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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 |
http://blog.gbrueckl.at/2015/06/calculating-pearson-correlation-coefficient-dax/?share=google-plus-1 | 1,558,500,295,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256763.42/warc/CC-MAIN-20190522043027-20190522065027-00473.warc.gz | 29,610,017 | 20,868 | # Calculating Pearson Correlation Coefficient using DAX
The original request for this calculation came from one of my blog readers who dropped me a mail asking if it possible to calculated the Pearson Correlation Coefficient (PCC or PPMCC) in his PowerPivot model. In case you wonder what the Pearson Correlation Coefficient is and how it can be calculated – as I did in the beginning – these links What is PCC, How to calculate PCC are very helpful and also offer some examples and videos explaining everything you need to know about it. I highly recommend to read the articles before you proceed here as I will not go into the mathematical details of the calculation again in this blog which is dedicated to the DAX implementation of the PCC.
UPDATE 2017-06-04:
Daniil Maslyuk posted an updated version of the final calculation using DAX 2.0 which is much more readable as it is using variables instead of separate measures for every intermediate step. His blog post can be found at https://xxlbi.com/blog/pearson-correlation-coefficient-in-dax/
Anyway, as I know your time is precious, I will try to sum up its purpose for you: “The Pearson Correlation Coefficient calculates the correlation between two variables over a given set of items. The result is a number between -1 and 1. A value higher than 0.5 (or lower than –0.5) indicate a strong relationship whereas numbers towards 0 imply weak to no relationship.”
The two values we want to correlate are our axes, whereas the single dots represent our set of items. The PCC calculates the trend within this chart represented as an arrow above.
The mathematical formula that defines the Pearson Correlation Coefficient is the following:
The PCC can be used to calculate the correlation between two measures which can be associated with the same customer. A measure can be anything here, the age of a customer, it’s sales, the number of visits, etc. but also things like sales with red products vs. sales with blue products. As you can imagine, this can be a very powerful statistical KPI for any analytical data model. To demonstrate the calculation we will try to correlate the order quantity of a customer with it’s sales amount. The order quantity will be our [MeasureX] and the sales will be our [MeasureY], and the set that we will calculate the PCC over are our customers. To make the whole calculation more I split it up into separate measures:
1. MeasureX := SUM(‘Internet Sales’[Order Quantity])
2. MeasureY := SUM(‘Internet Sales’[Sales Amount])
Based on these measures we can define further measures which are necessary for the calculation of our PCC. The calculations are tied to a set if items, in our case the single customers:
1. Sum_XY := SUMX(VALUES(Customer[Customer Id]), [MeasureX] * [MeasureY])
2. Sum_X2 := SUMX(VALUES(Customer[Customer Id]), [MeasureX] * [MeasureX])
3. Sum_Y2 := SUMX(VALUES(Customer[Customer Id]), [MeasureY] * [MeasureY])
4. Count_Items := DISTINCTCOUNT(Customer[Customer Id])
Now that we have calculated the various summations over our base measures, it is time to create the numerator and denominator for our final calculation:
1. Pearson_Numerator :=
2. ([Count_Items] * [Sum_XY]) – ([MeasureX] * [MeasureY])
3. Pearson_Denominator_X :=
4. ([Count_Items] * [Sum_X2]) – ([MeasureX] * [MeasureX])
5. Pearson_Denominator_Y :=
6. ([Count_Items] * [Sum_Y2]) – ([MeasureY] * [MeasureY])
7. Pearson_Denominator :=
8. SQRT([Pearson_Denominator_X] * [Pearson_Denominator_Y])
Having these helper-measures in place the final calculation for our PCC is straight forward:
1. Pearson := DIVIDE([Pearson_Numerator], [Pearson_Denominator])
This [Pearson]-measure can then be used together with any attribute in our model – e.g. the Calendar Year in order to track the changes of the Pearson Correlation Coefficient over years:
For those of you who are familiar with the Adventure Works sample DB, this numbers should not be surprising. In 2005 and 2006 the Adventure Works company only sold bikes and usually a customer only buys one bike – so we have a pretty strong correlation here. However, in 2007 they also started selling Clothing and Accessories which are in general cheaper than Bikes but are sold more often.
This has impact on our Pearson-value which is very obvious in the screenshots above.
As you probably also realized, the Grand Total of our Pearson calculation cannot be directly related to the single years and may also be the complete opposite of the single values. This effect is called Simpson’s Paradox and is the expected behavior here.
[MeasuresX] and [MeasureY] can be exchanged by any other DAX measures which makes this calculation really powerful. Also, the set of items over which we want to calculated the correlation can be exchanged quite easily. Below you can download the sample Excel workbook but also a DAX query which could be used in Reporting Services or any other tool that allows execution of DAX queries.
Sample Workbook (Excel 2013): Pearson.xlsx
DAX Query: Pearson_SSRS.dax
## 12 Replies to “Calculating Pearson Correlation Coefficient using DAX”
1. Dear Gerhard,
Great job!!!!
Thank very much for your post, it has been very useful to me.
I have combined your correlation calculation with a calculation of the linear regression slope, I created myself, with great benefits for my analysis.
I wonder if there is any way to select in DAX the two variable you want to calculate and the n term of the correlation from a Pivot Table/ Slicer/ Parametric table to make the correlation calculation more flexible and dynamic.
Thank very much
• Hi Alberto,
I am afraid there is not much we can do here in order to make it more flexible – you always need two fixed measures and a fixed column for the calculation.
However, you could create a a dummy-table for which holds all your measures and use SWITCH() in order to select the value of the measure which you want to return
something similar to this:
http://blogs.msdn.com/b/analysisservices/archive/2010/04/12/time-intelligence-functions-in-dax.aspx
=IF( COUNTROWS( VALUES( DimPeriod[Period]))=1,
IF( VALUES( DimPeriod[Period]) = “Current”, [Sales],
IF( VALUES( DimPeriod[Period]) = “MTD”, [Sales](DATESMTD(DimDate[Datekey])),
IF( VALUES( DimPeriod[Period]) = “QTD”, [Sales](DATESQTD(DimDate[Datekey])),
IF( VALUES( DimPeriod[Period]) = “YTD”, [Sales](DATESYTD(DimDate[Datekey])),
IF( VALUES( DimPeriod[Period]) = “LastYear”, [Sales](DATEADD(DimDate[Datekey],-1,YEAR)),
IF( VALUES( DimPeriod[Period]) = “PriorYearMTD”, [Sales](DATEADD(DATESMTD(DimDate[Datekey]),-1,YEAR)),
IF( VALUES( DimPeriod[Period]) = “PriorYearQTD”, [Sales](DATEADD(DATESQTD(DimDate[Datekey]),-1,YEAR)),
IF( VALUES( DimPeriod[Period]) = “PriorYearYTD”, [Sales](DATEADD(DATESYTD(DimDate[Datekey]),-1,YEAR)),
BLANK())))))))),[Sales])
you can then use this measure to drive your [MeasureX]
hope that helps,
-gerhard
I will try it, as far I understand I will have to implement the Switch function in each of the calculations: MeasureX and MeasureY.
Thank very Much !!!!
• if you want both of them to be dynamic, and selectable via slicer – yes
-gerhard
I’ve noticed that the Pearson measure doesn’t work right if measures X and Y are not SUMs. For example, if you change SUM to AVERAGE, the correlation coefficient will be wrong.
To address that, I would introduce two new measures:
Sum_X := SUMX(VALUES(Customer[Customer Id]), [MeasureX])
Sum_Y := SUMX(VALUES(Customer[Customer Id]), [MeasureX])
I would also modify the existing measures slightly:
Pearson_Numerator:=([Count_Items] * [Sum_XY]) – ([Sum_X] * [Sum_Y])
Pearson_Denominator_X:=([Count_Items] * [Sum_X2]) – ([Sum_X] ^ 2)
Pearson_Denominator_Y:=([Count_Items] * [Sum_Y2]) – ([Sum_Y] ^ 2)
This way, the correlation coefficient will be calculated correctly even if measures X and Y are not SUMs.
• yes, you are of course right
i just wanted to show the general approach of how PCC can be calculated in DAX.
With DAX 2.0 and the support of variables, it would be even simpler to put everything into a single measure and only have to specify every input parameter (items, MeasureX, MeasureY) once
I might update the post in the future when I have some more time but for the time being I will leave it up to the reader to do this
thanks,
-gerhard
3. Thank you very much. Do you know if there is a way of calculating the P-Value from r on DAX?
• do you have an example for me?
Ideally an Excel file or something to show what you want to achieve? | 2,051 | 8,497 | {"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} | 3.96875 | 4 | CC-MAIN-2019-22 | latest | en | 0.94089 |
https://jeeneetqna.in/464/when-litres-mixed-with-litres-each-the-moles-hcl-formed-equal | 1,653,226,655,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662545548.56/warc/CC-MAIN-20220522125835-20220522155835-00102.warc.gz | 400,318,374 | 9,662 | # When 22.4 litres of H2(g) is mixed with 11.2 litres of Cl2(g), each at S.T.P, the moles of HCl(g) formed is equal to
more_vert
When 22.4 litres of H2(g) is mixed with 11.2 litres of Cl2(g), each at S.T.P, the moles of HCl(g) formed is equal to
(a) 1 mol of HCl(g)
(b) 2 mol of HCl(g)
(c) 0.5 mol of HCl(g)
(d) 1.5 mol of HCl(g).
more_vert
done_all
Answer - (a) 1 mol of HCl(g)
Solution -
H+ Cl2 ⇒ 2HCl
no. of moles of H2 = 22.4/22.4 = 1 mole
no. of moles of Cl2 = 11.2/22.4 = 0.5 mole
thus limiting reagent is Cl2
for 1 mole Cl2 → 2 mole HCl
for 0.5 mole Cl2 → 1 mole HCl | 255 | 593 | {"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-2022-21 | latest | en | 0.721021 |
https://goprep.co/ex-3.b-q11-2x5y-8-3-3x-2y-5-6-solve-for-x-and-y-i-1njyuk | 1,611,550,615,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703564029.59/warc/CC-MAIN-20210125030118-20210125060118-00276.warc.gz | 371,705,071 | 59,036 | # Solve for x and y:
We have,
…eq.1
…eq.2
Let us first simplify eq.1 & eq.2 by taking LCM of denominators,
Eq.1
6x + 15y = 8 …eq.3
Eq.2
18x – 12y = 5 …eq 4
To solve these equations, we need to make one of the variables (in both the equations) have same coefficient.
Lets multiply eq.3 by 18 and eq.4 by 6, so that variable x in both the equations have same coefficient.
Recalling equations 3 & 4,
6x + 15y = 8 [×18]
18x – 12y = 5 [×6]
108x + 270y = 144
108x – 72y = 30
On solving these two equations we get,
342y = 114
Substitute in eq.3/eq.4, as per convenience of solving.
Thus, substituting in eq.3, we get
6x + = 8
6x + 5 = 8
6x = 8 – 5
6x = 3
Hence, we have and
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RELATED QUESTIONS : | 461 | 1,589 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2021-04 | latest | en | 0.84912 |
https://www.answers.com/Q/How_high_does_percent_error_have_to_be_considered_accurate | 1,556,217,265,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578732961.78/warc/CC-MAIN-20190425173951-20190425195951-00289.warc.gz | 628,590,543 | 51,119 | # How high does percent error have to be considered accurate?
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That depends a lot on the application. In some cases, a 10% error (or even more) may be acceptable, in other, 1%, in others, you need a much higher precision.
Generally, a %Error of approximately 5% is regarded as accurate. However, this is only a guideline for small experiments or data sets. Also, one should ensure that the relative standard deviation is less than 5% also. This ensures that the data set is precise.
1 person found this useful
# Can percent error be negative?
Sometimes you will take the absolute value of the percent error because your estimated number could be less than the theoretical, meaning the calculation is negative. But an absolute value is always positive. A percent error can be left as a negative though, and this would be perfectly acceptable (o ( Full Answer )
# How do you find percent error?
To determine the Percent value , you must have two variables: True value , the Recorded or Measured value , and it is represented by the following equation: . ( Recorded/Measured value - True value ) / True value * 100% . e.g (20.000 - 21.571) / 21.751*100% = -7.223%.
# What is the percent error formula?
Percent Error is the difference between the true value and the estimate divided by the true value and the result is multiplied by 100 to make it a percentage. The percent error obviously can be positive or negative; however, some prefer taking the absolute value of the difference. The formula is t ( Full Answer )
# Percent error formula?
Divide the calculated or estimated error by the magnitude of the measurement. Take the absolute value of the result, that is, if it is negative, convert to positive. This would make the percent error = | error / measurement |.
# Is there a difference percent error and percentage error?
The difference: - age (hey, it's not wrong...) In general, probably not - percent and percentage are often used interchangeably. The context of use may warrant a difference though, if strict semantics are being followed: "Percent error" would refer to the the maximum potential difference be ( Full Answer )
# What is the formula for percent error?
The formula of percent error is percent error= Your value/accepted value x 100 ------------ The definition of error is: difference between the accepted true value and the measured value of a quantity or parameter. But this is the absolute error . The relative (percent error) is: (m ( Full Answer )
# What if the percent error is negative?
9f you are using the equation %error =[(oberved value - true value)/true value]x100 a negative percent indicates the observed value is less than the true value by the calculated percent.
# Is a lie detector 99.9 percent accurate?
No. They can be fooled which is why they are not admissible as evidence in a court of law (at least in the US)
# How do you calculate percent error in physics?
Take the difference between your experimental value and your known value, and divide that difference by your know value. Say you experimentally found the force of gravity to be 8m/s 2 , and you know that the true/known value is 9.8 m/s 2 , the percent error would be l 8 - 9.8 l = 0.1836 or 18. ( Full Answer )
# Absolute percent error?
Absolute percent error is the Percent Difference between two values. Applying the equation for Absolute error . For example: 21.571 is the True value 20.000 is the Recorded Value . Thus: ( Recorded Value ) - ( True value ) = Absolute error ( 21.571 ) - ( 20.000 ) = 1.571 W ( Full Answer )
# Is an error considered an at bat?
Yes. If there is an error on a hit ball, the at bat counts against the batter as if it was an out.
# What is a percent error?
It is a measure measurement of the amount of error made in an experiment. It is obtained by comparing the actual result, with the result gotten from the experiment. % error = [(experimental value - true value) / true value] x 100
# What is the reason for percent error?
Percent error is used when you are comparing your result to a knownor accepted value. It is the absolute value of the difference ofthe values divided by the accepted value, and written as apercentage. Percent error is equal to the difference divided by theknown times 100 percent.
# Why do you calculate percent error?
you take the actual minus the theoretical number and divide that by the actual number and multiply it by 100.
# What is the definition of percent error?
the difference between the true value and the estimate divided by the true value. The result is multiplied by 100 to make a percentage * * * * * Nearly correct. It is the estimated (or measured) minus the true value, the difference divided by the true value and then the result multiplied by 100 ( Full Answer )
# Can the answer of a percent error problem be 300?
Yes, your percent error can be over 100%. This means that somewhere during your experiment you made a big error.
# Why are there percent errors?
i want ot say 1 in 8 but the codes have changed since all the brigde failures and i do not know
# What Percent error is an acceptable range?
An acceptable error range depends on the application. Forexample, a 5-10% error range on political pollingis commonly accepted as reasonable. A similar rate for surgicalerror would be appaling and targets tend to be in the 0.1-1% range. In general, an error range of 5%-35% is acceptable, with 0- ( Full Answer )
# What does a high standard error mean?
It means theres a high amount of variation between the results used to calculate the mean value for a particular sample or experiment
# How are the error and the percent error of a measurement calculated?
Take the correct value, subtract the value you got, and then divide that figure by the correct value. Then take the absolute value of that and multiply by 100. For example, say I weighed something and got that it was 2.5 grams, but it really was 2.7 grams. 2.7-2.5=.2. .2/2.7=.074. .074*100=7.4. Thus ( Full Answer )
# What is a good value of percent error?
That depends a lot on the application. In electronic circuits, an error of 5-10% or even more is often acceptable; for some applications, you need a much greater precision - even a millionth or less in some cases.\n\n
# A high percent error refers to?
A high percent error indicates that a certain value is very farfrom the accepted value. Percent error is the comparison of anestimated value to an exact one.
# How do you get percent error?
The definition of error is: difference between the accepted true value and the measured value of a quantity or parameter. But this is the absolute error . The relative (percent error) is: (measured value - accepted true value) . 100/accepted true value This value is exprssed as a percent ( Full Answer )
# Is 100 percent accurate the sonogram?
Well an sonogram use waves that are sent out an the bounce back off surfaces that it is directed to a more accurate machine would be a CAT scan.
# Percent error of O in MgO?
If you think to magnesium oxide and to concentration of oxygen in MgO the answer is 40,007 33 % oxygen.
# Why are parallax errors considered systematic errors?
simply speaking, systematic errors are those you can improve on( so if you have a systematic error, its probably your fault). Random errors are unpredictable and cannot be corrected. A parallax error can be corrected by you and if there is a parallax error, its probably your fault.
# Are pregnsncy tests 100 percent accurate?
no, i don't believe they are, because there are such things called "False pregnancies" that can have similar symptoms, but usually they are correct.
# Is a pap smear 100 percent accurate?
No it is not. There are various places for errors. The specimen may not be perfect, and there are possible errors in the processing too.
# What is the percent error of 2.5 pounds?
This error depends on the type of your balance; read the error of the balance on the scale, label or user manual.
# Is a passed ball from a catcher considered an error?
If the play is scored as a passed ball it is an error. It should only be scored a passed ball/wild pitch if a runner advances, or on the third strike the batter reaches first base safely. The scoring is sometimes a judgment call. Some statisticians are more strict on calling it a passed ball or wild ( Full Answer )
# Why does your poptropica say an error accured while connecting to this database?
There is an error with the poptropica site if you keep going back on it will work
# Are pee tests 99.9 percent accurate?
I believe they are fairly accurate, but probably not 99.9%. I've gotten false positives on urine tests before, but they're usually pretty accurate.
# What is the percent error of an error of 1 meter in a kilometer?
Since 1 kilometre = 1000 metres, then if an error is 1 m in 1 km, then that would translate to 1/1000 = 0.001 x 100 = 0.1% error.
# Why should documents for publishing should be accurate and error free?
Published documents should be accurate and error-free. Poorlywritten, inaccurate documents reflect badly on the writer and thebusiness associated with the article. When a document hasinaccuracies, the reader tends to distrust the entire article,including statistics, opinions, and facts.
# Is the internet news 100 percent accurate?
Nope, that's impossible, everything has some form of bias thrown in it, nothing is 100% accurate in this world. Take this "fact" for example: the sky is blue. No, the sky is not blue, it's blue sometimes, red/pink/purple other times, and sometimes it has white on it. Nothing is 100% accurate in this ( Full Answer )
# What is the difference between the low percent error and the high percent error?
If it's high that means it's very off and away from the actual value. If you find a low percent error it is very close or close to the true value.
# What is the percent error for 56-56.6?
If 56.0 is the true accepted value and 56.6 is the measured value the relative error 1,07 %.
# Why is zero error not considered in spherometer?
No, you have not any knowledge about spherometer. In spherometer also, similar to other measuring instruments zero error should be considered every time; in order to achieve accurate value.
# Are DNA Test 100 percent accurate?
No laboratory test is 100 percent accurate, but DNA testsare pretty close (a commonly cited figure is 99.5%, meaning there'sa 1 in 200 chance that a given individual will test as a positivematch for a DNA sample from someone else).
# What is the percent error of the measurement 7cm?
what is the number supposed to be if you git 7 cm? Percent error is the percentage that you're incorrect by, so you need another measurement. to do it, you find the difference and then divide by the original so (7-x)/true value then multiply by 100 to get percentage.
# How do you compute the percent error or formula?
Actual results divided by your expected results. Subtract that number from 1. Multiply that by 100. For example: Actual result : 9 grams Expected result: 10 grams 9/10 = .9 1 - .9 = .1 .1 x 100 = 10 10% error.
# What do you do when your sim's 3 game says error accured?
This will depend on when the error message comes up. If it is before the game pulls up the launcher, you need to re-install your game. If it is after the launcher but before you actually get into a household, it could be a few things. First, if you have custom content, you need to ensure that ther ( Full Answer )
# What does it mean to have a high percentage error?
It means that, relative to the true value of whatever you are trying to measure, the estimated (or calculated) value is quite a long way off.. If the real value of something is 5 but is measure as 7 the absolute error is 7 - 5 = 2, but the percentage error is 100*2/5 = 40%. If the true value is 10 ( Full Answer )
# How do you compute a percent error?
% error = |experime ntal value - theoretical value| /theoretical value * 100% . It is the absolute value of the differe nce betwee n the experime ntal a nd theoretical values divided by the theoretical value multiplied by 100%. .
# What does the term percent error refer to?
An error is the difference between a predicted value and the actual, observed, value. The percent error tells the user how close or how far off one was from the actual value in the form of a percentage.
# What does percent of error mean?
Is a term used to describe the proportion of audit adjustmentsfound in a sample of transactions. E.g: "Error": SubtractApproximate value from Exact value. Ignore any minus sign. Example: I estimated 260 people, but 325 came. 260 - 325 = -65, ignore the "-" sign, so my error is 65
# What is the difference between low percent error and high percent error?
The difference between low percent error and high percent error isone is low and the other is high
# What would be considered a source of error in an experimen?
The mice in the experimental group did not always drink the orangejuice that they were given
# How do I calculate what percent of a given forecast was accurate?
The answer depends on what the forecast was about and how you interpret accurate. Consider weather forecasts. If the weather forecast was "sunny", how much cloud cover is acceptable before the forecast is considered wrong? If the forecast was "dry" would a 5 second drizzle mean a total *FAIL*? Does ( Full Answer )
# What is the percent error of 4cm?
A percent error depends on the size of the measurement as well asthe error itself. It's very intuitive to think about: If you'remeasuring a piece of paper and you're off by 4 cm, you'll haveproblems; if you're measuring the moon, that's nothing. A biggerpercent error is a bigger deal to an engineer. ( Full Answer ) | 3,199 | 13,984 | {"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-2019-18 | latest | en | 0.925583 |
https://learn.k20center.ou.edu/lesson/3280 | 1,713,997,434,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819971.86/warc/CC-MAIN-20240424205851-20240424235851-00319.warc.gz | 333,589,860 | 10,652 | # Edible Oasis (Geometry)
## Arc Length and Sector Area of a Circle
Lydia Baker, Shayna Pond | Published: September 27th, 2023 by K20 Center
• Grade Level 9th, 10th, 11th, 12th
• Subject Mathematics
• Course Geometry
• Time Frame 80-90 Minutes
• Duration 2-3 Class Periods
### Summary
Through this lesson, students will use math and equity to discover the equation solving for any circle's arc length and sector area while investigating food deserts and how they impact a community.
### Essential Question(s)
How do you calculate the sector area and arc length of any circle?
### Snapshot
Engage 1
Students define food deserts, contemplate their impacts on a community, and use a map and tools to determine if their school is in a food desert.
Explore
Students work in pairs to discover the equation to solve the arc length and sector area for any circle.
Engage 2
Students calculate how much land would be needed to feed every student in the class for a year.
Explain
Students and the teacher review the Explore activity and practice solving for arc length and sector area.
Extend
Students solve various problems for arc length and sector area through a scavenger hunt.
Evaluate
Students decide where a grocery store should be placed to reduce a food desert and write a letter to the city requesting the store.
### Materials
• Lesson Slides
• Map of local town (provided by teacher; 1 per student)
• Arc Length and Sector Area Discovery Activity (attached; 1 page; 1 per student)
• Arc Length and Sector Area Discovery Activity Teacher Resource (attached; 1 page)
• Circle Practice Scavenger Hunt PPT (attached; 9 Pages front side only; 1st page answer key)
• Scavenger Hunt Answer Sheet (attached; 1 page; ½ pager per student)
• Optional: Norman Food Desert (attached; 1 page; 1 per person)
• Ruler
• Compass
• Calculator
• Paper (3 pages per student)
• Pencil
### Engage 1
25 Minute(s)
Use the attached lesson slides to guide the lesson.
Move through slides 2-4 to introduce the lesson, essential question, and objectives to the class. Before playing the video on slide 5, ask the class if they have ever heard of the term “food desert.” Spend a few minutes allowing students in the class to give their definitions of the term or hypothesize what the term means. Once the class has discussed, play the clip on the slide to give a formal definition.
Transition to slide 6, and ask every student to get out a piece of paper and a writing utensil.
Remind students that the United States classifies food deserts as areas where people are a mile or more away from affordable and nutritious food. This most often occurs in low-income urban and rural areas where there are not as many grocery stores and farmers markets available. Food deserts are not, however, an indication of poverty. Some individuals may be surprised to know they live more than a mile away from a grocery store because they are easily able to travel to one.
Ask students to think of the struggles people who do not have access to fresh and affordable food would have. Tell the students to independently list as many things they can think of on their paper in the 1 minute 30 seconds provided. Then start the timer.
When the time ends, bring the whole class back together and introduce the Stand Up, Sit Down strategy on slide 7. Explain to the class that they will be sharing their list of struggles. Each student will take turns voicing one struggle out loud. If something is said that is on a student’s list, they will mark it off. Students will only say something that has not been said out loud. Once all of the items on a student’s list have been said, they will sit down. This will continue until every student in the classroom is sitting.
After all students have shared their list, hold a whole class discussion about the challenges people in food deserts might face. Consider using the following guiding questions:
• How many challenges could you think of?
• What items did you have in common with other students?
• What do you consider affordable and nutritious food?
• Where can you buy affordable and nutritious food?
• What item (your own or someone else’s) do you think is the most severe implication of living in a food desert?
Move to slide 8, and play the clip to show the perspective of the documentary on the challenges those in food deserts face.
After the video clip, show slide 9 and introduce the Magnetic Statements strategy. Ask students to predict if their school is in a food desert. They will move to the side of the classroom that indicates their prediction. Ask a couple of students from each group to share why they chose ‘yes’ or ‘no.’ Use the following reflection questions when students share their reasoning.
Reflection questions:
• What store are you considering as a grocery store?
• Do they sell fresh fruits, vegetables, and meat?
• Are they frequently out of any of these foods?
• Is the food affordable for people in our town?
After discussing, have students return to their seats, move to slide 10, and hand students the map of their town (see teacher note at the beginning of this lesson for instructions), ruler, and compass. Using the steps on the slide, students will determine if their school is in a food desert.
After everyone in the class has determined if they live in a food desert, ask the class if anyone was surprised by the answer. Explain to students that circles and their properties can be used by people like city planners and food banks to determine where grocery stores and food resource centers need to be placed, just as the class used a circle to discover if they live in a food desert.
### Explore
15 Minute(s)
Move to slide 11, and inform students to put their town map away in a safe location to be used later in the lesson.
Introduce Elbow Partners to the class, and give each student one copy of the Arc Length and Sector Area Discovery Activity. Ask the partners to work together to complete this handout.
While pairs are working, float around the room to answer questions as necessary and keep students on track. Use the Arc Length and Sector Area Discovery Activity Teacher Resource as needed.
### Engage 2
10 Minute(s)
On slide 12, introduce the think-pair-share strategy while each student gets out a piece of paper and a writing utensil. Read the prompt on the slide out loud before asking the class to begin working on the task individually:
It would take 72,843.48 square feet to produce enough vegetables for 1 person for 1 year. How much land would be needed to feed every student in your classroom for one year? Design a circular plot of land to represent the amount of area needed to feed the students in your class. Share this circular plot of land equally, and allocate each student their slice of the circle. Calculate the sector area and arc length of each slice the student needs to feed themselves for 1 year.
While students work, walk around the room to check pacing and ensure that students create products similar to the example on slide 13.
Once the students have had an opportunity to try the problem on their own, instruct them to turn and talk to their elbow partner about the problem and either finish the problem or come to a consensus on the answer.
After pairs have worked for a few minutes, move to slide 13 to review the answer with the whole class.
Transition to slide 14, and introduce the S-I-T strategy. Ask students to write anything they found surprising, interesting, and troubling about the facts shared in the problem and the solution they have for the problem posed above. When the students have had enough time to write down their responses, ask a few students to share their responses for each bullet point.
Sample responses:
Surprising - It is surprising that someone can eat that much food!
Interesting - It is interesting that people know the exact square footage needed to feed a person.
Troubling - Where is all of the land to grow this food? Is there enough land for the population?
### Explain
15 Minute(s)
Ask all students to have their completed Arc Length and Sector Area Discovery Activity out and on their desks, move to slide 15, and bring the entire class together to review vocabulary terms and allow students to share what they learned in the Explore phase.
Ask for volunteers from the class to define the terms on the slide and share what equations they created in the Explore activity. Either write directly on the slide, or move to slide 16 to reveal all the definitions. Encourage students to use the back side of their discovery activity to take notes on as a later reference.
On slide 17, introduce the formal equations for arc length and sector area. Allow time for students to ensure they have the correct equations written on their discovery activity.
Conduct a whole class discussion about the three questions on the slide, so students can reflect on the equations.
1. What is that symbol in the numerator?
That is the symbol Theta (?); it is a variable in this equation that stands for the central angle of the part of the circle we are measuring. This symbol serves the same purpose as any other variable like x.
1. What do you notice about the equations?
They both have the ratio of part of the circle to the whole. They both are multiplied by the equation of circumference and area.
1. How are they similar to the area and circumference equations of a circle?
The arc length equation is multiplied by the circumference equation. The sector area equation is multiplied by the area equation.
Move through slides 18-27 to practice arc length and sector area questions together. As a whole class, work through the two examples together by asking students in the classroom to describe how to complete the problem while the teacher progresses through slides 18-21 to show example 1 and slides 22-27 to show example 2. Encourage students to practice working out these problems on the back side of their discovery activity to review later in the lesson if needed.
### Extend
15 Minute(s)
Move to slide 28 and pass out the Scavenger Hunt Answer Sheet to each student.
Explain to the students that they will be answering each question that is hanging around the room and record their answers on their answer sheets. All solutions are listed on the handout, so if a student does not see their answer, they will need to try the problem again to come up with the correct answer. Once the student has solved the problem and located the answer on their handout, they will write the capital letter from the question page on the line above the solution. The students will know they are finished because they have solved the riddle.
Emphasize to students that the questions can be answered in any order, and they are free to choose which questions to solve first.
Once students understand the instructions, dismiss them to begin working on the scavenger hunt.
While students are working, float around the room to answer questions and keep students on track.
Instruct students to turn in their answer sheets and any scratch paper before returning to their seats when they have completed the Scavenger Hunt.
### Evaluate
10 Minute(s)
Once all students have returned to their seats, if your class found that your school is in a food desert, transition to slide 29 and ask students to get out the map they completed in the Engage phase that shows their school’s location compared to nearby stores.
If your class found that your school is not in a food desert, transition to slide 30 and pass out the Norman Food Desert map to each student.
Read the instructions on the slide to the class to explain the activity. Allow students time to work independently to write the letter with the criteria written on the slide.
To end the class, have students turn in their letters, or allow time for students to share their letters with the whole class. | 2,467 | 11,929 | {"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} | 4.125 | 4 | CC-MAIN-2024-18 | latest | en | 0.844066 |
https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10B_Problems/Problem_22&oldid=69035 | 1,638,711,281,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363189.92/warc/CC-MAIN-20211205130619-20211205160619-00378.warc.gz | 190,286,169 | 11,367 | 2015 AMC 10B Problems/Problem 22
Problem
In the figure shown below, $ABCDE$ is a regular pentagon and $AG=1$. What is $FG + JH + CD$? $[asy] pair A=(cos(pi/5)-sin(pi/10),cos(pi/10)+sin(pi/5)), B=(2*cos(pi/5)-sin(pi/10),cos(pi/10)), C=(1,0), D=(0,0), E1=(-sin(pi/10),cos(pi/10)); //(0,0) is a convenient point //E1 to prevent conflict with direction E(ast) pair F=intersectionpoints(D--A,E1--B)[0], G=intersectionpoints(A--C,E1--B)[0], H=intersectionpoints(B--D,A--C)[0], I=intersectionpoints(C--E1,D--B)[0], J=intersectionpoints(E1--C,D--A)[0]; draw(A--B--C--D--E1--A); draw(A--D--B--E1--C--A); draw(F--I--G--J--H--F); label("A",A,N); label("B",B,E); label("C",C,SE); label("D",D,SW); label("E",E1,W); label("F",F,NW); label("G",G,NE); label("H",H,E); label("I",I,S); label("J",J,W); [/asy]$
Solution
Triangle $AFG$ is isosceles, so $AG=AF=1$. Using the symmetry of pentagon $FGHIJ$, notice that $\triangle IGF \cong \triangle JHG \cong \triangle DIJ \cong AFG$. Therefore, $JH=AF=1$.
Since $\triangle AJH \sim \triangle AFG$, $\frac{JH}{AF+FJ}=\frac{1}{1+FG}=\frac{FG}{FI}=\frac{FG}1$. From this, we get $FG=\frac{\sqrt{5} -1}{2}$.
Since $\triangle DIJ \cong \triangle AFG$, $DJ=DI=AF=1$. Since $\triangle AFG \sim ADC$, $\frac{AF}{AF+FJ+JD}=\frac1{2+FG} = \frac{FG}{CD}=\frac1{CD}$. Solving for $CD$, we get $CD = \frac{\sqrt{5} +1}{2}$
Therefore, $FG+JH+CD=\frac{\sqrt5-1}2+1+\frac{\sqrt5+1}2=\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }$
See Also
2015 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 20 Followed byProblem 22 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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Login to AoPS | 738 | 1,846 | {"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": 19, "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} | 4.5 | 4 | CC-MAIN-2021-49 | latest | en | 0.385794 |
https://www.hackmath.net/en/math-problem/48131 | 1,632,828,566,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060677.55/warc/CC-MAIN-20210928092646-20210928122646-00220.warc.gz | 832,359,272 | 13,608 | # A machine
A machine produces steel rods of normally distributed length; the mean length and the standard deviation being 50.0 cm and 0.5 cm respectively. The rods do not conform to safety standards if they are either shorter than 49.1 cm or longer than 50.7 cm in length. Find the number of rods from a consignment of 5000 rods that do not conform to safety standards.
x = 584
### Step-by-step explanation:
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https://www.jiskha.com/display.cgi?id=1363308854 | 1,503,530,993,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886124662.41/warc/CC-MAIN-20170823225412-20170824005412-00434.warc.gz | 911,905,069 | 4,087 | # algebra 2
posted by .
3n/(n2 -7n + 10) - 2n/(n2-8n+15)
• algebra 2 -
First factor the denominators.
3n/(n^2 -7n + 10) - 2n/(n^2-8n+15)
= 3n/[(n-5)(n-2)] -2n/[(n-5)(n-3)]
= [n/(n-5)] *{[3/(n-2)]-[2/(n-3)]}
= [n/(n-5)] *{[3(n-3)-2(n-2)/(n-2)(n-3)]}
= [n/(n-5)]*(n+5)/[(n-2)(n-3)]
This does not seem to be getting any simpler.
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y^2-x-4=0 a)find all x and y intercepts of the graph implied by this equation. (use an appropriate equation). b) test the graph for symmetry using algebra. (explain reasoning through algebra)
More Similar Questions | 705 | 2,320 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2017-34 | latest | en | 0.937162 |
https://flashman.neocities.org/Courses/m106f00.html | 1,638,195,205,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358774.44/warc/CC-MAIN-20211129134323-20211129164323-00032.warc.gz | 328,089,761 | 9,724 | ## Martin Flashman's Courses
Math 106 Calculus for Business and Economics
Fall, '00
Final Examination: Wednesday 12-20-00 12:40-14:40 or self scheduled
See Prof. Flashman
Checklist of topics for Final Exam
MTRF 1300-1350 SH 128
Last updated: 8/29/00
```Fall, 2000 Problem Assignments(Tentative as of 8-29-00) M.FLASHMAN
Section Problems (*= interesting but optional; SC= Self-Check)
------- --------------------------------------
Assignments and Recommended problems I```
```8/29 -> rev. sheet Problems 1,2,4,6
8/31 -> 1.1: 1-23 odd; 45-51 odd;111,112,115,116, 121-124
8/31 -> 1.2: (i)1-4,13-15,23-25,33-36
9/1 -> (ii)53-55,65-67,89,95,99
8/31 -> 1.3: 1-19 odd
9/1 -> 1.4:(i)1-14;17,23,27-30,45,51,53
9/5 -> (ii)55-57,69,71,73,75,76,78
9/7 -> rev. sheet Problems 5,8
9/7 -> 2.1: (i)SC:1-3; 1-5,9 (also draw T-figs to illustrate these functions);13,15,17,33
9/8 -> (ii)35-45 odd (also sketch T-figs);59,62,63,67,69,73,*77
9/11-> 2.1T: 1-5 odd; 11, *43
9/11 -> 2.3:(i)15-18;23,28,*30,31
9/12 -> (ii)36,37,40,41,45,51,52```
```Assignments and recommended problems II
9/14 -> 2.6: 2,3,5, *8
9/15 -> Read p 155-166. Do 2.6 19,13,21,27, 29, 31, 34, 51
9/18 -> 2.6: 15, 16, 22, 30, 33, 39-41,51
*2.6T: 1-9 odd
9/21 -> 3.1 (i) 1-21 odd, 22, 27,30, 35, 36, *72
9/22 -> (ii) 41-43,46,49; 55,57, 62, 63
9/22 ->3.2 (i) 1-9 odd, 12, 31, 32, 39
9/25 -> (ii) 15-23 odd, 33, 37, 46
9/25 -> (iii) 49,51,54, 58
9/26 ->3.3(i) 1-9 odd;24,47
9/28 (ii) 29,31,,59,63,65,71,*75
9/28 ->3.4 (i)1-5
9/29 -> (ii) 11, 13,16, 17, 19
10/2 -> Read handout plus pp 221-225.
->(iii) 23-27 odd,29,30
9/28 ->3.5 1-13 odd, 21-24, 29, 31, 32
9/29 ->3.6 (implicit diff'n) (i) 1,3,5, 9-11, 31
10/2 -> (related rates) (ii) 15, *29, 39, 43-45,48
10/2 -> (iii) 51,60
10/3 ->3.7 (i) 1-9 odd, 15-17, 27,29
10/5 -> (ii) 34,36-39, 41,44
10/6 -> 2.4 1-9 odd
10/6 ->2.5 (i)1-19 odd,39,41
10/12 -> (ii) 43-49 odd, 63, 71, 73, 76, 80
10/12 -> pp 141-144 IVT & bisection (iii) 87, 88, 93
10/16 -> 2.6 pp 167-169 "Diff implies cont" *57
Assignments and recommended problems III```
```10/9 ->4.1 (i)1-7 odd, 13-17,21, 23, 44-47
10/10 -> (ii) 37-40,25, 27, 49- 57 odd
10/12 -> (iii) 60-64, 72, 73, 77
10/13 -> 4.4 (i) 2,5,7,8,15,17,19,39, 41
10/16 -> (ii) 19-31 odd ; 45,49, 52, *55
10/16 -> 4.5 (i) 1,3, 15
10/17 -> (ii) 5
10/20 -> Read Example 5 (iii) 22
10/20 ->4.2 (i) 1-11 odd, 16, 23-27 odd
10/23 -> (ii) 26, 45-49 odd, 75, 81
10/23 ->4.3 (i) 29, 30, 32,33,35, 37,43
10/24 -> (ii) 40,45,65
10/26-> (iii) asymptotes: 1-15 odd,20-22, 61, 68```
``` Assignments and recommended problems IV
10/24 -> 5.1 READ 374-376 (i) 1,4,7,10,13,16,19,22,25
10/27 -> (ii)27,29,31,32
10/30 -> 5.2 (i) 1-25 odd
10/31 -> (ii) 20,22,27,29,31, 33-37 odd
10/31 -> 5.3 (i) 1-11 odd
11/2 -> (ii) 13-23 odd
11/3 -> 5.4 (i) 1-17 odd,6,18, 29
11/6 -> (ii) 14,23,27,31,34,35,41, 45, 46
11/7 -> (iii) 49, 54
11/6 -> 5.5 (i) 1-17 odd, 6,18, 29
11/7 -> (ii) 4, 14, 35, 37, 47, 49, 53, 55
Optional -> Handout problems on logs and exponentials.
11/9 -> 5.6 (i) 3,7,*9, 11, 15
11/10 -> (ii) 9,12
11/13 6.1 -> (i) 1-19 odd
11/13 -> (ii) 23-30, 51-57 odd, 61
11/17 -> (iii)65, 67, 69, 79
11/14 ->IV.E 1a,c; 3a,c; 5a,b; 13 a,b; 21
Assignment for 11/27: Read IV.F and 6.3 pp 466-471.
Assignments and recommended problems V
11/28-> IV.F 1, 3, 7,9, 19, 21.
11/28 -> 6.4 (i) 5-11 odd
11/30 -> (ii)10, 12, 23-29 odd
12/1 -> (iii) 19-22, 31-37 odd, 41-44```
```11/30 -> 6.2 (i) 1-13 odd
12/1 -> (ii) 19-27 odd, 6,8, 51,53
12/4 -> (iii) 45-47, 57, 59, 63```
```12/1 ->6.5 -> (i) [sub.] 1-11 odd
12/4 -> (ii) 2,4, 16, 29-33 odd
12/5 -> (iii) 41, 42, 43```
```12/5 ->6.6 area (i) 1-7
12/7 -> (ii) 9-23 odd, 35-37
12/8-> (iii) 27-30, 46
12/14 (i) 1-7 odd
12/11 -> value (ii) 9-17 odd```
```12/12 7.4 ->(i) 1-7 odd,15, 17, 19, 50
-> (ii) 35, 37, 39```
```12/5-> 8.1 (i)1-7 odd
12/7 -> (ii) 19, 20,25, 28,29, 35
12/8 -> 8.2 1-5,11-17 odd; 23-29 odd, 41,43
12/15->8.3 -> 21, 23, 25
12/15-> -> 1-7 odd
8.4 -> 1,3,*16, *19```
Monday Tuesday Thursday Friday Week 1 8/28 Course Introduction 8/29 Numbers, Variables, Algebra Review 1.1&1.2 8/31 More Algebra review and The coordinate plane1.3 Begin Functions? 9/1 More Algebra review. Lines 1.4 Begin Functions. 2.1 Week 2 9/ 4 No Class. Labor Day 9/5 Begin Functions and models. 2.1 & 2.3 9/7 More Functions and Models. The fence problem: functions, graphs, technology. 9/8 Slopes, rates and estimation. Week 3 9/11 The Derivative I 2.6 Motivation: Marginal cost, rates and slopes. 9/12 The Derivative II 2.6 9 /14 The Derivative II 2.6 9/15 More on the Derivative. Begin the Derivative Calculus I 3.1 Week 4 Summary I due 9/22 9/18 Calculus II 3.1 9/19 3.1 Justify Powers & Sums. 9/21 3.2 product rules 3.2 Justify product 9/22 3.2 quotient rule. 3.3 The Chain Rule Week 5 POW I due 9/29 9/25 3.3 Chain Rule 9/26 3.4 Marginal Applications 3.5 Higher order Derivatives 9/28 3.6 Implicit Differentiation Related Rates 3.6 9/29 More related rates. Week 6 Summary II due:10/5 10/2 Differentials 3.7 10/3 More on differentials. relative error. 10/5 Back-up: limits and continuity 2.4 & 2.5 IVT? 10/6 Begin First Derivative Analysis 4.1 Week 7 POW II due:10/13 10/9 More First Derivative analysis. 4.1 . 10/10 IVT.2.5 Bisection Method 10/12 Optimization I 4.4 10/13 Review Optimization I 4.4 Optimization II 4.5 Week 8 Summary III due: 10/16 10/16 More optimization. 10/17 Second Derivative Analysis 4.2 More Optimization 10/19 Examination I (covers through 10/13) 10/20 More optimization and Curves III 4.3 Week 9 POW III due: 10/28 10/23 More on Concavity. Horizontal Asymptotes. 10/25 Vertical Asymptotes 10/27 Start Exponential and Logarithmic functions 5.1 10/28 Logarithmic functions 5.2 Week 10 : Summary IV due:11/3 10/30 Interest and value 5.3 10/31 more on interesr.Start derivative og exp. 11/ 2 Derivatives of exponentials 5.4 11/3 Derivatives of Logarithms 5.5 Week 11 POW IV 11/6 Logarithmic differentiation. 5.5 11/7 Models using exponentials 5.6 11/9 More models. 11/10 6.1 Begin Differential equations and integration Week 12 Summary V 11/13 Euler's Method IV.E 11/14 Euler's Method and Area 11/16 Examination II (covers from 4.1 to 6.1 ) 11/17 Finding area by estimates and using anti-derivatives. Week 13 Thanksgiving Break 11/20 11/21 11/23 Thanksgiving 11/24 Week 14 11/27 The definite integral. 6.3 FT of calculus I 6.4 11/28 Substitution 6.2 Applications 6.5 11/30 Substitution in definite integrals. interpreting defintie integrals. 12/1 More area/ and applications 6.5 &.6.6 Week 15 POW V Due 12/5 Summary V Due 12/4 12/4 Intro to functions of 2 or more 12/5 More on areas. Functions of 2 variables: level curves, graphs. 12/7 Partial derivatives. 1st order . 12/8Value 6.7 2nd order partial derivatives. 8.2 Week 16 (last week of classes) 12/11 Improper integrals and value. 7.4 12/12 Surplus 6.7 12/14 Extremes 8.3 (Critical points) Least Squares. 12/15 Misc.OtherApplications LAST CLASS :) Week 17 Final Exam Week 12/18 12/19 12/20 12/21
`*Final examination may be self-scheduled M,T,W, or R. Contact Professor Flashman.`
I. Differential Calculus: A. *Definition of the Derivative Limits / Notation Use to find the derivative Interpretation ( slope/ velocity ) B. The Calculus of Derivatives * Sums, constants, x n, polynomials *Product, Quotient, and Chain rules *logarithmic and exponential functions Implicit differentiation Higher order derivatives C. Applications of derivatives *Tangent lines *Velocity, acceleration, marginal rates (related rates) *Max/min problems *Graphing: * increasing/ decreasing concavity / inflection *Extrema (local/ global) Asymptotes The differential and linear approximation D. Theory *Continuity (definition and implications) *Extreme Value Theorem *Intermediate Value Theorem E. Several Variable Functions Partial derivatives. (first and second order) Max/Min's and critical points. II. Differential Equations and Integral Calculus: A. Indefinite Integrals (Antiderivatives) *Definitions and basic theorem *Simple properties [ sums, constants, polynomials] *Substitution *Simple differential equations with applications B. Euler's Method, etc. Euler's Method *Simple differential equations with applications Tangent (direction) fields/ Integral Curves C. The Definite Integral Definition/ Estimates/ Simple Properties / Substitution *Interpretations (area / change in position/ Net cost-revenues-profit) *THE FUNDAMENTAL THEOREM OF CALCULUS - evaluation form Infinite integrals D. Applications *Recognizing sums as the definite integral *Areas (between curves). Average value of a function. Present Value. Consumer Savings.
```
```
Fall, 2000 COURSE INFORMATION M.FLASHMAN
MATH 106 : Calculus for Business and Economics MTRF 1300-1350 SH 128
OFFICE: Library 48 PHONE:826-4950
Hours (Tent.): MTWR 10:15-11:30 AND BY APPOINTMENT or chance!
E-MAIL:[email protected] WWW: http://www.humboldt.edu/~mef2/
***Prerequisite: HSU MATH 42 or 44 or 45 or math code 40.
• TEXT: Required: Calculus for the Managerial, Life, and Social Sciences, Fifth Edition by S.T. Tan, 2000.
• Excerpts from Sensible Calculus by M. Flashman as available on the web from Professor Flashman.
• Catalog Description: Logarithmic and exponential functions. Derivatives, integrals; velocity, curve sketching, area; marginal cost, revenue, and profit, consumer savings; present value.
• SCOPE: This course will deal with the theory and application to Business and Economics of what is often described as "differential and integral calculus." Supplementary notes and text will be provided as appropriate.
• TESTS AND ASSIGNMENTS: There will be several tests in this course. There will be several reality check quizzes, two midterm exams and a comprehensive final examination.
• Homework assignments are made regularly. They should be done neatly and passed in on the due date. Homework is graded Acceptable/Unacceptable with problems to be redone. Redone work should be returned for grading promptly.
• Exams will be announced at least one week in advance.
• THE FINAL EXAMINATION WILL BE SELF SCHEDULED.
• The final exam will be comprehensive, covering the entire semester.
• MAKE-UP TESTS WILL NOT BE GIVEN EXCEPT FOR VERY SPECIAL CIRCUMSTANCES! It is the student's responsibility to request a makeup promptly.
• *** DAILY ATTENDANCE SHOULD BE A HABIT! ***
• Team Activities: Every two weeks your team will be asked to submit a summary of what we have covered in class. (No more than two sides of a paper.) These may be organized in any way you find useful but should not be a copy of your class notes. I will read and correct these before returning them. Team participants will receive corrected photocopies.
• Your summaries will be allowed as references at the final examination only.
Every week (with some exceptions) teams will submit a response to the "problem/activity of the week." All cooperative problem work will be graded +(5 well done), ü(4 for OK), -(3 acceptable), or unacceptable(1) and will be used in determining the 80 points allocated for cooperative assignments.
• GRADES: Final grades will be determined taking into consideration the quality of work done in the course as evidenced primarily from the accumulation of points from tests and various assignments.
• Reality Quizzes 100 points 2 Midterm Examinations 200 points Homework 70 points Cooperative work 80 points Final Examination 200 points Total 650 points
• Cooperative problem assignments and summaries will be used to determine 80 points.
• The total points available for the semester is 650.
• Notice that only 400 of these points are from examinations, so regular participation is essential to forming a good foundation for your grades as well as your learning.
• MORE THAN 4 ABSENCES MAY LOWER THE FINAL GRADE FOR POOR ATTENDANCE.
• ** See the course schedule for the dates related to the following:
• No drops will be allowed without "serious and compelling reasons" and a fee.
• Students wishing to be graded with either CR or NC should make this request to the Adm & Rec office in writing or by using the web registration procedures.
• No drops will be allowed.
• Technology: The computer or a graphing calculator can be used for many problems. We will use Winplot and Microsoft Xcel.
• Winplot is freeware and may be downloaded from Rick Parris's website or directly from this link for Winplot . | 4,694 | 13,398 | {"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} | 3.75 | 4 | CC-MAIN-2021-49 | latest | en | 0.412765 |
http://cdn.physlink.com/education/askexperts/ae675.cfm | 1,726,212,538,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651510.65/warc/CC-MAIN-20240913070112-20240913100112-00090.warc.gz | 6,225,107 | 8,898 | QUESTION #675
# If an orbiting artificial satellite were to slow down, what would happen?
Asked by: Christopher Alec Maquiling
### Answer
Artificial satellites slow down all the time, and uniformly, they are no longer able to resist the pull of gravity and fall back to Earth. Satellites like the Space Shuttle do it intentionally and can control their descent. The Mir space station was a controlled crash in the South Pacific, and countless other satellites have been accidentally lost. The most common cause of what is known as orbital decay is the atmosphere. As you can imagine, there is no solid "border" to the atmosphere. Gravity holds most of it near the Earth, but out near the border of space there are very few particles, but certainly enough to slow down a satellite that ventures too close to the Earth. Satellites carry fuel to help maintain their orbits, but when that fuel runs out... sooner or later they will fall back to the ground. Hopefully it can be in a controlled manner. If you have any more questions, feel free to ask us, or hit up NASA.
Answered by: Frank DiBonaventuro, B.S., Physics grad, The Citadel, Air Force officer
It would fall to the planet's surface. Here's why: Imagine a car driving around in a perfect circle. This is a centripetal acceleration described by a = v^2 / r where v is the velocity of the car and r is the distance of its revolution. This equation of acceleration also applies to the motion of the satellite as it moves around the planet. (Here we are not considering the planet's rotational motion.) Now if we want to consider centripetal force, we multiply the centripetal acceleration by the mass of the satellite; this gives: F = m a Next consider that the weight of the artificial satellite is given by the equation: W = m g Now g is the acceleration due to gravity. In the physics classroom its value is usually given at sea level, but in reality g is a variable: closer to the center of mass it is strong, farther away it is weak. For the artificial satellite to orbit the planet without falling in, F and W must be equal. So then: F = W
m a = m g
a = g
v^2 / r = g Now if the satellite slows down, v decreases in value so that: v^2 / r < g This means that the acceleration due to gravity is greater than the centripetal acceleration. Gravity wins the fight and pulls the satellite to the planet's surface.
Answered by: Frederick Herrmann, B.A. | 537 | 2,410 | {"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-2024-38 | latest | en | 0.957233 |
http://www.slideshare.net/MOHOTASIMANIK/presentation-based-on-truss-and-frame | 1,484,802,712,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280483.83/warc/CC-MAIN-20170116095120-00543-ip-10-171-10-70.ec2.internal.warc.gz | 668,895,020 | 36,344 | Upcoming SlideShare
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### presentation based on Truss and Frame
1. 1. TRUSS & FRAME Course no-CE 416 course title- Prestress Concrete Design Sessinonal Presented by MD. Mohotasimur Rahman ID NO. 10.01.03.040 Course Teachers Munshi Galib Muktadir & Sabreena Nasrin Lecturer of Civil Engineering Department Ahsanullah University Of Science And Tecnology Dhaka, Bangladesh
2. 2. TRUSS - INTRODUCTION A truss is a structure composed of members fastened together in such a way to resist change in shape and it is rigid structure. Triangular unit A truss is a structure comprising one or more triangular units constructed with straight members whose ends are connected at joints referred to as nodes. Its purpose is to support a larger load or span a greater distance than any individual member from which the truss may be built
3. 3. TRUSS – INTRODUCTION CONTINUE Node External forces and reactions to those forces are considered to act only at the nodes. Moments (torques) are explicitly excluded because, and only because, all the joints in a truss are treated as pin joint or hinge joint . Result in forces in the members which Tie strut are either tensile or compressive forces.
4. 4. PLANE TRUSS VS SPACE TRUSS Plane Truss All member of truss and applied load lie in a same plane. Space Truss An elementary space truss consists of 6 members connected at 4 joints to form a tetrahedron. In a simple truss, m = 2n - 3 where m In a simple space truss, m = 3n - 6 is the total number of members and n is the where m is the number of members and n number of joints. is the number of joints.
5. 5. ROOF TRUSS TERMINOLOGY
6. 6. ROOF TRUSS TYPE
7. 7. BRIDGE TRUSS TERMINOLOGY
8. 8. BRIDGE TRUSS TYPE
9. 9. METHOD OF TRUSS ANALYSIS Joint Method Determine the Support Reaction. Apply Fx = 0 and Fy = 0 to every node and determine member force Dismember the truss and create a free-body diagram for each member and pin.
10. 10. METHOD OF TRUSS ANALYSIS Section method Determine the Support Reaction. To determine the force in member BD, pass a section through the truss as shown and create a free body diagram for the left side. With only three members cut by the section, the equations for static equilibrium may be applied to determine the unknown member forces, including FBD.
11. 11. FRAME (INTRODUCTION) Contain at least one multi-force member, i.e., member acted upon by 3 or more forces. Frames are designed to support loads and are usually stationary.
12. 12. ANALYSIS OF FRAME A free body diagram of the complete frame is used to determine the external forces acting on the frame. Internal forces are determined by dismembering the frame and creating free-body diagrams for each component. | 816 | 3,181 | {"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} | 3.75 | 4 | CC-MAIN-2017-04 | latest | en | 0.896467 |
https://physicsteacher.in/2020/09/03/scalar-product-formula-with-an-example/ | 1,720,843,104,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514484.89/warc/CC-MAIN-20240713020211-20240713050211-00218.warc.gz | 364,599,376 | 28,310 | High School Physics + more
# Scalar product formula with an example
Last updated on April 14th, 2021 at 05:17 pm
Scalar multiplication of two vectors yields a scalar product. Scalar multiplication is also known as the dot product. In the next section, we will see how the scalar product formula or equation is written.
## Scalar product formula | equation of dot product
The scalar product of 2 vectors A and B is expressed by the following equation:
A.B = AB cos φ, where φ is the angle between the vectors, A is the magnitude of vector A and B is the magnitude of vector B.
The scalar product is also called the dot product because of the dot notation that indicates it.
In figure (b), vector A is projected along the direction of vector B, as a result, a projected component of vector A is generated along the direction of vector B. And this component is A cos φ.
Similarly, in figure (c), the vector B is projected along the direction of vector A, as a result, a projected component of vector B is generated along the direction of vector A. And this component is B cos φ.
In both cases, you can see how the cos φ is generated as we are working to find out a scalar product of 2 vectors A and B.
## Numerical problem solving using the scalar product or dot product
Two vectors A and F are shown in the above 2 diagrams. Find the scalar or dot product of A and F.
Solution
From Figure, the magnitudes of vectors A and F are A = 10.0 and F = 20.0.
Angle θ , between them, is the difference: θ = φ − α = 110° − 35° = 75° .
Substituting these values into Equation of scalar product gives the scalar product.
A straightforward calculation gives us A.F = AF cos θ = (10.0)(20.0) cos 75° = 51.76.
## Work is scalar or vector?
Work is a scalar. It’s a scalar product of 2 vectors, force, and displacement. As work is scalar, only the Magnitude of work counts.
## Energy is scalar or vector?
Energy is a scalar. only the Magnitude of energy counts.
## Power is scalar or vector?
Power is a scalar quantity. It has no direction attached. Magnitude of power only matters. | 494 | 2,085 | {"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} | 4.71875 | 5 | CC-MAIN-2024-30 | latest | en | 0.939643 |
https://cheeksbay.com/2019/08/06/the-elephant-in-the-pajamas/ | 1,685,729,175,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648850.88/warc/CC-MAIN-20230602172755-20230602202755-00552.warc.gz | 199,891,143 | 24,858 | # The Elephant in the Pajamas
I’m not much on Internet mems, but this one caught my eye. What is the answer to the equation 8÷2(2+2)? If you’ve not already done so, try it and see what you get. Did you come up with 16 or 1? What answer we get says a lot not only about how we learned math, but about mathematics itself.
Mathematics, based on rigorous proof of concepts, doesn’t lend itself to ambiguity. It’s only when we learn quadratic equations that we enter the world of two equally valid mathematical answers. If mathematics was ambiguous, it would make for a poor tool, indeed. But while mathematics isn’t ambiguous, language can be. Though we
don’t think of equations (called algebraic notation) as a language, they are. It’s a rigorously constructed language, to be sure, but still a language. What an equation means depends as much on an agreed upn set of rules as any sentence in any language./p>
This sort of ambiguity in language shows up in a Groucho Marx joke from Animal Crackers: “One morning I shot an elephant in my pajamas. What it was doing in my
pajamas, I’ll never know.” The gag is the unexpected, but equally valid, interpretation of the first sentence. What it means is all in how you look at it.
Simple equations, like simple sentences, aren’t much of a problem.
“Bill hit the ball” is as clearly understood as 5 + 3. It’s only when our sentences and equations become more complex that the need for an agreed upon set of rules becomes important. In language, this is called grammar. In mathematics, it’s order of operation. What is the answer to the equation 5 + 3 ˣ 2? If you said “11,” you’re correct, because 5+3 ˣ 2 = 5+6 = 11. But why should that be the case? Why not 5+3 ˣ 2 to 8 ˣ 2 to 16? Because, by the order of operation, we multiply 3 by 2 before we add the result to 5. The order of operation is such a basic thing, we’re taught it relatively early in school, and use it unconsciously. If you went to school in the United States, you likely learned the mnemonic “Please excuse my dear Aunt Sally.” That stands for parenthesis, exponents, mathematics, division, addition, subtraction,” which is how we remember how to
evaluate an equation. This is known as PEMDAS. If you went to school somewhere else in the world, you probably learned brackets, operators, division, multiplication, addition, subtraction, or BODMAS. Someone who learned BODMAS would likely evaluate 8÷2(2+2)
as 8÷2(2+2) = 8÷2(4) = 4(4) = 16, while someone who learned PEDAS would likely evaluate the same equation as 8÷2(2+2) = 8÷2(4) = 8÷8 = 1. This is because BODMAS names division before multiplication, and PEMDAS names multiplication before division.
Having learned PEMDAS, I came up with 1. The idea that it could be 16 provoked a vicersal reaction. Nor was I alone. This equation provoked considerable, sometimes heated, discussion across the Internet. So, what is the correct answer?
That led to some casual research into orders of operation, and through a surprising foggy patch of mathematics. It seems that the question of whether multiplication should always be performed before division has been going on for a long time. In our era, the concensus is that multiplication and division are equally ranked in the order of
operation, as are addition and subtraction. This means you evaluate multiplication and division as you go through the equation from left to right. 8÷2(2+2) should
be solved by first evaluating what’s in the parenthesis, giving 8÷2(4). Then, working from left to right, 8÷2(4) = 4(4). Lastly, multiplying 4 by 4 gives the answer of 16. It doesn’t
matter if you learned BODMAS or PEMDAS: Since multiplication and division are on the same level of operation, you should come up with the same result.
If that goes against the grain, as it does to all of us who thought multiplication came before division, consider this: Is 8 ÷ 2 different than 8 ˣ ½? Is 4 – 2 different from 4 + (-2)? In both cased, they’re the same.
To those of us who thought the order of operation ranked multiplication before division (or division before multiplication, for that matter), giving division equal ranking with multiplication might seem like a new thing. The surprising answer is “Maybe.” While there’s some evidence that the order of operation we know predates algebraic notation itself, the question of whether multiplication should take precedence before division rocked on into the 20th Century. Opinions seemed to mostly hold to the same understanding we have now, but the multiplication prior to division opinion also shows up in at least one textbook, while there are some that side-step the issue by calling it a gray area. The picture we get, looking at textbooks from a century ago, is that placing multiplication and division on the same level was already winning out.
Regardless of how we reached the current understanding of the order of operation, placing multiplication and division on the same level is how it’s understood now. PEMDAS still stands for parenthesis, exponents, multiplication, division, addition, subtraction, but can perhaps be more clearly understood as parenthesis, exponents,
multiplication and division, addition and subtraction. We have to keep in mind multiplication and division are on the same level, worked as we come to them from left to right, just like addition and subtraction.
If that still seems a little confusing, it might help to think of it like this:
First Parenthesis
Then Exponents
Then Multiplication and Division
Then Addition and Subtraction
For some of us, that might take some getting used to, but that’s the way it’s done.
Here’some simple problems. Answers given below.
1. 8 + 5 – 4 + 3
2. 4 ÷ 2 ˣ 8 + 1
3. 6(4 – 2)
4. (4 + 1)2 – (5 – 2)2
5. 9 ÷ 3(5 – 2) | 1,397 | 5,748 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2023-23 | latest | en | 0.955009 |
http://sciencehq.com/physics/vectors.html | 1,582,398,283,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145713.39/warc/CC-MAIN-20200222180557-20200222210557-00465.warc.gz | 129,580,067 | 3,769 | # Vectors Index
## Vector product ( Multiplying vectors )
Vector Product: If we multiply a vector with any other vector or scalar quantity then the result is called vector product. Multiplying a vector by a scalar: If we multiply a vector by a scalar , then the result is a new vector. The magnitude of the vector formed by multiplying and , is the magnitude of multiplied by and the direction of the vector is same as vector if is positive and exactly oppsite of is negative. And to divide the vector with we can simply multiply the vector with Multiplying a vector by a...
Two vector can be easily summed by using vector sum by geometrical method but it is not practical way to add vectors. For a simpler and more practical way of adding two or more vectors then we can use component method of vector sum or add vectors by components. Let there be three vector , and and if vector is the sum of vectors and or , then, The x component of vector r is the sum of x components of vector a and b , the y component of vector r is the sum of y components of vector a and b and the z component of vector r is the sum...
## Components of vectors
Components of a vector: The components of a vector are two or more vectors ( Usually vectors along the x , y , z axes) whose vector sum is equal to the given vector. For ease in doing vector calculations Vectors are expressed in the form of two or three components ; Two if the vector is two dimensional and three if the vector is three dimensional . Components of a Two dimensional vector: Consider a two dimensional vector whose initial point is the point “o” the rectangular coordinate system and final point “A”. If we...
## Unit Vector
Unit Vector: A unit vector is a vector that has a magnitude of exactly 1 unit and points in a particular direction. A unit vector lacks both unit and dimension , But it have a magnitude of a unit and a certain direction. It’s sole purpose is to point toward a direction. A unit vector is denoted by a letter above which a hat ( ) is placed. The unit vector along any vector ( say vector a ( ) ) is calculated using the formula: Or by dividing the vector by it’s magnitude. Unit Vector along x , y & z axes: The unit... | 476 | 2,216 | {"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} | 4.46875 | 4 | CC-MAIN-2020-10 | latest | en | 0.920261 |
https://mypages.iit.edu/~smile/ch8715.html | 1,642,644,507,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301670.75/warc/CC-MAIN-20220120005715-20220120035715-00133.warc.gz | 468,447,179 | 2,074 | ```The MoleStringer, S. C. Morgan Park High School 881-5050 Objectives:
Resolution of Problems:
The role of a unifying hypothesis (law) and the reasoning
Application of Hypothesis:
Diatomic and monoatomic gaseous elements. Avogadro's
hypothsis. Determination of molecular weights, atomic
weights, etc.
Fundamental Skills:
Solving problems by use of the gas constants:
D(gas) = g mol wt(g)/22.4 l
g mol wt(g) = D(gas) * 22.4 l
D(gas) = (g/ml)*(1000ml/l)
Topic Statement:
The data below will apply to Avogadro's hypothesis. This
exercise preceded the review of the chapter on the gas laws:
Boyle's, Charles', Gay-Lussac's, indicating similarities in the
physical behavior of gases. The question of whether gases
display any similarities in chemical behavior may then be raised.
Discussion of the experimental findings of Gay-Lussac and Proust
used to explain Gay-Lussac's Law, and the inconsistency with
Dalton's Theory may be resolved. Also, consider the historical
treatment and significance as in the basic chapter.
The diagram of the mole concept and a work sheet follows below:
divide by 22.4 l -->| M |--> multiply by GAW/GMW
volume (l) <---------------| O |----------------------> mass (grams)
multiply by 22.4 l <--| L |<-- divide by GAW/GMW
| E |
divide by 6.02 * 1023 -->| |--> multiply by 6.02 * 1023
|
V
number of
particles
Diatomic elements: Define: Br, I, N, Cl, H, O, F Density = M/unit volume Volume = L * W * H Liquids = cylinder (3l) GAW = atoms Gas = displacement GMW = molecules Weight = grams D = M/V /--- mass at 1 mole GAW/m wt.MOLE -------<---- 6.02 * 1023 particles \--- at STP vol of 1 mole of gas = 22.4 l Substance Density Volume of 1 mole
1. Mercury | 13.55 g/cm3 |2. Iron | 7.85 g/cm3 |3. Ethanol (C2H5OH) | 0.80 g/ml |4. Propanol (C3H8OH) | 0.78 g/ml |5. Carbon tetrachloride | 1.59 g/ml | etc. Chemistry
all of these contain 1 mole
SUBSTANCE NUMBER OF MOLECULES FORMULA MASS(g) ML ESTIMATE VOL.
Copper 8.92 | | | | 63.5/8.92 =
Aluminum 2.70| | | | 27.0/27.0 = 10.0
Iron 7.86 | | | | 55.8/7.86 =
Lead 11.34 | | | |
Zinc 7.14 | | | |
Tin 5.75 | | | |
etc. | | | |
``` | 758 | 2,779 | {"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} | 3.578125 | 4 | CC-MAIN-2022-05 | latest | en | 0.772209 |
https://www.gradesaver.com/textbooks/math/prealgebra/prealgebra-7th-edition/chapter-4-review-page-319/33 | 1,532,223,572,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592875.98/warc/CC-MAIN-20180722002753-20180722022753-00261.warc.gz | 904,773,115 | 14,884 | Prealgebra (7th Edition)
Published by Pearson
Chapter 4 - Review: 33
Answer
$\frac{9}{x^2}$
Work Step by Step
We can simplify by canceling out common factors of both fractions and then multiplying. Multiplying 2 negatives yields a positive product. $-\frac{24x}{5}\times-\frac{15}{8x^3}=-\frac{3\times/\!\!8\times /\!\!x}{/\!\!5}\times-\frac{3\times/\!\!5}{/\!\!8\times /\!\!x\times x^2}=\frac{3\times3}{x^2}=\frac{9}{x^2}$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 191 | 590 | {"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} | 4.03125 | 4 | CC-MAIN-2018-30 | latest | en | 0.771118 |
https://justaaa.com/chemistry/87610-when-0109-g-of-zns-combines-with-enough-hcl-to | 1,695,946,554,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510462.75/warc/CC-MAIN-20230928230810-20230929020810-00389.warc.gz | 366,018,244 | 9,569 | Question
# When 0.109 g of Zn(s) combines with enough HCl to make 55.7 mL of HCl(aq) in...
When 0.109 g of Zn(s) combines with enough HCl to make 55.7 mL of HCl(aq) in a coffee cup calorimeter, all of the zinc reacts, which increases the temperature of the HCl solution from 23.2 °C to 24.8 °C: Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) Calculate the enthalpy change of the reaction ΔHrxn in J/mol. Insert your answer in kJ, but do not write kJ after the number. (Assume the density of the solution is 1.00 g/mL and the specific heat capacity of solution is 4.184 J/g°C.)
The amount of heat absorbed by solution , q = mcdt
Where
m = mass of solution = volume x density = 55.7 mL x 1.00 g/mL = 55.7 g
c = pecific heat capacity of solution = 4.184 J/g°C
dt = change in temperature = 24.8 - 23.2 = 1.6 oC
Plug the values we get q = 372.9 J
This amount of heat was liberated by 0.109 g of Zn
Thatmeans 0.109 g of Zn liberates 372.9 J
1 mol = 65.4 g of Zn liberates (65.4x372.9) / 0.109 = 223.7x103 J /mol
Therefore the enthalpy change of the reaction 223.7x103 J /mol
#### Earn Coins
Coins can be redeemed for fabulous gifts. | 385 | 1,130 | {"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-2023-40 | latest | en | 0.804557 |
http://www.sosmath.com/algebra/solve/solve4/s44/s4411/s441101/s441101.html | 1,550,582,776,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247490107.12/warc/CC-MAIN-20190219122312-20190219144312-00566.warc.gz | 429,380,896 | 4,397 | Note:
• A quadratic equation is a polynomial equation of degree 2.
• The ''U'' shaped graph of a quadratic is called a parabola.
• A quadratic equation has two solutions. Either two distinct real solutions, one double real solution or two imaginary solutions.
• There are several methods you can use to solve a quadratic equation:
1. Factoring
2. Completing the Square
4. Graphing
Solve for x in the following equation.
Problem 4.4a:
Solution:
Set the equation equal to zero by subtracting 2 from both sides.
If you have forgotten how to manipulate fractions, click on Fractions for a review.
Remove all the fractions by writing the equation in an equivalent form without fractional coefficients. In this problem, you can do it by multiplying both sides of the equation by 3.
Method 1:Factoring
The equation is not easily factored. Therefore, we will not use this method.
Method 2:Completing the square
Add 36 to both sides of the equation
Add to both sides of the equation:
Factor the left side and simplify the right side:
Take the square root of both sides of the equation:
Add 6 to both sides of the equation:
In the equation ,a is the coefficient of the term, b is the coefficient of the x term, and c is the constant. Substitute for a , for b, and for c in the quadratic formula and simplify.
Method 4:Graphing
Graph the equation, (formed by subtracting the right side of the equation from the left side of the equation). Graph (the x-axis). What you will be looking for is where the graph of crosses the x-axis. Another way of saying this is that the x-intercepts are the solutions to this equation.
You can see from the graph that there are two x-intercepts, one at 14.485281 and one at -2.485281.
The answers are and These answers may or may not be solutions to the original equations. You must verify that these answers are solutions.
Check these answers in the original equation.
Check the solution x=14.485281 by substituting 14.485281 in the original equation for x. If the left side of the equation
equals the right side of the equation after the substitution, you have found the correct answer.
• Left Side:
• Right Side:
Since the left side of the original equation is equal to the right side of the original equation after we substitute the value 14.485281 for x, then x=14.485281 is a solution.
Check the solution x=-2.485281 by substituting -2.485281 in the original equation for x. If the left side of the equation
equals the right side of the equation after the substitution, you have found the correct answer.
• Left Side:
• Right Side:
Since the left side of the original equation is equal to the right side of the original equation after we substitute the value -2.485281 for x, then x=-2.485281 is a solution.
The solutions to the equation are and
Comment:You can use the exact solutions to factor the left side of the original equation minus the right side of the original equation:
Since :
Since :
The product
Since and
then we could say
However the product of the first terms of the factors does not equal
Multiply by
Let s check to see if
The factors of are and
If you would like to review the solution to problem 4.4b, click on Problem
If you would like to test yourself by working some problems similar to this example, click on Problem
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Author: Nancy Marcus | 805 | 3,535 | {"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} | 4.71875 | 5 | CC-MAIN-2019-09 | latest | en | 0.926066 |
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[Marks]
[4]
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MATH 3705B Test 1 Solutions
January 23, 2012
1. L e cos(3 t) s 2
(a)
( s 2) 2 Answer: (a)
2t
=
+ 9
(b)
s
( s 2) 2 + 9
s + 2
(c) ( s + 2) 2 + 9
s 2
(d) s 2 + 9
(e) None of these
2.
3.
4.
1
3 s + 1
s
2
1
L
(a) 3 cos(t) + sin(t) (b) e t 2 e t (c) 2 e t + e t (d) 2 e t + e t (e) None of these
=
L u ( t 3)e
(a)
e 6 e 3s
2t
=
(b) e 3s
s
+ 2
(c)
e
3s
s +
2
(d) e 6 e 3s
s + 2
(e) None of these
L
1
(a)
(b)
(c)
(d)
(e)
2s
s 2 2 s + 4
se
=
1
e t cos( 3t) + 3 e 3t sin( 3 t)
u ( t 2)e t 2 cos[ 3(t 2)] + sin[ 3(t 2)]
u ( t 2)e t 2 cos[ 3( t 2)] + 3 sin[ 3( t 2)]
1
u ( t 2)e t 2 cos[ 3( t 2)] + 3 sin[ 3( t 2)]
1
None of the above
5. Solve the initial-value problem y + y 6 y = 0 , y (0) = 1 , y Solution:
(0) = 12.
[s 2 Y ( s ) sy (0) y (0)] + [sY ( s ) y (0)] 6 Y ( s )=0
6.
s 2 + s 6 = ( s 2)(s + 3) = s 2 s + 3 by
s + 13
s + 13
3
2
(0) = 0.
( s 2 + s 6)Y ( s ) s 13 = 0 Y ( s ) =
partial fractions. Hence, y ( t)=3 e 2t 2 e 3t .
Solve the initial-value problem y + 6 y + 13 y = 0 , y (0) = 1 , y
Solution:
[s 2 Y ( s ) sy (0) y (0)] + 6[sY ( s ) y (0)] + 13 Y ( s )=0
2
( s 2 + 6 s + 13)Y ( s ) s 6=0 Y ( s ) =
s + 6
( s + 3) + 3
s 2 + 6 s + 13 = ( s + 3) 2 + 4 =
3
( s + 3) 2 + 4 + 2
s + 3
( s + 3) 2 + 4 y ( t) = e 3t cos(2 t) + 3 e 3t sin(2 t) .
2
2 | 787 | 1,401 | {"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} | 4.25 | 4 | CC-MAIN-2020-24 | latest | en | 0.322758 |
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A086881 a(n) = (2*n)!*Sum[Sum[1/(i+j),{i,1,n}],{j,1,n}] 6
1, 34, 1788, 146256, 17485920, 2894002560, 635331029760, 178910029670400, 62920533840998400, 27042268338763776000, 13950701922125574144000, 8509745665997194493952000, 6059691013778107566981120000 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 LINKS Eric Weisstein's World of Mathematics, Hilbert Matrix. Eric Weisstein's World of Mathematics, Harmonic Number. FORMULA a(n) = (2*n)!*((2*n+1)*Psi(2*n+2)-(2*n+2)*Psi(n+2)+1-gamma). limit(a(n)/(2*n)!/n, n=infinity)=2*ln2. - Vladeta Jovovic, Aug 24 2003 Sum of all matrix elements M(i, j) = 1/(i+j) multiplied by (2*n)! (i, j = 1..n) or Sum of all matrix elements M(i, j) = 2*i/(i+j)^2 multiplied by (2*n)! (i, j = 1..n). a(n) = (2*n)!*Sum[Sum[2*i/(i+j)^2, {i, 1, n}], {j, 1, n}] - Alexander Adamchuk, Oct 24 2004 a(n) = (2n)! * ((2n+2)*H(2n+2) - 2(n+1)*H(n+1) - H(2n+1)), where H(n) is HarmonicNumber[n] = Sum[1/i, {i, 1, n}] = A001008(n)/A002805(n). - Alexander Adamchuk, Nov 01 2004 EXAMPLE a(2) = 4!*(1/(1+1)+1/(1+2)+1/(2+1)+1/(2+2)) = 24*(1/2+1/3+1/3+1/4) MATHEMATICA Table[((2*n)!)*Sum[Sum[1/(a+b), {i, 1, n}], {j, 1, n}], {n, 1, 20}] CROSSREFS Cf. A098118, A001008, A002805. Sequence in context: A212033 A202297 A296673 * A212023 A056566 A335090 Adjacent sequences: A086878 A086879 A086880 * A086882 A086883 A086884 KEYWORD nonn AUTHOR Alexander Adamchuk, Aug 21 2003 STATUS approved
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# Normalised Floating-Point Binary
This page shows you how to normalise floating-point binary numbers, which is a requirement of A level Computer Science courses.
Normalised floating-point binary numbers are the binary equivalent of denary standard form. Very large or very small numbers have their digits shifted left or right so that they start immediately to the right of the binary point - this forms the mantissa - and the number of places it's been shifted to the right becomes the exponent (with a negative number indicating that the mantissa was shifted to the left). The only slight difference between normalised binary and denary standard form is that, in denary, the mantissa is greater than or equal to one and less than ten - i.e. there are digits to the left of the decimal point - whereas in binary, because we use two's complement form for negative numbers, the bit before the point is used as the sign, with all of the actual number coming after the point.
This page is a normalised binary calculator (using the two's complement method) - you can enter a denary number in the textbox below and the required mantissa and exponent will be displayed, along with an explanation of the process. If you aren't familiar with the use of binary for fractions and negative numbers, click here.
Note that there seems to be some confusion over the method used for negative numbers, with some textbooks normalising first and some finding the complement first. This page shows you the result if you find the complement first - the preferred method for A level Computer Science. There are also other methods for storing floating-point binary numbers, including IEEE 754.
Input a
8
4
## Floating Point Representation
### Explanation
If you're interested to see how this page is coded you can view the Python prototype on Replit. | 380 | 1,890 | {"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-2024-26 | latest | en | 0.915432 |
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Jam Pots
Answer : Two of the pots, 13 and 19, were in their proper places. As every interchange may result in a pot being put in its place, it is clear that twenty-two interchanges will get them all in order. But this number of moves is not the fewest possible, the correct answer being seventeen. Exchange the following pairs: (3-1, 2-3), (15-4, 16-15), (17-7, 20-17), (24-10, 11-24, 12-11), (8-5, 6-8, 21-6, 23-21, 22-23, 14-22, 9-14, 18-9). When you have made the interchanges within any pair of brackets, all numbers within those brackets are in their places. There are five pairs of brackets, and 5 from 22 gives the number of changes required : 17.
Math Genius | 215 | 676 | {"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-2018-05 | latest | en | 0.902941 |
http://www.mathcasts.org/mtwiki/Sg/Sec1 | 1,542,219,747,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039742253.21/warc/CC-MAIN-20181114170648-20181114192648-00152.warc.gz | 453,860,466 | 6,187 | # Singapore Mathematics - Secondary One
Topic/Sub-topic Content Learning Outcomes
1 Numbers and Algebra
1.1 Numbers and the four operations Include:
• primes and prime factorisation
• finding HCF and LCM, squares, cubes, square roots and cube roots by prime factorisation
• negative numbers, integers, rational numbers, real numbers and their four operations
• calculations with the use of a calculator
• representation and ordering of numbers on the number line
• use of the symbols <, >, ≤, ≥
• approximation and estimation (including rounding off numbers to a required number of decimal places or significant figures, estimating the results of computation, and concepts of rounding and truncation errors)
1.2 Ratio, rate and proportion Include:
• ratios involving rational numbers
• writing a ratio in its simplest form
• average rate
• problems involving ratio and rate
1.3 Percentage Include:
• expressing one quantity as a percentage of another
• comparing two quantities by percentage
• percentages greater than 100%
• increasing/decreasing a quantity by a given percentage
• reverse percentages
• problems involving percentages
1.4 Speed Include:
1.5 Algebraic representation and formulae Include:
• using letters to represent numbers
• interpreting notations:
• ab as a \times b
• {a \over b} as a \div b
• a^2 as a \times a , a^3 as a \times a \times a , a^2b as a \times a \times b...
• 3y as y + y + y or 3 /times y
• {{3 \pm y} \over 5} as {{(3 \pm y)} \div 5} or {1 \over 5}\times {(3 \pm y)}
• evaluation of algebraic expressions and formulae
• translation of simple real-world situations into algebraic expressions
• recognising and representing number patterns (including finding an algebraic expression for the nth term)
1.6 Algebraic manipulation Include:
• addition and subtraction of linear algebraic expressions
• simplification of linear algebraic expressions, e.g.
• − 2(3x − 5) + 4x
• {2x \over 3}-{3(x-5) \over 2}
• factorisation of linear algebraic expressions of the form
• ax + ay (where a is a constant)
• ax + bx + kay + kby (where a , b and k are constants)
1.7 Functions and graphs Include:
1.8 Solutions of equations and inequalities Include:
• solving linear equations in one unknown (including fractional coefficients)
• solving simple inequality (e.g. 3x \le 5 )
• solving simple fractional equations that can be reduced to linear equations, e.g.
• {x \over 3}+{x-2 \over 4}=3
• {3 \over x-2}=6
• formulating a linear equation in one unknown to solve problems
2 Geometry and Measurement
2.1 Angles, triangles and polygons Include:
• right, acute, obtuse and reflex angles, complementary and supplementary angles, vertically opposite angles, adjacent angles on a straight line, adjacent angles at a point, interior and exterior angles
• angles formed by two parallel lines and a transversal: corresponding angles, alternate angles, interior angles
• properties of triangles and special quadrilaterals
• classifying special quadrilaterals on the basis of their properties
• angle sum of interior and exterior angles of any convex polygon
• properties of regular pentagon, hexagon, octagon and decagon
• properties of perpendicular bisectors of line segments and angle bisectors
• construction of simple geometrical figures from given data (including perpendicular bisectors and angle bisectors) using compasses, ruler, set squares and protractors, where appropriate
2.2 Mensuration Include:
• area of parallelogram and trapezium
• problems involving perimeter and area of composite plane figures (including triangle and circle)
• volume and surface area of cube, cuboid, prism and cylinder
• conversion between cm2 and m2 , and between cm3 and m3
• problems involving volume and surface area of composite solids
3 Statistics and Probability
3.1 Data handling Include:
• data collection methods such as:
• taking measurements, conducting surveys, classifying data, reading results of observations/outcomes of events
• construction and interpretation of:
• tables, bar graphs, pictograms, line graphs, pie charts, histograms
• purposes and use, advantages and disadvantages of the different forms of statistical representations
• drawing simple inference from statistical diagrams
Exclude:
• histograms with unequal intervals. | 977 | 4,262 | {"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} | 4.625 | 5 | CC-MAIN-2018-47 | latest | en | 0.834107 |
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# Linear Regression and Correlation: The Regression Equation
Summary: This module provides an overview of Linear Regression and Correlation: The Regression Equation as a part of Collaborative Statistics collection (col10522) by Barbara Illowsky and Susan Dean.
Note: You are viewing an old version of this document. The latest version is available here.
Data rarely fit a straight line exactly. Usually, you must be satisfied with rough predictions. Typically, you have a set of data whose scatter plot appears to "fit" a straight line. This is called a Line of Best Fit or Least Squares Line.
## Optional Collaborative Classroom Activity
If you know a person's pinky (smallest) finger length, do you think you could predict that person's height? Collect data from your class (pinky finger length, in inches). The independent variable, xx, is pinky finger length and the dependent variable, yy, is height.
For each set of data, plot the points on graph paper. Make your graph big enough and use a ruler. Then "by eye" draw a line that appears to "fit" the data. For your line, pick two convenient points and use them to find the slope of the line. Find the y-intercept of the line by extending your lines so they cross the y-axis. Using the slopes and the y-intercepts, write your equation of "best fit". Do you think everyone will have the same equation? Why or why not?
Using your equation, what is the predicted height for a pinky length of 2.5 inches?
## Example 1
A random sample of 11 statistics students produced the following data where xx is the third exam score, out of 80, and yy is the final exam score, out of 200. Can you predict the final exam score of a random student if you know the third exam score?
The third exam score, xx, is the independent variable and the final exam score, yy, is the dependent variable. We will plot a regression line that best "fits" the data. If each of you were to fit a line "by eye", you would draw different lines. We can use what is called a least-squares regression line to obtain the best fit line.
Consider the following diagram. Each point of data is of the the form (x,y)(x,y)and each point of the line of best fit using least-squares linear regression has the form ( x , y ^ ) (x, y ^ ).
The y^y^ is read "y hat" and is the estimated value of yy. It is the value of yy obtained using the regression line. It is not generally equal to yy from data.
The term |y0-y^0|=ε0|y0-y^0|=ε0 is called the "error" or residual. It is not an error in the sense of a mistake, but measures the vertical distance between the actual value of yy and the estimated value of yy.
εε = the Greek letter epsilon
For each data point, you can calculate, |yi-y^i|=εi|yi-y^i|=εi for i=1, 2, 3, ..., 11i=1, 2, 3, ..., 11.
Each εε is a vertical distance.
For the example about the third exam scores and the final exam scores for the 11 statistics students, there are 11 data points. Therefore, there are 11 εε values. If you square each εε and add, you get
( ε 1 ) 2 + ( ε 2 ) 2 + ... + ( ε 11 ) 2 = Σ i = 1 11 ε 2 ( ε 1 ) 2 +( ε 2 ) 2 +...+( ε 11 ) 2 = Σ i = 1 11 ε 2
This is called the Sum of Squared Errors (SSE).
Using calculus, you can determine the values of aa and bb that make the SSE a minimum. When you make the SSE a minimum, you have determined the points that are on the line of best fit. It turns out that the line of best fit has the equation:
y ^ = a + bx y ^ =a+bx
(1)
where a=y¯-bx¯a=y¯-bx¯ and b=Σ(x-x¯)(y-y¯) Σ(x-x¯)2b=Σ(x-x¯)(y-y¯) Σ(x-x¯)2.
x¯x¯ and y¯y¯ are the averages of the xx values and the yy values, respectively. The best fit line always passes through the point (x¯,y¯)(x¯,y¯).
The slope bb can be written as b=r( sy sx)b=r( sy sx) where sysy = the standard deviation of the yy values and sxsx = the standard deviation of the xx values. rr is the correlation coefficient which is discussed in the next section.
## Note:
Many calculators or any linear regression and correlation computer program can calculate the best fit line. The calculations tend to be tedious if done by hand. In the technology section, there are instructions for calculating the best fit line.
The graph of the line of best fit for the third exam/final exam example is shown below:
Remember, the best fit line is called the least squares regression line (it is sometimes referred to as the LSL which is an acronym for least squares line). The best fit line for the third exam/final exam example has the equation:
y ^ = -173.51 + 4.83x y ^ =-173.51+4.83x
(2)
The idea behind finding the best fit line is based on the assumption that the data are actually scattered about a straight line. Remember, it is always important to plot a scatter diagram first (which many calculators and computer programs can do) to see if it is worth calculating the line of best fit.
The slope of the line is 4.83 (b = 4.83). We can interpret the slope as follows: As the third exam score increases by one point, the final exam score increases by 4.83 points.
## Note:
If the scatter plot indicates that there is a linear relationship between the variables, then it is reasonable to use a best fit line to make predictions for yy given xx within the domain of x-values in the sample data, but not necessarily for x-values outside that domain.
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1. Calculate to 4 decimal places the new exchange rate after a 2% increase in the value of the dollar given a current exchange rate of EURUSD 1.1
2. Given the following exchange rates calculate the GBPEUR rate to 4 decimal places.
EURUSD 1.1043
GBPUSD 1.2970
3. Explain the relationship between your answer to part b of this question and triangular arbitrage.
## Exchange Rate Calculation with 2% Increase in US Dollar Value
If the new exchange rate would go up to 2%. Then, the new EURUSD would be:
Existing EURUSD: 1.1
New EURUSD: 1.1+ [1.1*2%] =1.1220
Hence, the Four decimal places in New EURUSD is .1220
(b):
We have given,
1 Euro = \$1.1043
1 GBP = \$1.2970
And we want to calculate one unit of GBP in terms of Euro, for this we have to apply following formula -
One unit of GBP in terms of USD
One GBP = _______________________________
One unit of EURO in terms of USD
1.2970
GBPEUR Rate = __________
1.1043
= 1.1745(EURO)
For understanding relationship between answer to part b of this question (stated above) and triangular arbitrage, firstly, we have to understand the meaning of triangular arbitrage.
Triangular Arbitrage: It is a process of finding an arbitrage opportunity from three different currencies in the foreign exchange market due to price difference in these currencies. It is also referred as cross currency arbitrage or three-point arbitrage (Reynolds, et. al., 2018).
Process of Triangular arbitrage: In this process we simultaneously convert one currency into another and so on and reach back with original currency which we have convert first. There are two ways to initiate this process which are -
1.Clockwise Arbitrage
2 Anticlockwise Arbitrage
Relationship between answer to part b of this question and triangular arbitrage: In both our answer and triangular arbitrage, there are 3 currencies involved. In our answer we calculate exchange rate through cross currency rate formula and also this formula is required to convert one currency to other (Reynolds, et. al., 2018).
Therefore, process used in answer to part b is same in all aspect as used in triangular arbitrage.
Home currency is US Dollar
Foreign currency is EUR
1. a) Exchange rate - January 2017 1 EUR=1.1 US\$
December 2017 1 EUR=1.144 US\$
In accordance to Purchasing Price Parity assumption, there is only one price across the world and difference between exchange rate of different countries is due to difference in inflation rate of those countries (McKinnon and Ohno, 2016).
As exchange rate of US\$ of December 2017 is higher than the exchange rate of January 2017 as compare to EUR that means that EUR is at forward premium.
By applying Purchasing Price Parity assumption in the given problem, it can have concluded that difference in exchange rate is due to inflation and inflation rate of EUR is higher than the inflation rate of US\$ in year 2017 because for purchase of 1EUR it required 1.1US\$ in January
2017 and for purchase of 1EUR it required 1.144US\$ in December 2017.
Difference between forward and spot rate *100 * 365
Rate of inflation= _________________________________ __________
Spot Rate No. of days
1.144-1.1
Rate of inflation= ________ *100 =4% per annum
1.1
Exchange rate of January 2017 is 1EUR=1.1US\$
If inflation rate of US is taken 3% higher than the euro, then real value of dollar will be calculated as follows:
Assume inflation rate of euro be 5%p.a.
Therefore, inflation rate of US will be 5+3= 8%
1+inflation rate of domestic country
## Calculating GBPEUR Rate to 4 Decimal Places
Forward Rate= ___________________________________ * Spot Rate
1+inflation rate of foreign country
1.05
Spot Rate =_________ * 1.144 = 1.1122
1.08
Therefore, real value of dollar will be 1EUR=1.1122 US\$
Change in Real value=1.1122-1.1=0.122
Change in Real value (%)=.0122/1.1*100=1.109%
(c)Change in real exchange rate can affect the currency in two ways: -
1. Positively
2. Negatively
1.Positively-If home currency is appreciated as compare to foreign currency then it will increase the value of home currency in foreign exchange market.
2.Negatively-If home currency is depreciated as compare to foreign currency then it will decrease the value of home currency in foreign exchange market.
The foreign exchange rate can be increase or decrease due to following factors-
1.Rate of inflation- If inflation rate of a country increases as compare to other country then exchange rate of first mentioned country will decrease. Higher the inflation rate lowers the exchange rate (Hodrick, 2014).
2.Interest Rate- If interest rate of a country increase then it will allow more inflows of foreign investment in the country which will increase the value of foreign exchange. But when we repay the invested amount then our currency will have depreciated (Cairns, 2018).
3.National Income- If national income of person in a country increase then it will result in outflows of home currency it will negatively impact the home currency and home currency will depreciated (Blanchard and Johnson, 2017).
4.Balance of payment-If total receipt of foreign currency exceeds total payment of foreign currency during the year then it will have positive impact on home currency
5.Speculation- It can affect in both positively and negatively
6.Scarcity of resource- If there is scarcity of any resource in a country then it will result into outflows of home currency and therefore depreciate the home currency.
(d) Concept and effect of interest rate in foreign exchange can be understand with the help of Interest rate parity theory:
According to Interest Rate Parity, the forward rate of one currency will contain a premium (or discount) that is determined by the differential in interest rates between the two countries. As a result, covered interest arbitrage will provide a return that is no higher than a domestic return.
And higher interest rate increases the value of a country 's currency. Higher interest rates tend
to attracts foreign investments, increasing the demand for and value of home country 's currency. Conversely, lower interest rates tend to be unattractive for foreign investment and decrease the currency 's relative value (Du, et. al., 2018).
Effect on foreign exchange rates due to change in interest rate can be understand by following steps: -
Step1. Increase in US interest rate.
Step2. Better return on US saving account
Step3. 'Hot Money Flows' in to take advantage of high interest rate.
Step4. Increased demand of US \$
Step5. Appreciation in value of US\$
Therefore, it can be concluded that if the interest rate of US (Home Currency) is higher than Euro area (Foreign
Currency), then the value of US currency will also be in increased so it always be expected that interest rate of US
## The Relationship Between Answer and Triangular Arbitrage
currency will increase so that value of US currency will be increased as compare to EURO.
(e) In European counties rate of interest of government bond is determined by the European Central Bank and national government which are same with other countries of Europe. But due to following factors practically the interest rate of government bond is different: -
1.The government directives to the central bank to accomplish the government's goal
2.The currency of the principal sum lent or borrow.
3.The term to maturity of the investment.
4.The perceived default probability of the borrower
5.The supply and demand in the market.
6.Interest rate can fluctuate accordance to status of economy.
1. People prefer to have their resource available in a form that can immediately be exchanged.
For understanding comparison between country risk and exchange rate volatility, we have understood the following terms -
Country Risk: Country risk is a risk related to country's economic and/or political risk which may affect the businesses and investment opportunities. For example, investment in one country, let us say, in United States is not same as investment in Australia, because of separate risks which are depending country to country (Brooke, 2016). It can be divided into two major groups, which are -
1.Political Risk
2.Economic Risk
Political Risk: Risk which is associated with country's political environment, their politicians and impact of political decisions on businesses. For example, desperate politicians may impose a risk to investors in certain strategic industries because of supportive nature for nation (McKellar, 2017).
Economic Risk: It is a risk which is associated with country's financial condition and country's ability to repay its debt. For example, high debt-to-GDP ratio of a country may not easily allow to a country to raise funds(money) in international market and this makes a country risky for its domestic economy (Lin, et. al., 2018).
For understanding the exchange rate volatility, firstly we should understand the following term -
Exchange Rate: It is a rate at which one currency can be converted into another currency. Exchange Rate allows us to determine how much of one currency is required to gain another currency (Boon and Hook, 2017).
Exchange Rate Volatility: Exchange rate volatility refers to the tendency for foreign currencies to appreciate or depreciate in value, thus affecting the profitability of foreign exchange trades. The volatility is the measurement of the amount that these rates change and the frequency of those changes. There is a vast amount of studies on the impact of exchange rate volatility on international trade, as a result we may consider that volatility in exchange rate for short period of time has no such impact on the international trade but exchange rate volatility in long period of time has very vast impact on international trade (Brooke, 2016).
Why there is a change in Exchange Rate? This is very important question and this is depending on following factors -
• Interest rate
• Inflation rate
• Balance of payment
• Government debt
• Political stability & performance
• Recession
• Speculation
Interest Rate: There is a positive correlation between country's interest rate and its exchange rate. This mean if interest rate of one country is more in relation to other country than its currency is appreciated as compared to other currency higher interest rates provide higher rates to lenders, thereby attracting more foreign capital, which causes a rise in exchange rates (Hajilee and Al Nasser, 2014).
Inflation Rate: Changes in market inflation cause changes in currency exchange rates. A country with a lower inflation rate than another, will see an appreciation in the value of its currency. The prices of goods and services increase at a slower rate where the inflation is low. Relationship between inflation rate and exchange rate can be understand through purchasing power parity (PPP) (Brooke, 2016).
Balance of Payment: It means country's current account balance. In other words, it reflects balance of trade & investment in foreign countries. If a country spends its currency more on imports than compared to exports, then it results in decline in its own currency. Balance of payments fluctuates exchange rate of its domestic currency (Gibson and Thirlwall, 2016).
Government Debt: Govt. debt means national debt owned by central government. If the market predicts government debt within a certain country, then investors will sell their bonds in the open market. This results in a decrease in the value of its exchange rate will follow (Greenwood, et. al., 2015).
Political Stability & Performance: A country's political state and economic performance can affect its currency strength. If there is a stability in political environment of a country, then it is more attractive to foreign investors, as a result, drawing investment away from other countries and invest with more political and economic stability. Increase in foreign capital, in turn, leads to an appreciation in the value of its domestic currency (Tunc, et. al., 2018).
Techniques for hedging exchange rate volatility risk: These can be categorised in two groups which are -
1. Internal Technique:
Invoicing
Netting
Outsourcing
2.External Technique:
Forward Contract
Money market Cover
Futures
Options
Comparison Between Country Risk Analysis and Exchange Rate Volatility: It is very important to analysis the country risk to identify the related aspect which are necessary for determination of exchange rate volatility. If there is more country risk, then there should be more volatile exchange rate (Brooke, 2016). It can be summarised in following points -
If there is no political stability, then there is more country risk. As a result, there should be more volatile exchange rate
In country risk analysis, we analysis the risk as a whole related to nation but in exchange rate volatility, we consider the risk related to volatility of exchange rate which is covered only in forex market (den Bossche, et. al., 2017).
In analysis of country risk, we take various measurement to overcome from more country risk which negatively affect the country but in exchange rate volatility, there are various techniques which are used to hedge currency risk exposure (Alper, 2017).
If there is an appreciation in the currency due to positive exchange rate volatility, then it means that there are less country risk in that aspect (Hirst, et. al., 2015).
Exchange rate volatility has a positive impact on one currency and negative impact on other currency. It means that one country has low risk as compared to other country (Matsushita, et. al., 2015).
The exchange rate is of great importance and can affect the overall growth and development of the country’s trade and economy. Thus, the country should try to make their currency strong and try to stabilize their exchange rate so as to improve the country’s trade and economics. As a result, there should be stable political and economic condition in the country (Boon and Hook, 2017).
When a country experiences a recession, its interest rates are likely to fall, decreasing its chances to acquire foreign capital. As a result, its currency weakens in comparison to that of other countries, therefore lowering the exchange rate. Recession is a country risk, we should analysis this to overcome from this situation (Tunc, et. al., 2018).
Globalization is the speedy segregation of the entire economy which helps the companies in creating the interlinks in several ways. Globalization enhances betterment in the international trade practices and communication of different countries. Since last 25-35 years there is the marvellous increment in the globalization accordance with the increment in number of multinational corporations which expands their businesses in several number of countries and operates from their home country (Steger, 2017). MNC' s used to develop their production centres at different places in different countries but generally at those places where there is the availability of cheap labour, markets are close, where the government policies face later their interests, where there is availability of raw materials etc. this kind of places are beneficial for MNC' s in order to reduce the costs of goods and for maximising the profits. After the production of goods in huge quantity the MNC' s sells their products in different countries. This creates the interlinks between MNC' s and different economies of different nations and promotes the globalization (Baylis, et. al., 2017).
The benefits for the developing interlink helps in “homogenization of global culture” as this statement is given by several authors in their articles. In present times due to increasing role of multinational corporations in the contemporary world economy, leads to availability of same products with the same logos, same movies, same brands for different items, everybody eats at the same restaurants and same drinks. In this scenario, the logos and brands efforts of MNC' s symbolises the globalization across the different nations and throughout the entire world economy (Ashford and Hall, 2018). The multinational corporations ensure enhancement of the Foreign Direct Investment (FDI), the value-added activities in several countries are controlled and managed by these MNC' s. In present time, the multinational corporations give the priority to manufacturing and services related with the extraction of raw materials and commodities. FDI scheme also beneficial for the funding form the international markets. These MNC 's focuses on the adaptation of the advanced technologies as it is essential for maintaining the quality of products and reducing the costs increasing the production within time and costs effective manner. This helps in maximisation of their profitability (Tarasenko, et. al., 2018).
The World Trade Organisation and the Organisation for Economies Cooperation and Development framed some rules and regulations for promoting the fair-trade practices throughout the entire world economy. Besides this, the World Trade Organisation also focuses on betterment of the role transportation and communications technologies that leads to reduction in costs incurred in managing the far-flung economic operations. The MNC 's promotes the advancement of the computing and telecommunication technologies, these technologies help in managing the operations of the business in different countries and in different time zones. As the main ordeals of globalization is distances, this all advancement in transportations and communications are made in order to reduce the effects of this obstacle (Light, 2018). There some advantages of the expansion of globalization or the 'MNC' s promotes the globalization is good for the entire economy” for supporting this statement some advantages are as follows.
• Empowerment to businesses:as the production activities of the MNC 's requires a lot of things according to the product line of the MNC. The availability of all things at one place or region is not possible as some regions have minerals, from iron ore to Bauxite. Some regions have rich oil and coal, natural gas or gold and diamond. These regions are more beneficial for the MNC 's as globalisation ensures every region to offers the best products or services to other regions. These regions help the MNC 's in procurement of the cheap technology, cheap steel, and better quality of raw material for the production of better-quality goods (Baylis, et. al., 2017).
• Facilitates the large market for Businesses: no doubt in the statement that 'globalization facilitates large market for MNCs’.Globalisation enhances the MNC 's to promote their businesses across different countries throughout the world. By facilitating the different countries for setting the production centres and for selling the products and services in different markets of several countries helps the MNC 's in acquiring the large market for promoting the sales in other words it helps in targeting large number of customers by facilitating the international markets different countries. It also helps in incrementing the revenues of the MNC 's (Beck, 2018).
• Helpful for consumers:The globalization also helps the consumers in getting goods at cheap prices, better services, acquiring advanced technological products. Along with this, it also provides of brands and large varieties of products according to the desire and budget of customers (Steger, 2017). Not only this, it helps in maintaining the status of high-class customers by providing them products of high level or international brands. Overall, the globalization helps in improving the living standard of the society by providing them advanced technological products and high-quality standard services at the outlets of the MNC 's that are available in different regions of the different countries (Tarasenko, et. al., 2018).
• Creation of Employment: the globalization allows the MNC 's to set up their production centres or factories in different regions of different countries. Generally,the MNC 's selects rural areas for setting the production centres and performing their manufacturing or production activities as well as selling activities performed by setting of various outlets in different market of a country. The setup of factory or production centres and outlets helps in creating the employment opportunities for the general public of that region as well as for people of that nation. This all helps in reducing the unemployment rates of economy by facilitating several employment opportunities to different people according to their knowledge and experience in particular stream (Ashford and Hall, 2018).
• Better relations:Globalization help in developing the better relations between two or more trading countries, who are trading with each other. This reduces the chances of war between those countries by developing good relations with the help of international trading or globalisation. As these countries enjoys the benefits of globalised economy and helps in developing the better relation with different countries and reduces the chance of war against those countries and in case of world war and financial crisis or any other major problem, there may be chances of getting support from those countries (Baylis, et. al., 2017).
• Technological Innovation:Emerging number of MNC' s leads to increase in level of competition helps in stimulating new technology development, along with the growth and progress in FDI, which is beneficial in betterment of economic output by making processes more effective and efficient (Tarasenko, et. al., 2018).
• Contributes to National Income: No doubt the globalization contributes to national income of country and countries, by providing thousands of employmentsin different regions of the country and tax paid by these MNC 's to the tax authorities of different countries in which the MNC 's are practising their business activities this, all leads to increment or contribution in national income of nation and nations (Beck, 2018).
• Economies of scale:Globalization ensures large companies to recognize economies of scale it helps in reduction in costs as well as in prices, which promote the economic growth, although this can hurt many small businesses attempting to compete domestically (Atack,2018).
Books and Journals:
Alper, A.E., 2017. Exchange Rate Volatility and Trade Flows. Fiscaoeconomia, 1(3), pp.14-39.
Ashford, N.A. and Hall, R.P., 2018. Globalization: Technology, trade regimes, capital flows, and the global financial system. In Technology, Globalization, and Sustainable Development (pp. 264-333). Routledge.
Atack, J., 2018. Estimation of economies of scale in nineteenth century United States manufacturing. Routledge.
Baylis, J., Smith, S. and Owens, P. eds., 2017. The globalization of world politics: an introduction to international relations. Oxford University Press.
Beck, U., 2018. What is globalization? John Wiley & Sons.
Blanchard, O. and Johnson, D.R., 2017. Macroeconomics. London usw.: Prentice-Hall International Inc.
Boon, T.H. and Hook, L.S., 2017. Real exchange rate volatility and the Malaysian exports to its major trading partners. In ASEAN in an Interdependent World: Studies in an Interdependent World (pp. 95-117). Routledge.
Brooke, M.Z., 2016. Handbook of international financial management. Springer.
den Bossche, V., Henry, P.L. and Zdouc, W., 2017. The Law and Policy of the World Trade Organization: Text, Cases and Materials. Cambridge University Press.
Du, W., Tepper, A. and Verdelhan, A., 2018. Deviations from covered interest rate parity. The Journal of Finance, 73(3), pp.915-957.
Gibson, H.D. and Thirlwall, A.P., 2016. Balance-of-payments theory and the United Kingdom experience. Springer.
Greenwood, R., Hanson, S.G. and Stein, J.C., 2015. A Comparative?Advantage Approach to Government Debt Maturity. The Journal of Finance, 70(4), pp.1683-1722.
Hajilee, M. and Al Nasser, O.M., 2014. Exchange rate volatility and stock market development in emerging economies. Journal of Post Keynesian Economics, 37(1), pp.163-180.
Hirst, P., Thompson, G. and Bromley, S., 2015. Globalization in question. John Wiley & Sons.
Hodrick, R., 2014. The empirical evidence on the efficiency of forward and futures foreign exchange markets. Routledge.
Light, A., 2018. The development of world trade organization law: Examining change in international law [Book Review]. Australian Year Book of International Law, 35, p.224.
Lin, S., Shi, K. and Ye, H., 2018. Exchange rate volatility and trade: The role of credit constraints. Review of Economic Dynamics.
Matsushita, M., Schoenbaum, T.J., Mavroidis, P.C. and Hahn, M., 2015. The World Trade Organization: law, practice, and policy. Oxford University Press.
McKellar, R., 2017. A short guide to political risk. Routledge.
McKinnon, R.I. and Ohno, K., 2016. 7 Purchasing power parity as a monetary. The Future of the International Monetary System: Change, Coordination of Instability? Change, Coordination of Instability? p.42.
Reynolds, J., Sögner, L., Wagner, M. and Wied, D., 2018. Deviations from Triangular Arbitrage Parity in Foreign Exchange and Bitcoin Markets.
Steger, M.B., 2017. Globalization: A very short introduction (Vol. 86). Oxford University Press.
Tarasenko, O., Yermakov, Y. and Pchelin, V., 2018. FORMATION OF COMPETITIVE ADVANTAGES OF ENTERPRISES IN TERMS OF GLOBALIZATION: COMPETITIVE DYNAMICS AND AN INTELLECTUAL COMPONENT. Baltic Journal of Economic Studies, 3(5), pp.469-473.
Tunc, C., Solakoglu, M.N., Babuscu, S. and Hazar, A., 2018. Exchange rate risk and international trade: The role of third country effect. Economics Letters, 167, pp.152-155.
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# FINA0301 Assignment 3 - 1 a The ini tial investment equals...
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1) a.The initial investment equals to (-800+75-45=)-770. This position yields \$815 after one year and so the rate of return is 0.05844. b. In a), we have to pay more that the risk-free rate so we are in risk- less position. And the arbitrage implied is that we should first borrow the money at 5% and then buy the position stated by a). As a result, we yield a sure return at 0.68%. c. From the above calculation, the initial investment would be \$775.25 and the yield would be \$815 one year later with the risk-less rate of 5%. So the difference between the call and put prices that would eliminate arbitrage is \$24.748. d. By using the method of c), we can see that for \$780, the difference is \$58.04105; for \$800, it is \$39.01646; for \$820, it is \$19.99187; and for \$840, it is 0.967283. 2) a.By using the put-call parity for the currency options which is: + P (K, T ) = e rf T x 0 + C (K, T ) + e rT K = e 0 .01 0.009 + 0.0006 + e 0 .05 0.009 = 0 .00891045 + 0.0006 +0.00856106 =0.00025 b.The option price is higher than expected, so we sell the put option and create a long put option synthetically to offset the risk perfectly. The following demonstrates the arbitrage process.
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null | null | null | null | null | null | Name that dwarf planet moonlet!
The debate about Pluto’s planetary status has simmered down, although I was always partial to the way Alan Stern, the Director of New Horizons mission to Pluto, explained it to me: it’s just a dwarf planet. We don’t call a small dog a different species just because they’re small, we just call it a miniature, something like that makes sense for Pluto and its cohorts like Sedna or Ceres. But we now know Pluto has two new moons, or dwarf moons or moonlets if we want to extend the debate, and SETI wants your help naming them! [Read more…]
For my missing friend
I mentioned a good friend of mine died last week from sudden cardiac arrest. His funeral is today and I probably won’t feel much like blogging. His best friend and I were given the honor of writing the service program obit, what mourners will read at the funeral. We were tasked by the director to hit certain points and there was a narrow word count, but we wanted to give an organic feel to it so that friends and family would know this wasn’t done by a paid writer who didn’t know the deceased, it was done by two people who love him and miss him. That rough draft is below with some details removed or altered out of respect for his privacy. [Read more…]
Updated: Comet ISON orbit and viewing possibilities
Comet ISON will pass within 40 million mile of earth on Christmas of this year and be in decent position for viewing by eye or small scope throughout the fall of 2013. Predicting the brightness of a comet is real dicey, it depends on what the comet is made of and how many times it has been close enough to the sun to vaporize some of those volatile substances. But the scant evidence we have is that ISON could be relatively new to the inner solar system and may be made off the traditional ices found in many comets.
Astronomers believe comets are lumps of exotic and water ices, some with rocky or metal compound inclusions left over from the formation of the solar system. They are, typically between several meters and several hundred kilometers in diameter. The average really big comet we see during a normal lifetime is in the five to 20 mile diameter range.. Periodic comets like Halleys observed up close also show a thin layer of black, tarry coating, possibly produced by repeated solar encounters acting on organic ices over time.
ISON was imaged by NASA’s Deep Impact spacecraft a few weeks ago producing the video above, and already indicated a faint tail 40,000 miles long. It’s still as far away as the planet Jupiter, beyond what solar astronomers call the frost line, where water and carbon dioxide are solid rocks, even methane is at least a slushy liquid. But other frozen gases, nitrogen and oxygen for example, are now well within their respective melting and boiling ranges and if those low boiling point substances exist in enough quantity on the surface to produce a faint tail at that distance in those frigid reaches of deep space, it bodes well for viewing later this year. It suggests this particular object may have never visited the inner system since its creation almost 5 billion years ago, or that it has visited only a few times over millennia.
That would mean the tiny nucleus could be chock full of pristine CO2, water, methane, and ammonia. If so, ISON could develop an enormous tail over a million miles long or longer as it passes Mars on the way to earth’s orbit, making it a once in a lifetime celestial sight!
While it won’t come real close to earth by some standards, it should be in a fair position relative to earth and sun for several months, meaning we’ll have a good long look at the full length of any tail that develops. It will be moving toward the sun when the earth is about 60 to 70 millions away and will appear high in the night sky during the course of a clear November evening later this year. ISON is a sun grazing comet, it will be heated to many hundreds of degrees Celsius for several days as it whips around Sol, enough that it might calve into smaller bodies or disintegrate. But if it survives, ISON will then swing back, tail leading the way, making its closest approach to earth on December 26, 2013, when it will be about 39 million miles away as shown by the blue portion of the cometary orbit above.
There is zero chance ISON or any sizable fragments of it will strike earth. But in the event an object similar in size and composition does hit us one day the results will be catastrophic. I made modest assumptions, assuming you were 1000 miles way from the impact site, that the object is mostly low density ice, that long period comets like ISON hit at a leisurely 20 miles per second. The fireball as it streaked over head and struck the earth would be so bright it would instantly blind anyone unfortunate enough to see it. The final crater would be almost a 100 miles wide and 20 miles deep.
The seismic wave would arrive within a few minutes, magnitude 10.2, greater than any recorded earthquake in recorded history. Even 1000 miles away ejecta would rain down like artillery for hours at near super sonic speeds. An hour and a half after impact the atmospheric shockwave would arrive to the tune of 400 mph windblasts. In short, it would be as devastating as the K-T impact, the world would catch on fire, followed by a nuclear winter and massive climatic shifts that would permanently alter weather patterns.
The good news is that won;t happen. The somewhat bad news is we have lousy experience with comets living up to their potential over the last 50 years. But after the letdown of Comet Kohoutek in 1974, and Halley’s Comet being in just about the least favorable position for viewing on record in 1986, it seems like we’re about due for a beauty. Regardless, with all the scopes on earth, in space, and space craft like Deep Impact trainined on ISON, this will be the most studied and well photographed comet in history.
I’ll be on the air shortly
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Yesterday I was near desperate, today I’m saved
UPDATE: Guys and Gals, I’m fighting off tears in my eyes as I write this, Just over a hundred bucks more now, and even if we don’t get it, the donations so far are enough that I’m basically saved the humiliation and risk that had me so freaked out. I’m going to leave this up through mid-morning, while I go to cardio rehab, and I’m sure that will do it. I can’t thank you enough, my stress level is dropping noticeably. I’ll announce as soon as we hit the target. Thank you all so much!
I’m just going to keep this short and sweet: I need several hundred dollars to cover some medical expenses from a massive unplanned heart attack I suffered last month. If that’s all you need to know you can donate here, StevenAndrew-at-Paypal. Any amount no matter how small is greatly appreciated. Email address of the account is DarkSydOThemoon -at- domain A.O.L plus the dot and the com if you’re set up that way.
I have great health insurance, I loaded up on extra insurance including extra short-term disability insurance out of my own pocket. That disability is enough for me to live on and it has been approved. I’ve literally been told the first check is now in the mail. But I haven’t had a pay check in over a month now, I’m tapped out from constant deductibles which started anew this year, big copays for heart surgery and MD support staff, plus hospital stay, follow-up appointments with a cardiologist and a rheumatologist — the autoimmune inflammatory condition provides significant risk of serious complications and had to be treated with Humira and other very expensive drugs — and endless Rx copays for all that. I had a bit of an anxiety meltdown today when I realized I don’t even have enough in checking to cover the copays for my appointments, rehab, and meds this week (The two most important meds are called Carvediol and Brilinta respectively if anyone’s interested). That first disability check probably isn’t going to show up this week, and it’s drawn on an out of state bank anyway.
I am so sorry, I feel like a heel, I never thought I would be in this situation, even planned for years and sacrificed to sock money away during better times for years and years. But I don’t know what else to do. I sure don’t want to depend on my 84-year-old retired father, his health is declining and he’s done enough for me over the last two years. But those meds and those appointments are absolutely critical. If I’m wrong and the checks hit tomorrow or Thurs, I may be able to offer to reverse some of the donations. But for now I’m beyond worried, I’m scared and desperate. | null | null | null | null | null | null | null | null | null |
http://ndsportspage.com/the-coming-mpu/homothetic-function-calculator-f67a50 | 1,720,897,931,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514512.50/warc/CC-MAIN-20240713181918-20240713211918-00344.warc.gz | 23,107,506 | 12,082 | But we have de ned previously that a function is homothetic if it is a monotone transformation of a homogeneous function. Show that the CES function is homothetic. Consumer’s surplus Mattias has quasilinear preferences and his demand function for books is B = 15 – 0.5p. Code to add this calci to your website Then, it is homothetic if and only if j j j j x u x 1 ( ) ( ) 1 • With homothetic preferences all indifference curves have the same shape. Information and translations of homothetic in the most comprehensive dictionary definitions resource on the web. utility functions which are increasing transformations of functions with this property. Explore anything with the first computational knowledge engine. Homothetic functions have been applied in a number of empirical studies. E. Common Functions E.3 Homothetic functions Definition: Homothetic function A function f x x( , ) 12 is homothetic if, for any x0 and 1, and any r! The agent’s effort, a affects current profits, q1 = a + #q 1, and future profits, q2 = a + # q2, where #qt are random shocks, and they are i.i.d with normal distribution N(0,s2). Homothetic Functions Afunctionishomothetic if it is a monotonic transformation of a linearly homogeneous function. Why? Typically economists and researchers work with homogeneous production function. How does the MRS depend on the ratio y/x? The wikipedia article about Homothetic preferences however defined a preference to be homothetic, if they can be represented by a utility function and the following is true: $$u(kx_1, kx_2) = k \cdot u(x_1, x_2)$$ And I am pretty sure, that this is not true for Cobb Douglas preferences: The agent’s effort, a affects current profits, q1 = a + #q 1, and future profits, q2 = a + # q2, where #qt are random shocks, and they are i.i.d with normal distribution N(0,s2). c. Show that the MRS is strictly diminishing for all values of δ . Quickly find the codes you need to include just about any kind of mathematical relationship in your web documents. Homoge-neous implies homothetic, but not conversely. <> The demand for x 2 = βm/p 2 which is also a linear function of m, p 2 remaining constant. The agent retires at the end of the first period, and … • Homothetic: Cobb-Douglas, perfect substitutes, perfect complements, CES. Show that your results from part (a) agree with our discussion of the cases d = 1 (perfect substitutes) and d = 0 (Cobb-Douglas). Do not proceed further unless the check box for homogeneous function is automatically checked off. �p =h�rx}J�ǘL��� The following shows that, in additively separable utility functions, any deviation from CES would give us non-homothetic preferences. How does the MRS depend on the ratio y/x? A Startling Fact about Inverse Calculator Uncovered . Homothetic Functions Afunctionishomothetic if it is a monotonic transformation of a linearly homogeneous function. But we have de ned previously that a function is homothetic if it is a monotone transformation of a homogeneous function. function fis homothetic. ... delta -1 since the mrs depends only on the ratio of the quantities x and y, the utility function is homothetic. b. b) d = 1 MRS is equal to alpha/ beta i.e a constant which is always the case for perfect substitutes. a. The following shows that, in additively separable utility functions, any deviation from CES would give us non-homothetic preferences. The Class of Homothetic Isoquant Production Functions' There is a wide choice of algebraic forms which can be used to represent and estimate the production function [23, 19]. 1. d. 1. d. function fis homothetic. Meaning of homothetic. Problem 4. A transformation of the variables of a tensor functions. Homothetic polygons in circle. ii. Homothetic production functions have the property that f(x) = f(y) implies f(λx) = f(λy). Why? (Homogenous function and homothetic function) Identify the following functions are either homogenous or homothetic (or both). Homogeneous production functions have the property that f(λx) = λkf(x) for some k. Homogeneity of degree one is constant returns to scale. The function wx is continuous. Calculate the MRS for this function when y/x = 0.9 and y/x = 1.1 for the two cases δ = 0.5 and δ = -1. triangle center functions are homogeneous What does homothetic mean? What does homothetic mean? Calculate the MRS for this function when y/x = 0.9 and y/x = 1.1 for the two cases δ = 0.5 and δ ... delta -1 since the mrs depends only on the ratio of the quantities x and y, the utility function is homothetic. Inverse function calculator helps in computing the inverse value of any function that is given as input. This happens with production functions. R is called homothetic if it is a mono-tonic transformation of a homogenous function, that is there exist a strictly increasing function g: R ! Some Examples •Perfect substitutes u(q 1,q 2) = aq 1 + bq 2: The MRS is −a/b and is constant. • Not homothetic: Quasilinear. So 4. changes the tensor into another whose components are linear (1) f(x1, X2, x3) = x1 x2 + x3; (2) f(x,y) = 24; (3) f(x, y) = xy2 + 2x3. The constant function f(x) = 1 is homogeneous of degree 0 and the function g(x) = x is homogeneous of degree 1, but h is not homogeneous of any degree. Indifference curves are parallel straight lines. ���K��Q,��:��i�1��p�^��e/�����E�BO�\��v2�֤�68�! In this paper, we classify the homothetic production functions of varibles 2 whose stream We also offer a table of HTML number codes . 2.5 Homogeneous functions Definition Multivariate functions that are “homogeneous” of some degree are often used in economic theory. Once a given algebraic form is chosen, certain key parameters are then estimated to determine the empirical functional relationship between the factor inputs and value-added. Such a function is an equation showing the relationship between the input of two factors (K and L) into a production process, and the level of output (Q), in which the elasticity of substitution between two factors is equal to one. 39 The Many-Good Case • Suppose utility is a function of n goods given by utility = U(x1, x2,…, xn) On the other hand, the quadratic and the CARA class are not homothetic. iv. Since the demand functions for both x 1 and x 2 are linear functions of m, the ICC will be a straight line through the origin as shown in Fig. Homothety and uniform scaling. The Steiner area formula and the polar moment of inertia were expressed during one-parameter closed planar homothetic motions in complex plane. Homothetic functions are functions whose marginal technical rate of substitution (the slope of the isoquant, a curve drawn through the set of points in say labour-capital space at which the same quantity of output is produced for varying combinations of the inputs) is homogeneous of degree zero. Author: Daniel Mentrard. Hints help you try the next step on your own. Show that your results from part (a) agree with our discussion of the cases δ = 1 (perfect substitutes) and δ = 0 (Cobb–Douglas). Since his monotonic, it remains to show that h 1 fis homogeneous. Ordinary Differential Equations. a. Flexibility and Non-Separable CES functions We let denote the user price of the ith input, and let be the cost-minizing demand for the ith input. "l�?b��%�]�$�����$sd�1F�f����,�gj_;՝|����Ge 0, if f x f x( ) ( )01d then f rx f rx( ) ( )01d. By using this website, you agree to our Cookie Policy. To recall, an inverse function is a function which can reverse another function. Economists have at different times examined many actual production functions and a famous production function is the Cobb-Douglas production function. It is clear that homothetiticy is … Calculate the compensated income, m´. 6. a, b. 18.2.c To make 18 guns, firm 1 needs 9 barrels of oil. A function is homothetic if it is a monotonic transformation of a homogenous function. c��}��@C�h�Һ������������ӭ��qB�d�e�Q4{ֶe��+��� >���GQU)ݫVGqfҔ��%�b�����"�G�7��Yk��מ��֑�D���&��J�5. We start with a look at homogeneity when the numerical values themselves matter. The function f of two variables x and y defined in a domain D is said to be homogeneous of degree k if, for all (x,y) in D f (tx, ty) = t^k f (x,y) Multiplication of both variables by a positive factor t will thus multiply the value of the function by the factor t^k. Now let us calculate the intertemporal elasticity of substitution for a homothetic utility function. Homogeneous Differential Equations Calculator Online calculator is capable to solve the ordinary differential equation with separated variables, homogeneous, exact, linear and Bernoulli equation, including intermediate steps in the solution. Weierstrass elliptic function, and Homothetic paradigm can be helpful for enhancing a collection of matrix updating methods based on constrained minimization of the distance functions. utility functions which are increasing transformations of functions with this property. By using this website, you agree to our Cookie Policy. A beautiful, free online scientific calculator with advanced features for evaluating percentages, fractions, exponential functions, logarithms, trigonometry, statistics, and more. It is denoted as: f(x) = y ⇔ f − 1 (y) = x. I know that homothetic production function implies that cost function is multiplicatively separable in input prices and output, and it can be written as C(w,y)=h(y)C(w,1). Show that the CES function is homothetic. 20. �u �o���g�H#�i�J��Sa}�y��@�3��_p�\PȐ����P}��ضwt� 4Te�h��eg�7��W�J~JB��Ş0e$h)'9��(1ؙ�E�j%?V,�S��u�={M�H���vfx^��#��EN�CԃI��2d����G"B��������H�� � M��閼 �vFn�6I�������o�e�?�0������gx��*Ca���7mS�k^Z �8 e�T^�������ě��V��z�]�;!7yvx�Z�NnӢK:��@�ה�2����bW�>��� Information and translations of homothetic in the most comprehensive dictionary definitions resource on the web. If the homothetic center S happens to coincide with the origin O of the vector space (S ≡ O), then every homothety with ratio λ is equivalent to a uniform scaling by the same factor, which sends → ↦ →. For any scalar a, the inverse of h, as noted prior, tells us … %PDF-1.4 Production functions may take many specific forms. You should be familiar with the idea of returns to scale. 4. b = 0. To make 33 units of butter, firm 2 needs 11 barrels of oil. An agent can work for a principal. • Any monotonic transformation of a homothetic function is homothetic. It is also called an anti function. Calculate the person´s demand for x and y at the new price. Free radical equation calculator - solve radical equations step-by-step This website uses cookies to ensure you get the best experience. Proposition: Suppose that the utility function, U RJ R: , is quasi-concave, increasing, and separable, J j U x u j x j 1 ( ) ( ). Homothetic production functions have the property that f(x) = f(y) implies f(λx) = f(λy). b. For any scalar a, the inverse of h, as noted prior, tells us … Meaning of homothetic. Let the multi-product production function be a homothetic closed form as in Equation (1), which is a generalization of the Cobb-Douglas homogeneous function. It's possible to use calculator … 2.5 Homogeneous functions Definition Multivariate functions that are “homogeneous” of some degree are often used in economic theory. Inverse function calculator helps in computing the inverse value of any function that is given as input. Define the reference cost, and reference value share for ith input by and , where Problem 4. 7.7(a), Fig 7.7(b) shows that the Engel curve for x … Proof of the results can be viewed by clicking on the "Proof" button. Homoge-neous implies homothetic, but not conversely. Definition of homothetic in the Definitions.net dictionary. Show that the CES function is homothetic. homogeneous functions of the components of the original tensor. �3��弅)S�\�k� -�7��o/vv/kt ���%@��5�#!����������7Ovg?/jwv�������8�f�d���2ʱ"n�tP��*Hzy3��"8́g|�/]����_ To recall, an inverse function is a function which can reverse another function. Knowledge-based programming for everyone. 2. A widely used class of homothetic function is the CRRA class. Can some one help me derive the functional form of profit function in case of homothetic production functions? Homogeneous Differential Equations. Practice online or make a printable study sheet. Write functions in terms of "m" and "n", where m=x and n=y. Proposition: Suppose that the utility function, U RJ R: , is quasi-concave, increasing, and separable, J j U x u j x j 1 ( ) ( ). �H�W? Given familiarity with the applications of homothety, we can identify a homothetic transformation and apply the results. 1. m 2 + n 2 = 1. It also provides visualization of solution on the slope field of the DE. There are lots of graphing calculator applications to be found on the internet that you may download on your smartphone. Use Refresh button several times to 1. The constant function f(x) = 1 is homogeneous of degree 0 and the function g(x) = x is homogeneous of degree 1, but h is not homogeneous of any degree. This point is called the external homothetic center, or external center of similitude (Johnson 1929, pp. An agent can work for a principal. Free functions composition calculator - solve functions compositions step-by-step This website uses cookies to ensure you get the best experience. This GeoGebra applet solves shows how to solve a homogeneous DE. Definition of homothetic in the Definitions.net dictionary. Homothetic functions are functions whose marginal technical rate of substitution (the slope of the isoquant, a curve drawn through the set of points in say labour-capital space at which the same quantity of output is produced for varying combinations of the inputs) is homogeneous of degree zero. ]����٫�\��^�T~�ԏGW˽l���"o����rP-�9��!K/W�_V�?���n���Q��V�?�dD4n��W"�h���� ��*��^��U�8�>���듽�)e�k�Šddk��4!�o�篠�S��o�A%�C����cl0JH8�*�"�BJV�U,�yM&J����26j_�$��_��c�G�f�O�?�ﴍ䶃J"�����p�w{D Decompose the change in demand for good x into a substitution and an income effect. R and a homogenous function u: Rn! How does the MRS depend on the ratio y/x? • With homothetic preferences all indifference curves have the same shape. The reference price and quantities are and .One can think of set i as {K,L,E,M} but the methods we employ may be applied to any number of inputs. 2 "a" and "b" are "x" and "y" coordinates of homothetic center, whereas "k" is a coefficient 3. a = − 4. Join the initiative for modernizing math education. 39 The Many-Good Case • Suppose utility is a function of n goods given by utility = U(x1, x2,…, xn) Ascertain the equation is homogeneous. Can some one help me derive the functional form of profit function in case of homothetic production functions? Weisstein, Eric W. "Homogeneous Function." Why? 19-20 and 41).If radii are drawn parallel but instead in opposite directions, the extremities of the radii pass through a fixed point on the line of centers which divides that line internally in the ratio of radii (Johnson 1929, pp. It is denoted as: f(x) = y ⇔ f − 1 (y) = x. A first order Differential Equation is homogeneous when it can be in this form: In other words, when it can be like this: M(x,y) dx + N(x,y) dy = 0. Topic: Circle, Polygons, Rotation Explore thousands of free applications across science, mathematics, engineering, technology, business, art, finance, social sciences, and more. From MathWorld--A Wolfram Web Resource. c. Show that the MRS is strictly diminishing for all values of δ . • If fis a homogeneous function of degree α6=0 ,thenfis homothetic. I know that homothetic production function implies that cost function is multiplicatively separable in input prices and output, and it can be written as C(w,y)=h(y)C(w,1). The cost function exists because a continuous function on a nonempty closed bounded set achieves a minimum in the set (Debreu [6, p. 16]). %�쏢 Homothetic preference functions yield income elasticities of demand equal to 1 for all goods across all possible levels of income because all level sets (i.e., indifference curves) are radial expansions of each other when a function is homothetic. Problem 1. Collection of teaching and learning tools built by Wolfram education experts: dynamic textbook, lesson plans, widgets, interactive Demonstrations, and more. Homothetic production functions additively separable utility functions which are increasing transformations of functions this! A famous production function exhibits constant returns to scale the other hand, the quadratic the! M, p 2 remaining constant y ) = y ⇔ f − 1 k 8 depend... Have the same shape that, in additively separable utility functions which are homothetic and quasilinear for Demonstrations... 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Functions Definition Multivariate functions that are “ homogeneous ” of some degree are used! You get the best experience applied in a number of empirical studies 1 Answer to a closed form production can...
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A107317 Semiprimes of the form 2*(m^2 + m + 1) (implying that m^2 + m + 1 is a prime). 2
6, 14, 26, 62, 86, 146, 314, 422, 482, 614, 842, 926, 1202, 1514, 2246, 2966, 3446, 5102, 5942, 6614, 7082, 7814, 8846, 9662, 10226, 11402, 12014, 12326, 12962, 16022, 16382, 19802, 20606, 22262, 24422, 24866, 27614, 28562, 34586, 38366, 40046 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS Twice A002383. Also semiprimes n such that 2*n - 3 is a square. - Giovanni Teofilatto, Dec 29 2005. This coincidence was noticed by Andrew S. Plewe. Proof that this is the same sequence: If X is n^2+(n+1)^2+1, then 2X-3 is 4n^2+4n+1 = (2n+1)^2. And if 2X-3 is a square, then since it's odd, 2X-3 = (2n+1)^2 and X = n^2+(n+1)^2+1. - Don Reble, Apr 18 2007 LINKS Vincenzo Librandi, Table of n, a(n) for n = 1..300 FORMULA a(n) = 2*A002383(n). a(n) = 2*(A002384(n)^2+A002384(n)+1). EXAMPLE a(1)=6 because 1^2 + 2^2 + 1 = 6 = 2*3; a(2)=14 because 2^2 + 3^2 + 1 = 14 = 2*7; a(3)=26 because 3^2 + 4^2 + 1 = 26 = 13*2. MATHEMATICA 2(#^2 + # + 1) & /@ Select[ Range[144], PrimeQ[ #^2 + # + 1] &] (* Robert G. Wilson v, May 28 2005 *) fQ[n_] := Plus @@ Last /@ FactorInteger@n == 2 && IntegerQ@Sqrt[2n - 3]; Select[ Range@43513, fQ[ # ] &] (* Robert G. Wilson v *) PROG (PARI) for(n=2, 100000, if(bigomega(n)==2&&issquare(2*n-3), print1(n, ", "))) /* Lambert Herrgesell */ CROSSREFS Cf. A002383, A002384. Sequence in context: A131951 A168648 A093776 * A071776 A063590 A128806 Adjacent sequences: A107314 A107315 A107316 * A107318 A107319 A107320 KEYWORD easy,nonn AUTHOR Giovanni Teofilatto, May 21 2005 EXTENSIONS Edited by Robert G. Wilson v, May 28 2005 Re-edited by N. J. A. Sloane, Apr 18 2007 STATUS approved
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Level 6 - NCEA Level 1
# Formulae and Substitution
Lesson
Out in the real world all sorts of amazing relationships play out every day: Air temperatures change with ocean temperatures. Populations of species rise and fall depending on seasons, food availability and the number of predators. The surface area of a human body can even be measured fairly accurately according to your height and weight.
One of the most powerful things about mathematics is its ability to describe and measure these patterns and relationships exactly. Given a mathematical formula for the relationship between, say, the weight of a patient and how much medication they should be given, we can find one quantity by substituting a value for the other.
We have come across so many different formulae in mathematics that allow us to measure quantities such as Area, Volume, Speed etc. Let's have a look at the process of substituting values into these formulae to find a particular unknown.
#### Examples
##### Question 1
The perimeter of a triangle is defined by the formula $P=x+y+z$P=x+y+z. Find $P$P if the length of each of its three sides are $x=5$x=5 cm, $y=6$y=6 cm and $z=3$z=3 cm.
Solution:
By inserting the number values of $x$x, $y$y and $z$z we have a new equation that we can use to find the value of $P$P:
$P=5+6+3$P=5+6+3
$P=14$P=14 cm
##### Question 2
The area of a square with side $a$a is given by the formula $A=a^2$A=a2. Find $A$A if $a=6$a=6 cm.
Solution:
From the information above, we know that we are finding the area of a square where each side measures $6$6cm.
Substituting our value for $a$a into the formula:
$A=6^2$A=62
$A=36$A=36 $cm^2$cm2
Now let's have a look at a worked video solution.
##### Question 3
The simple interest generated by an investment is given by the formula $I=\frac{P\times R\times T}{100}$I=P×R×T100.
Given that $P=1000$P=1000, $R=6$R=6 and $T=7$T=7, find the interest generated.
### Outcomes
#### NA6-5
Form and solve linear equations and inequations, quadratic and simple exponential equations, and simultaneous equations with two unknowns
#### 91027
Apply algebraic procedures in solving problems | 587 | 2,167 | {"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} | 4.8125 | 5 | CC-MAIN-2022-05 | longest | en | 0.873006 |
https://www.vedantu.com/iit-jee/3d-geometry | 1,610,920,862,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703513194.17/warc/CC-MAIN-20210117205246-20210117235246-00752.warc.gz | 1,080,582,275 | 87,681 | # 3d Geometry
View Notes
## Introduction to 3 Dimensional Geometry
3 Dimensional geometry involves the mathematics of shapes in 3D space and involves 3 coordinates in the XYZ plane which are x-coordinate, y-coordinate and z-coordinate. The shapes that occupy space are called 3D shapes. 3D shapes can also be defined as the solid shapes having three dimensions length, width, and height. Three-dimensional space is a geometric three-parameter model in which (there are three axes x,y, and z-axes) all known matter exists. These three dimensions are chosen from the terms length, width, height, depth, and breadth.
In a 3d space, three parameters are required to find the exact location of a point. 3-dimensional geometry plays a major role in JEE exams as a lot of questions are included in the exam. In this article, we will study the basic concepts of geometry involving 3-dimensional coordinate geometry which will help to understand different operations on a point in 3d plane.
### Coordinate System in 3D Geometry
In 3 dimensional coordinate geometry, a coordinate system refers to the process of identifying the position or location of a point in the coordinate plane. To understand more about coordinate planes and systems, refer to the coordinate geometry lesson which covers all the basic concepts, theorems, and formulas related to coordinate or analytic geometry.
The Properties of Three-dimensional Space:
• A point is represented by an ordered triple (x, y, and z) that consists of three numbers, an x-coordinate, a y-coordinate, a z-coordinate.
• In the two-dimensional xy-plane, these coordinates indicate the signed distance along the coordinate axes,
• The x-axis, y-axis, and z-axis, respectively, originate from the origin, denoted by O, which has coordinates (0, 0, and 0).Â
### Rectangular Coordinate System in Space
The coordinate system defines the position of a vector. In the rectangular coordinate in space, we refer to the three-dimensional space. To demonstrate the position of a vector mark a point as the origin, represented by the point ‘O’. The distance of any vector is now measured from this standard point.
Let ‘O’ be any point in space called origin and X’OX, Y’OY and Z’OZ be three lines perpendicular to each other and these three lines denote the coordinate X, Y, and Z-axis. The planes XY, YZ, and ZX are called the coordinate planes in space.  Â
### Coordinates of a Point in Space
Consider a point P in space. The position of the point P is given by the coordinates (x, y, z) where x, y, z denote the perpendicular distance from YZ-plane, ZX-plane, and XY-plane respectively. If the vectors i, j, k are assumed to be the unit vectors along OX, OY, OZ respectively, then the position vector of point P is xi + yj + zk or simply (x, y, z).
If ‘O’ is the origin and P is any point with coordinates (x, y, z) from the origin then the distance vector OP by the distance formula is given by OP = √x2 + y2 + z2.
### Some Key Points to Remember
• The formula for distance between the points P(x1, y1, z1) and Q (x2, y2, z2) is √(x1 – x2)2 + (y1 – y2)2 + (z1 – z2)2
• The point dividing the line joining the points P(x1, y1, z1) and Q (x2, y2, z2) in m : n ratio isÂ
(mx2 – nx1)/(m + n),Â
(my2 – ny1)/(m + n),
(mz2 – nz1)/(m + n) where m + n ≠0.Â
• The coordinates of the centroid of a triangle having vertices A (x1, y1, z1), B (x2, y2, z2) and C (x3, y3, z3) is G ((x1 + x2 + x3) / 3, (y1 + y2 + y3) / 3, (z1 + z2 + z3)/ 3).Â
### Direction Cosines of a Line
The cosines of the angles made by a directed line segment with the coordinate axes are called the direction cosines of that line.
As shown in the figure above, if α, β, and γ are the angles made by the line segment with the coordinate axes then these angles are said as direction angles; and the cosines of these directional angles are the direction cosines of the line.Â
Also, cos α, cos β, and cos γ are called the direction cosines and are denoted by l, m, and n respectively.
l = cos α,Â
m = cos β andÂ
n = cos γ
If three numbers are in proportion with the direction cosines of a line then they are called the direction ratios. Hence, if ‘a’, ‘b’, and ‘c’ are the direction ratios, and l, m, n are the direction cosines then, we must have,
a/l = b/m = c/n.
Some Key Points of Direction Cosines
• As we know that l = cos α, m = cos β, and n = cos γ also -1< cos x< 1 ∀ x ∈ R, so l, m, and n are real numbers with values varying between -1 to 1. So, direction cosine’s ∈ [-1,1].
• The angles between the x-axis and the coordinate axes are 0°, 90°, and 90°. So the direction cosines are cos 0°, cos 90° and cos 90° i.e. 1, 0, 0 respectively.
• The direction cosine of the x, y, and z axes are (1,0,0), (0,1,0) and (0,0,1).
• The direction cosines of a line parallel to any coordinate axis are equal to the direction cosines of the corresponding axis.
• The direction cosines are associated by the relation l2 + m2 + n2 =1.
• If the given line is inverted, then the dc’s will be cos (Ï€ − α), cos (Ï€ − β), cos (Ï€ − γ) or − cos α, − cos β, − cos γ.
• Thus, a line can have two sets of direction cosines according to its direction.
• If the two lines are parallel then their direction cosines are always the same. Â
• Direction ratios and direction cosines are proportional to each other and hence for a given line, there can be infinitely many direction ratios.
### Perpendicular Distance of a Point from a Line
Let us assume AB be the straight line passing through the point A (a, b, c) and having direction cosines l, m, and n. Now, if AN is the projection of line AP on the straight line AB then we have,
AN = l(x – a) + m(y – b) + n(z – c),
and AP = √(x–a)2 + (y–b)2 + (z–c)2Â
∴ Perpendicular distance of point P
PN = √(AP2 – AN2) | 1,669 | 5,898 | {"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} | 4.65625 | 5 | CC-MAIN-2021-04 | latest | en | 0.912775 |
http://www.educator.com/mathematics/calculus-ii/murray/trapezoidal-rule-midpoint-rule-left_right-endpoint-rule.php | 1,511,052,676,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805242.68/warc/CC-MAIN-20171119004302-20171119024302-00543.warc.gz | 396,304,394 | 51,001 | Start learning today, and be successful in your academic & professional career. Start Today!
• ## Related Books
### Trapezoidal Rule, Midpoint Rule, Left/Right Endpoint Rule
Main formula:
Trapezoidal Rule:
Midpoint Rule:
where,
Left/Right Endpoint Rules:
where for the Left Endpoint Rule,
For the Right Endpoint Rule,
Hints and tips:
• These formulas may seem long and complicated. Instead of memorizing them, remember the geometry on which they’re based. If you can draw the pictures of the trapezoids and rectangles, you can probably reconstruct the formulas quickly.
• The Midpoint Rule is more accurate than the Trapezoid Rule, even though it is simpler and requires fewer function evaluations.
• By graphing the curve that you’re estimating the area under, you can often tell whether the estimates from the various formulas will be higher or lower than the true area.
• Sometimes you will not have a graph of the function or an explicit formula. Instead, you will use one of these Rules to estimate integrals based on data from charts.
### Trapezoidal Rule, Midpoint Rule, Left/Right Endpoint Rule
Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.
• Intro 0:00
• Trapezoidal Rule 0:13
• Graphical Representation
• How They Work
• Formula
• Why a Trapezoid?
• Lecture Example 1 5:10
• Midpoint Rule 8:23
• Why Midpoints?
• Formula
• Lecture Example 2 11:22
• Left/Right Endpoint Rule 13:54
• Left Endpoint
• Right Endpoint
• Lecture Example 3 15:32
### Transcription: Trapezoidal Rule, Midpoint Rule, Left/Right Endpoint Rule
We are going to do a couple more examples here.0000
I want to get some more practice with the midpoint rule.0003
We are going to estimate the integral from 1 to 2 of x × ln(x).0006
Now, here b = 2, a = 1.0011
So, δx, which is b-a/n, is 2-1/4, so that is 1/4.0020
The midpoint rule says the integral is approximately equal to 1/4 × f(the midpoint of these 4 intervals).0030
So if we take the interval from 1 to 2 and split it into 4 pieces, that is 1 and 1/4, 1 and 1/2, 1 and 3/4.0042
Now we want the midpoints of those 4 integrals.0050
The midpoint of the first one is 1 and 1/8.0056
The midpoint of the second one between 1 and 1/4 and 1 and 1/2 is 1 and 3/8.0064
The midpoint of the third one is 1 and 5/8.0072
The last one is 1 and 7/8.0076
That is 1/4.0081
Now the function here, we get from the integral.0084
That is x ln(x), so that is 1 and 1/8, we want to plug that in there.0088
1 and 1/8 is 1.125 × ln(1.125).0095
We plug each one of these values in there and I will not write them all down, but the last one here is 1 and 7/8.0102
Plug that in, and we get 1.875 × ln(1.875).0110
That is an expression now that you could plug into your calculator.0123
You just take these numbers and these expressions and plug them all into your calculator.0128
What you come up with is .634493.0135
So, that tells us that our integral is approximately equal to .634493.0143
Notice there that we never actually solve the integral as we would have using some of our earlier techniques in Calculus 2.0154
We just picked different points and plugged them into the function and got an approximation of the area.0160
So the next example I would like to do is using the right endpoint rule.0000
We are going to find the integral of sin(x) from 1 to 2.0007
It says to use n = 4, so we will split that into 4 pieces.0013
δx is b-a/n, so that is 2/1/4, which is 1/4.0019
Then, the right endpoint rule says you take the width of these rectangles, which is 1/4, that is the δx...0028
... times, now you plug in for the height, you plug in the right endpoint.0037
So we are not going to look at the left endpoint of the first interval.0040
We are going to look at the right endpoint of those four intervals.0047
So we look at f(1 and 1/4) + f(1 and 1/2) + f(1 and 3/4) +f(2).0050
f(2) represents the right endpoint of the last interval there.0065
Now the function f here is sin(x), so we will be doing sin(1.25) + sin(1.5) + sin(1.75) + sin(2).0073
Again, this is something you can plug into your calculator.0093
When I plugged it into my calculator and simplified it down,0100
I got .959941 as an answer.0103
So, that is the approximate area under sin(x) using the right endpoint rule without actually doing any actual integration.0112
That is the end of the lecture on approximate integration.0126
We covered the trapezoid rule, the midpoint rule, and the left and right endpoint rules.0128
Those were all brought to you by educator.com.0138
This is educator.com, and today we are going to discuss three methods of integration approximation, the trapezoidal rule, the midpoint rule, and the left and right end-point rules.0000
The idea here is that you are trying to approximate the area under a curve0014
The function here is f(x) and we are trying to approximate that from x=a to x=b and we are trying to find that area.0027
What you have done so far in your calculus class, is you just take the integral of f and then you plug in the endpoints.0044
The point is that there are a lot of functions that you will not be able to take the integral of directly.0050
So, what we are going to try to do is find approximation techniques that do not rely on us being able to take the integral.0055
The idea for all of these techniques is that you start out by dividing the region between a to b into n equal partitions.0062
Then we are going to look at the area on each one of those.0082
That is the first part of the formula here.0086
Each one of these partitions is δx y, and δx comes from b - a, that is the total width, divided by n because there are n of these segments.0093
Now on the trapezoidal rule, what we are going to do is label each one of these points on the x axis.0108
This is x0, x1, x2 all the way up to xn is b.0115
That is where the next part of this formula comes from.0124
x0 is a, x1 is a + δX because it is a and then you go over δx, x2 is a + 2 δX all the way up to xn is a + n δx.0127
But, of course a + n δx is a + b - a/n.0147
So, that is a + b -a which is b, so xn is the same as b0156
We have labelled these points on the x axis, and what we are trying to do is approximate the area.0162
What we do to approximate the area is we are going to use several different rules.0172
The first one is called the trapezoidal rule.0179
The trapezoidal rule means that you draw little trapezoids on each of these segments, and then you find the area of these trapezoids.0183
The area of a trapezoid is = 1/2(base1 × base2 × height).0200
That area of the trapezoid is reflected in this formula right here.0218
The 1/2 gives you that 2 right there, the height of a trapezoid, that is the height, and that is the width of one of those trapezoids, and that is δX0225
Then you have base1 + base2 is, I am going to show this in red, base1 + base2, that is for the first trapezoid.0243
For the next trapezoid, base1 + base2, and so on, up to the last trapezoid, base1 + base2.0260
What you are doing is you are plugging each of these x0, x1, x2 into f to get these heights0279
But you only have 1 of the end one and 2 of each of the middle ones.0287
That is why you get 1 here, and two of each of the middle ones, and one of the end one.0297
So, that is where the formula for the trapezoidal rule comes from.0305
Let us try it out on an example.0308
Example 1 is we are going to try to estimate the integral from 1 to 2 of sin(x) dX.0313
Here is 1, and here is 2, and we are going to try to estimate that using n = 4.0322
That means we are going to divide the region from 1 to 2 into 4 pieces.0330
Using the formula for the trapezoid rule, we have δx/2.0336
Well δx is (b - a)/n, so that is 1/4.0341
δx/2 is 1/8, so we are going to have 1/8 times f(x0) + 2f(x1) + 2f(x2) + 2f(x3) + only 1 of f(x4).0352
These x0's are the division points in between 1 and 2, so this x0 is 1, so that is sin(1) + 2, now that is 1, x1 × δx is 1/4.0385
x2 can go over another unit of δx, so that is 1 and 1/2, 1 and 3/4 and finally, sin(2).0410
We are going to take all of that, and multiply it by 1/8.0430
At this point it is simply a matter of plugging all of these values into a calculator.0433
I have a TI calculator here, and I am going to plug in 1/8 sin(1) is 0.01745 + 2sin(1.25), which is 0.0281, and so on.0437
You can plug the rest of the values into a calculator.0468
What you get at the end simplifies down to 0.951462.0476
So, we say that the integral from 1 to 2 of sin(x) dX is approximately = 0.9514620486
Next rule we are going to learn is the midpoint rule.0503
It is the same idea, where you have a function that you want to integrate from a to b, and you break the region up into partitions.0508
So, you have x0 = a, xn = b and a bunch of partitions in between, each partition is δx y.0522
Except, in each partition, instead of building trapezoids, we are going to build rectangles.0533
We are going to build rectangles on the height of the middle of the partition.0542
Here, we are going to look at the middle of the partition,0548
We see how tall the function is at the middle of the partition and we build a rectangle that is that height.0555
We do that on every rectangle.0564
The formula we get in total is δx, that is the width of the rectangles, times the height of these rectangles,0575
I have labelled the midpoints of those rectangles x1* and x2* and xn*.0596
Those represent these midpoints, so that is x1*, there is x2*, and so on.0616
Those are the midpoints so x1* is just x0 + x1/2, x2* is just x2 + x2/2,0622
And so on and those just represent the midpoints of each of these intervals.0632
We plug those midpoints in to find the heights of the rectangles and estimate the area.0640
What we are going to do for the last rule today, is we are going to use instead of the midpoints, we will use the left endpoints of each rectangle.0650
Instead of having to find the midpoints, the x1× and the x2× and so on will be the left endpoint of each interval.0663
We will use those to get the heights.0677
We will see those in the second.0680
First we will do an example with the midpoint rule using the same integral as before.0682
Sin(x), there is 1, there is 2, again we are using n = 4 so we will break it up into 4 partitions.0686
Except we are going to use a slightly different formula to solve it.0698
Again, δx = 1/4, because that is the width of each of these rectangles, but now we are going to look at the midpoints of those 4 rectangles to find the heights.0700
The midpoints are, well, this is 1 right here, that is 1 and 1/4, the midpoint there is 1 and 1/8.0715
The next midpoint is halfway between 1 and 1/4 and 1 and 1/2 and that is 1 and 3/8.0731
The formula that we get is δX(f(1 + 1/8) + f(1 + 3/8) + f(1 + 5/8) + f(1 + 7/8) and that is just 1/4.0740
Now the function here is sin(x) so we will be doing sin(1 + 1/8), that is 1.125, + sin(1 + 3/8), that is 1.375, sin(1 + 5/8) is 1.625, and sin(1 + 7/8) is 1.875.0771
Now it is just an expression that we can plug into our calculator.0800
I worked this out ahead of time, I got 0.9589440805
That is our best approximation for the integral using the midpoint rule.0827
The next formula we want to learn is the right and left endpoint rule.0835
We will talk about the left endpoint rule first.0840
It is pretty much the same as the midpoint rule.0845
Again, you are drawing these rectangles except instead of using the midpoint to find the height of the rectangle, you are using the left endpoint.0849
So that means you start out with the exact same formula,0862
Except that the star points that you choose to plug in to find the heights are just the left endpoints, x0, x1, up to xn-1.0865
You do not go up to xn.0876
For the right endpoint rule it is the same formula except you use the right endpoints.0879
The right endpoint would be x1, x2, up to xn.0885
You do not have x0 anymore because that is the first endpoint of the left formula.0890
Let me draw these in different colors here.0896
The left endpoints give you x0, x1, up to xn-1, so that is the left endpoint rule.0902
The right endpoint, I will do that one in blue, is x1, x2, up to xn-1, and xn.0912
You are using the right endpoints so that is the right endpoint rule.0928
We will do another example.0932
Again we are going to figure out the integral, or estimate the integral of sin(x) from 1 to 2.0934
We are going to use the left endpoint rule so that means the key points that we plug in for the heights are 1, 1 + 1/4, 1 + 1/2, and 1 + 3/4.0947
Again δx = 1/4 and our formula says δx × f(left endpoint).0964
So, f(1) + f(1 + 1/4) + f(1 + 1/2) + f(1 + 3/4).0975
We do not go to 2 because that was the right endpoint of the left interval.0990
The integral is approximately equal to that.1000
That is sin(1) + sin(1 + 1/4) + sin(1 + 1/2) + sin(1 + 3/4).1010
That is something that you can plug into your calculator, and when I did that ahead of time I got 0.942984.1022
That is our estimation of that integral using the left endpoint rule.1038 | 3,909 | 13,226 | {"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} | 4.40625 | 4 | CC-MAIN-2017-47 | latest | en | 0.865162 |
https://dsp.stackexchange.com/questions/72338/how-to-interpret-frequency-response-of-moving-average-system-in-discrete-time-si/72339 | 1,627,582,643,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153892.74/warc/CC-MAIN-20210729172022-20210729202022-00588.warc.gz | 241,094,060 | 39,245 | # How to interpret frequency response of moving average system in discrete time signal processing?
Excerpt from Discrete Time Signal Processing by Alan Oppenheim:
The text says this can be considered as an approximation of a Low Pass Filter. A low pass filter in continuous signal system is something that would block frequencies above a cutoff. But here in the discrete system, how is this to be understood? In a discrete time system, "frequency" is just angle/sample. Does the picture above mean "the system allows samples taken at the rate (0 to $$2\pi/5$$) / sample and blocks the rest" ?
EDIT:
Is it like this: say there is signal with a frequency range 20Hz to 50Hz. The sampling frequency in 100Hz. The lower frequency components get smaller $$\omega$$ and for higher frequency (50Hz) it gets close to $$\pi$$. If I have a low pass filter, it would block out those samples with large $$\omega$$. And this functionality now resembles a continuous time low pass filter. Is this a correct interpretation?
In discrete time, a frequency of zero is just a constant sequence, in complete analogy with continuous time. The main difference between discrete time and continuous time is that in discrete time you have a maximum frequency, whereas in continuous time you (theoretically) don't have one. That maximum frequency in discrete time is achieved for a sequence with alternating values, i.e., the sequence $$(-1)^n$$ has a maximum frequency of $$\omega=\pi$$. There can be no larger frequency.
If the sequence was obtained from sampling a continuous-time signal with a sampling frequency $$f_s$$, then $$\omega=\pi$$ translates to an actual frequency in Hertz via $$f=\omega/(2\pi f_s)$$.
Returning to the example of the discrete-time lowpass filter, from the frequency response you see that higher frequencies (up to the maximum frequency $$\omega=\pi$$) are attenuated. That means that a discrete-time sinusoid's amplitude is changed according to the magnitude of the filter's frequency response at the sinusoid's frequency.
So if the input signal equals
$$x[n]=A\sin(n\omega_0)\tag{1}$$
the amplitude of the corresponding output signal is given by $$A|H(e^{j\omega_0})|$$. If $$\omega_0$$ happens to be a multiple of $$2\pi/5$$ then the input is completely suppressed because these frequencies are the zeros of the filter's frequency response.
• Thank you. I have made an edit to the question. Could you please have a look? – Satheesh Paul Dec 31 '20 at 15:15
• @SatheeshPaul: Your interpretation sounds ok, except for the fact that you don't block out "samples with large $\omega$", but signal components with large frequencies. The notion of samples with a large frequency doesn't make sense. Just like in the continuous-time case, a lowpass filter attenuates signal components with higher frequencies. – Matt L. Dec 31 '20 at 15:54
You are right in your concerns here; as is obvious from the magnitude of the frequency response, a moving-average system does not block any portion of the spectrum except at those $$M-1$$ null frequencies, where $$M$$ is the length of the moving-average impulse response.
However, it's also quite evident from the shape of the spectrum that high frequencies are attenuated while low frequencies are passed better through. From this it's justfied to call the system as a low pass filter, which attenuates high frequencies while letting low frequencies.
Nevertheless, it's not a best kind of lowpass filter. The main advantage this filter is that it requires a very simple summation to implement. It doesn't need multiplications, whereas other better filter characteristics require multiplications (as the weights) to implement.
And also historically, for some similar reasons, the MA filter finds extensive usage in reducing noise variations on time-series data. | 837 | 3,817 | {"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": 16, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2021-31 | latest | en | 0.929176 |
http://wikieducator.org/Thermodynamics/FirstLaw-simple | 1,542,748,937,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039746800.89/warc/CC-MAIN-20181120211528-20181120233528-00490.warc.gz | 361,331,868 | 9,858 | # Work, Heat, Energy, and the First Law (simplified)
Note: This is a simplified version, for a version including derivatives go HERE.
## Work
Work, W
Force acting through a distance
In most applications of thermodynamics we are mainly interested in mechanical work due to pressure of a fluid. Then the work is
$W=-P\Delta V=-P(V_2-V_1)$
where P is the pressure and V is the volume. The symbol Δ means change - in this case change in volume. The reason for the minus sign is explained below.
## Energy
Energy
Capacity to do work
Internal Energy, U
The total energy of the system[1]
## Heat
Heat
Energy transferred due to a temperature difference
No heat transfer between a system and its surroundings
Exothermic process
A process which releases heat
Endothermic process
Heat is denoted by the symbol, Q
## Signs
Heat and work are considered positive if they are transferred from surroundings to the system.
## Observations
The laws of thermodynamics are based on observations of the natural world. The first law is based on two observations concerning energy:
1. Energy can be transferred between a system and its surroundings by only two ways: work and heat
2. The total energy of a system and its surroundings is always constant (The conservation of energy)
## First Law
These two observations can be combined into the First Law of Thermodynamics:
The internal energy of a system is constant unless changed by doing work or by heating
## Enthalpy
Often in thermodynamics we use a quantity known as the Enthalpy, H. It is defined as: $H=U+PV$
Enthalpy is probably the most common thermodynamic quantity used, especially in chemistry and engineering. Especially common is the use of the "enthalpy of reaction". Often this is called just the "heat of reaction"[2]. There are likewise heats of solution, mixing, vaporization, etc.
## Heat Capacity
Heat Capacity
The change in heat per unit temperature
We usually use two specific types of heat capacity:
• Heat Capacity at constant pressure, CP.
• Heat Capacity at constant volume, CV.
## Notes
1. Note that some references say the internal energy is the energy due to the internal vibrations, etc. In other words that other than kinetic or potential energy. However, the definition used here is equivalent and is easier to understand.
2. it is actually the heat change at constant pressure. See here for details | 519 | 2,387 | {"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} | 3.84375 | 4 | CC-MAIN-2018-47 | latest | en | 0.942459 |
https://sozvyezdami.wordpress.com/2016/06/12/fermats-sum-of-squares-pythagoras-pegboards-fermats-primes-regular-polygons-part-1/ | 1,542,550,671,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039744381.73/warc/CC-MAIN-20181118135147-20181118161147-00457.warc.gz | 718,033,076 | 16,024 | Fermat’s sum of squares, Pythagoras, pegboards, Fermat’s primes, regular polygons, part 1
Daydreaming again. When do polygons remind me of grids? They seem to me to show interesting holes in the way we associate shapes with numbers.
For a grid, all of the hypoteneuses of the triangles fit a pattern:
Say that hypoteneuse length is n. Then the prime factors of n^2 that are of form 4k+3 must have only even powers in the factorization.
That’s because of Fermat’s pattern for the sum of two squares and Pythagoras’ theorem for the hypoteneuse of a right triangle. In fact this same pattern is linked to the patterns of Pythagorean Triples.
For a regular polygon with n sides (pentagon=5, heptagon=7, etc), the constructible ones, with algebraic coordinates, the number of sides fit a pattern.
The prime factors of n are of form $2^{2^k}+1$ and $2^{k}$
There is a nice explanation of the sum of two squares pattern here:
Su, Francis E., et al. “Sums of Two Squares.” Math Fun Facts. <http://www.math.hmc.edu/funfacts>
There is a nice explanation of Gauss and Wantzel applying Fermat primes to regular polygons here:
Constructing Regular Polygons, Math Teacher Link University of Illinois at Urbana-Champaign <http://mtl.math.uiuc.edu/node/29>
But is there not some relation between these two things? In both cases we have geometric ideas of objects, but they are limited by the patterns in prime factorizations of an integer.
For the grid, there are some Quadrances (distance squared) that we cannot represent on it. For regular polygons, there are some n-gons that we cannot represent with the Descartes’ coordinates being ‘constructible numbers’, in other words, square roots combined with addition, subtraction, multiplication, and addition. Or, in other language, they are not ‘constructible’ with a straightedge and compass.
But let me ask this… how does one build an angle x if one cannot build the n-gon made of wheel-like triangle-slices out of that angle x? Does this not limit which rational angles are representable in such a system? Because every n-gon is associated with a rational angle, as the triangles fanning out from the central point, fan-like.
To my amateur’s mind, these limits are fascinating. They represent a sort of ‘unevenness’ or partiality of the systems we have constructed in our minds, such that what is seemingly a smooth even pattern, a grid, in fact, in another perspective, is full of holes and things it does not cover. The constructible n-gons, as well, display an uneven pattern where only certain rational angles are representable.
Even if one admits the Real Numbers, and posits that all things are possible, such as each point on a grid, or each angle on a circle, including irrationals and transcendentals and all of the other fascinating animals in the mathematical universe, it doesn’t wipe away the fact that these animals are not uniformly distributed. There are little oceans and islands, little clusters lying about in different places, that I never expected when I was first introduced to these ideas many years ago. | 701 | 3,079 | {"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": 2, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2018-47 | latest | en | 0.900014 |
https://www.physicsforums.com/threads/what-is-magnitude.239733/ | 1,696,414,342,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511364.23/warc/CC-MAIN-20231004084230-20231004114230-00896.warc.gz | 1,031,906,456 | 16,968 | # What is magnitude?
• nehach
In summary, magnitude just means how big something is. Displacement is a vector quantity, which has direction and magnitude. Magnitude is always positive, or zero. To calculate it, you can use the formula posted by CompuChip.f
#### nehach
Actually, i have just been to class 11th so i m not clear about magnitude so please help me in knowing magnitude and magnitude of displacement...
what is magnitude in physics?
Last edited by a moderator:
Basically, magnitude just means how big it is.
If you have a vector (in, say, three dimensions) $$\vec x = (x_1, x_2, x_3)$$ then the magnitude of that vector is given by (Pythagoras): $$|\vec x| = \sqrt{x_1^2 + x_2^2 + x_3^2}$$.
still not very clear how it is related to distance and displacement...then wt is magnitude of displacement.
Welcome to PF!
Actually i have just been to class 11th so i m not clear about magnitude so please help me in knowing magnitude and magnitude of displacement...
Hi nehach ! Welcome to PF!
"magnitude" is a long word which means something very simple.
As CompuChip says, it just means how big something is.
For example, if something is displaced along the x-axis by 3, then the magnitude of its displacement is also 3.
But if something is displaced along the x-axis by -3, then the magnitude of its displacement is still 3.
(magnitude is always positive, or zero.)
still not very clear how it is related to distance and displacement...then wt is magnitude of displacement.
Displacement is a vector quantity defined as the change in position - from an initial point i to a final point f. Its magnitude is the length of the straight line between i and f and its direction is from i to f. The actual path taken is irrelevant.
Distance is a scaler quantity defined as the path length, i.e., it does depend on the actual path taken. Distance only equals the magnitude of the displacement for straight line paths.
Hope this hepls
if a person start from i and reach to f
distance between i and f is 5 km
i ----------5KM------------f
now kindly tell me what is magnitude and what is displacement
kindly tell me the formula how to calculate both of them
because i am confisued in this
Hi nehach!
Displacement: 5 km East.
Magnitude: 5 km.
thanks tiny-tim now its clear to me
hey could u please give me ur gmail id so that instead of wasting our time
we can discuss any problem directly
if u think ok
kindly give me ur gmail id
and kindly tell me the formula of calulating both displacement and magnitude
Displacement is a vector quantity. It has direction and magnitude. The absolute quantity of vector is its magnitude.
You can use the formula posted by CompuChip to calculate it
Sorry if I said something wrong
Simpler still, magnitude is the size or length (always positive or zero) of a vector, independent of the vector direction. "Displacement" in Newtonian physics, is a vector, and has two elements, a signed length (positive, negative, or zero) and a direction.
thanks jeff i got it | 700 | 3,017 | {"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} | 3.8125 | 4 | CC-MAIN-2023-40 | latest | en | 0.908045 |
http://www.enotes.com/homework-help/factor-x-2-12x-35-b-x-2-13x-14-431217 | 1,477,618,098,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988721415.7/warc/CC-MAIN-20161020183841-00024-ip-10-171-6-4.ec2.internal.warc.gz | 431,019,176 | 9,445 | # Factor x^2-13x-14
justaguide | College Teacher | (Level 2) Distinguished Educator
Posted on
The factors of x^2-13x-14 have to be determined.
x^2-13x-14
= x^2 - 14x + x - 14
= x(x - 14) + 1(x - 14)
= (x + 1)(x - 14)
The factored form of x^2-13x-14 = (x + 1)(x - 14) | 123 | 274 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2016-44 | latest | en | 0.774881 |
https://math.stackexchange.com/questions/3878962/some-basic-question-about-obtain-minimal-polynomial-of-algebraic-element | 1,716,521,265,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058677.90/warc/CC-MAIN-20240524025815-20240524055815-00360.warc.gz | 319,862,566 | 35,515 | # Some basic question about obtain minimal polynomial of algebraic element.
For example, let $$c=p^{1/2}+q^{1/3}$$ for some prime integers $$p$$ is not $$q$$. To find minimal polynomial of $$c$$ over $$\mathbb{Q}(p^{1/2})$$, use $$(c-p^{1/2})^3=q$$, then we can obtain $$f=x^3-3p^{1/2}x^2+3px-p^{3/2}-q$$. It is minimal polynomial of $$c$$ over $$\mathbb{Q}(p^{1/2})$$, but I don't know how to sure that this polynomial is minimal polynomial.
If I prove the polynomials which degree 0, 1, 2 cannot satisfy $$g(c)=0$$, then can I say that $$f$$ is minimal polynomial? I guess I also have to show that $$f$$ is irreducible.. I already know there are unique monic irreducible polynomial, so if I prove $$f$$ is irreducible, then we can say that minimal polynomial is $$f$$ which have degree 3?
Either of the following statements would show that $$f$$ is the minimal polynomial of $$c$$ in $$\mathbb Q(p^{1/2})$$:
1. There is no polynomial $$g\in\mathbb Q(p^{1/2})[x]$$ with $$\deg g<\deg f$$ and $$g(c)=0$$.
2. The polynomial $$f$$ is irreducible.
It is clear that the first condition implies the second since, if $$f=gh$$ with $$\deg g,\deg h>0$$, then either $$g(c)=0$$ or $$h(c)=0$$, but both have strictly smaller degree. The second also implies the first because we can take the greatest common divisor: if $$g$$ is a polynomial for which $$g(c)=0$$, then $$\gcd(f,g)$$ is also such a polynomial (this essentially follows from the fact that the minimal polynomial exists). This polynomial has degree strictly smaller than that of $$f$$, and must also divide $$f$$, so its existence would contradict the irreducibility. In general, the minimal polynomial of $$c$$ must divide such an $$f$$ -- this can actually be taken to be a definition of the term "minimal polynomial."
Condition 1 is useful if you want to show something has minimal polynomial of high degree, but can't be sure what its minimal polynomial should actually be. Condition 2 requires a lot less to be verified; in particular, $$f$$ is of degree $$3$$, so, if it is not irreducible, it has to have a linear factor. Proving that your $$f$$ has no linear factor is much easier than dealing with an arbitrary polynomial of degree $$\leq 2$$, so it's a much better path forward. | 621 | 2,244 | {"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": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 36, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2024-22 | latest | en | 0.925853 |
https://eguruchela.com/math/formulas/algebra-factor-and-product.php | 1,721,687,166,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517927.60/warc/CC-MAIN-20240722220957-20240723010957-00638.warc.gz | 193,127,342 | 3,312 | # List of formulas related to Product and Factor
## Factoring Formulas
$a^2 - b^2 = (a-b)(a+b)$ $a^3 - b^3 = (a-b)\left(a^2 + ab + b^2\right)$ $a^3 + b^3 = (a+b)\left(a^2 - ab + b^2\right)$ $a^4 - b^4 = (a-b)(a+b)\left(a^2 + b^2\right)$ $a^5 - b^5 = (a-b)\left(a^4 + a^3b + a^2b^2 + ab^3 + b^4\right)$
### Product Formulas
$(a + b)^2 = a^2 + 2ab + b^2$ $(a - b)^2 = a^2 - 2ab + b^2$ $(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$ $(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$ $(a + b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4$ $(a - b)^4 = a^4 - 4a^3b + 6a^2b^2 - 4ab^3 + b^4$ $(a + b + c)^2 = a^2 + b^2 + c^2 +2ab + 2ac + 2bc$ $(a + b + c + ...)^2 = a^2 + b^2 + c^2 + ... + 2(ab + ac + bc + ... )$ | 393 | 688 | {"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} | 4.15625 | 4 | CC-MAIN-2024-30 | latest | en | 0.140768 |
https://oeis.org/A308095 | 1,722,648,367,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640353668.0/warc/CC-MAIN-20240802234508-20240803024508-00277.warc.gz | 356,167,910 | 4,442 | The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A308095 a(n) is the sum of sigma (i.e., A000203) over the totatives of n. 1
1, 1, 4, 5, 15, 7, 33, 19, 40, 26, 87, 27, 127, 50, 84, 82, 220, 59, 277, 90, 187, 140, 407, 103, 401, 193, 352, 207, 660, 127, 762, 309, 485, 339, 646, 244, 1098, 423, 677, 390, 1342, 268, 1480, 525, 758, 639, 1758, 416, 1666, 581, 1191, 770, 2250, 527, 1742, 821, 1527, 1016, 2786, 502, 3014 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,3 COMMENTS a(n) <= A024916(n-1) for n >= 2, with equality if and only if n is prime. LINKS Robert Israel, Table of n, a(n) for n = 1..10000 Robert Israel, Plot of a(n)/n^2 for 1 <= n <= 20000 FORMULA a(n) = Sum_{1<=k<=n; gcd(k,n)=1} A000203(k). EXAMPLE a(3) = sigma(1) + sigma(2) = 4. MAPLE f:= proc(n) local k; add(numtheory:-sigma(k), k=select(t -> igcd(t, n)=1, [\$1..n])) end proc; map(f, [\$1..100]); PROG (PARI) a(n) = sum(k=1, n, if (gcd(n, k)==1, sigma(k))); \\ Michel Marcus, May 13 2019 CROSSREFS Cf. A000203, A024916, A307997. Sequence in context: A222501 A225204 A363817 * A321348 A257311 A369790 Adjacent sequences: A308092 A308093 A308094 * A308096 A308097 A308098 KEYWORD nonn,look AUTHOR J. M. Bergot and Robert Israel, May 12 2019 STATUS approved
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Last modified August 2 21:25 EDT 2024. Contains 374875 sequences. (Running on oeis4.) | 653 | 1,683 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2024-33 | latest | en | 0.593308 |
https://www.physicsforums.com/threads/bearnoullis-equation.60939/ | 1,544,698,712,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376824675.15/warc/CC-MAIN-20181213101934-20181213123434-00505.warc.gz | 994,833,967 | 15,686 | Bearnoulli's equation
1. Jan 24, 2005
vinter
Consider a U- tube filled with water. The water in one arm of it is pushed down and left, result- the full water column will start oscillating.
Now, consider an instant when the water levels in the two arms are not same and water in one arm is going up with a velocity v, water in the other arm is going down with the same velocity. Apply Bernoulli's equation to the points on the two water surfaces open to air, one on each, i.e, one point on the surface of water column in the left arm and one point on the surface of water column in the right arm.
You will have
pressure + half * (rho) * (v^2) + (rho ) *g * h = a similar expression for the second point.
This creates all the problem. The pressures at the two points are same, so that term will cancel out. the velocities are same, so the half rho v squared term will go. the remaining term is the rho*g*h term which cannot go since the heights are different in the two columns, that means the above inequality cannot hold in such a situation. But that is the Bernoulli's thrm!
What's wrong here?
Is Bernoulli's equation wrong, or we have not yet learnt to apply it properly?
2. Jan 24, 2005
vincentchan
you should not apply your whatever equation in this problem... check your text book in what situation your equation hold and tell me why can't you use it in this problem
3. Jan 24, 2005
arildno
You can certainly use the NON-STATIONARY version of Bernoulli's equation here.
But it is rather tricky..
4. Jan 24, 2005
vinter
Why is the stationary, non-viscous, non-rotational version not applicable here?
(I hope you all understood that equation I have written. Or do I need to use latex? I have not learnt it's tags and stuff yet, so for now, I will just tell you that rho was the famous greek letter that represents density of water here.)
-vinter
5. Jan 24, 2005
Q_Goest
Interesting question... Bernoulli's equation is for steady state, whereas the example you gave has an accelerating fluid.
6. Jan 24, 2005
arildno
Hi, vinter:
Let us write the Euler equations as:
$$\frac{\partial\vec{v}}{\partial{t}}+\nabla(\frac{1}{2}\vec{v}^{2})+\vec{c}\times\vec{v}=-\frac{1}{\rho}\nabla{p}-g\vec{k}$$
I've used the vector identities:
$$(\vec{v}\cdot\nabla)\vec{v}=\nabla(\frac{1}{2}\vec{v}^{2})+\vec{c}\times\vec{v},\vec{c}\equiv\nabla\times\vec{v}$$
Now, let us multiply our equation with the streamline tangent $$d\vec{s}$$.
Since the streamline tangent is parallell to the velocity field, we gain, by integrating between two points:
$$\oint_{s_{0}}^{s_{1}}\frac{\partial\vec{v}}{\partial{t}}\cdot{d}\vec{s}+\oint_{s_{0}}^{s_{1}}\nabla(\frac{1}{2}\vec{v}^{2}+\frac{p}{\rho}+gz)\cdot{d}\vec{s}=0$$
(s is a scalar variable running along the streamline, (x,y,z) are therefore functions of s)
The second integral is easy to evaluate, since we have a gradient field:
$$(\frac{1}{2}\vec{v}^{2}+\frac{p}{\rho}+gz)\mid_{s=s_{1}}-(\frac{1}{2}\vec{v}^{2}+\frac{p}{\rho}+gz)\mid_{s=s_{0}}=-\oint_{s_{0}}^{s_{1}}\frac{\partial\vec{v}}{\partial{t}}\cdot{d}\vec{s}$$
Note that in the stationary case, the right-hand side is zero; i.e, you get the usual Bernoulli equation.
However, in your case the right-hand-side is NOT zero.
The non-stationary version resolves your dilemma:
The height difference equals the curve integral on your right-hand side.
Last edited: Jan 24, 2005
7. Jan 25, 2005
vinter
Thanks Arildno.
I am not very much familiar with the del operators and the advanced fluid dynamics. But considering what you just said and what I know, I still have a doubt..
Let's consider a simple Bernoulli equation application example - water coming out of a container through a hole in it. This example is there in almost all the books on elementary fluid dynamics(eg = Sears and Zemansky "University Physics" by Addison-Wesly) They readily use the stationary version of Bernoulli's equation. And they use the continuity equation (amount of fluid crossing through any cross section is the same) to get different velocities at different points depending on the area of cross section. But since the velocities ARE different at different points, the motion has to be accelerated. And if it is, how can the integral on the right hand side of your equation be zero? In that case, are we allowed to use the stationary Bernoulli's equation? If yes and if it doesn't give any absurd results there, it shouldn't do that here too.
8. Jan 25, 2005
arildno
An emptying tank should, strictly speaking, always be regarded as a case of non-stationary flow.
However, if we think of a container in which the top area of the container (i.e, where the fluid surface is) is a lot bigger than the drain hole (a rather typical situation), two features follows:
1) The fluid layer by the top area is practically "at rest" (at least over quite some time)
2) The height difference between the surface area and the drain hole remains practically constant.
This justifies the stationary treatment of the situation, in which we say that at a fixed spatial position, the velocity profile does not change with time.
I don't know which specific examples Spears&Zemansky uses; so I can't say whether they have told which approximations they've used.
Note:
So, it is the area ratio which makes the problems different:
In the emptying tank case, the surface area is a lot bigger than the exit area;
in the U-tube case, the "inlet"/"outlet" areas are equal.
Last edited: Jan 25, 2005
9. Jan 26, 2005
vinter
OK thanks for the help | 1,494 | 5,518 | {"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} | 3.53125 | 4 | CC-MAIN-2018-51 | latest | en | 0.939948 |
http://www.physicsforums.com/showthread.php?t=308870 | 1,369,150,110,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368700132256/warc/CC-MAIN-20130516102852-00058-ip-10-60-113-184.ec2.internal.warc.gz | 645,998,593 | 7,759 | ## Control Engineering
1. The problem statement, all variables and given/known data
If the steady-state error of the proportional-plus-derivative-plus acceleration system is to be less than 10% determine a suitable value for Kp.
2. Relevant equations
The input of the system is: v
The output of the system is: y
The transfer function of the system including the controller is:
(23(Kp+4s+s^2)) / (s^3+19s^2+48s+96+23Kp)
3. The attempt at a solution
With steady state: s = 0
Therefore: y steady state = v x (23Kp / (96 + 23Kp))
The error of the system is: v-y
Therefore: error = v - (23Kp / (96+23Kp))v
error = 96v / (96 + 23Kp)
How can I correctly substitute the error of 10% now in this equation.
I think (v-y)/y = 0.1 [error]
But this will leave 3 variables in the equation...
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Hi, The error is a ratio. So when you get to error = ( 96 / ( 96 + 23 Kp ) ) v, The error in percentage is: %error = ( 96 / ( 96 + 23 Kp ) ) < 10% There you can then solve for the minimum Kp.
Cheers. So that means that: error = v - y error [%] = ((v-y)/v)*100% So: error = ( 96 / ( 96 + 23 Kp ) ) v error [%] = ( 96 / ( 96 + 23 Kp ) ) * 100% ( 96 / ( 96 + 23 Kp ) ) * 100% < 10% ( 96 / ( 96 + 23 Kp ) ) < 0.1 Then I get: Kp > 37.565 Is that correct? Thanks in advance!
## Control Engineering
Hi, you must have done something wrong in the algebra, because Kp should be positive.
( 96 / ( 96 + 23 Kp ) ) < 0.1
Conceptually, when the denominator is 960, you will have exactly 0.1 error.
As Kp increases, the error decreases further. When you solve it using algebra
you can verify that Kp to give you a denominator that is greater than 960. | 578 | 1,851 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2013-20 | latest | en | 0.794478 |
http://www.docstoc.com/docs/2135914/free-fraction-worksheets | 1,438,083,161,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042981856.5/warc/CC-MAIN-20150728002301-00292-ip-10-236-191-2.ec2.internal.warc.gz | 399,857,527 | 40,501 | # free fraction worksheets by tdelight
VIEWS: 10,616 PAGES: 8
• pg 1
``` Title: Fraction Action
! Problem Solving Students will demonstrate the ability to rename and simplify
fractions
to solve problems.
! Communication The students will read, write, and discuss fractions.
! Reasoning Students will deduce reasonable methods of renaming fractions and
mixed numbers.
! Connections Students will apply fractions to their interest in basketball.
! Number Sense Students will see simplest form as an equivalent fraction.
& Relationships
Brief Overview:
Students will rename mixed numbers and improper fractions using basketball as a
theme. They will also put fractions and mixed numbers in simplest form. They will
use counters, fraction circles, and small basketball hoops to accomplish these tasks.
This lesson is intended for Grade 5.
Duration/Length:
This lesson take three 40 minute class periods over 3 days.
Prerequisite Knowledge:
The ability to identify fractions, find equivalent fractions, identify mixed numbers,
identify improper fractions, multiples, and common denominators, and familiarity with
coordinate geometry.
Objectives:
! Students will be able to change an improper fraction to a mixed number.
! Students will be able to change a mixed number to an improper fraction.
! Students will be able to put fractions in simplest form and show the method used.
Materials/Resources/Printed Materials:
! Fraction Circles
! Fraction Bars
! Construction paper
! Multiplication chart
! Counters
! Scissors
! Paper plates
! Small basketball hoop and nerf ball
Development/Procedures:
! Give students one paper plate and construction paper. Have them trace the plate
five times to make five basketballs. Have students cut them out. Fold each
basketball into eighths and cut one apart. Review identification of mixed
numbers and fractions to eighths.
! Begin to change mixed numbers to fractions using basketballs. Students work in
groups of three or four. Instructor shows a mixed number on the board. The
students must work together to change to an improper fraction using the
! After discussion of pictures and written work students will deduce a mathematical
method for renaming mixed numbers.
! Students will practice several problems using basketballs. Students complete the
top of worksheet # 1 at this time.
! After practice page, students work in groups to come up with a way to do the
reverse (solve improper to mixed).
! After a discussion of methods the instructor reviews the proper method for
completing a worksheet (finish Worksheet # 1).
! Review renaming of mixed numbers and improper fractions. Instructor divides
students into four teams. Each team is given a flash card with an improper
fraction or a mixed number to rename. If it is correct , that team shoots the
! Pass out bags of basketball counters for groups of two students each.
! Instructor will name a fraction and show it on the overhead. The procedure for
finding the simplest form with counters will be shown.
! Do “Think-Pair-Share” with Worksheet # 2.
! Each student is given a fraction card and must find its simplest form by walking
around the room looking at other cards. Students must work together and share
how they found their match using the overhead and counters.
! Complete “Free-Throw “ problem solving evaluation.
Evaluation:
Teacher will observe students making mixed numbers and improper fractions in a group
situation.
Students will play “Fraction Action Basketball.”
Students will play “Fraction Action Match Up.”
Students will complete “Free -Throw” problem solving (Worksheet # 3) and a given journal
topic (Worksheet # 4) explaining the procedures for renaming mixed numbers and fractions.
Problem solving is to be used throughout the unit.
Have students make a list of their activities during a given day. Estimate the number of
hours spent doing each activity. Represent each activity as a fractional part of a 24-hour day.
Label sections of the circle graph for each activity. Color the correct number of parts for
each activity. Put all the fractions in simplest form. Find which activity on which they spent
most of their time. Share graphs with the class.
Tangram Coordinates: students must use coordinates to plot points forming tangram pieces.
Students will find area in fraction form, and make sure their fractions are represented in
simplest form. ( See attached Tangram Coordinates Worksheet ).
Authors:
Sandy Landon Stephanie Ellsworth
Occohannock Elementary Kiptopeke Elementary
Northampton Northampton
Name: Date: #1
Fraction Action Worksheet # 1
Renaming Mixed Numbers and Improper Fractions
Rename each mixed number as an improper fraction.
1. 6 1/3 =____ 2. 11 3/9 = ____
3. 9 4/8 =____ 4. 6 5/9 = ____
5. 10 3/7 =____ 6. 4 7/10 = ____
7. 8 2/5 =____ 8. 2 9/12 = ____
9. 3 3/6 =____ 10. 5 1/8 =____
--------------------------------------------------------------------------------------------------
Rename each improper fraction as a mixed number.
11. 34/6=____ 12. 13/7=____
13. 97/8=____ 14. 43/5=____
15. 62/10=____ 16. 22/11=____
17. 55/4=____ 18. 31/3=____
19. 73/8=____ 20. 25/6=____
Name: Date: #2
Fraction Action Worksheet # 2
Simplest Form
Put each fraction in simplest form. Draw Pictures to help find the answer.
Example: 2/4= **/**** = one group of two in the numerator,
Two groups of two in the denominator
2/4=1/2
1. 6/14 =____ 2. 4/6 =____
3. 12/16 =____ 4. 15/25 =___
5. 3/15 =____ 6. 7/9 =____
7. 6/9 =____ 8. 10/18 =____
9. 16/36 =____ 10. 9/12 =____
_____________________________________________________________
Write yes if the fraction is in simplest form. Write no if it is not in simplest form.
1. 8/12 ______ 2. 3/5 ______
3. 7/11 ______ 4. 3/12 ______
5. 8/16 ______ 6. 12/30 ______
7. ½ ______ 8. 5/14 _____
9. 28/42 ______ 10. 9/16 ______
Name: Date: #3
Fraction Action Worksheet # 3
Action Fractions
Free Throw Problem Solving
The Lakers are practicing free throws. Each player takes 10 shots. Player #1 makes 3
of his throws. Player #2 makes 6 of his throws. Player #3 makes 5 of his throws. Player
#4 makes 2 of his throws, and player 5 makes 4 of his throws. Find the fraction for the
number of throws each player made. Then put each fraction into simplest form. Draw a
chart to show your data and the method you used to solve this problem.
Draw Chart Here:
Write your own problem using fractions. Include putting fractions in simplest form and
renaming. Give your paper to a friend to solve.
______________________________________________________________
______________________________________________________________
______________________________________________________________
______________________________________________________________
______________________________________________________________
Name: Date: ___________________ #4
Fraction Action Worksheet # 4
Journal Entry:
Explain procedures for renaming mixed numbers and fractions. Draw pictures and show
______________________________________________________________
______________________________________________________________
______________________________________________________________
______________________________________________________________
_____________________________________________________________
______________________________________________________
______________________________________________________
______________________________________________________
______________________________________________________
______________________________________________________
______________________________________________________
______________________________________________________
______________________________________________________
______________________________________________________
______________________________________________________
______________________________________________________
______________________________________________________
______________________________________________________
______________________________________________________
______________________________________________________
______________________________________________________
______________________________________________________
______________________________________________________
______________________________________________________
______________________________________________________
______________________________________________________
Tangram Coordinates
Form the Tangram by connecting the coordinates.
Find the area of each Tangram piece in fraction form. Simplify each fraction.
1. (8,0) and (0,8) Lift pencil.
2. (4,0) and (0,4) Lift pencil.
3. (2,2) and (8,8) Lift pencil.
4. (2,2) and (2,6) Lift pencil.
5. (4,0) and (6,2) Lift pencil.
```
To top | 1,931 | 9,834 | {"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-2015-32 | longest | en | 0.887861 |
https://www.infoapper.com/unit-converter/volume/us-gal-to-cm3/ | 1,600,554,118,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400192887.19/warc/CC-MAIN-20200919204805-20200919234805-00302.warc.gz | 912,311,778 | 8,707 | Want to Enable Key Features?
# Convert US Gallon to Centimeter Cube (us gal to cm3)
In next fields, kindly type your value in the text box under title [ From: ] to convert from US gallon to centimeter cube (us gal to cm3). As you type your value, the answer will be automatically calculated and displayed in the text box under title [ To: ].
From:
To:
Definitions:
US Gallon (abbreviation: US gal): is a unit of measurement for liquid capacity in the US customary units. Three significantly different sizes are in current use: the imperial gallon defined as 4.54609 litres, which is used in the United Kingdom, Canada, and some Caribbean nations; the US gallon defined as 231 cubic inches (3.785 L), which is used in the US and some Latin American and Caribbean countries; and the least-used US dry gallon defined as 1/8 US bushel (4.405 L).
Centimeter Cube (abbreviations: cm3, or cm cube): is the SI derived unit of volume. It is the volume of a cube with edges one centimeter in length.
## How to Convert US Gallons to Centimeters Cube
### Example: How many centimeters cube are equivalent to 84.13 US gallons?
As;
1 US gallons = 3785 centimeters cube
84.13 US gallons = Y centimeters cube
Assuming Y is the answer, and by criss-cross principle;
Y equals 84.13 times 3785 over 1
(i.e.) Y = 84.13 * 3785 / 1 = 318432.05 centimeters cube
Answer is: 318432.05 centimeters cube are equivalent to 84.13 US gallons.
### Practice Question: Convert the following units into cm3:
N.B.: After working out the answer to each of the next questions, click adjacent button to see the correct answer.
( i ) 10.72 us gal
( ii ) 72.4 us gal
( iii ) 21.98 us gal
References
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Results: | 558 | 2,257 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2020-40 | longest | en | 0.881764 |
https://stackoverflow.com/questions/1817143/upper-bound-lower-bound | 1,723,401,930,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641008125.69/warc/CC-MAIN-20240811172916-20240811202916-00044.warc.gz | 426,399,639 | 43,503 | # upper bound, lower bound
What does it mean to prove an upper bound or lower bound to an algorithm?
Proving an upper bound means you have proven that the algorithm will use no more than some limit on a resource.
Proving a lower bound means you have proven that the algorithm will use no less than some limit on a resource.
"Resource" in this context could be time, memory, bandwidth, or something else.
• By the way, the proof can be done by induction, contradiction or pigeonhole principle i guess, is that right? Commented Nov 30, 2009 at 0:01
• Could you please also explain about how upper/lower bound is related with BIG O, OMEGA and THETA notation? Commented Apr 9, 2015 at 5:35
Upper and lower bounds have to do with the minimum and maximum "complexity" of an algorithm (I use that word advisedly since it has a very specific meaning in complexity analysis).
Take, for example, our old friend, the bubble sort. In an ideal case where all the data are already sorted, the time taken is f(n), a function dependent on `n`, the number of items in the list. That's because you only have to make one pass of the data set (with zero swaps) to ensure your list is sorted.
In a particularly bad case where the data are sorted in the opposite to the order you want, the time taken becomes f(n2). This is because each pass moves one element to the right position and you need `n` passes to do all elements.
In that case, the upper and lower bounds are different, even though the big-O complexity remains the same.
As an aside, the bubble sort is much maligned (usually for good reasons) but it can make sense in certain circumstances. I actually use it in an application where the bulk of the data are already sorted and only one or two items tend to be added at a time to the end of the list. For adding one item, and with a reverse-directional bubble sort, you can guarantee the new list will be sorted in one pass. That illustrates the lower bound concept.
In fact, you could make an optimization of the bubble sort that sets the lower bound to f(1), simply by providing an extra datum which indicates whether the list is sorted. You would set this after sorting and clear it when adding an item to the end.
• Wouldn't insertion sort make more sense in your case (only a few items added at the end of the list)? Commented Jun 2, 2011 at 4:20
Whatever the bound (upper or lower), we are always talking about the worst-case input that we can consider. For example, in sorting, we assume that the worst-case is an unsorted input list.
My understanding is that problems have a lower bound. For example, we say that the lower bound of comparison-based sorting is \Omega(n log n); we are making no assumptions about what particular comparison-based sorting algorithm we use. Whatever the algorithm (merge sort, quick sort, etc), we cannot do better than this bound of \Omega(n log n). Lower bounds tell us, intuitively, how hard a particular problem is.
When we talk about a specific algorithm, then we talk about upper bounds. For example, we say that the upper bound of bubble sort is O(n^2) and the upper bound of merge sort is O(n log n). Upper bounds, intuitively, tell us how good a particular algorithm is at solving the problem.
• I don't agree. Algorithms can have lower bounds. Several sorting algorithms have a linear lower bound (even on sorted input, you need to visit each element to verify that it is sorted). For example, this means that quicksort can never run in less than time linear to its input, even in the ideal case. Commented May 24, 2012 at 8:50
• There you are making a distinction between the 'best-case' and 'worst-case' running time of an algorithm. That is different from the lower-bound of a problem itself (which is the worst-case lower bound). Commented May 24, 2012 at 9:01 | 871 | 3,818 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2024-33 | latest | en | 0.940137 |
https://www.cut-the-knot.org/arithmetic/rapid/Specifics/MultiplyBy25.shtml | 1,701,376,015,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100232.63/warc/CC-MAIN-20231130193829-20231130223829-00277.warc.gz | 831,378,939 | 4,578 | # Division/Multiplication by 25
It is easy to remember that 100/25 = 4, and therefore 1000/25 = 40, 10000/25 = 400, etc.
For example, what is 4725 / 25? This is 4×40 + 7×4 + 1 = 160 + 28 + 1 = 189.
Equivalenetly, sometimes it is easier to multiply by 4 than to divide by 25. For example, divide 4732 by 25. We already know that 4725/25 = 189. What remains is to divide 7 by 25. Multiply 7 by 4 and divide by 100 to obtain 0.28 = 7/25. The final result is then 4732/25 = 189.28.
Assume now we wish to find 48×25:
48×25 = 48×100/4 = (48/4)×100 = 1200.
This trick is useful even when the division by 4 is less straightforward:
63×25 = 63×100/4 = 6300/4 = 3150/2 = 1575.
As you see, dividing by 4 can be done in two steps.
• Multiplication by 9, 99, 999, (Multiply + Subtract) etc.
• Squaring 2-Digit Numbers
• Division by 5
• Multiplication by 2
• Multiplication by 5
• Multiplication by 9, 99, 999, etc. (Something Special)
• Product of 10a + b and 10a + c where b + c = 10
• Product of numbers close to 100
• Product of two one-digit numbers greater than 5
• Product of 2-digit numbers
• Squaring Numbers in Range 26-50
• Squaring Numbers in Range 51-100
• Squares of Numbers That End in 5
• Squares Can Be Computed Squentially
• How to Compute Fast Any Square
• Adding a Long List of Numbers
• | 452 | 1,302 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2023-50 | latest | en | 0.877939 |
http://cs.roanoke.edu/Fall2002/CPSC120A/lab5/lab5in.html | 1,696,042,527,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510575.93/warc/CC-MAIN-20230930014147-20230930044147-00438.warc.gz | 12,640,432 | 6,372 | # Lab 5 In-Class: Exploring Data Representation & Conditional (if) Statements
Log onto the Linux system and create a lab5 subdirectory of your cpsc120/labs directory for today's work. As usual, you will need to have three windows open: an xterm, emacs, and Netscape.
### A Base Conversion Program
In class we learned an algorithm for converting a base 10 number to another base by repeatedly dividing by the base. Each time a division is performed the remainder and quotient are saved. At each step, the number used in the division is the quotient from the preceding step. The algorithm stops when the quotient is 0. For example to convert the base 10 number 1878 to base 8 you would do the following:
``` Quotient Remainder
1878 divided by 8 --> 234 6
234 divided by 8 --> 29 2
29 divided by 8 --> 3 5
3 divided by 8 --> 0 3
```
The number in the new base is the sequence of remainders in reverse order (the last one computed goes first; the first one goes last). In this example, the base 8 answer is 3526 (that is 187810 = 35268). In this exercise you will use this algorithm to write a program that converts a base 10 number to a 4-digit number in another base (you don't know enough programming yet to be able to convert any size number). The base 10 number and the new base (2 - 9) will be input to the program. The start of the program is in the file BaseConvert.java. Open the file in Netscape, save it to your lab5 subdirectory, then open it in emacs. Modify the program one step at a time as follows:
1. The program will only work correctly for base 10 numbers that fit in 4 digits in the new base. We know that in base 2 the maximum unsigned integer that will fit in 4 bits is 11112 which equals 15 in base 10 (or 24 - 1). In base 8, the maximum number is 77778 which equals 4095 in base 10 (or 84 - 1). In general, the maximum base 10 number that fits in 4 base b digits is b4 - 1. Add an assignment statement to the program to compute this value for the base that is input and assign it to the variable maxNumber. Add a statement that prints out the result (appropriately labeled). Compile and run the program to make sure it is correct so far.
2. Now it is time to add the code to do the conversion. The comments below guide you through the calculations -- replace them with the appropriate Java statements. Note that you will need to declare a new variable to hold the quotient.
``` // First compute place0 -- the units place. Remember this comes
// from the first division so it is the remainder when the
// base 10 number is divided by the base (HINT %).
// Then compute the quotient (integer division / will do it!) and
// store the result into the quotient variable.
// Now compute place1 -- this is the remainder when the quotient
// from the preceding step is divided by the base.
// Then compute the new quotient and store the result into
// the quotient variable.
// Repeat the idea from above to compute place2 and the next quotient
// Repeat again to compute place3
```
3. So far the program does not print out the answer. Recall that the answer is the sequence of remainders written in reverse order -- note that this requires concatenating the four digits that have been computed. Since they are each integers, if we just add them the computer will perform arithmetic instead of concatenation. Instead, we can use a variable of type String and then use the + operator to do string concatenation. (Remember that if either parameter to + is a string, it does concatenation instead of addition.) Near the top of the program a variable named baseBNum has been declared as an object of type String and initialized to an empty string. Add statements to the program to concatenate the digits in the new base to baseBNum and then print the answer. Compile and run your program. Test it using the following values: Enter 2 for the base and 13 for the base 10 number -- the program should print 1101 as the base 2 value; enter 8 for the base and 1878 for the number -- the program should print 3526 for the base 8 value; enter 3 for the base and 50 for the number -- the program should print 1212.
4. Don't print your program yet! You are going to add something to it later.
### Color Codes
The basic scheme for representing a picture in a computer is to break the picture down into small elements called pixels and then represent the color of each pixel by a numeric code (this idea is discussed in section 1.5 of the text). In most computer languages, including Java, the color is specified by three numbers -- one representing the amount of red in the color, another the amount of green, and the third the amount of blue. These numbers are referred to as the RGB value of the color. In Java, each of the three primary colors is represented by an 8-bit code. Hence, the possible base 10 values for each have a range of 0-255. Zero means none of that color while 255 means the maximum amount of the color. Pure red is represented by 255 for red, 0 for green, and 0 for blue, while magenta is a mix of red and blue (255 for red, 0 for green, and 255 for blue). In Java you can create your own colors. So far in the graphics programs we have written we have used the pre-defined colors, such as Color.red, from the Color class. However, we may also create our own Color object and use it in a graphics program. One way to create a Color object is to declare a variable of type Color and instantiate it using the constructor that requires three integer parameters -- the first representing the amount of red, the second the amount of green, and the third the amount of blue in the color. For example, the following declares the Color object myColor and instantiates it to a color with code 255 for red, 0 for green, and 255 for blue.
```
Color myColor = new Color(255, 0, 255);
```
The statement page.setColor(myColor) then will set the foreground color for the page to be the color defined by the myColor object. The file Colors.java contains an applet that defines myColor to be a Color object with color code (200, 100, 255) - a shade of purple. Save the program and its associated HTML file Colors.html to your lab5 directory, compile and run it using the appletviewer. Now make the following modifications:
1. Change the constructor so the color code is (0,0,0) --- absence of color. What color should this be? Run the program to check.
2. Try a few other combinations of color codes to see what you get. Page 793 of the text shows you the codes for the pre-defined colors in the Color class.
3. Now we will modify the program to generate random colors. Notice on page 793 of the text that there is a constructor for the Color class that takes a single integer as an argument. The first 8 bits of this integer are ignored while the last 24 bits define the color -- 8 bits for red, 8 for green, and the last 8 bits for blue. Hence, the bit pattern
``` 00000000000000001111111100000000
```
should represent pure green. Its base 10 value is 65280. Change the declaration of the myColor object to
``` Color myColor = new Color (65280);
```
Compile and run the program. Do you see green?
4. Now add the following statements to the program:
• declare generator to be an object of type Random (the import statement is already there);
• declare colorCode to be a variable of type int;
• assign colorCode a random integer value (use the nextInt() method);
• replace the number 65280 in the Color constructor above with the variable colorCode
Compile and run the program -- reload the program several times so you can see the different random colors generated.
5. The Color class has methods that return the individual color codes (for red, green, and blue) for a Color object. For example,
``` redCode = myColor.getRed();
```
returns the code for the red component of the myColor object (redCode should be a variable of type int). The methods that return the green and blue components are getGreen and getBlue, respectively. Add statements to the program, similar to the above, to get the three color codes (you need to declare some variables). Then add statements such as
``` page.drawString("Red: " + redCode, ____ , ____ );
```
to label the rectangle with the three color codes (fill in the blanks with appropriate coordinates so each string is drawn inside the rectangle -- you also need to set the drawing color to something such as black so the strings will show up). Compile and run the program to make sure it works. Reload several times to see the different colors and their corresponding codes displayed.
6. Print the final version of your applet.
### A Program Using If Statements (Conditionals)
A conditional statement (also called a selection statement) in a programming language is one that allows a choice as to which statement is executed next. That choice is based on some condition -- a boolean expression that is either true or false. In Java (and most other programming languages) two of the conditional statements are if and if ... else... . If and if ... else... statements are discussed in Section 3.2 of the text. An if statement is used when you want to do something when some condition is true but do nothing otherwise. Some examples are:
``` if (grade >= 90)
System.out.println ("Congrats! You made an A!");
```
In this example the condition is grade >= 90 (grade would be a variable given a value earlier in the program). If it is true (if the value of the variable grade is greater than or equal to 90 when this statement is executed) the congratulation statement would be printed; otherwise it will not be printed.
``` if (numStudents != 0)
testAverage = sumOfGrades / numStudents;
```
In this example, the condition is numStudents != 0. != is the "not equal" operator. If the value of the variable numStudents is not equal to 0 when the condition is evaluated, the assignment statement will be evaluated.
An if ... else ... statement is used when you want to do one thing when a condition is true but something else when it is false. Some examples are
``` toss = Math.abs (generator.nextInt()) % 2;
if (toss == 0)
System.out.println ("Tails");
else
System.out.println ("Heads");
```
In this example the condition is toss == 0, where == is the "equal" operator. (Remember a single = mark is the assignment operator.)
``` if (numStudents > 0)
testAverage = sumOfGrades / numStudents;
else
System.out.println ("You must have at least one student!!");
```
The file WaterBill.java contains a program to compute a customer's water and sewer bill. The charge for water is based on consumption -- a customer who uses no more than 7500 gallons in a month pays \$0.002 per gallon but a customer who uses more pays \$0.002 per gallon for the first 7500 gallons plus \$0.0035 per gallon for each gallon over 7500. Note that the charge depends on a condition -- whether or not the amount of water used is over 7500 gallons. Hence an if is needed to compute the charge for the water. Do the following:
1. Save the file to your directory and open it in emacs.
2. Study the code and find the if... else... that computes the charge for the water -- there are two different formulas used for the calculation depending on whether numGals is less than or equal to CUTOFF (7500) or not. Also notice near the bottom of the program an if that prints a message for customers who use less than half of 7500.
3. Run the program several times. Enter numbers greater than 7500 (such as 10000), numbers between 7500 and half of 7500 (such as 6000), and numbers less than half of 7500 (such as 3000).
4. Suppose the charge for sewer services is also based on water usage. A person who uses less than 7500 gallons of water pays a flat rate of \$7.50 for sewer but a person who uses more pays \$0.001 per gallon. Add an if ... else... to the program to compute the sewer charge. Update other parts of the program to take this into account -- update the calculation of the total bill and add a statement to print the sewer charge in the bill.
5. Suppose senior citizens get a break on the utility tax. The tax rate for senior citizens is 8.5% but it is 12% for all others. Add statements to the program to compute the tax. This requires that you find out if the customer is a senior citizen! Use a variable of type char for this. You need to do the following:
1. Add a prompt that asks the user to answer the question "Are you a senior citizen?" by entering a y or n. Use the Keyboard class to read in the answer (as a char not a String); store the answer in the variable senior.
2. Write an if to compute the utility tax.
3. Add the utility tax to the total bill.
4. Print the utility tax as part of the bill.
6. Add an if statement that prints a warning to customers who use more than 3 times the cut off (7500) gallons. Warn them that they are subject to a \$500 fine if their excessive water use continues (just print a message -- don't add the fine to their bill)!
7. Be sure your program works correctly, then print it.
### Adding an if to the Base Conversion Program
Get your base conversion program back (note -- instead of re-opening it, look under buffers on the emacs menu bar, then click on BaseConversion.java). Remember that this program only computes 4 digits in the new base so if the user enters a number that is too large to fit in 4 digits the answer produced by the program is incorrect (it is incomplete -- it only gives the rightmost 4 digits). Unfortunately a lot of people believe the computer so we need to be sure the computer doesn't print out incorrect information. Replace the statement that prints the answer with an if ... else ... If the number is too large to fit in 4 digits in the new base print a message saying so (remember you calculated the maximum number that will fit); otherwise, print the answer. To test the program, use the following input: 4 for the base and 375 for the base 10 number (the program should say 375 is too big to fit in a 4-digit base 4 number); 7 for the base and 5341 for the number (again, too big); 7 for the base and 537 for the base 10 number (the program should print 1365).
Print a copy of your program.
### Hand In
• The three programs (BaseConvert.java, Colors.java, and WaterBill.java).
• Tar your directory and email it to your instructor at roanoke.edu with the subject cpsc120 lab5. | 3,335 | 14,448 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2023-40 | latest | en | 0.896779 |
http://www.varsitytutors.com/ap_physics_b-help/using-ohm-s-law | 1,484,604,201,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560279368.44/warc/CC-MAIN-20170116095119-00232-ip-10-171-10-70.ec2.internal.warc.gz | 760,845,228 | 24,185 | AP Physics B : Using Ohm's Law
Example Questions
Example Question #1 : Using Ohm's Law
A student assembles a circuit made up of a voltage source and two resistors. All three circuit elements are connected in series. If the voltage across the voltage source is and the resistance of the resistors are and respectively, what is current through the second resistor?
Explanation:
Recall that the current through resistors connected in series is the equal for all components. The current through the first resistor and the current through the second resistor should be the same.
The first step in solving involves calculating the equivalent resistance of the circuit. Since the resistors are in series the equivalent resistance of this circuit is the sum of the two resistances.
Now we can use Ohm’s law to solve for the current. | 168 | 833 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2017-04 | latest | en | 0.915803 |
https://www.physicsforums.com/threads/puck-collision-problem.212543/ | 1,477,608,631,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988721405.66/warc/CC-MAIN-20161020183841-00404-ip-10-171-6-4.ec2.internal.warc.gz | 969,412,337 | 17,959 | # Puck Collision Problem
1. Feb 1, 2008
### davidphysics
Hey. This is my first post... heres my question [Diagram Attached]:
The drawing shows a collision between two pucks on an air-hockey table. Puck A has a mass of 0.028 kg and is moving along the x axis with a velocity of +5.5 m/s. It makes a collision with puck B, which has a mass of 0.068 kg and is initially at rest. After the collision, the two pucks fly apart with the angles shown in the drawing.
Find the final speed of
(a) puck A and
(b) puck B. [both in m/s]
I am having trouble with this and although I know to use conservation of momentum, should I assume its elastic?
If you could post an explanation that would be great, but I would also like a final answer in numbers please.
File size:
11.6 KB
Views:
68
2. Feb 1, 2008
### tony873004
No one is interested in doing your homework for you, but rather in helping you understand. So don't be too disappointed with only an explanation.
Have you tried using momentum formulas and a little trig? Put in a little effort so people can see where you're having trouble.
3. Feb 1, 2008
### davidphysics
Ok sorry let me explain what I've gotten to... Basically I've set up both horizontal and vertical equations for both Po=Pf and Ko=Kf. Before I go any further, am I correct to assume that it is elastic and that the kinetic energy is conserved?
To be honest I don't wan't to use up my 4 trys on webassign, which I'm sure your familiar with.
By the way, thanks for the quick response.
4. Feb 1, 2008
### davidphysics
And when I plug in the values for "v" in those formulas I should use for example Vacos65 and Vasin65 for the horiz and vert respectively, right?
For some reason I keep getting the Vert for a and b=0, because both the initial momentum and kin energy=0. So when I solve I get Vert: Va=Vb=0... I feel like I'm doing something very stupid.
5. Feb 1, 2008
### tony873004
Yes, I'm familiar with the 4 attempts :)
Unless energy is being carried away in sound waves or by other means, I would guess your assumption is correct. Your picture isn't approved yet, so I can't be sure.
6. Feb 1, 2008
### davidphysics
O didn't realize that, please read my post that I just posted a second ago, and I'll put the image on imageshack.
POSTED:
And when I plug in the values for "v" in those formulas I should use for example Vacos65 and Vasin65 for the horiz and vert respectively, right?
For some reason I keep getting the Vert for a and b=0, because both the initial momentum and kin energy=0. So when I solve I get Vert: Va=Vb=0... I feel like I'm doing something very stupid.
Last edited: Feb 1, 2008
7. Feb 1, 2008
Bump (sorry)
8. Feb 1, 2008
### tony873004
Keep in mind your total momentum before the collision was completely on the x-axis. There was no momentum on the y-axis. So afterwards, the y-components of the momentums of your two objects should also equal zero since it is conserved.
I'm on my way out the door, so I can't help you further. But others are sure to jump in. Good luck, and welcome to the forum.
9. Feb 1, 2008
### davidphysics
Wow, then that image is very deceiving... thanks for your help -Everyone else JUMP IN! :)
Just thinking of it that makes perfect sense, now I see how these pictures mess with your mind.
Disregard everything above, yea so the Vert momentum up = vert momentum down
10. Feb 1, 2008
### hage567
But the initial momentum is not zero. It is in the vertical direction, but not in the horizontal. The initial kinetic energy is not zero either, since A is moving.
11. Feb 1, 2008
### davidphysics
I mean the initial VERT momentum. But thanks for "The initial kinetic energy is not zero either, since A is moving." Can't believe I screwed that up. I don't have any more time now, but I'll work on that tomorrow and will come back to this forum later.
Thanks! and cya
12. Feb 1, 2008
### usman27
You need to apply conservation of momentum and energy. Initial conditions do not require you to resolve the velocity vector into components. You need to resolve the velocity vector into x and y components for the final condition.
13. Feb 3, 2008
### davidphysics
Hi. I am still having trouble and would like help. Its not working out.
Thanks
14. Feb 3, 2008
### hage567
Show how you set up your equations. We can't find what you're doing wrong if you don't show us your work.
15. Feb 3, 2008
### davidphysics
Nevermind - I figured it out. I was making it much too difficult. All I had to do was Po=Pf for the horiz and vert, and solve that set of equations. | 1,191 | 4,568 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2016-44 | longest | en | 0.949147 |
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Soda is often sold in 20 ounce bottles. A pack of sugar (the kind they have at restaurants for your coffee or tea) typically contains 4 grams of sugar. If an 8 ounce bottle of cream soda has 30 grams of sugar, then drinking a 20 ounce bottle of cream soda would be equivalent to eating how many packets of sugar?
18.75 packs of sugar
Step-by-step explanation:
20/8 = 2.5 (how many 8 ounce bottles in 20 oz)
2.5 * 30 = 75 (to find how many grams of sugar in the 20 oz bottle)
75/4 = 18.75 (how many packs of sugar it’s equal to)
you can also use a ratio to solve:
8/30 = 20/x; cross multiply
600 = 8x; divide 8 on both sides
x = 75
75/4 = 18.74 (how many packs of sugar it’s equal to) | 208 | 703 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2023-23 | latest | en | 0.938137 |
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We'll provide some tips to help you choose the best Algebra help free step by step for your needs. Math can be a challenging subject for many students.
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This will usually result in a quadratic equation which can be solved using standard methods. In some cases, it may be possible to solve the equations directly without first solving for one of the variables. However, this is usually more difficult and it is often easier to use substitution. Whether or not to use substitution depends on the form of the equations and the preference of the person solving them. | 491 | 2,517 | {"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-2022-49 | longest | en | 0.955726 |
https://forhinhexes.blogspot.com/2022/04/ | 1,656,719,467,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103947269.55/warc/CC-MAIN-20220701220150-20220702010150-00313.warc.gz | 313,219,982 | 13,775 | ## April 21, 2022
### Trade VI: Simplifying to Complexity
I've been working on my algorithms to generate trade networks and manufactured goods. Although probably several years out of date now, Tao's Trade System remains required reading for anyone wanting to do something similar.
This iteration was a much-needed rework of the way a lot of the prices were generated. I had gone through Alexis' work linked above and simplified it to my needs. He is using references based in the real world, whereas I am trying to generate from scratch.
I found that of all the equations, the final price of an object simplified to just a few variables. The first is the local value of a unit (oz) of gold (in cp, all prices are expressed in cp and can then be abstracted back up to sp or gp as needed). I'll call this $g$. To find this number, we need the local availability $g_\ell$ and the total availability $g_t$. $g_t$ is the total number of references reachable from the point in question. This is easy to do with network algorithms. Local availability is a weighted distance calculation over all locations $i$ reachable from $\ell$: $g_\ell = \sum_i \frac{g_i}{\textrm{dist}\left(\ell, i\right) + 1}$ By this calculation, if $i$ has 2 gold references, but is 4 days away, it contributes $\frac{2}{4+1} = 0.4$ references to $g_\ell$.
Right now, I am treating each hex as a node in the graph, but if there is no defined settlement there, all its resources are given to the closest city hex for pricing purposes.
We need a rarity factor $r$ that scales with the size of the network. As the network grows larger, a smaller $r$ is needed to balance things out. I'm trying this out for now: $r = -\frac{1}{n}$ where $n$ is the number of nodes in the network.
The last two constants are the number of gold pieces per oz ($p = \frac{1\textrm{ gp}}{0.48\textrm{ oz}}$) and ratio of cp to gp ($c = \frac{100}{1}$). These are easy to change. A half-ounce gold piece is somewhat hefty; many coins in history would have been much smaller amounts. Gold is valuable enough that a small bit is worth a lot, and hence a great deal of value can be expressed in quite tiny coins. I like the idea of a gp being a weightier coin. I also try to base the sizes of my coins on current analogs that I can actually show to my players.
A thousand of these is no joke
So the final equation comes together: $g = \left(r \cdot \frac{g_t}{g_\ell} + 1\right) \cdot p \cdot c$
For an individual resource $q$, the equation is similar. First, we have to define the base cost $b$ of a unit of $q$. I found that this was the most important factor; it essentially represents the ideal economy where everything is in perfect supply. Whether a board-foot of wood is defined as 1 cp or 18 cp will have approximately a 18x effect on the final price of wood no matter where you are in the world. This becomes the key object of research when sketching out the system. It is then an easy matter to determine how many units of $q$ are equivalent to one reference of gold (assuming a gold ref = 1500 oz): $\mathit{ref}_q = \frac{b}{\mathit{ref}_g = 1500\textrm{ oz}}$
We obtain the rarity $r_q$ in a similar way as above with gold, using the distance weighted availability. The final price (in cp) of an item at location $\ell$ is then: $\_q = \frac{g}{\mathit{ref}_q} \cdot \left(r \cdot \frac{q_t}{q_\ell} + 1\right) \mathit{ref}_g$
There are some other ways to view this equation. It can simplify again to: $\_q = \frac{g}{b} \cdot \left(r \cdot \frac{q_t}{q_\ell} + 1\right)$
The next step is to determine the availability of labor references. I haven't quite decided how to assign these so I'll save that for a future post. We obtain the available references by once again iterating on the network: $L_\ell = \sum_i \frac{L_i}{\textrm{dist}\left(\ell, i\right) + 1}$
The cost of a material $m$ at a given stage (eg, hematite $\to$ iron ore) is then the cost of the raw materials (the unit cost $\$_m$times the number of units$n_m\$) plus the labor cost, which is raw material cost divided by the labor references: $\_m = \_q \cdot n_m + \frac{\_q \cdot n_m}{\mathit{ref}_L}$
This step is repeated for each stage of the process, which can be quite complex. I developed a JSON schema to represent each manufactured material. To raise an auroch from a calf to weaned, you require the following.
{
"item": "auroch (weaned)",
"unit": "hd",
"stage": 1,
"tech": 7,
"weight": 200,
"recipe": {
"materials": {
"auroch (calf)": 1,
"min": [
{
"maize": 483,
"oats": 483,
"barley": 483,
"cassava": 483,
"rice": 483,
"wheat": 483
}
]
},
"labor": "herdsman"
}
}
Our raw materials are 1 auroch calf + whichever is cheaper between 483 lbs of feed, plus the labor of a herdsman. As long as every manufactured item "downstream" exists, this "item" will be available for purchase.
## April 8, 2022
### Resources XXIII: Crops and Climates
I've recently completed a little side project that will help resource placement significantly.
I cross-referenced the crop yields from EarthStat with Koppen climate data to get the prevalence for 175 different categories. The data isn't perfect - it doesn't take politics or demographics into account, for example - but I think it will be useful either in automation (as I plan to use it) or in beginning to think about where crops should be placed on a world map.
More details and the data itself can be found here in its own git repo. | 1,446 | 5,427 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2022-27 | latest | en | 0.939981 |
http://www.wyzant.com/resources/answers/10859/earns_hourly_rate_for_35h_work_and_increased_rate_for_overtime_he_worked_39h_and_received_787_20_next_worked_41h_and_received_844_80_hourly_and_overtime | 1,397,877,666,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609535745.0/warc/CC-MAIN-20140416005215-00381-ip-10-147-4-33.ec2.internal.warc.gz | 770,029,415 | 10,453 | Search 72,499 tutors
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## Earns hourly rate for 35h work and increased rate for overtime. He worked 39h and received \$787.20, next worked 41h and received \$844.80. hourly and overtime?
what is the rate for hourly and overtime
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# 2 Answers
Given: earns regular hourly rate for 35 hours and earns overtime hourly rate for working extra hours
received a total of \$787.20 for working 39 hours
==> earns regular rate for 35 hours and overtime rate for 4 hours
received a total of \$844.80 for working 41 hours
==> earns regular rate for 35 hours and overtime rate for 6 hours
Let 'x' represent the regular rate earned per hour and 'y' represent the overtime rate earned per hour. That is,
x = regular rate per hour
y = overtime rate per hour
With this, we arrive at the following system of linear equations:
(1) 35x + 4y = 787.20
(2) 35x + 6y = 844.20
To solve by substitution, solve one of the above equations for one of the variables (either x or y) then substitute this into the other equation. For instance, let's solve equation (1) for y:
(1) 35x + 4y = 787.20
subtract 35x from both sides
4y = 787.20 - 35x
divide both sides by 4
4y/4 = 787.20/4 - 35x/4
y = 196.80 - 8.75x
(2) 35x + 6y = 844.80
35x + 6(196.80 - 8.75x) = 844.80
distribute the 6 into each term inside the parenthesis
35x + 1180.80 - 52.50x = 844.80
combine like terms
-17.50x + 1180.80 = 844.80
subtract 1180.80 from both sides
-17.50x = -336
divide both sides by -17.50
x = 19.20
Now that we've solved for x, solve for y by substituting the value found for x into the equation we determined for y:
y = 196.80 - 8.75x
y = 196.80 - 8.75(19.20)
y = 196.80 - 168
y = 28.80
Thus,
x = regular hourly rate ==> \$19.20/hr
y = overtime hourly rate ==> \$28.80/hr
let x be regular pay rate and y be overtime pay rate.
first case: worked for 39 hours, so get paid regular rate for 35 hours and overtime of 4 hours
35x + 4y = 787.20 eq. 1
second case: worked 41 hours, so get paid regular rate for 35 hours and overtime of 6 hours
35x + 6y = 844.80 eq. 2
so figure out x and y from the given equations. | 730 | 2,265 | {"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} | 4.40625 | 4 | CC-MAIN-2014-15 | latest | en | 0.904843 |
https://www.acemywork.com/statistics-assignment-help-24-7/ | 1,675,795,559,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500628.77/warc/CC-MAIN-20230207170138-20230207200138-00368.warc.gz | 626,505,107 | 22,562 | ## Statistics Assignment Help
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Statistics Assignment Help | 1,595 | 9,178 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2023-06 | latest | en | 0.922083 |
https://www.gradesaver.com/textbooks/math/algebra/college-algebra-10th-edition/chapter-6-section-6-3-exponential-functions-6-3-assess-your-understanding-page-434/15 | 1,531,961,070,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590362.13/warc/CC-MAIN-20180718232717-20180719012717-00412.warc.gz | 887,012,178 | 12,991 | ## College Algebra (10th Edition)
(a) $2^{3.14} \approx 8.815$ (b) $2^{3.141} \approx 8.821$ (c) $2^{3.1415} \approx 8.824$ (d) $2^{\pi} \approx 8.825$
Use a calculator to evaluate each expression, and then round-off the answer to three decimal places: (a) $2^{3.14} \approx 8.815$ (b) $2^{3.141} \approx 8.821$ (c) $2^{3.1415} \approx 8.824$ (d) $2^{\pi} \approx 8.825$ | 163 | 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": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2018-30 | latest | en | 0.480046 |
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Brilliant’s Composite Mathematics Class 8 Solutions Chapter 9 Shares and Debentures
Welcome to NCTB Solutions. Here with this post we are going to help 8th class students for the Solutions of Brilliant’s Composite Mathematics Class 8 Math Book, Chapter 9, Shares and Debentures. Here students can easily find step by step solutions of all the problems for Shares and Debentures, Exercise 9.1 and 9.2 Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily.
Shares and Debentures Solution Exercise 9.2
Question no – (1)
Solution :
Investment = 80 rupees per share
Income = 4 rupees per income
Income percentage
= 4/30 × 100
= 25%
Thus, the income percent will be 25%
Question no – (2)
Solution :
Investment = 75 rupees per share
Income = 125 × 21/100
= 105/4
= 26.25 rupees per share .
Annual yield,
= 26.25/75 × 100
= 35%
Hence, the annual yield percent will be 35% .
Question no – (3)
Solution :
Sonu received = (750 × 2)
= 1500 rupees in a whole year
Since, the rate of interest is 10%,
Total face value of all the debentures is,
= 1500 × 100/10
= 15000 rupees
The value of 1 debenture
= 15000/300
= 50 rupees
Therefore, the face value of one debenture will be Rs 50.
Question no – (4)
Solution :
The cost price of a debentures worth Rs 100 = 90
Return from a debentures,
= 100 × 9/100
= 9 rupees
Yield percent,
= (9/90 × 100)%
= 10%
Question no – (5)
Solution :
Face value of 1 debenture = 100 rupees
Market value of 1 debenture = 112 rupees
Rate per cent that Murlidhar earns,
= (8/112 × 100)%
= 7.14%
Hence, the rate percent will be 7.14%
Question no – (6)
Solution :
Since premium is 10% the total face value is,
= 5500 × 100/110
= 5000 rupees
Income derived at 18%
= 5000 × 18/100
= 900 rupees
Question no – (7)
Solution :
Let the face value be 100 rupees in each case
1st case : Investment at 10% discount = 90 rupees
Income at 10% return,
= 90 × 10/100
= 9 rupees
2nd case : Investment at 10% premium = 110 rupees
Income at 20% return
= 110 × 20/100
= 22 rupees
Return on investment per cent in the first case
= 9/90 × 100
= 10% and in the second,
Case = 22/100 × 100
= 20%
Therefore, the second investment is better.
Question no – (8)
Solution :
Let the face value be 100 rupees in each case
1st case :
Investment at 8% premium = 108 rupees Income at 15% interest
= 108 × 15/100
= 16.2 rupees
2nd case :
Investment at 4% premium = 104 rupees
Income 14% return,
= 104 × 14/100
= 14.56 rupees
Thus, the first investment is better.
Question no – (9)
Solution :
Market Value at each debentures at 15% Premium 115 rupees. The value of each debenture including brokerage,
= 115 × 101/100
= 116.15 rupees
No of debentures he bought,
= 5200 ÷ 100
= 52.
His total investment
= 52 × 116.15
= 6039.80 rupees
Question no – (10)
Solution :
Total face value of the debentures = 7500 rupees.
Total market value,
= 7500 × 105/100
= 7875 rupees
Brokerage,
= 7875 × 2/100
= 157.50 rupees
= 7875 – 157.50
= 7717.50 rupees.
Question no – (11)
Solution :
Harish’s income in 1 debenture = 15 rupees.
Total number of shares,
= 37500/15
= 2500.
His total investment,
= 2500 × 80 × 101/100
= 20200 rupees
Question no – (14)
Solution :
Vineet gets = 10 × 8/100
= 0.8 rupees per share.
So, his income per cent on investment
= 0.8/7.5 × 100
= 32/3%
= 10.67%
Question no – (15)
Solution :
Ratio of return from a debenture and a share is
= 40/3 : 20
= 2 : 3.
Since, the ratio of total return from debentures and shares is = 1 : 1,
Therefore, the ratio of number of debentures and shares is = 3 : 2.
The ratio of investment per debenture and per share is 11 : 10.
The ratio of invest on debenture and on shares is,
= (3 × 11) : (2 × 10)
= 33 : 20
He invested on debentures,
= 33/53 × 5300
= 3300
He invested on shares,
= 20/52 × 5300
= 2000 Rs
Question no – (16)
Solution :
Number of debentures,
= 6000/100
= 60.
Market Value
= 60 × 105
= 6300 rupees.
Dividend = 6000 × 20/100
= 1200 rupees.
Brokerage = 6300 × 1/100
= 63 rupees.
Sona will get,
= 6300 + 1200 – 63
= 7337 rupees.
Previous Chapter Solution :
Updated: May 29, 2023 — 6:32 am | 1,516 | 4,594 | {"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} | 4.65625 | 5 | CC-MAIN-2024-10 | longest | en | 0.800763 |
http://www.devmathrevival.net/?p=2706?shared=email&msg=fail | 1,519,290,492,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814079.59/warc/CC-MAIN-20180222081525-20180222101525-00250.warc.gz | 437,892,677 | 10,413 | ## Problem Solving Skills
This is the story of what a student did when confronted with a procedural problem for which she did not ‘remember’ the standard procedure.
One of the in-class assessments I used is a ‘worksheet’; it’s like an open-note quiz over a set of material. During our last class (intermediate algebra), the worksheet is longer than usual because it ‘covers’ the entire course. Item 6 on this worksheet is:
Rationalize the denominator
As I said, she could not remember ‘what to do’. However, she did a great thing … she recognized that both numbers could be written as a power of 2:
I was very pleased that she did this, but the student was frustrated … she then could not see what to do. This is pretty typical when novices dive in to the world of ‘non-standard problems’ — problems for which we lack a remembered process.
Of course, it was pretty easy to guide her through the remainder of the work:
Obviously, the expectation (this is our traditional intermediate algebra course) was that students would apply the standard procedure (multiplying top & bottom by the cube root of 4). Students do not like that procedure, and I tell them that the procedure itself is seldom needed.
The alternate method worked only because there was a common base between numerator and denominator, and I doubt if the student will gain any long-term benefit from this experience. This was more of a positive thing for me, as a teacher and problem-solver: Noticing a special pattern within a problem is a critical problem solving skill.
I’m sharing this story just because I had fun with it!
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### 1 Comment
• By schremmer, December 30, 2016 @ 12:21 pm
There is an interesting shift in emphasis here: on the first line, we have “did not ‘remember’ the standard procedure” while, thereafter, we now have remember ‘what to do’ (my boldface).
I believe that this shift is the core of the problem: We do realize that the problem is with remember (something totally anti-learning) but we immediately shift to what to do (something unfortunately just as anti-learning).
What to do ought to proceed from the question. (As John Hold wrote once upon a time, in How Children Fail, we force the students to be “answer oriented” instead of letting them remain “question oriented”.)
The issue above was purely a linguistic one. The student did not know the meaning of the “code”. It is as if I asked an English speaker to do something in, say, Norwegian. Sure, even languages can be superficially forgotten but most of the time they come back very quickly.
To use a related example: After I have coerced students into reading 3x^{+4} as “3 multiplied by 4 copies of x” and 3x^{-4} as “3 divided by 4 copies of x”, I immediately ask them to “do” 2x^{+4}•5x^{+3}, 2x^{+4}•5x^{-3}, 2x^{-4}•5x^{+3}, 2x^{-4}•5x^{-3}, … and … they “do it”. (Without me showing them anything of course. Just insisting that they go back to reading the code.) Then, after they have made their peace with “division of fractions”, I can ask them without further ado to “do” 2x^{+4}÷5x^{+3}, 2x^{+4}÷5x^{-3}, 2x^{-4}÷5x^{+3}, 2x^{-4}÷5x^{-3}, … and … they “do it”. | 789 | 3,178 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2018-09 | latest | en | 0.970558 |
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