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http://www.matrixlab-examples.com/xor.html | 1,561,615,324,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628000894.72/warc/CC-MAIN-20190627055431-20190627081431-00536.warc.gz | 276,708,367 | 5,054 | # XOR - Logical EXCLUSIVE OR
A B x o r (A,B)
0 0 0
0 1 1
1 0 1
1 1 0
For the logical exclusive OR, XOR(A,B), the result is logical 1 (TRUE) where either A or B, but not both, is nonzero. The result is logical 0 (FALSE) where A and B are both zero or nonzero.
A and B must have the same dimensions (or one can be a scalar).
This is the common symbol for the 'Exclusive OR'
Example:
If matrix A is:
A =
0 0 1 1
1 1 0 0
0 0 0 0
1 1 1 1
and matrix B is:
B =
0 0 0 0
1 1 1 1
0 1 0 1
1 0 1 0
Then, the logical EXCLUSIVE OR between A and B is:
>> xor(A,B)
ans =
0 0 1 1
0 0 1 1
0 1 0 1
0 1 0 1
>>
Example:
If vector x is:
x =
0 1 2 3 0
and vector y is:
y =
1 2 3 0 0
Then, the logical exclusive OR between x and y is:
ans =
1 0 0 1 0
From 'Exclusive OR' to home
From 'XOR' to 'Boolean Algebra Menu'
Top Logical AND Logical OR De Morgan's Laws | 406 | 1,098 | {"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-2019-26 | latest | en | 0.153533 |
https://www.slideserve.com/oshin/vibration-and-waves | 1,591,215,923,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347435987.85/warc/CC-MAIN-20200603175139-20200603205139-00420.warc.gz | 885,981,456 | 13,624 | # L 21 – Vibration and Waves [ 2 ] - PowerPoint PPT Presentation
L 21 – Vibration and Waves [ 2 ]
1 / 24
L 21 – Vibration and Waves [ 2 ]
## L 21 – Vibration and Waves [ 2 ]
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##### Presentation Transcript
1. L 21 – Vibration and Waves[ 2 ] • Vibrations (oscillations) • resonance • clocks – pendulum • springs • harmonic motion • Waves • mechanical waves • sound waves • musical instruments
2. VIBRATING SYSTEMS • Mass and spring on air track • Mass hanging on spring • Pendulum • Torsional oscillator All vibrating systems have one thing in common restoring force
3. Springs obey Hooke’s Law spring force (N) elastic limit of the spring amount ofstretchingor compressing (m) • the strength of a spring is measured by how much • force it provides for a given amountof stretch • we call this quantity k, the spring constant in N/m • magnitude ofspring force = k amount of stretch Fspring = k x
4. X=0 x kx mg m example • A mass of 2 kg is hung from a spring that has a spring constant k = 100 N/m. By how much will it stretch? • The downward weight of the mass is balanced by the upward force of the spring. • w = mg = k x2 kg × 10 m/s2 = (100 N/m) × x20 N = 100 N/m ×x x = 0.2 m or 20 cm
5. simple harmonic oscillatormass and spring on a frictionless surface equilibrium position frictionless surface k m spring that can be stretched or compressed A A 0 k is the spring constant, which measures the stiffness of the spring in Newtons per meter
6. Terminology • AMPLITUDE A: maximum displacement from equilibrium (starting position) • PERIODT: time for one complete cycle • FREQUENCYf : number of complete cycles per unit time; one cycle per second = 1 Hertz (Hz)
7. position -A 0 +A time follow the mass – position vs. time + A - A T T T
8. Period (T) and frequency (f ) of the mass-spring system Newton’s 2nd Law: F = ma = k x a = (k/m) x Units: (k/m) (N/m)/kg (kg m/s2 /m)/kg 1/s2, So has units of time (s). If the mass is quadrupled, the period is doubled.
9. L x mg Frestoring The pendulum: T and f Frestoring = mg (x/L) F = m a = mg (x/L) a = (g/L) x (g/L) (m/s2)/m 1/s2 Does NOT depend on m
10. PE KE PE -A 0 +A KE+PE KE+PE KE GPE GPE KE+GPE KE+GPE Energy in the simple harmonic oscillator A stretched or compressed spring has elastic Potential Energy Etotal = KE + PE = constant The pendulum is driven by Gravitational potential energy
11. Waves vibrations that move • What is a wave?A disturbance that moves (propagates) through something • Due to the elastic nature of materials • The “wave” - people stand up then sit down, then the people next to them do the same until the standing and sitting goes all around the stadium. • the standing and sitting is the disturbance • notice that the people move up and down but the disturbance goes sideways !
12. Why are waves important? • a mechanical wave is a disturbance that moves through a medium ( e.g. air, water, strings) • waves carry and transmit energy • they provide a means to transport energy from one place to another • electromagnetic waves (light, x-rays, UV rays, microwaves, thermal radiation) are disturbances that propagate through the electromagnetic field, even in vacuum (e.g. light from the Sun)
13. Types of waves • Mechanical waves: a disturbance that propagates through a medium • waves on strings • waves in water • ocean waves • ripples that move outward when a stone is thrown in a pond • sound waves – pressure waves in air • Electromagnetic waves • Light waves • Radio waves
14. transverse wave on a string • jiggle the end of the string to create a disturbance • the disturbance moves down the string • as it passes, the string moves up and then down • the string motion in vertical but the wave moves in the horizontal (perpendicular) direction transverse wave • this is a single pulse wave (non-repetitive) • the “wave” in the football stadium is a transverse wave
15. Wave speed: How fast does it go? • The speed of the wave moving to the right is not the same as the speed of the string moving up and down. (it could be, but that would be a coincidence!) • The wave speed is determined by: • the tension in the string more tension higher speed • the mass per unit length of the string (whether it’s a heavy rope or a light rope) thicker rope lower speed
16. Slinky waves • you can create a longitudinalwave on a slinky • instead of jiggling the slinky up and down, you jiggle it in and out • the coils of the slinky move along the same direction (horizontal) as the wave
17. S N the diaphragm of The speaker moves in and out SOUND WAVES • longitudinal pressure disturbances in a gas • the air molecules jiggle back and forth in the same direction as the wave • Sound waves cannot propagate in a vacuum DEMO Patm
18. I can’t hear you! Since sound is a disturbance in air, without air (that is, in a vacuum) there is no sound. vacuum pump
19. Sound – a longitudinal wave
20. The pressure waves make your eardrum vibrate • we can only hear sounds between about 30 Hz and 20,000 Hz • below 30 Hz is called infrasound • above 20,000 is called ultrasound The eardrum is a very sensitive membrane Capable of responding to displacements on the order of the size of an atom
21. Sound and Music • Sound pressure waves in a solid, liquid or gas • The speed of sound vs • Air at 20 C: 343 m/s = 767 mph 1/5 mile/sec • Water at 20 C: 1500 m/s • copper: 5000 m/s • Depends on density and temperature 5 second rule for thunder and lightning
22. Why do I sound funny whenI breath helium? • The speed of sound depends on the mass of the molecules in the gas • Sound travels twice as fast in helium, because Helium is lighter than air • The higher sound speed results in sounds of higher pitch (frequency)
23. Acoustic resonance tuning fork resonance shattering the glass | 1,590 | 5,925 | {"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} | 3.78125 | 4 | CC-MAIN-2020-24 | latest | en | 0.738592 |
https://quizizz.com/en-in/mixed-numbers-and-improper-fractions-worksheets-class-8?page=1 | 1,723,354,510,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640975657.64/warc/CC-MAIN-20240811044203-20240811074203-00347.warc.gz | 378,517,344 | 27,004 | Fractions and Decimals
17 Q
7th - 8th
12 Q
6th - 8th
Multiplying Fractions and Mixed Numbers
20 Q
6th - 8th
6.4 Converting between Improper Fractions and Mixed Numbers
15 Q
8th
Rules for Multiplying Fractions
13 Q
5th - 8th
Improper Fractions and Mixed Numbers
15 Q
8th
21 Q
7th - 8th
General Mathematics
10 Q
8th - Uni
16 Q
8th - 9th
Improper Fractions and Mixed Numbers
10 Q
6th - 8th
Dividing Decimals, Fractions, and Mixed Numbers
16 Q
6th - 8th
Improper & Mixed Numbers
18 Q
6th - 8th
The Number System
10 Q
8th
Fractions Review
21 Q
5th - 8th
Improper Fractions and Mixed Numbers
16 Q
5th - 8th
Converting Improper Fractions to Mixed Numbers
10 Q
6th - 8th
Improper fractions and measuring length, U4L2.LVL2
11 Q
6th - 8th
Rational and Irrational Numbers
15 Q
7th - 8th
Fraction Vocabulary
15 Q
6th - 8th
Multiplying Fractions Review
12 Q
6th - 8th
Fraction Vocabulary
12 Q
5th - 8th
Fractions and Mixed Numbers quiz
12 Q
6th - 8th
Fractions, decimals and integers (Basics)
20 Q
6th - 8th
Dividing Fractions Rev
13 Q
6th - 8th
## Explore printable Mixed Numbers and Improper Fractions worksheets for 8th Class
Mixed Numbers and Improper Fractions worksheets for Class 8 are essential tools for teachers looking to help their students master the challenging concepts of fractions in Math. These worksheets provide a variety of problems that allow students to practice converting between mixed numbers and improper fractions, simplifying fractions, and performing operations with fractions. By incorporating these worksheets into their lesson plans, teachers can ensure that their Class 8 students develop a strong foundation in fractions, which is crucial for success in more advanced Math courses. Furthermore, these worksheets can be easily tailored to meet the individual needs of each student, making them a valuable resource for both in-class instruction and homework assignments. Mixed Numbers and Improper Fractions worksheets for Class 8 are an indispensable resource for teachers who want to help their students excel in Math.
In addition to Mixed Numbers and Improper Fractions worksheets for Class 8, Quizizz offers a wide range of resources that can help teachers enhance their Math lessons. Quizizz is a platform that allows teachers to create interactive quizzes, polls, and presentations that can be used in the classroom or assigned as homework. With Quizizz, teachers can easily track student progress and identify areas where students may need additional support. The platform also offers a vast library of pre-made quizzes and worksheets, covering a wide range of topics and grade levels. This means that teachers can quickly find and customize resources that align with their curriculum and learning objectives. By incorporating Quizizz into their teaching strategies, teachers can create engaging and effective Math lessons that cater to the diverse needs of their Class 8 students. | 710 | 2,979 | {"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-2024-33 | latest | en | 0.938407 |
https://oneclass.com/textbook-notes/ca/utsc/sta-actb/stab-22h3/128410-chapter-5docx.en.html | 1,621,377,032,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989874.84/warc/CC-MAIN-20210518222121-20210519012121-00631.warc.gz | 450,980,511 | 81,749 | STAB22H3 Chapter Notes - Chapter 5: Box Plot, Equalize
23 views2 pages
29 Jan 2013
School
Department
Course
Stats: Data and Models Canadian Edition
Chapter 5 Understanding and Comparing Distributions
Boxplot and 5-Number Summaries
- 5-number summary can be displayed in a boxplot
- Making the boxplot:
o Vertical axis spanning the extent of the data
o Indicate Q1, median, and Q3 with short horizontal lines, and connect them with vertical
lines to form a box
o (in class) indicate the minimum and maximum values with an asterisk (*), connect the *s
to the box with a straight line
o Erect “fences” around the main part of the data to identify outliers, shown with a dotted
line place the upper fence 1.5 IQRs above Q3 (Q3 + 1.5IQR) and the lower fence 1.5
IQRs below Q1 (Q1 1.5IQR)
o If a data value falls outside the fences, it does not get connect with a whisker (the line
from the box connecting the min. and max. values)
If the min. or max. value falls outside of the fence, the most extreme value within
the fences is also marked (because it is the highest/lowest non-outlier)
- The height of the box made by the Q1, median, and Q3 is equal to the IQR
- If the median is roughly at the centre of the box, the middle half of the data is roughly symmetric,
if it is not centred, the distribution is skewed
- The length of the whiskers also indicate distribution symmetry/skewness
Comparing Groups with Boxplots
- Boxplots offer a balance of information and simplicity; hide the details while displaying the
overall summary information
- Looking at boxplots side-by-side, we can see which groups have higher medians, greater IQRs,
and greater ranges
Outliers
- Cases that stand out from the rest of the data almost always deserve our attention
- An outlier is a value that doesn’t fit in with the rest of the data
- Firstly, try to understand the outlier(s) in the context of the data
- Look at the gap between the case and the rest of the data when considering whether it is an outlier
- Some outliers are just errors, many are just different
- Report summaries and analyses with and without the outlier to see its influence and then decide
what to think about the data
- Never ignore an outlier or drop it from analysis without comment
- A timeplot is a display of values against time
- Smoothing Timeplots
o Timeplots often show a lot of point-to-point variation that we want to see past so that we
can see underlying smooth trends and how the values around that trend vary (timeplot
o A smooth trace can highlight long-term patterns and help us see them through the local
variation; helps us see the main trend and points that don’t fit in the overall pattern
Looking Into the Future
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Class Notes
Textbook Notes | 706 | 2,908 | {"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-2021-21 | latest | en | 0.865064 |
http://wiki.computationalthinkingfoundation.org/wiki/Fractris_Design | 1,566,503,457,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027317359.75/warc/CC-MAIN-20190822194105-20190822220105-00265.warc.gz | 206,280,161 | 6,591 | # Fractris Design
In the Fractris game, you will be trying to finish as many rows as possible using different combinations of fractions. The computer will start you off with a random fraction and then it will be up to you to add your fractions to fill up the row exactly to score points. You will gain more points if you can avoid using the same fraction twice.
To start playing, click on the Run button. You will see a block fall immediately on the left of the grid. To add to that block, select a fraction on the right side by clicking on it. You should then see a block fall into place. Keep adding blocks until you have filled the row. If you fill your row up without using any fraction more than once, it will disappear so you can start a new row. Otherwise, your row will remain and you will start on the next row up. You'll have 250 seconds to finish as many rows as possible!
## Explorations
• If the computer sends down a 1/3 block, how can you finish the row with the fewest number of blocks and without using the same size block twice?
• If the computer sent down 1/5, would you be able to fill the row? If so, how could you do it with the fewest blocks? If not, explain why not and tell how close you could get to completing the row.
• What do all the fractions in the Fractris game (1/2, 1/3, 1/4, 1/6, 1/12, 5/12) have in common?
• Bonus: What are all the different combinations of the fractions 1/2, 1/3, 1/4, 1/6. 1/12, and 5/12 that will sum to 1 without using any fraction twice? Explain how you know that you have found all the ways.
## Standards
• Number & Operations
• work flexibly with fractions, decimals, and percents to solve problems
• Problem Solving
• solve problems that arise in mathematics and in other contexts
• Communication
• communicate mathematical thinking coherently and clearly to peers, teachers, and others
• use the language of mathematics to express mathematical ideas precisely
• Connections
• recognize and apply mathematics in contexts outside of mathematics | 485 | 2,012 | {"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-2019-35 | latest | en | 0.90621 |
https://openstax.org/books/college-physics-2e/pages/12-3-the-most-general-applications-of-bernoullis-equation | 1,723,079,615,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640713903.39/warc/CC-MAIN-20240808000606-20240808030606-00570.warc.gz | 336,912,613 | 100,506 | College Physics 2e
# 12.3The Most General Applications of Bernoulli’s Equation
College Physics 2e12.3 The Most General Applications of Bernoulli’s Equation
## Learning Objectives
By the end of this section, you will be able to:
• Calculate using Torricelli’s theorem.
• Calculate power in fluid flow.
## Torricelli’s Theorem
Figure 12.8 shows water gushing from a large tube through a dam. What is its speed as it emerges? Interestingly, if resistance is negligible, the speed is just what it would be if the water fell a distance $hh$ from the surface of the reservoir; the water’s speed is independent of the size of the opening. Let us check this out. Bernoulli’s equation must be used since the depth is not constant. We consider water flowing from the surface (point 1) to the tube’s outlet (point 2). Bernoulli’s equation as stated in previously is
$P1+12ρv12+ρgh1=P2+12ρv22+ρgh2.P1+12ρv12+ρgh1=P2+12ρv22+ρgh2.$
12.29
Both $P1P1$ and $P2P2$ equal atmospheric pressure ($P1P1$ is atmospheric pressure because it is the pressure at the top of the reservoir. $P2P2$ must be atmospheric pressure, since the emerging water is surrounded by the atmosphere and cannot have a pressure different from atmospheric pressure.) and subtract out of the equation, leaving
$12ρv12+ρgh1=12ρv22+ρgh2.12ρv12+ρgh1=12ρv22+ρgh2.$
12.30
Solving this equation for $v22v22$, noting that the density $ρ ρ$ cancels (because the fluid is incompressible), yields
$v22=v12+2g(h1−h2).v22=v12+2g(h1−h2).$
12.31
We let $h=h1−h2h=h1−h2$; the equation then becomes
$v22=v12+2ghv22=v12+2gh$
12.32
where $hh$ is the height dropped by the water. This is simply a kinematic equation for any object falling a distance $hh$ with negligible resistance. In fluids, this last equation is called Torricelli’s theorem. Note that the result is independent of the velocity’s direction, just as we found when applying conservation of energy to falling objects.
Figure 12.8 (a) Water gushes from the base of the Studen Kladenetz dam in Bulgaria. (credit: Kiril Kapustin; http://www.ImagesFromBulgaria.com) (b) In the absence of significant resistance, water flows from the reservoir with the same speed it would have if it fell the distance $hh$ without friction. This is an example of Torricelli’s theorem.
Figure 12.9 Pressure in the nozzle of this fire hose is less than at ground level for two reasons: the water has to go uphill to get to the nozzle, and speed increases in the nozzle. In spite of its lowered pressure, the water can exert a large force on anything it strikes, by virtue of its kinetic energy. Pressure in the water stream becomes equal to atmospheric pressure once it emerges into the air.
All preceding applications of Bernoulli’s equation involved simplifying conditions, such as constant height or constant pressure. The next example is a more general application of Bernoulli’s equation in which pressure, velocity, and height all change. (See Figure 12.9.)
## Example 12.5
### Calculating Pressure: A Fire Hose Nozzle
Fire hoses used in major structure fires have inside diameters of 6.40 cm. Suppose such a hose carries a flow of 40.0 L/s starting at a gauge pressure of $1.62×106N/m21.62×106N/m2$. The hose goes 10.0 m up a ladder to a nozzle having an inside diameter of 3.00 cm. Assuming negligible resistance, what is the initial water pressure at the base of the hose?
### Strategy
Here we must use Bernoulli’s equation to solve for the pressure, since depth is not constant.
### Solution
Bernoulli’s equation states
$P 1 + 1 2 ρv 1 2 +ρ gh1 =P2+ 12 ρv22 +ρ gh2 , P 1 + 1 2 ρv 1 2 +ρ gh1 =P2+ 12 ρv22 +ρ gh2 ,$
12.33
where the subscripts 1 and 2 refer to the initial conditions at ground level and the final conditions inside the nozzle, respectively. We must first find the speeds $v1v1$ and $v2v2$. Since $Q = A 1 v 1 Q = A 1 v 1$ , we get
$v1=QA1=40.0×10−3m3/sπ(3.20×10−2m)2=12.434m/s.v1=QA1=40.0×10−3m3/sπ(3.20×10−2m)2=12.434m/s.$
12.34
Similarly, we find
$v2=56.588 m/s.v2=56.588 m/s.$
12.35
(This rather large speed is helpful in reaching the fire.) Now, taking $h1h1$ to be zero, we solve Bernoulli’s equation for $P2P2$:
$P2=P1+12ρ v 1 2 − v 2 2 −ρgh2.P2=P1+12ρ v 1 2 − v 2 2 −ρgh2.$
12.36
In the proposed solution, $P2=0P2=0$, so
$P1–P2=P1 = 12 (1000 kg/m3 )[ (12.434m/s )2 −(56.588m/s )2] −(1000 kg/m3 )(9.80 m/s2 )(10.0 m) ≈1.62×106N/m2. P1–P2=P1 = 12 (1000 kg/m3 )[ (12.434m/s )2 −(56.588m/s )2] −(1000 kg/m3 )(9.80 m/s2 )(10.0 m) ≈1.62×106N/m2.$
12.37
### Discussion
This value is a gauge pressure, since the initial pressure was given as a gauge pressure. Thus the nozzle pressure is very close to atmospheric pressure, as it must because the water exits into the atmosphere without changes in its conditions.
## Power in Fluid Flow
Power is the rate at which work is done or energy in any form is used or supplied. To see the relationship of power to fluid flow, consider Bernoulli’s equation:
$P+12ρv2+ρgh=constant.P+12ρv2+ρgh=constant.$
12.38
All three terms have units of energy per unit volume, as discussed in the previous section. Now, considering units, if we multiply energy per unit volume by flow rate (volume per unit time), we get units of power. That is, $(E/V)(V/t)=E/t(E/V)(V/t)=E/t$. This means that if we multiply Bernoulli’s equation by flow rate $QQ$, we get power. In equation form, this is
$P+12ρv2+ρghQ=power.P+12ρv2+ρghQ=power.$
12.39
Each term has a clear physical meaning. For example, $PQPQ$ is the power supplied to a fluid, perhaps by a pump, to give it its pressure $PP$. Similarly, $12ρv2Q12ρv2Q$ is the power supplied to a fluid to give it its kinetic energy. And $ρghQρghQ$ is the power going to gravitational potential energy.
## Making Connections: Power
Power is defined as the rate of energy transferred, or $E/tE/t$. Fluid flow involves several types of power. Each type of power is identified with a specific type of energy being expended or changed in form.
## Example 12.6
### Calculating Power in a Moving Fluid
Suppose the fire hose in the previous example is fed by a pump that receives water through a hose with a 6.40-cm diameter coming from a hydrant with a pressure of $0.700×106N/m20.700×106N/m2$. What power does the pump supply to the water?
### Strategy
Here we must consider energy forms as well as how they relate to fluid flow. Since the input and output hoses have the same diameters and are at the same height, the pump does not change the speed of the water nor its height, and so the water’s kinetic energy and gravitational potential energy are unchanged. That means the pump only supplies power to increase water pressure by $0.92×106N/m20.92×106N/m2$ (from $0.700×106N/m20.700×106N/m2$ to $1.62×106N/m21.62×106N/m2$).
### Solution
As discussed above, the power associated with pressure is
$power = PQ = 0.920×106N/m2 40.0×10−3m3/s. = 3.68×104W=36.8kW . power = PQ = 0.920×106N/m2 40.0×10−3m3/s. = 3.68×104W=36.8kW .$
12.40
### Discussion
Such a substantial amount of power requires a large pump, such as is found on some fire trucks. (This kilowatt value converts to about 50 hp.) The pump in this example increases only the water’s pressure. If a pump—such as the heart—directly increases velocity and height as well as pressure, we would have to calculate all three terms to find the power it supplies. | 2,190 | 7,323 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 41, "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} | 4.625 | 5 | CC-MAIN-2024-33 | latest | en | 0.893728 |
https://magedkamel.com/7a-solved-problem-12-1-part-2 | 1,723,634,936,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641107917.88/warc/CC-MAIN-20240814092848-20240814122848-00645.warc.gz | 293,790,006 | 33,721 | # 7A- Solved problem 12-1-part 2-connection nominal load.
Last Updated on August 6, 2023 by Maged kamel
## Solved problem 12-1-part 2-connection nominal load.
We can find the bearing Nominal load from Table 7-4 for the available bearing strength for the external bolts. No provision in Table 7-5 for external bolts of the outer distance of the bolt with the value of le=3″. Later we will estimate the nominal shear of fasteners by using Table J3.2
Finally, we find out the design strength of the connection after comparing the least load of the different nominal loads for tensile failure, bearing strength, and shear strength.
We continue our discussion to the Solved problem 12-1-part 2 for bearing connection, quoted from Prof. McCormack’s book.
Determine the design strength φ*Pn and the allowable strength Pn/Ω for the bearing type connection shown in Fig.12-5. The steel is A36-Fy=36 KSi and Fu=58 ksi.
The bolts are 7/8″ in A325, the bolts are standard sizes, and the threads are excluded or X from the shear plane. Assume that deformations at bolt holes are a design consideration.
### Solved problem 12-1-part 2-Bearing Failure nominal load by table 7-4.
There is a table 7-4 for the available bearing strength at the bolt, based on the diameter of the bolt and type of bolt,Fu, and inner spacing.
We have 7/8″ and Fult=58 ksi, and a standard hole STD, the spacing is 3″ between bolts.
Move horizontally at the spacing of 3″, intersecting with the vertical line from the nominal diameter of 7/8″ at Φ*rn . We get the value of Φ*rn= 91.40 kips/inch of the plate thickness. Since the plate thickness is 1/2″. Then multiply by (1/2).
We get the value the design value ASD value of rn. from the same table when selecting Move horizontally at the spacing of 3″, intersects with the vertical line from the nominal diameter of 7/8″ at (1/Ω) *rn.
We get the value of (1/Ω) *rn =60.90 kips/inch of the plate thickness. Since the plate thickness is 1/2″. Then multiply by (1/2). For a single bolt but/inch of plate thickness, our plate thickness is 1/2″.
#### LRFD value for bearing.
From Table 7-4 we have Φ*rn=91.4 kips/inch, then multiply by 1/2″, the final Φ*Rn=45.70 kips for one bolt.
#### ASD value for bearing.
For the (1/Ω)*rn=60.90 kips/inch, then multiply by 1/2″, the final (1/Ω)*Rn=30.45 kips for one bolt. The calculations are very close to the previous calculations done by using equations.
### Solved problem 12-1-part 2. Shear Failure nominal load by table J3.2.
The next step will be the shear estimation. For shear, we need table J3.2. For ASTM A325, we have two figures the first one is for shear which is 54 ksi, and 68 kips, which value to choose from?
The solved problem mentioned that bolts are excluded, and the thread will not help in the shear strength Fnv=68 ksi, to estimate the shear. The shear calculation is checking how many planes and calculating the area, which is given. For shear estimation.
We have two methods In the first method, we have table 7-1 available shear strength for bolts.
We have N and X, we check the diameter, and also the table gives the area of the bolt check diameter. The table will give the LRFD and ASD values.
We have d=7/8″ for x, we have S and D, where s is the single shear and D is the double shear Φ*Fnv=51 ksi/inch2 of diameter area or Fnv/Ω =34.0 ksi/inch2 of diameter area for LRFD, which can give for 7/8″bolt Φ*Fnv=51*0.601=30.65 kips for single shear. While for 7/8″ bolt. Fnv/Ω*Fnv=34*0.601=20.43 kips for a single shear.
### Solved problem 12-1-part 2.Shear Failure nominal load by table 7-1.
We can go to Table- 7-1 for the available shear strength for bolts. Select the column of 7/8″ with the thread condition x as a horizontal line. The intersected value will be Φ*Rn =30.70 kips for one bolt. For the ASD we have (1/Ω)*Rn =20.40 kips.
How many bolts do we have? we have 4 bolts, then multiply by 4 For Φ*Rn*n=4*30.70 =122.80 kips.
For the ASD we have (1/Ω)*Rn*n=20.40*4=81.60 kips.ASD value=81.60 kips. The block shear was not included in the given solved problem 12-1-part 2.4*0.601=20.43 kips for a single shear.
### Solved problem 12-1-part 2.The final summary for the nominal load.
We take all the estimated values and put these values in a form of a table, which is shown herewith. This is the summary for yielding for LRFD Φ*Rn=194.40 kips, for fracture Φ*Rn=261 kips. In the case of shear Φ*Rn=122.60 kips. , you can double-check between the values by diving the LRFD /1.5 to get the corresponding ASD values. Ω =1.50. Which are the lowest values? the lowest value is 122.60 kips for the LRFD from shearing.
The final value of Φ*Rn=122.60 kips and the values for THe ASD design nominal loads are added for each limit state and the final ASD value selected is 81.74 kips.
Thanks a lot and see you in the next post.
This is a link for the previous post-A solved problem 12-1 for bearing connections.
This is a very useful source for the design of various Steel elements, A Beginner’s Guide to the Steel Construction Manual, 15th ed, Chapter 4 – Bolted Connections.
This is a very useful link: A Beginner’s Guide to the Steel Construction Manual, 15th ed, Chapter 4 – Bolted Connections
Scroll to Top | 1,464 | 5,217 | {"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} | 3.984375 | 4 | CC-MAIN-2024-33 | latest | en | 0.874327 |
https://www.gradesaver.com/textbooks/math/algebra/algebra-1/chapter-7-exponents-and-exponential-functions-7-6-exponential-functions-practice-and-problem-solving-exercises-page-450/19 | 1,534,278,423,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221209562.5/warc/CC-MAIN-20180814185903-20180814205903-00603.warc.gz | 885,903,738 | 14,118 | ## Algebra 1
There will be $4800$ foxes after $45$ years.
There are 3 15 year periods in 45 years, so $x = 3$. According to the order of operations, we simplify inside parentheses, then we simplify powers, then we multiply and divide, and finally, we add and subtract. When we do this, we find: $y=75 \cdot 4^3 = 75 \cdot 64 = 4800$ | 102 | 333 | {"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} | 4.21875 | 4 | CC-MAIN-2018-34 | longest | en | 0.824538 |
https://www.tutorialspoint.com/statistics/harmonic_number.htm | 1,716,308,050,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058504.35/warc/CC-MAIN-20240521153045-20240521183045-00460.warc.gz | 946,464,109 | 26,608 | # Statistics - Harmonic Number
Harmonic Number is the sum of the reciprocals of the first n natural numbers. It represents the phenomenon when the inductive reactance and the capacitive reactance of the power system becomes equal.
## Formula
${ H = \frac{W_r}{W} \\[7pt] \, where\ W_r = \sqrt{ \frac{1}{LC}} } \\[7pt] \, and\ W = 2 \pi f$
Where −
• ${f}$ = Harmonic resonance frequency.
• ${L}$ = inductance of the load.
• ${C}$ = capacitanc of the load.
## Example
Calculate the harmonic number of a power system with the capcitance 5F, Inductance 6H and frequency 200Hz.
Solution:
Here capacitance, C is 5F. Inductance, L is 6H. Frequency, f is 200Hz. Using harmonic number formula, let's compute the number as:
${ H = \frac{\sqrt{ \frac{1}{LC}}}{2 \pi f} \\[7pt] \implies H = \frac{\sqrt{ \frac{1}{6 \times 5}} }{2 \times 3.14 \times 200} \\[7pt] \, = \frac{0.18257}{1256} \\[7pt] \, = 0.0001 }$
Thus harmonic number is ${ 0.0001 }$. | 320 | 949 | {"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} | 4.21875 | 4 | CC-MAIN-2024-22 | latest | en | 0.74285 |
https://www.physicsforums.com/threads/final-exam-questions.120165/page-5 | 1,628,026,184,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154471.78/warc/CC-MAIN-20210803191307-20210803221307-00125.warc.gz | 978,910,060 | 17,233 | # Final Exam Questions
Okay so for a problem, say i wana find how far must u bring ur hands across the string to get a frequency of 350. How would u go about solving that
Doc Al
Mentor
Alt+F4 said:
A satellite is in circular orbit at a fixed radius from the center of the earth and with a constant speed. Which one of the following statements is correct about the satellite?
(a) The acceleration is constant but the velocity is not.
(b) Both the acceleration and the velocity are constant.
(c) Neither the acceleration nor the velocity are constant.
Ans:C
WHy so? Constant Speed = Constant Velocity
Careful! Velocity and acceleration are both vectors--direction counts.
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam2/sp99 [Broken] Question 18
(5)(6) + (2) ( 20) - (x)(25) = 0
X = 2.8 why is it 2
Last edited by a moderator:
well
Px = 0
Py = 0
Doc Al
Mentor
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam2/sp99 [Broken] Question 18
(5)(6) + (2) ( 20) - (x)(25) = 0
X = 2.8 why is it 2
First things first. Momentum is a vector. Start by finding the total momentum of the 5-kg and 2-kg pieces.
Last edited by a moderator:
okay so Total Mom = (5)(6) for X, For Y it is (2)(20)
so for X = 30
so for Y = 40
so the vecotr of the third one
u get an angle of 53.13
Doc Al
Mentor
Alt+F4 said:
okay so Total Mom = (5)(6) for X, For Y it is (2)(20)
so for X = 30
so for Y = 40
so the vecotr of the third one
u get an angle of 53.13
So what's the magnitude of the total momentum (of those two pieces)?
50 = MV
got it thanks
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam2/sp04 [Broken]
Question 22
15*9.8*50 / 80 = 91.8 ~92
What exactly is the formula since i dont get why you do that, i just have it memorized Why multiply by radii?
Last edited by a moderator:
Doc Al
Mentor
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam2/sp04 [Broken]
Question 22
15*9.8*50 / 80 = 91.8 ~92
What exactly is the formula since i dont get why you do that, i just have it memorized Why multiply by radii?
For the cylinder to be in equilibrium, the torques must balance. Torque depends on the moment arm.
Last edited by a moderator:
Okay i need this straighen out before exam, An elevator going downward say it is Accelerating. Does that Mean Acc is - or is it postive?
So can u just explain to me When say Gravity is negative etc..
Does - Acceleration mean that ur slowing down? Deacc
Alt+F4 said:
Okay i need this straighen out before exam, An elevator going downward say it is Accelerating. Does that Mean Acc is - or is it postive?
So can u just explain to me When say Gravity is negative etc..
Does - Acceleration mean that ur slowing down? Deacc
Think about it this way. In general, g is only the magnitude of the acceleration in the y direction. Or, $$a_y=a_/smallfreefall=-g$$
Now, this is true when you choose the positive y-direction to point vertically upward. It is $$a_y$$ that is negative, not g. Does that help?
Last edited:
Doc Al
Mentor
Alt+F4 said:
Okay i need this straighen out before exam, An elevator going downward say it is Accelerating. Does that Mean Acc is - or is it postive?
Acceleration is the rate of change of velocity. It's a vector and has direction. Calling acceleration + or - is just a sign convention to indicate direction. Usually, + means up and - means down. Note that acceleration and velocity are not necessarily in the same direction. Just because the elevator is moving down, doesn't mean it's acceleration is downward. (It could be slowing down, for instance.)
So can u just explain to me When say Gravity is negative etc..
All that means is that the force of gravity (and the acceleration due to gravity) acts downward, the negative direction.
Does - Acceleration mean that ur slowing down?
Not at all. Acceleration just means that your velocity is changing. Toss a ball straight up into the air. The acceleration is always downward (negative). As the ball rises, it slows; as it falls, it speeds up. But it's always accelerating down.
"Deacceleration" is a confusing term; I would avoid it. (It sometimes is used to mean "slowing down", but it's more helpful to say that your acceleration is negative (compared to your velocity).)
Note that you can have acceleration without speeding up or slowing down. (Think circular motion.) | 1,202 | 4,484 | {"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} | 3.84375 | 4 | CC-MAIN-2021-31 | latest | en | 0.872407 |
http://mathhelpforum.com/algebra/index33.html | 1,527,430,865,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794868316.29/warc/CC-MAIN-20180527131037-20180527151037-00035.warc.gz | 179,960,490 | 15,173 | # Algebra Forum
Pre-Algebra and Basic Algebra Help Forum: Basic calculations, order of operations, variables solving, exponential and logarithmic equations
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# maths algebra forum
Click on a term to search for related topics.
Use this control to limit the display of threads to those newer than the specified time frame.
Allows you to choose the data by which the thread list will be sorted. | 987 | 2,576 | {"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} | 3.765625 | 4 | CC-MAIN-2018-22 | latest | en | 0.686724 |
https://studylib.net/doc/11379628/ph.d.-exam-in-topology-august-18--1993 | 1,709,512,707,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476409.38/warc/CC-MAIN-20240304002142-20240304032142-00265.warc.gz | 541,746,166 | 12,270 | # Ph.D. Exam in Topology August 18, 1993
```Ph.D. Exam in Topology
August 18, 1993
Instructions: For a passing grade you must work at least five of the following problems.
1. Suppose that f : X → Y is a continuous and 1-1 map from X onto Y . If X is compact and
Y is Hausdorff, prove that f is a homeomorphism.
2. Use the homotopy or homology structure of the circle to prove that every continuous map
on the closed disc f : D → D admits a point p ∈ D for which f (p) = p.
3. Let f : (0, 1] → [0, 1] denote a continuous function with the property that
f
1
2n
=0
and f
1
2n − 1
= 1,
for all n = 1, 2, 3, . . . . Prove that the set
X = Graph(f ) ∪ ({0} × [0, 1])
is connected but not path connected.
4. Let X be path connected. Prove that the following are equivalent:
a) X is simply connected.
b) Every map of the unit circle S1 into X extends to a map from the closed unit disc into X.
c) If f and g are paths in X such that f (0) = g(0) and f (1) = g(1) then f ' g (endpoint
homotopic).
5. Suppose that A is a subset of the topological space X. Suppose that x ∈ A and y lies in the
set complement X\A of A. If ϕ denotes a path in X joining x = ϕ(0) to y = ϕ(1), show that
there exists a “time” t∗ at which ϕ(t∗ ) lies in the boundary ∂A.
6. Prove that every continuous open mapping from R to R is a homeomorphism onto its
image. Recall that an open mapping f from X to Y has the property that if V is an open set
in X, then f (V ) is an open set in Y .
7. In this problem you may use the fact that the homotopy group of the figure eight, denoted
“8”, in the plane is the free group F2 (α, β) on two symbols. Let f : 8 → S be the map which
folds the figure eight at the intersection point to make a circle. Determine exactly the induced
homotopy map f∗ : π1 (8) → π1 (S1 ). Identify the kernel of this homomorphism.
``` | 582 | 1,826 | {"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-2024-10 | latest | en | 0.946017 |
http://mathhelpforum.com/algebra/43081-fractions.html | 1,480,721,253,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540698.78/warc/CC-MAIN-20161202170900-00231-ip-10-31-129-80.ec2.internal.warc.gz | 174,512,397 | 11,679 | 1. ## fractions
I need someone to help me understand how to do these problems.
2. Your images are hard to read. You might want to try to find another method to present your work. This is what I think you have. Please confirm and others will be glad to assist.
a) $\frac{-2(18-2)}{4(5-6)}-\frac{3(27-37)}{5(7-4)}$
b) $\frac{\frac{3}{4}-7}{\frac{1}{2}-\frac{4}{5}}$
c) $\frac{\frac{3}{8}}{\frac{1}{4}-\frac{7}{16}}$
d) $4 \frac{3}{5}+7 \frac{1}{10}$
e) $9\left(\frac{15}{81}\right)\left(\frac{27}{15}\rig ht)$
3. your pics are kinda blurry, so make sure i got the problem right.
$\frac{\frac{3}{4} - 7 }{\frac{1}{2} - \frac{4}{5}}$
Get like denominators on both the top and bottom of the big fraction.
$\frac{\frac{3}{4} - \frac{28}{4}}{\frac{5}{10} - \frac{8}{10}}$
Simplify: $\frac{\frac{-25}{4}}{\frac{-3}{10}}$
Multiply by the reciprocal: $\frac{-25}{4}\cdot\frac{-10}{3}$
$= \frac{250}{12}$
$= \frac{125}{6}$
C is very similar.
On A, do what's in parenthesis first, get a like denominator and simplify.
For: $4 \frac{3}{5}+7 \frac{1}{10}$, get improper fractions, then a like denominator and finally simplify.
For: $9\left(\frac{15}{81}\right)\left(\frac{27}{15}\rig ht)$
Multiply straight across; numerator by numerator; denominator by denominator.
You can think of it as: $\bigg(\frac{9}{1}\bigg)\left(\frac{15}{81}\right)\ left(\frac{27}{15}\right)$
Look for cross canceling. The 15 is an obvious one.
Then simplify as much as possible.
all of what masters has is correct
5. ## Thanks
I want to thank both of you. I still have one question how do you write out the problems on the computer so i dont have to take pictures.
6. Originally Posted by MistaMista
I want to thank both of you. I still have one question how do you write out the problems on the computer so i dont have to take pictures.
This forum allows you to enter $\text{\LaTeX}$ (LaTeX) code directly: simply enclose it in $$and$$ tags. To learn the basics, visit the LaTex Help forum.
7. Originally Posted by MistaMista
I want to thank both of you. I still have one question how do you write out the problems on the computer so i dont have to take pictures.
While you're learning to post using Latex, you could simply express your problems using brackets and parentheses like this:
a) [-2(18 - 2)] / [4(5 - 6)] - [3(27 - 37)] / [5(7 - 4)]
b) (3/4 -7) / (1/2 -4/5)
etc... | 756 | 2,371 | {"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": 15, "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} | 4.375 | 4 | CC-MAIN-2016-50 | longest | en | 0.836637 |
https://quizgecko.com/quiz/mathematical-concepts-quiz-polynomials-equations-algebra-triangles-txzb90 | 1,716,039,296,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057422.43/warc/CC-MAIN-20240518121005-20240518151005-00779.warc.gz | 438,440,001 | 30,826 | # Mathematical Concepts Quiz: Polynomials, Equations, Algebra, Triangles
UnequivocalAnemone
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## 12 Questions
Scalene Triangle
### Which type of triangle has at least two sides of equal length and two equal interior angles?
Isosceles Triangle
60 degrees
### What type of triangle has all sides of equal length?
Equilateral Triangle
Scalene Triangle
### What is the definition of an equilateral triangle?
All sides of equal length
### Which of the following is a polynomial expression?
$x^2 - 4x + 4$
### What is the general form of a linear equation?
$ax + b = c$
### Which of the following is an algebraic expression?
$3x - 7y$
0
### Which of the following is the solution to the linear equation $2x - 3 = 7$?
$x = 5$
4
## Mathematical Concepts
Mathematics is a broad field of study encompassing various concepts such as geometry, algebra, arithmetic, calculus, and statistics. These mathematical concepts form the foundation of many scientific, engineering, and technological applications. Let's delve into some of these important areas:
### Polynomials
A polynomial is a mathematical expression consisting of variables and coefficients combined using the operations of addition, subtraction, multiplication, and non-negative integer powers. For example, x^2 - 4x + 4 and 3x^2 - 2x + 1 are polynomial expressions. Polynomials are often used in algebra, calculus, and various other mathematical fields.
### Linear Equations
A linear equation is a mathematical statement that describes a relationship between two quantities, where each term could involve only itself, one power of another variable, and constant terms. Linear equations typically have the general form ax + b = c, where a, b, and c are constants, and x represents a variable. Solving linear equations involves finding the value of the variable x that satisfies the equation.
### Algebraic Expressions
Algebraic expressions are mathematical combinations of numbers and symbols representing unknown values or variables. They can be used to represent relationships between different variables and perform operations on them. Examples of algebraic expressions include 3x - 7y, (4x^2 + 5) / (2x - 1), and sqrt(x^2 + y^2). These expressions can be simplified or manipulated using algebraic rules to solve problems involving numerical values.
### Triangles
In geometry, a triangle is a polygon with three edges and three vertices. It is one of the basic shapes in Euclidean geometry. Classification of triangles is based on their side lengths and angles. There are several types of triangles:
#### Scalene Triangle
A scalene triangle has all three sides of different lengths. In other words, there are no congruent sides. An example of a scalene triangle is shown below:
______________
| |
| A |
| |
| |
| B |
| |
|____________|
#### Isosceles Triangle
An isosceles triangle has at least two sides of equal length and at least two equal interior angles. An example of an isosceles triangle looks like this:
*
***
*****
*******
*********
#### Equilateral Triangle
Equilateral means all sides are of equal length. Thus, an equilateral triangle is one whose three sides are of equal length, and all internal angles are of 60 degrees. Here is an example of an equilateral triangle:
*
***
*****
******
*****
***
*
These mathematical concepts, including triangles, polynomials, linear equations, and algebraic expressions, provide fundamental building blocks for understanding more complex mathematical theories and applications.
Test your knowledge of fundamental mathematical concepts such as polynomials, linear equations, algebraic expressions, and triangles with this quiz. Explore various areas of mathematics that serve as the foundation for scientific, engineering, and technological applications.
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Error: | 865 | 4,120 | {"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} | 4.59375 | 5 | CC-MAIN-2024-22 | latest | en | 0.88904 |
http://mathhelpforum.com/geometry/15817-rectangular-prism.html | 1,527,049,075,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865411.56/warc/CC-MAIN-20180523024534-20180523044534-00450.warc.gz | 180,539,342 | 9,276 | 1. ## Rectangular Prism
A right rectangular prism has the following dimensions, x, y, z, where x + y + z = 15. Suppose that the prism's main diagonal has a length of 11. Then, determine the total surface area of the prism.
2. Originally Posted by Ideasman
A right rectangular prism has the following dimensions, x, y, z, where x + y + z = 15. Suppose that the prism's main diagonal has a length of 11. Then, determine the total surface area of the prism.
So,
$\displaystyle x+y+z=15$
Square,
$\displaystyle (x+y+z)^2 = 15^2 = 225$
Open,
$\displaystyle x^2+y^2+z^2+2xy+2xz+2yz = 225$
But,
$\displaystyle \sqrt{x^2+y^z+z^2}=11$
Thus,
$\displaystyle 121 + 2xy+2xz+2yz = 225$
Thus,
$\displaystyle 2xy+2xz+2yz = 104$ | 232 | 713 | {"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 | 0.793911 |
http://mathhelpforum.com/algebra/204180-trying-figure-out-cost.html | 1,480,762,012,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540915.89/warc/CC-MAIN-20161202170900-00492-ip-10-31-129-80.ec2.internal.warc.gz | 183,431,409 | 10,868 | # Thread: Trying to figure out cost
1. ## Trying to figure out cost
Alright, I am new here, I am going to try and explain this as well as I can.
I have a product that I paid 22.50 for. It was 280 watts with 1 rail at an efficiency level of 85. Now I have deduced the cost per watt, that was easy. However I am trying to find a constant (I guess). If I am paying .08036 per watt, how can I find if I am paying to much for a product. There are multiple levels of efficiency (85, 88, 90, etc) and you can have any number of rails (1,2,3, etc) that would come apart of the equation.
So, is there a constant that I can say, with x rail and at efficiency level x and x watts, this is what the price should be at.
I hope I explained this well. If you have any questions please ask. Thanks for your time.
2. ## Re: Trying to figure out cost
Hey onlyname.
For the efficiency, does this mean that 100% effeciency uses only the needed power, but 50% efficiency requires twice as much power to actually do what's meant to do?
If this is the case, then it would make sense to take into account efficiency is to take the per watt rating and divide by the efficiency where this is a number from 0 to 1 inclusive. If 100%, then nothing changes: if its 50% you double the figure, 25% you quadruple and 0% then you can't supply enough to get the job done.
To convert a percentage to a fraction simply divide by 100.
3. ## Re: Trying to figure out cost
Can you explain this for me in more detail.
"If this is the case, then it would make sense to take into account efficiency is to take the per watt rating and divide by the efficiency where this is a number from 0 to 1 inclusive. If 100%, then nothing changes: if its 50% you double the figure, 25% you quadruple and 0% then you can't supply enough to get the job done."
I am lost on the number 0 to 1 inclusive.
4. ## Re: Trying to figure out cost
The 0 corresponds to 0% and the 1 corresponds to 100%, that's all.
5. ## Re: Trying to figure out cost
Okay, so 50% would be .5, 80% .8, correct?
6. ## Re: Trying to figure out cost
Yes that's correct, and remember to divide your per watt figure by your efficiency (i.e. pw = per watt, e = effeciency then pw/e). | 584 | 2,214 | {"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-2016-50 | longest | en | 0.969565 |
http://shakespir.com/ebook/mathematics-with-creative-designs-on-a-graphing-ti-calculator-the-beginner-s-wor-8225 | 1,618,288,144,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038072082.26/warc/CC-MAIN-20210413031741-20210413061741-00143.warc.gz | 91,092,856 | 8,472 | # Mathematics with Creative Designs on a Graphing TI - Calculator, The Beginner’
The Beginner’s Worksheet
(Linear functions in the slope-intercept form)
A design can be created on most of the TI graphing Calculators by using the following steps.
The linear function design can be created using the following steps:
· Setting the viewing window: Press the key [ZOOM]. Select 6: Standard. Press the key [ZOOM] again and select 5; this will set the window in square settings. Be sure that your calculator is in degree mode.
· Be sure that the axes are off: Press the key [2ND] then press [ZOOM] key and follow the screen.
· Enter the following equations: Press the key [Y =] and enter the following equations.
· Press the [GRAPH] key once the equations are entered.
The graphs on screen will look like:
Sketch the graph on a paper and label the lines with their equations. Labeling the graphs helps beginners in writing the shading directions.
Labeled graphs
To shade the area between the curves press the [2ND] key and then the [DRAW] and select 7: Shade(.
· Write the shading directions in the order as follows:
.
· Shading with different pattern and patres: Four different patterns can be created by entering one of the numbers, either 1 or 2 or 3 or 4. Enter 1 for vertical lines, 2 for horizontal lines, 3 for lines with negativeslope, and 4 for lines with positive slope.
· The last parameter in the shading command, the patres, defines the spacing between the pixels. For distinct effect the spacing between the pixels can be adjusted by using 2, 3, 4, 5, 6, 7, or 8. [The number _]1[ is not recommended as number ]1[ will shade the area totally dark_].
· Enter the following sequence of shade commands, one at a time. Press [ENTER] each time to execute the command. To be able to write, for example~,~ press [VARS], move to Y-VARS, press [ENTER] and select one of the “y” to be posted. Now try the following shadings.
The design, after each shading, will _] look like:
The final design on the screen will look like:
· Writing Text: Press the [2ND] key and then the [DRAW]. Select 0: Text. Press the [2ND] key and then the [ALPHA] key to clock the Text option, position the cursor where you want your name to begin and write your name.
The Zero Key works as the space key as well as the key to erase any portion of the graph or the text while the [ALPHA] key is locked. Move the cursor to the portion of the unwanted graph and press Zero Key. Repeat this step until the unwanted portion of the graph is erased.
· Store the design: Press the [2ND] key and then the [DRAW] key and move the cursor to STO. Select [1: StorePic *]and[ *]enter a number from 1 to 9, or 0 (zero). The picture will be stored by pressing [ENTER]
· Recall the design: Press the [2ND] key and then the [PRGM] key and move the cursor to STO. The picture can be recalled by selecting 2: ReclPic and entering the same number where the picture was stored
## Mathematics with Creative Designs on a Graphing TI - Calculator, The Beginner’
One of the major challenges of the CCSS is to engage and motivate students for deeper conceptual understanding of mathematics. The natural way to achieve this challenge will be the best practice, which can be done by using technology providing visual approaches. The Designing approach provides teachers a unique tool to teach algebraic concepts by creating designs on a graphing calculator using equations of mathematical functions. The creative design helps students to enhance visual thinking. It helps students to understand and to apply Algebra, Geometry and Trigonometry concepts in creative and enjoyable way.The visual illustrations inspire and motivate students in the learning process. You can learn how to create a mathematical design on a TI-83, TI-84, TI-84 Plus, TI-Inspire, and more TI- Graphing Calculators.
• Author: husains4ever
• Published: 2015-10-09 18:05:06
• Words: 556 | 939 | 4,015 | {"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-2021-17 | latest | en | 0.873264 |
https://nrich.maths.org/public/topic.php?code=-778&cl=3&cldcmpid=786 | 1,582,610,101,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146033.50/warc/CC-MAIN-20200225045438-20200225075438-00112.warc.gz | 483,558,447 | 9,327 | # Resources tagged with: Resourceful
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##### Age 11 to 14 Challenge Level:
The clues for this Sudoku are the product of the numbers in adjacent squares.
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How many solutions can you find to this sum? Each of the different letters stands for a different number.
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##### Age 11 to 14 Challenge Level:
Imagine you have an unlimited number of four types of triangle. How many different tetrahedra can you make?
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##### Age 7 to 14 Challenge Level:
There are nasty versions of this dice game but we'll start with the nice ones...
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Use the differences to find the solution to this Sudoku.
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Can you find a relationship between the number of dots on the circle and the number of steps that will ensure that all points are hit?
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Find a cuboid (with edges of integer values) that has a surface area of exactly 100 square units. Is there more than one? Can you find them all?
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The Tower of Hanoi is an ancient mathematical challenge. Working on the building blocks may help you to explain the patterns you notice.
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Play this game and see if you can figure out the computer's chosen number.
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Here is a machine with four coloured lights. Can you develop a strategy to work out the rules controlling each light?
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Can you find the values at the vertices when you know the values on the edges of these multiplication arithmagons?
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Charlie likes tablecloths that use as many colours as possible, but insists that his tablecloths have some symmetry. Can you work out how many colours he needs for different tablecloth designs?
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How many moves does it take to swap over some red and blue frogs? Do you have a method?
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A game for 2 or more people, based on the traditional card game Rummy. Players aim to make two `tricks', where each trick has to consist of a picture of a shape, a name that describes that shape, and. . . .
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Players take it in turns to choose a dot on the grid. The winner is the first to have four dots that can be joined to form a square.
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Can you find the values at the vertices when you know the values on the edges?
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Who said that adding, subtracting, multiplying and dividing couldn't be fun?
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How many different symmetrical shapes can you make by shading triangles or squares?
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My two digit number is special because adding the sum of its digits to the product of its digits gives me my original number. What could my number be?
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A game in which players take it in turns to turn up two cards. If they can draw a triangle which satisfies both properties they win the pair of cards. And a few challenging questions to follow...
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Can you find a reliable strategy for choosing coordinates that will locate the treasure in the minimum number of guesses?
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Draw some isosceles triangles with an area of $9$cm$^2$ and a vertex at (20,20). If all the vertices must have whole number coordinates, how many is it possible to draw?
### Where Can We Visit?
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Charlie and Abi put a counter on 42. They wondered if they could visit all the other numbers on their 1-100 board, moving the counter using just these two operations: x2 and -5. What do you think?
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If you are given the mean, median and mode of five positive whole numbers, can you find the numbers?
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Move your counters through this snake of cards and see how far you can go. Are you surprised by where you end up?
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Semi-regular tessellations combine two or more different regular polygons to fill the plane. Can you find all the semi-regular tessellations?
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How many winning lines can you make in a three-dimensional version of noughts and crosses?
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If you move the tiles around, can you make squares with different coloured edges?
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A game in which players take it in turns to try to draw quadrilaterals (or triangles) with particular properties. Is it possible to fill the game grid?
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Gabriel multiplied together some numbers and then erased them. Can you figure out where each number was?
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Imagine you were given the chance to win some money... and imagine you had nothing to lose...
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Do you know a quick way to check if a number is a multiple of two? How about three, four or six?
### How Much Can We Spend?
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A country has decided to have just two different coins, 3z and 5z coins. Which totals can be made? Is there a largest total that cannot be made? How do you know?
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Six balls are shaken. You win if at least one red ball ends in a corner. What is the probability of winning?
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In 15 years' time my age will be the square of my age 15 years ago. Can you work out my age, and when I had other special birthdays?
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What is the smallest number of answers you need to reveal in order to work out the missing headers?
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Generate three random numbers to determine the side lengths of a triangle. What triangles can you draw?
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Here is a machine with four coloured lights. Can you make two lights switch on at once? Three lights? All four lights?
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Why not challenge a friend to play this transformation game?
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A spider is sitting in the middle of one of the smallest walls in a room and a fly is resting beside the window. What is the shortest distance the spider would have to crawl to catch the fly?
### Funny Factorisation
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Using the digits 1 to 9, the number 4396 can be written as the product of two numbers. Can you find the factors?
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Seven balls are shaken. You win if the two blue balls end up touching. What is the probability of winning?
### Substitution Cipher
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Find the frequency distribution for ordinary English, and use it to help you crack the code.
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A hexagon, with sides alternately a and b units in length, is inscribed in a circle. How big is the radius of the circle?
### Which Solids Can We Make?
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Interior angles can help us to work out which polygons will tessellate. Can we use similar ideas to predict which polygons combine to create semi-regular solids?
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Can you find the hidden factors which multiply together to produce each quadratic expression?
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Can you work out which spinners were used to generate the frequency charts?
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Imagine a room full of people who keep flipping coins until they get a tail. Will anyone get six heads in a row?
### In a Box
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Chris and Jo put two red and four blue ribbons in a box. They each pick a ribbon from the box without looking. Jo wins if the two ribbons are the same colour. Is the game fair? | 2,102 | 9,059 | {"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": 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} | 3.53125 | 4 | CC-MAIN-2020-10 | longest | en | 0.831293 |
https://math.stackexchange.com/questions/1715484/find-lim-x-to0-frac-tan-frac-pi4x-frac1x-e2x2 | 1,575,859,224,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540517156.63/warc/CC-MAIN-20191209013904-20191209041904-00239.warc.gz | 448,012,053 | 31,656 | # Find $\lim_{x\to0}\frac{(\tan(\frac{\pi}{4}+x))^{\frac{1}{x}}-e^2}{x^2}$
Find $\lim_{x\to0}\frac{(\tan(\frac{\pi}{4}+x))^{\frac{1}{x}}-e^2}{x^2}$
My attempt:
$\lim_{x\to0}(\tan(\frac{\pi}{4}+x))^{\frac{1}{x}}=e^2$
So the required limit is in $\frac{0}{0}$ form.
Then i used L hospital form.
$\lim_{x\to0}\frac{e^2(\frac{2}{x\cos 2x}-\frac{1}{x^2}\log(\frac{1+\tan x}{1-\tan x}))}{2x}$
I am stuck here.
• @OlivierOloa yeah sure... :) – tired Mar 27 '16 at 11:23
You may observe that, by the Taylor expansion, you get, as $x \to 0$, \begin{align} \tan x&=x+\frac{x^3}3+O(x^5) \tag1 \\ \frac{1+\tan x}{1-\tan x}&=1+2 x+2 x^2+\frac{8 x^3}{3}+O(x^5) \tag2 \end{align} giving \begin{align} \log\left(\frac{1+\tan x}{1-\tan x}\right)&=2 x+\frac{4 x^3}{3}+O(x^5) \tag3 \\\\\frac1x\log\left(\frac{1+\tan x}{1-\tan x}\right)&=2 +\frac{4 x^2}{3}+O(x^4) \tag4. \end{align} Then $$\left(\tan\left(\frac{\pi}{4}+x\right)\right)^{\large \frac1x}-e^2=e^{\large \frac1x\log\left(\frac{1+\tan x}{1-\tan x}\right)}-e^2 \tag5$$ rewrites $$\left(\tan\left(\frac{\pi}{4}+x\right)\right)^{\large \frac1x}-e^2=e^2 \left(e^{\large \frac{4 x^2}{3}+O(x^4)}-1\right)=e^2 \left(\frac{4 x^2}{3}+O(x^4)\right)\tag6$$ yielding, as $x \to 0$,
$$\lim_{x\to 0}\frac{\left(\tan\left(\frac{\pi}4+x\right)\right)^{\large \frac1x}-e^2}{x^2} = \frac{4e^2}3. \tag7$$
Note that $$\tan\left(\frac{\pi}{4} + x\right) = \frac{1 + \tan x}{1 - \tan x}$$ and therefore \begin{align} L &= \lim_{x \to 0}\dfrac{\left(\tan\left(\dfrac{\pi}{4} + x\right)\right)^{1/x} - e^{2}}{x^{2}}\notag\\ &= \lim_{x \to 0}\dfrac{\left(\dfrac{1 + \tan x}{1 - \tan x}\right)^{1/x} - e^{2}}{x^{2}}\notag\\ &= \lim_{x \to 0}\dfrac{\exp\left(\dfrac{1}{x}\log\left(\dfrac{1 + \tan x}{1 - \tan x}\right)\right) - e^{2}}{x^{2}}\notag\\ &= e^{2}\lim_{x \to 0}\dfrac{\exp\left(\dfrac{1}{x}\log\left(1 + \dfrac{2\tan x}{1 - \tan x}\right) - 2\right) - 1}{x^{2}}\notag\\ &= e^{2}\lim_{x \to 0}\dfrac{\exp\left(\dfrac{1}{x}\log\left(1 + \dfrac{2\tan x}{1 - \tan x}\right) - 2\right) - 1}{\dfrac{1}{x}\log\left(1 + \dfrac{2\tan x}{1 - \tan x}\right) - 2}\cdot\dfrac{\dfrac{1}{x}\log\left(1 + \dfrac{2\tan x}{1 - \tan x}\right) - 2}{x^{2}}\notag\\ &= e^{2}\lim_{t \to 0}\dfrac{\exp(t) - 1}{t}\cdot\lim_{x \to 0}\dfrac{\dfrac{1}{x}\log\left(1 + \dfrac{2\tan x}{1 - \tan x}\right) - 2}{x^{2}}\notag\\ &= e^{2}\lim_{x \to 0}\dfrac{\dfrac{1}{x}\log\left(1 + \dfrac{2\tan x}{1 - \tan x}\right) - 2}{x^{2}}\notag\\ &= e^{2}\lim_{x \to 0}\dfrac{\log\left(\dfrac{1 + \tan x}{1 - \tan x}\right) - 2x}{x^{3}}\notag\\ &= e^{2}\lim_{x \to 0}\dfrac{\log\left(\dfrac{1 + \tan x}{1 - \tan x}\right) - 2\tan x + 2\tan x - 2x}{x^{3}}\notag\\ &= e^{2}\left(\lim_{x \to 0}\dfrac{\log\left(\dfrac{1 + \tan x}{1 - \tan x}\right) - 2\tan x}{x^{3}} + 2\lim_{x \to 0}\frac{\tan x - x}{x^{3}}\right)\notag\\ &= e^{2}\left(\lim_{x \to 0}\dfrac{\log\left(\dfrac{1 + \tan x}{1 - \tan x}\right) - 2\tan x}{\tan^{3}x}\cdot\frac{\tan^{3}x}{x^{3}} + 2\lim_{x \to 0}\frac{\sec^{2} x - 1}{3x^{2}}\right)\text{ (via LHR)}\notag\\ &= e^{2}\left(\lim_{u \to 0}\dfrac{\log\left(\dfrac{1 + u}{1 - u}\right) - 2u}{u^{3}}\cdot 1 + \frac{2}{3}\lim_{x \to 0}\frac{\tan^{2}x}{x^{2}}\right)\text{ (putting }u = \tan x)\notag\\ &= e^{2}\left(\lim_{u \to 0}\dfrac{2\left(u + \dfrac{u^{3}}{3} + o(u^{3})\right) - 2u}{u^{3}}\cdot 1 + \frac{2}{3}\right)\notag\\ &= e^{2}\left(\frac{2}{3} + \frac{2}{3}\right)\notag\\ &= \frac{4e^{2}}{3}\notag \end{align} The simplifications done above are obvious and are aimed to simplify the use of series expansions and L'Hospital Rule. Also note that the variable $$t = \frac{1}{x}\log\left(1 + \dfrac{2\tan x}{1 - \tan x}\right) - 2 = \dfrac{\log\left(1 + \dfrac{2\tan x}{1 - \tan x}\right)}{\dfrac{2\tan x}{1 - \tan x}}\cdot\dfrac{\tan x}{x}\cdot\frac{2}{1 - \tan x} - 2$$ tends to $1\cdot1\cdot 2 - 2 = 0$ as $x \to 0$ and this fact has been used in the solution provided.
As a rule I prefer to use series expansions only when they are available via memory (like binomial, exponential, logarithmic and sine/cosine series) and try to use algebraic simplification so as to limit the use of series expansion to these simple series. If I need to do complicated stuff with series (like multiplying/dividing series or finding series of composite functions) I show the evaluation of series coefficients.
In the same spirit as Olivier Oloa's answer.
By Taylor $$\tan(\frac{\pi}{4}+x))=1+2 x+2 x^2+\frac{8 x^3}{3}+\frac{10 x^4}{3}+\frac{64 x^5}{15}+O\left(x^6\right)$$ $$\log\Big(\tan(\frac{\pi}{4}+x)) \Big)=2 x+\frac{4 x^3}{3}+\frac{4 x^5}{3}+O\left(x^6\right)$$ $$\frac 1x\log\Big(\tan(\frac{\pi}{4}+x)) \Big)=2+\frac{4 x^2}{3}+\frac{4 x^4}{3}+O\left(x^5\right)$$ $$\exp\Big(\frac 1x\log\Big(\tan(\frac{\pi}{4}+x)) \Big)\Big)=\Big(\tan(\frac{\pi}{4}+x)) \Big)^{\frac 1x}=e^2+\frac{4 e^2 x^2}{3}+\frac{20 e^2 x^4}{9}+O\left(x^5\right)$$ $$\frac{(\tan(\frac{\pi}{4}+x))^{\frac{1}{x}}-e^2}{x^2}=\frac{4 e^2}{3}+\frac{20 e^2 x^2}{9}+O\left(x^3\right)$$ which shows the limit and how it is approached.
To show how good is the approximation : using $x=\frac 1{10}$, the expression is $\approx 10.0191$ while the approximation gives $\frac{61 e^2}{45}\approx 10.0163$ that is to say a relative error of less than $0.03$%. | 2,385 | 5,234 | {"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": 3, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2019-51 | latest | en | 0.399896 |
https://savvycalculator.com/parking-lot-size-calculator/ | 1,713,493,484,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817253.5/warc/CC-MAIN-20240419013002-20240419043002-00885.warc.gz | 457,703,751 | 45,829 | # Parking Lot Size Calculator
## Introduction
Are you in need of a simple and efficient way to calculate the size of a parking lot? Look no further! In this guide, we will walk you through the process of creating an HTML Parking Lot Size Calculator. This tool will allow you to determine the number of parking spaces based on the area per space (in square feet) and the overall parking lot size. Whether you’re designing a parking facility or just curious about the space required, our step-by-step instructions, formula, example, and frequently asked questions will provide you with all the information you need.
## How to Use
Using our HTML Parking Lot Size Calculator is straightforward. Follow these steps:
1. Open a text editor or HTML development environment of your choice.
2. Create an HTML form to collect input from users.
3. Implement a JavaScript function to calculate the parking lot size based on the formula: PLS = S * AS.
4. Add an interactive button to trigger the calculation.
5. Display the results to the user.
Now, let’s dive into the details.
## Formula
The formula for calculating the Parking Lot Size (PLS) is:
PLS = S * AS
Where:
• PLS represents the Parking Lot Size (in square feet).
• S stands for the number of parking spaces.
• AS denotes the area per space (in square feet).
## Example
Let’s illustrate the usage of our HTML Parking Lot Size Calculator with an example:
Suppose you want to build a parking lot with 50 spaces, and each parking space requires 200 square feet of area.
1. Enter the number of spaces (S): 50
2. Enter the area per space (AS): 200
Click the “Calculate” button, and the calculator will display the Parking Lot Size (PLS) as follows:
Parking Lot Size (PLS): 10,000 sq. ft
## FAQs
1. Why do I need a Parking Lot Size Calculator?
• A Parking Lot Size Calculator helps you determine the space required for your parking facility, aiding in efficient planning and cost estimation.
2. Can I use this calculator for irregularly shaped parking lots?
• This calculator assumes a uniform layout. For irregularly shaped lots, consult a professional architect or designer.
3. How do I make the button clickable in my HTML form?
• You can achieve this by using the <button> HTML element with an onclick attribute calling your JavaScript function.
4. Can I customize the calculator’s design and layout?
• Yes, you can style and format the HTML form and the calculator’s output to match your website’s aesthetics.
## Conclusion
Creating an HTML Parking Lot Size Calculator is a useful addition to your website, helping users quickly estimate the space needed for parking. By following the steps outlined in this guide and understanding the formula (PLS=S∗AS), you can provide a valuable tool for your audience. Feel free to customize the design and functionality to suit your specific needs. | 609 | 2,861 | {"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-2024-18 | longest | en | 0.843516 |
https://web2.0calc.com/questions/please-help_9696 | 1,601,376,626,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401641638.83/warc/CC-MAIN-20200929091913-20200929121913-00654.warc.gz | 659,116,820 | 6,987 | +0
0
87
3
Solve the inequality (x - 1)/(x + 1) > 2.
May 10, 2020
#1
+744
+1
^m^, whymenotsmart
May 10, 2020
#2
+21953
+1
Solve the inequality: (x - 1)(x + 1) > 2
Multiply out: x2 - 1 > 2
Get one side to be zero: x2 - 3 > 0
Factor: [ x + sqrt(3) ] · [ x - sqrt(3) ] > 0
The number line is now broken into 5 regions:
x < - sqrt(3) x = - sqrt(3) - sqrt(x) < x < sqrt(3) sqrt(3) x > sqrt(3)
Test each of these regions, one at a time:
-- for x < - sqrt(3) choose a number smaller than - sqrt(3)
I'm going to choose -10.
Does -10 work in the inequality x2 - 3 > 0 --> (-10)2 - 3 > 0 ---> 100 - 3 > 0
Yes, that's true! So the region x < - sqrt(3) is part of the answer.
-- for x = - sqrt(3) This doesn't work because there is no equal sign in the original problem.
-- for - sqrt(x) < x < sqrt(3) choose a number in this region
I'm going to choose 0.
Does 0 work in the inequality x2 - 3 > 0 --> (0)2 - 3 > 0 ---> 0 - 3 > 0
No, that's not true ... so this region is not part of the answer.
-- for x = sqrt(3) This doesn't work because there is no equal sign in the original problem.
-- for x > sqrt(3) choose a number greater than sqrt(3)
I'm going to choose 10.
Does 10 work in the inequality x2 - 3 > 0 --> (10)2 - 3 > 0 ---> 100 - 3 > 0
Yes, that's true! So the region x > sqrt(3) is also part of the answer.
So, the answer is: either x < - sqrt(3) or x > sqrt(3)
May 10, 2020
#3
0
Hello Geno,
I believe he meant $$\frac{(x-1)}{(x+1)}>2$$ Since there is "/"
Guest May 11, 2020 | 673 | 1,661 | {"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} | 4.25 | 4 | CC-MAIN-2020-40 | latest | en | 0.426664 |
https://puzzling.stackexchange.com/questions/3447/from-knights-to-kings-on-a-rectangle/3507 | 1,713,731,829,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817819.93/warc/CC-MAIN-20240421194551-20240421224551-00773.warc.gz | 423,378,799 | 45,404 | # From Knights to Kings on a rectangle
In From knights to kings, we were asked
On a NxN chess board $N^2$ knights are placed (one per cell). Each pair of knights, who control each other (i.e. one move away from each other) are friends. One day all knights are promoted to kings. Is it possible to put them on the board (in different order) that all friends will be similarly one move away from each other?
We found out that for a square board, we can't put all the knights next to their friends when $N>3$. What if we are using a $N\times M$ board ($N\ge M$) instead? For example, if $N=3,M=2$:
A B C
D E F
You only need to swap D and F to get a solution
A B C
F E D
It's fairly obvious that all $N\times 1$ and $N\times 2$ boards will work. What other board sizes will work under these conditions?
• Are they allowed to be 1 move away from someone they aren't friends with? Oct 31, 2014 at 16:49
• @kaine Yes. As long as a piece is next to all of its friends, it doesn't care what other pieces it is next to. Oct 31, 2014 at 18:24
Assuming "castle" maneuvers are not a part of this drill: Also, symmetry is not drawn out; you can always flip these boards around.
Anything 4x4 and larger doesn't have a valid solution. A 4x4 board will generate knights with 4 friends. You can't place 5 friendly kings on a board where each is one move away from the other 4 (neglecting "castle" maneuvers!). That was incorrect reasoning. The "friendliness" is not contagious.
Anything 6x7 and larger definitely doesn't have a valid solution. This size board will result in knights with 8 friends, some of whom have 6 or more non-mutual friends. You can't arrange these 2 such that the first has access to all 8 and the second has access to 6 or more additional friends.
Square 6x6 not allowed.
Can't do 5x6 or larger. You'll end up with two knights that have 8 friends, 0 mutual friends. x has friends a-h, X has friends A-H:
. a . b .
c A . B d
C . x . D
e . X . f
E g . h F
. G . H .
They must go here on the King Board:
. . 0 0 0 . 0 0 0 . 0 0 0 . .
. . 0 x 0 . 0 x 0 . 0 x 0 . .
. . 0 0 0 . 0 0 0 . 0 0 0 . .
0 0 0 . . . 0 0 0 . 0 0 0 . .
0 X 0 . . . 0 X 0 . 0 X 0 . .
0 0 0 . . . 0 0 0 . 0 0 0 . .
Each of the friends of x and X (meaning a-h and A-H) have at least 2 non-mutual friends with the x's. This means none of them can go on the corners of the King Board, since they won't have access to anyone else, nor can they be boxed-in on an edge, so we can't solve this one.
Square 5x5 is not allowed.
Nor can you do 5x4:
a b c d
e f g h
i j k l
m n o p
q r s t
j and k each have 6 friends. Each of j's friends have at least 1 friend that isn't mutual with j (same goes for k). Also, since j and k have 0 friends in common, they must be diagonal from each other. These are the only places on the King Board where j and k could go and have access to 6 friends each. All of them puts one of their friends in a spot where they can't reach all of their own friends:
. . . . . . . . . . . .
. . j . . . j . . . j .
. k . . . . . . . . . .
. . . . . k . . . . k .
. . . . . . . . . . . .
Square 4x4 is not allowed.
3x4 is the largest that's good to go I've solved explicitly so far:
1 2 3
4 5 6
7 8 9
10 11 12
which means (knight: friends)
1: 6,8
2: 7,9
3: 4,8
4: 3,9,11
5: 10,12
6: 1,7,11
7: 2,6,12
8: 1,3
9: 2,4,10
10: 5,9
11: 4,6
12: 5,7
and valid solution:
3 8 1
4 11 6
9 2 7
10 5 12
• I don't understand your argument making 4x4 impossible. A knight can have 4 friends but the 4 friends aren't friends. So they don't need to be neighbours to each other. Oct 31, 2014 at 20:42
• @FlorianF You're totally right. My original understanding of the question involved "friend groups", and I never discarded this conclusion after I updated my understanding of friendliness! I'll remove that comment. Oct 31, 2014 at 20:46
For a 3x3 board we can do
A B C A H C
D E F -> F E D
G H I G B I
For a 3x4 board we can do
A B C D A J C L
E F G H -> G H E F
I J K L I B K D
I can't find others.
• ... and it's exactly the same solution for 3x4 as anregen's. Maybe it's the only one? Oct 31, 2014 at 20:35
• can you change the 3x3? there are 2 D's in the solved part but it's too small of an edit for me to do. Oct 31, 2014 at 20:53
• Thanks. Fixed. It was a test to see if you are paying attention. :-) Oct 31, 2014 at 22:09
I realize the 4x4 case has already been shown to not work, but let me try explaining it a bit differently -- a bit more "locally". Consider a large board, and we'll only look at the 4x4 corner of it.
In an actual 4x4 board, by symmetry there are only three kinds of pieces, a corner piece (like $A$), and edge piece (like $B$), and a 'central' piece (like $C$).
$$\begin{matrix} A & B & . & c \\ c & . & a & b \\ b & a & C & . \\ c & .& . & . \end{matrix}$$
In a larger board there may be more room initially (and thus more friends to place), which gives more space/options to place pieces after promotion to kings, but I will show this extra room cannot help.
Now the corner piece $A$ has two friends $(a,a)$, the edge piece $B$ has three friends $(b,b,C)$ and the center piece $C$ has four friends $(c,c,c,B)$.
Regardless of how we rearrange the board, some piece needs to go in the corner.
Can $A$,$B$, or $C$ be placed in the corner after promotion to kings?
• As a king in the corner only has room for three friends, it is clear a central piece like $C$ cannot be put there.
• If we put $B$ in the corner, we have two edge friends $(b,b)$ and one central friend $(C)$ to place. So this gives us only two symmetry distinct layouts to consider
$$\begin{matrix} B & b & . \\ C & b & . \\ . & . & . \end{matrix} \quad\quad or \quad\quad \begin{matrix} B & b & . \\ b & C & . \\ . & . & . \end{matrix}$$
The first doesn't leave enough room for the three other friends of $C$, so can be eliminated.
The second, there is no way to place the three other friends of $C$ without reducing at least one $b$ to only having room for one more friend.
$$\begin{matrix} B & b & c \\ b & C & c \\ . & . & c \end{matrix} \quad\quad , \quad\quad \begin{matrix} B & b & . \\ b & C & c \\ . & c & c \end{matrix}$$
Therefore there isn't room for the friends of the $b$ pieces, and this layout will not work.
So an edge piece like $B$ cannot be placed in the corner after promotion to king.
• That leaves $A$. Can a corner piece be placed in the corner after promotion?
Well, if we put $A$ in the corner, this constrains placement of many pieces. Let me rename the starting positions to make the discussion more clear:
$$\begin{matrix} A & f & \bar{e} & . \\ \bar{f} & . & \bar{a} & e \\ e & a & . & \bar{f} \\ . & \bar{e} & f & k \end{matrix}$$
The overbars are used to allow making it clear two positions (such as $a$ and $\bar{a}$) are symmetry equivalent, even if we need to refer to them as separate pieces in the discussion.
$A$ has friends $(a,\bar{a})$, the $a,\bar{a}$ each have two distinct $(e,\bar{e})$ and two common $(A,k)$ friends, the two $e$ have two common friends $(f,f)$ and the $\bar{e}$ have two common friends $(\bar{f},\bar{f})$.
If we put $A$ in the corner, this gives us only two symmetry distinct layouts to consider
$$\begin{matrix} A & \bar{a} & . \\ a & . & . \\ . & . & . \end{matrix} \quad\quad or \quad\quad \begin{matrix} A & \bar{a} & . \\ . & a & . \\ . & . & . \end{matrix}$$
Looking at the first, forces placement of quite a few pieces
$$\begin{matrix} A & \bar{a} & . & . \\ a & . & . & . \\ . & . & . & . \\ . & . & . & . \end{matrix} \quad\quad \rightarrow \quad\quad \begin{matrix} A & \bar{a} & \bar{e} & . \\ a & k & e & . \\ \bar{e} & e & . & . \\ . & . & . & . \end{matrix}$$
While the bar's over the $e$ could be changed, it doesn't change the issue that there is no way to place the $(f,f)$ and $(\bar{f},\bar{f})$ so that they are common friends with the $e$ and $\bar{e}$ respectively. So this layout is not possible.
The second layout offers more possibilities for where to place the $e,\bar{e}$ but the same problem arises. There is no way to place the $e$ pieces such that they can simultaneously be by their $a$ and $f$ friends.
• So no piece can be placed in the corner. Thus 4x4 is not possible.
That was explained "locally" in the sense that having extra room on the board would not help matters. It should be clear that any board containing a 4x4 corner as a subset will not work.
As the question states 1xN and 2xN are not interesting, this then only leaves 3xN for N$\ge$3 to consider.
Let's first look at the 3x3 case.
$$\begin{matrix} a & d & g \\ f & x & b \\ c & h & e \end{matrix}$$
The friends can be written as cyclic graphs. $x$ stands alone, and $$a \rightarrow b \rightarrow c \rightarrow d \rightarrow e \rightarrow f \rightarrow g \rightarrow h \rightarrow a$$
This is a completely "local" observation, in that any board having a 3x3 sub board in it will have this requirement. The unconstrained nature of $x$ along with the cyclic permutations for placement of $a...h$ gives quite a bit of freedom and many solutions for the 3x3 case.
A 3x4 board has two 3x3 sub-boards in it, and thus two cycles:
$$\begin{matrix} a & d,1 & g,4 & 7 \\ f & 6 & b & 2 \\ c & h,3 & e,8 & 5 \end{matrix}$$
There are no restrictions beyond the two cycles $a...h$ and $1...8$. So just shuffling to maintain the two cycles after promotion to kings gives
$$\begin{matrix} a & h,3 & g,4 & 5 \\ b & 2 & f & 6 \\ c & d,1 & e,8 & 7 \end{matrix}$$
There are of course other solutions, swapping $(f,6)$ or $(b,2)$ or $(2,f)$, or symmetry related solutions, etc. Note however that this is already constrained enough that once a corner for $a$ is chosen, all but the middle row is already fixed. These two cycles are so tightly interwoven that even if there was more room on the board, besides freedom in the middle row (or by symmetry rotating to a 4x3 solution), no additional solutions would be gained. So this is a local observation and can be applied to limit larger boards containing 3x4 subboards.
That proves by construction that the 3x3 and 3x4 case are possible. Now to finish this we can attempt to prove any board containing a 3x5 sub-board is not possible.
When we go to 3x5, we can either consider this three 3x3 sub-boards, or two 3x4 sub-boards. The 3x3 board has a lot of freedom (one piece was completely constrained) which turned out to have enough freedom to simultaneously solve the two 3x3 subboards in a 3x4 board. The resulting 3x4 solution still has some freedom, but is much more restrictive.
It turns out that it is not possible to simultaneously solve the two 3x4 subboards in a 3x5 board.
$$\begin{matrix} a & d,1 & g,4,A & 7,D & G\\ f & 6 & b,F & 2 & B\\ c & h,3 & e,8,C & 5,H & E \end{matrix}$$
There are no restrictions beyond the cycles $a...h$, $1...8$, and $A...H$.
Selecting $a$ to be in the same corner, the solutions from the 3x4 subboards gives the top and bottom rows as:
$$\begin{matrix} a & h,3 & g,4,A & 5,H & G \\ . & . & . & . & . \\ c & d,1 & e,8,C & 7,D & E \end{matrix}$$
There piece $b,F$ cannot simultaneously be by $a$ and $c$ and $G$ and $E$. So there is no solution.
One could also go back and prove the 4x4 board is not possible this way, by looking at it as two 3x4 subboards. | 3,578 | 11,320 | {"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": 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-2024-18 | latest | en | 0.96172 |
https://www.coursehero.com/file/p41aksu/Ben-believes-that-the-appropriate-analysis-is-to-calculate-the-future-value-of/ | 1,603,615,957,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107888402.81/warc/CC-MAIN-20201025070924-20201025100924-00498.warc.gz | 659,954,587 | 112,786 | Ben believes that the appropriate analysis is to calculate the future value of
# Ben believes that the appropriate analysis is to
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4)Ben believes that the appropriate analysis is to calculate the future value of each option. How would you evaluate this statement? 1. If Ben decides not to attend any MBA program, the FV will be: FV= PV x ( 1 + r ) 40 = 935,283.4855 ( 1 + 0.065 ) 40 = 11,612,549.466 2. If Ben decides to attend Wilton University, the FV will be: FV= PV x ( 1 + r ) 40 = 1806116.4 ( 1.065 ) 38 + 17633.57 ( 1.065 ) 2 - 132860 ( 1.065 ) 2 + 82076.508 ( 1.065 ) 2 = 19,771,099.95 + 20000.43 – 150693.13 + 93,093.23 = 19,733,500.4
3. If Ben decides to attend MBA at Mount Perry College, FV will be: FV= PV x ( 1 + r ) 40 = 1463821.2 ( 1.065 ) 39 + 15870.22 ( 1.065 ) 2 - 84033.34 ( 1.065 ) 1 + 44400 ( 1.065 ) 1 = 17,065,646.13 +18,000.40 – 89,495.50 + 47,286 = = 17,041,437.03 FV(Wilton) is the highest, so again, we can claim that MBA at Wilton University would be Ben’s best choice. 5)What initial salary would Ben need to receive to make him indifferent between attending Wilton University and staying in his current position? 6)Suppose, instead of being able to pay cash for his MBA, Ben must borrow the money. The current borrowing rate is 5.4%. How would this affect his decision? 1. If Ben decides to attend Wilton University, the total cost = \$146,000, r = 5.4%, n=5, so the PMT= \$34,096.06
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Want to read all 4 pages? | 530 | 1,540 | {"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} | 3.84375 | 4 | CC-MAIN-2020-45 | latest | en | 0.895375 |
http://math.stackexchange.com/questions/27959/quicksort-with-trivalued-logic/28022 | 1,469,515,990,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824756.90/warc/CC-MAIN-20160723071024-00085-ip-10-185-27-174.ec2.internal.warc.gz | 157,775,284 | 19,768 | # Quicksort with Trivalued Logic
Does anyone know a way to do a quick sort with trivalued logic?
The problem I’m trying to solve is this: I’m trying to display a view of a complex 3d object from a given viewing angle. I’ve broken the object into many 2d surfaces that I can draw separately, but to display the image properly, I need to determine the z-order of the surfaces – a classic computer drawing problem. It’s guaranteed that none of the surfaces intersect, so the problem is solvable. It would be simple if on comparing any two surfaces, I could always determine which one is in front – then a simple mergesort would suffice. But very often, if I compare two surfaces, it’ll turn out that, with the angle I’m viewing from, there’s no overlap at all. One surface is over here, and the other surface is over there, so it’s impossible to say which one is in front.
In mathematical terms, what I’m trying to do is sort a set of entities - call them a, b, c, etc. Transitivity is guaranteed: If a < b is true and b< c is true then a < c is always true. But the complicating factor is the trivalued logic: a < b could be unknown. A consequence is the final sorted list may contain small sets of elements within which the order doesn’t matter, eg. The result may be a < (b, c) < d etc.
Note that even if a < b is unknown, other comparisons may indirectly force a certain ordering for a, b. Eg. If a < b is unknown, but it turns out that a < c = true and b < c is false, then the sorted order must be a < c < b.
I can solve the problem with a bubble sort, but that’s bad because O(N^2) comparisons, and each comparison is very expensive (since it involves figuring out whether two surfaces can block each other when viewed from a certain angle). Is there a way to solve this with a faster sort? (eg. Some adaptation of a mergesort)?
-
– joriki Mar 19 '11 at 13:27
BTW, what you call "associativity" is usually called "transitivity". – joriki Mar 19 '11 at 13:41
If your polygons are convex, you can probably use a sweep algorithm. You can move your reference frame pretty simply, there are camera formulas for that, and then sort your polygons by biggest and smallest z-value, and sweep far to near. Just keep track of the set of elements you can see at each time, and that should be enough. en.wikipedia.org/wiki/Sweep_line_algorithm – leif Mar 19 '11 at 17:22
@joriki - well spotted on the transitivity, thanks. I've edited to fix that. Topological sorting definitely looks interesting. – Simon Robinson Mar 19 '11 at 17:48
@leif - unfortunately my polygons are not guaranteed to be convex. – Simon Robinson Mar 19 '11 at 17:51 | 654 | 2,635 | {"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} | 3.6875 | 4 | CC-MAIN-2016-30 | latest | en | 0.917835 |
http://www.docstoc.com/docs/72399520/Interaction | 1,432,693,801,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207928865.24/warc/CC-MAIN-20150521113208-00285-ip-10-180-206-219.ec2.internal.warc.gz | 407,953,223 | 43,650 | # Interaction by dfsdf224s
VIEWS: 5 PAGES: 29
• pg 1
``` Interaction
shared variables
can be read and written by any process (most interaction)
difficult to implement
interactive variables
can be read by any process, written by only one process (some interaction)
easier to implement
boundary variables
can be read and written by only one process (least interaction)
but initial value can be seen by all processes
easiest to implement
1/29
Interactive Variables
boundary variable var a: T· S = ∃a, a′: T· S
interactive variable ivar x: T· S = ∃x: time→T· S
The value of variable x at time t is x t
But sometimes we write x for x t , x′ for x t′ , x′′ for x t′′ , ...
a:= a+x
is really
a:= a + x t
Most laws still work but not the Substitution Law
2/29
Interactive Variables
suppose boundary a , b ; interactive x , y ; time t
ok = a′=a ∧ b′=b ∧ t′=t
x′=x ∧ y′=y means x t′ = x t ∧ y t′ = y t
3/29
Interactive Variables
suppose boundary a , b ; interactive x , y ; time t
ok = a′=a ∧ b′=b ∧ t′=t
a:= e = a′=e ∧ b′=b ∧ t′=t
x:= e = a′=a ∧ b′=b ∧ x′=e ∧ (∀t′′· t≤t′′≤t′ ⇒ y′′=y)
∧ t′ = t+(the time required to evaluate and store e )
P. Q = ∃a′′, b′′, t′′· (substitute a′′, b′′, t′′ for a′, b′, t′ in P )
∧ (substitute a′′, b′′, t′′ for a, b, t in Q )
P||Q = ∃tP, tQ· (substitute tP for t′ in P )
∧ (substitute tQ for t′ in Q )
∧ t′ = max tP tQ
∧ (∀t′′· tP≤t′′≤t′ ⇒ xt′′=x(tP)) interactive variables of P
∧ (∀t′′· tQ≤t′′≤t′ ⇒ yt′′=y(tQ)) interactive variables of Q
4/29
Interactive Variables
example boundary a , b ; interactive x , y ; extended integer time t
(x:= 2. x:= x+y. x:= x+y) || (y:= 3. y:= x+y) x left, y right, a left, b right
= (a′=a ∧ xt′=2 ∧ t′=t+1. a′=a ∧ xt′= xt+yt ∧ t′=t+1. a′=a ∧ xt′= xt+yt ∧ t′=t+1)
|| (b′=b ∧ yt′=3 ∧ t′=t+1. b′=b ∧ yt′= xt+yt ∧ t′=t+1)
= (a′=a ∧ x(t+1)=2 ∧ x(t+2)= x(t+1)+y(t+1) ∧ x(t+3)= x(t+2)+y(t+2) ∧ t′=t+3)
|| (b′=b ∧ y(t+1)=3 ∧ y(t+2)= x(t+1)+y(t+1) ∧ t′=t+2)
= x(t+1)=2 ∧ x(t+2)= x(t+1)+y(t+1) ∧ x(t+3)= x(t+2)+y(t+2)
∧ y(t+1)=3 ∧ y(t+2)= x(t+1)+y(t+1) ∧ y(t+3)=y(t+2)
∧ a′=a ∧ b′=b ∧ t′=t+3
= x(t+1)=2 ∧ x(t+2)=5 ∧ x(t+3)=10 ∧ y(t+1)=3 ∧ y(t+2)=y(t+3)=5 ∧ a′=a ∧ b′=b ∧ t′=t+3
5/29
Thermostat
thermometer || control || thermostat || burner
inputs to the thermostat:
• real temperature , which comes from the thermometer and indicates the actual
temperature.
• real desired , which comes from the control and indicates the desired temperature.
• boolean flame , which comes from a flame sensor in the burner and indicates whether
there is a flame.
outputs of the thermostat:
• boolean gas ; assigning it T turns the gas on and ⊥ turns the gas off.
• boolean spark ; assigning it T causes sparks for the purpose of igniting the gas.
6/29
Heat is wanted when the actual temperature falls ε below the desired temperature, and not
wanted when the actual temperature rises ε above the desired temperature, where ε is
small enough to be unnoticeable, but large enough to prevent rapid oscillation. To obtain
heat, the spark should be applied to the gas for at least 1 second to give it a chance to ignite
and to allow the flame to become stable. But a safety regulation states that the gas must not
remain on and unlit for more than 3 seconds. Another regulation says that when the gas is
shut off, it must not be turned on again for at least 20 seconds to allow any accumulated gas
to clear. And finally, the gas burner must respond to its inputs within 1 second.
thermostat = (gas:= ⊥ || spark:= ⊥ ). GasOff
GasOff = if temperature < desired – ε
then ((gas:= T || spark:= T || t+1 ≤ t′ ≤ t+3). spark:= ⊥ . GasOn )
else (((frame gas, spark· ok) || t < t′ ≤ t+1). GasOff)
GasOn = if temperature < desired + ε ∧ flame
then (((frame gas, spark· ok) || t < t′ ≤ t+1). GasOn)
else ((gas:= ⊥ || (frame spark· ok) || t+20 ≤ t′ ≤ t+21). GasOff)
7/29
Communication Channels
Channel c is described by
message script Mc string constant
time script Tc string constant
read cursor rc extended natural variable
write cursor wc extended natural variable
M = 6 ; 4 ; 7 ; 1 ; 0 ; 3 ; 8 ; 9 ; 2 ; 5 ; ...
T = 3 ; 5 ; 5 ; 20 ; 25 ; 28 ; 31 ; 31 ; 45 ; 48 ; ...
↑ ↑
r w
8/29
Input and Output
c! e= Mw = e ∧ Tw = t ∧ (w:= w+1)
c! = Tw = t ∧ (w:= w+1)
c? = r:= r+1
c = M r–1
√c = Tr ≤ t
M = 6 ; 4 ; 7 ; 1 ; 0 ; 3 ; 8 ; 9 ; 2 ; 5 ; ...
T = 3 ; 5 ; 5 ; 20 ; 25 ; 28 ; 31 ; 31 ; 45 ; 48 ; ...
↑ ↑
r w
9/29
Input and Output
c! e= Mw = e ∧ Tw = t ∧ (w:= w+1)
c! = Tw = t ∧ (w:= w+1)
c? = r:= r+1
c = M r–1
√c = Tr ≤ t
if √key
then ( key?.
if key="y"
then screen! "If you wish."
else screen! "Not if you don't want.")
else screen! "Well?"
10/29
Input and Output
Repeatedly input numbers from channel c , and output their doubles on channel d .
S = ∀n: nat· Md wd+n = 2 × Mc rc+n
S ⇐ c?. d! 2×c. S
proof
c?. d! 2×c. S
= rc:= rc+1. Md wd = 2 × Mc rc–1 ∧ (wd:= wd+1). S
= Md wd = 2 × Mc rc ∧ ∀n: nat· Md wd+1+n = 2 × Mc rc+1+n
= ∀n: nat· Md wd+n = 2 × Mc rc+n
= S
11/29
Communication Timing
real time need to know implementation
transit time input and output take time 0
communication transit takes time 1
input c? becomes t:= max t (Tc rc + 1). c?
check √c becomes Tc rc + 1 ≤ t
12/29
Communication Timing
W = t:= max t (Tr + 1). c?
= wait (if necessary) for input and then read it
W ⇐ if √c then c? else (t:= t+1. W)
proof
if √c then c? else (t:= t+1. W)
= if Tr + 1 ≤ t then c? else (t:= t+1. t:= max t (Tr + 1). c?)
= if Tr + 1 ≤ t then (t:= t. c?) else (t:= max (t+1) (Tr + 1). c?)
= if Tr + 1 ≤ t then (t:= max t (Tr + 1). c?) else (t:= max t (Tr + 1). c?)
= W
13/29
Recursive Communication
dbl = c?. d! 2×c. t:= t+1. dbl
weakest solution
∀n: nat· Md wd+n = 2 × Mc rc+n ∧ Td wd+n = t+n
strongest implementable solution
(∀n: nat· Md wd+n = 2 × Mc rc+n ∧ Td wd+n = t+n)
∧ rc′=wd′=t′=∞ ∧ wc′=wc ∧ rd′=rd
strongest solution
⊥
∀n: nat· Md wd+n = 2 × Mc rc+n ∧ Td wd+n= t+n ⇐ dbl
dbl ⇐ c?. d! 2×c. t:= t+1. dbl
14/29
Recursive Construction
dbl0 = T
dbl1 = c?. d! 2×c. t:= t+1. dbl0
= rc:= rc+1. Md wd = 2 × Mc rc–1 ∧ Td wd = t ∧ (wd:= wd+1). t:= t+1. T
= Md wd = 2 × Mc rc ∧ Td wd = t
dbl2 = c?. d! 2×c. t:= t+1. dbl1
= rc:= rc+1. Md wd = 2 × Mc rc–1 ∧ Td wd = t ∧ (wd:= wd+1). t:= t+1.
Md wd = 2 × Mc rc ∧ Td wd = t
= Md wd = 2 × Mc rc ∧ Td wd = t ∧ Md wd+1 = 2×Mc rc+1 ∧ Td wd+1 = t+1
dbl∞ = ∀n: nat· Md wd+n = 2 × Mc rc+n ∧ Td wd+n = t+n
15/29
Monitor
x0in x0req
x0ack x0out
x
x1in x1req
x1ack x1out
monitor = (√x0in ∨ Tx0in rx0in = m) ∧ (x0in?. x:= x0in. x0ack!)
∨ (√x1in ∨ Tx1in rx1in = m) ∧ (x1in?. x:= x1in. x1ack!)
∨ (√x0req ∨ Tx0req rx0req = m) ∧ (x0req?. x0out! x)
∨ (√x1req ∨ Tx1req rx1req = m) ∧ (x1req?. x1out! x).
monitor
16/29
Monitor
x0in x0req
x0ack x0out
x
x1in x1req
x1ack x1out
monitor ⇐ if √x0in then (x0in?. x:= x0in. x0ack!) else ok.
if √x1in then (x1in?. x:= x1in. x1ack!) else ok.
if √x0req then (x0req?. x0out! x) else ok.
if √x1req then (x1req?. x1out! x) else ok.
t:= t+1. monitor
17/29
Communicating Processes
c! 2 || (c?. x:= c)
= Mw = 2 ∧ (w:= w+1) || (r:= r+1. x:= Mr–1 )
= Mw = 2 ∧ w′ = w+1 ∧ r′ = r+1 ∧ x′ = Mr
c! 1. (c! 2 || (c?. x:= c)). c?
channel declaration
chan c: T· P
= ∃Mc: ∞*T· ∃Tc: ∞*xnat· var rc , wc: xnat := 0· P
18/29
ignoring time
chan c: int· c! 2 || (c?. x:= c)
= ∃M: ∞*int· ∃T: ∞*xnat· var r, w: xnat := 0·
x′ = Mr ∧ Mw = 2 ∧ r′ = r+1 ∧ w′ = w+1 ∧ (other variables unchanged)
= ∃M: ∞*int· ∃T: ∞*xnat· var r, w: xnat·
x′ = M0 ∧ M0 = 2 ∧ r′=1 ∧ w′=1 ∧ (other variables unchanged)
= x′=2 ∧ (other variables unchanged)
= x:= 2
including time
chan c: int· c! 2 || (t:= max t (Tr + 1). c?. x:= c)
= x′=2 ∧ t′ = t+1 ∧ (other variables unchanged)
19/29
chan c: int· t:= max t (Tr + 1). c?. c! 5
= ∃M: ∞*int· ∃T: ∞*xnat· var r, w: xnat := 0·
t:= max t (Tr + 1). r:= r+1. Mw = 5 ∧ Tw = t ∧ (w:= w+1)
= ∃M: ∞*int· ∃T: ∞*xnat· ∃r, r′, w, w′: xnat·
r:= 0. w:= 0. t:= max t (Tr + 1). r:= r+1.
Mw = 5 ∧ Tw = t ∧ r′=r ∧ w′ = w+1 ∧ t′=t
= ∃M: ∞*int· ∃T: ∞*xnat· ∃r, r′, w, w′: xnat·
M0 = 5 ∧ T0 = max t (T0 + 1) ∧ r′=1 ∧ w′=1 ∧ t′ = max t (T0 + 1)
= t′=∞
20/29
chan c, d: int· (c?. d! 6) || (d?. c! 7)
chan c, d: int· (t:= max t (Tc rc + 1). c?. d! 6) || (t:= max t (Td rd + 1). d?. c! 7)
= ∃Mc, Md: ∞*int· ∃Tc, Td: ∞*xnat· ∃rc, rc′, wc, wc′, rd, rd′, wd, wd′: xnat·
Md 0 = 6 ∧ Mc 0 = 7 ∧ rc′ = wc′ = rd′ = wd′ = 1
∧ Tc 0 = max t (Td 0 + 1) ∧ Td 0 = max t (Tc 0 + 1)
∧ t′ = max (max t (Td 0 + 1)) (max t (Tc 0 + 1))
= t′=∞
21/29
Power Series Multiplication
Input on channel a : a0 a1 a2 ... A = a0 + a1×x + a2×x2 + ...
Input on channel b : b0 b1 b2 ... B = b0 + b1×x + b2×x2 + ...
Output on channel c : c0 c1 c2 ... C = c0 + c1×x + c2×x2 + ...
A1 = a1 + a2×x + a3×x2 + ... B1 = b1 + b2×x + b3×x2 + ...
A2 = a2 + a3×x + a4×x2 + ... B2 = b2 + b3×x + b4×x2 + ...
C = A × B = a0×b0 + (a0×b1 + a1×b0)x + (a0×B2 + A1×B1 + A2×b0)×x2
〈!c: rat → C = A×B〉 c ⇐ (a? || b?). c! a×b.
var a0: rat := a· var b0: rat := b· chan d: rat·
〈!c: rat → C = A×B〉 d
|| ((a? || b?). c! a0×b + a×b0. C = a0×B + D + A×b0)
C = a0×B + D + A×b0 ⇐ (a? || b? || d?). c! a0×b + d + a×b0. C = a0×B + D + A×b0
22/29
Review
Boolean Theory laws proof
Number Theory Character Theory
Bunches Sets Strings Lists
Functions Quantifiers
Specification Refinement exact precondition exact postcondition
Program Development Time Calculation real time recursive time
Space Calculation maximum space average space
Scope variable declaration frame
Data Structures array element assignment
Control Structures while loop loop with exit for loop
23/29
Review
Time Dependence wait
Assertions checking backtracking
Subprograms function procedure
Probabilistic Programming random number generator
Functional Programming refinement timing
Recursive Data Definition construction induction
Recursive Program Definition construction induction
Theory Design and Implementation data theory program theory
Data Transformation
Independent Composition sequential to parallel transformation
Interactive Variables Communication Channels
24/29
Disjoint Composition
Independent composition P||Q requires that P and Q have no variables in common, although
each can make use of the initial values of the other's variables by making a private copy. An
alternative, let's say disjoint composition, is to allow both P and Q to use all the variables with
no restrictions, and then to choose disjoint sets of variables v and w and define
P |v|w| Q = (P. v′=v) ∧ (Q. w′=w)
(a) Prove that if P and Q are implementable specifications, then P |v|w| Q is
implementable.
Application Law 〈v→b〉 a = (substitute a for v in b )
Let the remaining variables (if any) be x .
25/29
Disjoint Composition
P. v′=v expand dependent composition
= ∃v′′, w′′, x′′· 〈v′, w′, x′ → P〉 v′′ w′′ x′′ ∧ v′=v′′ one-point v′′
= ∃w′′, x′′· 〈v′, w′, x′ → P〉 v′ w′′ x′′ rename w′′, x′′ to w′, x′
= ∃w′, x′· 〈v′, w′, x′ → P〉 v′ w′ x′ apply
= ∃w′, x′· P
Q. w′=w
= ∃v′, x′· Q
P |v|w| Q = (P. v′=v) ∧ (Q. w′=w) = (∃w′, x′· P) ∧ (∃v′, x′· Q)
26/29
Disjoint Composition
( P |v|w| Q is implementable) definition of implementable
= ∀v, w, x· ∃v′, w′, x′· P |v|w| Q use previous result
= ∀v, w, x· ∃v′, w′, x′· (∃w′, x′· P) ∧ (∃v′, x′· Q) identity for x′
= ∀v, w, x· ∃v′, w′· (∃w′, x′· P) ∧ (∃v′, x′· Q)
= ∀v, w, x· ∃v′· ∃w′· (∃w′, x′· P) ∧ (∃v′, x′· Q) distribution (factoring)
= ∀v, w, x· ∃v′· (∃w′, x′· P) ∧ (∃w′· ∃v′, x′· Q) distribution (factoring)
= ∀v, w, x· (∃v′· ∃w′, x′· P) ∧ (∃w′· ∃v′, x′· Q)
= ∀v, w, x· (∃v′, w′, x′· P) ∧ (∃v′, w′, x′· Q) splitting law
= (∀v, w, x· ∃v′, w′, x′· P) ∧ (∀v, w, x· ∃v′, w′, x′· Q) definition of implementable
= ( P is implementable) ∧ ( Q is implementable)
27/29
Disjoint Composition
Independent composition P||Q requires that P and Q have no variables in common, although
each can make use of the initial values of the other's variables by making a private copy. An
alternative, let's say disjoint composition, is to allow both P and Q to use all the variables with
no restrictions, and then to choose disjoint sets of variables v and w and define
P |v|w| Q = (P. v′=v) ∧ (Q. w′=w)
(b) Describe how P |v|w| Q can be executed.
Make a copy of all variables. Execute P using the original set of variables and in parallel
execute Q using the copies. Then copy back from the copy w to the original w . Then throw
away the copies.
28/29
Disjoint Composition
Independent composition P||Q requires that P and Q have no variables in common, although
each can make use of the initial values of the other's variables by making a private copy. An
alternative, let's say disjoint composition, is to allow both P and Q to use all the variables with
no restrictions, and then to choose disjoint sets of variables v and w and define
P |v|w| Q = (P. v′=v) ∧ (Q. w′=w)
(b) Describe how P |v|w| Q can be executed.
P |v|w| Q ⇐ var cv:=v· var cw:=w· var cx:=x·
(P || 〈v, w, x, v′, w′, x′→ Q〉 cv cw cx cv′ cw′ cx′). w:= cw
29/29
```
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As spectators at the show walk by the cage they look quizzically at the cat and say, “It looks like a Persian but it has short hair. The sign says Exotic. Exotic what?!”
They are bred to meet the Persian standard in every way with one very special exception: the coat has a thick, dense, plush, short coat. The Exotic coat is unique to the breed and gives them a soft, rounded, teddy bear look. Their wonderful coat requires much less combing than a Persian’s and will not mat or tangle. Because of the ease of grooming for this special breed, Exotics are sometimes affectionately referred to as the lazy man’s Persian.
What is it like living with an Exotic? Are they like Persians, or do they resemble their shorthaired ancestors? Over the years, as the type and coat have changed, so has the personality of the Exotic. As the Exotic’s line of Persian ancestors became longer and longer, their temperament has become more and more Persian like. Indeed, there is no longer much difference in the temperament of the two breeds. Exotics have a quiet, endearing nature. Their voices are seldom heard.
The Exotic is an ideal breed that produces a quiet, sweet, peaceful and loyal companion. They are easy going and not much seems to disturb them. In general, they are extremely affectionate. They quietly beg for your attention by just sitting in front of you with an irresistible look focused on your eyes. They will jump in your lap to curl up for a nap or push their wet nose right into your face. Some like to sit on your shoulder and hug you when you pet them. They may or may not sleep with you as some prefer cooler places like the bricks on the hearth or the tiled floor. An Exotic is very comfortable to have in your home. They give you privacy and are not constantly demanding attention. They will, They are just as playful and fun loving as other breeds. They will jump until exhausted trying to catch a toy on a stick, or they will sit and carefully study how to get the toy down from the top of the bookcase where it was placed when you stopped playing with them.
When people call for a pet kitten, they almost always ask for a female, thinking that a girl will be sweeter and more loving. Many also believe that males will be more aggressive and prone to spray. However, neither assumption is correct. Male Exotics are, in general, more affectionate than females. Females can be somewhat more aloof. They always seem to have more important things to do than cuddle with their owner. Exotics mature later than most other breeds, and since all pets should be neutered and spayed at an early age, problems related to spraying and other adult urges need never be a concern.
Exotic kittens exhibit the same level of activity as do Persian kittens. Some breeders say that the Exotic kittens do everything first: open their eyes, climb out of the box, start eating, etc. Adult Exotics enjoy simple pleasures, like watching water drip from a faucet or chasing paper balls around the house.
The easy going nature of the Exotic allows it to fit into your home at any age. Exotics stay playful as adults and bring pleasure for many years. All things considered, the Exotic is a wonderful addition to any family. Adorable to look at, peaceful and clean, what more could you ask for the perfect pet. The Exotic is really the “best of two worlds.”
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Unformatted text preview: ized below. 2 As we shall see shortly, when solving equations involving secant and cosecant, we usually convert back to cosines and sines. However, when solving for tangent or cotangent, we usually stick with what we’re dealt. 638 Foundations of Trigonometry Tangent and Cotangent Values of Common Angles θ(degrees) θ(radians) 0◦ 0 45◦ 60◦ 90◦ undefined √ 3 1 √ 3 √ undefined π 6 π 4 π 3 π 2 cot(θ) √ 0 30◦ tan(θ) 0 3 3 1 3 3 Coupling Theorem 10.6 with the Reference Angle Theorem, Theorem 10.2, we get the following. Theorem 10.7. Generalized Reference Angle Theorem. The values of the circular functions of an angle, if they exist, are the same, up to a sign, of the corresponding circular functions of its reference angle. More specifically, if α is the reference angle for θ, then: cos(θ) = ± cos(α), sin(θ) = ± sin(α), sec(θ) = ± sec(α), csc(θ) = ± csc(α), tan(θ) = ± tan(α) and cot(θ) = ± cot(α). The choice of the (±) depends on the quadrant in which the terminal side of θ lies. We put Theorem 10.7 to good use in the following example. E...
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Jill Tulane University ‘16, Course Hero Intern | 629 | 2,333 | {"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} | 4.0625 | 4 | CC-MAIN-2018-22 | latest | en | 0.871379 |
http://mathhelpforum.com/calculus/84551-maclaurin-series-check-my-work-print.html | 1,505,844,365,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818685912.14/warc/CC-MAIN-20170919164651-20170919184651-00563.warc.gz | 212,786,194 | 2,769 | # Maclaurin series - check my work
• Apr 19th 2009, 06:03 PM
mollymcf2009
Maclaurin series - check my work
Cn someone check my Maclaurin series for:
ln(1+x)
I got:
$\sum^{\infty}_{n=0} (-1)^n \frac{(n-1)!}{(1+x)^n}$
• Apr 19th 2009, 06:07 PM
TheEmptySet
Quote:
Originally Posted by mollymcf2009
Cn someone check my Maclaurin series for:
ln(1+x)
I got:
$\sum^{\infty}_{n=0} (-1)^n \frac{(n-1)!}{(1+x)^n}$
$f(x)=\ln(1+x)$
Then
$f'(x)=\frac{1}{1+x}=\frac{1}{1-(-x)}=\sum_{n=0}^{\infty}(-1)^nx^n$
Now integrating both sides we get
$f(x)=C+\sum_{n=0}^{\infty}(-1)^n\frac{x^{n+1}}{n+1}$
And since $f(0)=\ln(1)=0=C$ we get
$f(x)=\sum_{n=0}^{\infty}(-1)^n\frac{x^{n+1}}{n+1}$ | 318 | 682 | {"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": 0, "codecogs_latex": 7, "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-2017-39 | longest | en | 0.449862 |
https://www.queryhome.com/puzzle/41769/mohan-covered-distance-between-cities-taking-hours-travel | 1,720,897,595,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514512.50/warc/CC-MAIN-20240713181918-20240713211918-00578.warc.gz | 853,027,642 | 25,180 | # Mohan covered a distance of 360 km between two cities, taking............How many hours did he travel at 60 km per hour?
203 views
Mohan covered a distance of 360 km between two cities, taking a total of 13 hours 30 minutes.
If part of the distance was covered at 50 km per hour speed and the rest at 60 km per hour speed.
How many hours did he travel at 60 km per hour?
posted Jan 17, 2022
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https://nationwriters.com/problem-1-j-mcwilliams-swim-club-is-trying-to-calibrate-their-chlorine-pump-to-ensure-that-the-right-amount-of-chlorine-1-0a%C2%A2a%C2%ACaeoe1-5-ppm-of-free-chlorine-is-mixed-into-the-wate/ | 1,652,688,228,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662510097.3/warc/CC-MAIN-20220516073101-20220516103101-00170.warc.gz | 498,956,005 | 18,008 | # Problem 1 J. McWilliams Swim Club is trying to calibrate their chlorine pump to ensure that the right amount of chlorine (1.0–1.5 ppm of free chlorine) is mixed into the water. Thirty samples of fou
Problem 1
J. McWilliams Swim Club is trying to calibrate their chlorine pump to
ensure that the right amount of chlorine (1.0–1.5 ppm of free chlorine) is mixed into the water. Thirty samples of four readings at random times during the week
were taken. These data can be found in the worksheet Prob. 8-26 in the C08Data.xlsx file.
a. Compute the mean and range of each sample, calculate control limits, and plot them on and R control charts.
b. Does the process appear to be in statistical control? Calculate descriptive statistics that may help you to determine the answer to this question. What evidence is there for your conclusion?
Problem 2
The Chair of the Management Department at UB wants to construct a p-chart for determining whether the five (5) faculty members teaching the course are under control with regards to the number of students who fail in the course. Accordingly, the Chair sampled 200 final grades from last year for each instructor and computed the proportion of failures per instructor as 0.01, 0.05, 0.02, 0.04, and 0.03 respectively. Calculate the centerline and the control limits.
Problem 3
A process at Printwright, Inc.’s largest facility is used to make plastic gears for a computer printer. The data found in the worksheet Prob. 9-15 were gathered by a quality analyst. The gears were designed to be 3.57 ± 0.05centimeters (cm) in diameter.
A. Construct a histogram of the data.
b.What can you observe about the shape of the distribution?
c. What would you recommend to the production manager based on your analysis?
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https://statsidea.com/en/r-learn/the-way-to-carry-out-tough-regression-in-r-step-by-step/ | 1,726,100,938,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651405.61/warc/CC-MAIN-20240911215612-20240912005612-00806.warc.gz | 495,627,968 | 19,068 | # The way to Carry out Tough Regression in R (Step-by-Step)
Tough regression is a form we will be able to importance as an extra to familiar least squares regression when there are outliers or influential observations within the dataset we’re operating with.
To accomplish tough regression in R, we will be able to importance the rlm() serve as from the MASS bundle, which makes use of refer to syntax:
Refer to step by step instance presentations the right way to carry out tough regression in R for a given dataset.
### Step 1: Build the Knowledge
First, let’s form a faux dataset to paintings with:
```#form information
df <- information.body(x1=c(1, 3, 3, 4, 4, 6, 6, 8, 9, 3,
11, 16, 16, 18, 19, 20, 23, 23, 24, 25),
x2=c(7, 7, 4, 29, 13, 34, 17, 19, 20, 12,
25, 26, 26, 26, 27, 29, 30, 31, 31, 32),
y=c(17, 170, 19, 194, 24, 2, 25, 29, 30, 32,
44, 60, 61, 63, 63, 64, 61, 67, 59, 70))
#view first six rows of information
x1 x2 y
1 1 7 17
2 3 7 170
3 3 4 19
4 4 29 194
5 4 13 24
6 6 34 2
```
### Step 2: Carry out Familiar Least Squares Regression
Then, let’s are compatible an familiar least squares regression style and form a plot of the standardized residuals.
In observe, we incessantly believe any standardized residual with an absolute price more than 3 to be an outlier.
```#are compatible familiar least squares regression style
ols <- lm(y~x1+x2, information=df)
#form plot of y-values vs. standardized residuals
plot(df\$y, rstandard(ols), ylab='Standardized Residuals', xlab='y')
abline(h=0)```
From the plot we will be able to see that there are two observations with standardized residuals round 3.
This is a sign that there are two doable outliers within the dataset and thus we might get pleasure from appearing tough regression in lieu.
### Step 3: Carry out Tough Regression
Then, let’s importance the rlm() serve as to suit a powerful regression style:
```library(MASS)
#are compatible tough regression style
tough <- rlm(y~x1+x2, information=df)```
To decide if this tough regression style do business in a greater are compatible to the information in comparison to the OLS style, we will be able to calculate the residual usual error of every style.
The residual usual error (RSE) is a method to measure the usual diversion of the residuals in a regression style. The decrease the worth for RSE, the extra carefully a style is in a position to are compatible the information.
Refer to code presentations the right way to calculate the RSE for every style:
```#to find residual usual error of ols style
abstract(ols)\$sigma
[1] 49.41848
#to find residual usual error of ols style
abstract(tough)\$sigma
[1] 9.369349
```
We will be able to see that the RSE for the tough regression style is way not up to the familiar least squares regression style, which tells us that the tough regression style do business in a greater are compatible to the information.
### Backup Sources
The way to Carry out Easy Unbending Regression in R
The way to Carry out A couple of Unbending Regression in R
The way to Carry out Polynomial Regression in R | 878 | 3,103 | {"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-2024-38 | latest | en | 0.818578 |
https://oursland.edublogs.org/2019/01/28/picture-problem-painted-cube-jeff-stack/ | 1,632,067,550,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056892.13/warc/CC-MAIN-20210919160038-20210919190038-00320.warc.gz | 479,496,195 | 10,430 | # Picture Problem Painted Cube 6.EE
The painted cube problem is deceptively complex, and has multiple levels of understanding applied to it. Students are shown a cube broken up into smaller sections. The cube has been painted with a particular pattern based on what faces are visible from the outside of the cube. Sections with one face exposed are painted red on that exposed side. Sections with two exposed sizes are painted pink. Sections with three exposed sides are painted green. Sections with no exposed sides are unpainted, or are clear. Student’s jobs are to count how many of each painted cube there are in a given cube, as well as to devise functions so as to quickly count out how many of a particular section there are. Their answers would be written out in table form; said tables would be given to them. Students can also be asked to find out how much surface area there is in these cubes, assuming that each section is 1 inch on each side.
CCSS.Math.Content.6.EE.A.1
Write and evaluate numerical expressions involving whole-number exponents.
CCSS.Math.Content.6.EE.A.2
Write, read, and evaluate expressions in which letters stand for numbers.
CCSS.Math.Content.6.EE.B.5
Understand solving an equation or inequality as a process of answering a question: which values from a specified set, if any, make the equation or inequality true? Use substitution to determine whether a given number in a specified set makes an equation or inequality true.
CCSS.Math.Content.6.EE.B.6
Use variables to represent numbers and write expressions when solving a real-world or mathematical problem; understand that a variable can represent an unknown number, or, depending on the purpose at hand, any number in a specified set.
CCSS.Math.Content.8.EE.A.2
Use square root and cube root symbols to represent solutions to equations of the form x2 = p and x3 = p, where p is a positive rational number. Evaluate square roots of small perfect squares and cube roots of small perfect cubes. Know that √2 is irrational.
2x2x2 cube Green Pink Red unpainted How many? What’s the equation?
3x3x3 cube Green Pink Red unpainted How many? 12 What’s the equation?
4x4x4 cube Green Pink Red unpainted How many? What’s the equation?
5x5x5 cube Green Pink Red unpainted How many? 54 What’s the equation?
NxNxN Green Pink Red unpainted How many? 4n 6(n-2)^2 What’s the equation? | 541 | 2,377 | {"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} | 4.375 | 4 | CC-MAIN-2021-39 | latest | en | 0.921821 |
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Question:
# As a member of FEMA, you’re required to set up a contingency plan to supply meals to residents of a town devastated by a tornado. A breakfast ration weighs 12 ounces and the lunch and dinner rations weigh 18 ounces each. Assuming a food truck can carry 3 tons and that each resident will receive 3 meals per day, how many residents can you feed from one truck during a 10-day period?
A 200 residents
explanation
One ton = 2,000 pounds, so one truck can carry 6,000 pounds. There are 16 ounces in a pound, so one truck can carry 96,000 ounces. The total daily ration for each resident is 12 ounces + 18 ounces + 18 ounces, or 48 ounces. The number of daily rations supplied can be expressed as 96,000 ÷ 48 = 2,000. Dividing 2,000 by 10 days results in 200 residents who can be fed by one truck during this 10-day period.
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# help
0
110
2
Given ABCD slove for X ?
Apr 27, 2020
#1
+21953
+1
Since it is a parallelogram, consecutive angles are supplementary.
Apr 27, 2020
#2
+1
A + B = 180
(30 + 5X) + (15 + 10X) = 180
45 + 15X = 180
15X = 180 – 45
15X = 135
x = 9
Apr 27, 2020 | 140 | 280 | {"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-2020-40 | latest | en | 0.648961 |
https://www.jiskha.com/display.cgi?id=1285717840 | 1,516,611,214,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891196.79/warc/CC-MAIN-20180122073932-20180122093932-00364.warc.gz | 915,656,796 | 3,488 | # Physics
posted by .
a girl jumps off a high diving platform with a horizontal velocity of 2.77m/s and lands in the water 1.2s later the acceleration of gravity is 9.8m/s squared
how high is the platform and how far from the base of the platform does she land?
• Physics -
first do the vertical problem
We will have to assume she has no initial speed up, just horizontal
h = h initial + Vi t - 4.9 t^2
0 = h initial + 0 (1.2) - 4.9 (1.2)^2
solve for h initial
her horizontal speed never changed so horizontal distance is 2.77*1.
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More Similar Questions | 629 | 2,280 | {"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-2018-05 | latest | en | 0.869394 |
https://stats.stackexchange.com/questions/122430/whats-the-difference-between-estimating-equations-and-method-of-moments-estimat | 1,653,168,538,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662541747.38/warc/CC-MAIN-20220521205757-20220521235757-00051.warc.gz | 610,094,001 | 67,502 | # What's the difference between estimating equations and method of moments estimators?
From my understanding, both are estimators that are based on first providing an unbiased statistic $T(X)$ and obtaining the root to the equation:
$$c(X) \left( T(X) - E(T(X)) \right) = 0$$
Secondly both are in some sense "nonparametric" in that, regardless of what the actual probability model for $X$ may be, if you think of $T(\cdot)$ as a meaningful summary of the data, then you will be consistently estimating that "thing" regardless of whether that thing has any probabilistic connection with the actual probability model for the data. (e.g. estimating the sample mean from Weibull distributed failure times without censoring).
However, method of moments seems to insinuate that the $T(X)$ of interest must be a moment for a readily assumed probability model, however, one estimates it with an estimating equation and not maximum likelihood (even though they may agree, as is the case for means of normally distributed random variables). Calling something a "moment" to me has the connotation of insinuating a probability model. However, supposing for instance we have log normally distributed data, is the method of moments estimator for the 3rd central moment based on the 3rd sample moment, e.g. $$\hat{\mu_3} = \frac{1}{n}\sum_{i=1}^n \left( X_i - \bar{X} \right)^3$$
Or does one estimate the first and second moment, transform them to estimate the probability model parameters, $\mu$ and $\sigma$ (whose estimates I will denote with hat notation) and then use these estimates as plug-ins for the derive skewness of lognormal data, i.e.
$$\hat{\mu_3} = \left( \exp \left( \hat{\sigma}^2 \right) + 2\right) \sqrt{\exp \left( \hat{\sigma}^2-1\right)}$$
The most common justification of the method of moments is simply the law of large numbers, which would seem to make your suggestion of estimating $\mu_3$ by $\hat{\mu}_3$ "method of moments" (and I'd be inclined to call it MoM in any case).
However, a number of books and documents, such as this for example (and to some extent the wikipedia page on method of moments) imply that you take the lowest $k$ moments* and estimate the required quantities for given the probability model from that, as you imply by estimating $\mu_3$ from the first two moments.
*(where you need to estimate $k$ parameters to obtain the required quantity)
--
Ultimately, I guess it comes down to "who defines what counts as method of moments?"
Do we look to Pearson? Do we look to the most common conventions? Do we accept any convenient definition? --- Any of those choices has problems, and benefits.
The interesting bit, to me, is whether one can always or almost always reparameterize a parametric family to characterize an estimation problem in EE as the solution to the moments of a (possibly bizarre) distribution function?
Clearly there are large classes of distribution for which method of moments would be useless.
For an obvious example, the mean of the Cauchy distribution is undefined.
Even when moments exist and are finite, there could be a large number of situations where the set of equations $f(\mathbf{\theta},\mathbf{y})=0$ has 0 solutions (think of some curve that never crosses the x-axis) or multiple solutions (one that crosses the axis repeatedly -- though multiple solutions aren't necessarily an insurmountable problem if you have a way to choose between them).
Of course, we also commonly see situations where a solution exists but doesn't lie in the parameter space (there may even be cases where there's never a solution in the parameter space, but I don't know of any -- it would be an interesting question to discover if some such cases exist).
I imagine there can be more complicated situations still, though I don't have any in mind at the moment.
• The wikipedia article seems to say that MoM is used when you are interested in estimating the DF of iid data, $X_1, X_2, \ldots, X_n \sim_{iid} F$ where $F$ has a known parametric distribution and by estimating the necessary $k$ moments for solving systems of equations for the $p$ parameters, $p < k???$ you get the appropriate estimator. The interesting bit, to me, is whether one can always or almost always reparameterize a parametric family to characterize an estimation problem in EE as the solution to the moments of a (possibly bizarre) distribution function? Nov 3, 2014 at 19:18
• Hi AdamO -- I wrote a comment but it started to get too long so I moved it up to the end of my answer. If you can expand on where that interest in your last sentence lies, I think that might make for some stimulating areas to explore (I don't claim to know any more than you on this, but I'd love to try to learn more). Nov 3, 2014 at 21:06
Estimating equations is a more general method, it doesnt specify from where you get the estimating equation. Maximum likelihood is also an example of estimating equations, as it leads to the score equation. Various forms of quasi (or pseudo)-likelihood are other examples! as are method of moments.
• Likelihoods with local extrema do not have unique solutions when solving their corresponding score functions (but directly maximizing the likelihood would address this), so to me, there are some pathological cases where ML is more general than EE... but for somewhat regular parametric models, their ML solution can be obtained with score equations as EEs. I am not as familiar with MoM. Nov 3, 2014 at 23:10 | 1,252 | 5,471 | {"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} | 3.5625 | 4 | CC-MAIN-2022-21 | longest | en | 0.908225 |
https://infinitylearn.com/questions/physics/speed-particle-executing-shm-given-by-equation-v | 1,702,260,376,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679103464.86/warc/CC-MAIN-20231211013452-20231211043452-00536.warc.gz | 351,454,577 | 12,918 | Simple hormonic motion
Question
# The speed of a particle executing SHM is given by the equation ${\mathrm{v}}^{2}=144-9{\mathrm{x}}^{2}$ then the wrong statements among the following is
Moderate
Solution
## maximum velocity in SHM is at x = 0 that is when there is no displacement. Thus putting x = 0, the maximum velocity comes out to be = 12 units $\begin{array}{l}\mathrm{V}=\sqrt{144-9{\mathrm{x}}^{2}}\\ =3\sqrt{{4}^{2}-{\mathrm{x}}^{2}}\end{array}$ Comparing with the equation $\mathrm{V}=\mathrm{\omega }\sqrt{{\mathrm{A}}^{2}-{\mathrm{x}}^{2}}$ we get As a result T = $2\mathrm{\pi }/3$ Also we know acceleration = $-{\mathrm{\omega }}^{2}\mathrm{x}$ in the case of SHMFor maximum acceleration $\mathrm{x}=\mathrm{A}=4$ which gives us maximum acceleration as 36 units, Hence option 3 is wrong.
Get Instant Solutions | 266 | 835 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 7, "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-2023-50 | latest | en | 0.637106 |
https://www.jiskha.com/questions/568084/use-alaytical-and-graphical-methods-to-solve-the-inequality-3x-6-7x-2-5-0 | 1,582,003,190,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875143505.60/warc/CC-MAIN-20200218025323-20200218055323-00115.warc.gz | 797,848,509 | 4,677 | # algebra II
use alaytical and graphical methods to solve the inequality.
(3x+6/7x^2+5) > 0
1. 👍 0
2. 👎 0
3. 👁 151
1. The denominator is always +
so the real question is when is the numerator >0
3x + 6 >0
3 x > -6
x > -2
1. 👍 0
2. 👎 0
posted by Damon
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# Bollinger Bands
Posted in Indicators of technical analysis
Bollinger bands indicator was created by a financial analyst John Bollinger in 1980s. Bollinger bands are a complex technical analysis indicator which calculation is based on the volatility of the market (similar to Keltner channel indicator or Donchian channel indicator). The main idea behind the indicator is that it highlights the overbought and oversold conditions on the market but yet some other traders use it to identify strong trends (so as a matter of fact they use it as a trend-following indicator).
The Bollinger bands indicator is quite simple to calculate and display it in the price chart. As its name says the indicator consists of one moving average and two bands that surround him – one band from the upper side and the other from below. As the main moving average is used 20-days Simple moving average but many traders use to change it for Exponential moving average (as the EMA can adapt faster to changing conditions on the market). The bands are calculated as the standard deviations of the last 20 days prices. The calculation of the standard deviation lies quite behind this article, but for people that like statistics the standard deviation formula looks like:
And for people who are not interested in statistical formulas at all the following table that really makes it quite simple should work just fine. So you can calculate the standard deviation like:
Price Average price Pi - PAVG (Pi – PAVG)2 (Pi – PAVG)2 x 1/N 100 70 +30 900 300.00 80 70 +10 100 33.33 30 70 -40 1600 533.33 ∑ 866.66
The number 866.66 is the variance of these prices. You can calculate the standard deviation as the square root of the variance, so in this case the standard deviation equals to 29.44. If you still don’t understand the calculation, don’t worry it is also one of the Excel functions.
Next we are going to explain a 3-day Bollinger Bands calculation. The formula for Bollinger bands calculation then looks like this:
1. Calculate moving average of chosen period of time. John Bollinger used simple moving average, so you would just count the days and divide it by the number of days. In our case (the table above) the simple moving average of the last three days’ prices is 70 USD.
2. Calculate the Standard deviation of the chosen days’ prices and double it. In our case the standard deviation is 29.44 USD so doubled it would be (29.44 * 2) = 58.88 USD.
3. Calculate the upper band and lower band by adding the doubled standard deviation to the average price and by subtracting it from the price.
Middle curve = Simple moving average
Upper band = SMA + (2 * Standard deviation)
Lower band = SMA – (2 * Standard deviation)
Now we have 3 curved that follow and surround the price of the underlying asset. The middle curve for the most actual day is set to 70 USD. The upper level is set to 128.88 USD and the lower band is set to 11.12 USD. These bands are valid for the most actual day. When the next (new) trading day ends, the newest close price is included in the calculation instead of the oldest one - so the simple moving average changes a bit and the bands as well. It is also important to realize that the bands width depends on the market volatility - the more volatile the prices are the wider the bands become and vice versa (only hypothetically - should the price remain unchanged all the time then the bands would equal to simple moving average so instead of three curves we would get just one).
How to use the indicator in technical trading:
The theory that lies behind this indicator says that in 68.2 % of the time the price should move inside the bands when they are created from SMA ± (1 * Standard deviation). It also applies that if you double the standard deviation the price should move inside the bands in 95.4 % of the time. If you triple the standard deviation the price should move inside the Bollinger bands in 99.6 % of all the time.
Now it is quite easy to figure out how to use the Bollinger bands indicator. We use double the standard deviation so 95.4 % of the prices should move inside our bands. Only 4.6 % of prices should move outside them. If they do so it is considered to be an extraordinary situation that can’t be lasting long and the prices should get back into the limits soon. All this probability theory and statistics rules applies to normal distribution. There is a discussion whether the underlying assets prices and the low number of days (used in the Bollinger bands calculation) can be considered for normal distribution. Well, the truth is that various practical researches proved that the probability of staying prices inside the bands can be lower – e.g. not 95.4 % of the time, but just 85 % - 90 % of the time. Maybe that is the reason why some traders prefer not to choose double the Standard deviation but even triple it. Then they can get much more reliable signals to go short or long. In later years John Bollinger introduced three other variations of Bollinger bands. They are called: Impulse BB, percent Bandwidth (%B) and Bandwidth delta. The next image shows how the Bollinger bands looks in a candlestick chart.
Signals when to buy and sell:
The use of Bollinger bands indicator varies a lot. It is commonly used as an oscillator that should identify the oversold and overbought conditions on the market. That means when the price gets above (or even touches) the upper band it gives a signal that the price is too high (the market is overbought) and the time has come to go short. There is increased probability that the price will get back below the upper band and even go lower. In other words, after touching or overcoming the upper limit the price starts to decline and later it even touches the middle Bollinger band (simple moving average). This is a form of price correction that can come either in ranging market or in strongly trending market, doesn’t matter. But after all there is a difference. While in strongly trending market the price will probably go back to the Middle Bollinger band and then it bounces again upwards, in ranging market there is more probable that the price will break through the Middle band and even continue to the other (lower) band. We can simplify this and say that in trending market the price will probably move between the Simple moving average and one of the bands. In ranging market the price will probably move between both bands, sometimes it bounces back from the middle, sometimes it breaks through it.
As mentioned at the beginning of this article, some traders don’t use it for identifying overbought and oversold market conditions but as a trend-following indicator. They wait while price breaks through the upper or lower band (in ranging market) and take it as a signal that a new trend has just begun (in other words that the long-lasting war between bulls and bears has just been won by one of these groups and that is the reason why the price has broke the upper or lower band).
As there is many ways how to use the Bollinger bands and some of them are opposed, traders who don’t want to rely just on this indicator use to combine it with some other indicators. There were some trading systems tested with ADX indicator (ADX identified the strength of the trend while Bollinger bands provided signals to buy or sell at the right time). Some other people prefer to combine it with chart patterns like supports and resistances, price channels and others.
Despite Bollinger bands indicator is already one of the most popular indicator among traders in our opinion it offers great opportunities to develop it even further. E.g. some traders already adapted the indicator when they used Exponential moving average instead of the basic Simple moving average. But it would be even more interesting to combine and test it with KAMA indicator. Unlikely the Simply moving average or Exponential moving average the KAMA indicator is based on the market volatility as well (as the Bollinger bands is) so it could generate very interesting signals. VIDYA indicator or Variable index dynamic average is another moving average that uses market volatility for its calculation. The opportunities how to develop the Bollinger bands indicator are almost unlimited.
As with almost all of the technical indicators the best thing every trader can do is to test his own data, his own settings, and his own rules how to trade. Surprisingly, sometimes the best result can be achieved with settings that are not common and rules that are quite strange at a first glance – the more things a trader can change and experiment with the better for him and his trading results. If you are interested in a deeper study of the indicator the following link leads to the technical indicators in Excel files for download. | 1,868 | 8,824 | {"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-2019-22 | latest | en | 0.928799 |
https://www.jiskha.com/questions/22269/you-and-a-friend-play-the-following-game-you-pay-your-friend-3-each-turn-and-then-flip-a | 1,627,457,623,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153531.10/warc/CC-MAIN-20210728060744-20210728090744-00685.warc.gz | 874,218,970 | 5,272 | # Probability
You and a friend play the following game: You pay your friend \$3 each turn and then flip a fair coin. It it’s tails, your friend pays you \$(2^n), where n is the number of times you’ve flipped the coin, and the game ends. If it’s heads, you have the choice of stopping and continuing. If you have m dollars to start with, and you play the game either until you win or until you have no money left, what will you win on the average?
Ah the good old St Petersburg Paradox.
Well, if you and your friend each have a infinite amount of money, then you can expect to win and infinite amount, on average.
1. 👍
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posted by on .
I am so confused with some word problems math is not my strong point here is the problem if you know how to solve it please tell me how to solve but just don't give me the answer
Austin has 42 paperback books that are either Mysteries or science fiction. If he has 16 more Mysteries than science fiction books, how many mysteries does he have?
Let x be the number of mysteries. Then there will be x - 16 science fiction books.
Write
x + (x -16) = 2x - 16 = 42
which reduces to
2x = 58
Solve for x.
i don't know how you got that but i talked to a real tutor and 2x doesnt equal 58 it equals 26
youre right danielle; drwls made a slight mistake in his algebra; check the solution that i posted
M = MYSTERY BOOKS
S = SCIENCE BOOKS
M - 16 = S because he was 16 more mysteries than scifis.
M + S = 42 because he has a total of 42 mysteries and scifis
M + S = 42
M - 16 = S
M + (M - 16) = 42
M + M - 16 = 42
2M - 16 = 42
2M = 26
M = 13
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Sample edHelper.com - Exponents Worksheet
Name _____________________________ Date ___________________
Exponents
(Answer ID # 0367577)
Complete each of the equations. Explain your answer.
1.
5 (7 1) =
2.
5h (7h h) =
3
Complete each of the equations. Explain your answer.
4.
(2 8) = 144
5.
(2j 8j) = 144j3
Complete each of the equations. Explain your answer.
6.
e e e e e =
7.
8 (5 6) =
8
Complete each of the equations. Explain your answer.
9.
+ c + c + c + c = 5c
10.
9j + 9j + 9j + 9j + = 45j
11
Complete each of the equations. Explain your answer.
12.
5 (4 + 8) =
13.
5e (4e + 8e) =
14
Evaluate to a single number.
15 23
16 62
17 50
Evaluate to a single number.
18.
50 + 83
19.
23 + 74
Simplify.
20.
4-8 × 4-5
21.
34 × 36
Simplify.
22 (16x)3 (32x)2
23 (-5x)0
Simplify.
24.
98(-9)-6
÷ 9-6
25.
512(-5)3
Simplify.
26 (-4x) ÷ (-9x)3
27 (32x)2 ÷ (22x)0
Simplify.
28 (-9h3q6)(-5h5)
29 (-l4)(12l3n5)
Z represents a missing exponent. What value should Z be?
30.
(-2q4)(-8qZo6) = 16q10o6
31.
(-8l4a6)(10l2)(3lZ) = -240l9a6
Simplify.
32 (-8g-6o6)(-10g3)(3g4o-4)
33 (-8d-3v-3)(-4d-5v-5)(d-3)
Z represents a missing exponent. What value should Z be?
34.
(6y3t3)(-8yZt-2) = -48y5t
35.
(-9o-2w2)(-9o3w3)(5oZ) = 405o-3w5
Simplify.
36 (-5s-5q5)5(-s-6q6)5
37 (2y3)4
Z represents a missing exponent. What value should Z be?
38.
(-z-2w2)4(z-2)Z = z-16w8
39.
(2p-4)Z = 16p-16
Simplify.
40.
-5v5b2-8v6b3
41.
-5d2o37d3o6
Z represents a missing exponent. What value should Z be?
42.
-4z6-12zZf6
=
z3f6
43.
6pZs37p6s2
=
6s7
Simplify.
44.
(-10r6o2)(-5r5o3)2r2
45.
-d5(-11d3)(-2d2t3)
Z represents a missing exponent. What value should Z be?
46.
-7w3x59w6
-9w2-8wZ
=
-7x58w5
47.
-10t5v6(11t3vZ)(11t3v6)
=
-10121tv3
Simplify.
48.
(-q3o5)(-3)2-4q5o3
49.
(-12m5i5)(2)55m6
Z represents a missing exponent. What value should Z be?
50.
9jZ(-3j2x2)5
=
-127j7x10
51.
-11a4(-9a2)(2aZe6)3
=
1172a7e18
Rewrite the number in scientific notation.
52 3,000,000
53 9.52e-12
Rewrite the number in decimal notation.
54 4.64 x 1012
55 8.2 x 10-12
Simplify and write the answer in scientific notation.
56 (4.1 x 10-7) - (3.4 x 10-8)
57 (3.6 x 107)(6 x 10-5)
Simplify and write the answer in scientific notation.
58.
5 x 1054 x 10-5
59.
(7 x 105)(8 x 10-6)4 x 107
Sample
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Answer key also includes questions | 1,158 | 2,620 | {"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-2017-17 | latest | en | 0.528615 |
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