square, to the nearest hundredth. Identify smaller shapes within the composite figure. This figure consists of a circular region and a square. If you discover the area that each, girlfriend can find the area of the entire figure. Find the area that the square. . Find the area of the one region. The radius is 2 feet. Note the the region is of a entirety circle, therefore you should multiply the area the the one by . Use 3.14 as an approximation because that . 4 feet2 + feet2 = around 13.42 feet2 Add the two areas together. Because your measure up of the circular’s area is approximate, the area of the figure will be an approximation also. Answer The area is around 13.42 feet2. | 2,288 | 9,781 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2022-21 | latest | en | 0.932157 |
https://worksheetgenius.com/unit-converter/convert-813-ft-to-ft-us/ | 1,713,938,882,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819067.85/warc/CC-MAIN-20240424045636-20240424075636-00507.warc.gz | 557,771,952 | 6,493 | # Convert 813 ft to ft-us
In this article I will show you how to convert 813 feet into us survey feet. Throughout the explanation below I might also call it 813 ft to ft-us. They are the same thing!
## How to Convert Feet to Us survey feet
A foot is smaller than a us survey foot. I know that a ft is smaller than a ft-us because of something called conversion factors.
Put very simply, a conversion factor is a number that can be used to change one set of units to another, by multiplying or dividing it. So when we need to convert 813 feet into us survey feet, we use a conversion factor to get the answer.
The conversion factor for ft to ft-us is:
1 ft = 0.999998 ft-us
Now that we know what the conversion factor is, we can easily calculate the conversion of 813 ft to ft-us by multiplying `0.999998` by the number of feet we have, which is 813.
813 x 0.999998 = 812.998374 ft-us
So, the answer to the question "what is 813 feet in us survey feet?" is 812.998374 ft-us.
## Feet to Us survey feet Conversion Table
Below is a sample conversion table for ft to ft-us:
Feet (ft) Us survey feet (ft-us)
0.010.01
0.10.1
11
22
33
55
1010
2020
5050
100100
10001000
## Best Conversion Unit for 813 ft
Sometimes when you work with conversions from one unit to another, the numbers can get a little confusing. Especially when dealing with really large numbers.
I've also calculated what the best unit of measurement is for 813 ft.
To determine which unit is best, I decided to define that as being the unit of measurement which is as low as possible, without going below 1. Smaller numbers are more easily understood and can make it easier for you to understand the measurement.
The best unit of measurement I have found for 813 ft is fathoms and the amount is 135.5 fm.
## Link to Us / Reference this Page
Please use the tool below to link back to this page or cite/reference us in anything you use the information for. Your support helps us to continue providing content!
• "Convert 813 ft to ft-us". WorksheetGenius.com. Accessed on April 24, 2024. http://worksheetgenius.com/unit-converter/convert-813-ft-to-ft-us/.
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• Convert 813 ft to ft-us. WorksheetGenius.com. Retrieved from http://worksheetgenius.com/unit-converter/convert-813-ft-to-ft-us/. | 607 | 2,407 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2024-18 | latest | en | 0.917456 |
null | null | null | null | null | null | Saturday, May 24, 2008
YouTube weekend
Political gaffe edition
This one may just end Hillary Clinton's campaign.
“My husband did not wrap up the nomination in 1992 until he won the California primary somewhere in the middle of June, right? We all remember Bobby Kennedy was assassinated in June in California. I don’t understand it,” she said, dismissing calls to drop out.
What Does RFK's Assassination Have to Do with It?
I can usually forgive a stupid statement from a Presidential candidate because they have to talk so much. Obama 's bitter comment or McCain's 100 years war response to a question come to mind.
But this latest one from Hillary is not the first time she has said something similar. Time reports that in March she responded to a question asking if she would drop out of the race:
NBC News is reporting that she included similar remarks in her stump speech in West Virginia and Kentucky.
Big Dan said...
She's saying, "I'm hanging around in a mathematically impossible race...maybe Obama will get assassinated, that's why I'm not dropping out".
And Steve "whoring for Hillary" Corbett is immediately defending her, even though the implication is OBVIOUS!
Corbett: YOU STINK!
pj the wb lefty said...
Even if her historical context was correct, it wasn't a good example. Bobby Kennedy and 1968 are remembered because he was asassinated, not because the campaign ran into the month of June (which it did).
Add to that the crack Hucksterbee made recently at the NRA convention about someone shooting at Obama (for which he actually apologized to his credit at least), and you can see why there'd be some heightened sensitivity. Just because something is accurate doesn't mean you should cite it even if used in context.
Big Dan said...
Obama will not pick Hillary as his VP, because you get "Billary" Bill Clinton along with that, and he wants to purge his administration from the Clinton cabal and their devisive Rovian tactics, and wants to divorce himself from the "old Democrats", and break up the Bush/Clinton's being in the White House since 1980.
1980 - Bush I VP
1984 - Bush I VP
1988 - Bush I president
1992 - Bill Clinton president.
1996 - Bill Clinton president.
2000 - Shrub president.
2004 - Shrub president.
If Hillary was president in 2008, it would be 32 straight years of Bush's & Clintons, like 2 Royal Families! I thought that's what we had the Revolution about!
Big Dan said...
You can tune in to Corbett, and he has "whoring for Hillary" all lined up with the latest: Obama smokes, Obama hired non-union, Obama is associated with Casey who disrespected a soldier, and like clockwork - defending the in defensible Hillary/RFK statement!
Tune in to Corbett, for "whoring for Hillary", there's something going on there, very suspicious, with Corbett's show anymore!
Big Dan said...
OH! I forgot: SEXISM! Sorry, Corbett!
Gee! According to Corbett, Hillary hasn't done anything wrong yet!
Like this RFK statement and making up that Bosnia sniper LIE!
Of course, he finds it in his heart to "forgive her" for those things, though. She "mis-spoke", "mis-remembered", "didn't mean it", mis-everthing!
Big Dan said...
I thought Hillary was "Rocky", too! When she was here, she compared herself to Rocky Balboa! When she went to Indiana, she had on a NASCAR suit and an Amish beard & corncob pipe when she was talking to the Amish. And I think she dressed up as a horse in Kentucky. What happened to Rocky and NEPA? And growing up shooting a gun on Lake Winola?
I have to give her credit, for reinventing herself from state to state, that takes guts.
Big Dan said...
She's the "David Bowie" of politicians!
Anonymous said...
Corbett is whoring for Hillary because his strong personality wife is making him. He is whipped and wrapped. I wonder is it sexist if you support someone just because they are a woman???
D.B. Echo said...
You know, it would be far more reasonable to interpret what she said as "The only way to stop me in this Primary race is to shoot me." Though the more obvious and reasonable way to interpret what she said was "It ain't over 'til it's over, and lots of Primaries have run until June, and just because Barack Obama and his media lapdogs want to anoint him and move on already and break out the fancty new graphics they've already designed doesn't mean I'm about to quit."
Hillary Hate is in full swing, as usual. The Obamanites are doing Karl Rove's work for him. Remember: Obama cannot win without the support of the people who are currently backing Hillary Clinton. Team Obama had better reign in its minions and start working on a strategy for winning in November. 'Cause they can still very easily pry Defeat out of the jaws of Victory.
Anonymous said...
How do you spell CUNT?
Anonymous said...
ANON 11:28
You spell it B U S H.
Anonymous said...
also she lied about Bill Clinton. He was far ahead by March of that year. Tsongas had dropped out and it was just him and Jerry Brown......
I know Hillary lie, go figure
Anonymous said...
Hillary lie, no never!?!
Anonymous said...
Vince Foster death, that was a suicide wasn't it? Clean shirt, left-handed with a gun in the right, and spunk in his drawers. Yah suicide.
D.B. Echo said...
Anonymous, where do you get your Vincent Foster facts? Rush Limbaugh or Karl Riove?
NEPAExpat said...
We all know the Junior Senator from NY has to stay in the race to have any chance to reimburse herself for the debts accumulated by the campaign. If she doesn't raise that money by x date, she cannot reimburse herself for the $10 million of her own money she planted into the campaign.
To make up for this money shortfall, after the campaign finally ends, there is an unsubstantiated rumor of a Talking Pantsuit Barbie. Unfortunately, one of the phrases will be "delegate math is hard". | null | null | null | null | null | null | null | null | null |
https://www.knowpia.com/knowpedia/Exsecant | 1,723,542,000,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641075627.80/warc/CC-MAIN-20240813075138-20240813105138-00774.warc.gz | 636,109,407 | 32,596 | BREAKING NEWS
Exsecant
## Summary
The external secant function (exsecant, symbolized exsec) is a trigonometric function defined in terms of the secant function:
${\displaystyle \operatorname {exsec} \theta =\sec \theta -1={\frac {1}{\cos \theta }}-1.}$
It was introduced in 1855 by American civil engineer Charles Haslett, who used it in conjunction with the existing versine function, ${\displaystyle \operatorname {vers} \theta =1-\cos \theta ,}$ for designing and measuring circular sections of railroad track.[3] It was adopted by surveyors and civil engineers in the United States for railroad and road design, and since the early 20th century has sometimes been briefly mentioned in American trigonometry textbooks and general-purpose engineering manuals.[4] For completeness, a few books also defined a coexsecant or excosecant function (symbolized coexsec or excsc), ${\displaystyle \operatorname {coexsec} \theta ={}}$${\displaystyle \csc \theta -1,}$ the exsecant of the complementary angle,[5][6] though it was not used in practice. While the exsecant has occasionally found other applications, today it is obscure and mainly of historical interest.[7]
As a line segment, an external secant of a circle has one endpoint on the circumference, and then extends radially outward. The length of this segment is the radius of the circle times the trigonometric exsecant of the central angle between the segment's inner endpoint and the point of tangency for a line through the outer endpoint and tangent to the circle.
## Etymology
The word secant comes from Latin for "to cut", and a general secant line "cuts" a circle, intersecting it twice; this concept dates to antiquity and can be found in Book 3 of Euclid's Elements, as used e.g. in the intersecting secants theorem. 18th century sources in Latin called any non-tangential line segment external to a circle with one endpoint on the circumference a secans exterior.[8]
The trigonometric secant, named by Thomas Fincke (1583), is more specifically based on a line segment with one endpoint at the center of a circle and the other endpoint outside the circle; the circle divides this segment into a radius and an external secant. The external secant segment was used by Galileo Galilei (1632) under the name secant.[9]
## History and applications
In the 19th century, most railroad tracks were constructed out of arcs of circles, called simple curves.[10] Surveyors and civil engineers working for the railroad needed to make many repetitive trigonometrical calculations to measure and plan circular sections of track. In surveying, and more generally in practical geometry, tables of both "natural" trigonometric functions and their common logarithms were used, depending on the specific calculation. Using logarithms converts expensive multiplication of multi-digit numbers to cheaper addition, and logarithmic versions of trigonometric tables further saved labor by reducing the number of necessary table lookups.[11]
The external secant or external distance of a curved track section is the shortest distance between the track and the intersection of the tangent lines from the ends of the arc, which equals the radius times the trigonometric exsecant of half the central angle subtended by the arc, ${\displaystyle R\operatorname {exsec} {\tfrac {1}{2}}\Delta .}$ [12] By comparison, the versed sine of a curved track section is the furthest distance from the long chord (the line segment between endpoints) to the track[13] – cf. Sagitta – which equals the radius times the trigonometric versine of half the central angle, ${\displaystyle R\operatorname {vers} {\tfrac {1}{2}}\Delta .}$ These are both natural quantities to measure or calculate when surveying circular arcs, which must subsequently be multiplied or divided by other quantities. Charles Haslett (1855) found that directly looking up the logarithm of the exsecant and versine saved significant effort and produced more accurate results compared to calculating the same quantity from values found in previously available trigonometric tables.[3] The same idea was adopted by other authors, such as Searles (1880).[14] By 1913 Haslett's approach was so widely adopted in the American railroad industry that, in that context, "tables of external secants and versed sines [were] more common than [were] tables of secants".[15]
In the late-19th and 20th century, railroads began using arcs of an Euler spiral as a track transition curve between straight or circular sections of differing curvature. These spiral curves can be approximately calculated using exsecants and versines.[15][16]
Solving the same types of problems is required when surveying circular sections of canals[17] and roads, and the exsecant was still used in mid-20th century books about road surveying.[18]
The exsecant has sometimes been used for other applications, such as beam theory[19] and depth sounding with a wire.[20]
In recent years, the availability of calculators and computers has removed the need for trigonometric tables of specialized functions such as this one.[21] Exsecant is generally not directly built into calculators or computing environments (though it has sometimes been included in software libraries),[22] and calculations in general are much cheaper than in the past, no longer requiring tedious manual labor.
## Catastrophic cancellation for small angles
Naïvely evaluating the expressions ${\displaystyle 1-\cos \theta }$ (versine) and ${\displaystyle \sec \theta -1}$ (exsecant) is problematic for small angles where ${\displaystyle \sec \theta \approx \cos \theta \approx 1.}$ Computing the difference between two approximately equal quantities results in catastrophic cancellation: because most of the digits of each quantity are the same, they cancel in the subtraction, yielding a lower-precision result.
For example, the secant of is sec 1° ≈ 1.000152, with the leading several digits wasted on zeros, while the common logarithm of the exsecant of is log exsec 1° ≈ −3.817220,[23] all of whose digits are meaningful. If the logarithm of exsecant is calculated by looking up the secant in a six-place trigonometric table and then subtracting 1, the difference sec 1° − 1 ≈ 0.000152 has only 3 significant digits, and after computing the logarithm only three digits are correct, log(sec 1° − 1) ≈ −3.818156.[24] For even smaller angles loss of precision is worse.
If a table or computer implementation of the exsecant function is not available, the exsecant can be accurately computed as ${\textstyle \operatorname {exsec} \theta =\tan \theta \,\tan {\tfrac {1}{2}}\theta {\vphantom {\Big |}},}$ or using versine, ${\textstyle \operatorname {exsec} \theta =\operatorname {vers} \theta \,\sec \theta ,}$ which can itself be computed as ${\textstyle \operatorname {vers} \theta =2{\bigl (}{\sin {\tfrac {1}{2}}\theta }{\bigr )}{\vphantom {)}}^{2}{\vphantom {\Big |}}={}}$ ${\displaystyle \sin \theta \,\tan {\tfrac {1}{2}}\theta \,{\vphantom {\Big |}}}$ ; Haslett used these identities to compute his 1855 exsecant and versine tables.[25][26]
For a sufficiently small angle, a circular arc is approximately shaped like a parabola, and the versine and exsecant are approximately equal to each-other and both proportional to the square of the arclength.[27]
## Mathematical identities
### Inverse function
The inverse of the exsecant function, which might be symbolized arcexsec,[6] is well defined if its argument ${\displaystyle y\geq 0}$ or ${\displaystyle y\leq -2}$ and can be expressed in terms of other inverse trigonometric functions (using radians for the angle):
${\displaystyle \operatorname {arcexsec} y=\operatorname {arcsec}(y+1)={\begin{cases}{\arctan }{\bigl (}\!{\textstyle {\sqrt {y^{2}+2y}}}\,{\bigr )}&{\text{if}}\ \ y\geq 0,\\[6mu]{\text{undefined}}&{\text{if}}\ \ {-2}
the arctangent expression is well behaved for small angles.[28]
### Calculus
While historical uses of the exsecant did not explicitly involve calculus, its derivative and antiderivative (for x in radians) are:[29]
{\displaystyle {\begin{aligned}{\frac {\mathrm {d} }{\mathrm {d} x}}\operatorname {exsec} x&=\tan x\,\sec x,\\[10mu]\int \operatorname {exsec} x\,\mathrm {d} x&=\ln {\bigl |}\sec x+\tan x{\bigr |}-x+C,{\vphantom {\int _{|}}}\end{aligned}}}
where ln is the natural logarithm. See also Integral of the secant function.
### Double angle identity
The exsecant of twice an angle is:[6]
${\displaystyle \operatorname {exsec} 2\theta ={\frac {2\sin ^{2}\theta }{1-2\sin ^{2}\theta }}.}$
• Chord (geometry) – A line segment with endpoints on the circumference of a circle, historically used trigonometrically
• Exponential minus 1 – The function ${\displaystyle x\mapsto e^{x}-1,}$ also used to improve precision for small inputs
## Notes and references
1. ^ Cajori, Florian (1929). A History of Mathematical Notations. Vol. 2. Chicago: Open Court. §527. "Less common trigonometric functions", pp. 171–172.
2. ^ The original conception of trigonometric functions was as line segments, but this was gradually replaced during the 18th and 19th century by their conception as length ratios between sides of a right triangle or abstract functions; when the exsecant was introduced, in the mid 19th century, both concepts were still common.
Bressoud, David (2010). "Historical Reflections on Teaching Trigonometry" (PDF). Mathematics Teacher. 104 (2): 106–112. doi:10.5951/MT.104.2.0106.
Van Sickle, Jenna (2011). "The history of one definition: Teaching trigonometry in the US before 1900". International Journal for the History of Mathematics Education. 6 (2): 55–70.
3. ^ a b Haslett, Charles (1855). "The Engineer's Field Book". In Hackley, Charles W. (ed.). The Mechanic's, Machinist's, and Engineer's Practical Book of Reference; Together with the Engineer's Field Book. New York: James G. Gregory. pp. 371–512.
As the book's editor Charles W. Hackley explains in the preface, "The use of the more common trigonometric functions, to wit, sines, cosines, tangents, and cotangents, which ordinary tables furnish, is not well adapted to the peculiar problems which are presented in the construction of Railroad curves. [...] Still there would be much labor of computation which may be saved by the use of tables of external secants and versed sines, which have been employed with great success recently by the Engineers on the Ohio and Mississippi Railroad, and which, with the formulas and rules necessary for their application to the laying down of curves, drawn up by Mr. Haslett, one of the Engineers of that Road, are now for the first time given to the public." (pp. vi–vii)
Charles Haslett continues in his preface to the Engineer's Field Book: "Experience has shown, that versed sines and external secants as frequently enter into calculations on curves as sines and tangents; and by their use, as illustrated in the examples given in this work, it is believed that many of the rules in general use are much simplified, and many calculations concerning curves and running lines made less intricate, and results obtained with more accuracy and far less trouble, than by any methods laid down in works of this kind. [...] In addition to the tables generally found in books of this kind, the author has prepared, with great labor, a Table of Natural and Logarithmic Versed Sines and External Secants, calculated to degrees, for every minute; also, a Table of Radii and their Logarithms, from 1° to 60°." (pp. 373–374)
Review: Poor, Henry Varnum, ed. (1856-03-22). "Practical Book of Reference, and Engineer's Field Book. By Charles Haslett". American Railroad Journal (Review). Second Quarto Series. XII (12): 184. Whole No. 1040, Vol. XXIX.
4. ^ Kenyon, Alfred Monroe; Ingold, Louis (1913). Trigonometry. New York: The Macmillan Company. p. 5.
Hudson, Ralph Gorton; Lipka, Joseph (1917). A Manual of Mathematics. New York: John Wiley & Sons. p. 68.
McNeese, Donald C.; Hoag, Albert L. (1957). Engineering and Technical Handbook. Englewood Cliffs, NJ: Prentice-Hall. pp. 147, 315–325 (table 41). LCCN 57-6690.
Zucker, Ruth (1964). "4.3.147: Elementary Transcendental Functions - Circular functions". In Abramowitz, Milton; Stegun, Irene A. (eds.). Handbook of Mathematical Functions. Washington, D.C.: National Bureau of Standards. p. 78. LCCN 64-60036.
5. ^ Bohannan, Rosser Daniel (1904) [1903]. "\$131. The Versed Sine, Exsecant and Coexsecant. §132. Exercises". Plane Trigonometry. Boston: Allyn and Bacon. pp. 235–236.
6. ^ a b c Hall, Arthur Graham; Frink, Fred Goodrich (1909). "Review Exercises". Plane Trigonometry. New York: Henry Holt and Company. § "Secondary Trigonometric Functions", pp. 125–127.
7. ^ Oldham, Keith B.; Myland, Jan C.; Spanier, Jerome (2009) [1987]. An Atlas of Functions (2nd ed.). Springer. Ch. 33, "The Secant sec(x) and Cosecant csc(x) functions", §33.13, p. 336. doi:10.1007/978-0-387-48807-3. ISBN 978-0-387-48806-6. Not appearing elsewhere in the Atlas [...] is the archaic exsecant function [...].
8. ^ Patu, Andræâ-Claudio (André Claude); Le Tort, Bartholomæus (1745). Rivard, Franciscus (Dominique-François) [in French] (ed.). Theses Mathematicæ De Mathesi Generatim (in Latin). Paris: Ph. N. Lottin. p. 6.
Lemonnier, Petro (Pierre) (1750). Genneau, Ludovicum (Ludovico); Rollin, Jacobum (Jacques) (eds.). Cursus Philosophicus Ad Scholarum Usum Accomodatus (in Latin). Vol. 3. Collegio Harcuriano (Collège d'Harcourt), Paris. pp. 303–.
Thysbaert, Jan-Frans (1774). "Articulus II: De situ lineæ rectæ ad Circularem; & de mensura angulorum, quorum vertex non est in circuli centro. §1. De situ lineæ rectæ ad Circularem. Definitio II: [102]". Geometria elementaria et practica (in Latin). Lovanii, e typographia academica. p. 30, foldout.
van Haecht, Joannes (1784). "Articulus III: De secantibus circuli: Corollarium III: [109]". Geometria elementaria et practica: quam in usum auditorum (in Latin). Lovanii, e typographia academica. p. 24, foldout.
9. ^ Galileo used the Italian segante.
Galilei, Galileo (1632). Dialogo di Galileo Galilei sopra i due massimi sistemi del mondo Tolemaico e Copernicano [Dialogue on the Two Chief World Systems, Ptolemaic and Copernican] (in Italian).
Galilei, Galileo (1997) [1632]. Finocchiaro, Maurice A. (ed.). Galileo on the World Systems: A New Abridged Translation and Guide. University of California Press. pp. 184 (n130), 184 (n135), 192 (n158). ISBN 9780520918221. Galileo's word is segante (meaning secant), but he clearly intends exsecant; an exsecant is defined as the part of a secant external to the circle and thus between the circumference and the tangent.
Finocchiaro, Maurice A. (2003). "Physical-Mathematical Reasoning: Galileo on the Extruding Power of Terrestrial Rotation". Synthese. 134 (1–2, Logic and Mathematical Reasoning): 217–244. doi:10.1023/A:1022143816001. JSTOR 20117331.
10. ^ Allen, Calvin Frank (1894) [1889]. Railroad Curves and Earthwork. New York: Spon & Chamberlain. p. 20.
11. ^ Van Brummelen, Glen (2021). "2. Logarithms". The Doctrine of Triangles. Princeton University Press. pp. 62–109. ISBN 9780691179414.
12. ^ Frye, Albert I. (1918) [1913]. Civil engineer's pocket-book: a reference-book for engineers, contractors and students containing rules, data, methods, formulas and tables (2nd ed.). New York: D. Van Nostrand Company. p. 211.
13. ^ Gillespie, William M. (1853). A Manual of the Principles and Practice of Road-Making. New York: A. S. Barnes & Co. pp. 140–141.
14. ^ Searles, William Henry (1880). Field Engineering. A hand-book of the Theory and Practice of Railway Surveying, Location, and Construction. New York: John Wiley & Sons.
Searles, William Henry; Ives, Howard Chapin (1915) [1880]. Field Engineering: A Handbook of the Theory and Practice of Railway Surveying, Location and Construction (17th ed.). New York: John Wiley & Sons.
15. ^ a b Jordan, Leonard C. (1913). The Practical Railway Spiral. New York: D. Van Nostrand Company. p. 28.
16. ^ Thornton-Smith, G. J. (1963). "Almost Exact Closed Expressions for Computing all the Elements of the Clothoid Transition Curve". Survey Review. 17 (127): 35–44. doi:10.1179/sre.1963.17.127.35.
17. ^ Doolittle, H. J.; Shipman, C. E. (1911). "Economic Canal Location in Uniform Countries". Papers and Discussions. Proceedings of the American Society of Civil Engineers. 37 (8): 1161–1164.
18. ^ For example:
Hewes, Laurence Ilsley (1942). American Highway Practice. New York: John Wiley & Sons. p. 114.
Ives, Howard Chapin (1966) [1929]. Highway Curves (4th ed.). New York: John Wiley & Sons. LCCN 52-9033.
Meyer, Carl F. (1969) [1949]. Route Surveying and Design (4th ed.). Scranton, PA: International Textbook Co.
19. ^ Wilson, T. R. C. (1929). "A Graphical Method for the Solution of Certain Types of Equations". Questions and Discussions. The American Mathematical Monthly. 36 (10): 526–528. JSTOR 2299964.
20. ^ Johnson, Harry F. (1933). "Correction for inclination of sounding wire". The International Hydrographic Review. 10 (2): 176–179.
21. ^ Calvert, James B. (2007) [2004]. "Trigonometry". Archived from the original on 2007-10-02. Retrieved 2015-11-08.
22. ^ Simpson, David G. (2001-11-08). "AUXTRIG" (Fortran 90 source code). Greenbelt, MD: NASA Goddard Space Flight Center. Retrieved 2015-10-26.
van den Doel, Kees (2010-01-25). "jass.utils Class Fmath". JASS - Java Audio Synthesis System. 1.25. Retrieved 2015-10-26.
"MIT/GNU Scheme – Scheme Arithmetic" (MIT/GNU Scheme source code). v. 12.1. Massachusetts Institute of Technology. 2023-09-01. exsec function, arith.scm lines 61–63. Retrieved 2024-04-01.
23. ^ In a table of logarithmic exsecants such as Haslett 1855, p. 417 or Searles & Ives 1915, II. p. 135, the number given for log exsec 1° is 6.182780, the correct value plus 10, which is added to keep the entries in the table positive.
24. ^ The incorrect digits are highlighted in red.
25. ^ Haslett 1855, p. 415
26. ^ Nagle, James C. (1897). "IV. Transition Curves". Field Manual for Railroad Engineers (1st ed.). New York: John Wiley and Sons. §§ 138–165, pp. 110–142; Table XIII: Natural Versines and Exsecants, pp. 332–354.
Review: "Field Manual for Railroad Engineers. By J. C. Nagle". The Engineer (Review). 84: 540. 1897-12-03.
27. ^ Shunk, William Findlay (1918) [1890]. The Field Engineer: A Handy Book of Practice in the Survey, Location, and Track-Work of Railroads (21st ed.). New York: D. Van Nostrand Company. p. 36.
28. ^ "4.5 Numerical operations". MIT/GNU Scheme Documentation. v. 12.1. Massachusetts Institute of Technology. 2023-09-01. procedure: aexsec. Retrieved 2024-04-01.
"MIT/GNU Scheme – Scheme Arithmetic" (MIT/GNU Scheme source code). v. 12.1. Massachusetts Institute of Technology. 2023-09-01. aexsec function, arith.scm lines 65–71. Retrieved 2024-04-01.
29. ^ Weisstein, Eric W. (2015) [2005]. "Exsecant". MathWorld. Wolfram Research, Inc. Retrieved 2015-11-05. | 5,188 | 18,895 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 19, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2024-33 | latest | en | 0.897647 |
http://mathhelpforum.com/trigonometry/206691-trigonometry-identities-help.html | 1,526,916,418,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864405.39/warc/CC-MAIN-20180521142238-20180521162238-00121.warc.gz | 193,325,189 | 10,338 | 1. ## Trigonometry identities help
Section B
Number 9) II
What i did
________
cos2x=1-2sin275
cos 2x = cos (x+x)
cos (x+x) = cosA*CosB-SinA*SinB
cos( 60 + 90)= cos60*cos90 - sin60*cos90
section B part III number 9
____________________
cos 2x = cos2x - sin2x
cos( x + x) = cosA*CosB-SinA*SinB
cos(90 + 45)= cos90*cos45-sin90*sin45
I attempted this one similarly.. wrong answer... help!
2. ## Re: Trigonometry identities help
9. II: Note that $\displaystyle 1 - 2 \sin^2 {75^{\circ}} = \cos(150^{\circ})$.
3. ## Re: Trigonometry identities help
I know, sorry i made a mistake in labelling and typing , now fixed,
4. ## Re: Trigonometry identities help
Oh okay, I was a little confused with the 45/60 thing.
Also, $\displaystyle \sin(60^{\circ}) = \frac{\sqrt{3}}{2}$, not $\displaystyle \frac{2}{\sqrt{3}}$. The sine/cosine of an angle can only be from -1 to 1 inclusive. Since $\displaystyle -\frac{2}{\sqrt{3}}$ is outside the set [-1,1], this answer cannot be correct.
5. ## Re: Trigonometry identities help
got it! thanks ! | 359 | 1,046 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2018-22 | latest | en | 0.679871 |
null | null | null | null | null | null | Taking out the trash
Turned out, not so much.
Removing the trash from TotC didn't make it feel epic. Partially, this could be due to the other design change of TotC, namely that it takes place in a single round room with no real distinguishing features. Combined, these two facts make the bosses actually start to feel like trash pulls. I did Faction Champions my first time in a kind of bored daze, wondering why it was taking so long to kill them. Ironically, instances like Black Morass and Violet Hold, despite basically being the same (big area, you don't go anywhere) managed to be more compelling than TotC because they used variations in the terrain and trash spawning in waves to break up the tedium of the big round room. Perhaps TotC would have been better with a few naval battles just like actual coliseums had.
There's a fine line in trash difficulty and, more importantly, trash respawn times. Trash difficulty should be high enough to engage your group and get them focused, and this difficulty can often be used by designers to give you a sense of what the boss mechanics are going to be. This is a useful role for trash, to get players looking forward to the boss and what it may or may not do to them. If the trash respawn is too long, once you've cleared it, then it may as well not matter at all. If it's too short, it ends up chewing up limited time you may have to learn a boss's full suite of mechanics and work on defeating him.
Take two bosses, Halfus Wyrmbreaker and Magmaw, and consider their trash and timers. Magmaw has three trash mobs, one with a straightforward suite of abilities and two with a charge/switch mechanic. These trash pulls don't take long to deal with once you know the mechanics, but they don't particularly educate you about the boss and serve no real purpose. However, they're at least easy to deal with, so their two-hour respawn time is not onerous. (Although I'm not discussing them here, the trash for the Omnotron Defense System is superior in that it is educational in nature.)
The trash pulls on the way to Halfus Wyrmbreaker, on the other hand, are excruciatingly tedious. They involve mechanics based on other sections of trash (the glowing orbs that do massive damage to anyone) that require a group to learn on the fly how to deal with them. They teach you nothing at all about the Halfus fight. And there are a lot of these pulls. In this situation, there is absolutely no reason for a two-hour respawn time for this trash. Forcing people to reclear this trash every two hours when first learning the boss is, in effect, working in a "We don't want you to spend your three hours a night working on this boss; if you can't get him in two, we want you to call it" mode -- which is fine, if that's what you're going for. But it's a punitive design choice, one that punishes you for being new to an encounter.
Getting a boss to 20% over the course of a night, being near the end of your raid and wanting to make that one more attempt that you think can push you to victory is a fun part of raiding. Being unable to do so because all the trash just respawned and you only have 20 minutes left in your raid is not a fun part of raiding, and I don't think it's a good design choice to force time-outs on us. The two-hour respawn, in this case, basically penalizes your raid for taking a raid break, for having a tank AFK because her husband called from work or what have you.
Honestly, the best trash I've seen in the past three years of raiding was Ulduar trash. It often taught you what to expect in upcoming fights, it was numerous and complex enough to be entertaining without being so numerous that it felt like a slog, and it engaged players in how to overcome it without becoming more daunting than the bosses. So far, to my mind, Cataclysm trash is somewhat above vanilla and BC but below the bar set by Ulduar. You definitely need trash, but right now, we maybe need a bit of a cleaning service, since trash is actually serving as an artificial and pointless gatekeeper to content.
This article was originally published on WoW Insider. | null | null | null | null | null | null | null | null | null |
https://cosmolearning.org/courses/what-every-5th-grader-should-know-about-math/ | 1,611,523,475,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703557462.87/warc/CC-MAIN-20210124204052-20210124234052-00307.warc.gz | 276,378,618 | 11,151 | ### Course Description
There is one common denominator to success in school, at university, in life and science and it is having a firm grasp of what 5th Graders should know. Michel van Biezen will give you all you need to know to fill the gaps in your math education, in this thoughtfully organized series called “What a Fifth Grader Should Know."
Michel van Biezen presents every basic mathematical concept that all 5th graders (and adults) should know to be successful
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### Video Lectures & Study Materials
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# Lecture Play Lecture
1 Series Content Overview Play Video
2 Addition of 2-Single Digit Play Video
3 Addition of 2-Double Digits Play Video
4 Addition of Double Digit Numbers: 1 Play Video
5 Addition of Double Digit Numbers: 2 Play Video
6 Addition of Double Digit Numbers: 3 Play Video
7 Addition of Triple Digit Numbers Play Video
II. Subtraction
8 Subtraction of Double Digit Numbers: 1 Play Video
9 Subtraction of Double Digit Numbers: 2 Play Video
10 Subtraction of Double Digit Numbers: 3 Play Video
11 Subtraction of Double Digit Numbers: 4 Play Video
12 Subtraction of Double Digit Numbers: 5 Play Video
III. Multiplication
13 Multiplication: Introduction Play Video
14 Multiplication: Multiply by "0" Play Video
15 Multiplication: Multiply by "1" Play Video
16 Multiplication: Multiply by "2" Play Video
17 Multiplication: Multiply by "3" Play Video
18 Multiplication: Multiply by "4" Play Video
19 Multiplication: Multiply by "5" Play Video
20 Multiplication: Multiply by "6" Play Video
21 Multiplication: Multiply by "7" Play Video
22 Multiplication: Multiply by "8" Play Video
23 Multiplication: Multiply by "9" Play Video
24 Memorizing Multiplication Table MADE EASY! Play Video
25 Multiplication: Multiply by "10" Play Video
26 Multiplication: Squaring a 1-Digit Number Play Video
27 Multiplication: Doubling Numbers Play Video
28 Multiplication of 1 and 2 Digit Numbers Play Video
29 Multiplication of 2 2-Digit Numbers: 1 Play Video
30 Multiplication of 2 2-Digit Numbers: 2 Play Video
31 Multiplication of 2 2-Digit Numbers: 2 Play Video
32 How to Square 2-Digit Number Ending in 5 Play Video
33 How to Square Any 2-Digit Number Play Video
34 How to Square 2-Digit Number Ending in 1 Play Video
35 How to Multiply 2-Digit Numbers by 11 Play Video
36 How to Multiply Any Number by 25 Play Video
37 How to Multiply Any Number by 5, 50, 500 Play Video
38 Great Trick for Checking Multiplication! Play Video | 658 | 2,589 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2021-04 | longest | en | 0.772824 |
https://www.effortlessmath.com/math-topics/how-to-find-volume-by-spinning-disk-method/ | 1,723,768,069,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641319057.20/warc/CC-MAIN-20240815235528-20240816025528-00041.warc.gz | 582,753,587 | 12,115 | # How to Find Volume by Spinning: Disk Method
The Disk Method calculates volumes by revolving a region around an axis, forming disks. Integrating these disk areas, perpendicular to the axis of revolution, yields the solid's volume. It's ideal for solids with simple, symmetric shapes, where the radius varies along the axis, allowing precise volume calculation through continuous addition of infinitesimally thin disk volumes.
This method leverages the symmetry of revolving shapes, transforming a 2D area into a 3D object. Each disk’s volume is $$\pi r^2 \Delta x$$, where $$r$$ is the radius function. By accumulating these volumes along the axis, complex volumetric problems are simplified. Especially useful in engineering and design, the Disk Method provides a foundational tool for calculating storage capacities, structural volumes, and more.
Here are the steps to use the method:
1. Define the Radius: The radius $$r$$ of each disk is a function of $$x$$ (or $$y$$, depending on the axis of revolution), denoted as $$r(x)$$ or $$r(y)$$.
2. Disk Volume: The volume of each infinitesimally thin disk is given by the formula $$dV = \pi [r(x)]^2 dx$$ (if revolving around the y-axis) or $$dV = \pi [r(y)]^2 dy$$ (if revolving around the x-axis), where $$dV$$ is the differential volume of the disk.
3. Set Up the Integral: The total volume $$V$$ of the solid is the integral of these differential volumes over the interval of revolution. Formally, if revolving around the y-axis:
$$V = \int_{a}^{b} \pi [r(x)]^2 dx$$
And if revolving around the x-axis:
$$V = \int_{c}^{d} \pi [r(y)]^2 dy$$
Here, $$a$$ and $$b$$ (or $$c$$ and $$d$$) are the bounds of the region being revolved, determined by the intersection points of the curve and the axis of revolution or the specified limits of integration.
4. Evaluate the Integral: Compute the integral to find the total volume of the solid of revolution. This step may require analytical techniques or numerical methods depending on the function defining the radius.
By applying this method, you can calculate the volume of a wide variety of shapes with rotational symmetry, providing a powerful technique for analyzing physical objects and theoretical constructs in mathematics, physics, and engineering.
Let’s calculate the volume of a solid formed by revolving the region under the curve $$y = \sin(x)$$ from $$x = 0$$ to $$x = \pi$$ around the x-axis.
Step-by-Step Solution:
• The radius of each disk is given by the function $$y = \sin(x)$$.
1. Disk Volume:
• The volume of each disk is $$dV = \pi [r(x)]^2 dx = \pi [\sin(x)]^2 dx$$.
1. Set Up the Integral:
• The total volume $$V$$ is the integral of $$\pi [\sin(x)]^2$$ from $$x = 0$$ to $$x = \pi$$:
$$V = \int_{0}^{\pi} \pi [\sin(x)]^2 dx$$
1. Simplify the Integral Using a Trigonometric Identity:
• Use the identity $$\sin^2(x) = \frac{1 – \cos(2x)}{2}$$ to simplify the integral:
$$V = \pi \int_{0}^{\pi} \frac{1 – \cos(2x)}{2} dx$$
The volume of the solid formed by revolving the region under the curve $$y = \sin(x)$$ from $$x = 0$$ to $$x = \pi$$ around the x-axis is approximately $$4.93$$ cubic units.
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SAVE $40 It was$89.99 now it is \$49.99 | 886 | 3,365 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2024-33 | latest | en | 0.836993 |
https://www.sarthaks.com/2828087/income-25-more-than-bs-expenditure-less-then-find-the-income-of-if-savings-of-and-is-12-each | 1,675,831,495,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500671.13/warc/CC-MAIN-20230208024856-20230208054856-00488.warc.gz | 952,470,337 | 15,874 | # A’ income 25% more than B. B’s expenditure 2/6 less then A. find the Income of A if savings of A and B is 12 of each.
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A’ income 25% more than B. B’s expenditure 2/6 less then A. find the Income of A if savings of A and B is 12 of each.
1. 40
2. 58
3. 30
4. 80
5. None
by (24.2k points)
selected
Correct Answer - Option 3 : 30
Calculation:
Let be assume Income of B = I then the income of A = 5I/4
Expenditure of A = E, then the expenditure of B = (2/3) × E
Savings of A = (5I/4) – E = (5I – 4E)/4 = 12 … (1)
Savings of B = I – (2/3)E = (3I – 2E)/3 = 12 … (2)
When we will solve the equation (1) and (2) then
⇒ I = 24, A’s income = (5/4) × 24 = 30
∴ The required result will be 30. | 280 | 710 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2023-06 | latest | en | 0.89853 |
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# 410Hw04ans - STAT 410 Fall 2009 Homework #4 (due Friday,...
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STAT 410 Fall 2009 Homework #4 (due Friday, September 25, by 3:00 p.m.) 1. Let X, Y, and Z be i.i.d. Uniform [ 0 , 1 ] random variables Find the probability distribution of W = X + Y + Z. That is, find ( 29 w f W . Hint: If V = X + Y, we know the p.d.f. of V, f V ( v ) ( see Examples for 09/18/2009 ): f V ( v ) = v if 0 < v < 1, f V ( v ) = 2 – v if 1 < v < 2, f V ( v ) = 0 otherwise. Now use convolution formula to find the p.d.f. of W = V + Z. There will be 5 possible cases; two of them are “boring”, two of them are “exciting”, and one is “really exciting”. ( 29 < < = otherwise 0 1 0 1 Z z z f ( 29 < < - = < - < = - otherwise 0 1 1 otherwise 0 1 0 1 Z w v w v w v w f ( 29 w f W = ( 29 ( 29 ( 29 - = - + dv v w f v f w f Z V Z V (convolution) Case 1 : w < 0. ( 29 w f W = 0.
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Case 2 : 0 < w < 1. Then w – 1 < 0. ( 29 ( 29 2 1 2 0 W w dv v w f w = = . Case 3 : 1 < w < 2. Then 0 < w – 1 < 1. ( 29 ( 29 ( 29 ( 29 2 3 3 1 2 1 2 1 1 1 W - + - = - + = - w w dv v dv v w f w w ( 29 ( 29 2 1 2 1 + - - = w w . Case 4 : 2 < w < 3. Then 1 < w – 1 < 2. ( 29 ( 29 ( 29 ( 29 2 3 2 9 3 2 1 2 2 2 2 1 W w w w dv v w f w - = + - = - = - . Case 5 : w > 3. Then w – 1 > 2. ( 29 w f W = 0.
2. Let X and Y be two independent random variables, with probability density functions f X ( x ) and f Y ( y ) , respectively. ( 29 = otherwise 0 1 0 3 2 X x x x f ( 29 = otherwise 0 1 0 2 Y y y y f Find the p.d.f. f W ( w ) of W = X + Y. f W ( w ) = ( 29 ( 29 - - dx x w f x f Y X . ( 29 x w f - Y = ( 29 - - otherwise 0 1 0 if 2 x w x w = ( 29 - - otherwise 0 1 if 2 w x w x w Case 1 . w < 0. ( 29 w f Y X + = 0. Case 2 . 0 < w < 1. ( 29 w f Y X + = ( 29 - w dx x w x 0 2 2 3 = 0 4 3 2 3 2 = = - x w x x w x = 4 2 1 w .
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Case 3 . 1 < w < 2. ( 29 w f Y X + = ( 29 - - 1 1 2 2 3 w dx x w x = 1 1 4 3 2 3 2 - = = - w x x x w x = ( 29 ( 29 4 3 1 2 3 1 2 2 3 2 - + - - - w w w w = w w w 2 3 2 1 2 4 - + - . Case 4 . w > 2. ( 29 w f Y X + = 0. ( 29 w f Y X + = < < - + - < < otherwise 0 2 1 2 3 2 1 1 0 2 1 2 4 4 w w w w w w
OR Case 1 . w < 0. F X + Y ( w ) = 0. ( 29 w f Y X + = F X ' + Y ( w ) = 0. Case 2 . 0 < w < 1. F X + Y ( w ) = - w x w dx dy y x 0 0 2 2 3 = … ( 29 w f Y X + = F X ' + Y ( w ) = … Case 3 . 1 < w < 2. F X + Y ( w ) = … ( 29 w f Y X + = F X ' + Y ( w ) = … Case 4 . w > 2. F X + Y ( w ) = 1. ( 29 w f Y X + = F X ' + Y ( w ) = 0.
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3. Let X and Y have the pdf f ( x , y ) = 1, 0 < x < 1, 0 < y < 1, zero elsewhere. Find the
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## 410Hw04ans - STAT 410 Fall 2009 Homework #4 (due Friday,...
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Ask a homework question - tutors are online | 1,401 | 3,153 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2016-50 | longest | en | 0.714088 |
https://nrich.maths.org/2375 | 1,679,927,479,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948632.20/warc/CC-MAIN-20230327123514-20230327153514-00033.warc.gz | 520,776,528 | 4,524 | ### Integral Equation
Solve this integral equation.
### Integral Sandwich
Generalise this inequality involving integrals.
### Integral Inequality
An inequality involving integrals of squares of functions.
# Area L
##### Age 16 to 18Challenge Level
Let $f(x)$ be a continuous increasing function in the interval $a\leq x\leq b$ where $0 < a < b$ and $0\leq f(a) < f(b)$.
Can you prove the following formula with the help of a sketch? $$\int_{f(a)}^{f(b)} f^{-1}(t) \,dt + \int_a^bf(x) \,dx = bf(b) - af(a).$$
Why must $f(x)$ be increasing in the interval $a\leq x\leq b$?
How could you evaluate a similar integral if $f(x)$ is decreasing?
Once you've proved the formula, find the value of $\int _1^4 \sqrt t \,dt$, in two different ways; firstly by evaluating the integral directly, and secondly by using the formula above with $f(x)=x^2$.
Have a go at using the formula to evaluate $\int_0^1\sin^{-1}t \,dt.$
Can you find other functions that you can integrate more easily using this formula than by other means? | 289 | 1,025 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2023-14 | latest | en | 0.810177 |
hillby.wordpress.com | 1,498,683,663,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128323801.5/warc/CC-MAIN-20170628204133-20170628224133-00538.warc.gz | 766,312,359 | 37,839 | # Credit Cards & Recursive Geometric Formulas
This is my first lesson post with materials, so please comment if you need more information, or if things don’t work.
Learning Targets:
1. Students will be able to write a recursive formula with a percent increase or decrease.
2. Students will know how a credit card balance is calculated (hindsight: they need to know how a credit card works)
3. Students will know about credit card fees, APR, interest and monthly payments.
Background: Our textbook (Discovering Advanced Algebra) uses recursion to build into linear functions and exponential functions. Our district decided at the beginning of the year that we could take the geometric recursive sequences and teach them at the intro to exponential functions. The student teacher I’m working with had the idea that we could use credit cards to make a real-world connection. Unfortunately he got saddled with all of the lesson planning for PreCalculus and never got to flesh out the idea. So I hopped on that train and made it into a groupwork jigsaw activity.
What makes it a groupworthy task: credit card calculations involve a lot of separate pieces of information, and you don’t need to be an expert on every aspect to grasp the whole, but you do need the basics. By dividing up the resources of content knowledge, the final group has a reason to seek a contribution from each member.
Activate Prior Knowledge: We opened with the warm-up to make sure students were all clear on percentages, and finding increase and decrease. In years past this was tricky, but it was smooth sailing this year. The discussion questions at the end of the warm-up helped a ton to prepare them for presenting their work to the rest of the class.
Grouping: Each group of four students were assigned individual expert tasks. They split to meet with their matching experts from other groups & work through their sheets. The FEES experts have an actual credit card offer on the back of their sheet.
Products: After completing their requisite tasks, they head back to their original groups, and get the overall Task Card. In a 100 minute period, this is where we had to stop and we’ll continue on Monday.
The final product will be a public vote tally on which credit card is “better.” I haven’t figure out how to guide that into a mathematical argument. Perhaps each vote needs to put up 2 reasons for the vote, and discussion will ensue? I have to see it play out once and then decide what can be added.
We devoted an entire 100 minute class to practicing this process (it’s on three chart papers on the wall now). It was myself, a student teacher and a literacy coach in a class of 32 students. We chose a problem in a future section, and told students to base their reading off solving this problem.
We gave all of the students 3 minutes of private quiet think time to read the problem and assess their level of understanding. They held up fingers (1-4) and we grouped the students by their level of understanding. I re-explained the rest of the process and how they should expect to repeat it multiple times.
I sat down and worked with the small group of 1’s to get them started on understanding the question. I pretty much kept repeating the two questions: “What do you understand?” “What information are you looking for?” Even the IDKers could say what information they were looking for. If it was from before the section, I directed them to the index after they couldn’t find it in the index.
1. Read the problem (or example)
2. Assess your level of understanding (this is on a second poster)
1. I don’t understand the question –> I’m reading to understand the question (you may need help from a classmate or teacher)
2. I kind of understand the question, but can’t start the problem -> I’m reading to find an example
3. I understand the question, but I need an example –> I’m reading to see possible solutions
4. I understand the question and know what to do –> Go for it!
3. Prepare to read (this is on a third poster)
1. What information do you know?
2. What information do you need? (just one thing at a time)
3. Run your eyes over the section of the textbook and find several places where that might be located
4. Choose one location
5. Did that answer your question? Yes – solve the problem. No – repeat!
We (the student teacher and I) need to devote another 40-50 minutes in a month or so to reinforce this process. But already we’ve learned a lot about where students get stuck reading and locating information. Just breaking down the “prepare to read” was a huge eye opening experience for me.
# First Week
Although this blog is for my own personal reflection, I have to pretend I have a vast audience that is chomping at the bit for regular postings. Perhaps this can be another avoidance strategy to lesson planning on Sunday nights.
What’s different and wonderful about this first week of the school year?
To begin, I know nearly all of my students already. Those that I don’t know, know a fair amount about me and how demanding and strict I can be. I gave an exit slip on the first day of each class asking for questions about me. I didn’t give any spiel about who I am and how I teach. I figured if they were interested, they would ask. When the question, “are you strict?” came up, I was in a great position to explain how I am strict, but I’ve gotten very good about not being a jerk. Yes, I will hold you to a high standard. No, I will not let you pass just because you’re sucking up. But there’s just no good reason I would need to be mean while enforcing my rules and standards.
Also, It’s wonderful to know the curriculum I’m teaching. I taught Algebra II last year, I taught PreCalc two years ago, and this is my third year teaching Physics.
I’m very excited about standards based grading. The concept is not new to my students. Most of them have had SBG in their language arts classes for the past two years. The shift for them is from pieces of evidence to make the grade, to multiple assessments. I’m not grading homework, and I won’t accept homework as evidence of learning. There has been a strong culture of copying for the past several years and I’m not going to forget.
Finally, it’s wonderful to have a student teacher to work with. I couldn’t help but step in when he was leading an activity on the first day, but I do think that we’ve established to the students an expectation that we are partner teachers. As time goes on, I’m sure I’ll need to step in less, and I think that’s appropriate. Planning the first days of lessons was incredibly productive. We’re both tossing out ideas, we’re both weighing in on what we think will be effective. I still have the final say, which again will change when he takes over the class.
Goals for curriculum:
– Make something cohesive out of the first semester of Precalc (before trig)!
– Document the heck out of Physics, so someone else can step in and take over next year. I just realized that video is going to be important there!
Right. Enough procrastinating. And check out Baths – Cerulean, excellent working music.
# Being cool sucks, or everyone’s a geek
I have spent my entire summer learning. I have been reading a ton of blogs, taking classes, reading books, visiting the radiation department in a hospital. It has been great! I get in this mode where I binge on information. I’ll read one article, which will set me off on a several hour journey where my mind is in overdrive and I can’t read fast enough to satisfy my now apparent hunger for information.
What I like even more about this is how much more fun and easy this is as an adult. Let me explain by contrasting with a teenage mind.
You can find this in the brain research, or if you’re very introspective you can recall this about yourself. As a teenager, when I learned something new it went into my brain as “this new thing.” There were no connections between “this new thing” and what I’ve learned before, unless it was made explicit for me. That could be from a teacher pointing out the connection, or a question leading me to that connection, or an experience. Did you notice how “this new thing” has dink in common with “that old thing?” Point is, as a teenager I would learn something new and LATER connect to previous knowledge. (it’s pretty obvious that this is our jobs as teachers)
But as an adult, I don’t learn brand new things! Everything I learn is a branch from something I know already. Many times this is due to exploring something deeper based on my own curiosity, which will immediately make anything I learn a branch off previous knowledge. I am made aware of this when I am talking to people at parties about what they do. There is an awkward first minute or two where I am struggling to keep up, then I’ll ask or they’ll volunteer the connection. “Well it’s similar to BLANK.” “Oh! I know about blank! How is it different?”
In both cases, once I had a starting place, I could keep adding on new information. Even as a teenager, if I had a starting place I could tack on a ton of new knowledge, so long as I followed the path I saw ahead. As an adult the path forward a little clearer (i know where to look now) and I know where I am! Everything has a connection, and I’m never more than an arm’s length from comfortable knowledge.
Okay, so about how being cool sucks.
Every time I’ve tried to be cool, it meant either proudly proclaiming a lack of knowledge (“what is dungeons and dragons”) or pretending to know something I didn’t (“yeah, they do sound like sonic youth mixed with velvet underground”). And that makes it very difficult to learn what I want to learn. If it’s not cool to learn too much about something, it’s like trying to go for a walk and having a cop stop you 1 block from your house in every direction. Every time that I pretended that I knew something, I spent the entire time in an awkward space. Like knowing that I’m in a pine forest, but not knowing if I’m in North America or Europe.
I want to communicate this problem in the clearest way possible to my students. I feel like I had to actively fight the tendency of the school system to make me fit in, to make me learn what they wanted and not “waste my time” learning more. And all of my classmates made it very uncool to geek-out and learn something really in depth. But EVERYONE geeks-out on something. Just because you geek-out on football and I geek-out on music and math doesn’t make you better than me. It doesn’t make me smarter than you.
This is our* model for humans and learning. Everyone wants to geek out on something. So my job is to create a culture that removes most** of the inhibitions to geeking out.
our* – I can’t take all the credit, this has been an ongoing discussion with my wife
most** – well I don’t want everyone totally absorbed in their little world.
# What does that final grade mean?
I have control on how a final grade is calculated – but not how it is used.
When my students leave my class, they’ll get one of five letters on their transcript. It’s a secret code that is distilled from my teaching, their learning, their performance and the heavy thumb of my value system (do I count zeros, allow re-takes, grade homework?).
The really crazy part about this whole process is how that code gets shipped off to some college, scholarship, university or whatever. They ship the code – but NOT THE KEY! How crazy is that?
And to make it worse, they do their best to break the code based purely on assumptions! Some people count a B in an AP class as more than an A in a regular class. How do they know that? I assume they treat all Algebra II grades as the same. How do they know that? They don’t. It’s a huge assumption that leads to a loophole for students to game the system. Some students have caught on to this, and they know it’s all about the letter grade until they finish their undergraduate degree.
I don’t think the problem is grade inflation – the problem is the grade is meaningless.
next – what this means for my grades
# Educational perspective
I’m going to be working with a teaching candidate (student teacher) this coming year. I’m already psyched, because the two conversations that we’ve had have left us both wondering where the time went.
As I automatically reflected on the second meeting, I realized that I’m already starting to, sort of…. groom him into my philosophy of education. That sounds sort of sinister, but once I explain a bit more it might not.
Over the past two years of teaching and from psychology classes before that, I’ve come to a couple of tenets of dispositions that I think are essential to being a good teacher.
1) Every student can get it, and getting it is only dependent on their previous experiences (inside and outside the classroom).
2) People are logical creatures. They may have false premises, or those premises may be simply emotions, but in the end what they are doing makes sense to them on some level. I feel that this is critical for empathy.
3) Every answer to any question is right on some level. I mean even silence is an answer (it could be “I’m scared of you” or “I have two conflicting possibilities” or “I’m bored”)
3a) People can tell when their incorrect answer doesn’t match up. If they ask, point out what parts of their thinking are correct, and they will be able to identify the rest. Let them know help is on offer, but do not help unless you are asked. Respect them enough to let them think for themselves.
4) I can control my behaviors. Through these, and only through these can I do anything about my attitudes. The same is true for everyone else in the world, so it’s pointless to talk about changing attitudes or intentions unless it is through changing behaviors.
5) More important than relevance for getting students interested is success. If they don’t feel successful on some level, then no amount of relevance is going to make them try your subject.
I know there are a few people already reading this blog. What needs to be added to the list? What needs to be clarified? What are your fundamental dispositions?
# thank you teaching for…
totally stolen from dan meyer.
Teaching has taught me when to be a leader, and to be extremely good at sussing out true motivations of behavior.
Outside of the classroom, I used to take charge simply for the sake of leading. This has lead to some hostile situations. Now I’m much better about hanging back, waiting until the moment when someone taking the lead is actually necessary, rather than simply what I want. (not perfect, but much better)
Because of teaching education, I had a fairly deep theoretical background in adolescent psychology and counseling. But in the classroom, I had to figure out something to do for the millions of times “i don’t know” comes up. Now I’m pretty good at asking “is [blank] what’s really going on” without being a total dick about it.
There is plenty more, but time management definitely isn’t one of them yet. Sorry teaching, but I’m good with the first two. | 3,490 | 15,152 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2017-26 | longest | en | 0.950633 |
http://betterlesson.com/lesson/reflection/17555/density-proof | 1,488,121,263,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501172017.60/warc/CC-MAIN-20170219104612-00303-ip-10-171-10-108.ec2.internal.warc.gz | 29,165,839 | 23,114 | ## Reflection: High Quality Task Density & The Earth - Section 3: Density "Proof"
This was an idea derived from a Japanese "lesson study" reading that I conducted, in which the math teacher slowly piece-mewled information that allowed students to solve a complex mathematics problem. In a sense, there are two things working here: students need to figuratively (and sometimes literally) put their heads together to solve the task, but they also need to actively select the information they need to do so. This ultimately pushes the cognitive load more onto the students in that they select the tools, procedures, and steps necessary to solve the problem.
Another element in what I wanted to do here was to have my students struggle with a problem. In thinking about how this went, I think it was interesting to let students struggle with this. I feel at times that class time was "wasted" in letting them grapple with a problem, and some students in particular groups seemed to reach frustration level quicker than I thought. I think a potential work around here would be to do this as a summative/post-lesson group assignment. We did it after an exceedingly brief introduction - something to the effect of "let's remind ourselves what density is - here's the formula, and here's what I was thinking..." and then launching into the problem noted in the video - how can an aircraft carrier float despite its massive weight? Can we prove its density? If I moved this until after students had "re-familiarized" themselves with the content and process, they might have built up a bit more intrinsic perseverance. Even so, I'm excited to try this again and continue to take a few instructional risks in my classroom.
Density Proof
# Density & The Earth
Unit 2: The Dynamic Earth
Lesson 3 of 16
## Big Idea: Students try to solve a unique challenge problem showing how an aircraft carrier can float in water.
Print Lesson
3 teachers like this lesson
Standards:
60 minutes
### Kane Koller
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Environment: Urban | 633 | 3,009 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2017-09 | latest | en | 0.966296 |
https://myriverside.sd43.bc.ca/linap2016/2017/11/ | 1,701,672,579,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100525.55/warc/CC-MAIN-20231204052342-20231204082342-00343.warc.gz | 485,719,473 | 17,531 | # EAL Skills：Charles Journal
Hello，Ms G
Here is my Doc for Journal Writing
Thank you
Lina Pan
A independent moment in my life-1aytttg
Nov.28,2017 Lina Pan
Journal Writing: a independent moment in your life
Learn to survive, learn to be independent, learn to be strong. When we are born, we have embarked on a path of life. This path has setbacks and ups and downs, but as long as you learn to survive, you will definitely take this path.
I remember one afternoon, my mother and I went shopping, shopping malls clothes can be so beautiful ah, a wide range of clothes is very attractive to everyone’s attention! I saw a favorite dress, involuntarily released her mother’s hand, intend To touch it. May be back, my mother is gone. Here is crowded, how can I do? I am anxious to turn around and start looking for a person in the mall. I found the top floor from the first floor has not found a mother’s figure, getting dark, I am more and more anxious. This is a timed broadcast coming, a thought in my mind, I can find the mall’s announcer aunt to help! I crowded from the crowd to the radio, aunt explained the situation, Aunt will broadcast a search for inspiration, and soon, my mother came. I pounced toward her mother’s arms, “Mom, scared me, I thought I could never see you again!” Said, grievances shed tears, but my mother smiled and touched my head, “children, You can find a way to overcome the difficulties, which shows that you have grown up, what are you crying? “On the way back, I clutched my mother’s hand for fear of losing again.
We must learn to survive, learn to be strong, we can’t over-rely on money, rely on our parents, and whenever something emerges, we rely on ourselves. This memorable experience, let me know to learn to be independent!
# Week 11 in Math 10
This is the last week to study Factoring Polynomial Expressions Lesson.At this week, we review about Polynomial Operations Lesson and Factoring Polynomial Expressions Lesson. We did two classes’ prastice text. And after Finished found that some defects exist. Let me know, this unit is a very living unit, learning points of knowledge can be combined with the knowledge of the first few units,so not only review the knowledge of this unit point, but also indirectly review the knowledge of the first few units.This also sounded the alarm for the review of my midterm test,Review is not a unit or two things, but all the knowledge points learned before. So to review the comprehensive words, you should then sort through the knowledge of previous learning points, and to do the same mistakes before doing the same, let your brain start to recall those points of knowledge.Here are some of the questions I did not quite understand, did wrong and hesitated when I did my review exercises on Friday.
# EAL Skills: Complex and Compound-Complex Sentences.
Compound – Complex Sentence
1）After I finish the meeting, my coffee was too cold, so I heated it in the microwave.
Complex Sentence
1)Because my coffee was too cold, I heated it in the microwave.
2)I heated my coffee in the microwave because it was too cold.
Compound – Complex Sentence
1)Sarah cried when her rabbit got sick , but she soon got better.
Complex Sentence
1)Sarah cried but she soon got better.
2) Sarah soon got better although she cried.
Lina Pan
# Week 10 in Math 10
This week we learned factoring Polynomial Expression Lesson.Listen to the name to know the content of the previous unit to learn, because the last unit called Polynomial Operations Lesson.This unit compared to the previous unit, the difference is not not particularly large, but added some new knowledge in it.
I think this module is mainly to test the common factor of our previous study,having mastered this is the same calculation as the previous one. And I learned something new .
Notice that :
1) If the product is positive then the two integers must be either both positive or both negative.
2) If the product is negative, then one integer is positive and the other is negative.
3) In a prefect square trinomial, the first and last terms must be perfect squares and the middle term must be twice the product of the square roots of the first and last terms.
4) The method of factoring by splitting the value of b into two integers whose product is ac and whose sum is b is called the method of decomposition.
# Week 9 in Math 10
This week we learned something about Polynomial Operations Lesson, This is a week of challenges and fun.
Say he is mainly fun to use different drawing methods to solve and confirm the accuracy of the answer or not.Say he has the challenge mainly lies in the back to the more complicated questions we want to calculate, almost the things we just learned are concentrated in a problem, so the operation will be more prone to deviation.
The following is a post-experience for me:When doing the questions do not worry about removing the brackets, especially in the face of the square and the cubic power of the demolition of the brackets as soon as possible, will affect the accuracy of the back of the operation.So if you want to calculate the square or cubic can be written as the following form:(?x-?)(?x+?) or (?x-?)(?x-?)(?x-?),instead of directly into(?x2-?x+?x….) ,this will make it clear that you know what to do next. However, if this is done, there are still discrepancies with the answer, it is quite possible that even if the calculation is not passed, so my small solution can not be guaranteed.
(This is where I made the question is wrong, I hope this kind of question is not very clear or troublesome students help). # The picture write by myself and choose on the math book.
1.Read page 23 about Food and Housing. What kind of foods do the Arctic People eat? Would you eat the same food? Describe what their housing looks like? Could you live in the Arctic?
Inuit people have a variety of diets, including three seals, three major fish, including white salmon, arctic and trout, walrus, white whale and some reindeer. They are hunting for Canadian and blue goose, ducks, sea pigeons, seagulls, terns, collecting eggs and berries. But there are almost no birds, reindeer or fish in Grise Fiord and Resolute Bay; instead, their diet is mainly limited to seals, walruses and polar bears.
The high Arctic Islands were built in 1926 in the Arctic Islands protected area, and RCMP banned Inuit hunting musk cows. I wouldn’t eat the same food.
Live in the tent in the winter, because there is not enough snow to build a snow house, every morning up like a ball rolled up.No,I couldn’t, it’s too cold.
2.Look at page 32. Why are people relocating all the time?
There are not many wild animals for hunting. Since hunting and fishing was their main source of food, they were forced to move around, following the seasonal migration patterns of area animals.
3.Look at page 43. Why are there so many Indigenous peoples of the Arctic countries? Did the number surprise you? Yes or No? Why or Why not?
Indigenous peoples lived in the Arctic for thousands of years. It is estimated that the proportion of indigenous population accounts for about 10 per cent of the total population of the Arctic. The Arctic has more than 40 different races. The Arctic is now living in 20 ethnic groups, with a total population of about 2 million, with a very wide geographical distribution around the Arctic Ocean. It’s surprise me, because I did not expect the Arctic in the cold, food shortage areas, but also with so many people in this hard place to survive, and people still a lot.
Because they live here for at least 4,000 years,they are thousands of years ago from Asia through the Bering Strait ice bridge to the Americas, because it was the century glacier, the straits frozen, you can go directly to North America.
# Week 8 in Math 10
This week we use knowledge about Trigonometry to solving triangles and problems.During I solving problems , I have a little trouble on it. Here is a question I will not do. I think I will do it, but my answer is inconsistent with the correct answer later. I really have no clue, so I raised the question. After teaching the students to explain, I know how to do it.So I used to do wrong because I used 9 years of age on the vertex, as well as parallel theory, so the answer is very different from the true answer In fact this question or should be used Trigonometry to solve the problem.When he explained half of the time he re-painted a triangle and just the answer we had just written, I was very puzzled. He explained to me that the original is not every question of the map is accurate. So when faced with such a type of problem, in order to do the problem can be re-painted, so that will not be due to the number of too many wrong.I think his little trick is good, so I think when I encounter a similar problem the next time you can also use this trick.
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1B METHODS LECTURE NOTES Richard Jozsa, DAMTP Cambridge [email protected] October 2011
PART III: Inhomogeneous ODEs; Fourier transforms
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64
THE DIRAC DELTA FUNCTION
The Dirac delta function and an associated construction of a so-called Green’s function will provide a powerful technique for solving inhomogeneous (forced) ODE and PDE problems. To motivate the introduction of the delta function consider the set F of (suitably well behaved) real functions f : R → R. Any such function g may be viewed as a map G from F to R viz. for any f we set Z ∞ g(x)f (x) dx (1) G(f ) = −∞
i.e. g appears as the kernel of an integral with variable f (and here we are just assuming that the integrals exist). Such “functions of a function” are called functionals. These functionals are clearly linear in their argument f and we ask: do we get all linear maps from F to R in this way? Well, consider the following (clearly linear) functional L(f ) = f (0). What should the kernel be? If we call it δ(x) we would need Z ∞ δ(x)f (x) dx = f (0) for all f .
(2)
−∞
If we’re thinking of δ as being a bonafide continuous function then if c 6= 0, we should have δ(c) = 0 – because if δ(c) 6= 0 we could choose an f that is nonzero R ∞only very near to c and eq. (2) could be made to fail. Next choosing f (x) = 1 we need −∞ δ(x) dx = 1 yet δ(x) = 0 for all x 6= 0! Thus intuitively the graph of δ(x) must enclose unit area but over a vanishingly small base i.e. we need an “infinite spike” at x = 0. Such objects that are not functions in the usual sense but have well defined properties such as eq. (2) are called generalised functions or distributions. It is possible to develop a mathematically rigorous theory of them and below we’ll outline how to do this. But for this “methods” course we will use them only via a set of rules (cf below) motivated intuitively (but the rules can be justified in the rigorous theory). One possible approach to making sense of the delta function is to view it as the limit of a sequence of (ordinary, integrable) functions Pn (x) that should have the property Z ∞ lim Pn (x)f (x) dx = f (0) (3) n→∞
−∞
R∞ and we also normalise them to have −∞ Pn (x) dx = 1 for all n. Such sequences are not unique – it can be shown (not obvious in all cases – we’ll see eqs. (5,6) in Fourier transforms later!) that each of the following sequences has the required properties:
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20
20
20
15
15
15
10
10
10
5
5
5
0
0
0
−5 −1
0
1
−5 −1
0
1
−5 −1
0
1
Figure 1: Pn (x) for the top-hat, the Gaussian, and sin(nx)/(πx), showing convergence towards the delta function, for n = 2, 4, 8, 16, 32 (thin solid, dashed, dotted, dot-dashed, thick solid).
|x| < n1 ; 0 otherwise, n 2 2 Pn (x) = √ e−n x π
Pn (x) =
n 2
Pn (x) =
sin(nx) . πx
(4) (5) (6)
These are shown in the figure. These examples also show that the limit in eq. (3) cannot be taken inside the integral ? to form a “limit kernel” δ(x) = limn→∞ Pn (x) since Pn (0) → ∞ in all cases! Note that the sequence in eq. (6) does not even have Pn (c) → 0 for c 6= 0 e.g. Pn (π/2) = sin(nπ/2) π 2 /2 2 which is ±2/π for all odd n. Indeed in eq. (3) contributions from f (c) for c 6= 0 are eliminated by Pn oscillating faster and faster near x = c and thus contributing zero effect by increasingly exact cancellations of the oscillations with envelope f for any suitably continuous f .
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Physical significance of delta functions In physics the delta function models point sources in a continuum e.g. suppose we have a unit point charge at x = 0 (in R ∞one dimension). Then its charge density ρ(x) should satisfy ρ(x) = 0 for x 6= 0 and −∞ ρ(x) dx = total charge = 1 i.e. ρ(x) = δ(x) and the physical intuition is well modelled by the sequence eq. (4). In mechanics delta functions model impulses e.g. for a particle in oneR dimension with momentum p = mv, Newton’s t law gives dp/dt = F so p(t2 ) − p(t1 ) = t12 F dt. If a particle is struck impulsively (e.g. hammer blow) at t = 0 within interval (t1 , t2 ), with force acting over aR vanishingly small t time ∆t, and resulting in a finite momentum change ∆p = 1 say, then t12 F dt = 1 and F is nonzero only very near t = 0. So in the limit of vanishing time interval ∆t, F (t) = δ(t) models the impulsive force. The delta function was introduced by P. A. M. Dirac in the 1930s from considerations of position and momentum in quantum mechanics.
6.1
Properties of the delta function
• Note that in the basic defining property eq. (2) we can take the range of integration to be any interval [a, b] that contains x = 0, because we can replace f by the function f˜ that agrees with f on [a, b] and is zero outside [a, b] (or alternatively we can use a sequence such as eq. (4)). If [a, b] does not contain x = 0 then the integral is zero. Properties of the delta function can be intuitively obtained by manipulating the integral in eq. (2) in standard ways, manipulating δ(x) as though it were a genuine function. (You are asked to do some of these on exercise sheet 3). The validity of this procedure can be justified in a rigorous theory of generalised functions. • Substituting x0 = x − c in the integral below we get the so-called sampling property: Z b f (c) a < c < b; f (x)δ(x − c) dx = (7) 0 otherwise. a (i.e. δ(x − c) has the spike at x = c). • Substituting x0 = ax in the integral with kernel δ(ax) we get the scaling property: if a 6= 0 then
δ(ax) =
1 δ(x) |a|
By this equality (and those below) we mean that both sides in the same way if R∞ R ∞ behave 1 used as the kernel of an integral viz. −∞ δ(ax)f (x) dx = −∞ |a| δ(x)f (x) dx. • If f (x) has simple zeroes at n isolated points x1 , . . . , xn then δ(f (x)) will have spikes at these x values and n X δ(x − xk ) δ(f (x)) = . 0 (x )| |f k k=1 • If g(x) is integrable and continuous around x = 0 then g(x)δ(x) = g(0)δ(x).
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• Using the first bullet point above, we can see that the integral of the delta function is the Heaviside step function H(x): Z x 0 for x < 0 δ(ξ) dξ = H(x) = 1 for x > 0 −∞ (and usually we set H(0) = 1/2). More generally, delta functions characteristically appear in derivatives of functions with jump discontinuities e.g. if f on [a, b] has a jump discontinuity of size K at a < c < b and is differentiable for x < c and x > c, then f 0 (x) = g(x) + Kδ(x − c) where g(x) for x 6= c is the derivative of f (x) (and g(c) may be given any value). Indeed integrating this f 0 to a variable upper limit x, we regain f (x) with the delta function introducing the jump as x crosses x = c. • We can develop the notion of derivative of the delta function by formally applying integration by parts to δ 0 and then using the defining properties of δ itself: Z ∞ Z ∞ ∞ 0 δ(x − c)f 0 (x) dx δ (x − c)f (x) dx = [δ(x − c)f (x)]−∞ − −∞
−∞
0
= 0 − f (c) Similarly, provided f (x) is sufficiently differentiable, we can formally derive: Z ∞ f (x)δ (n) (x) dx = (−1)n f (n) (0). −∞
Example. Compute I =
R∞ 0
δ 0 (x2 − 1)x2 dx.
The substitution u = x2 − 1 gives dx = Z
√du 2 u+1
and
√ d 1 u+1 u+1 δ (u) du = − ( ) =− . 2 du 2 4 u=0 0
I= −1
Towards a rigorous mathematical theory of generalised functions and their manipulation (optional section). A (complex valued) function φ on R is called a Schwartz function if φ, φ0 , φ00 , . . . are all defined and lim xm φ(n) (x) = 0 for all m, n = 0, 1, 2, . . .. (8) x→±∞ 2
(e.g. φ(x) = p(x)e−x for any polynomial p(x)). These conditions requiring φ to be extremely well behaved asymptotically, will be needed to guarantee the convergence (existence) of integrals from −∞ to ∞ involving modified versions of the φ’s. Let S be the set of all Schwartz functions. Now let g be any continuous function on R that is “slowly growing” in the sense that g(x) =0 x→±∞ xn lim
for some n = 0, 1, 2, . . . .
(9)
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(e.g. x3 + x, e−x , sin x, x ln |x| but not ex , , e−x etc.) Then g defines a functional i.e. linear map, from S to C via Z ∞ g(x)φ(x) dx any φ ∈ S (10) g{φ} = −∞
(noting that the integral is guaranteed to converge by our restrictions on g and φ). Next note that the functional corresponding to g 0 can be related to that of g using integration by parts: Z ∞ Z ∞ 0 ∞ 0 g(x)φ0 (x) dx = g{−φ0 }. g (x)φ(x) dx = [gφ]−∞ − g {φ} = −∞
−∞
The boundary terms are guaranteed to be zero by eqs. (8,9). Thus g 0 {φ} = g{−φ0 }
(11)
which is derived (correctly) by standard manipulation of integrals of honest functions. Note that now the action of g 0 on any φ is defined wholly in terms of the action of the original g on a suitable modified argument viz. −φ0 . Next recall that there are more functionals on S than those arising from g’s as above. def For example we argued intuitively that the functional δ{φ} = φ(0) does not arise from any such g. Now suppose we have any given functional G{φ} and we want to establish a notion of derivative G0 (or any other modified version of G) for it. The key new idea here is the following procedure: we first consider G’s that arise from honest functions g and then express the action of g 0 (or any other modified version of g) on φ, entirely in terms of the action of g itself on a suitable modified argument (as we did above in eq. (11) for the case of derivatives). We then use the latter formula with g replaced by any G as the definition of G0 i.e. we define G0 {φ} to be G{−φ0 } for any G. For example we get def δ 0 {φ} = δ{−φ0 } = −φ0 (0) where the last equality used the (given) definition of δ itself. The finalR result is the same as what we would get (non-rigorously) if we “pretended” that ∞ δ{φ} = −∞ δ(x)φ(x) dx for some imagined (“generalised”) function δ(x) on R, and then R∞ formally manipulated it the integral −∞ δ 0 (x)φ(x) dx. Similarly the translation (x → x − a) and scaling (x → ax) and other rules that we listed previously can all be rigorously justified using the above Schwartz function formalism. This formalism also extends in a very natural way to make rigorous sense of Fourier transforms of generalised functions that we’ll encounter in chapter 8. If you’re interested in learning more about generalised functions, I’d recommend looking at chapter 7 of D. Kammler, “A first course in Fourier analysis” (CUP). - warning. If we think of functionals G as generalised functions γ via G{φ} = RExample ∞ γ(x)φ(x) dx ≡ Gγ {φ} then there is a temptation to add and multiply the γ’s just −∞ like we do routinely for ordinary functions. For linear combinations this works fine i.e. Gc1 γ1 +c2 γ2 = c1 Gγ1 + c2 Gγ2 but products are problematic: Gγ1 {φ}Gγ2 {φ} = 6 Gγ1 γ2 {φ}!
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This is because the γ’s are being used as kernels in integrals and using γ1 γ2 as a kernel is not the same as the product of using the two γ’s separately as kernels (even if they are bonafide g’s!). A dramatic example is γ1 (x) = x and γ2 (x) = δ(x). We have def (xδ){φ} = δ{xφ} but the latter is zero for all φ! (as xφ(x) = 0 at x = 0) i.e. we have (xδ) being identically zero as a generalised function even though neither x nor δ are zero as generalised functions.
Fourier series of the delta function For f (x) = δ(x) on −L ≤ x ≤ L we can formally consider a Fourier series (in complex form for convenience) f (x) =
∞ X
cn e
inπx L
n=−∞
with cn Thus eq.(2) gives cn =
1 2L
1 = 2L
Z
L
f (x)e
−inπx L
dx.
−L
for all n and ∞ 1 X inπx δ(x) = e L . 2L n=−∞
RL Indeed using the RHS expression in −L δ(x)f (x) dx we regain f (0) as the sum of the complex Fourier coefficients cn of f . Extending periodically to all R we get ∞ 1 X inπx e L . δ(x − 2mL) = 2L n=−∞ m=−∞ ∞ X
(12)
(which is an example of a more general result called Poisson’s summation formula for Fourier transforms).
Eigenfunction expansion of the delta function Revisiting SL theory, let Yn (x) be an orthonormal family of eigenfunctions for a SL problem on [a, b] with weight function w(x). For ξ ∈ (a, b), δ(x−ξ) satisfies homogeneous boundary conditions and we expect to be able to write δ(x − ξ) =
∞ X
Cn Yn (x)
n=1
with Z
b
w(x)Ym (x)δ(x − ξ) dx = w(ξ)Ym (ξ)
Cm = a
so δ(x − ξ) = w(ξ)
∞ X n=1
Yn (x)Yn (ξ).
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70
w(x) δ(x w(ξ)
− ξ) = δ(x − ξ) we can alternatively write δ(x − ξ) = w(x)
∞ X
Yn (x)Yn (ξ).
(13)
n=1
These expressions are consistent with the sampling property since if g(x) = is the eigenfunction expansion of g then Z
b
g(x)δ(x − ξ)dx = a
=
∞ X ∞ X m=1 n=1 ∞ X
Z Dm Yn (ξ)
P∞
m=1
Dm Ym (x)
b
w(x)Yn (x)Ym (x)dx, a
Dm Ym (ξ) = g(ξ),
m=1
by orthonormality. The eigenfunction expansion of the delta function is also intimately related to the eigenfunction expansion of the Green’s function that we introduced in §2.7 (as we’ll see later) and our next task is to develop a theory of Green’s functions for solving inhomogeneous ODEs.
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GREEN’S FUNCTIONS FOR ODEs
Using the concept of a delta function we will now develop a systematic theory of Green’s functions for ODEs. Consider a general linear second-order differential operator L on [a, b], i.e. Ly(x) = α(x)
d2 d y + β(x) y + γ(x)y = f (x), 2 dx dx
(14)
where α, β, γ are continuous, f (x) is bounded, and α is nonzero (except perhaps at a finite number of isolated points), and a ≤ x ≤ b (which may be ±∞). For this operator L, the Green’s function G(x; ξ) is defined as the solution to the problem LG = δ(x − ξ)
(15)
satisfying homogeneous boundary conditions G(a; ξ) = G(b; ξ) = 0. (Other homogeneous BCs may be entertained too, but for clarity will will treat only the given one here.) The Green’s function has the following fundamental property (established below): the solution to the inhomogeneous problem Ly = f (x) with homogeneous boundary conditions y(a) = y(b) = 0 can be expressed as Z y(x) =
b
G(x; ξ)f (ξ)dξ.
(16)
a
i.e. G is a the kernel of an integral operator that acts as an inverse to the differential operator L. Note that G depends on L, but not on the forcing function f , and once G is determined, we are able to construct particular integral solutions for any f (x), directly from the integral formula eq. (16). We can easily establish the validity of eq. (16) as a simple consequence of (15) and the sampling property of the delta function: Z b Z b L G(x; ξ)f (ξ)dξ = (LG) f (ξ) dξ, a a Z b = δ(x − ξ)f (ξ) dξ = f (x) a
and Z y(a) =
b
Z G(a; ξ)f (ξ) dξ = 0 =
a
b
G(b; ξ)f (ξ) dξ = y(b) a
since G(x; ξ) = 0 at x = a, b.
7.1
Construction of the Green’s function
We now give a constructive means for determining the Green’s function (without recourse to eigenfunctions, although the two approaches must provide equivalent answers). Our
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construction relies on the fact that for any x 6= ξ away from ξ, LG = 0 so G can be constructed in terms of solutions of the homogeneous equation. We proceed with the following steps: (1) Construct a general homogeneous solution for x < ξ from two linearly independent solutions y1 (x) and y2 (x) to the homogeneous problem Ly = 0, and so G(x; ξ) = A(ξ)y1 (x) + B(ξ)y2 (x), a ≤ x < ξ.
(17)
Here A and B are independent of x, but typically will later be chosen to be dependent on ξ, the upper endpoint of the interval being considered. (2) Construct a general solution for x > ξ from two linearly independent solutions Y1 (x) and Y2 (x) to the homogeneous problem Ly = 0: G(x; ξ) = C(ξ)Y1 (x) + D(ξ)Y2 (x), ξ < x ≤ b.
(18)
Note: Y1 and Y2 will generally not be the same as y1 and y2 (as other choices will be more convenient, see steps (3) and (4) below), but of course Y1 and Y2 must at least be linear combinations of y1 and y2 . Just like A and B above, C and D are independent of x, but typically will depend on ξ. There are thus four constants, (A, B, C, and D) and we need four further conditions to determine G uniquely. (3) Apply the homogeneous boundary condition at x = a to eliminate either A or B, using the fact that G(a; ξ) = 0 = Ay1 (a) + By2 (a). (19) (4) Similarly, apply the homogeneous boundary condition at x = b to eliminate either C or D, using G(b; ξ) = 0 = CY1 (b) + DY2 (b). (20) Note: we generally choose the forms of y1 , y2 , Y1 , Y2 to make the application of these two BCs simple. (5) The two remaining conditions that we need are associated with the properties at the point x = ξ. Firstly G(x, ξ) must be continuous there, (as we will argue below) and so Ay1 (ξ) + By2 (ξ) = CY1 (ξ) + DY2 (ξ). (6) The fourth and final condition is the jump condition, which is the requirement (as we will argue below) that x=ξ+ 1 dG = dx x=ξ− α(ξ) i.e. lim+
x→ξ
dG dG 1 − lim− = dx x→ξ dx α(ξ)
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and so CY10 (Ξ) + DY20 (Ξ) − Ay10 (Ξ) − By20 (Ξ) =
1 , ι(Ξ)
where Îą(x) is the coefficient of the second derivative in the operator L as defined in eq. (14). Note: often, but not always, the operator L has Îą(x) = 1. Also everything here assumes that the right hand side of eq. (15) is +1 Ă— δ(x − Ξ) whereas some applications have an extra factor Kδ(x − Ξ) which must first be scaled out before applying our formulae from here. (7) With the four conditions in (3), (4), (5) and (6), the Green’s function is uniquely determined, and then for any given forcing function f (x), the solution to the forced problem with homogeneous boundary conditions y(a) = 0 = y(b) is Z b y(x) = G(x; Ξ)f (Ξ)dΞ a Z x Z x = Y1 (x) C(Ξ)f (Ξ)dΞ + Y2 (x) D(Ξ)f (Ξ)dΞ a a Z b Z b B(Ξ)f (Ξ)dΞ. A(Ξ)f (Ξ)dΞ + y2 (x) + y1 (x) x
x
Rb
Rx Rb Note that the Ξ integral a is separated at x into two parts (i) a and (ii) x . In the range of (i) we have Ξ < x so the formula eq. (18) for G(x; Ξ) with coefficients C(Ξ), D(Ξ) is applicable even though this expression incorporated the BC at x = b. For (ii) we have x > Ξ so we use the G(x; Ξ) expression from eq. (17), that incorporated the BC at x = a.
The conditions on G(x; Ξ) at x = Ξ We are constructing G(x; Ξ) from pieces of homogeneous solutions (as required by eq. (15) for all x 6= Ξ) which are restrictions of well behaved (differentiable) functions on (a, b). At x = Ξ eq. (15) imposes further conditions on how the two homogeneous pieces must match up. The continuity condition: first suppose there was a jump discontinuity at x = Ξ. d2 d2 d 0 G âˆ? δ(x − Ξ) and dx Then dx 2 G âˆ? δ (x − Ξ). However eq. (15) shows that dx2 G cannot involve a generalised function of the form δ 0 (but only δ). Hence G cannot have a jump discontinuity at x = Ξ and it must be continuous there. Note however that it is fine for G0 to have a jump discontinuity at x = Ξ, and indeed we have: The jump condition: eq. (15) imposes a constraint on the size of any prospective jump discontinuity in dG/dx at x = Ξ, as follows – integrating eq. (15) over an arbitrarily small interval about x = Ξ gives: 2 Z Ξ+ Z Ξ+ Z Ξ+ Z Ξ+ dG dG Îą(x) dx + β(x) dx + Îł(x)G dx = δ(x − Ξ)dx = 1. dx2 dx Ξ− Ξ− Ξ− Ξ− Writing the three LHS integrals as T1 + T2 + T3 = 1.
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we have T3 → 0 as → 0 (since G is continuous and γ is bounded); T2 → 0 as → 0 (as dG/dx and β are bounded). Also ι is continuous and ι(x) → ι(Ξ) as → 0 so Z
Ξ+
T1 → ι(Ξ) lim →0
Ξ−
x=Ξ+ d2 G dG dx = Îą(Ξ) , 2 dx dx x=Ξ−
giving the jump condition.
Example of construction of a Green’s function Consider the problem Ly = −y 00 − y = f (x)
y(0) = y(1) = 0.
We follow our algorithm: 1. For 0 ≤ x < Ξ, G00 + G = 0, which suggests G = A cos x + B sin x (especially noting the x = 0 BC). 2. For Ξ < x ≤ 1, G00 + G = 0, which suggests (cf x = 1 BC!) G = C cos(1 − x) + D sin(1 − x). These expressions are motivated by looking at the BCs. G here could of course be expressed in terms of the basic sin x and cos x homogeneous solutions i.e. G = (C cos 1 + D sin 1) cos x + (C sin 1 − D cos 1) sin x, ˆ sin x. = Cˆ cos x + D but then see step 4 below! 3. Applying the x = 0 boundary condition G(0, Ξ) = 0 gives A = 0. 4. Applying the x = 1 boundary condition G(1, Ξ) = 0 gives simply C = 0. Note that ˆ D ˆ form we would have got Cˆ = −D ˆ tan 1 and things just get more messy for the C, in steps 5 and 6 below. 5. Therefore, G(x; Ξ) =
B sin x 0 < x < Ξ; D sin(1 − x) Ξ < x < 1.
Applying the continuity condition, B=D
sin(1 − Ξ) . sin Ξ
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6. Therefore, G(x; ξ) =
sin x D sin(1−ξ) 0 < x < ξ; sin ξ D sin(1 − x) ξ < x < 1.
In our operator L we have α(x) = −1 and so the jump condition is that sin(1 − ξ) cos x D [− cos(1 − x)]ξ+ − D = −1 sin ξ ξ− and we get sin ξ , sin 1 which then means that the Green’s function is: ( D=
G(x; ξ) =
sin(1−ξ) sin x sin 1 sin(1−x) sin ξ sin 1
0 < x < ξ; ξ < x < 1.
7. And so we are able to construct the complete solution to −y 00 − y = f (x) as (taking care to use the second G formula in the first integral, and the first G formula in the second integral!): Z Z sin x 1 sin(1 − x) x f (ξ) sin ξdξ + f (ξ) sin(1 − ξ)dξ. (21) y(x) = sin 1 sin 1 x 0
General solution for self adjoint Ls If L is self adjoint i.e. β(x) = dα/dx (and we have homogeneous BCs y(a) = y(b) = 0) then we can give a general formula for the Green’s function. Let y1 and y2 be two independent solutions of the homogeneous equation Ly = 0 which satisfy the BCs at x = a and x = b respectively (so not the same as y1 , y2 used above!) Then consider G(x; ξ) =
y1 (x)y2 (ξ)H(ξ − x) + y2 (x)y1 (ξ)H(x − ξ) J(y1 , y2 )
where J(y1 , y2 ) = α(x)W (y1 , y2 ) = α(x)[y1 (x)y20 (x) − y2 (x)y10 (x)] is called the conjunct (and W is the Wronskian that you met last year). Since L is self adjoint J(y1 , y2 ) is necessarily a (non-zero) constant (exercise – show this e.g. consider dJ/dx and use Ly1 = Ly2 = 0 so y1 Ly2 − y2 Ly1 = 0). Note that the H(ξ − x) and H(x − ξ) factors above imply that for x < ξ (resp. x > ξ) only the first (resp. the second) terms appear. Now it is straightforward to verify that the G(x; ξ) formula above satisfies the continuity and jump conditions at x = ξ (exercise – e.g. consider x → ξ from above and below separately, and for the jump condition remember that J(y1 , y2 ) is a constant). Thus the above formula provides a neat way of immediately writing down the Green’s function.
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This formula also shows that the Green’s function is symmetric in variables x and ξ i.e. G(x; ξ) = G(ξ; x) as featured in our example above, and also seen in our previous expression (in § 2.7) of G(x; ξ) in terms of a series of eigenfunctions Yn viz. G(x; ξ) =
∞ X Yn (x)Yn (ξ) n=1
λn
.
Green’s functions for inhomogeneous BCs Homogeneous BCs were essential to the notion of a Green’s function (since in eq. (16) the integral represents a kind of “continuous superposition” of solutions for individual ξ values). However we can also treat problems with inhomogeneous BCs using a standard trick to reduce them to homogeneous BC problems: (1) First find a particular solution yp to the homogeneous equation Ly = 0 satisfying the inhomogeneous BCs (usually easy). (2) Then using the Green’s function we solve Lyg = f
with homogeneous BCs y(a) = y(b) = 0.
(3) Then since L is linear the solution of Ly = f with the inhomogeneous BCs is given by y = yp + yg . Example. Consider −y 00 − y = f (x) with inhomogeneous BCs y(0) = 0 and y(1) = 1. The general solution of the homogeneous equation −y 00 − y = 0 is c1 cos x + c2 sin x and the (inhomogeneous) BCs require c1 = 0 and c2 = 1/ sin 1 so yp = sin x/ sin 1. Using the Green’s function for this L calculated in the previous example, we can write down yg (solution of Ly = f with homogeneous BCs) as given in eq. (21) and so the final solution is Z Z sin x sin(1 − x) x sin x 1 + f (ξ) sin ξdξ + f (ξ) sin(1 − ξ)dξ. y(x) = sin 1 sin 1 sin 1 x 0
Equivalence of eigenfunction expansion of G(x; ξ) For self adjoint Ls (with homogeneous BCs) we have two different expressions for the Green’s function G(x; ξ) =
y1 (x)y2 (ξ)H(ξ − x) + y2 (x)y1 (ξ)H(x − ξ) J(y1 , y2 )
and G(x; ξ) =
∞ X Yn (x)Yn (ξ) n=1
λn
.
(22)
(23)
The first is in terms of homogeneous solutions y1 and y2 (Ly = 0) whereas the second is in terms of eigenfunctions Yn and eigenvalues λn of the corresponding SL system LYn = λn Yn w(x).
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In § 2.7 we derived eq. (23) without any mention of delta functions, but it may also be quickly derived using the eigenfunction expansion of δ(x − ξ) given in eq. (13) viz. δ(x − ξ) = w(x)
∞ X
Yn (x)Yn (ξ).
n=1
Indeed viewing ξ as a parameter and writing the Green’s function as an eigenfunction expansion ∞ X G(x; ξ) = Bn (ξ)Yn (x) (24) n=1
then LG = δ(x − ξ) gives LG =
∞ X
Bn (ξ)LYn (x) =
∞ X
n=1
Bn (ξ)λn w(x)Yn (x)
n=1
= δ(x − ξ) = w(x)
∞ X
Yn (x)Yn (ξ).
n=1
Multiplying the two end terms by Ym (x) and integrating from a to b we get (by the weighted orthogonality of the Ym ’s) Bm (ξ)λm = Ym (ξ) and then eq. (24) gives the expression eq. (23). Remark. Note that the formula eq. (23) requires that all eigenvalues λ be nonzero i.e. that the homogeneous equation Ly = 0 (also being the eigenfunction equation for λ = 0) should have no nontrivial solutions satisfying the BCs. Indeed the existence of such solutions is problematic for the concept of a Green’s function, as providing an inverse operator for L: if nontrivial y0 with Ly0 = 0 exist, then the inhomogeneous equation Ly = f will not have a unique solution (since if y is any solution then so is y + y0 ) and then the operator L is not invertible. This is just the infinite dimensional analogue of the familiar situation of a system of linear equations Ax = b with non-invertible coefficient matrix A (and indeed a matrix is non-invertible iff it has nontrivial eigenvectors belonging to eigenvalue zero). Example. As an illustrative example of the equivalence, consider again Ly = −y 00 − y on [a, b] = [0, 1] with BCs y(0) = y(1) = 0. The normalised eigenfunctions and corresponding eigenvalues are easily calculated to be √ Yn (x) = 2 sin nπx λn = n2 π 2 − 1 so by eq. (23) we have G(x; ξ) = 2
∞ X sin nπx sin nπξ . 2π2 − 1 n n=1
(25)
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On the other hand we had in a previous example the expression constructed from homogeneous solutions: G(x; ξ) =
sin(1 − x) sin ξ H(x − ξ) + sin x sin(1 − ξ) H(ξ − x) sin 1
and using trigonometric addition formulae we get G(x; ξ) = cos x sin ξ H(x − ξ) + sin x cos ξ H(ξ − x) − cot 1 sin x sin ξ .
(26)
Now comparing eqs (25,26) and viewing ξ as a parameter and x as the independent variable, we see that the equivalence of the two expressions amounts to the Fourier sine series in eq. (25) of the function (for each ξ) in eq. (26) f (x) = P cos x sin ξ H(x − ξ) + sin x cos ξ H(ξ − x) − cot 1 sin x sin ξ ∞ = n=1 bn (ξ) sin nπx. √ The Fourier coefficients are given as usual by (noting that sin nπx has norm 1/ 2): Z 1 f (x) sin nπx dx bn (ξ) = 2 0
and a direct (but rather tedious) calculation (exercise?...) gives bn (ξ) =
2 sin nπξ n2 π 2 − 1
as expected. (In this calculation note that the Heaviside functions H(x − ξ) and H(ξ − x) R1 R1 Rξ merely alter the integral limits from 0 to ξ and 0 respectively).
7.2
Physical interpretation of the Green’s function
We can think of the expression b
Z y(x) =
G(x; ξ) f (ξ) dξ a
as a ‘summation’ (integral) of individual point source effects, of strength f (ξ) with G(x; ξ) characterising the elementary effect (as a function of x) of a unit point source placed at ξ. To illustrate with a physical example consider again the wave equation for a horizontal elastic string with gravity and ends fixed at x = 0, L. If y(x, t) is the (small) transverse displacement we had T
∂ 2y ∂ 2y − µg = µ ∂x2 ∂t2
0≤x≤L
y(0) = y(L) = 0.
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Here T is the (constant) tension in the string and µ is the mass density per unit length which may be a function of x. We consider the steady state solution ∂y/∂t = 0 i.e. y(x) is the shape of a (non-uniform) string hanging under gravity, and d2 y µ(x)g ≡ f (x) = 2 dx T
(27)
is a forced self adjoint equation (albeit having a very simple Ly = y 00 ). Case 1 (massive uniform string): if µ is constant eq. (27) is easily integrated and setting y(0) = y(L) = 0 we get the parabolic shape y=
µg x(x − L). 2T
Case 2 (light string with point mass at x = ξ): consider a point mass concentrated at x = ξ i.e. the mass density is µ(x) = mδ(x − ξ) with µ = 0 for x 6= ξ. For x 6= ξ the string is massless so the only force acting is the (tangential) tension so the string must be straight either side of the point mass. To find the location of the point mass let θ1 and θ2 be the angles either side. Then resolving forces vertically the equilibrium condition is mg = T (sin θ1 + sin θ2 ) ≈ T (tan θ1 + tan θ2 ) where we have used the small angle approximation sin θ ≈ tan θ (and y hanging, is negative). Thus mg ξ(ξ − L) y(ξ) = T L for the point mass at x = ξ. Hence from physical principles we have derived the shape for a point mass at ξ: ( x(ξ−L) mg 0 ≤ x ≤ ξ; L y(x) = × (28) ξ(x−L) T ξ ≤ x ≤ L; L which is the solution of eq. (27) with forcing function f (x) =
mg δ(x T
− ξ).
Next let’s calculate the Green’s function for eq. (27) i.e. solve LG =
d2 G = δ(x − ξ) dx2
with G(0; ξ) = G(L; ξ) = 0.
Summarising our algorithmic procedure we have successively: (1) For 0 < x < ξ, G = Ax + B. (2) For ξ < x < L, G = C(x − L) + D. (3) BC at zero implies B = 0. (4) BC at L implies D = 0. (5) Continuity implies C = Aξ/(ξ − L). (6) The jump condition gives A = (ξ−L) so finally L ( x(ξ−L) 0 ≤ x ≤ ξ; L G = ξ(x−L) ξ ≤ x ≤ L; L
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and we see that eq. (28) is precisely the Green’s function with a multiplicative scale for a point source of strength mg/T . Case 3 (continuum generalisation): if in case 2 we had several point masses mk at xk = ξk we can sum the solutions to get X mk g G(x; ξk ). (29) y(x) = T k For the continuum limit we imagine (N − 1) masses mk = µ(ξk )∆ξ placed at equal intervals xk = ξk = k∆ξ with ∆ξ = L/N and k = 1, . . . , N − 1. Then by the Riemann sum definition of integrals, as N → ∞, eq. (29) becomes Z L g µ(ξ) G(x; ξ) dξ. y(x) = T 0 If µ is constant this function reproduces the parabolic result of case 1 as you can check by direct integration (exercise, taking care with the limits of integration).
7.3
Application of Green’s functions to IVPs
Green’s functions can also be used to solve initial value problems. Consider the problem (viewing now the independent variable as time t) Ly = f (t), t ≥ a, y(a) = y 0 (a) = 0.
(30)
The algorithm for construction of the Green’s function is very similar to the previous BVP method. As before, we want to find G such that LG = δ(t − τ ). 1. Construct G for a ≤ t < τ as a general solution of the homogeneous equation: G = Ay1 (t) + By2 (t). 2. But now, apply both boundary (i.e. initial) conditions to this solution: Ay1 (a) + By2 (a) = 0, Ay10 (a) + By20 (a) = 0. Since y1 and y2 are linearly independent the determinant of the coefficient matrix (being the Wronskian) is non-zero and we get A = B = 0, and so G(t; τ ) = 0 for a ≤ t < τ! 3. Construct G for τ ≤ t, again as a general solution of the homogeneous equation: G = Cy1 (t) + Dy2 (t). 4. Now apply the continuity and jump conditions at t = τ (noting that G = 0 for t < τ ). We thus get Cy1 (τ ) + Dy2 (τ ) = 0, Cy10 (τ ) + Dy20 (τ ) =
1 , α(τ )
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where α(t) is as usual the coefficient of the second derivative in the differential operator L. This determines C(τ ) and D(τ ) completing the construction of the Green’s function G(t; τ ). For forcing function f (t) we thus have b
Z y(t) =
f (τ )G(t; τ ) dτ a
but since G = 0 for τ > t (by step 2 above) we get Z t y(t) = f (τ )G(t; τ ) dτ a
i.e. the solution at time t depends only on the input (forcing) for earlier times a ≤ τ ≤ t. Physically this is a causality condition. Example of an IVP Green’s function Consider the problem d2 y + y = f (t), y(0) = y 0 (0) = 0. 2 dt
(31)
Following our procedure above we get (exercise) 0 0 ≤ t ≤ τ; G(t; τ ) = C cos(t − τ ) + D sin(t − τ ) t ≥ τ. Note our judicious choice of independent solutions for t ≥ τ : continuity at t = τ implies simply that C = 0, while the jump condition (α(τ ) = 1) implies that D = 1, and so 0 0 ≤ t ≤ τ; G(t; τ ) = sin(t − τ ) t ≥ τ, which gives the solution as Z
t
f (τ ) sin(t − τ )dτ.
y(t) =
(32)
0
7.4
Higher order differential operators
We mention that there is a natural generalization of Green’s functions to higher order differential operators (and indeed PDEs, as we shall see in the last part of the course). If Ly = f (x) is a nth -order ODE (with the coefficient of the highest derivative being 1 for simplicity, and n > 2) with homogeneous boundary conditions on [a, b], then Z b y(x) = f (ξ)G(x; ξ)dξ, a
where
1B Methods • G satisfies the homogeneous boundary conditions; • LG = δ(x − ξ); • G and its first n − 2 derivatives continuous at x = ξ; • d(n−1) G/dx(n−1) (ξ + ) − d(n−1) G/dx(n−1) (ξ − ) = 1. See example sheet III for an example.
82
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8 8.1
83
FOURIER TRANSFORMS From Fourier series to Fourier transforms
Recall that Fourier series provide a very useful tool for working with periodic functions (or functions on a finite domain [0, T ] which may then be extended periodically) and here we seek to generalise this facility to apply to non-periodic functions, on the infinite domain R. We begin by imagining allowing the period to tend to infinity, in the Fourier series formalism. Suppose f has period T and Fourier series (in complex form) f (t) =
∞ X
cr eiwr t
(33)
r=−∞
. We can write wr = r∆w with frequency gap ∆w = 2π/T . The where wr = 2πr T coefficients are given by 1 cr = T
Z
T /2 −iwr u
f (u)e −T /2
∆w du = 2π
Z
T /2
f (u)e−iwr u du.
(34)
−T /2
R∞ For this integral to exist in the limit as T → ∞ we’ll require that −∞ |f (x)| dx exists, but then the 1/T factor implies that for each r, cr → 0 too. Nevertheless for any finite T we can substitute eq. (34) into eq. (33) to get Z ∞ X ∆w iwr t T /2 f (t) = e f (u)e−iwr u du 2π −T /2 r=−∞
(35)
Now recall the Riemann sum definition of the integral of a function g: ∞ X
Z
∆w g(wr ) →
g(w) dw −∞
r=−∞
with w becoming a continuous variable. For eq. (35) we take eiwr t g(wr ) = 2π
Z
T /2
f (u)e−iwr u du
(36)
−T /2
(with t viewed as a parameter) and letting T → ∞ we get Z ∞ Z ∞ 1 iwt −iwu dw e f (u)e du . f (t) = 2π −∞ −∞
(37)
(Actually here the function g in eq. (36) changes T → ∞ i.e. as ∆w → 0 but it R ∞ too, as −iwu has a well defined limit g∆w (wr ) → g0 (wr ) = −∞ f (u)e du so the limit of eq. (35) is still eq. (37) as given).
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Introduce the Fourier transform (FT) of f , denoted f˜: Z ∞ ˜ f (t)e−iwt dt f (w) =
(38)
−∞
and then eq. (37) becomes 1 f (t) = 2π
Z
f˜(w) eiwt dw
(39)
−∞
which is Fourier’s inversion theorem, recovering f from its FT f˜. Thus (glossing over issues of when the integrals converge) the FT is a mapping from functions f (t) to functions g(w) = f˜(w) where we conventionally use variable names t and w respectively. Warning:R different books give slightly different definitions of FT. Generally we can have ∞ f˜(w) = A −∞ f (t)e±iwt dt and then the inversion formula is Z ∞ 1 f˜(w)e∓iwt dt f (t) = 2πA −∞ (pairing opposite ± signs). Quite common is A = √12π , symmetrically giving the same constant for the FT and inverse FT formulas, but in this course we will always use A = 1 as above. Some other books also use e±2πiwt in the integrals. Remarks. • The dual pair of variables t and w above are referred to by different names in different applications. If t is time, w is frequency; if t is space x then w is often written as k (wavenumber) or p (momentum, especially in quantum mechanics) and k-space may be called spectral space. • the Fourier transform is an example of an integral transform with kernel e−iwt . There are other important integral transforms such as the Laplace transform with kernel e−st (s real, and integration range 0 to ∞) that you’ll probably meet later. • if f has a finite jump discontinuity at t then (just as for Fourier series) the inversion (t− ) . formula eq. (39) returns the average value f (t+ )+f 2
8.2
Properties of the Fourier transform
A duality property of FT and inverse FT Note that there is a close structural similarity between the FT formula and its inverse viz. eqs. (38,39). Indeed if a function f (t) has the function f˜(w) as its FT then replacing t by −t in eq. (39) we get Z ∞ 1 f (−t) = f˜(w) e−iwt dw. 2π −∞ Then from eq. (38) (with the variable names t and w interchanged) we obtain the Duality property: if g(x) = f˜(x) then g˜(w) = 2πf (−w).
(40)
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85
This is a very useful observation for obtaining new FTs: we can immediately write down the FT of any function g if g is known to be the FT of some other function f . Writing F T 2 for the FT of the FT we have (F T 2 f )(x) = 2πf (−x) or equivalently f (x) = 1 ˜˜ (f )(−x), and then F T 4 (f ) = 4π 2 f i.e. iterating FT four times on a function is just 2π the identity operation up to a constant 4π 2 .
Further properties of FT Let f and g have FTs f˜ and g˜ respectively. Then the following properties follow easily from the integral expressions eqs (38,39). (Here λ and µ are real constants). • (Linearity) λf + µg has FT λf˜ + µ˜ g; • (Translation) if g(x) = f (x − λ) then g˜(k) = e−ikλ f˜(k); • (Frequency shift) if g(x) = eiλx f (x) then g˜(k) = f˜(k − λ). Note: this also follows from the translation property and the duality in eq. (40). 1 ˜ f (k/λ); • (Scaling) if g(x) = f (λx) then g˜(k) = |λ| • (Multiplication by x) if g(x) = xf (x) then g˜(k) = if˜0 (k) (applying integration by parts in the FT integral for g). • (FT R ∞ of a derivative) dual to the multiplication rule (or applying integration by parts to −∞ f 0 (x)e−ikx dx) we have the derivative rule: if g(x) = f 0 (x) then g˜(k) = ik f˜(k) i.e. differentiation in physical space becomes multiplication in frequency space (as we had for Fourier series previously). The last property can be very useful for solving (linear) differential equations in physical space – taking the FT of the equation can lead to a simpler equation in k-space (algebraic equation for an ODE, or an ODE from a PDE if we FT on one or more variables cf. exercise sheet 3, question 12). Thus we solve the simpler problem and invert the answer back to physical space. The last step can involve difficult inverse-FT integrals, and in some important applications techniques of complex methods/analysis are very effective.
Example. (Dirichlet’s discontinuous formula) Consider the “top hat” (English) or “boxcar” (American) function 1 |x| ≤ a, f (x) = 0 |x| > a for a > 0. Its FT is easy to calculate: Z a Z −ikx ˜ f (k) = e dx = −a
a
−a
cos(kx) dx =
2 sin(ka) . k
so by the Fourier inversion theorem we get Z 1 ∞ ikx sin ka 1 |x| < a, e dk = 0 |x| > a. π −∞ k
(41)
(42)
(43)
Now setting x = 0 (so the integrand is even, and rewriting the variable k as x) we get Dirichlet’s discontinuous formula: π Z ∞ a > 0, 2 sin(ax) 0 a = 0, dx = (44) π x 0 − 2 a < 0,
1B Methods
86 =
π sgn(a), 2
(45)
(To get the a < 0 case we’ve simply used the fact that sin(−ax) = − sin(ax).) The above integral (forming the basis of the inverse FT of f˜) is quite tricky to do without the above FT inversion relations. It can be done easily using the elegant methods of complex contour integration (Cauchy’s integral formula) but a direct “elementary” method is the following: introduce the two parameter(!) integral Z ∞ sin(λx) −αx e dx (46) I(λ, α) = x 0 with α > 0 and λ real. Then
∂I = ∂λ
Z
∞ −αx
Z
∞ −(α+λi)x
dx ,
e dx = < 0 −(α+λi)x ∞ 1 −e α =< =< = 2 . α + λi 0 α + λi α + λ2 cos(λx)e
0
Now integration w.r.t. λ gives λ I(λ, α) = arctan + C(α) α with integration constant C(α) for each α. But from eq. (46) I(0, α) = 0 so C(α) = 0. Then considering I(λ, 0) = limα→0 I(λ, α) we have I(λ, 0) = limα→0 arctan(λ/α) which gives eq. (44).
8.3
Convolution and Parseval’s theorem for FTs
In applications it is often required to take the inverse FT of a product of FTs i.e. we ˜ want to find h(x) such that h(k) = f˜(k)˜ g (k) where f˜ and g˜ are FTs of known functions f and g respectively. Applying the definitions of the Fourier transform and its inverse, we can write Z ∞ 1 f˜(k)˜ g (k)eikx dk, h(x) = 2π −∞ Z ∞ Z ∞ 1 ikx −iku = g˜(k)e f (u)e du dk. 2π −∞ −∞ Now (assuming the f and g˜ are absolutely integrable) we can change the order of integration to write Z ∞ Z ∞ 1 ik(x−u) h(x) = f (u) g˜(k)e dk du, 2π −∞ −∞ Z ∞ h(x) = f (u)g(x − u) du = f ∗ g, (47) −∞
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87
This integrated combination h = f ∗ g is called the convolution of f and g. Convolution is called faltung (or ‘folding’) in German, which picturesquely describes the way in which the functions are combined – the graph of g is flipped (“folded”) about the variable vertical line u = x/2 and then integrated against f . By exploiting the dual structure of eq. (40) of FT and inverse FT, the above result readily shows (exercise) that a product of functions in physical (x) space has a FT that is the convolution of the individual FTs (with a 2π factor): Z ∞ 1 ˜ f˜(u)˜ g (k − u)du. (48) h(x) = f (x)g(x) → h(k) = 2π −∞ Parseval’s theorem for FTs An important simple corollary of (47) is Parseval’s theorem for FTs. Let g(x) = f ∗ (−x). Then Z ∞ g˜(k) = f ∗ (−x)e−ikx dx −∞ ∗ ∗ Z ∞ Z ∞ −iky ikx f (y)e dy = f˜∗ (k). f (−x)e dx = = −∞
−∞
Thus by eq. (47) the inverse FT of f˜(k)f˜∗ (k) is the convolution of f (x) and f ∗ (−x) i.e. Z ∞ Z ∞ 1 ∗ f (u)f (u − x)du = |f˜(k)|2 eikx dk. 2π −∞ −∞ Setting x = 0 gives the Parseval formula for FTs: Z ∞ Z ∞ 1 2 |f (u)| du = |f˜(k)|2 dk. 2π −∞ −∞
(49)
Thus the FT as a mapping from functions f (x) to functions f˜(k) preserves the squared norm of the function (up to a constant 1/2π factor).
8.4
The delta function and FTs
There is a natural extension of the notion of FT to generalised functions. Here we will indicate some basic features relating to the Dirac delta function, using intuitive arguments based on formally manipulating integrals (without worrying too much about whether they converge or not). The results can be rigorously justified using the Schwartz function formalism for generalised functions that we outlined in chapter 7, and we will indicate later how this formalism embraces FTs too (optional section below). Writing f (x) as the inverse of its FT we have Z ∞ Z ∞ 1 ikx −iku f (x) = e f (u)e du dk, 2π −∞ −∞ Z ∞ Z ∞ 1 ik(x−u) = f (u) e dk du. 2π −∞ −∞
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88
Comparing this with the sampling property of the delta function, we see that the term in the square brackets must be a representation of the delta function: Z ∞ 1 eik(x−u) dk. (50) δ(u − x) = δ(x − u) = 2π −∞ R∞ 1 1eikx dk and we get the FT In particular setting u = 0 we have δ(−x) = δ(x) = 2π −∞ pair f (x) = δ(x) ↔ f˜(k) = 1. (51) Note that this is with directly putting f (x) = δ(x) into the basic FT R ∞also consistent −ikx ˜ formula f (k) = −∞ f (x)e dx and using the sampling property of δ. Eq. (51) can be used to obtain further FT pairs by formally applying basic FT rules to it. The dual relation eq. (40) gives Z ∞ ˜ f (x) = 1 ↔ f (k) = e−ikx dx = 2πδ(k). −∞
From the translation property of FTs (or applying the sampling property of δ(x − a)) we get, Z ∞ ˜ f (x) = δ(x − a) ↔ f (k) = δ(x − a)e−ikx dx = e−ika −∞
and the dual relation f (x) = eiax ↔ f˜(k) = 2πδ(k − a). Then e±iθ = cos θ ± i sin θ gives f (x) = cos(ωx) ↔ f˜(k) = π [δ(k + ω) + δ(k − ω)] , f (x) = sin(ωx) ↔ f˜(k) = πi [δ(k + ω) − δ(k − ω)]
(52)
So, a highly localised signal in physical space (e.g. a delta function) has a very spread out representation in spectral space. Conversely a highly spread out (yet periodic) signal in physical space is highly localised in spectral space. This is illustrated too in exercise sheet 3, in computing the FT of a Gaussian bell curve.
Towards a rigorous theory of FTs of generalised functions (optional section) Many of the FT integrals above actually don’t technically converge! (e.g. FT(1) which we identified as 2πδ(k)). Here we outline how to make rigorous sense of the above results, extending our previous discussion of generalied functions in terms of Schwartz functions. If f is any suitably regular ordinary function on R and φ is any Schwartz function then it is easy to see from the FT definition eq. (38) that Z ∞ Z ∞ ˜ ˜ dx. f (x)φ(x) dx = f (x)φ(x) −∞
−∞
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Here as usual the tilde denotes FT and we have written the independent variables as x rather than t or k. Hence if F and F˜ denote the functionals on S (space of Schwartz functions) associated to f and f˜ then we have ˜ F˜ {φ} = F {φ} and we now define(!) the FT of any generalised function by this formula. For example for the delta function we get Z ∞ (a) (b) (c) ˜ ˜ ˜ 1.φ(t) dt. δ{φ} = δ{φ} = φ(0) = −∞
(where (a) is by Rdefinition, (b) is the action of the delta function and (c) is by setting ∞ ˜ w = 0 in φ(w) = −∞ φ(t)e−iwt dt). Thus comparing to the formal notation of generalised function kernels Z ∞ ˜ ˜ δ{φ} ≡ δ(x)φ(x) dx −∞
˜ we have δ(x) = 1 as expected. Similarly all our formulas above may be rigorously established. Example (FT of the Heaviside function H). 1 x>0 1 x=0 H(x) = 2 0 x<0 We begin by noting that H(x) = 21 (sgn(x) + 1) and recalling Dirichlet’s discontinuous formula Z ∞ Z ∞ sin kx sin kx dk = 2 dk = πsgn(x). k k −∞ 0 Thus for the FT of sgn(x) we have Z
sgn{φ} f = = = = =
˜ ds sgn(s) φ(s) Z−∞ ∞ Z ∞ ˜ sin us du ds φ(s) πu −∞ −∞ Z ∞Z ∞ ius e − e−ius ˜ φ(s) du ds 2πiu −∞ Z−∞ ∞ 1 [φ(u) − φ(−u)] du −∞ iu Z ∞ 1 2 φ(x) dx. −∞ ix
(In the second last line we have used the Fourier inversion formula, and to get the last line we have substituted u = x and u = −x respectively in the two terms of the integral). Thus comparing with Z ∞
sgn{φ} f =
−∞
sgn(k)φ(k) f dk
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we get sgn(k) f =
2 . ik
Finally (using FT(1) = 2πδ(k)) we get 1 1 ˜ H(k) = FT (sgn(x) + 1) = + πδ(k). 2 ik It is interesting and instructive here to recall that H 0 (x) = δ(x) so from the differentiation rule for FTs we get ˜ ˜ ik H(k) = δ(k) =1 ˜ even though H(k) is not ik1 but equals ik1 + πδ(k)! However this is not inconsistent since ˜ using the correct formula for H(k), we have ˜ ik H(k) = ik(
1 + πδ(k)) = 1 + iπkδ(k) ik
and as noted previously (top of page 67) xδ(x) is actually the zero generalised function, ˜ so we can correctly deduce that ik H(k) = 1!
8.5
FTs and linear systems, transfer functions
FTs are often used in the systematic analysis of linear systems which arise in many engineering applications. Suppose we have a linear operator L acting on input I(t) to give output O(t) e.g. we may have an amplifier that can in general change the amplitude and phase of a signal. Using FTs we can express the physical input signal via an inverse FT as Z ∞ 1 iwt ˜ I(w)e dw I(t) = 2π −∞ which is called the synthesis of the input, expressing it as a combination of components with various frequencies each having amplitude and phase given by the modulus and ˜ argument respectively, of I(w). The FT itself is known as the resolution of the pulse (into its frequency components): Z ∞ ˜ I(w) = I(t)e−iwt dt. −∞
˜ Now suppose that L modifies the amplitudes and phases via a complex function R(w) to produce the output Z ∞ 1 iwt ˜ ˜ O(t) = R(w) I(w)e dw. (53) 2π −∞ ˜ R(w) is called the transfer function of the system and its inverse FT R(t) is called the response function.
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˜ Warning: in various texts both R(t) and R(w) are referred to as the response or transfer function, in different contexts. Thus ˜ R(w) =
Z
R(t)e
−iwt
1 R(t) = 2π
dt
−∞
Z
∞ iwt ˜ R(w)e dw
−∞
and looking at eq. (53) we see that R(t) is the output O(t) of the system when the input ˜ has I(w) = 1 i.e. when the input is δ(t) i.e. LR(t) = δ(t). Hence R(t) is closely related to the Green’s function when L is a differential operator. ˜ ˜ Eq. (53) also shows that O(t) is the inverse FT of a product R(w) I(w) so the convolution theorem gives Z ∞
I(u)R(t − u) du.
O(t) =
(54)
−∞
Next consider the important extra physical condition of causality. Assume that there is no input signal before t = 0 i.e. I(t) = 0 for t < 0, and suppose the system has zero output for any t0 if there was zero input for all t < t0 (e.g. “the amplifier does not hum..”) so R(t) = 0 for t < 0 (as R(t) is output for input δ(t) which is zero for t < 0). Then in eq, (54) the lower limit can be set to zero (as I(u) = 0 for u < 0) and the upper limit to t (as R(t − u) = 0 for u > t): Z t O(t) = I(u)R(t − u) du (55) 0
which is formally the same as our previous expressions in chapter 7.3 with Green’s functions for IVPs.
General form of transfer functions for ODEs In many applications the relationship between input and output is given by a linear finite order ODE: ! ! m n X X dj di Lm [I(t)] = bj j [I(t)] = ai i [O(t)] = Ln [O(t)]. (56) dt dt j=0 i=0 For simplicity here we will consider the case m = 0 so the input acts directly as a forcing term. Taking FTs we get ˜ ˜ I(ω) = a0 + a1 iω + . . . + an−1 (iω)n−1 + an (iω)n O(ω) 1 ˜ R(ω) = . [a0 + a1 iω + . . . + an−1 (iω)n−1 + an (iω)n ] So the transfer function is a rational function with an nth degree polynomial as the denominator. ˜ The denominator of R(w) can be factorized into a product of roots of the form (iω − cj )kj P for j = 1, . . . , J (allowing for repeated roots) where kj ≥ 1 and Jj=1 kj = n). Thus ˜ = R
(iω − c1
)k1
1 . . . (iω − cJ )kJ
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and using partial fractions, this can be expressed as a simple sum of terms of the form Γmj , 1 ≤ m ≤ kj (iω − cj )m where Γmj are constants. So we have ˜ R(ω) =
XX j
m
Γmj (iω − cj )m
and to find the response function R(t) we need to Fourier-invert a function of the form ˜ m (ω) = h
1 , m ≥ 0. (iω − α)m+1
We give the answer and check (by Fourier-transforming it) that it works. Consider the function h0 (t) = eαt for t > 0, and zero otherwise i.e. h0 (t) = 0 for t < 0. Then Z ∞ ˜ e(α−iω)t dt h0 (ω) = 0 (α−iω)t ∞ e = α − iω 0 1 = iω − α provided <(α) < 0. So h0 (t) is identified. Remark: If <(α) > 0 then the above integral is divergent. Indeed recalling the theory of linear constant coefficient ODEs, we see that the cj ’s above are the roots of the auxiliary equation and the equation has solutions with terms ecj t which grow exponentially if <(cj ) > 0. Such exponentially growing functions are problematic for FT and inverse FT integrals so here we will consider only the case <(cj ) < 0 for all j, corresponding to ‘stable’ ODEs whose solutions do not grow unboundedly as t → ∞. Next consider the function h1 (t) = teαt for t > 0 and zero otherwise, so h1 (t) = th0 (t). Recalling the “multiplication by x” rule for FTs viz: if f (x) → f˜(w) then xf (x) → if˜0 (w), we get ˜ 1 (w) = i d h dw
1 iw − α
=
1 (iw − α)2
(which may also be derived directly by evaluating the FT integral
R∞ 0
teαt e−iwt dt).
Similarly (or proof by induction) we have (for Re(α) < 0) tm eαt 1 t > 0; m! hm (t) = ↔ h˜m (ω) = , m≥0 0 t≤0 (iω − α)m+1
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and so it is possible to construct the output from the input using eq. (55) for such stable systems easily. Physically, see that functions of the form hm (t) always decay as t → ∞, but they can increase initially to some finite time maximum (at time tm = m/|α| if α < 0 and real for example). We also see that R(ξ) = 0 for ξ < 0 so in eq. (54) the upper integration limit can be replaced by t, so O(t) depends only on the input at earlier times, as expected. Example: the forced damped oscillator The relationship between Green’s functions and response functions can be nicely illustrated by considering the linear operator for a damped oscillator: d d2 (57) y + 2p y + (p2 + q 2 )y = f (t), p > 0. 2 dt dt Since p > 0 the drag force −2py 0 acts opposite to the direction of velocity so the motion is damped. We assume that the forcing term f (t) is zero for t < 0. Also y(t) and y 0 (t) are also zero for t < 0 and we have initial conditions y(0) = y 0 (0) = 0. Taking FTs we get
so
(iω)2 y˜ + 2ipω y˜ + (p2 + q 2 )˜ y = f˜ f˜ ˜ f˜ = = y˜ R −ω 2 + 2ipω + (p2 + q 2 )
and taking inverse FTs we get Z
t
R(t − τ )f (τ )dτ, Z ∞ 1 eiω(t−τ ) dω. R(t − τ ) = 2π −∞ p2 + q 2 + 2ipω − ω 2 y(t) =
0
with
Now consider LR(t − τ ), using this integral formulation, and assuming that formal differentiation within the integral sign is valid d d2 R(t − τ ) + 2p R(t − τ ) + (p2 + q 2 )R(t − τ ) dt2 dt Z ∞ 1 (iω)2 + 2ipω + (p2 + q 2 ) iω(t−τ ) = e dω 2π −∞ p2 + q 2 + 2ipω − ω 2 Z ∞ 1 = eiω(t−τ ) dω = δ(t − τ ), 2π −∞ using (50). Therefore, the Green’s function G(t; τ ) is the response function R(t − τ ) by (mutual) definition. On sheet 3, question 1 you are asked to fill in the details, computing both R(t) and the Green’s function explicitly. Example. FTs can also be used for ODE problems on the full line R so long as the functions involved are required to have suitably good asymptotic properties for the FT integrals to exist. As an example consider d2 y − A2 y = −f (x) 2 dx
−∞<x<∞
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such that y → 0, y 0 → 0 as |x| → ∞, and A is a positive real constant. Taking FTs we have f˜ y˜ = 2 , (58) A + k2 and we seek to identify g(x) such that g˜ =
A2
1 . + k2
Consider h(x) =
e−µ|x| , µ > 0. 2µ
Since h(x) is even, its FT can be written as Z ∞ ˜h(k) = < 1 exp [−x(µ + ik)] dx µ 0 1 1 1 = < = 2 µ µ + ik µ + k2 so we have identified g(x) =
e−A|x| . 2A
The convolution theorem then gives Z ∞ 1 y(x) = f (u) exp(−A|x − u|)du. 2A −∞
(59)
This solution is clearly in the form of a Green’s function expression. Indeed the same expression may be derived using the Green’s function formalism of chapter 7, applied to the infinite domain (−∞, ∞) (and imposing suitable asymptotic BCs on the Green’s function for |x| → ∞).
8.6
FTs and discrete signal processing
Another hugely important application of FTs is to the theory and practice of discrete signal processing i.e. the manipulation and analysis of data that is sampled at discrete times, as occurring for example in any digital representation of music or images (CDs, DVDs) and all computer graphics and sound files etc. Here we will give just a very brief overview of a few fundamental ingredients of this subject, which could easily justify an entire course on its own. Consider a signal h(t) which is sampled at evenly spaced time intervals ∆ apart: hn = h(n∆)
n = . . . − 2, −1, 0, 1, 2, . . .
The sampling rate or sampling frequency is fs = 1/∆ (samples per unit time) and introduce the angular frequency ws = 2πfs (samples per 2π time). Consider two complex exponential signals with pure frequencies h1 (t) = eiw1 t = e2πif1 t
h2 (t) = eiw2 t = e2πif2 t .
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These will have precisely the same ∆ samples when (w1 − w2 )∆ is an integer multiple of 2π i.e. when (f1 − f2 )∆ is an integer. We can avoid this possibility by choosing ∆ so that |f1 − f2 |∆ < 1. A general signal h(t) is called wmax - bandwidth limited if its ˜ ˜ is supported entirely in [−wmax , wmax ]. Writing FT h(w) is zero for |w| > wmax i.e. h fmax = wmax /2π we see that we can distinguish different fmax -bandwidth limited signals with ∆-sampling if 1 (60) 2fmax ∆ < 1 i.e. ∆ < 2fmax which is called the Nyquist condition (with Nyquist frequency f = 1/2∆ for sampling interval ∆). For a given sampling interval ∆ violation of eq. (60) leads to the phenomenon of aliasing i.e. an inability to distinguish properties of signals with frequencies exceeding half the sampling rate. Via the relation (f1 − f2 )∆ = integer, such high frequencies f1 will be “aliased” onto lower frequencies f2 within [−fmax , fmax ]. Reconstruction of a signal from its samples – the sampling theorem Suppose g is a wmax -bandwidth limited signal and write fmax = wmax /2π. From the definition of the Fourier transform we have Z ∞ g(t)e−iwt dt, g˜(w) = −∞
and the inversion formula with the bandwidth limit gives Z ∞ Z wmax 1 1 iwt g˜(w)eiwt dw. g˜(w)e dw = g(t) = 2π −∞ 2π −wmax Now let us set the sampling interval to be ∆ = 1/(2fmax ) = π/wmax so samples are taken at tn = n∆ = nπ/wmax giving Z wmax 1 iπnw g(tn ) = gn = g˜(w) exp dw. 2π −wmax wmax Now recalling our formulas for complex Fourier series coefficients of a function on [a, b] = [−wmax , wmax ] (cf page 12 of lecture notes) we see that the g−n are wmax /π times the complex Fourier coefficients cn for a function g˜p (w) which coincides with g˜(w) on (−wmax , wmax ) and is extended to all R as a periodic function with period 2wmax i.e. ∞ π X −inπw g˜p (w) = gn exp . wmax n=−∞ wmax Therefore the actual (bandwidth limited) Fourier transform of the original function is ˜ the product of the periodically repeating g˜p (ω) and a box-car function h(w) as defined in (41): 1 |w| ≤ wmax , ˜h(w) = , 0 otherwise. ˜ g˜(w) = g˜p (w)h(w) " # ∞ π X −inπw ˜ = gn exp h(w), wmax n=−∞ wmax
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This is an exact equality: the countably infinite discrete sequence of samples g(n∆) = gn of a bandwidth-limited function g(t), completely determines its Fourier transform g˜(w). We can now apply the Fourier inversion formula to exactly reconstruct the full signal g(t) for all t, from its samples at t = n∆. Assuming that swapping the order of integration and summation is fine, we have Z ∞ 1 g˜(w)eiwt dw, g(t) = 2π −∞ Z wmax ∞ X nπ 1 exp iw t − = gn dw, 2wmax n=−∞ wmax −wmax ∞ X 1 exp(i[wmax t − πn]) − exp(−i[wmax t − πn]) = gn nπ 2wmax n=−∞ i t− wmax
= =
∞ X n=−∞ ∞ X
gn
sin(wmax t − πn) , wmax t − πn
g(n∆) sinc[wmax (t − n∆)]
n=−∞
where sinc(t) =
sin t . t
In summary, the bandwidth-limited function can be represented (for continuous time) exactly by this representation in terms of its discretely sampled values, with the continuous values being filled in by this expression which is known as the Shannon-Whittaker sampling formula. This result is called the (Shannon) sampling theorem. Such full reconstruction of bandwidth-limited functions, by multiplying by the ‘Whittaker sinc’ function or cardinal function sinc(t) centred on the sampling points and summing, is at the heart of all digital music reproduction.
8.7
The discrete Fourier transform
For any natural number N the discrete Fourier transform (mod N ) denoted DFT is defined to be the N by N matrix with entries 2πi
[DFT]mn = e− N
mn
m, n = 0, 1, 2, . . . , (N − 1).
(61)
Note that here we number rows and columns from 0 to N − 1 rather than the more conventional 1 to N . Thus the first row and √ column are always all 1’s√and DFT is a symmetric matrix. Some books include a 1/ N pre-factor (just like a 1/ 2π factor that we did not use in our definition of FT). In further pure mathematical theory (group representation theory) there is a general notion of a Fourier transform on any group G. It may be shown that the Fourier transform of functions on R, the Fourier series of periodic finctions on R, and DFT, are all examples of this single construction, just using different groups viz. respectively R with addition,
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the circle group {eiθ : 0 ≤ θ < 2π} with multiplication, and the group ZN of integers mod N with addition. Correspondingly DFT enjoys a variety of nice properties analogous to those we’ve seen for Fourier series and transforms. We now derive some of these from the definition of DFT above. So far we have that DFT is a linear mapping from CN to CN . The inverse DFT √1 DFT N
is actually a unitary matrix i.e. the inverse is the adjoint (conjugate transpose):
−1 † 1 1 √ DFT = √ DFT N N
so
† 1 1 √ DFT √ DFT = I N N
so
1 DFT† . N To see this write w = e−2πi/N and note that each row or column of DFT is a geometric sequence 1, r, r2 , . . . , rN −1 with ratio r being a power of w (depending on the choice of row or column). Now recall the basic property of such series of N th roots of unity: if r = wa for any a 6= 0 (or more generally a is not a multiple of N ) then r 6= 1 but rN = 1 so 1 − rN 1 + r + r2 + . . . + rN −1 = =0 1−r but if a = 0 (or is a multiple of N ) then r = 1 so DFT−1 =
1 + r + r2 + . . . + rN −1 = 1 + 1 + . . . + 1 = N. √ Using this, it is easy to see (exercise) that the set of rows (or set of columns) of DFT/ N is an orthonormal set of vectors in CN and hence that ( N1 DFT† )(DFT) = I. A Parseval equality (also known as Plancherel’s theorem) For any a = (a0 , . . . , aN −1 ) ∈ CN , if a ˜ = (˜ a0 , . . . , a ˜N −1 ) = DFT a then a·a=
1 a ˜·a ˜. N
This follows immediately from the general relationship of adjoints to inner products viz. (Au, v) = (u, A† v). Hence (˜ a, a ˜) = (DFT a , DFT a) = (a , DFT† DFT a) = N (a , a). Convolution property For any a = (a0 , . . . , aN −1 ) and b = (b0 , . . . , bN −1 ) introduce the convolution c = a ∗ b defined by N −1 X ck = am bk−m k = 0, . . . , N − 1 m=0
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(where the subscript k − m is computed mod N to be in 0, . . . , N − 1). Then DFT maps convolutions into straightforward component-wise products: c˜k = a ˜k ˜bk for each k = 0, . . . , N − 1 . To see this, use the DFT and convolution definitions to write c˜k =
N −1 X l=0
cl wkl =
N −1 N −1 X X
am bl−m wkl
m=0 l=0
and then substitute for index l introducing p = l − m mod N to get a ˜k ˜bk directly on RHS (exercise). Remark (FFT) (Optional). DFT has a wide variety of applications in both pure and applied mathematics. One of its most important properties is the existence of a so-called fast Fourier transform (FFT) which refers to the following idea: suppose we want to compute DFT a for some a ∈ CN and N is large. Direct matrix multiplication will require O(N 2 ) basic arithmetic operations (additions and multiplications) – for each of the N components of DFT a we need O(N ) operations to compute the inner product of a row of DFT and the column vector a. Now because of the very special structure of the DFT matrix entries (as a pattern of roots of unity) it can be shown that if N = 2M is a power of 2, then there is a faster algorithm to compute DFT a, requiring only O(N log N ) basic arithmetic operations! – a dramatic exponential (N → log N ) saving in run time. This is the so-called FFT algorithm (cf wikipedia for an explicit description, which we omit here), often attributed to a paper by Cooley and Tukey from 1965, but it already appears in the notebooks of Gauss from 1805 – he invented it on the side, as an aid to speed up his by-hand calculations of trajectories of asteroids! An important application of FFT is to provide a “super-fast” algorithm for integer multiplication. If a = an−1 . . . a0 and b = bn−1 . . . b0 are two n digit numbers (written as usual in terms of decimal or binary digits) then the direct calculation of the product c = ab by long multiplication takes O(n2 ) elementary arithmetic (additions and multipliPn−1 steps m cations of single digits). However if we write a = m=0 am 10 and similarly for b, then P2n−2 a simple calculation shows that c = m=0 cm 10m with C = (c0 , c1 , . . . , c2n−2 ) being the convolution A ∗ B of the (2n − 1)-dimensional vectors A = (a0 , . . . , an−1 , 0, . . . , 0) and B = (b0 , . . . , bn−1 , 0, . . . , 0) of the digits extended by zeroes. With this observation we can construct a fast multiplication algorithm along the following lines: represent a and b as extended vectors A and B of their digits; take DFTs (O(n log n) steps with FFT) to ˜ form the entry-wise product (O(n) steps); finally take inverse DFT to get get A˜ and B; the components of A ∗ B i.e. the desired result c (O(n log n) steps again). Thus we have O(n log n) steps in all compared to O(n2 ) for standard long multiplication. For example (assuming all the big-O constants to be 1) if the numbers have 1000 digits, then we get (only!) about n log n = 3000 steps compared to about n2 = 1 million steps! Example (finite sampling) The sampling theorem showed how in some circumstances (viz. bandwidth limited signals) we can reconstruct a signal h(t) and hence also its FT exactly, from a discrete
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but infinite set of samples h(n∆). But realistically in practice we can obtain only a finite number N of samples say hm = h(tm ) for tm = m∆ and m = 0, 1, . . . , N − 1, and ˜ at some points. The DFT arises in the hence expect only an approximation to its FT h, computation of such approximations. Suppose h(t) is negligible outside [0, T ] and we sample N points as above with ∆ = T /N . ˜ at w with frequency f = w/2π. By approximating the FT integral Now consider the FT h as a Riemann sum using our sampled values hm in [0, T ], we have Z ∞ ˜ h(t)e−2πif t dt (62) h(w) = −∞ N −1 X
≈ ∆
hm e−2πif (m∆) .
(63)
m=0
Since FT is an invertible map, from our data of N samples of h(t) we’d like to estimate ˜ n ) of the FT at frequencies fn = wn /2π. We know that if |f | exceeds the N samples h(w Nyquist frequency fc = 1/2∆ associated to the sampling rate ∆, then features with such frequencies become indistinguishably aliased into frequencies with |f | < fc . Hence we’ll choose our N discrete frequency values fn to be equally spaced in [−fc , fc ] i.e. fn =
n N∆
n = −N/2, . . . , 0, . . . , N/2 − 1.
(64)
Subsitiuting these into the above approximation we get ˜ n) = ∆ h(f
N −1 X
2πi
hm e− N
mn
m=0
˜ and h, ˜ and h whose components are the discrete values of h so if we introduce vectors h then the above gives ˜ = −∆ DFT(h) h (the minus sign arising from the fact that in eq. (64) n runs from −N/2 to N/2 − 1 rather than from 0 to N − 1). Remark (quantum computing) (optional) √ DFT (or more precisely the unitary matrix QFT = DFT/ N ) has a fundamental significance in quantum mechanics. In quantum theory, states of physical systems are described by complex vectors (finite dimensional for many physical properties) and physical time evolution is represented by a unitary operation on the vectors. Thus QFT represents an allowable quantum physical state transformation and further to FFT (as a mathematical transformation of a list of N complex numbers), it can be shown that QFT as a quantum physical transformation on a physical quantum state can be physically implemented in O((log N )2 ) elementary quantum steps i.e. exponentially faster than the O(N log N )-time FFT on classical data. This is a very remarkable property with momentous consequences: if we had a “quantum computer” that could implement elementary quantum operations on quantum states as its computational steps (rather than the conventional Boolean operations on classical bit values) then the superduper-fast QFT could
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be exploited (as a consequence of a little bit of number theory...) to provide an algorithm for integer factorisation that can factorise any integer of n digits in O(n3 ) steps, whereas the best known classical factoring algorithm has a profoundly slower run time of about 1/3 O(en ) steps. Because of this time speed up, such a quantum computer would be able to efficiently (i.e. realistically in actual practice) crack currently used cryptosystems (e.g. RSA) whose security relies on the hardness of factoring, and decipher much encrypted secret information that exists the public domain. Because all known classical factoring algorithms are so profoundly slower, such decryption is in practice impossible on a regular computer - it can in principle be done but would typically need an estimated running time well exceeding the age of the universe to finish the task. The tripos part III course titled “Quantum computation� provides a detailed exposition of QFT and the amazing quantum factoring algorithm.
IB Mathematical Methods 1c (Cambridge)
IB Mathematical Methods 1c (Cambridge), astronomy, astrophysics, cosmology, general relativity, quantum mechanics, physics, university degre... | 22,192 | 68,164 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2017-34 | latest | en | 0.918832 |
http://mathoverflow.net/feeds/question/83023 | 1,369,144,826,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368700074077/warc/CC-MAIN-20130516102754-00060-ip-10-60-113-184.ec2.internal.warc.gz | 170,322,183 | 2,839 | How to do integrals involving two Bessel functions and another function? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T14:00:26Z http://mathoverflow.net/feeds/question/83023 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/83023/how-to-do-integrals-involving-two-bessel-functions-and-another-function How to do integrals involving two Bessel functions and another function? renphysics 2011-12-09T02:56:54Z 2011-12-10T03:17:04Z <p>I often encounter the integrals in the following form:</p> <p>$\int_0^\infty{\rm Bessel}(ax)\cdot{\rm Bessel}(bx)\cdot f(cx)dx$,</p> <p>where Bessel can be $J$, $N$, $H^{(1)}$, $H^{(2)}$, $I$, or $K$; and $f(x)$ can be $\sin(x)$, $e^x$, etc. For example,</p> <p>$\int_0^\infty K_\nu(ax)I_\nu(bx)\cos(cx)dx=\frac{1}{2\sqrt{ab}}Q_{\nu-1/2}(\frac{a^2+b^2+c^2}{2ab})$,$\qquad{\rm Re}(a)>|{\rm Re}(b)|$, $c>0$, ${\rm Re}(\nu)>-1/2$</p> <p>I already know the result of this integral because it is in Gradshtein & Ryzhik's book. Sometimes the integration is with respect to the order of the Bessel function.</p> <p>Both Mathematica 8 and Maple 15 cannot do this kind of integrals. When the integral involves two Bessel functions or two other special functions, Mathematica and Maple usually cannot do even if the integral has a closed-form result. My questions are as follows:</p> <ol> <li><p>Is there any general theory about how to do this kind of integrals? (I wonder how the authors of that book did the above integral.)</p></li> <li><p>I know there is a Mathematica package "<a href="http://www.risc.jku.at/research/combinat/software/HolonomicFunctions/" rel="nofollow">HolonomicFunctions</a>." It seems that this package can help, but it does not seem very straightforward to obtain the final result. This package can verify the integrals with already known results, but can it do new integrals as above? Are there any better ways to deal with these integrals by computer?</p></li> </ol> http://mathoverflow.net/questions/83023/how-to-do-integrals-involving-two-bessel-functions-and-another-function/83053#83053 Answer by Jacques Carette for How to do integrals involving two Bessel functions and another function? Jacques Carette 2011-12-09T13:01:59Z 2011-12-09T13:01:59Z <p>The best tool for trying to deal with such integrals is Fredéric Chyzak's <a href="http://algo.inria.fr/chyzak/mgfun.html" rel="nofollow">MGfun package</a> (available as part of the <a href="http://algo.inria.fr/libraries/" rel="nofollow">Algolib</a> library).</p> <p>For your example, you should get a system of differential equations (for the integrand) for $a,b,c$ and $x$; you can leave $\nu$ as a parameter, or get a (mixed) difference equation for it. Then using this package, you can try to do elimination, which will give you a new system for the answer. </p> <p>Note that this method is a vast generalization of using the Meijer G-function as an intermediary (since MeijerG is the most general function whose series expansion / asymptotic expansion at 0 has coefficients satisfy a recurrence of order <strong>1</strong>). </p> http://mathoverflow.net/questions/83023/how-to-do-integrals-involving-two-bessel-functions-and-another-function/83102#83102 Answer by Tom Dickens for How to do integrals involving two Bessel functions and another function? Tom Dickens 2011-12-10T03:17:04Z 2011-12-10T03:17:04Z <p>Take a look at "A Treatise on the Theory of Bessel Functions" by Watson. There is a long chapter on integrating Bessel functions over the infinite range $0-\infty$. </p> <p>In addition, I think a Mellin transform approach could very well get you what you want. The idea is to find Mellin transforms of the functions in the integrand (most often separated into two parts) and use Parseval's theorem to write the integral as a contour integral. Then one can often move the contour over the poles of the integrand and generate a series representation of the integral, which can sometimes be identified as some known special function. </p> <p>You could look at the paper <a href="http://www.risc.jku.at/publications/download/risc_3924/AlgorithmicMellinTransform.pdf" rel="nofollow">http://www.risc.jku.at/publications/download/risc_3924/AlgorithmicMellinTransform.pdf</a> which briefly describes the method and shows a computer algebra technique for getting the final result. There is also a great, simple book by Fikoris called "Mellin Transform Method for Integral Evaluation." </p> <p>Good luck, Tom</p> | 1,251 | 4,490 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2013-20 | latest | en | 0.73238 |
http://slideplayer.com/slide/7512065/ | 1,713,349,607,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817146.37/warc/CC-MAIN-20240417075330-20240417105330-00028.warc.gz | 32,448,959 | 18,679 | # 8.1 Solving Systems of Linear Equations by Graphing
## Presentation on theme: "8.1 Solving Systems of Linear Equations by Graphing"— Presentation transcript:
8.1 Solving Systems of Linear Equations by Graphing
To solve by graphing, graph both linear equations. This gives an approximate solution. Algebraic methods are more exact (next 2 sections). If the graphs intersect at one point the system is consistent and the equations are independent.
8.1 Solving Systems of Linear Equations by Graphing
If the graphs are parallel lines, there is no solution and the solution set is . The system is inconsistent. If the graphs represent the same line, there are an infinite number of solutions. The equations are dependent.
8.2 Solving Systems of Linear Equations by Substitution
Solving by substitution: Solve for a variable Substitute for that variable in the other equation Solve this equation for the remaining variable Put your solution back into either of the original equations to solve for the other variable Check your solution with the other equation
8.2 Solving Systems of Linear Equations by Substitution
Example: From the first equation we get y=2x-7, so substituting into the second equation:
8.2 Solving Systems of Linear Equations by Substitution
If when using substitution both variables drop out and you get something like: 10=6 The system inconsistent and there is no solution (parallel lines) If when using substitution both variables drop out and you get something like: 10=10 The system dependent and every solution of one line is also on the other (same lines)
8.3 Solving Systems of Linear Equations by Elimination
Solving systems of equations by elimination: Write equations in standard form (variables line up) Multiply one of the equations to get coefficients of one of the variables to be opposites Add (or subtract) equations – so that one variable drops out Solve for the remaining variable. Plug you solution back into one of the original equations and solve for the other variable.
8.3 Solving Systems of Linear Equations by Elimination
Example: Multiply the second equation by 3 to get: Adding equations you get:
8.3 Solving Systems of Linear Equations by Elimination
If when using elimination both variables drop out and you get something like: 10=6 The system inconsistent and there is no solution (parallel lines) If when using elimination both variables drop out and you get something like: 10=10 The system dependent and every solution of one line is also on the other (same lines)
8.4 Linear Systems of Equations in Three Variables
Linear system of equation in 3 variables: Example:
8.4 Linear Systems of Equations in Three Variables
Graphs of linear systems in 3 variables: Single point (3 planes intersect at a point) Line (3 planes intersect at a line) No solution (all 3 equations are parallel planes) Plane (all 3 equations are the same plane)
8.4 Linear Systems of Equations in Three Variables
Solving linear systems in 3 variables: Eliminate a variable using any 2 equations Eliminate the same variable using 2 other equations Eliminate a different variable from the equations obtained from (1) and (2)
8.4 Linear Systems of Equations in Three Variables
Solving linear systems in 3 variables: Use the solution from (3) to substitute into 2 of the equations. Eliminate one variable to find a second value. Use the values of the 2 variables to find the value of the third variable. Check the solution in all original equations.
8.5 Applications of Linear Systems of Equations
Solving an applied problem by writing a system of equations: Determine what you are to find – assign variables Draw a diagram, figure or make a chart of information. Write the system of equations Solve the system using substitution or elimination Answer the question from the problem.
8.5 Applications of Linear Systems of Equations
Mixture problem: How many ounces of a 5% solution must be added to a 20% solution to get 10 ounces of 12.5% solution. Let x = # ounces of 5% solution Let y = # ounces of 20% solution
8.5 Applications of Linear Systems of Equations
Solution to mixture problem in 2 variables: | 900 | 4,140 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2024-18 | latest | en | 0.891565 |
null | null | null | null | null | null | April 19, 1 9 9 6
BU's Scientology Connection
| article (part 1) | article (part 2) | secrets | tangled web | scientology's letter | dan kennedy's response | more responses |
The secrets of Scientology
by Dan Kennedy
(Hypertext links by Ron Newman and Thor Iverson)
To Scientologists, Steven Fishman is an apostate who's spread vicious lies about the church to which he used to belong. To church critics, he's a hero who's exposed the truth about Scientology.
Evaluating Fishman's credibility is difficult, to say the least. He's an ex-convict who served a prison term for financial crimes that he claims he was ordered to commit. Church officials deny there was any such order, and they deny just about everything else Fishman has said about his old organization -- including his allegation that they ordered Fishman to murder his psychiatrist and commit suicide.
In 1991, the church sued Fishman for defamation in connection with comments he made to Time magazine for a cover story on Scientology. (The church also sued Time, and the case has yet to be resolved.) In response, Fishman introduced into the court records a number of secret, copyrighted church documents pertaining to top-level, "Operating Thetan" ( or OT) teachings. It was these documents that Arnaldo Lerma excerpted and uploaded to alt.religion.scientology in 1995.
Church critics say Fishman's "declaration" -- portions of which appear below -- is an excellent summary of those documents. The full text of the declaration can be found on "The Real Steve Fishman Home Page" (http://www.xs4all.nl/~fishman).
Who is Steven Fishman?
Was he really the biological father of Jesus Christ?
The Church of Scientology said he was. His auditors, Nancy Witkowski, Catherine Fox, Leah Abady, Ann Glushakow, Margaret Supak, Richard Reese, John Eastment, Hans Stahli, and Ray Mithoff all checked Steven Fishman on the e- meter over a period of years and told him over and over again that he was the biological father of Jesus Christ, and that it was Steve Fishman's responsibility to de-Christianize the planet by exposing the lie and the myth of the immaculate conception, and thereafter bring all of Christianity into Scientology as the largest FSM (Field Staff member), or conversion movement, of planet earth.
You see, the Church of Scientology is an anti-Christian religion. On Saint Hill Special Briefing Course Tape #112, L. Ron Hubbard, the founder of Dianetics and Scientology who died a fugitive from justice in 1986, said, "Christ died for his own sins." In a confidential student briefing in 1980, Hubbard described Christ as a "pedophile" and as "lover of young boys." [Note: church officials have denied that Hubbard ever called Jesus a pedophile.]
I was told by Fred Hare, the Organizational Executive Secretary (OES) of the Mission of Fort Lauderdale, in 1987, that Christ was a later life cycle of the evil Emperor "Xenu," who, according to Hubbard, freeze-dried clusters of thetans or souls and transported them from an alien planet, Helatrobus, to Earth, which Hubbard called "Teegeeack." The word "Teegeeack," according to Russell Means, the leader of the American Indian Movement, is the name of a tribe of American Indians who settled and lived in Oregon in the early nineteenth century. The word "Teegeeack" means "tribe." Yet, Hubbard tried to pass off false definitions and concepts in his poorly written, illogically contrived "Advanced Technology," known also as the "Upper Level Materials," to unsuspecting Scientologists. In the "Advanced Technology," Hubbard talked about Xenu exploding clusters or freeze-dried packages of thetans inside volcanoes located in Las Palmas and Hawaii. Scientific evidence refutes completely that there was any explosion seventy-five million years ago in Las Palmas, because there was no volcanic activity present there at that point in time.
Scientology is a Satanic cult which has its origins in the work of Aleister Crowley, a well-known Satanist. Hubbard was a disciple and student of Crowley between 1947 and 1949.
The "Advanced Technology" refers to "Body Thetans," or "BTs." After the purported volcanic explosions, and all of the thetans were released into the atmosphere, some attached themselves to and occupied the bodies of "genetic entities," and these were the bodies of animals, plants, and fish. Hubbard's cosmology does not quarrel with evolution, but rather supports it. According to the "Advanced Technology," each living thing, or "genetic entity," is occupied and controlled by a thetan, and since there are far more thetans in the atmosphere than live organisms on earth, there are many unattached thetans, and some of these, in their attempt to occupy a body, attach themselves instead to a body part, such as a nose hair or a toe nail, and these are called "Body Thetans" by the Church.
The scam of the "Advanced Technology" involves the removal of the Body Thetans one by one, at great expense, because the exorcism of each body thetan requires certain precise actions on the e-meter, which in turn has to be checked by a course supervisor.
The process of "intending away body thetans" is identical to the Satanic ritual of demonic exorcism. There is no difference, other than the Church's use of an e-meter, a rudimentary galvanic skin-response device which is a crude imitation of a lie detector....
[E]ven if a Scientologist doing the OT levels is told by his Case Supervisor that he has to get rid of a trillion body thetans, and he has managed to beg, borrow, or steal the "donations" needed to audit out the body thetans and to have his auditing supervised (at that point the Scientologist is "self-auditing" on the e-meter or auditing himself by holding the two soup cans together in one hand and writing down the readings of the e-meter needle with the other), he has the logistical question to deal with: "What if the body thetans I have sent away ever come back?"
After all, what guarantee is there that after you "intend away" a body thetan (still assuming for the ease of understanding that the body thetans are like "germs"), it will not come back to re-attach itself to you?
What this causes is the most destructive kind of paranoia imaginable. A fear so devastating to the human mind that it has resulted in countless people going insane; or in Scientology's own words, "spinning in." Scientology even has named a condition for it: PTS Type III, where a person goes totally out of control in an induced paranoid psychosis.
It is for that reason why it is so vital and important for the upper levels be free and available on the Internet. Everyone, Scientologists and non-Scientologists alike, should have the right to "informed consent in spiritual matters," which is a religious freedom and basic religious right which is just as important as free speech on the Internet.
Click for the Church of Scientology's response to this article
Kennedy's sidebar: Scientology's Tangled Web | null | null | null | null | null | null | null | null | null |
http://www.topperlearning.com/forums/ask-experts-19/how-many-terms-of-ap-16-14-12-are-needed-to-give-the-mathematics-arithmetic-progression-48969/reply | 1,487,623,483,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501170609.0/warc/CC-MAIN-20170219104610-00131-ip-10-171-10-108.ec2.internal.warc.gz | 633,008,177 | 37,536 |
Question
Tue July 26, 2011 By:
# How many terms of A.P. 16,14,12,....are+ needed to give the sum 60? Explain why we get double answer?
Tue July 26, 2011
16,14,12
sum=n/2{2a+(n-1)d}
60=n/2{2*16+(n-1)(-2)}
120=n{34-2n}
120=34n-2n2
n2 -17n+60=0
n2 -12n-5n+60=0
n(n-12)-5(n-12)=0
n=5,12
here,the common difference is negative therefore terms go on diminishing and 9th term becomes zero.All terms after 9th term are negative.These negative terms when added to positive terms from 6th to 8th term,they cancel out each other and the sum remains same.
Related Questions
Fri February 17, 2017
# The 19th term of an A. P. is equal to three times its sixth term. If its 9th term is 19, find the A.P.
Mon February 13, 2017
| 267 | 721 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2017-09 | longest | en | 0.934128 |
https://math.answers.com/other-math/What_does_this_mean_write_the_place_value_and_the_value_of_each_underlined_digit | 1,674,914,688,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499634.11/warc/CC-MAIN-20230128121809-20230128151809-00229.warc.gz | 401,408,786 | 47,706 | 0
What does this mean write the place value and the value of each underlined digit?
Wiki User
2012-09-18 20:13:14
It sounds like this:
The place value is the power of 10 that is connected to the column of a given digit. In the number 12345.67 the digit 5 is in the ones column, the digit 4 is in the tens column, the digit 3 is in the hundreds column, the digit 2 is in the thousands column, and the 1 is in the ten thousands column. 6 is in the tenths column, and 7 is in the hundredths column.
So, if the 2 above is underlined, you should realize that the value of it is 2,000, or two thousand. It's 2, but it is in the thousands column.
Wiki User
2012-09-18 20:13:14
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1984 Reviews | 218 | 735 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2023-06 | longest | en | 0.873605 |
https://marshallbrain.com/science/hair-strength.htm | 1,606,393,768,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141188146.22/warc/CC-MAIN-20201126113736-20201126143736-00416.warc.gz | 393,787,086 | 3,541 | ## How strong is a strand of your hair?
By Marshall Brain
In the video, we used a stack of books, a strand of hair and some pennies to see how strong hair is. By taping pennies onto a strand of hair one by one, we were able to measure the strength of a single strand of hair.
We used pennies because they are small, standardized and easy to obtain. In the video we calculated that a penny weighs 2.5 grams, and that the strand of hair we used could hold 23 pennies before breaking. Which is pretty amazing!
If you want to try some hair experiments at home, here's what you need:
• A stack of books
• A pencil
• Some tape
• A handful of pennies
• A strand of hair
Step 1: Securely tape one end of the strand of hair onto the pencil
Step 2: Wedge the pencil into the stack of books so that the pencil is sticking out and the hair is hanging down (see the video).
Step 3: Securely tape one penny onto the strand of hair and see if the hair can hold it.
Step 4: Keep taping on pennies until the strand breaks.
In our experiment, we managed to tape 23 pennies on before the strand broke. 23 times 2.5 grams is 57.5 grams. In the video, when we actually weighed the stack of pennies, it weighed 62 grams. The tape actually weighed 4.5 grams!
The fact that the tape had weight is important. It means that this might not have been the best way to perform this experiment if you don’t have a scale handy. It would have been better to hang a small cup or bag on the hair and add the pennies one by one to the cup, so that there is no “extra” weight being added.
There are lots of experiments you could try at home. Is straight hair stronger, or curly? Is blond hair stronger, or brown or black? If you get hair wet, does it get stronger or weaker? What about if you dip it in a weak acid like vinegar? If you braid three strands of hair together, is it 3 times stronger? What if you tie a knot in the middle? Have fun performing your own science experiments! | 478 | 1,959 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2020-50 | longest | en | 0.944993 |
https://math.stackexchange.com/questions/3248720/30-languages-spoken-how-many-participants-are-at-the-conference | 1,563,598,717,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526446.61/warc/CC-MAIN-20190720045157-20190720071157-00238.warc.gz | 464,638,789 | 38,219 | # 30 languages spoken, how many participants are at the conference?
At the conference there are 30 languages spoken. We know that every triple of participants finds common language but there is no language that more than half of the participants speaks.
1) What is the smallest number of participants?
2) What is the largest number of participants?
3) What is the largest number of participants when every triple uses different language?
4) What is the largest number of participants if every participant uses different language in every triple he is part of.
The idea is to use double counting but I am not sure how.
When $$p$$ is the number of participants then $${p \choose 3}$$ Is number of triples. And when $$l_i$$ is number of participants speaking language $$i=1,..,30$$ then $$l_i \le {p \over 2}$$
So 1) and 3) Are both 6 participants what about 2) and 4)-any ideas?
• Does anyone have an idea? – KarmaL Jun 10 at 18:28
• There must be at least $6$ participants, since three people with a common language are at most half the participants. I suggest you stat by seeing if you can work out a way to do part 1) with $6$ participants. (I would guess this is possible, but I haven't tried.) I find it usually helps my thinking to have a concrete case to work on. – saulspatz Jun 10 at 18:47
• Similarly the answer to $3)$ would also be $6$ as with $7$ participants, you'd need at least ${{7}\choose{3}} = 35$ number of languages and this number would just keep rising as the number of participants grew – Sauhard Sharma Jun 10 at 18:51
• And also 2) a 4) are the same, aren't they? Because criterium in 4 is the case of the most participants in general. Does anyone know how to country thé largest number? – KarmaL Jun 11 at 5:05
There must be at least $$6$$ participants, since the common language spoken by the members of a triple is spoken by no more than half the participants. If there are exactly $$6$$ participants, then there are $${6\choose3}=20$$ triples, and we can assign a different language to each triple. That is, the members of $$\{1,2,3\}$$ speak language $$1$$, and no other participants do; the members of $$\{1,2,4\}$$ and no other participants speak language $$2$$, and so on. We may assume that the other $$10$$ languages are spoken only by participant $$1$$. This shows that the answer to part $$1)$$ is $$6$$.
For part 2) notice that we can add another $$6$$ participants, if number $$7$$ speaks exactly the same languages as number $$1$$, number $$8$$ exactly the same languages as number $$2$$ and so on. In fact, we can add as many participants as we like provided that participants $$j$$ and $$k$$ speak exactly the same languages precisely when $$j\equiv k\pmod{6}$$. So for part $$2),$$ there is no maximum.
This also shows, in response to the OP's comment that parts $$2)$$ and $$4)$$ are not the same. For part $$4)$$, an individual cannot be a member of more than $$30$$ triples, since he must use a different language in each triple, so if there are $$p$$ participants, we must have $${p-1\choose2}\leq30\implies p\leq9$$. Note that in the solution to part $$1)$$ given above with $$6$$ participants, each person uses a different language in every triple. You have to determine whether there are solutions with $$7$$, $$8$$ or $$9$$ participants.
Part $$3)$$ is trivial.
• typo? in part (4) ${8 \choose 2} = 28 \le 30$ so $p \le 9$ instead of $p < 9$? – antkam Jun 11 at 18:54
• @antkam Some kind of mistake anyway; thanks, I'll fix it – saulspatz Jun 11 at 19:18
Suppose there are $$p$$ participants. Then the following must hold:
• Each participant belongs to $${p-1 \choose 2}$$ triplets and must speak (at least) that many languages. So the total number of (language, speaker) pairs $$\ge p {p-1 \choose 2}$$.
• Each language can be spoken by at most $$\lfloor p/2 \rfloor$$ speakers, so the total number of (language, speaker) pairs $$\le 30 \lfloor p/2 \rfloor$$.
• Combining: $$p {p-1 \choose 2} \le 30 \lfloor p/2 \rfloor$$, i.e. $$p(p-1)(p-2) \le 60 \lfloor p/2 \rfloor$$.
It is easy to verify that $$p \ge 7$$ fails the above requirement. Thus the maximum is still $$6$$, which @saulspatz has achieved. | 1,155 | 4,168 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 46, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2019-30 | latest | en | 0.931447 |
https://www.thestudentroom.co.uk/showthread.php?t=5041662 | 1,643,388,660,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320306301.52/warc/CC-MAIN-20220128152530-20220128182530-00204.warc.gz | 1,074,517,157 | 32,342 | # maths C2 trig help!
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#1
Hi, I'm stuck on a few trig eq. Q out the textbook;
Solve the equation 3sin^2x=1+5cosx in the range 0-180
Solve the equation cos(3x-pi)=-1
I understand everything else I just need help in rearranging the equations (possibly using the identities) to find x! I can't quite get it right, help would be appreciated, ty.
-A
0
4 years ago
#2
(Original post by anton6846)
Hi, I'm stuck on a few trig eq. Q out the textbook;
Solve the equation 3sin^2x=1+5cosx in the range 0-180
Solve the equation cos(3x-pi)=-1
I understand everything else I just need help in rearranging the equations (possibly using the identities) to find x! I can't quite get it right, help would be appreciated, ty.
-A
If you want help rearranging, just post your attempt at the rearrangement and we'll tell you exactly where the errors lie.
1
4 years ago
#3
(Original post by anton6846)
Hi, I'm stuck on a few trig eq. Q out the textbook;
Solve the equation 3sin^2x=1+5cosx in the range 0-180
Solve the equation cos(3x-pi)=-1
I understand everything else I just need help in rearranging the equations (possibly using the identities) to find x! I can't quite get it right, help would be appreciated, ty.
-A
For the first one, have you tried expressing sin2x in terms of cosx? Think about identities.
0
#4
(Original post by RDKGames)
If you want help rearranging, just post your attempt at the rearrangement and we'll tell you exactly where the errors lie.
Well I've moved pi to the RHS and divided by 3 but I know this is wrong?
(Original post by TheMindGarage)
For the first one, have you tried expressing sin2x in terms of cosx? Think about identities.
o ya thanks ill give it a go.
0
4 years ago
#5
(Original post by anton6846)
Well I've moved pi to the RHS and divided by 3 but I know this is wrong?
Seems like a valid portion of the full procedure. Must be going wrong elsewhere if you can't get the correct answer.
0
4 years ago
#6
you must learn these:
https://mathsteaching.wordpress.com/...ic-identities/
0
4 years ago
#7
(Original post by anton6846)
Hi, I'm stuck on a few trig eq. Q out the textbook;
Solve the equation 3sin^2x=1+5cosx in the range 0-180
Solve the equation cos(3x-pi)=-1
I understand everything else I just need help in rearranging the equations (possibly using the identities) to find x! I can't quite get it right, help would be appreciated, ty.
-A
The first one needs to be converted into a quadratic in cos(x) first. You can inverse cos both sides of the second but you will need to extend the interval to get all solutions. There are two examples just like this at this link:
http://studywell.com/maths/pure-math...ric-equations/
Hope that helps.
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3.18% | 929 | 3,309 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2022-05 | latest | en | 0.938897 |
https://enigmaticcode.wordpress.com/2011/12/04/enigma-1662-red-face/ | 1,521,831,437,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257648431.63/warc/CC-MAIN-20180323180932-20180323200932-00431.warc.gz | 581,122,874 | 17,668 | # Enigmatic Code
Programming Enigma Puzzles
## Enigma 1662: Red face?
From New Scientist #2828, 3rd September 2011 [link]
Our local craft shop stocks an ornament made of a piece of carved wood. Its surface consists of a number of faces, all being different colours but the same-sized regular polygon, and with the same number of faces meeting at each vertex. I bought a number of these ornaments, all absolutely identical, and placed them on my mantelpiece.
My nephew saw them and noted that three of them had the same colour face resting on the mantelpiece and that in all other cases the colours were different. “That’s an unnatural set-up,” he commented, and he arranged them so that all the colours resting on the surface were different.
However, he was embarrassed when I proved to him that the previous situation was more natural because, if the ornaments were placed at random, then there was 50 per cent more chance of having three the same (and the rest different) than having them all different.
How many ornaments did I have, and what shape were they?
[enigma1662]
### 5 responses to “Enigma 1662: Red face?”
1. Jim Randell 4 December 2011 at 12:08 pm
The following Python program uses the mathematical combination function C(n, k) to compute the permutations where all displayed faces are different, and where exactly three objects display the same face. And then determines the solution where these numbers are in the ratio of 2:3. It runs in 30ms.
```from enigma import C, factorial, printf
# consider m objects with n faces. m <= n
# number of faces in the platonic solids
for n in (4, 6, 8, 12, 20):
for m in range(3 + 1, n + 1):
# pd = number of permutations where all displayed faces are different
pd = factorial(n, n - m)
# p3 = number of permutations where exactly 3 objects display same face
p3 = factorial(n, n - m + 2) * C(m, 3)
s = ('***SOLUTION***' if (2 * p3 == 3 * pd) else '')
printf("objects={m} faces={n} pd={pd} p3={p3} [{r:.2f}] {s}", r=float(p3) / float(pd))
```
Solution: There are 11 icosahedral ornaments.
• jimrandell 4 December 2011 at 12:09 pm
And here’s an even simpler version:
```# consider m objects with n faces. m <= n
# pd = number of permutations where all displayed faces are different
# p3 = number of permuations where exactly 3 objects display same face
# 2 * p3 == 3 * pd when m * (m - 1) * (m - 2) == 9 * (n - m + 2) * (n - m + 1)
# number of faces in the platonic solids
for n in (4, 6, 8, 12, 20):
for m in range(3 + 1, n + 1):
if m * (m - 1) * (m - 2) == 9 * (n - m + 2) * (n - m + 1):
print("objects={m} faces={n}".format(n=n, m=m))
```
2. Hugh Casement 6 May 2016 at 9:06 am
What I can’t work out is the most likely configuration. Is it two faces the same and all the others different? Or two of one colour and two of another, with the rest different? Or what? I usually get in a muddle with permutations and combinations, so would welcome any help there.
• Jim Randell 31 May 2016 at 3:50 pm
@Hugh: I’m never too confident that I’ve calculated such counting problems correctly, but I wrote a program to do 100 million random trials, selecting 11 numbers from 1 to 20.
It would seem that the mostly likely configuration is (2, 2, 1, 1, 1, 1, 1, 1, 1), i.e. one pair of objects sharing one value (colour), another pair sharing another value and all the rest having their own unique value. | 953 | 3,372 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2018-13 | latest | en | 0.938135 |
null | null | null | null | null | null | Cape Code Community Readies to Take Down $10 Million Wind Turbines
People say the wind turbines cause too much noise, make them sick
The people living in the town of Falmouth are now getting ready to rid their community of some wind turbines which they claim cause nothing but noise pollution and health issues.
Although the money spent on installing these wind turbines amounts to $10 million (€7.65 million), this Cape Cod community maintains that the negative aspects of having these devices near them more than outweigh any benefits that they might get from them.
According to Daily Mail, the people of Falmouth are first and foremost concerned about the fact that, when up and running, these wind turbines produce tremendous amounts of noise, and disrupt the sleeping patterns of both people and whatever wildlife is present nearby.
They also claim that the wind turbine's movements impact on one's overall wellbeing. As pointed out by one of the town's residents, “Every time the blade has a downward motion it gives off a tremendous energy, gives off a pulse.”
As was to be expected, these two wind turbines are also blamed for diminishing properly values, both because they ruin local landscapes and because of their affecting people and natural ecosystems in said ways.
The same source informs us that, in order to tear down these wind turbines, the town's officials would have to spend an additional $5 million - $15 million (€3.82 million - €11.47 million), meaning that Falmouth should expect some serious financial losses.
Still, as one local argues, “You can't put a monetary value on people's health and that's what's happened here. A lot of people are sick because of these.”
Falmouth's residents and officials are expected to hold a town meeting this coming April.
The purpose of this meeting is that of deciding once and for all whether or not taking down the wind turbines is the wisest thing to do.
Hot right now · Latest news | null | null | null | null | null | null | null | null | null |
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