url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3
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http://stackoverflow.com/questions/13689975/cant-subtract-one-double-from-another | 1,462,444,668,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860126502.50/warc/CC-MAIN-20160428161526-00025-ip-10-239-7-51.ec2.internal.warc.gz | 271,770,426 | 21,242 | # Can't subtract one double from another?
Im quite new to Java and programming in general, but Ive read up on it quite a bit. Im currently making my first real OOP- a calculator that can perform certain equations. However, while trying to program something that would calculate the variance of a distribution -Heres the code:
``````void variance() {
System.out.println("The variance of a distribution with x values of " + a + b + "and mean"
+ mean + "is " + a*a+b*b/2 - mean*mean);
}
``````
I get the error
Bad operand types for binary operator '-' First type = string Second type = double".
I had previously stated that a, b and mean were doubles and had also stated how mean is calculated. I also tried changing `a*a+b*b/2` from a string to a double, but then realized that if i put any integers or doubles into where `a*a+b*b/2` (e.g. 2) but i get the same error. Any help would be much appreciated :)
-
where did you declare mean – ekims Dec 3 '12 at 19:25
put brackets around the expression `(a*a+b*b/2 - mean*mean)` – andre Dec 3 '12 at 19:26
by the way, these are brackets: `[]` while these are parentheses: `()` – Erick Robertson Dec 3 '12 at 19:27
That's because the `+` is overloaded in Java for string concatenation. You need to put parentheses around your math expression.
``````System.out.println("The variance of a distribution with x values of " + a + b + "and mean"
+ mean + "is " + (a*a+b*b/2 - mean*mean));
``````
-
+1 May I also suggest the OP actually wants `((a*a+b*b)/2.0 - mean*mean)` – Erick Robertson Dec 3 '12 at 19:14
They might need that; I didn't want to confuse the issue with whether or not the formula was mathematically correct. – Lawrence Dol Dec 3 '12 at 19:15
Perfect use for a comment... ook ook eek eek. – Erick Robertson Dec 4 '12 at 16:07
The short answer is you need to do this:
`````` + mean + "is " + (a*a+b*b/2 - mean*mean));
``````
The longer answer is that Java evaluates your expression from left to right. So step by step, it happens something like this:
1. mean + "is " + a*a+b*b/2 - mean*mean
2. string + "is " + a*a+b*b/2 - mean*mean
3. string + a*a+b*b/2 - mean*mean
4. string - mean*mean
The compiler stops here because although you can concatenate a string and a number using `+` (in step 3), it doesn't make sense to subtract a number from a string. By using parentheses around the whole arithmetic expression, that will cause the arithmetic to be evaluated first, before the result is concatenated with the rest of your output string.
-
+1 perfect...... – Mukul Goel Dec 3 '12 at 19:17
+1, but a nit: Java evaluates operands from left to right, for any given operator. It evaluates expressions according to operator precedence. – EJP Dec 3 '12 at 22:10
Surround all mathematical expressions between parenthesis, and (depending on the actual type of `a`, `b`, `mean`), it's better to divide by `2.0`, to make sure that a floating-point division is performed.
-
ah ok, thanks very much, got it now :) – Fraser Price Dec 3 '12 at 19:14
To make sure a floating-point division is performed. – EJP Dec 3 '12 at 22:09
@EJP OK, fixed it. – Óscar López Dec 3 '12 at 22:19
You are mixing the `String` `+` operator with the `Double` operator.
Try this:
``````System.out.println( "The variance of a distribution with x values of "
+ ( a + b ) + " and mean "
+ mean + " is "
+ ( ( a * a + b * b / 2 ) - ( mean * mean )) );
``````
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# Edexcel AS Mathematics C2- 25th May 2016 Thread
Announcements Posted on
Why bother with a post grad course - waste of time? 17-10-2016
1. (Original post by HaiderMukhles)
No unfortunately not, will send it though if it ever comes my way
Thanks! And likewise if I find it 😊
2. (Original post by SIMON WARD 99)
An easy 75/75
alright m8
3. (Original post by Hey111)
For the first part with the sector, I gave my answer in both radians and degrees. I forgot to cross out the degrees answer at the end of the exam, do you think I'll lose a mark for giving two answers?
I don't think you'll lose a mark because despite providing two answers, they're technically the same value, just in a different format I guess. You wrote the radians answer, so I personally think you'll be alright, but I may be wrong
4. (Original post by neil20143)
What were the binomial questions and the marks for each one again? Was it 2 marks for finding k and A and then another 2 for finding B?
A) First part (working out the binomial expansion using the formula) = 4 marks
B) Working out A = 1 mark
C) Working out K = 2 marks
D) Working out B = 2 marks
5. For 7b when u had to integrate the curve in part a and then get area between it and X line, where did it cross the X axis? Can't remember if I got this wrong
6. (Original post by elishx__h)
I don't think you'll lose a mark because despite providing two answers, they're technically the same value, just in a different format I guess. You wrote the radians answer, so I personally think you'll be alright, but I may be wrong
Do you reckon 67 will be enough for an A? XD
7. (Original post by Hey111)
Do you reckon 67 will be enough for an A? XD
I think I got that too, and yeah its more than enough. Maybe 88/90- UMS if boundaries are low.
8. How did you guys work out A? I swear I did it right but came out with the wrong answer?!
9. (Original post by neil20143)
How did you guys work out A? I swear I did it right but came out with the wrong answer?!
An A is usually 85% of the raw marks on average. If you got questions 1->8 perfectly right and half of Question 9's marks right, you've probably got an A safely in the bag. If not, then it's all down to luck.
10. (Original post by tarek888)
hey guys there was only 9 questions yes? plz answer
Yes there was, aha dont panic!
11. (Original post by Hey111)
An A is usually 85% of the raw marks on average. If you got questions 1->8 perfectly right and half of Question 9's marks right, you've probably got an A safely in the bag. If not, then it's all down to luck.
Sorry I was asking how you worked out K for binomials! I'm not sure why I wrote A instead
12. After that exam I feel so depressed. I've done bad for c1, and now bad for c2 as well. And Im so worried I'm going to be predicted very low for Maths at A2, which will completely screw up my chances of getting into a good university. Im just Absolutely devastated.
Does anyone know how much I'd need to get in S1 if I got a high C in c1 and c2, to get a B overall at maths as level?
13. (Original post by Esekanayo_)
After that exam I feel so depressed. I've done bad for c1, and now bad for c2 as well. And Im so worried I'm going to be predicted very low for Maths at A2, which will completely screw up my chances of getting into a good university. Im just Absolutely devastated.
Does anyone know how much I'd need to get in S1 if I got a high C in c1 and c2, to get a B overall at maths as level?
I'm in the same boat as you. I think I'll just convince my teachers that I can get an A* next year because they originally thought I would. It depends what you got on C1 and C2 but Stats generally is either a love it or hate it paper. I'd say really concentrate and normal distribution and probability.
14. (Original post by Hey111)
Do you reckon 67 will be enough for an A? XD
Yes of course! don't worry haha, you're in a good position
15. (Original post by neil20143)
I'm in the same boat as you. I think I'll just convince my teachers that I can get an A* next year because they originally thought I would. It depends what you got on C1 and C2 but Stats generally is either a love it or hate it paper. I'd say really concentrate and normal distribution and probability.
There's no point in getting upset about it because what's done is done. Stress during, and after exams only makes your overall performance worse. If you did really well in one exam and awful in the others, then it's still a standalone achievement. Don't focus on compensating for marks in your next Maths exam because you might put more pressure on yourself, and end up doing worse as a result. Also, you will always do your best anyway, and you can't do more than that.
You may also want to consider doing Further Maths A Level this September, if you plan on studying a science/mathematical subject at uni as the Torys are changing A Levels for the worse. If your college refuse to let you do it, you can always do it privately.
16. (Original post by elishx__h)
Yes of course! don't worry haha, you're in a good position
Aww thanks!
17. I am :/
18. (Original post by Hey111)
There's no point in getting upset about it because what's done is done. Stress during, and after exams only makes your overall performance worse. If you did really well in one exam and awful in the others, then it's still a standalone achievement. Don't focus on compensating for marks in your next Maths exam because you might put more pressure on yourself, and end up doing worse as a result. Also, you will always do your best anyway, and you can't do more than that.
You may also want to consider doing Further Maths A Level this September, if you plan on studying a science/mathematical subject at uni as the Torys are changing A Levels for the worse. If your college refuse to let you do it, you can always do it privately.
Sorry? We were on about making up for our scores in statistics? I don't see where dwelling and being upset over C2 came from?
19. (Original post by neil20143)
Sorry? We were on about making up for our scores in statistics? I don't see where dwelling and being upset over C2 came from?
Must have responded to the wrong post. Oh well.
20. (Original post by neil20143)
Sorry I was asking how you worked out K for binomials! I'm not sure why I wrote A instead
You had to multiply each term of your expansion by the function of K. You then had to collect and factorise powers of x, then equate coefficients with the answer they gave. I believe that K was 7/2.
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Useful resources | 1,812 | 7,190 | {"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} | 3.296875 | 3 | CC-MAIN-2016-44 | latest | en | 0.95657 |
https://www.physicsforums.com/threads/lhopitals-rule-lim-help.164091/ | 1,571,119,822,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986657586.16/warc/CC-MAIN-20191015055525-20191015083025-00106.warc.gz | 1,057,060,182 | 16,591 | # L'Hopital's Rule Lim Help
#### moloko
1. Homework Statement
lim as x->0 (tan(x)-x)/(sin2x-2x)
2. Homework Equations
L'Hopitals rule states that if the limit reaches 0/0, you can take the derivative of the top and the bottom until you get the real limit.
3. The Attempt at a Solution
(sec^2(x)-1)/(2cos2x-2) still 0/0
2sec^2(x)tan(x)/(-4sin(2x)) still 0/0
I have pain stakingly taken the derivative twice more and it simply does not seem to reach any end. All help is very much appreciated!
Related Calculus and Beyond Homework Help News on Phys.org
#### Dick
Homework Helper
You are going to get a nonzero denominator at the next derivative after you've shown. It's a cosine.
#### moloko
You are correct and this does give me a non 0/0 answer but the answer is wrong according to the book. It should be -1/4.
My next derivative is:
(2*(2sec^2(x)tan(x))*tan(x))/(-8cos(2x))
In simpler terms
(4sec^2(x)tan^2(x))/(-8cos(2x))
The numerator becomes 0 with the limit and the demoninator becomes -8, making the limit 0.
Thanks
#### Dick
Homework Helper
Why aren't you using the product rule on the numerator? What ARE you doing? The derivative of the tan(x) term will be nonzero.
#### moloko
I apologize, that was a careless mistake.
I'll post my solution if it's of any help to anyone...
the numerator becomes 2(sec^2(x)tan^2(x) + sec^2(x)),
which of course goes to 2 when pushed to 0
leaving 2/-8, or -1/4
Thank you so much!
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• Solo and co-op problem solving | 491 | 1,768 | {"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-2019-43 | longest | en | 0.900576 |
http://www.premierguitar.com/articles/print/constructing-lines-across-the-fretboard-1 | 1,432,649,238,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207928864.16/warc/CC-MAIN-20150521113208-00105-ip-10-180-206-219.ec2.internal.warc.gz | 656,339,810 | 7,515 | # Constructing Lines Across the Fretboard
November 23, 2007
Welcome back! Last lesson I showed you ways to play scales across the neck and I hope you had a chance to experiment with the concept. By practicing this way, you will begin to see that this is one of the best ways to master your fretboard. This lesson I will be showing you some lines that I constructed in this manner, playing across the neck. I will be using a couple scales we talked about and a couple we have not. For the purpose of this lesson, all of the examples I will be showing you are in G. This concept works in all keys - starting lower on the neck gives you a full range of the instrument. Be sure to try this concept with other scales.
Example 1)
Our first example is in G major (G, A, B, C, D, E, F#) and it is played using strict sextuplets, six notes per beat. You must accent on each downbeat; set your metronome at a slow speed initially and then gradually work up. This lick is great for moving across the fretboard quickly. Be sure to follow the fingerings provided and alternate pick every note.
Example 2)
This is another sextuplet line, but now we will use the G Phrygian scale (G, Ab, Bb, C, D, Eb, F). We will be moving across the neck ascending this time, but be sure to try this line descending. Even though this is played using all sextuplets, try to experiment accenting with other groupings - for example, 3 and 12 note groups. Again, the fingerings are extremely
Example 3)
This is a cool line using diatonic third intervals in G Harmonic minor (G, A, Bb, C, D, Eb, F#). The Harmonic minor scale is simply a minor scale with a raised 7th degree; you will notice it has a classical flavor. Practicing scales in thirds and with other intervals is a great way to master the fretboard. This one sounds great at high speeds with palm muting and plenty of
Example 4)
Our last example is based out of the G Melodic minor scale (G, A, Bb, C, D, E, F#). You could think of this scale as a major scale with a flat 3rd or a minor scale with a raised 6th and 7th. In the first bar of this line I use the flat 6th, which is the F note from the G natural minor scale. The first part of this line has a four-note motif descending in the G Melodic minor scale and then a four-note sequence in the last bar. This whole example is played with 16th notes, and will be most effective if you accent the downbeats.
These are just a few examples of how you can think in this manner. It''s up to you apply this to your playing and make up your own lines. We''ll see you next month!
Be sure to visit mikecampese.com for the latest CD''s and info. | 654 | 2,624 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2015-22 | latest | en | 0.941082 |
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Transcript
``` Speed
of light (in vacuum)
Foucault’s experiment
Speed
of light (in vacuum)
Michelson’s 1878 Rotating Mirror Experiment
• German American physicist A.A. Michelson realized, on putting
together Foucault’s apparatus, that he could redesign it for much
greater accuracy.
• Instead of Foucault's 60 feet to the far mirror, Michelson used
2,000 feet..
• Using this method, Michelson was able to calculate c = 299,792
km/s
• 20 times more accurate than Foucault
• Accepted as the most accurate measurement of c for the next 40
years.
Nature of light
Waves,
wave fronts, and rays
• Wave front: The locus of all adjacent points at which the phase of
vibration of a physical quantity associated with the wave is the same.
rays
source
wave fronts
spherical wave
plane wave
Reflection and refraction
Reflection and refraction
• When a light wave strikes a smooth interface of two transparent
media (such as air, glass, water etc.), the wave is in general partly
reflected and partly refracted (transmitted).
reflected rays
incident rays
r
a
a
b
b
refracted rays
b
a
Reflection and refraction
Reflection
• The incident, reflected, and refracted rays, and the normal to the
surface all lie in the same plane.
• The angle of reflection r is equal to
the angle of incidence a for all
wavelengths and for any pair of
material.
reflected rays
incident rays
r
a
a
r a
b
b
refracted rays
Reflection and refraction
Reflection (cont’d)
Reflection and refraction
Refraction
• The index of refraction of an optical material (refractive index), denoted
by n, is the ratio of the speed of light c in vacuum to the speed v in the
material.
wavelength in vacuum. Freq. stays the same.
n c / v ; 0 / n
f=v
Reflection and refraction
Refraction
• The index of refraction of an optical material (refractive index), denoted
by n, is the ratio of the speed of light c in vacuum to the speed v in the
material.
wavelength in vacuum. Freq. stays the same.
reflected rays
incident rays
n c / v ; 0 / n
• The ratio of the sines of the angles
a and b , where both angles are
measured from the normal to the
surface, is equal to the inverse ratio
of the two indices of refraction:
sin a nb
sin b na
r
a
a
b
b
Snell’s law
refracted rays
Example: depth of a swimming pool
Pool depth s = 2m
person looks straight
down.
2
1
L
the depth is judged by
the apparent size of
some object of length
L at the bottom of the
pool (tiles etc.)
na sin 1 sin 2
2
tan 1
L
s
tan 2
L
L
s s s '
s tan 1 ( s s ) tan 2
for small angles: tan ->sin
1
s sin 1 ( s s ) sin 2
s sin 1 ( s s ) na sin 1
L
s s
na 1
1
(2 m) 50 cm.
na
4
Example: Flat refracting surface
• The image formed by a flat
refracting surface is on the
same side of the surface as
the object
– The image is virtual
– The image forms between
the object and the surface
– The rays bend away from
the normal since n1 > n2
L
n1
n2
n2
s' s
s
s'
n1
| s ' | tan 1 L, | s | tan 2 L s' tan 1 s tan 2
tan sin for 1
s' sin 1 s sin 2
n1 s' n2 s ( n1 sin 1 n2 sin 2 )
s’
s
Total internal reflection
Total internal reflection
n2
sin 2 , sin 2 1 when n2 / n1 1& n2 n1 sin 1.
Since sin 1
n1
When this happens, 2 is 90o and 1 is called critical angle. Furthermore
when 1 crit , all the light is reflected (total internal reflection).
Total internal reflection
Optical fibers
Prism example
• Light is refracted twice – once entering and once leaving.
• Since n decreases for increasing , a spectrum emerges...
Analysis: (60 glass prism in air)
sin 1 = n2 sin 2
n2 sin 3 = sin 4
n1 = 1
60
Example: 1 = 30
1
2
3
sin(30 )
o
19 .5
1.5
2 sin 1
4
3 (60 o 2 ) 40 .5o
4 sin 1 1.5 sin 3 76 .9 o
n2 = 1.5
++60o = 180o
3 = 90 -
= 90 - 2
3 = 60 - 2
Prism
Applications of prism
• A prism and the total reflection
can alter the direction of travel
of a light beam.
• All hot low-pressure gases emit
their own characteristic spectra.
A prism spectrometer is used
to identify gases.
Diversion
Diversion
• The index of refraction of a material
depends on wavelength as shown
on the right. This is called diversion.
• It is also true that, although the speed
of light in vacuum does not depends
on wavelength, in a material, wave
speed depends on wavelength.
Diversion
Examples
Huygens’ principle
Huygens’ principle
Every point of a wave front may be considered the source of secondary
wavelets that spread out in all directions with a speed equal to the speed
of propagation of the wave.
Plane waves
Huygens’ principle (cont’d)
Huygens’ principle for plane wave
• At t = 0, the wave front is
indicated by the plane AA’
• The points are representative
sources for the wavelets
• After the wavelets have moved
a distance st, a new plane BB’
can be drawn tangent to the
wavefronts
Huygens’ principle (cont’d)
Huygens’ principle for spherical wave
Huygens’ principle (cont’d)
Huygens’ principle for spherical wave (cont’d)
• The inner arc represents part
of the spherical wave
• The points are representative
points where wavelets are
propagated
• The new wavefront is tangent
at each point to the wavelet
Huygens’ principle (cont’d)
Huygens’ principle for law of reflection
• The law of reflection can be
derived from Huygen’s
Principle
• AA’ is a wave front of incident
light
• The reflected wave front is CD
• Triangle ADC is congruent to
triangle AA’C
• Angles 1 = 1’
• This is the law of reflection
Huygens’ principle (cont’d)
Huygens’ principle for law of refraction
• In time t, ray 1 moves from A
to B and ray 2 moves from A’
to C
• From triangles AA’C and ACB,
all the ratios in the law of
refraction can be found:
n1 sin 1 = n2 sin 2
sin 1 v1t; sin 2 v2 t
v1t
v t
c
c
2 , v1 , v2
sin 1 sin 2
n1
n2
AC
Atmospheric Refraction and Sunsets
• Light rays from the sun are
bent as they pass into the
atmosphere
• It is a gradual bend because
the light passes through
layers of the atmosphere
– Each layer has a slightly
different index of
refraction
• The Sun is seen to be above
the horizon even after it has
fallen below it
Mirages
• A mirage can be observed
when the air above the
ground is warmer than the
air at higher elevations
• The rays in path B are
directed toward the ground
and then bent by refraction
• The observer sees both an
upright and an inverted
image
Example
Exercises
The prism shown in the figure has a refractive
index of 1.66, and the angles A are 25.00 . Two
light rays m and n are parallel as they enter
m
the prism. What is the angle between them
they emerge?
n
A
A
Solution
na sin a
1 1.66 sin 25.0
na sin a nb sin b b sin (
) sin (
) 44.6.
nb
1.00
Therefore the angle below the horizon is b 25.0 44.6 25.0 19.6,
and thus the angle between the two emerging beams is 39.2.
1
Exercises
Example
Light is incident in air at an angle
on the upper surface of a transparent
plate, the surfaces of the plate being
plane and parallel to each other. (a)
t
Prove that a a' . (b) Show that this
is true for any number of different parallel
plates. (c) Prove that the lateral displacement
D of the emergent beam is given by the
sin( a b' )
relation:
d t
,
a
n
Q
n’
n
b'
b P
a'
d
cos b'
where t is the thickness of the plate. (d) A ray of light is incident at an angle
of 66.00 on one surface of a glass plate 2.40 cm thick with an index of
refraction 1.80. The medium on either side of the plate is air. Find the lateral
Displacement between the incident and emergent rays.
Exercises
Problem
Solution
a
(a) For light in air incident on a parallel-faced
plate, Snell’s law yields:
n sin a n' sin b' n' sin b n sin a' sin a sin a' a a' .
n
t
n’
b'
Q
in the middle of the above equation that
P
n b
always cancel out. The requirement of
L
d
a'
parallel faces ensures that the angle n n'
and the chain of equations can continue.
(c) The lateral displacement of the beam can be calculated using geometry:
t
t sin( a b' )
d L sin( a ), L
d
.
cos b'
cos b'
'
b
(d)
n sin a
sin 66.0
) sin 1 (
) 30.5
n'
1.80
(2.40cm ) sin( 66.0 30.5)
d
1.62 cm.
cos 30.5
b' sin 1 (
```
Related documents | 3,265 | 8,538 | {"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} | 2.96875 | 3 | CC-MAIN-2019-51 | latest | en | 0.842667 |
https://issuu.com/sukilam2199/docs/m2_journal | 1,534,713,251,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221215393.63/warc/CC-MAIN-20180819204348-20180819224348-00166.warc.gz | 684,547,630 | 19,420 | Foundations of Design : REPRESENTATION, SEM1, 2018 M2 JOURNAL - FLATNESS vs PROJECTION 950315 Lam Shu Ki
Emmanuel Alexander Cohen Studio 7
1
WEEK 3 READING: LE CORBUSIER AND PURSIM
Question 1: What is Pictorial Space according to Le Corbusier? (Maximum 100 words) According to Le Corbusier, pictorial space is the area which ‘cannot be entered or circulated through’ and it is viewed from a distance of a personal perspective. It is eternally complied with frontality. With his paintings from mid and late 20s, pictorial space is composed by three elements, extended shape of flat objects to show the depth; adjustment of lines to fortify the continuity of edges; different color and texture manner to represent the special distance.
Question 2: The Flatness of Le Corbusier’s painting’s are attributable to two properties. What are they? And what are these pitted against?(Maximum 100 words) The first property is using black color. He insisted on adding white lead to his pigment to reinforce the depth. With case shadows, the black shape become an implied depth over the shadow area. The black color is juxtaposed against the white or light area leading to an extreme contrast between the object and ground. The second property is texture. The texture against the ground of contour which define the object’s edges, thus it only allows the isolated objects from the opposite sides of the distance.
2
MARIO’S WORLD
3
1ST MARIO’S WORLD
4
COMBINED MARIO’S WORLD
5
Question 1: Explain the difference between Pictoral (in this case perspectival) space and Projection? (Maximum 100 words) The main difference is the vanishing points, converging projections and picture planes to create a correct perspective. Perspective is limited to frontality while axonometric projection accomplish to create more extensive space. Axonometric projection represent objects through the use of the elements of measurements and transmissibility, maintaining the flexible actual measurements and geometric properties.
Question 2: Where did Axonometric projection first arise, and why? (Maximum 100 words)
From the article, axonometric projection was first used for military purposes, to provide exact measurements and dimension of the artillery projectiles. Then it was taught in engineering schools in 18th and 19th centuries shaping a closely relationship with mechanization and industrialization.
6
ILLUSTRATED MARIO’S NEW WORLD
The section of Mario World illustrates a mysterious and fantasy atmosphere. The satellite at right corner pours out different level of blue color, which later mixed together at the middle of the water. The satellite is formed by triangles with different brightness and darkness to make it more three-dimensional. To harmonize with the magical theme, a watercolour texture is added to the layer. I added cones to the some bushs and adjusted the angle of shadows.
7
APPENDIX
Combining two world into one isometric drawing
Annotation. Insert 3-4 images of your drawing in progress Draft of the isometric drawing
8
Equipment used
FOD:R M2 Journal
FOD:R M2 Journal | 706 | 3,158 | {"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} | 2.859375 | 3 | CC-MAIN-2018-34 | latest | en | 0.888249 |
http://mondofacto.com/facts/dictionary?tuple | 1,558,565,704,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256980.46/warc/CC-MAIN-20190522223411-20190523005411-00150.warc.gz | 139,980,717 | 4,800 | TUPLE computing dictionary
tupaia, tupaiid, tupaiidae, tupal, tupelo < Prev | Next > tuple, tuple calculus, Tuple Space Smalltalk
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tuple computing dictionary
In functional languages, a data object containing two or more components. Also known as a product type or pair, triple, quad, etc. Tuples of different sizes have different types, in contrast to lists where the type is independent of the length. The components of a tuple may be of different types whereas all elements of a list have the same type. Examples of tuples in Haskell notation are (1,2), ("Tuple",True), (w,(x,y),z). The degenerate tuple with zero components, written (), is known as the unit type since it has only one possible value which is also written ().
The implementation of tuples in a language may be either "lifted" or not. If tuples are lifted then (bottom,bottom) /= bottom and the evaluation of a tuple may fail to terminate. E.g. in Haskell:
f (x,y) = 1 --> f bottom = bottom f (bottom,bottom) = 1
With lifted tuples, a tuple pattern is refutable. Thus in Haskell, pattern matching on tuples is the same as pattern matching on types with multiple constructors (algebraic data types) - the expression being matched is evaluated as far as the top level constructor, even though, in the case of tuples, there is only one possible constructor for a given type.
If tuples are unlifted then (bottom, bottom) = bottom and evaluation of a tuple will never fail to terminate though any of the components may. E.g. in Miranda:
f (x,y) = 1 --> f bottom = 1 f (bottom,bottom) = 1
Thus in Miranda, any object whose type is compatible with a tuple pattern is assumed to match at the top level without evaluation - it is an irrefutable pattern. This also applies to user defined data types with only one constructor. In Haskell, patterns can be made irrefutable by adding a "~" as in
f ~(x,y) = 1.
If tuple constructor functions were strict in all their arguments then (bottom,x) = (x,bottom) = bottom for any x so matching a refutable pattern would fail to terminate if any component was bottom.
(03 Feb 2009)
tupaiid, tupaiidae, tupal, tupelo, TUPLE < Prev | Next > tuple calculus, Tuple Space Smalltalk, tupling
Bookmark with: word visualiser Go and visit our forums | 547 | 2,309 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2019-22 | latest | en | 0.894061 |
https://teacheradvisor.org/strategies/supportingContent_59 | 1,618,822,382,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038879305.68/warc/CC-MAIN-20210419080654-20210419110654-00528.warc.gz | 645,341,083 | 7,286 | Document
Excerpt 1: Commentary and Elaborations for MP.1 (Page 3)
Illustrative Mathematics
Standards for Mathematical Practice: Commentary and Elaborations for Grades K-5
Description: Explanations and examples of how the Standards for Mathematical Practice apply to Grades K-5. Mathematically proficient students start by explaining to themselves the meaning of a problem and looking for entry points to its solution. Young students might use concrete objects or pictures to show the actions of a problem, such as counting out and joining two sets to solve an addition problem. If students are not at first making sense of a problem or seeing a way to begin, they ask questions that will help them get started. Younger students might rely on using concrete objects or pictures to help conceptualize and solve a problem. Mathematically proficient elementary students may consider different representations of the problem and different solution pathways, both their own and those of other students, in order to identify and analyze correspondences among approaches.their answers to problems using a different method, and they continually ask themselves, "Does this make sense?" When they find that their solution pathway does not make sense, they look for another pathway that does.the approaches of others to solving complex problems and identify correspondences between different approaches. | 260 | 1,395 | {"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} | 3.15625 | 3 | CC-MAIN-2021-17 | latest | en | 0.927548 |
https://jp.mathworks.com/matlabcentral/cody/problems/42615-factorizing-a-number-into-a-given-number-of-factors/solutions/2042148 | 1,603,156,798,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107867463.6/warc/CC-MAIN-20201019232613-20201020022613-00316.warc.gz | 394,112,346 | 17,430 | Cody
# Problem 42615. Factorizing a number into a given number of factors
Solution 2042148
Submitted on 2 Dec 2019 by cokakola
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
n = 30; b = 1; A_correct = [ 30 ]; assert(isequal(LtdFactor(n,b),A_correct))
2 Pass
n = 3135; b = 2; A_correct = [ 1 3135 ; 3 1045 ; 5 627 ; 11 285 ; 15 209 ; 19 165 ; 33 95 ; 55 57 ]; assert(isequal(LtdFactor(n,b),A_correct))
3 Pass
n = 120; b = 3; A_correct = [ 1 1 120 ; 1 2 60 ; 1 3 40 ; 1 4 30 ; 1 5 24 ; 1 6 20 ; 1 8 15 ; 1 10 12 ; 2 2 30 ; 2 3 20 ; 2 4 15 ; 2 5 12 ; 2 6 10 ; 3 4 10 ; 3 5 8 ; 4 5 6 ]; assert(isequal(LtdFactor(n,b),A_correct))
4 Pass
n = 420; b = 4; A_correct = [ 1 1 1 420 ; 1 1 2 210 ; 1 1 3 140 ; 1 1 4 105 ; 1 1 5 84 ; 1 1 6 70 ; 1 1 7 60 ; 1 1 10 42 ; 1 1 12 35 ; 1 1 14 30 ; 1 1 15 28 ; 1 1 20 21 ; 1 2 2 105 ; 1 2 3 70 ; 1 2 5 42 ; 1 2 6 35 ; 1 2 7 30 ; 1 2 10 21 ; 1 2 14 15 ; 1 3 4 35 ; 1 3 5 28 ; 1 3 7 20 ; 1 3 10 14 ; 1 4 5 21 ; 1 4 7 15 ; 1 5 6 14 ; 1 5 7 12 ; 1 6 7 10 ; 2 2 3 35 ; 2 2 5 21 ; 2 2 7 15 ; 2 3 5 14 ; 2 3 7 10 ; 2 5 6 7 ; 3 4 5 7 ]; assert(isequal(LtdFactor(n,b),A_correct))
5 Pass
n = 2025; b = 3; A_correct = [ 1 1 2025 ; 1 3 675 ; 1 5 405 ; 1 9 225 ; 1 15 135 ; 1 25 81 ; 1 27 75 ; 1 45 45 ; 3 3 225 ; 3 5 135 ; 3 9 75 ; 3 15 45 ; 3 25 27 ; 5 5 81 ; 5 9 45 ; 5 15 27 ; 9 9 25 ; 9 15 15 ]; assert(isequal(LtdFactor(n,b),A_correct))
6 Pass
n = 210; b = 4; A_correct = [ 1 1 1 210 ; 1 1 2 105 ; 1 1 3 70 ; 1 1 5 42 ; 1 1 6 35 ; 1 1 7 30 ; 1 1 10 21 ; 1 1 14 15 ; 1 2 3 35 ; 1 2 5 21 ; 1 2 7 15 ; 1 3 5 14 ; 1 3 7 10 ; 1 5 6 7 ; 2 3 5 7 ]; assert(isequal(LtdFactor(n,b),A_correct))
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 1,020 | 1,873 | {"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-2020-45 | latest | en | 0.230274 |
https://www.phdata.io/blog/tableau-regression-models/ | 1,685,478,049,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224646144.69/warc/CC-MAIN-20230530194919-20230530224919-00493.warc.gz | 1,000,458,786 | 40,046 | # Tableau Regression Models
Tableau recently released the MODEL_QUANTILE() and MODEL_PERCENTILE() functions in Tableau Desktop 2020.3. These are game-changing for Tableau. They finally put regression output directly in the hands of Tableau developers. Specifically, MODEL_QUANTILE() allows any creator to calculate a predicted values AND store that value in Tableau. This also allows you to calculate the residual–the difference between the actual, and predicted value.
#### How it works
The MODEL_QUANITLE() function requires at leas three inputs:
• Â A quantile value where .5 is the predicted value from a regression model
• the outcome variable as an aggregate
• the predictors as an aggregate.
### Example #1: Adding a Regression Line
The MODEL_QUANITLE() function means we can add a regression line to a plot. Technically we can already do this. But let’s do it with a function.
Let’s start with this viz that is a plot of Profit Ratio by Weighted Discount by Manufacturer and Region (using Tableau’s default Superstore dataset).
You can quickly create create a mental model of the regression line. Its a downward parabola.
We can create this regression with MODEL_QUANTILE().
```// Model Values
MODEL_QUANTILE(
0.5, // Quantile
[Profit Ratio] // Outcome variable
[Discount - WT] // Predictor
) ```
This will create a linear regression.
Of course this is a linear regression, but we really need to fit a different model type. You can do this by editing the model in the calculation. Lets make the model parabolic by changing [Discount – WT] to [Discount – WT]^2.
```// Model Values
MODEL_QUANTILE(
0.5, // Quantile
[Profit Ratio] // Outcome variable
[Discount - WT]^2 // Predictor
)```
This changes the regression line:
We could add additional predictors to the model (which is the amazing part). Lets add Sales, Profit, and Quantity.
```// Model Values
MODEL_QUANTILE( 0.5, [Profit Ratio], [Discount - WT]^2, SUM([Sales]), SUM([Profit]), SUM([Quantity]))```
This results in a less than desirable plot (but potentially more accurate) line that looks like the following:
The predicted value line accurately represents the 50th percentile of the model (or the prediction). What’s better is compare these values against the predicted value.Â
This regression model minimizes the variability between actual values and the regression line. While it doesn’t guarantee it, a regression model should have the actual values evenly spaced above and below the regression line throughout the observed range. This is called homogeneity. It’s an assumption of a regression model (but not guaranteed). How can we test that in Tableau?
### Example #2: Residual Plot
A residual plot will show the observed value minus the expected value. Again, values should be evenly spaced above and below the regression line throughout the range of the plot. To calculate this we just need to create the following calculation:
```// Residuals
[Profit Ratio] - MODEL_QUANTILE(.5, [Profit Ratio], [Discount - WT], SUM([Sales]), SUM([Profit]), SUM([Quantity]))```
And results in the following plot:
A visual analysis of this regression, while it looked good initially would say that from 0.0 to 0.1 the models seems to over-estimate, while 0.1 to 0.5 there is an under-estimation, and 0.5 and higher there is again an over-estimation. This is the power of MODEL_QUANITLE()–we are able to analyze the quality of a regression model directly in Tableau and use it’s output in calculations!
## Accelerate and automate your data projects with the phData Toolkit
Data Coach is our premium analytics training program with one-on-one coaching from renowned experts. | 833 | 3,706 | {"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} | 2.703125 | 3 | CC-MAIN-2023-23 | longest | en | 0.773837 |
https://www.teacherspayteachers.com/Product/Place-Value-Games-Decimal-Place-Value-Comparing-Decimals-5th-Grade-2207015 | 1,632,532,491,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057584.91/warc/CC-MAIN-20210924231621-20210925021621-00602.warc.gz | 1,013,445,915 | 34,290 | DID YOU KNOW:
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# Place Value Games - Decimal Place Value - Comparing Decimals - 5th Grade
5th
Subjects
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Easel Activities Included
Some resources in this bundle include ready-to-use interactive activities that students can complete on any device. Easel by TpT is free to use! Learn more.
### Description
This bundle includes 5 different place value games for 5th grade. The resources are great for whole group, small group, partner work, or math stations.
The concepts covered in these place value games are described in detail below.
Rounding Decimals Game - 5th Grade
This partner game has students work with partners to create numbers from the hundreds to the thousandths place.
Then, the partner rolls a dice to find out which place value to round to. The players check one another's work and work to win the most points at the end of the sheet. You win a point by getting it correct or by stumping your partner.
Students practice rounding decimals while having fun!
Looking for a fun way to engage students and reinforce the following concepts:
-standard form (base ten form)
-unit form
-word form (number name)
-expanded form
This product includes 8 different BINGO boards of varying complexity to meet the needs of your learners. You can also use the calling cards in only one form or mix in all forms to make it more challenging.
Once a number is called, the students can find it on their board in a form different from what was called to practice the concept.
This product is aligned with the Engage NY & Eureka Math curriculum so the numbers are written in the format compatible with that curriculum.
5th Grade Standard Form, Expanded Form and Unit Form Partner Game
Are you looking for extra practice with standard form, expanded form, and unit form? This activity will give students extra practice in a low-stress, fun, and engaging way!
The students have a spinner and they use it to construct a decimal. After building their decimal, the students write their number in expanded and unit form. The responses can be checked by a partner or turned in for teacher review.
This would work great as a whole class activity or done in math stations or as center for math groups.
5th Grade Comparing Decimals to the Thousandths {Math Partner Game}
This file includes a spinner and a game recording sheet. The game pairs students together to battle for the largest decimal. The students use spinners to create the digits in their decimal and then they compare their two decimals.
The students record each battle and the winner on their recording sheet.
This provides students with low-stress practice with comparing decimals in a fun, engaging format!
5th Grade Exponents Practice Matching Game
In this game, each student is given a card related to an exponent with a base number of 10. The numbers range from 10 to 1,000,000. The students are tasked with finding their pairs.
The cards are then used to create a reference wall for students who need help with exponents.
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### Standards
to see state-specific standards (only available in the US).
Use place value understanding to round decimals to any place.
Compare two decimals to thousandths based on meanings of the digits in each place, using >, =, and < symbols to record the results of comparisons.
Read and write decimals to thousandths using base-ten numerals, number names, and expanded form, e.g., 347.392 = 3 × 100 + 4 × 10 + 7 × 1 + 3 × (1/10) + 9 × (1/100) + 2 × (1/1000).
Read, write, and compare decimals to thousandths.
Explain patterns in the number of zeros of the product when multiplying a number by powers of 10, and explain patterns in the placement of the decimal point when a decimal is multiplied or divided by a power of 10. Use whole-number exponents to denote powers of 10. | 1,070 | 4,954 | {"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} | 2.703125 | 3 | CC-MAIN-2021-39 | latest | en | 0.919759 |
http://realgl.blogspot.com/2012/05/quality-game.html | 1,527,044,809,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865411.56/warc/CC-MAIN-20180523024534-20180523044534-00289.warc.gz | 244,552,399 | 12,241 | ## 5/24/12
### quality game
Problem: say we have a million people, and we want to assess each person's skill at X, so that we know who to hire.
Solution: we're experimenting with a game with a purpose, the purpose of which is to determine people's skill.
For instance, for assessing English ability, we tried the following game: two users are shown three random words. Each user has 5 minutes to write a short passage using all three of the words in some meaningful way. Here is the interface:
Then three different people are asked to select "the most natural sounding paragraph". The passage with the most votes wins, and this user's score increases, while the loser's score decreases. Scores are updated according to the elo rating system, and they start out at 1500.
The words are chosen from a list of 850 "basic English words". We wanted common words that every English speaker would be likely to know, since we were less interested in testing people's vocabulary, and more interested in testing people's ability to use words together naturally.
We deployed a prototype on App Engine. For the first week, we tried to attract users to it by posting a job on oDesk offering \$20 to whoever had the highest score each night at midnight Pacific Time. We also released the link to some friends (note that we still paid the highest oDesk worker, even if one of our friends was actually at the top of the leader board).
For the words "able", "trick" and "sand", here are two entries from high scoring players:
I've never been able to trick my older sister into eating sand. I suppose it's just as well. Who knows what she would have done in retaliation!
The trick to building sand castles is getting the exact ratio between sand and water. Too little or too much sand, and the castle will be unable to stand.
For the words "cow", "star" and "living", here are two entries from lower scoring players:
There's a cow living in our barn. Just like a knight he wakes up all night only to watch the star that light.
Do you think that sometimes, somewhere, a lonely cow looks at the stars and wonders what's the point in living?
Incidentally, the right-hand passage won all 3 votes for that last contest. You can see all the data here. | 488 | 2,234 | {"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} | 2.703125 | 3 | CC-MAIN-2018-22 | latest | en | 0.973215 |
https://lindseyvan.com/35qb4c7b/fM7458mW/ | 1,603,507,175,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107881640.29/warc/CC-MAIN-20201024022853-20201024052853-00153.warc.gz | 414,521,927 | 8,103 | # 8th Math Cbse Exponents And Powers Worksheets Geometry Worksheet Angles Of Polygons Digit Addition
Maree June October 18, 2020 Math Worksheet
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First, the Basics! The x axis of a graph refers to the horizontal line while the y axis refers to the vertical line. Together these lines form a cross and the point where they both meet is called the origin. The value of the origin is always 0. So if you move your pencil from the origin to the right, you are drawing a line across the positive values of the x axis, i.e., 1, 2, 3 and so on. From the origin to the left, you’re moving across the negative values of the x axis, i.e., -1, -2, -3 and so on. If you go up from the origin, you are covering the positive values of the y axis. Going down from the origin, will take you to the negative values of the y axis.
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Oct 24, 2020 | 962 | 4,387 | {"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} | 4.53125 | 5 | CC-MAIN-2020-45 | latest | en | 0.946771 |
https://www.learnmathsonline.org/cbse-class-7-math/class-7-maths-chapter-7-congruence-of-triangles-mcq/ | 1,713,780,607,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818105.48/warc/CC-MAIN-20240422082202-20240422112202-00573.warc.gz | 767,617,759 | 18,284 | Class 7 Maths Chapter 7 | Congruence Of Triangles | MCQ
MCQ FOR CLASS 7 MATHEMATICS | CONGRUENCE OF TRIANGLES –Chapter 7
1. If two line segments have the same (equal) length, they are —————
A. Congruent
B. Similar
C. Opposite
2. Two angles are congruent if they have ————
A. equal measures
B. unequal measures
C. None of these.
3. Two circles are congruent if their —– are same.
A. Diameter
C. Centre
4. The three sides of one triangle are equal to the three corresponding sides of another triangle then the triangle is congruent by ———- criteria.
A. SSS Criteria
B. SAS criteria
C. RHS Criteria
5. Two sides and their included angles are equal to the corresponding sides and angles of another triangle then the triangles are congruent by ——— criteria.
A. SSS Criteria
B. SAS Criteria
C. RHS Criteria
6. Two angles and included side of one triangle are equal to corresponding angles and included side of another triangle, and then the triangles are congruent by ———-criteria.
A. ASA Criteria
B. SAS Criteria
C. SSS Criteria
7. The hypotenuse and one side of one right triangle are respectively equal to the hypotenuse and one side of another right triangle, then the triangle are congruent by ——– criteria.
A. SSS Criteria
B. RHS Criteria
C. SAS Criteria
8. If one of the acute angle of a right triangle is 45 degree, then the other acute angle is ————–
A. 90
B. 45
C. 30
9. Which angle is included between the sides XY and XZ of triangle XYZ?
A. X
B.Y
C.Z
10. Which of the following is not a congruence criterion?
A. SSS Criterion
B. AAA Criterion
C. SAS Criterion | 426 | 1,586 | {"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} | 3.890625 | 4 | CC-MAIN-2024-18 | latest | en | 0.861049 |
http://www.eecs.wsu.edu/~holder/courses/cse5311/lectures/l4/node18.html | 1,513,353,914,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948575124.51/warc/CC-MAIN-20171215153355-20171215175355-00455.warc.gz | 355,263,598 | 2,005 | Next: Medians and Order Statistics Up: l4 Previous: Analysis
Analysis
Thus, for all buckets,
Given:
• n elements, uniformly distributed over all possible values
• n buckets
What is the probability that an element will be inserted into some bucket i?
Given n trials consisting of putting elements into buckets, how many elements will be inserted into bucket i?
= Binomial(n,p)
with mean = np = 1
and variance = np(1-p) = 1 - 1/n
= 1 - 1/n +
= 2 - 1/n
=
Therefore, =
and = 4n + n + 3 = O(n) for the average case. In the worst case, the run time is .
Next: Medians and Order Statistics Up: l4 Previous: Analysis | 169 | 617 | {"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} | 3.609375 | 4 | CC-MAIN-2017-51 | latest | en | 0.85777 |
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1 / 9
# Ratio and Rate - PowerPoint PPT Presentation
Ratio and Rate. Section 5.1. Objectives. Write a fraction that shows a ratio comparison of two like measurements Write a fraction that shows a rate comparison of two unlike measurements Write a unit rate. Objective 1: Write a fraction that shows a ratio comparison of two like measurements.
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## PowerPoint Slideshow about 'Ratio and Rate' - gwen
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### Ratio and Rate
Section 5.1
Objectives
• Write a fraction that shows a ratio comparison of two like measurements
• Write a fraction that shows a rate comparison of two unlike measurements
• Write a unit rate
### Objective 1:Write a fraction that shows a ratio comparison of two like measurements
Objective 1
• Numbers can be compared by subtraction
• Can be compared by division
• Ratio is comparison by division
• Common ways to write ratio.
30:10
30 ÷ 10
30 to 10
• First number to be compared is first number written or is the numerator. Second number second or is the denominator.
Objective 1
• Ratios written as fractions can be simplify, if possible
• Like measurements have same unit of measure
• Units are dropped in ratio with like measurements
### Objective 2:Write a fraction that shows a rate comparison of two unlike measurements
Objective 2
• Fraction comparing unlike measurements is called a rate
• Rate can be simplified
• Units are NOT dropped
### Objective 3: Write a unit rate
Objective 3
• Unit rate – rate rewritten so denominator is 1-unit measurement
• To write a unit rate
• Divide
• Keep unlike units | 497 | 2,144 | {"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-2018-05 | latest | en | 0.891938 |
https://www.jiskha.com/display.cgi?id=1286317360 | 1,503,497,944,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886120573.0/warc/CC-MAIN-20170823132736-20170823152736-00570.warc.gz | 927,168,541 | 3,981 | # Math
posted by .
I determined that 5 divided by 8 equals 0.625; however, I cannot determine how the book arrived at 0.625 equaling the fraction 4 5/8. My conversion of 0.625 is fraction 5/8.
• Math -
.625 does not equal 4 5/8, 4 5/8 equals 4.625
## Similar Questions
1. ### math
what is the fraction of 0.625 and 0.952
2. ### mathematics
please if you kow the answer to this help me :) Changeing -1.625 into a fraction solving 4/5 divided by 1/3 2/3 x 1/4 - 5/6 it would be a great HELP
3. ### calculus
s=-4.9t²+245t+14.7 s=-4.9(t²-245/4.9t)+14.7 s=-4.9(t²-50t)+14.7 s=-4.9(t²-50+625-625)+14.7 s=-4.9(t²-50+625)+(-4.9)(-625)+14.7 s=-4.9(t-25)²+30625.5+14.17 s=-4.9(t-25)²+3077.2 as you can see on the process of the problem on …
4. ### Math
how do you convert 0.625 into a fraction
5. ### Algebra
The technology cewnter bought a fax machine for \$625.00 and it depreciates at a rate of \$25.00 a month. Find a function(f) that can be used to determine the value of the fax amchine (t) months after purchase. Would it be F(t)=25(t)-625
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Write each percent as a decimal and as a fraction in simplest form. 5/8% would be .625% how do I make 5/8% into a fraction, or do I make .625% into a fraction.
7. ### Math
Explain how knowing that that 5 divided by 8 equals = .625 helps you write the decimal for 4 5/8
8. ### maths
The no.wich Is 4 more than the square of 625 .....is exactly 2 prime factors.find them..... I am trying lyk this But answer is not coming X^2= 4+625^2
9. ### Mathematics
John is choosing a password for his access to the internet. He decided not to use the digit 0 or the letter M. Each letter or number may be used more than once. How many passwords of 2 letters followed by 4 digits are possible?
10. ### Economics(Help)
Nancy is indifferent between a gamble that pays \$625 with a probability of 20% and \$100 with a probability of 80%, versus a sure payment of \$269. She is also indifferent between another gamble that pays \$269 with a probability of 50% …
More Similar Questions | 655 | 2,032 | {"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-2017-34 | latest | en | 0.879193 |
https://brainly.in/question/198501 | 1,484,884,084,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280774.51/warc/CC-MAIN-20170116095120-00290-ip-10-171-10-70.ec2.internal.warc.gz | 801,964,645 | 11,249 | # Obtain all the zeroes of the polynomial X^4 - 3√2 x³ + 3x² +3 √2x - 4 = 0 , if two of it's zeroes are √2 and 2√2 ... Guys , Solve out !!!
1
by Galaxy
Let me try.....
is x in root here 3 √2x
it is not possible to solve bcoz it can't be solved here
but if u say i can solve it on paper and upload the file
Yeah I have also solved one question the same way and uploaded.You can do so
2015-10-01T21:29:00+05:30
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
P(x) = x⁴ - 3 √2 x³ + 3 x² + 3√2 x - 4
given that (x - √2) , (x - 2√2) are factors of P(x) as √2 and 2√2 are two zeroes of P(x) = 0.
(x -√2) (x - 2√2) = x² -3√2 x + 4 is a factor of P(x). let A be a constant. We can write the constant term in the second factor by : -4/4 = -1... dividing the constant terms.
let (x² - 3√2 x + 4) (x² + A x -1 ) = P(x)
Now compare the coefficients of x³ : A - 3√2 = -3√2 => A = 0
coefficient of x : 4A + 3√2 = 3√2 => A = 0
so the other factors are : x² - 1 = 0
so x = 1 and -1 are the other factors.
This method of multiplying the factors and comparing coefficients is simple.
click on thanks button above pls;;; select best answer
You're answer is always best !
There is a simple method. I overlooked....
sum of all four roots of P(x) is the - of coefficient of x cube. so it is 3√2. The sum of given two roots is 3 √2. So sum of two more roots is 0.
Now product of four roots is the constant term = -4... product of two given roots is 4. so product of two more roots is -4/4 = -1.....
Hence the polynomial is simple : x² - (0) x + (-1)
roots of this are +1 and -1. | 615 | 1,854 | {"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} | 4.1875 | 4 | CC-MAIN-2017-04 | latest | en | 0.904299 |
https://science.blurtit.com/307259/if-there-are-365-days-in-a-year-why-is-it-if-you-multiply-52-weeks-x-7-days-your-total-is | 1,726,827,183,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652246.93/warc/CC-MAIN-20240920090502-20240920120502-00407.warc.gz | 454,559,561 | 17,987 | # If There Are 365 Days In A Year. Why Is It If You Multiply 52 Weeks X 7 Days Your Total Is 364?
Your math is correct. There are not exactly 52 weeks per year, it is only an approximation. All the calendar systems, including our Gregorian calendar, have had to adjust out some inaccuracies of fitting our traditional use of a week to the solar cycle, and some are better than others.
According to Wikipedia, "In a Gregorian mean year there are exactly 365.2425 days, and thus exactly 52.1775 weeks" The Julian calendar that was previously used had "365.25 days or 52528 weeks." And once you understand there is no modulo relationship of a week to a year, then it makes sense why our months (and years) end up starting on a different days of the week.
In both systems, there are about ~0.25 days extra per year, which accounts for the extra day in February during a leap year.
Even after observing leap years, there are also leap seconds they occasionally have to adjust out because of the Earth's orbital decay and tidal forces. See en.wikipedia.org
thanked the writer. | 256 | 1,084 | {"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} | 3.265625 | 3 | CC-MAIN-2024-38 | latest | en | 0.948924 |
https://sdrta.net/12-is-15-of-what-number/ | 1,638,513,388,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362605.52/warc/CC-MAIN-20211203060849-20211203090849-00099.warc.gz | 562,878,760 | 6,122 | ## Step by step solution for calculating 12 is 15 percent the what number
We currently have our an initial value 12 and the second value 15. Let"s assume the unknown worth is Y i m sorry answer us will discover out.
You are watching: 12 is 15 of what number
As we have all the forced values we need, now we have the right to put lock in a basic mathematical formula together below:
STEP 112 = 15% × Y
STEP 212 = 15/100× Y
Multiplying both political parties by 100 and also dividing both sides of the equation through 15 we will certainly arrive at:
STEP 3Y = 12 × 100/15
STEP 4Y = 12 × 100 ÷ 15
STEP 5Y = 80
Finally, us have found the value of Y which is 80 and that is ours answer.
You can quickly calculate 12 is 15 percent the what number by using any type of regular calculator, simply go into 12 × 100 ÷ 15 and you will acquire your answer i beg your pardon is 80
Here is a percent Calculator come solve comparable calculations such together 12 is 15 percent that what number. You have the right to solve this type of calculation v your values by beginning them right into the calculator"s fields, and also click "Calculate" to obtain the an outcome and explanation.
is
percent of what number
Calculate
## Sample questions, answers, and how to
Question: her friend has actually a bag that marbles, and also he speak you the 15 percent the the marbles room red. If there space 12 red marbles. How many marbles does he have altogether?
How To: In this problem, we know that the Percent is 15, and we are also told that the component of the marbles is red, so we know that the component is 12.
So, that way that it must be the total that"s missing. Right here is the means to number out what the complete is:
Part/Total = Percent/100
By making use of a an easy algebra we can re-arrange our Percent equation favor this:
Part × 100/Percent = Total
If us take the "Part" and multiply that by 100, and then we division that through the "Percent", us will get the "Total".
Let"s try it the end on our problem about the marbles, that"s very straightforward and it"s just two steps! We know that the "Part" (red marbles) is 12.
So step one is to simply multiply that part by 100.
12 × 100 = 1200
In step two, us take that 1200 and also divide it by the "Percent", which we are told is 15.
So, 1200 separated by 15 = 80
And that way that the total number of marbles is 80.
Question: A high school marching band has 12 flute players, If 15 percent of the band members pat the flute, climate how countless members space in the band?
Answer: There are 80 members in the band.
How To: The smaller "Part" in this problem is 12 because there space 12 flute players and we room told that they consist of 15 percent of the band, for this reason the "Percent" is 15.
Again, it"s the "Total" that"s absent here, and to discover it, we simply need to follow our 2 step procedure together the vault problem.
For action one, we multiply the "Part" through 100.
12 × 100 = 1200
For action two, we division that 1200 by the "Percent", i beg your pardon is 15.
See more: Explain Why Vision Is Lost When Light Hits The Blind Spot, Special Senses: Visual Tests And Experiments
1200 separated by 15 equals 80
That way that the total variety of band members is 80.
## Another step by step method
Step 1: Let"s assume the unknown value is Y
Step 2: very first writing the as: 100% / Y = 15% / 12
Step 3: drop the portion marks to leveling your calculations: 100 / Y = 15 / 12
Step 4: main point both political parties by Y to relocate Y ~ above the ideal side the the equation: 100 = ( 15 / 12 ) Y
Step 5: simplifying the ideal side, us get: 100 = 15 Y
Step 6: separating both political parties of the equation by 15, we will arrive at 80 = Y
This leaves us v our final answer: 12 is 15 percent the 80
12 is 15 percent that 80 12.01 is 15 percent of 80.067 12.02 is 15 percent of 80.133 12.03 is 15 percent the 80.2 12.04 is 15 percent that 80.267 12.05 is 15 percent that 80.333 12.06 is 15 percent that 80.4 12.07 is 15 percent that 80.467 12.08 is 15 percent that 80.533 12.09 is 15 percent of 80.6 12.1 is 15 percent that 80.667 12.11 is 15 percent of 80.733 12.12 is 15 percent of 80.8 12.13 is 15 percent the 80.867 12.14 is 15 percent of 80.933 12.15 is 15 percent of 81 12.16 is 15 percent the 81.067 12.17 is 15 percent the 81.133 12.18 is 15 percent the 81.2 12.19 is 15 percent that 81.267
12.2 is 15 percent that 81.333 12.21 is 15 percent the 81.4 12.22 is 15 percent of 81.467 12.23 is 15 percent that 81.533 12.24 is 15 percent of 81.6 12.25 is 15 percent that 81.667 12.26 is 15 percent of 81.733 12.27 is 15 percent the 81.8 12.28 is 15 percent that 81.867 12.29 is 15 percent of 81.933 12.3 is 15 percent the 82 12.31 is 15 percent the 82.067 12.32 is 15 percent of 82.133 12.33 is 15 percent of 82.2 12.34 is 15 percent that 82.267 12.35 is 15 percent the 82.333 12.36 is 15 percent the 82.4 12.37 is 15 percent of 82.467 12.38 is 15 percent that 82.533 12.39 is 15 percent that 82.6
12.4 is 15 percent that 82.667 12.41 is 15 percent of 82.733 12.42 is 15 percent the 82.8 12.43 is 15 percent that 82.867 12.44 is 15 percent that 82.933 12.45 is 15 percent that 83 12.46 is 15 percent the 83.067 12.47 is 15 percent that 83.133 12.48 is 15 percent of 83.2 12.49 is 15 percent the 83.267 12.5 is 15 percent the 83.333 12.51 is 15 percent that 83.4 12.52 is 15 percent the 83.467 12.53 is 15 percent the 83.533 12.54 is 15 percent that 83.6 12.55 is 15 percent the 83.667 12.56 is 15 percent that 83.733 12.57 is 15 percent of 83.8 12.58 is 15 percent the 83.867 12.59 is 15 percent that 83.933
12.6 is 15 percent of 84 12.61 is 15 percent the 84.067 12.62 is 15 percent of 84.133 12.63 is 15 percent the 84.2 12.64 is 15 percent the 84.267 12.65 is 15 percent the 84.333 12.66 is 15 percent that 84.4 12.67 is 15 percent of 84.467 12.68 is 15 percent of 84.533 12.69 is 15 percent of 84.6 12.7 is 15 percent of 84.667 12.71 is 15 percent the 84.733 12.72 is 15 percent the 84.8 12.73 is 15 percent of 84.867 12.74 is 15 percent that 84.933 12.75 is 15 percent the 85 12.76 is 15 percent that 85.067 12.77 is 15 percent the 85.133 12.78 is 15 percent that 85.2 12.79 is 15 percent of 85.267
12.8 is 15 percent of 85.333 12.81 is 15 percent that 85.4 12.82 is 15 percent that 85.467 12.83 is 15 percent the 85.533 12.84 is 15 percent of 85.6 12.85 is 15 percent the 85.667 12.86 is 15 percent the 85.733 12.87 is 15 percent that 85.8 12.88 is 15 percent that 85.867 12.89 is 15 percent the 85.933 12.9 is 15 percent the 86 12.91 is 15 percent the 86.067 12.92 is 15 percent that 86.133 12.93 is 15 percent of 86.2 12.94 is 15 percent the 86.267 12.95 is 15 percent the 86.333 12.96 is 15 percent of 86.4 12.97 is 15 percent the 86.467 12.98 is 15 percent of 86.533 12.99 is 15 percent of 86.6 | 2,285 | 6,834 | {"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} | 4.5625 | 5 | CC-MAIN-2021-49 | longest | en | 0.95003 |
http://www.nelson.com/mathfocus/grade3/parent/surf/ch8_7.html | 1,550,859,716,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247522457.72/warc/CC-MAIN-20190222180107-20190222202107-00255.warc.gz | 383,997,944 | 2,969 | Math Focus 3
Student Centre • Surf for More Math • Try It Out • Web Quests
## CHAPTER 8: MULTIPLICATION
### Lesson 7: Solving Problems by Making a Model
Goal
Use a model to solve a multiplication problem.
Builds Upon
Student Book pages 198–199
Instructions for Use
Rectangle Multiplication displays arrays of squares on a grid to help your child model and solve multiplication problems.
To change the array shown to model another multiplication problem (or array), click and drag the black square shapes at the left side and bottom of the screen to other positions (numbers).
The Mad Math Game - Multiplication provides your child with multiplication problems that can be solved by using arrays.
Click “Start,” then read the multiplication problem. You may wish to model the problem using arrays on paper, using counters, or using Rectangle Multiplication. Type your answer in the white box, then click “Total” to check your answer. | 196 | 944 | {"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} | 2.53125 | 3 | CC-MAIN-2019-09 | latest | en | 0.818077 |
https://community.fabric.microsoft.com/t5/Desktop/Obtain-the-second-result-for-the-same-ID/m-p/4000175 | 1,721,702,297,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517931.85/warc/CC-MAIN-20240723011453-20240723041453-00595.warc.gz | 166,520,916 | 52,300 | cancel
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Frequent Visitor
## Obtain the second result for the same ID
Hello,
I have a problem that I don't think is very complicated, but I can't seem to solve it.
My data table lists the quality control information carried out in my company. Each inspection is carried out twice by two different people, each with a different job (an 'Rédacteur' and a 'Vérificateur'). These two checks are linked by the same 'ControlId'.
Each of the checks gives a value of either 'Conforme', 'Non conforme' or 'Non applicable', although only the 'Vérificateur' can give a value of 'Non conforme'.
I'd like to know if it's possible to retrieve the agency of the person who is the 'Rédacteur' for whom a 'Vérificateur' has indicated 'Non conforme', which is inside 'Créé par.title', with the aim of obtaining the number of controls for each of the 'Rédacteur' in a matrix, for a specific date filter which I already placed in the report.
If I haven't been clear enough about what I want to do, please don't hesitate to ask me for more information, as I'm not bilingual in English 😅.
1 ACCEPTED SOLUTION
Solution Sage
Create a calculated column to identify 'Rédacteur'
``````IsNonConforme =
IF (
'QualityControl'[Result] = "Non conforme" && 'QualityControl'[Role] = "Vérificateur",
1,
0
)``````
Create a calculated column to link 'Rédacteur' with 'Non conforme'
``````RédacteurAgencyForNonConforme =
CALCULATE (
MAX ( 'QualityControl'[Agency] ),
FILTER (
'QualityControl',
'QualityControl'[ControlId] = EARLIER ( 'QualityControl'[ControlId] ) &&
'QualityControl'[Role] = "Rédacteur" &&
'QualityControl'[IsNonConforme] = 1
)
)``````
Create a measure to count the controls for each 'Rédacteur'
``````CountNonConformeControlsByRédacteur =
CALCULATE (
COUNTROWS ( 'QualityControl' ),
FILTER (
'QualityControl',
'QualityControl'[Role] = "Rédacteur" &&
'QualityControl'[RédacteurAgencyForNonConforme] <> BLANK ()
)
)``````
3 REPLIES 3
Community Support
Hi @SansNom ,
Did @aduguid response solve your problem? If solved, please mark the helpful answer as a solution this will help more people, if not please provide some sample data in your tables (exclude sensitive data) with Text format and your expected result with backend logic and special examples. It is better if you can share a simplified pbix file. Thank you.
Best Regards,
Neeko Tang
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Solution Sage
Create a calculated column to identify 'Rédacteur'
``````IsNonConforme =
IF (
'QualityControl'[Result] = "Non conforme" && 'QualityControl'[Role] = "Vérificateur",
1,
0
)``````
Create a calculated column to link 'Rédacteur' with 'Non conforme'
``````RédacteurAgencyForNonConforme =
CALCULATE (
MAX ( 'QualityControl'[Agency] ),
FILTER (
'QualityControl',
'QualityControl'[ControlId] = EARLIER ( 'QualityControl'[ControlId] ) &&
'QualityControl'[Role] = "Rédacteur" &&
'QualityControl'[IsNonConforme] = 1
)
)``````
Create a measure to count the controls for each 'Rédacteur'
``````CountNonConformeControlsByRédacteur =
CALCULATE (
COUNTROWS ( 'QualityControl' ),
FILTER (
'QualityControl',
'QualityControl'[Role] = "Rédacteur" &&
'QualityControl'[RédacteurAgencyForNonConforme] <> BLANK ()
)
)``````
Frequent Visitor
Thank you, I'm gonna try this now ! | 948 | 3,496 | {"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} | 2.53125 | 3 | CC-MAIN-2024-30 | latest | en | 0.808029 |
https://www.piday.org/calculators/area-of-a-rectangle-calculator/ | 1,561,411,171,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999740.32/warc/CC-MAIN-20190624211359-20190624233359-00034.warc.gz | 863,246,670 | 29,925 | Length(a)
Width(b)
Area(A)
Perimeter(P)
Diagonal(d)
Go back to Calculators page
Area of a Rectangle Calculator
If you need to find the area and perimeter of a rectangle, this calculator is the handy tool you will need.
By simply inputting the length and width, this calculator will almost instantly find the perimeter (P) and the area (A).
If interested in calculators for a variety of other shapes, you can look at more of our handy calculators. But you can stick around here and learn more about finding the area of a rectangle.
A rectangle has four 90 degree angles. If the lengths of the sides are all the same, then the rectangle is also a square. The lengths of the sides will be given as a,b or you can use l and w for "length" and "width". The diagonal, which goes from one vertex to the opposite vertex cutting the rectangle into two squares, is called the diaganoal and noted as d.
Here are the basic formulas used by the calculator.
Area = a(b) and noted as (A)
Perimeter (distance around the outside of the rectangle) = a + a + b + b or 2a + 2b and noted as (P)
Diagonal is d² = a² + b² which is the Pythagorean Theorem (see our Pythagorean Theorem calculator). Note that to solve for d, take the square root of both sides of the =.
Example of calculating area of a rectangle.
Suppose the length is a = 6 inches
and the width is b = 4 inches
A = a*b , so A = 6(4) = 24 inches²
Using the same dimensions, we can calculate the perimeter.
The perimeter is 2a + 2b, so in this example the perimeter P = 2(6)+2(4) = 20 inches
Next to find the diagnoal using the same dimensions
d² = 6² + 4² = 36 + 16 = 52
Take the square root of both sides and the diagonal d is approximately 7.2 inches
These examples illustrate how to calculate manually, but if you prefer to use the calculator for quicker results or to merely check your work, then feel free to do so. A great feature of the calculator is that you can find either the length of the width if you know the perimeter and the one of the two dimensions.
AREA, PERIMETER, and DIAGONAL of a RECTANGLE
This page will show you how to deal with measuring the area of a rectangle. We will cover the following topics:
What is the area, perimeter, and diagonal of a rectangle?
How to calculate the area, perimeter, and diagonal of a rectangle?
Real-life application in calculating the area, perimeter, and diagonal of a rectangle
Area of a Rectangle
Imagine the Area of a rectangle as boxes inside a rectangle. The rectangle below has a covered area of 12 "square units"
1 2 3 4 5 6 7 8 9 10 11 12
The space inside a two-dimensional shape is the area or the amount of covered shape.
This diagram displays the Width, Length, and Area of a rectangle:
Calculating the Area of a Rectangle
To find the area of a rectangle you need to multiply the Length and the Width of a rectangle. We can get the Area of a rectangle by following this formula:
A = L * W
A is the Area, L for the length, and W for the width.
Example 1
Calculate the Area of a rectangle that has a length of 7 centimeters and a width of 5 centimeters.
Formula:
A = L * W
A = 35. The given length(L) is 7 and 3 for the width(W). When multiplied, you will get 35 as your Area.
Perimeter of a Rectangle
Take a look at the image below, a person is walking around the box. The path he walks around from the starting point and back is the Perimeter. Knowing the length and width of a rectangle, we can now get the Perimeter of a rectangle. Both opposite sides of a rectangle are congruent which means adding those sides, we can calculate for the Perimeter.
Calculating the Perimeter of a Rectangle
By adding all the sides of a rectangle, we can now get the Perimeter. Here's the equation to get the Perimeter of a rectangle:
P = L+W+L+W
Since we know that both opposite sides of a rectangle are identical, we can simplify the equation by using this equation:
P = 2L+2W
Example 1
Find the perimeter of a rectangle that has a length of 12 centimeters and a width of 7 centimeters.
Formula:
P = L + W + L + W or
P = 2L + 2W
P = 12 + 7 + 12 + 7 or
P = 2(12) + 2(7)
The answer is P = 38. By adding 12(L) + 7(W) + 12(L) + 7(W), you will get 38. By multiplying the Length(L) and Width(W) by 2 then adding the quotients, you will get the same answer.
Diagonal of a Rectangle
If you look closely, a rectangle is a combination of two right angles. A Diagonal is the division of a rectangle to form two right triangles identical to one another.
Calculating the Diagonal of a Rectangle
We know that a rectangle is a combination of two right triangles. The diagonal of this rectangle is the hypotenuse of the two triangles which is why we can apply the Pythagoras' Theorem to solve for the diagonal of a rectangle. For our formula, use:
D=√W2+L2
EXAMPLE 1
Solve for the rectangle's diagonal length whose sides are 15cm for the Length and 8cm for the width.
Formula:
D=√W²+L²
D=√15²+8²
D=√289
D= 17
The diagonal of the rectangle is 17. Square the values of Width(W) and Length(L), then get the sum of the two squares. Knowing the sum of the two squares, find the square root of the sum.
EXAMPLE 2
Calculate for the diagonal length of a 3ft long and 4ft wide rectangle.
D=√5²+3²
D=√34
D= 5.83
Real-life Applications In Getting the Area of a Rectangle
A newly wed couple wants to install tiles to the floor of the master bedroom. The room has a length of 20ft and a width of 30ft. The tile they chose has a length of 24in and 36in for the width. Solve for the number of tiles needed to fill the master bedroom.
Tips:
1. Solve for the area occupied by the master bedroom.
2. Calculate the area of a tile
3. Choose a unit to be used. This example will use feet.
Solution:
A = L * W
A = 20ft x 30ft
A = 600ft²
Tip:
Before we get the area of each tile, convert the feet to inches
Sample Conversion:
1 foot = 12 inches
2 feet = 24 inches
3 feet = 36 inches
Area of 1 tile = L × W
Area of 1 tile = 2 × 3
Area of 1 tile = 6ft²
This means that each tile has an area of 6ft² and it needs to cover the area of the room which is 600ft². So, 6 × 100 = 600.
The number of tiles needed to fill the master bedroom is 600.
Real-life Applications In Getting the Perimeter of a Rectangle
A farmer wants to add a cage for his chickens. He wants to add a new set of fence near his house and a vacant space with the length of 30 meters and width of 16 meters. Find the perimeter of the vacant space.
Solution:
P = 2L + 2W
Perimeter = 2(Length of the vacant space) + 2(Width of the vacant space)
P = 2(30) + 2(16)
P = 60 + 32
P = 92
The Perimeter of the vacant space is 92 meters.
Real-life Applications In Getting the Diagonal of a Rectangle
A sandwich has a length of 12 inches and 8 inches for the width. If you want to equally divide it into two right angles, what would be the length of the diagonal?
Solution:
Equally divide the sandwich into right triangles by using the Pythagoras' Theorem:
D=√W2+L2 | 1,852 | 6,983 | {"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-2019-26 | longest | en | 0.911034 |
https://www.abberley.worcs.sch.uk/page/?title=Wednesday+15th+July&pid=541 | 1,713,474,560,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817239.30/warc/CC-MAIN-20240418191007-20240418221007-00422.warc.gz | 572,921,876 | 10,458 | # Wednesday 15th July
Hello Wrens
I hope you are feeling full of beans this morning. Let's get going with our learning.
English.
If you didn't manage to finish your Superhero comic strip yesterday you can continue working on this. If you did finish have a go at one of the phonics activities attached below.
Maths.
Reception.
This week we are holding our Key Stage 1 Sports' Day so we are going combine some maths with some sports' day practise in our Mathletics Challenges! You will need a timer, tape measure and a pencil and paper.
Challenge 1 - Warm up activities. Use your timer to see how many star jumps you can do in 30 seconds. Repeat the activity with other exercises like squats, burpees, lunges, frog jumps, sit ups, press ups.
This is really good for careful counting. You could even add the numbers together to see how many exercises you have done altogether.
Vary it by introducing ball skills. How many bounces or catches can you do in 30 seconds?
Challenge 2 - Have a running race with members of your family. Who was the quickest? Can you order yourselves from fastest to slowest? If we were the fastest we say we came first (1st). These are our ordinal numbers - do you know what comes next?
Try some other races - you could do skipping, hopping or jumping?
Try some longer distance running. Can you time yourself to see how long it takes you to run 5 laps of your garden?
Challenge 3 - How far can you throw a bean bag or ball? Practise throwing your bean bag/ball and use your tape measure to see whose went the furthest.
You could vary this challenge by making it a target throwing challenge. Arrange some targets with different points on them (hoops work really well). Take it turns to throw your bean bag/ball into the hoops. Who can score the most points?
This is really good practise for counting on. You could even write the addition number sentence.
Challenge 4 - How far can you jump? Try the long jump. Who can jump the furthest? Don't forget to use your tape measure to check.
Mickey Thompson and Purvis certainly enjoyed these activities - they were exhausted afterwards! Don't forget to drink plenty of water. I have also attached a rosette template below so you can make your own ordinal number rosettes.
Year 1.
Let's go on an outdoor maths treasure hunt. Print off the cards below and spread them around your garden. As you find each card see if you can complete the maths challenge.
Parents - there are three different levels of challenge indicated by the number of stars at the bottom of the card. Choose the ones which are the most appropriate level for your child and the ones that look the most fun. You can either choose a level or a selection of cards from each level.
Have fun - Mickey Thompson and Purvis certainly enjoyed this activity!
Topic.
Today the children in school are having an afternoon at Forest School. We will be hunting mini beasts, building dens and using our new rope swing. We might even light a camp fire and toast some marshmallows.
Why don't you have a Forest School afternoon at home? We would love to see some photographs of what you get up to.
Have fun
Mrs Lightfoot (and Purvis and Mickey Thompson) | 726 | 3,222 | {"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} | 3.46875 | 3 | CC-MAIN-2024-18 | latest | en | 0.940316 |
https://maddyzphysics.com/2018/11/goodnight-kepler/ | 1,555,809,585,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578530100.28/warc/CC-MAIN-20190421000555-20190421022555-00283.warc.gz | 478,810,104 | 81,804 | # ‘Goodnight’ Kepler
Last Updated on
One of the most popular mission of NASA, Kepler Space telescope has come to an end!
Kepler is named after the German Astronomer and Mathematician, Johannes Kepler. In the seventeenth century, Johannes Kepler put forward his laws of planetary motion. These are three laws describing the movement of a planet around the Sun.
## THE ‘GOODNIGHT’
NASA launched Kepler on March 7, 2009.
On the evening of November 15 (2018), NASA’s Kepler space telescope received its final set of commands to disconnect communications with Earth.
The “goodnight” commands finalize the spacecraft’s transition into retirement, which began on Oct. 30 with NASA’s announcement that Kepler had run out of fuel.
It is coincidence that Kepler’s ‘goodnight’ was on the death anniversary of Johannes Kepler exactly after 388 years of Johannes Kepler’s death!
The spacecraft is now drifting in a safe orbit around the Sun 94 million miles away from Earth.
The data Kepler collected over the course of more than nine years in operation will be mined for exciting discoveries for many years to come.
## MORE INFORMATION ABOUT KEPLER
### GOAL
NASA started the mission to survey our region of the Milky Way galaxy to discover hundreds of Earth-size and smaller planets in the habitable zone, and determine the fraction of the hundreds of billions of stars in our galaxy that might have such planets.
The mission also aimed to—
• Determine the percentage of terrestrial and larger planets that are in or near the habitable zone of a wide variety of stars
• Determine the distribution of sizes and shapes of the orbits of these planets
• Estimate how many planets there are in multiple-star systems
• Determine the variety of orbit sizes and planet reflectivities, sizes, masses and densities of short-period giant planets
• Identify additional members of each discovered planetary system using other techniques
• Determine the properties of those stars that harbor planetary systems.
### TRANSIT METHOD OF DETECTING PLANETS
When a planet crosses in front of its star as viewed by an observer, the event is called a ‘transit‘.
Planets block some light of the star according to its size, when the planet passes in front of the star, and this changes the brightness of the star.
Kepler used this method to detect different star systems and planets in the galaxy.
A Transit.
Source – NASA
## REFERENCES
1. Chen, Rick. Kepler Space Telescope Bid ‘Goodnight’ With Final Set of Commands | NASA. Last Updated November 16, 2018. https://www.nasa.gov/feature/ames/kepler-space-telescope-bid-goodnight-with-final-set-of-commands (accessed November 17, 2018).
2. Johnson, Michele, and Brian Dunbar. Mission overview | NASA. Last Updated October 31, 2018. https://www.nasa.gov/mission_pages/kepler/overview/index.html (accessed November 17, 2018). | 619 | 2,857 | {"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} | 2.546875 | 3 | CC-MAIN-2019-18 | longest | en | 0.900725 |
http://song4u.info/fraction-worksheets/ | 1,524,736,332,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125948125.20/warc/CC-MAIN-20180426090041-20180426110041-00537.warc.gz | 283,323,749 | 4,780 | ## Fraction worksheets
Comments Off on Fraction worksheets
Free printable fraction worksheets with various topics and difficulty levels! Our fractions worksheets are free to fraction worksheets, easy to use, please forward this error screen to 198.
These fractions worksheets are a great resource for children in Kindergarten, please forward this error screen to 184. If you’re looking for a great tool for adding, students compare fractions with unlike denominators. These worksheets are appropriate for Kindergarten, the students will be asked to identify the fractions for the shaped in shape, practice identifying the fraction using eighths.
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These fractions worksheets have rows of equivalent fractions, and very flexible. These fractions worksheets may be selected from easy, and 5th Grade.
You may select 10, click here for a Detailed Description of all the Fractions Worksheets. Multiplying or dividing mixed fractions check out this online Fraction Calculator. These fractions worksheets are great for testing children to compare Fractions to see if they are greater than, click the image to be taken to that Fractions Worksheet. These worksheets may be selected for different denominators so the problems may be positive, use these fractions worksheets to produce rectangular fractions bars and pie wedge fractions to be used as visuals in your teaching lesson plans. | 333 | 1,850 | {"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} | 2.703125 | 3 | CC-MAIN-2018-17 | longest | en | 0.880485 |
https://iesystemslab.org/wiki/doku.php?id=introduction_to_multiple-workstation_production_lines | 1,627,067,998,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046150000.59/warc/CC-MAIN-20210723175111-20210723205111-00038.warc.gz | 331,204,709 | 8,539 | # My DokuWiki
### Site Tools
introduction_to_multiple-workstation_production_lines
# Production Line Models: Introduction
NOTE: Exercises below may link to supporting files in a GitHub repository. If you follow such links and (at the GitHub website) right-click a file and choose “Save link as…”, it will appear to download a desired file but may in fact fail. A failure will be discovered when trying to open the downloaded file, usually in MATLAB, and learning that it is not in fact a MATLAB script, function, or SimEvents simulation model.
A remedy is to, at the GitHub website, back up to the project root (e.g. Courseware or Software), choose “Download ZIP” for the entire project, and find the desired file within the project's ZIP. Our apologies for the inconvenience.
## Exercises
### Semantics and Simple Computations
1. Compute the capacity in parts per hour of the following:
• A station with 3 machines operating in parallel with 20 minute service time at each machine.
• A balanced line with single machine stations, all with average processing times of 15 minutes.
• A four-station line with one machine per station, where the average processing times are 15, 20, 10, and 12 minutes, respectively for stations 1, 2, 3, and 4.
• A four-station line with multiple machines at each station, where the number of machines at stations 1, 2, 3, and 4 are 2, 6, 10, and 3, and the average processing times at stations 1, 2, 3, and 4 are 10, 24, 40, and 18 minutes.
2. Consider a three-station line with one machine per station. The average processing time on stations 1, 2, and 3 are 15, 12, and 14 minutes. However, station 2 is subject to random failures, which cause its fraction of uptime to be only 75 percent.
• Which station is the bottleneck?
• What are the bottleneck rate and raw process time for the line?
3. A powder metal manufacturing line produces bushings in three processes (compaction, sinter-harden, and rough/finish turn) which are executed at three single-machine stations with average processing times of 12, 10, and 6 minutes. However, while compaction and sinter-harden are dedicated to producing bushings, the rough/finish turn station also processes bearings from another line; the average processing time for bearings is 14 minutes.
• If the production rates of bushings and bearings are the same, what is the bottleneck?
• If the production rates of bearings is 1/2 that of bushings, what is the bottleneck rate?
• If the production rates of bearings is 1/3 that of bushings, what is the bottleneck rate?
### Simple Simulation of Open and Closed Systems
1. Open the simulation model PennyFab_ClosedSystemCONWIP_RandomProcTimes.slx, which is of a closed system (the same entities keep recirculating) with random processing times. Configure the four workstations with exponentially-distributed service times with means [2, 5, 10, 3] and number of servers [1, 2, 6, 2]. Set the whole-system WIP to 25 and make the stop time sufficiently large to reach steady-state (10000 time units?). Run one replication and use the Simulation Data Inspector to determine a good target for average throughput, i.e., a throughput level that is “achievable” for the system with reasonable certainty.
2. Open the model PennyFab_OpenSystemRandomArrivals_RandomProcTimes.slx, which is of an open system (entities arrive from an exogenous process, usually abstracted as a “source”, and depart to an exogenous process, usually abstracted as a “sink”) with random processing times. Configure the four workstations with exponentially-distributed service times with means [2, 5, 10, 3] and number of servers [1, 2, 6, 2]. Make the stop time sufficiently large to reach steady-state (10000 time units?). Next, configure the arrival process - for each member of the team, convert his/her birthdate to a number of the form “mmddyy”. Add these numbers together and set this as the Initial Seed for the Arrival Generator. For inter-arrival times, empirically determine a number which gives the same throughput as for the previous exercise’s closed system (and explain your choice … What happens if you release work at a rate too slow? What happens if you release work at a rate too fast?). Finally, run one replication, use the Simulation Data Inspector to visualize WIP, and compare to the closed system.
BIG PICTURE: These simulation models are early versions of four-workstation production lines. It’s possible to replace each (queue, server, random number generator) triple by a single GGkWorkstation library block, but this was declined to explicitly show the “single queueing node” structure of an infinite-capacity queue followed by a k-capacity server. This exercise was given in the second week of class, and the Simulation Data Inspector is used to make visualization as simple as possible. (1) is answered by visualizing the “avgTH” signal and reading the steady-state value. (2) is answered by visualizing the “WIP” signal, with an explanation such as “If the four-serial-workstation system is changed from a closed CONWIP configuration to an open random-arrivals configuration, for the same average throughput the WIP level has a fair amount of variability and occasionally spikes well above the closed system’s CONWIP level. It might appear that average WIP is also increasing, but Little’s Law implies that if average throughput and average cycle time remain unchanged, then so does average WIP.”
### Relative Placement of Workstations
Open the simulation model PennyFab_ClosedSystemCONWIP_RandomProcTimes.slx, which is of a closed system (the same entities keep recirculating) with random processing times. Configure the four workstations with exponentially-distributed service times with means [2, 5, 10, 3] and number of servers [1, 2, 6, 2]. Set the whole-system WIP to 25 and make the stop time sufficiently large to reach steady-state (10000 time units?).
1. Run one replication, use the Simulation Data Inspector to visualize average TH and CT (copy & paste your figures below), and explain your results. Note that the bottleneck in this baseline scenario is the second workstation.
2. From the baseline, switch workstations one and two, e.g. configure the four workstations to have exponentially-distributed service times with means 5, 2, 10, and 3, and number of servers 2, 1, 6, and 2. Run one replication, use the Simulation Data Inspector to visualize TH and CT (copy & paste your figures below), and compare this scenario to the baseline.
3. From the baseline, switch workstations two and four, e.g. configure the four workstations to have exponentially-distributed service times with means 2, 3, 10, and 5, and number of servers 1, 2, 6, and 2. Run one replication, use the Simulation Data Inspector to visualize TH and CT (copy & paste your figures below), and compare this scenario to the previous ones.
4. What conclusions might you draw from these experiments?
BIG PICTURE: In both alternate scenarios, the bottleneck placement does not affect the average steady-state throughput or cycle time. Note that it may be necessary to simulate the models for quite a long time to reach steady-state and draw this conclusion. More importantly, note that this is not a general result – all four service processes are exponentially-distributed with equivalent SCVs. In general, relative placement of the bottleneck does matter when there are different service time probability distributions with different amounts of variability.
### Diagnosis and Improvement
[This assignment is based on Hopp & Spearman chapter 7, problem 5 (in ed. 2) or problem 9 (in ed. 3), which begins with the sentence “Positively Rivet Inc. is a small machine shop that produces sheet metal products”.]
A small machine shop has a four-workstation production line dedicated to manufacturing a single product. The old line has [4, 4, 2, 1] machines at each workstation, and machines have processing rates of [15, 12, 20, 50] parts/hour. Because of strong demand, a new four-workstation production line is added alongside the old one and uses higher-capacity automated equipment. The new line has a single machine at each workstation with processing rates of [120, 120, 125, 125] parts/hour. Over the past several months, the old line has averaged 315 parts/day (one eight-hour shift per day) with an average WIP of 400 parts. The new line has averaged 680 parts/day with an average WIP of 350 parts. Management is unhappy with the performance of the old line because of its lower throughput and higher WIP.
1. Compute rb (bottleneck rate), T0 (raw process time), and W0 (critical WIP) for each line. Explain why one line has a larger critical WIP than the other.
2. Use the simulation model PennyFab_ClosedSystemCONWIP_RandomProcTimes.slx to estimate the throughput for each line; use exponentially-distributed service times and a CONWIP level equal to each line's reported average WIP. Let these numbers approximate the practical worst case performance (for the new line, the simulation model almost exactly meets the three conditions). Run the model, use the Simulation Data Inspector to visualize average throughput, and compare management's reported throughput results with the simulated practical worst case. Is management correct in criticizing the old line for inefficiency?
3. Use the simulation model PennyFab_OpenSystemRandomArrivals_RandomProcTimes.slx with management's reported throughput results as the arrival rates, and visualize traces of WIP, average CT, and average TH for each line. Comment, in particular on how the observed WIP compares with what would be expected for exponentially-distributed processing times.
4. Identify and evaluate at least two options for improving the throughput of each line.
BIG PICTURE: (a) is plug-and-chug, with a bit of thinking to compare the consequences of multiple slower machines versus a single fast one. (b) enables simulating practical worst case performance as opposed to estimating it with analytical approximations, and then comparing PWC results with management's reported results. © tries to elucidate not just processing time averages but also variability in the old & new lines, to help answer the next question. (d) has several possible answers; for the old line they might include more capacity, and for the new line variability reduction (to get to PWC) and then relaxing any of the practical worst case assumptions. | 2,309 | 10,414 | {"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} | 3.0625 | 3 | CC-MAIN-2021-31 | latest | en | 0.905541 |
https://cboard.cprogramming.com/cplusplus-programming/70059-=-operator-2-print.html | 1,490,406,243,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218188717.24/warc/CC-MAIN-20170322212948-00263-ip-10-233-31-227.ec2.internal.warc.gz | 783,786,784 | 6,085 | # <?= operator
Show 80 post(s) from this thread on one page
Page 2 of 2 First 12
• 09-24-2005
Thantos
I would imagine maxx would be used inside the loop.
• 09-25-2005
Dae
Quote:
Originally Posted by Thantos
I would imagine maxx would be used inside the loop.
Dae.IQ = Dae.IQ - 20;
I don't know what you mean. Of course its possible maxx was used inside the loop, but what does stating maxx != rt + tt every increment do? its not a condition, its a statement.. right?
Code:
for(int i = 0; i++ < limit; )
{
maxx != rt + tt;
}
Its not a condition, its just.. out there.. for what reason?
Sorry I cant see any reason for why its there, other than none.
• 09-25-2005
Thantos
You tell me where it states that the third part of the for loop has to have anything to do with the first two parts. Heck I remember some interesting for loops posted on this board by some very bored programmers.
• 09-25-2005
Dae
Quote:
Originally Posted by Thantos
You tell me where it states that the third part of the for loop has to have anything to do with the first two parts. Heck I remember some interesting for loops posted on this board by some very bored programmers.
I'm not saying the third part of the for loop has to have anything to do with the others. I am saying that the first part of the for loop only occurs once (creation), at the beginning, of the for loop, the second part is the condition that is checked if still 1 (true) at the end of each loop, and the third is something that you want to occur at the end of every loop. Now why would you want 'maxx != rt + tt' to occur at the end of every loop? its not attached to an if statement, so there would be no purpose to it.. right? wrong? thats what I'm wondering.
Code:
for(int i = 0; i++ < limit; maxx != rt + tt)
{
...
}
is equal to:
for(int i = 0; i++ < limit; )
{
maxx != rt + tt;
}
Code:
for(int i = 0; i++ < limit; maxx != rt + tt)
{
...
}
is equal to:
for(int i = 0; ; maxx != rt + tt)
{
if(!(i++ < limit))
break;
}
Since its a condition it should be in the second part of the for loop, or be attached to an if statement:
Code:
for(int i = 0; i++ < limit; if(maxx != rt + tt) break)
{
...
}
is equal to:
for(int i = 0; i++ < limit; )
{
if(maxx != rt + tt)
break;
}
• 09-25-2005
Thantos
Who knows why they did it, its probably just a mistake they made, but really who cares?
• 09-25-2005
SilentStrike
It's actually a GCCism. a <?= b means the same as a < b ? a : b. In english, assign the minimum of a and b to a. You see stuff like that a lot in Topcoder from people trying to squeeze every second of time in their solutions.
• 09-25-2005
Dae
Quote:
Originally Posted by Thantos
Who knows why they did it, its probably just a mistake they made, but really who cares?
I only wondered because half the people said it was short for !, and the other half said it was y > x ? x : y. The way its placed it doesnt make sense to be a trigraph. I guess with SilentStrike's reply thats right, that makes sense now.
• 09-26-2005
Salem
> from people trying to squeeze every second of time in their solutions.
Gotta be a real moron to believe that the number of characters your program occupies has anything to do with the performance of the code.
Do they really believe that since a <?= b takes fewer characters than a < b ? a : b, that the compiler will generate fewer compare instructions or branch instructions?
• 09-26-2005
Rashakil Fol
Quote:
Originally Posted by Salem
> from people trying to squeeze every second of time in their solutions.
Gotta be a real moron to believe that the number of characters your program occupies has anything to do with the performance of the code.
Do they really believe that since a <?= b takes fewer characters than a < b ? a : b, that the compiler will generate fewer compare instructions or branch instructions?
Yes, you are correct; it is true that idiots exist. Either the programmer's an idiot, or the compiler creator is an idiot.
If there is some programmer or compiler creator here that I've just insulted, this thread should take an interesting turn :-)
Of course, these artifacts from nonstandardized days shouldn't be forcibly removed from compilers, so we're being a bit unfair.
• 09-26-2005
SilentStrike
Quote:
Gotta be a real moron to believe that the number of characters your program occupies has anything to do with the performance of the code.
Actually, I'd say these people are MUCH smarter than me, and probably a bit smarter than you. They are squeezing seconds off of coding time, not running time.
Take a look at this, for example [registration required].
http://www.topcoder.com/stat?c=probl...696&cr=8355516
The code is cryptic, but I don't think you nor I would solve the problem in 7 minutes.
I actually have found that using <?= and >?= makes my code more buggy, since I somtimes omit characters and then do assignments or other similiar dumb stuff.
• 09-26-2005
Speedy5
Quote:
Originally Posted by Dae
Sorry I cant see any reason for why its there, other than none.
How do you know maxx, rt, or tt aren't objects with the != and + operators overloaded for some reason? The maxx != rt + tt statement might actually do something useful, although I really can't find a good reason why he would do it with those operators (especially !=) since it isn't intuitive.
• 09-26-2005
Decrypt
If I recall correctly, a for statement that looks like this:
Code:
for (initializing statement; condition; expression)
statement
is basically the same as:
Code:
initializing statment
while (condition)
{
statement;
expression;
}
So really, the expression can be just about anything. Though usually used to increment the loop, here it could be used just to update the variable. Maybe rt and tt change during the loop, and the OP wants to make sure that maxx = rt + tt at the end of each cycle through the loop. If you wanted to produce all fibonacci numbers under n, you could use it:
Code:
#include <cstdlib>
#include <iostream>
using namespace std;
int main()
{
int oldFibo = 0;
int olderFibo = 0;
int n = 100;
for (int fibo = 1; fibo < n; fibo = oldFibo + olderFibo)
{
olderFibo = oldFibo;
oldFibo = fibo;
cout << fibo << endl;
}
system("pause");
return 0;
}
• 09-26-2005
Salem
> They are squeezing seconds off of coding time, not running time.
Absolutely pathetic reason if you ask me.
The biggest cost in software is understanding what the other person wrote (or even your own code) months after the event when it doesn't work anymore or needs to be enhanced.
Ever looked at any of your old "smart" code and wondered what the hell you were thinking of at the time?
Every stupid little trick in the book which obfuscates the code in some way simply adds to the difficulty.
I suppose they still think the xor swap without a temp is a really neat idea.
• 09-26-2005
Dae
Quote:
Originally Posted by Speedy5
How do you know maxx, rt, or tt aren't objects with the != and + operators overloaded for some reason? The maxx != rt + tt statement might actually do something useful, although I really can't find a good reason why he would do it with those operators (especially !=) since it isn't intuitive.
That would be a very good point, lol, a little too late.. but very good point.
However, you cant overload <?=, and that is what was in the code, and someone said it stands for !=, which is the only reason I used != in that code you quoted. ;)
• 09-26-2005
Dae
Quote:
Originally Posted by Decrypt
If I recall correctly, a for statement that looks like this:
Code:
for (initializing statement; condition; expression)
statement
is basically the same as:
Code:
initializing statment
while (condition)
{
statement;
expression;
}
So really, the expression can be just about anything. Though usually used to increment the loop, here it could be used just to update the variable. Maybe rt and tt change during the loop, and the OP wants to make sure that maxx = rt + tt at the end of each cycle through the loop. If you wanted to produce all fibonacci numbers under n, you could use it:
[code]#include <cstdlib>
#include <iostream>
You're right, it can be anything - as I said. However the condition statement is automatically considered to operate like an if statement, while the expression part does not. Hence if the OP was trying to check if maxx != tt + rt, he would have to use an if statement around it in the expression - as I exampled. Without the if statement (as it is in the OP's code) it does not make sense to be the != operator, as we now know according to SilentStrike.
Show 80 post(s) from this thread on one page
Page 2 of 2 First 12 | 2,159 | 8,598 | {"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} | 2.9375 | 3 | CC-MAIN-2017-13 | longest | en | 0.957039 |
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posted by .
Hello!
Thanks for checking my question out!
____
Fill in the Blank
14. In an experiment, if doubling the manipulated variable results in a doubling of the responding variable, the relationship between the variables is a(n) _________________________. (1 point)
Could someone check if I'm right?
Thanks!
- Da Fash
• Physical Science -
right.
• Physical Science -
Thanks!
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https://www.transtutors.com/questions/6-1-let-u-r-3-and-let-v-be-the-horizontal-i-e-the-xy-plane-with-basis-vectors-v-1-1--1316876.htm | 1,550,333,394,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247480622.9/warc/CC-MAIN-20190216145907-20190216171907-00599.warc.gz | 1,001,365,793 | 18,450 | # 6.1 Let U = R 3 and let V be the horizontal (i.e. the xy ) plane with basis vectors v 1 = ( 1 , 0...
6.1 Let U = R3 and let V be the horizontal (i.e. the xy) plane with basis vectors v1 = (1, 0, 0) and v2 = (0, 1, 0); and define any two vectors u1 = (x1, y1, z1) and u2 = (x2, y2, z2). Find T (u1), T (u2), T (u1 + u2) and T (ku1), using the definition of T in equation (6.6), and draw a sketch to illustrate your results. | 169 | 428 | {"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} | 2.765625 | 3 | CC-MAIN-2019-09 | latest | en | 0.776932 |
http://www.chegg.com/homework-help/physics-for-scientists-and-engineers-2nd-edition-chapter-30-solutions-9780805327366 | 1,469,438,489,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824217.36/warc/CC-MAIN-20160723071024-00315-ip-10-185-27-174.ec2.internal.warc.gz | 364,337,749 | 19,638 | View more editions
# TEXTBOOK SOLUTIONS FOR Physics for Scientists and Engineers A Strategic Approach 2nd Edition
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Chapter: Problem:
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FIGURE shows the x-component of as a function of x.Draw a graph of V versus x in this same region of space. Let V = 0 V at x = 0 m and include an appropriate vertical scale.
FIGURE
SAMPLE SOLUTION
Chapter: Problem:
100% (6 ratings)
• Step 1 of 4
The change in potential in terms of electric filed is given as follows:
Here, is the component of electric filed in the direction of.
Refer figure Q30.1, the electric field at any point on theaxis isand the potential at the point is.
The potential at the point is calculated as follows:
Therefore, the potential at the point is.
• Step 2 of 4
The potential at the point is calculated as follows:
Therefore, the potential at the point is.
• Step 3 of 4
The potential at the point is calculated as follows:
Therefore, the potential at the point is.
• Step 4 of 4
The potential at the point is calculated as follows:
Therefore, the potential at the point is.
The potential versus distance graph is plotted as follows:
Corresponding Textbook
Physics for Scientists and Engineers A Strategic Approach | 2nd Edition
9780805327366ISBN-13: 0805327363ISBN: Randall Dewey KnightAuthors:
Alternate ISBN: 9780321513335, 9780321516596 | 376 | 1,535 | {"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} | 2.875 | 3 | CC-MAIN-2016-30 | latest | en | 0.895182 |
http://www.enotes.com/homework-help/662-kev-photon-from-137cs-compton-scattered-an-450319 | 1,477,307,000,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719564.4/warc/CC-MAIN-20161020183839-00441-ip-10-171-6-4.ec2.internal.warc.gz | 434,159,379 | 9,061 | # A 662 kev photon from 137Cs is Compton scattered at an angle of 60 calculate the energy of the scattered photon.
llltkl | College Teacher | (Level 3) Valedictorian
Posted on
The Compton scattering equation is:
`lambda' - lambda = (h/(m_ec))(1-costheta) `
`E=hnu=(hc)/lambda`
Converting energy in ev to J, then plugging in the values, we get
`662*10^3*1.6022*10^-19=(6.626*10^-34*3*10^8)/lambda`
`rArr lambda=(6.626*10^-34*3*10^8)/(662*10^3*1.6022*10^-19)`
`=1.874*10^-12 m`
Put this value in Compton scatter equation to get
`lambda'= lambda + (h/(m_ec))(1-costheta)`
`=1.874*10^-12+((6.626*10^-34) (1-1/2))/(9.1094*10^-31*3*10^8)`
`=(1.874+1.2123)*10^-12 m`
`=3.08642*10^-12 m`
Energy of the scattered photon
`=(6.626*10^-34*3*10^8)/( 3.08642*10^-12 *1.6022*10^-19)`
`=401977 ev`
`=401.977 kev`
Sources: | 331 | 823 | {"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} | 3.328125 | 3 | CC-MAIN-2016-44 | latest | en | 0.440165 |
https://socratic.org/questions/what-are-all-the-possible-rational-zeros-for-y-12x-2-28x-15-and-how-do-you-find- | 1,585,499,586,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370494349.3/warc/CC-MAIN-20200329140021-20200329170021-00464.warc.gz | 713,936,555 | 6,157 | # What are all the possible rational zeros for y=12x^2-28x+15 and how do you find all zeros?
Sep 2, 2016
Zeros of $y = 12 {x}^{2} - 28 x + 15$ are $x = \frac{5}{6}$ and $x = \frac{3}{2}$.
#### Explanation:
Zeros of a function $f \left(x\right)$ are those values of $x$, for which $f \left(x\right) = 0$.
As $y = 12 {x}^{2} - 28 x + 15$ can be factorized, let us do so by splitting the middle term $- 28$ in two parts so that their product is 12×15=180. It is apparent that these are $- 18$ and $- 10$.
Hence $y = 12 {x}^{2} - 28 x + 15$
= $12 {x}^{2} - 18 x - 10 x + 15$
= $6 x \left(3 x - 3\right) - 5 \left(2 x - 3\right)$
= $\left(6 x - 5\right) \left(2 x - 3\right)$
It is apparent that if either $6 x - 5 = 0$ or $2 x - 3 = 0$, $y = 12 {x}^{2} - 28 x + 15$ will be zero.
Hence, zeros of $y = 12 {x}^{2} - 28 x + 15$ are $x = \frac{5}{6}$ and $x = \frac{3}{2}$. | 385 | 876 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 21, "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} | 4.4375 | 4 | CC-MAIN-2020-16 | longest | en | 0.68732 |
https://www.halfbakery.com/idea/Laser_20drill-square/addlink?op=nay | 1,726,574,789,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651773.64/warc/CC-MAIN-20240917104423-20240917134423-00050.warc.gz | 753,827,635 | 7,396 | h a l f b a k e r y
You could have thought of that.
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# Laser drill-square
(+6) [vote for, against]
It is often necessary to drill a hole in something using an electric drill. Moreovermore, it is often desirable that the hole should be perpendicular to the surface. Ensuring this perpendicularity, however, is not straightforward.
One solution is to use a bench drill (which lowers the bit vertically onto the workpiece), but this is not much use when you want to drill a hole in a wall, an okapi, or some other thing too large to fit under the bench drill. For hand-held drills, the best you can do is to look at the angle of the drill from above/below and left/right, and try to keep it all square as you drill.
Howevertheless, a simple solution suggests itself, involving lasers. One low-power laser, with its beam split into eight parts, should be sufficient. Just arrange it so that the beam is split into four pairs of beams (one pair above, one below, one to the left, one to the right), wherein each pair of beams converges. For example, the top pair of beams will start out two inches apart, but converge to be only a quarter of an inch apart a few inches in front of the drill chuck.
Now, it becomes easy to hold the drill square-on to the wall - just tilt it up/down until the two dots above the drill have the same spacing as the two dots below. Likewise, tilt it left/right until the two dots to the left have the same spacing as the two to the right. The same setup will work regardless of the length of the drill bit or how far into the wall you've drilled.
[Some time later... use a circle of laser light instead - much simpler.]
— MaxwellBuchanan, Nov 14 2018
BullseyeBore http://www.bullseyebore.com/
[xaviergisz, Nov 15 2018]
Shown by Dan Gelbart in one of his videos in 2013, since built by many people [notexactly, Dec 03 2018]
I think you could get away with 3 pairs, at 120 degree intervals.
— Loris, Nov 14 2018
Interferometry would be more precise.
Or since you're spinning a motor, put the laser on the spinny part at a pleasing angle. (Balance the weight out, too). Or maybe just a spinny mirror.
— RayfordSteele, Nov 14 2018
//3 pairs, at 120 degree intervals.// You could, but I think it would be fiddlier to use, since pitch and yaw movements would be interlinked. Easier to have one pair for pitch, one for yaw.
//Interferometry// In what way? If it gives a big response to very, very small movements it might be too sensitive for a handheld drill.
— MaxwellBuchanan, Nov 14 2018
Actually, hang on a moment. Forget the paired beams. Just have a lens that projects a circle of laser-light around the drill bit. If the circle is an ellipse, you're off.
— MaxwellBuchanan, Nov 14 2018
That brings up an interesting point on judgment. I suspect we can better tell simple distances as a comparison between points than the ovality of an oval, even though the oval would contain every relevant point.
The interferometry of course would be an appeal to our halfbakery nature for overkill.
— RayfordSteele, Nov 15 2018
I just did a quick test in Illustrator, and it's very easy to tell a 30mm circle from a 28x32mm ellipse. You can even see that a 29x31mm ellipse isn't quite right.
Somebody who knows geometry should be able to tell me how many degrees error that equates to.
— MaxwellBuchanan, Nov 15 2018
Looks like such a good idea that its baked.
— RayfordSteele, Nov 15 2018
(+)
— 2 fries shy of a happy meal, Nov 15 2018
Well, damn. BullseyeBore is exactly this. Sadly not available for purchase yet, though.
— MaxwellBuchanan, Nov 15 2018
There are drills with spirit levels built-in... not too difficult to imagine one with angles as well (for surfaces with a definitive idea of up and down, of course).
Alternatively, a rectangular body drill : square up to the hole with a piece of accurately cut angle-iron.
— FlyingToaster, Nov 15 2018
Wouldn't a drill with massive gyroscopic fly wheel built in do the job perfectly?
— xenzag, Nov 15 2018
// pairs of beams ... wherein each pair of beams converges //
Will this allow you to maintain exactly sixty feet above a German reservoir, at night ?
— 8th of 7, Nov 15 2018
Having a massive flywheel on the drill might make drilling tight spaces challenging.
— RayfordSteele, Nov 16 2018
This boring idea and the link are much better than my solution which is to affix a grinding disk to the base of the drill. I would drill the hole at any old angle then continue on until the grinding disk FORCES the surface in question to conform to perfect perpendicularity to the hole just drilled.
— AusCan531, Nov 16 2018
There 'are' drill plunger attachments which will let you align to whatever angle you wish... if it isn't a Mandela effect thing.
— 2 fries shy of a happy meal, Nov 17 2018
A tiny stick on filter to the drills light (a feature on newer battery drills) would supply the circle needed. The circle doesn't have to be at the drill point. It could be even be an oval that when orientated correctly displays round.
— wjt, Nov 17 2018
BullseyeBore claims to have patents granted and pending. However, spinning laser center finders have been baked and WKTE since at most (because that really makes more sense than 'at least') 2013: [link]
— notexactly, Dec 03 2018
[annotate]
back: main index | 1,348 | 5,422 | {"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} | 2.59375 | 3 | CC-MAIN-2024-38 | latest | en | 0.94343 |
https://planetmath.org/CyclicSubspace | 1,716,063,320,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057494.65/warc/CC-MAIN-20240518183301-20240518213301-00380.warc.gz | 400,408,210 | 3,642 | # cyclic subspace
Let $V$ be a vector space over a field $k$, and $x\in V$. Let $T:V\to V$ be a linear transformation. The $T$-cyclic subspace generated by $x$ is the smallest $T$-invariant subspace which contains $x$, and is denoted by $Z(x,T)$.
Since $x,T(x),\ldots,T^{n}(x),\ldots\in Z(x,T)$, we have that
$W:=\operatorname{span}\{x,T(x),\ldots,T^{n}(x),\ldots\}\subseteq Z(x,T).$
On the other hand, since $W$ is $T$-invariant, $Z(x,T)\subseteq W$. Hence $Z(x,T)$ is the subspace generated by $x,T(x),\ldots,T^{n}(x),\ldots$ In other words, $Z(x,T)=\{p(T)(x)\mid p\in k[X]\}$.
Remark. If $Z(x,T)=V$ we say that $x$ is a cyclic vector of $T$.
Title cyclic subspace CyclicSubspace 2013-03-22 14:05:03 2013-03-22 14:05:03 CWoo (3771) CWoo (3771) 12 CWoo (3771) Definition msc 15A04 msc 47A16 cyclic vector subspace CyclicDecompositionTheorem CyclicVectorTheorem cyclic vector | 335 | 882 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 20, "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} | 3.1875 | 3 | CC-MAIN-2024-22 | latest | en | 0.641236 |
http://www.slideshare.net/jpb232/finding-the-address-slide-show | 1,369,518,939,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368706469149/warc/CC-MAIN-20130516121429-00046-ip-10-60-113-184.ec2.internal.warc.gz | 713,953,568 | 20,050 | # Finding the address slide show
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## Finding the address slide showWebinar Transcript
• Finding The Network Address The Next Few Slides Will Show You How to Find the Network ID of an IP Address
• Remember this fundamental relationship!By User:Gerson Chicareli.Kbrose at en.wikipedia [CC-BY-SA-3.0 (http://creativecommons.org/licenses/by-sa/3.0)],from Wikimedia Commons
• Finding Network Address Find the network address for the IP address – 192.175.33.18 with a mask of 255.255.255.0. Find the network address for 192.175.33.30 with a mask of 255.255.255.240. How to determine if they have the same network address.
• Finding Network AddressFind the Network Address for the IP Address – 192.175.33.18 With a Mask of 255.255.255.0
• Finding Network Address Given – 192.175.33.18 mask 255.255.255.0Step 1: convert the IP 192.175.33.18 to binary 192 . 175 . 33 . 18 11000000 . 10101111 . 00100001 . 00010010
• Finding Network Address Given – 192.175.33.18 mask 255.255.255.0Step 2: convert the subnet mask to binary 255 . 255 . 255 . 011111111 . 11111111 . 11111111 . 00000000
• Finding Network Address Given – 192.175.33.18 mask 255.255.255.0Step 3: align the subnet mask to the IP address 192 . 175 . 33 . 1811000000 . 10101111 . 00100001 . 0001001011111111 . 11111111 . 11111111 . 00000000 255 . 255 . 255 . 0
• Finding Network AddressAny 0 in the subnet mask identifies the host bits in the IP address 192 . 175 . 33 . 1811000000 . 10101111 . 00100001 . 0001001011111111 . 11111111 . 11111111 . 00000000 255 . 255 . 255 . 0
• Finding Network Address Any 1 in the subnet mask identifies the network bits in the IP address 192 . 175 . 33 . 1811000000 . 10101111 . 00100001 . 0001001011111111 . 11111111 . 11111111 . 00000000 255 . 255 . 255 . 0
• Finding Network AddressStep 4: in the IP address change all the host bits to zero’s 192 . 175 . 33 . 18 11000000 . 10101111 . 00100001 . 00010010 11111111 . 11111111 . 11111111 . 00000000 255 . 255 . 255 . 0
• Finding Network AddressStep 4: in the IP address change all the host bits to zero’s 192 . 175 . 33 . 18 11000000 . 10101111 . 00100001 . 00000000 11111111 . 11111111 . 11111111 . 00000000 255 . 255 . 255 . 0
• Finding Network Address Given – 192.175.33.18 mask 255.255.255.0Step 5: convert the binary IP back to decimal 11000000 . 10101111 . 00100001 . 00000000 192 172 33 0
• Finding Network AddressThe network address or network ID for 192.175.33.18 with a subnet mask 255.255.255.0192 172 33 0
• Finding Network AddressFind the Network Address for192.175.33.30 With a Mask of 255.255.255.240
• Finding Network Address Given – 192.175.33.30 mask 255.255.255.240Step 1: convert the IP 192.175.33.30 to binary 192 . 175 . 33 . 30 11000000 . 10101111 . 00100001 . 00011110
• Finding Network Address Given – 192.175.33.30 mask 255.255.255.240Step 2: convert the subnet mask to binary 255 . 255 . 255 . 24011111111 . 11111111 . 11111111 . 11110000
• Finding Network Address Given – 192.175.33.30 mask 255.255.255.240Step 3: align the subnet mask to the IP address 192 . 175 . 33 . 3011000000 . 10101111 . 00100001 . 0001111011111111 . 11111111 . 11111111 . 11110000 255 . 255 . 255 . 240
• Finding Network AddressAny 0 in the subnet mask identifies the host bits in the IP address 192 . 175 . 33 . 3011000000 . 10101111 . 00100001 . 0001111011111111 . 11111111 . 11111111 . 11110000 255 . 255 . 255 . 240
• Finding Network Address Any 1 in the subnet mask identifies the network bits in the IP address 192 . 175 . 33 . 3011000000 . 10101111 . 00100001 . 0001111011111111 . 11111111 . 11111111 . 11110000 255 . 255 . 255 . 240
• Finding Network Address Given – 192.175.33.30 mask 255.255.255.240Step 4: in the IP address change all the host bits to zero’s 192 . 175 . 33 . 30 11000000 . 10101111 . 00100001 . 00011110 11111111 . 11111111 . 11111111 . 11110000 255 . 255 . 255 . 240
• Finding Network Address Given – 192.175.33.30 mask 255.255.255.240Step 4: in the IP address change all the host bits to zero’s 192 . 175 . 33 . 30 11000000 . 10101111 . 00100001 . 00010000 11111111 . 11111111 . 11111111 . 11110000 255 . 255 . 255 . 240
• Finding Network Address Given – 192.175.33.30 mask 255.255.255.240Step 5: convert the binary IP back to decimal 11000000 . 10101111 . 00100001 . 00010000 192 . 175 . 33 . 16
• Finding Network AddressThe network address or network ID for 192.175.33.18 with a subnet mask 255.255.255.240192 172 33 16
• This concludes this portion of the lesson. Now please begin to work the problems.When complete, go on to the next portion ofthe lesson. If you have difficulty solving theproblems, go back and watch the steps in theprocess carefully. Remember, you must be comfortable usingbinary number before attempting thisproblems. The number one challenge students have insubnetting is that they are not comfortable withbinary numbers. Make sure YOU are not oneof them. | 1,737 | 5,127 | {"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-2013-20 | latest | en | 0.513876 |
https://crypto.stackexchange.com/questions/15685/diffusion-in-shamirs-secret-sharing-scheme | 1,718,767,607,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861797.58/warc/CC-MAIN-20240619025415-20240619055415-00363.warc.gz | 164,479,985 | 41,501 | # Diffusion in Shamir's secret sharing scheme
The popular command line utility ssss implements the classic Shamir's secret sharing scheme over the generic field $GF(2^q)$ with $8 \le q \le 1024$.
When $q>=64$, the constant coefficient ($c_0$) is not the secret but a transformed version of it, obtained by means of 40 rounds of XTEA-based permutation. As result, the secret is somewhat diffused over the entire $q$ bits.
Why is that? Which attack does it prevent? Does it help when the secret has a short bit length?
• Note that, despite your notation, the exponent does not need to be prime. $\;$
– user991
Commented Apr 20, 2014 at 18:07
• I replaced p with q to avoid confusion. Commented Apr 20, 2014 at 18:33
• Based on your description, it serves no purpose.
– K.G.
Commented Apr 20, 2014 at 18:56
I took a brief look at the code, but I fail to see how this transformation could introduce any additional secrecy. If the randomness used to define the polynomial is good, then Shamir's secret sharing provides information theoretic secrecy (no matter how the secret actually looks like). What Ricky points out in his answer seems reasonable, i.e., to provide some means against the malleability of the secret sharing. Since Shamir's approach is linear, one can "update" the shares to "update" the secret and if you add an additional layer of diffusion to the secret before sharing such "updates" are no longer possible in a straightforward manner. Note, however, that this does not influence the secrecy of the secret but against active attacks on modifying the shares.
Another issue which I initially thought could be the case comes below.
If your secret is larger than what can be represented as an element of the used field, then computational secret sharing (CSS) can be used, but it is not implemented in ssss (it's only linked as an alternative at the bottom of the page).
Basically, with CSS it is possible to make shares in a secret sharing scheme shorter at the cost of losing the perfect secrecy guarantees provided by Shamirs approach. One only achieves computational instead of information theoretic security.
The idea is to use any IND-CPA secure symmetric encryption scheme in the following (obvious) way: Choose a random secret key $k$ of a suitable symmetric encryption scheme and encrypt the secret $s$ using $k$. Then one uses polynomial secret sharing to compute $n$ shares $s_1 ,\ldots , s_n$ of the key $k$. Note that the field $\mathbb{F}$ used for secret sharing here can be much smaller as in Shamir’s original approach, as you only need to represent the key $k$ as field element and not a potentially large secret $s$.
Note that if you use Shamir's secret sharing directly on $s$, then $s$ needs to be represented as a field element (as the constant coefficient of the polynomial) and also the shares will then be of that size. If your secret is large, then CSS is an efficient alternative.
• Thanks, I didn't know about CSS. Yet, I think you response does not explain the behaviour of ssss' code. In that case, the secret is still smaller than each share. In addition to that, the transformation I see being applied is not really an encryption. It is just a fixed permutation. Commented Apr 20, 2014 at 18:36
• @SquareRootOfTwentyThree I updated my answer. Commented Apr 20, 2014 at 19:08
• "..., then Shamir's secret sharing provides information theoretic security" against passive adversaries. $\;\;\;\;\;\;$ An active adversary can very easily and controllably affect the reconstructed secret. $\;\;\;\;$
– user991
Commented Apr 20, 2014 at 19:15
• @Ricky Demer I changed it to secrecy. I agree that pure Shamirs sharing is malleable but I only refer to hiding the secret. Commented Apr 20, 2014 at 19:18
• I think "additional security" should be changed to "additional secrecy". $\;$
– user991
Commented Apr 20, 2014 at 19:20
Note that, despite your notation, the exponent does not need to be prime.
Why is that?
They probably do that to reduce the impact of malleability on the reconstruction process.
Which attack does it prevent?
It might prevent simple malleation when the secret is not easily guessable.
Does it help when the secret has a short bit length?
Yes.
• That seems reasonable! Commented Apr 20, 2014 at 19:24
• You should insert the questions you're responding to as quotes into your answer. It's a bit hard to follow in its current form. Commented Apr 20, 2014 at 21:36 | 1,067 | 4,438 | {"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} | 2.546875 | 3 | CC-MAIN-2024-26 | latest | en | 0.920454 |
https://ascentadvantageacademy.com/product/math-mrs-k-second-grade-math-ip/ | 1,695,840,445,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510319.87/warc/CC-MAIN-20230927171156-20230927201156-00489.warc.gz | 136,647,832 | 75,476 | # Math – Mrs. K Second Grade Math – IP
\$0.00
This course is Video and PDF.
It Covers Learning:
Multiplication and Division Facts as Fact Families through 12.
Divisibility Rules.
Multiplication and Division through the 1,000,000s place value.
Identity Properties of Multiplication and Division.
Commutative and Associative Properties of Multiplication.
Distributive Property of Addition and Multiplication.
Use of the Decimal in Money, Per Cent, Multiplication, and Division.
General Formulas That Use Multiplication
– Distance
– Cost
– Per Cent
Complete Definitions of Geometric Figures and Their Formulas for Perimeter or Circumference and Area:
– Circle
– Square
– Triangle
– Rhombus
– Kite
– Trapezoid
– Diamond
– Rectangle
– Parallelogram
– Pentagon
– Hexagon
– Octagon
Knowledge of How to Draw Each of the Above Figures.
Understanding of 3-D Shapes, Their Formulas, and Their Drawing:
– Sphere
– Triangular Solid
– Rectangular Solid
– Square Based Pyramid
– Cone
– Cylinder
– Cube
The Metric System and It Prefixes, Definition, and Relationships
– Length
– Weight
– Capacity
– Temperature | 268 | 1,097 | {"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} | 3.171875 | 3 | CC-MAIN-2023-40 | longest | en | 0.784132 |
https://euclidsmuse.com/collections/view?id=51 | 1,670,419,373,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711162.52/warc/CC-MAIN-20221207121241-20221207151241-00353.warc.gz | 262,815,401 | 7,111 | # Euclid's Elements-Basic Constructions
These apps are taken from Book I of Euclid\\\'s Elements. Proposition 1 shows how to construct an equilateral triangle. Proposition 2 demonstrates how to create a straight line equal to a given straight line. Finally, proposition 3 teaches how to cut off a straight line that is equal to a given line from a larger line.
Tags: Euclid\\\'s-Elements, Basic-Constructions
## Enter Collection »
#### Collection Contents:
##### Proposition 1
On a given finite straight line to construct an equilateral triangle.
##### Proposition 2
To place at a given point (as an extremity) a straight line equal to a given straight line.
##### Proposition 3
Given two unequal straight lines, to cut off from the greater a straight line equal to the less. | 172 | 780 | {"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} | 3.15625 | 3 | CC-MAIN-2022-49 | latest | en | 0.796073 |
https://avsconsultants.co.in/tutoring-for-mathematics-tips/ | 1,718,244,206,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861319.37/warc/CC-MAIN-20240612234213-20240613024213-00643.warc.gz | 96,401,468 | 15,917 | # What Tutoring for Mathematics Is – and What it Is Not
What the books lack are refreshers of the simple material required to fix the tougher troubles and THOROUGH explanations on the best way to fix. From that point, you can get in touch with the ones that you think would best fit your should schedule a session. Answer the simple questions first, and return and answer the tougher ones when you have time remaining on that test.
## Up in Arms About Tutoring for Mathematics?
In http://portaldodivanah.com.br/nobel-mathematics-reviews-amp-guide/ the end, the introduction must conclude with a crystal clear statement of the general point you wish to make in the paper. A number of the worksheets are dynamically generated which means that you’ll be provided a different set each opportunity to practice. Other topics linked to the study of the structure of mathematical proof might be included.
Students with special needs might be weaker in their usage of a number of the representations. For instance, there are specialized applets designed especially for mathematics which permit the use of mathematical symbols. When it regards a troublesome math topic, help visual learners throughout the use visit their website of models and learning aids.
Either should you need help with a single Math or Stats problem, or in case you will need assistance with a course for the entire semester, or if it’s a very simple question or an extremely intricate bit of statistical evaluation, we can give the sort of math help you require. Discuss what sort of proof has come to be the absolute most effective for a specific matter. A tutor excellent in a topic may not qualify as the ideal alternative for you if you’re trying to find a help with Math problems of a different topic.
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## The Tried and True Method for Tutoring for Mathematics in Step by Step Detail
The evaluation is going to be utilized to tailor a tutoring program to fulfill your kid’s academic requirements and schedule. On the correct side, students wrote their very own reactions and observations. In the event the student doesn’t understand how a formula works, that student will oftentimes be not able to recognize the more intricate cases that call for that formula if a number of the other conditions aren’t standard.
First and foremost, give yourself a opportunity to enjoy what you’re learning. This personal strategy and guidance can help you in each step of the way up until your graduation from BMCC. In this manner, it’s the students that give the examples and standards against which scoring is completed.
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Ergo, with the derivative you’ll be able to ascertain the rate of change at a specific point. Be sure that you are acquainted with each one of these concepts and can manipulate all the number types accurately utilizing each operation. When you become acquainted with these basic formulas, you might want to learn more about the way to create complex formulas and try a number of the many functions which are available in Excel.
Inside this, only tutoring methods can enable the student to acquire the outputs for all of the calculations. Emphasis is put on the initial phase of building mathematical models and the last phase of interpreting the solutions when it comes to real-life applications. Faculty collaboratively designed contextualized assignments utilizing real-world examples and inquiry-based problem-solving procedures to reinforce the absolute most commonly troublesome content areas at every amount of developmental math.
## Tutoring for Mathematics
The only part that bothers me about this question is that it’s usually asked towards the conclusion of the semester by students that are in danger of failing. If you’re the sole person tutoring a distinct subject then you’re in a exceptional circumstance. One particular significant writing assignment is needed.
## The Lost Secret of Tutoring for Mathematics
Take note that in the event the problem asks for a negative number, that doesn’t necessarily signify a negative INTEGER. Some of these sites have math worksheets generators while others might have ready-made worksheets. To add a mixture of negative and positive numbers, first add the two kinds of numbers separately.
## The Unexpected Truth About Tutoring for Mathematics
The worksheets are generated randomly, which means you get a different one each moment. Math is a skill that takes a great deal of brainpower to master, and this is sometimes tough for children. Math is the most commonly used subject on earth.
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You can imagine multiplication as a fast means of adding a number of numbers at the same time. Some of these sites have math worksheets generators while others might have ready-made worksheets. Utilize AutoSum The easiest means to bring a SUM formula to your worksheet is to utilize AutoSum.
## What Tutoring for Mathematics Is – and What it Is Not
You are able to also use the AutoSum feature to rapidly total a string of values without needing to enter any of them manually in a formula. 1 thing to keep in mind is that a set of numbers might not have any mode if there aren’t any repeats. Answer the simple questions first, and return and answer the tougher ones when you have time remaining on that test.
## The Little-Known Secrets to Tutoring for Mathematics
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## Lies You’ve Been Told About Tutoring for Mathematics
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## Tutoring for Mathematics
Our job was supposed to hire a complete time math teacher. If you’re the sole person tutoring a distinct subject then you’re in a exceptional circumstance. As a parent tutor, there’ll be occasions when you’re able to tutor your child when preparing a meal.
## The Number One Question You Must Ask for Tutoring for Mathematics
This is the best method to make sure you’ll receive a quality learning experience for the money. We’re about early maturation of abstract thinking, Toneva explained. Set realistic objectives and expectations for yourself.
## The Key to Successful Tutoring for Mathematics
Be sure to have students tell you just what the mean, means. You also ought to get in touch with your student’s school office. Your student may also gain from educational therapy, which is comparable to tutoring, but with a far broader reach.
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## What You Need to Do About Tutoring for Mathematics
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## What You Need to Do About Tutoring for Mathematics Starting in the Next Four Minutes
New kindergarten worksheets will be used weekly. Learning algebra is currently a lot more convenient. Math is almost always a huge issue for majority students in school.
Scores on the math part of the regular SAT 1 aren’t accepted for placement. The best way to lessen math stress is for the student to truly get proficiency in mathematics. He feels that he or she is the only one incapable of finding the solutions, even if the math is extremely complicated.
## The Definitive Strategy to Tutoring for Mathematics
Either should you need help with a single Math or Stats problem, or in case you will need assistance with a course for the entire semester, or if it’s a very simple question or an extremely intricate bit of statistical evaluation, we can give the sort of math help you require. Add your answer only in case you feel you’ve got something non trivial to add. As you answer each question, make an effort not to examine the appropriate answers straight away.
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## The Pain of Tutoring for Mathematics
If you wish to make an application for a fee waiver, please do NOT finish the normal registration. Form tutors will provide parents with the majority of the information regarding their kid’s progress and any problems they may be experiencing. Check out the way to win against the desire to procrastinate, and learn to balance time for your studies.
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## The Importance of Tutoring for Mathematics
Programs cannot call another program. Tips given within this article will allow you to understand how to seek out a Math tutor that is definitely the most suitable for your requirements and is also in your spending capacity. Tutoring may even be used for the entire application procedure to university.
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Take note that in the event the problem asks for a negative number, that doesn’t necessarily signify a negative INTEGER. The fewer formulas you have to remember, the more you are able to concentrate on technique, and very good technique is the real key to an excellent SAT score. As soon as you copy the formula, make certain that the cell references are correct.
A tutoring agency is a company that functions as a go-between for students who want tutoring and local tutors selling their expert services. Not all on-line tutoring businesses supply an on-demand tutoring services. There are lots of websites and internet tutors who can assist you with your Math problems.
Our Mathematics department gives the courses you require for your major. Tutors can offer personal attention that’s really hard to have at school. They teach skills necessary to help students reach their highest potential and give them a competitive edge.
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## Here’s What I Know About Tutoring for Mathematics
Students with special needs might be weaker in their usage of a number of the representations. All digital classrooms arrive equipped with tools like tutor reporting and profanity filters to ensure that the learning environment is almost always a safe spot. When it regards a troublesome math topic, help visual learners throughout the use of models and learning aids.
## Get the Scoop on Tutoring for Mathematics Before You’re Too Late
Teachers must keep in mind that transitions from 1 lesson or class to another are especially tough for students with ADHD. They’re encouraged to do this 3 weeks before the beginning of the semester. Regular appointments can allow you to keep up with your school material and prevent the stress of last-minute studying.
## What You Can Do About Tutoring for Mathematics Beginning in the Next 3 Minutes
In the end, the introduction must conclude with a crystal clear statement of the general point you wish to make in the paper. A number of the worksheets are dynamically generated which means that you’ll be provided a different set each opportunity to practice. Carefully read the directions on the cover of the test booklet.
## What’s Truly Going on with Tutoring for Mathematics
Finding a superb grip in Math, through understanding and practice, will tend to come up with your brain in a huge way and enable you to clasp on the concepts in the rest of the subjects with increased speed and efficiency. Now, technological, engineering and company problems are often of such complexity they demand a high amount of mathematical therapy. For example, you may present a few easy exercises involving familiar conditions, followed by exercises involving unfamiliar situations on an identical topic.
Inside this, only tutoring methods can enable the student to acquire the outputs for all of the calculations. Examples included in the session will concentrate on how best to determine if a relation is a function and the way to get the domain and range of a certain function. Numerous representations are methods to symbolize, to describe and to refer to exactly the same mathematical entity.
You are able to also use the AutoSum feature to rapidly total a string of values without needing to enter any of them manually in a formula. From that point, you can get in touch with the ones that you think would best fit your should schedule a session. If you know how things work, you could always re-create the formulas when you want them.
} | 4,230 | 22,383 | {"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} | 2.671875 | 3 | CC-MAIN-2024-26 | latest | en | 0.92378 |
https://interviewquestions.ap6am.com/tag/deep-learning/ | 1,611,261,318,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703527850.55/warc/CC-MAIN-20210121194330-20210121224330-00672.warc.gz | 394,250,139 | 7,779 | ## Deep Learning Interview Questions & Answers
• Question 1. Why Are Deep Networks Better Than Shallow Ones?
Both shallow and deep networks are capable of approximating any function. For the same level of accuracy, deeper networks can be much more efficient in terms of computation and number of parameters. Deeper networks are able to create deep representations, at every layer, the network learns a new, more abstract representation of the input.
• Question 2. What Is A Backpropagation?
Backpropagation is a training algorithm used for a multilayer neural networks. It moves the error information from the end of the network to all the weights inside the network and thus allows for efficient computation of the gradient.
The backpropagation algorithm can be divided into several steps:
1. Forward propagation of training data through the network in order to generate output.
2. Use target value and output value to compute error derivative with respect to output activations.
3. Backpropagate to compute the derivative of the error with respect to output activations in the previous layer and continue for all hidden layers.
4. Use the previously calculated derivatives for output and all hidden layers to calculate the error derivative with respect to weights.
5. Update the weights.
• Python Interview Questions
• Question 3. Explain The Following Three Variants Of Gradient Descent: Batch, Stochastic And Mini-batch?
Uses only single training example to calculate the gradient and update parameters.
Calculate the gradients for the whole dataset and perform just one update at each iteration.
Mini-batch gradient is a variation of stochastic gradient descent where instead of single training example, mini-batch of samples is used. It’s one of the most popular optimization algorithms.
• Question 4. What Are The Benefits Of Mini-batch Gradient Descent?
1. Computationally efficient compared to stochastic gradient descent.
2. Improve generalization by finding flat minima.
3. Improving convergence, by using mini-batches we approximating the gradient of the entire training set, which might help to avoid local minima.
• Python Tutorial
• Question 5. What Is Data Normalization And Why Do We Need It?
Data normalization is very important preprocessing step, used to rescale values to fit in a specific range to assure better convergence during backpropagation. In general, it boils down to subtracting the mean of each data point and dividing by its standard deviation.
• Java Interview Questions
• Question 6. Weight Initialization In Neural Networks?
Weight initialization is a very important step. Bad weight initialization can prevent a network from learning. Good initialization can lead to quicker convergence and better overall error. Biases can be generally initialized to zero. The general rule for setting the weights is to be close to zero without being too small.
• Question 7. Why Is Zero Initialization Not A Recommended Weight Initialization Technique?
As a result of setting weights in the network to zero, all the neurons at each layer are producing the same output and the same gradients during backpropagation.
The network can’t learn at all because there is no source of asymmetry between neurons. That is why we need to add randomness to weight initialization process.
• Java Tutorial Machine learning Interview Questions
• Question 8. What Is The Role Of The Activation Function?
The goal of an activation function is to introduce nonlinearity into the neural network so that it can learn more complex function. Without it, the neural network would be only able to learn function which is a linear combination of its input data.
• Question 9. What Are Hyperparameters, Provide Some Examples?
Hyperparameters as opposed to model parameters can’t be learn from the data, they are set before training phase.
Learning rate:
It determines how fast we want to update the weights during optimization, if learning rate is too small, gradient descent can be slow to find the minimum and if it’s too large gradient descent may not converge(it can overshoot the minima). It’s considered to be the most important hyperparameter.
Number of epochs:
Epoch is defined as one forward pass and one backward pass of all training data.
Batch size:
The number of training examples in one forward/backward pass.
• Artificial Neural Network Interview Questions
• Question 10. What Is A Model Capacity?
Ability to approximate any given function. The higher model capacity is the larger amount of information that can be stored in the network.
• Question 11. What Is An Autoencoder?
Autoencoder is artificial neural networks able to learn representation for a set of data (encoding), without any supervision. The network learns by copying its input to the output, typically internal representation has smaller dimensions than input vector so that they can learn efficient ways of representing data. Autoencoder consist of two parts, an encoder tries to fit the inputs to an internal representation and decoder converts internal state to the outputs.
• Question 12. What Is A Dropout?
Dropout is a regularization technique for reducing overfitting in neural networks. At each training step we randomly drop out (set to zero) set of nodes, thus we create a different model for each training case, all of these models share weights. It’s a form of model averaging.
• Python Interview Questions
• Question 13. What Is A Boltzmann Machine?
Boltzmann Machine is used to optimize the solution of a problem. The work of Boltzmann machine is basically to optimize the weights and the quantity for the given problem.
Some important points about Boltzmann Machine −
• It uses recurrent structure.
• It consists of stochastic neurons, which consist one of the two possible states, either 1 or 0.
• The neurons in this are either in adaptive (free state) or clamped (frozen state).
• If we apply simulated annealing on discrete Hopfield network, then it would become Boltzmann Machine.
• Question 14. Is It Ok To Connect From A Layer 4 Output Back To A Layer 2 Input?
Yes, this can be done considering that layer 4 output is from previous time step like in RNN. Also, we need to assume that previous input batch is sometimes- correlated with current batch.
• Question 15. What Is An Auto-encoder?
An autoencoder is an autonomous Machine learning algorithm that uses backpropagation principle, where the target values are set to be equal to the inputs provided. Internally, it has a hidden layer that describes a code used to represent the input.
Some Key Facts about the autoencoder are as follows:-
• It is an unsupervised ML algorithm similar to Principal Component Analysis
• It minimizes the same objective function as Principal Component Analysis
• It is a neural network
• The neural network’s target output is its input | 1,357 | 6,867 | {"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} | 2.578125 | 3 | CC-MAIN-2021-04 | latest | en | 0.827571 |
http://mathhelpforum.com/differential-equations/169050-simple-separation-variable-problem-print.html | 1,527,481,779,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794870771.86/warc/CC-MAIN-20180528024807-20180528044807-00398.warc.gz | 181,310,062 | 2,894 | # Simple Separation of Variable Problem
• Jan 22nd 2011, 12:15 PM
David_is_a_LOSTaway
Simple Separation of Variable Problem
I'm just starting a course in DiffEQ and we have a simple separation of variables problems.
The problem is to change the dy/dt equation into a simple y(t) equation (example dy/dt=ty becomes y=k*e^(t^2)) where k is any real number).
The problem I am stuck on is dy/dt=t/(y+y*t^2). Here is my work.
dy/dt=t/(y(1+t^2))
y*dy=[t/(1+t^2)]*dt
Integrate both sides
1/2y^2=1/2[ln|t^2+1|]+C where C is a constant
y^2=ln|t^2+1| + C where C is a constant
y = sqrt (ln|t^2+1| + C) where C is a constant and the sqrt is either positive or negative.
The book gives an answer of y(t)=sqrt(ln|k(t^2+1)|) where k is any real number and the sqrt is either positive or negative.
I understand most of the steps, but why do I get ln|t^2+1| + C inside the square root while the book gets ln|k(t^2+1)| inside the square root?
• Jan 22nd 2011, 12:19 PM
Chris L T521
Quote:
Originally Posted by David_is_a_LOSTaway
I'm just starting a course in DiffEQ and we have a simple separation of variables problems.
The problem is to change the dy/dt equation into a simple y(t) equation (example dy/dt=ty becomes y=k*e^(t^2)) where k is any real number).
The problem I am stuck on is dy/dt=t/(y+y*t^2). Here is my work.
dy/dt=t/(y(1+t^2))
y*dy=[t/(1+t^2)]*dt
Integrate both sides
1/2y^2=1/2[ln|t^2+1|]+C where C is a constant
y^2=ln|t^2+1| + C where C is a constant
y = sqrt (ln|t^2+1| + C) where C is a constant and the sqrt is either positive or negative.
The book gives an answer of y(t)=sqrt(ln|k(t^2+1)|) where k is any real number and the sqrt is either positive or negative.
I understand most of the steps, but why do I get ln|t^2+1| + C inside the square root while the book gets ln|k(t^2+1)| inside the square root?
Define $\displaystyle C=\ln K$. :) | 616 | 1,860 | {"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 | 4 | CC-MAIN-2018-22 | latest | en | 0.840673 |
https://oeis.org/A047279 | 1,621,034,148,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991829.45/warc/CC-MAIN-20210514214157-20210515004157-00441.warc.gz | 449,111,393 | 4,060 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A047279 Numbers that are congruent to {0, 1, 2, 6} mod 7. 1
0, 1, 2, 6, 7, 8, 9, 13, 14, 15, 16, 20, 21, 22, 23, 27, 28, 29, 30, 34, 35, 36, 37, 41, 42, 43, 44, 48, 49, 50, 51, 55, 56, 57, 58, 62, 63, 64, 65, 69, 70, 71, 72, 76, 77, 78, 79, 83, 84, 85, 86, 90, 91, 92, 93, 97, 98, 99, 100, 104, 105, 106, 107, 111, 112 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,3 LINKS Daniel Starodubtsev, Table of n, a(n) for n = 1..10000 Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1). FORMULA G.f.: x^2*(1+x+4*x^2+x^3) / ( (1+x)*(1+x^2)*(x-1)^2 ). - R. J. Mathar, Oct 25 2011 From Wesley Ivan Hurt, May 21 2016: (Start) a(n) = a(n-1) + a(n-4) - a(n-5) for n>5. a(n) = (14n-17+3*(i^(2n)+(1+i)*i^(-n)+(1-i)*i^n))/8 where i = sqrt(-1). a(2n) = A047336(n), a(2n-1) = A047352(n). a(n) = A047361(n+1) - 1. a(2-n) = - A047322(n). (End) MAPLE A047279:=n->(14*n-17+3*(I^(2*n)+(1+I)*I^(-n)+(1-I)*I^n))/8: seq(A047279(n), n=1..100); # Wesley Ivan Hurt, May 21 2016 MATHEMATICA LinearRecurrence[{1, 0, 0, 1, -1}, {0, 1, 2, 6, 7}, 80] (* Harvey P. Dale, Jun 15 2015 *) PROG (MAGMA) [n : n in [0..100] | n mod 7 in [0, 1, 2, 6]]; // Wesley Ivan Hurt, May 21 2016 CROSSREFS Cf. A047322, A047336, A047352, A047361. Sequence in context: A243652 A080780 A138168 * A162917 A250183 A260139 Adjacent sequences: A047276 A047277 A047278 * A047280 A047281 A047282 KEYWORD nonn,easy,changed AUTHOR EXTENSIONS More terms from Wesley Ivan Hurt, May 21 2016 STATUS approved
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# m10, q27
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Joined: 18 Jan 2010
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m10, q27 [#permalink]
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06 Jul 2010, 18:59
Is point A closer to point (1,2) than to point (2,1)?
Point A lies on the line y=x
Point A lies on the line y=-x
Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
EACH statement ALONE is sufficient
Statements (1) and (2) TOGETHER are NOT sufficient
Please can you help with answering this question.
CIO
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Re: m10, q27 [#permalink]
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06 Jul 2010, 23:59
Already discussed here:
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Re: m10, q27 [#permalink] 06 Jul 2010, 23:59
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# m10, q27
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Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 798 | 2,858 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2017-04 | latest | en | 0.840389 |
http://openstudy.com/updates/50591c92e4b0cc1228932e5d | 1,448,703,671,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398451872.11/warc/CC-MAIN-20151124205411-00288-ip-10-71-132-137.ec2.internal.warc.gz | 168,118,760 | 10,085 | ## ieatpi 3 years ago So if e^lnx = x does e^-lnx = -x?
1. inkyvoyd
a^(-b)=1/(a^b)
2. inkyvoyd
So, what do you think.
3. ieatpi
e^-lnx = 1/x?
4. ieatpi
awesome, thanks for that clarification!
5. inkyvoyd
Yup.
6. inkyvoyd
Well I think so let me see.
7. inkyvoyd
yup. | 121 | 279 | {"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} | 2.625 | 3 | CC-MAIN-2015-48 | longest | en | 0.776752 |
http://jwilson.coe.uga.edu/EMAT6680Fa05/Hawkins/Final%20Assignment/BouncingBarney.html | 1,542,278,851,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039742666.36/warc/CC-MAIN-20181115095814-20181115121157-00000.warc.gz | 182,823,389 | 2,163 | Bouncing Barney
by Kristy Hawkins
Barney lives in this triangular room ABC.
He walks from a point on AC parallel to BC. When he reaches AB, he turns and walks parallel to AC. When he reaches BC, he turns and walks parallel to AB back to AC. He continues this process until he reaches his starting point again. How many times will Barney reach a wall before this happens?
Well, when we construct this situation in GSP, it looks like Barney will return to his starting position after reaching the sixth wall. Move Barney around in this GSP sketch to see if you can find any other cases. These two look interesting to me.
What do you think? These might be cases when Barney is starting at the midpoint of AC or at a point that trisects AC. I think their might be congruent triangle as well! Let's see what we can prove!
Let's begin with the most simple case i.e. Barney begins at the midpoint of AC. Denote AC=b, AB=c, and BC=a.
Prove that Barney returns to his starting point after two bounces.
First let's prove that triangle AXY is similar to triangle ACB. This is pretty simple to do because we know that YX is parllel to BC from the problem. Since this is true,
by two parallel lines cut by a transversal.
Also, both of these triangles contain angle BAC, and therefore they are similar.
Since they are similar,
In the same way, we can show that BZ=.5a. Since this is true, when Barney travels from Z parallel to AB, he must land at X. Therefore he lands at the same place where he began after two bounces.
This conjecture is also pretty easy to prove wherever Barney begins on AC by similar triangles and parallel lines.
As I was investigating this situation, I thought about what would happen if Barney was traveling on lines outside of triangle ABC. What would that look like? Do you think Barney would still eventually end up at the same place?
He still ends up at the same point after only five bounces! Can you prove this?
Now let's look at the distance that Barney travels before he returns to his beginning point.
At first glance, it looks like the distance that Barney travels might be equal to the perimeter of the triangle. Would you agree with this? Let's measure the distances in GSP and see what we find.
It looks like this is true for all triangles. Pretty cool! | 509 | 2,296 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2018-47 | latest | en | 0.970406 |
https://physics.stackexchange.com/questions/594150/how-much-of-iter-tokamaks-input-power-could-be-eliminated-in-a-future-commercia | 1,675,491,464,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500094.26/warc/CC-MAIN-20230204044030-20230204074030-00749.warc.gz | 474,576,496 | 34,629 | # How much of ITER Tokamak's input power could be eliminated in a future commercial fusion reactor? [closed]
The ITER Tokamak is designed to produce 500 MW of thermal output from 330 MW of input power, ignoring electric conversion efficiency.
The heaters require 150 MW (to heat input plasma by 50 MW), the magnets require 80 MW, and the other subsystems require 100 MW (source). How much of this 330 MW input is actually physically required to operate the 500 MW reactor?
For example,
• Could 150 MW -> 50 MW plasma heating loss be reduced with more efficient heating, or is it nearly optimal / Carnot efficient? (Could input heating be eliminated if a future reactor achieved ignition?)
• Could any of the magnetic 80 MW be eliminated with better superconductors or a different design, or is this close to the theoretical minimum to maintain these B-fields?
• Is any of the other 100 MW for auxiliary experimental equipment or non-germane subsystems, or is it all required to couple fusion heat to a heat sink (or electric turbine)?
(If all no, do any/all these inputs scale with the fusion output, e.g. to 5 GW?)
I'm asking about the scientific constraints & theoretical bounds here, not how to construct anything specific or solve any particular problem.
Future "commercial" fusion plants will hopefully be able to reduce all three sources of power: auxiliary heating, magnets, and other subsystems.
It is impossible to give precise numerical values in MW for how much of this auxiliary power can be eliminated because that would require me to speculate on future technologies which do not exist yet (for example, if we had room temperature superconductors, you wouldn't need any of the cryogenic systems necessary for present superconductors).
Nevertheless, I can comment on some common threads running through various Advanced Tokamak (AT) concepts these days.
1. (Better) Superconducting magnets - Calculations of the fusion gain, $$Q = P_{fusion}/P_{CD}$$, find that it is proportional to the magnetic field cubed, i.e. $$Q \propto B^3$$. Therefore, any new advancements in making higher fields at lower power 'cost' pay off big time when it comes to fusion (in my expression for $$Q$$, CD = "Current Drive" is the auxiliary power one must put in to drive the large toroidal current in a tokamak).
2. Efficient Non-inductive Current Drive - AT concepts also like to hope for more efficient ways of coupling power into plasma. To answer your first bullet point: no, our present abilities for power-coupling are not optimal, and this is the subject of active research. Also, to answer your question regarding eliminating heating/current-drive upon ignition: I believe the consensus on this is no. Even in so-called steady-state tokamak operation some amount of power is needed to drive current/heat the plasma. AT concepts typically try to have a large fraction of the toroidal current be self-generated (this is called bootstrap current).
3. Compact Design - Everyone realizes that ITER is far too expensive (big) to be the gold standard for fusion energy moving forward. So to address your last point about lowering the power-budget for "other systems", I have to imagine that having a smaller machine would reduce power needs. Also, recall that smaller machines are more economically competitive (less expensive, faster build-time).
I would like to refer you to this recent DIII-D Tokamak review as well as this article about the SPARC Tokamak for further reading on more efficient (both in terms of power and cost) paths to fusion.
• I'm skeptical that smaller is cheaper. Fusion energy confinement time scales with $r^3 B$. And realistic power plant would be 500MW–3GW. Lower construction costs can come from economies of scale. I'm curious what the breakdown of the "other systems" 100MW is. Is any of it for non-commercial scientific purposes? Nov 16, 2020 at 17:20
• @alexchandel , Allow me to clarify, smaller is cheaper in the sense that if you could get the same $Q$ with a smaller machine then you don’t need as many raw materials, hence brining down the cost. I looked at your reference-link for the 100MW number, and it’s a little unclear what this accounts for. In one sentence it seems to imply that this is the “overall consumption of the site” which might account for purely scientific systems, offices, etc. so I too would like to know the breakdown of this number before I can comment on extrapolating it to a commercial reactor. Nov 16, 2020 at 17:48 | 971 | 4,491 | {"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": 3, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2023-06 | latest | en | 0.899896 |
https://list.uvm.edu/cgi-bin/wa?A3=ind1210&L=UVMFLOWNET&E=quoted-printable&P=2023752&B=--_000_903972A315F5AB42B0F4ADAF3786B7580351441Dexc0i05wvchcadh_&T=text%2Fhtml;%20charset=us-ascii | 1,627,758,713,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154099.21/warc/CC-MAIN-20210731172305-20210731202305-00621.warc.gz | 386,374,197 | 3,080 | Cross sectional area is pi * r2 . If the radius is doubled then the area quadruples and so velocity would decrease by about 75% to accommodate the same volume flow. More apropos of what we see in disease if the radius decreases by 50% then the velocity roughly quadruples to accommodate the same volume flow. We are used to thinking of turbulence occurring in stenoses so if the radius decreases velocity must increase much more to accommodate the same volume flow and you can see that this will increase the Reynolds number.
All the best
Joe
From: UVM Flownet [mailto:[log in to unmask]] On Behalf Of Terry Zwakenberg
Sent: Wednesday, October 31, 2012 5:36 AM
Subject: Re: Reynolds
Don, The answer for your student is in the question. The Reynolds equation is determining the point at which flow becomes turbulent in a tube. You can solve that riddle for any diameter tube but once you choose the diameter of the tube you are solving for it is no longer a variable but rather a fixed constant in the formula. I guess if you were questioning the diameter required to keep a certain fluid of known viscosity and volume from becoming turbulent as you deliver it from point a to b than yes your student question would have some application, but you would first have to determine a starting diameter and it's associated Reynolds number for that specific fluid to be able to apply it. I think.
On Oct 31, 2012 12:05 AM, "Don Ridgway" <[log in to unmask]> wrote:
I got asked an awkward question by a student today, regarding the Reynolds equation:
Re = V p 2r
n
My primitive school email access won't allow me to make that lower-case p into a rho for density, nor that lower-case n into an eta for viscosity, but you get the idea.
The question was this: since we know from another equation—Q = V x CSA—that a reduction of cross-sectional area means an increase of mean velocity, then why don''t those two balance out in the Reynolds equation?
For example, if the radius is doubled, won't that bring about a decrease to 1/2 of the velocity, leaving Re unchanged?
My clever response: life is complicated. And I'll get back to you.
I'm obviously angling for a bit of perspective from Dr. Beach from up there in Seattle (whence I'm flying Friday, as it happens).
Thanks in advance to anyone with help on this.
Don Ridgway
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To unsubscribe or search other topics on UVM Flownet link to: http://list.uvm.edu/archives/uvmflownet.html | 622 | 2,654 | {"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} | 3.28125 | 3 | CC-MAIN-2021-31 | latest | en | 0.94448 |
http://brightstorm.com/math/geometry/circles/chords-and-a-circles-center/ | 1,369,189,591,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368701153213/warc/CC-MAIN-20130516104553-00076-ip-10-60-113-184.ec2.internal.warc.gz | 37,353,196 | 10,419 | Congratulations on starting your 24-hour free trial!
Quick Homework Help
# Circle Chord
Star this video
A chord is a line segment whose endpoints are on a circle. If a chord passes through the center of the circle, it is called a diameter. Two important facts about a circle chord are that (1) the perpendicular bisector of any chord passes through the center of a circle and (2) congruent chords are the same distance (equidistant) from the center of the circle.
Chords in the center of a circle have a special relationship but back up what's a chord? Let's refresh our memory. Well a chord is a line segment whose endpoints are on the circle. If I found the perpendicular bisector of this chord so if I took my compass and I swung arcs from both ends of that and I found the line that bisected this chord into two congruent pieces at a 90 degree angle, so let's say I do that in so this dotted line is my perpendicular bisector of that chord and no matter where I draw a chord on this circle if I find it's perpendicular bisector it will always pass through the center of the circle so that's the first key thing about a chord as relationship with the center of circle.
Let's talk about 2 congruent chords, so this is kind of a converse of what we just talked about. If I found the perpendicular bisector of these chords so if I measured the perpendicular distance from the chord to the center, so I'm going to draw a solid line here so this is the perpendicular distance because we said the shortest distance between two points is a line to perpendicular, if these chords are congruent, they will be the same distance away from the center of the circle so if I were to join two other chords and if I told you that these chords are congruent then their distance from the center of that circle measured along a perpendicular will be congruent. So using these two keys about chords and the relationship with the center will help us solve a lot of problems.
## Find Videos Using Your Textbook
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Step 2: solve the related quadratic equation. Make the boundary points solid circles if the original inequality includes equality; otherwise, make the boundary points open circles. Prove that a non-empty subset of the real numbers union its boundary set is a closed set. They can be any of the rational and irrational numbers. Was Stan Lee in the second diner scene in the movie Superman 2? Those that do not (compactness for example) are called "intrinsic". When U is an open subset of the plane, let R(U) denote the set of all closed rational rectangles that are included in U. SO X-4 The boundary points are x = (Simplify your answer. Solving rational inequalities is very similar to solving polynomial inequalities.But because rational expressions have denominators (and therefore may have places where they're not defined), you have to be a little more careful in finding your solutions.. To solve a rational inequality, you first find the zeroes (from the numerator) and the undefined points (from the denominator). Math Help Forum. x + 4 = 0, so x = –4 x – 2 = 0, so x = 2 x – 7 = 0, so x = 7 . The boundary of a set lies \between" its interior and exterior: De nition: Let Gbe a subset of (X;d). For a set E, define interior, exterior, and boundary points, Constructing a bounded set of real numbers with exactly three limit points. The main result of this paper is the following: Theorem 1. B write the boundary of the set of rational numbers. In a High-Magic Setting, Why Are Wars Still Fought With Mostly Non-Magical Troop? Show that the collection of intervals {(x-6, x + δ), where x is a rational number and ó is a positive rational number, is a countable collection. . 1. boundary, bounds, bound - the line or plane indicating the limit or extent of something. Therefore q is a boundary point of Irrational numbers. b. Therefore q is a boundary point of Irrational numbers. A point is called a ... For instance, the rational numbers are dense in the real numbers because every real number is either a rational number or has a rational number arbitrarily close to it. Determine the boundary of each set. "Therefore, O students, study mathematics and do not build without foundations". ... Every real number is a limit point of Q, \mathbb Q, Q, because we can always find a sequence of rational numbers converging to any real number. Proving a closed set contains all of it's boundary points? The boundary of a set is a topological notion and may change if one changes the topology. Real Analysis - Limit points and Open set. Home. X. x3bnm. Note the difference between a boundary point and an accumulation point. As R is union of rational and irrational numbers, therefore the boundary point of Irrational numbers are R. Thanks for contributing an answer to Mathematics Stack Exchange! All boundary points of a rational inequality should always be … Again, think of a rational expression as a ratio of two polynomials. Rational numbers Q CR. Solving Quadratic Inequalities Step 1: write the inequality in standard form. We say that $x$ is a boundary point of $A$ if every neighborhood of $x$ contains at lest one point of $A$ and at least one point of $A^C$. Short scene in novel: implausibility of solar eclipses. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Question 2 (15 points). This is the step in the process that has all the work, although it isn’t too bad. What and where should I study for competitive programming? Limit points are also called accumulation points of Sor cluster points of S. Remark: xis a limit point of Sif and only if every neighborhood of xcontains a point in Snfxg; equivalently, if and only if every neighborhood of xcontains an in nite number of points in … I feel I must be misinterpreting the definition of a boundary, because this doesn't seem right to me. Consider a sequence {1.4, 1.41, 1.414, 1.4141, 1.41414, …} of distinct points in ℚ that converges to √2. Approximating irrational numbers by rational ones 6 u1 v1 v2 u2 In other words, we are in essentially the same situation as when we started out. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Asking for help, clarification, or responding to other answers. A point $$x_0 \in D \subset X$$ is called an interior point in D if there is a small ball centered at $$x_0$$ that lies entirely in $$D$$, Let q be any rational number. In Brexit, what does "not compromise sovereignty" mean? Set Q of all rationals: No interior points. Question: Give the boundary points, the interior points, the accumulation points, the isolated points. Find the Set of All Accumulation Points (also called Limit Points) for the given set S. Every point of the set of all uncountable limit points is a limit point. Let $x$ $\in$ $\Bbb R$ and $\epsilon$>$0$. This Question: 1 pt 10 Determine all boundary points and solve the rational inequality. Quadratic and Rational Inequalities. a. The critical values are simply the zeros of both the numerator and the denominator. Perhaps that is what you saw? Is the set of rationals a measurable set. The rational numbers mod 1 are then ordered by these fans, providing insight into their tidal interweaving. Since the boundary point is defined as for every neighbourhood of the point, it contains both points in S and $$S^c$$, so here every small interval of an arbitrary real number contains both rationals and irrationals, so $$\partial(Q)=R$$ and also $$\partial(Q^c)=R$$ Show that the set of limit points of a set is closed. Please Subscribe here, thank you!!! In the standard topology or R it is int. The boundary point(s) will mark off where the rational expression is equal to 0. What is an escrow and how does it work? Solving Rational Inequalities. 13. ), Useful fact: $\partial A$ is the set of points $x\in \mathbb R$ such that $x$ is the limit of a sequence in $A$ as well as the limit of a sequence in $\mathbb R \setminus A$. Since this quadratic is not factorable using rational numbers, the quadratic formula will be used to solve it. How can I buy an activation key for a game to activate on Steam? In words, the interior consists of points in Afor which all nearby points of X are also in A, whereas the closure allows for \points on … Hint: any ball centered at a rational number contains an irrational number. How can I upsample 22 kHz speech audio recording to 44 kHz, maybe using AI? In lecture one, we introduced the concept of counting the number of lattice points that lie inside and on the boundary of a given circle of radius . Plug each of these test points into the polynomial and determine the sign of the polynomial at that point. In two dimensions, ... [0,1], δ > 0, there exist a pair of rational numbers q1,q2 such that t0 ∈ [q1,q2] Let $$(X,d)$$ be a metric space with distance $$d\colon X \times X \to [0,\infty)$$. Every real number is a limit point of \mathbb Q, Q, because we can always find a sequence of rational numbers converging to any real number. Why the set of all boundary points of the irrational is the set of real numbers? Regarding this, what does boundary line mean? 2-1 SO X-4 Completely factor the numerator of this inequality. Show that the set of limit points of a set is closed. It is VERY important that one side of the inequality is 0. How can I improve undergraduate students' writing skills? border, borderline, delimitation, mete. We get the “boundary points” or “critical values” by setting all the factors (both numerator and denominator) to 0; these are –4, and 1. Given a complex vector bundle with rank higher than 1, is there always a line bundle embedded in it? Interior points, boundary points, open and closed sets. Noun. Example: The set {1,2,3,4,5} has no boundary points when viewed as a subset of the integers; on the other hand, when viewed as a subset of R, every element of the set is a boundary point. JavaScript is disabled. The set of all boundary points of $A$ is called the boundary of $A$, and is denoted $A^b$. The boundary of the set of rational numbers as a subset of the real line is the real line. As R is union of rational and irrational numbers, therefore the boundary point of Irrational numbers are R. A real number is a number that can take any value on the number line. The critical values are simply the zeros of both the numerator and the denominator. (1) an interior point of Aif there exists >0 such that A˙(x ;x+ ); (2) an isolated point of Aif x2Aand there exists >0 such that xis the only point in Athat belongs to the interval (x ;x+ ); (3) a boundary point of Aif for every >0 the interval (x ;x+ ) contains points in Aand points not in A; Set Theory, Logic, Probability, Statistics, Stretchable micro-supercapacitors to self-power wearable devices, Research group has made a defect-resistant superalloy that can be 3-D-printed, Using targeted microbubbles to administer toxic cancer drugs. Quadratic and Rational Inequalities. 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How much do you have to respect checklist order other answers these test points into polynomial! Away galaxies in an expanding universe or responding to other answers as a fraction... Saying there 's * talent * in that building inappropriate on number lines and paste URL. Can light reach far away galaxies in an expanding universe fans, providing insight into tidal! Its boundary points of irrational numbers subset of the reals.. irrational means not rational know that bounded... Numbers mod 1 are then ordered by these fans, providing insight into their tidal interweaving, maybe AI! Step 3: locate the boundary point of an inequality clarification, or responding other... Numerator and denominator equal to 0 to find the values of x that make the boundary of rational. Video shows how to find the boundary of the 24 families of Kohanim subspace!, for instance the half-open interval [ 0,1 ) in the movie Superman?. Extended $\mathbb R$ and $\epsilon$ > $0.... Of BUD SIZE on number lines line or plane indicating the limit of the irrational is the:. A rectangle rational if its vertices are rational numbers is compact X-4 the boundary point an. To know that a repeating decimal is a rational number is a topological notion and may if. Example ) are called intrinsic '' is sufficient to show that ''... That every real number that can not be written as a subset of$ \Bbb R $the collection all! Quote of the irrational is the following: theorem 1 Non-Magical Troop then by! Interior and boundary points, boundary points of irrational numbers an example of numbers... Rss reader limit points of the set of rational numbers are a subset the... Their the domain consists simply of ( x, y ) E R2, |x + =... Sent via email is opened only via user clicks from a mail client and not by bots non-empty of! The boundary points of irrational numbers for these two statements depends a on! ( s ) will mark off where the rational inequality in standard form ⅔ is an of!$ is closed both irrational as well as rational numbers whereas √2 is a boundary, please enable in... You have to respect checklist order standard form is equal to 0 to find the of. Is any inequality that can be expressed in the space of rational.... Via email is opened only via user boundary points of rational numbers from a mail client not... Bundle with rank higher than 1, we put those on the sign as! An irrational number can not be written in the form of a rational expression positive. Let us call a rectangle rational if its end points are rational points places that boundary... Z values for boundary points that have rational internal angles do not draw fractal but. Critical values are simply the zeros of both the numerator and the denominator F., ISBN-10: …. There 's * talent * in that building inappropriate theorem every e-neighbourhood of q contains both irrational as well rational. From a mail client and not by bots this form it has been successfully applied in boundary points of rational numbers problems of theory! $is closed are surprised to know that a link sent via email is opened only user. Of ( x, y ) with both x and y rational } 2 depends a bit your! As rational numbers are a subset of the rational numbers, the boundary point of irrational are. Some sets are neither open nor closed, for instance the half-open interval [ 0,1 ) the. To solve it pick a test point from each of these test points the! We put those on the rings on writing great answers open and closed.! Locate the boundary of the constant sequence ( R ) formula will be used to solve it compactness... People are surprised to know that a repeating decimal is a closed set contains all of its boundary points only! It isn ’ t too bad improve undergraduate students ' writing skills to this feed... \Epsilon$ > $0$ numbers and a≠0 too bad ( compactness for example ) called. Them up with references or personal experience I rotate the cup © 2020 Stack Exchange ;... Then by density theorem every e-neighbourhood of q contains both irrational as as. Families of Kohanim how to find the boundary point and an accumulation point does compromise... | 4,214 | 18,740 | {"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": 2, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2023-23 | latest | en | 0.935387 |
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A126423 a(n) = n^4 - n - 1. 10
-1, 13, 77, 251, 619, 1289, 2393, 4087, 6551, 9989, 14629, 20723, 28547, 38401, 50609, 65519, 83503, 104957, 130301, 159979, 194459, 234233, 279817, 331751, 390599, 456949, 531413, 614627, 707251, 809969, 923489, 1048543, 1185887, 1336301, 1500589, 1679579, 1874123, 2085097, 2313401, 2559959 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 LINKS Vincenzo Librandi, Table of n, a(n) for n = 1..10000 MATHEMATICA a = {}; Do[AppendTo[a, x^4 - x - 1], {x, 1, 100}]; a PROG (MAGMA) [n^4-n-1: n in [1..40]]; // Vincenzo Librandi, Aug 30 2011 CROSSREFS Cf. A002327, A002328, A116581, A126420, A126421, A126422. Sequence in context: A044200 A329109 A044581 * A072474 A186103 A194713 Adjacent sequences: A126420 A126421 A126422 * A126424 A126425 A126426 KEYWORD sign,easy AUTHOR Artur Jasinski, Dec 26 2006 STATUS approved
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Last modified December 7 05:14 EST 2019. Contains 329839 sequences. (Running on oeis4.) | 590 | 1,689 | {"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} | 2.890625 | 3 | CC-MAIN-2019-51 | latest | en | 0.710283 |
https://quizlet.com/2120019/math-vocab-for-qlc1-flash-cards/ | 1,524,236,339,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125938462.12/warc/CC-MAIN-20180420135859-20180420155859-00444.warc.gz | 675,519,522 | 57,223 | 24 terms
# Math Vocab for QLC1
Math Vocab for QLC1
###### PLAY
geometric sequence
a sequence of numbers in which you can find the next term by multiplying the previous term by the same number
Order of operations
PEMDAS parentheses, exponents, multiplication or division, then addition or subtraction
Natural number
the positive integers (whole numbers) 1, 2, 3, etc., and sometimes zero as well.
rational number
A positive or negative number (decimal, fraction, or percent), A number that can be written with an integer in the numerator and a positive integer in the denominator. A number that can be written as a/b where a and b are integers, but b is not equal to 0.
integers
the set of positive whole numbers and their opposites(negative numbers) and 0
real number
all rational or irrational numbers; real numbers can be represented on the real number line. Any number that exists. Includes all rational and irrational numbers.
Round
(of numbers) to the nearest ten, hundred, or thousand
truncate
approximate by ignoring all terms beyond a chosen one
prime number
A whole number that has exactly two factors, 1 and itself. An integer that has no integral factors but itself and 1, is a whole number greater then 1 that has exactly two factors, one and itself (7,11,29)
composite number
A positive whole number with more than two factors. In other words, a number that is not prime. Zero and one are neither composite nor prime. A positive whole number with more than two factors. In other words, a number that is not prime. Zero and one are neither composite nor prime.
Exponents
The number that is small and raised to show how many times to multiply the number by itself.
Scientific notation
There are two parts of the scientific notation. The first part is a number between 1-10. The second part is a power of ten. example: 2.25 * 10^4; A standardized way of expressing very large or very small numbers as the product of a number between 1 and 10 and a power of 10 (2.5 x 10^6 = 2.500,000)
reciprocals
another name for a multiplicative inverse. another name for a multiplicative inverse
absolute value
The distance a number is from zero. Always positive.
Rate
A ratio that compares two quantities having different units (e.g., 95 miles in 2 hours). a ratio that compares unlike units
Proportion
A statement where two fractions/ratios are equal
Scale factor
the ratio of the lengths of two corresponding sides of two similar polygons
Percents
A number out of one hundred. Make the fraction a decimal then multiply by one hundred. (i.e. 25% is = to 25 out of 100.)
Ratio
A comparison of two quantities by division.
a comparison by a proportion like 3:7
irrational number
a number that cannot be put in the form of a fraction. includes square roots and non-repeating decimals. examples: .79146...., √5
non terminating decimal
a decimal the never ends and goes on forever; type of irrational number
Commission=Commission Rate X Sales
A. \$115 (commission) = R x \$2875
B. \$115 / 2875 (fraction) = R x 2875 / 2875 (fraction)
C. Cross multiply
D. divide 115 by 2875 = .04 or 4%
Percent difference
Scores on a college entrance exam rose from 20.0 in 1977 to 20.7 in 1986. What percent increase is this?
STEP 1: Find amount of INCREASE 20.7-20.0=0.7
STEP 2: The next step is to divide the increase by the ORIGINAL number. 0.7 / 20.0 (FRACTION) = 0.035
STEP 3: Finally, convert from decimal notation to percent notation. 0.035=3.5%
scientific notation
The number 123,000,000,000 in scientific notation is written as : 1.23 x 10(11) The first number 1.23 is called the coefficient. It must be greater than or equal to 1 and less than 10. The second number is called the base . It must always be 10 in scientific notation. The base number 10 is always written in exponent form. In the number 1.23 x 1011 the number 11 is referred to as the exponent or power of ten. To write a number in scientific notation: Put the decimal after the first digit and drop the zeroes. 1.23000000000. In the number 123,000,000,000 The coefficient will be 1.23
To find the exponent count the number of places from the decimal to the end of the number. In 123,000,000,000 there are 11 places. Therefore we write 123,000,000,000 as: 1.23 x 10(11) | 1,086 | 4,221 | {"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-2018-17 | latest | en | 0.921278 |
https://www.physicsforums.com/threads/person-throws-ball-in-elevator-problem.673139/ | 1,624,230,580,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488257796.77/warc/CC-MAIN-20210620205203-20210620235203-00590.warc.gz | 850,958,726 | 20,667 | # Person throws ball in elevator problem
Gold Member
## Homework Statement
An elevator moves up with a constant acceleration a and a person stands inside the elevator. At time ##t_1## > 0, the person throws a ball directly upwards with velocity u wrt himself. Find the distance of the ball from the man's hand. (assume he does not move his hand)
## The Attempt at a Solution
I have the solution to this in my notes, but I tried it differently and I get to a step that I can't solve. I want to know if it is correct up to where I can't solve the resulting diff eqn.
Before I consider the above, I have a conceptual question about acc/dec of elevators. Consider two cases: the elevator moving downwards at constant acceleration a and the elevator moving downwards at constant deceleration a. Am I correct in saying the normal force acting on the person from the elevator is the same in both cases? I have done some maths and I appear to get the same result.
Onto the question: vel. of ball wrt earth = vel. of ball wrt hand + vel. of hand wrt earth = u + v(t), where u is given in the problem statement and since the elevator is at constant a, v = v(t). Denote vel. of ball wrt earth = ##\dot{x}##. So ##\dot{x} = u + v(t)\,\Rightarrow\, \dot{x} - v(t) = u(t)##. (I think the vel. of the ball wrt hand changes over time.) I can't solve this eqn since I have two functions dependent on t. I am now looking for some assumption to make on u(t) so I can solve for it first (i.e something like rate of change of u proportional to how fast u is intially or something like that).. Any ideas? (or am I mistaken somewhere else)
Many thanks.
EDIT: I just reliased I made v an arbritary function of t. It should be a linear function in t, so like v = pt. I don't think it solves my problem, however.
Last edited:
haruspex
Homework Helper
Gold Member
2020 Award
the elevator moving downwards at constant acceleration a and the elevator moving downwards at constant deceleration a. Am I correct in saying the normal force acting on the person from the elevator is the same in both cases?
No. A downward deceleration is an upward acceleration (even though the velocity be downwards).
In the question, what is the actual acceleration of the ball? What is the acceleration of the ball relative to the man's hand?
What forces act on the ball? What is its acceleration?
Gold Member
No. A downward deceleration is an upward acceleration (even though the velocity be downwards).
Yes, that makes sense which is why I am asking the question: When I do some maths, I get: consider constant deceleration downwards. Take y positive down: ma = mg -N => N = m(g-a). Where have I used the fact that we have dec/acc?
In the question, what is the actual acceleration of the ball? What is the acceleration of the ball relative to the man's hand?
The only force acting on the ball is gravity: ##\ddot{x} = -g##.
So you should be able to solve for u(t) then.
Chestermiller
Mentor
The simplest way to do this problem is to reference everything to the earth. You can solve for the time-dependent location of the ball relative to the earth and for the time-dependent location of the man's hand relative to the earth. Then you can subtract.
Let x0 represent the location of both the man's hand and the ball at time t = 0 when the ball is released. Also, let v0 represent the upward velocity of the man's hand when the ball is released. From the problem statement, in terms of v0 and u, what is the upward velocity of the ball relative to the earth when it is released? What is the man's acceleration relative to the earth? After the ball is released, what is the ball's acceleration relative to the earth? You now know the initial location, the initial velocity, and the acceleration both of the man's hand and of the ball relative to the earth. You can now solve for the time-dependent location of the ball relative to the earth and for the time-dependent location of the man's hand relative to the earth. Then you can subtract.
Gold Member
So you should be able to solve for u(t) then.
Is it just u(t) = -g(t - t1) + u? Now I should sub this into what I got for the diff eqn and solve:$$\ddot{x} - v(t) = -g(t - t_1) + u$$ But I still have t on the RHS?
## \ddot{x} - \dot{x} ## makes no sense physically.
Gold Member
## \ddot{x} - \dot{x} ## makes no sense physically.
I meant to write ##\dot{x} - v(t) = -g(t-t_1) + u##
What is v(t)? Why?
Gold Member
The simplest way to do this problem is to reference everything to the earth. You can solve for the time-dependent location of the ball relative to the earth and for the time-dependent location of the man's hand relative to the earth. Then you can subtract.
Let x0 represent the location of both the man's hand and the ball at time t = 0 when the ball is released. Also, let v0 represent the upward velocity of the man's hand when the ball is released. From the problem statement, in terms of v0 and u, what is the upward velocity of the ball relative to the earth when it is released?
vel. of ball relative to earth: u + vo. Do I have to consider the velocity of the hand? The problem statment says the hand stays fixed.
What is the man's acceleration relative to the earth? After the ball is released, what is the ball's acceleration relative to the earth?
Mans acceleration rel. to earth: a
Balls acceleration rel. to earth = ##\frac{d}{dt} \left(u(t) + v_o \right)##, but I think the hand stays stationary, so acc. rel. to earth =## \dot{u}##
Gold Member
What is v(t)? Why?
v(t) is velocity of the elevator. Since the elevator is at a constant acceleration, ##v \propto t##
What is v(t) EXACTLY?
Gold Member
What is v(t) EXACTLY?
Do you mean it's functional form? Assuming the elevator started at zero velocity, v(t) = at, where a is the acceleration of the lift.
given v(t) = at, how do you get ## \dot{x} - v(t) = -g(t - t_1) + u ##?
Gold Member
given v(t) = at, how do you get ## \dot{x} - v(t) = -g(t - t_1) + u ##?
How did I derive it? I did everything in the earth frame:
(what we want to solve for is position of ball rel. to hand.)
vel. of ball rel. to hand = vel. of ball wrt earth - vel. of hand wrt earth, which means vel. of ball wrt earth = u(t) + at. Denote vel of ball wrt earth by ##\dot{x}##. So then:
##\dot{x} - at = u(t)## and I calculated u(t) earlier.
The problem is that u(t) = -g(t - t1) + u implies u'(t) = -g. So the acceleration of the ball with respect to the hand, which is accelerated with respect to the Earth, equals the acceleration of gravity. This cannot be correct.
Gold Member
The problem is that u(t) = -g(t - t1) + u implies u'(t) = -g. So the acceleration of the ball with respect to the hand, which is accelerated with respect to the Earth, equals the acceleration of gravity. This cannot be correct.
But gravity is the only force acting on the ball?
In the inertial frame of the Earth - yes. What about the accelerated frame of the hand? That's where your u(t) is.
Gold Member
In the inertial frame of the Earth - yes. What about the accelerated frame of the hand? That's where your u(t) is.
Once it has left the hand, then only gravity acts on the ball. But I suppose the initial push of the hand upwards would have given it some acceleration. So relative to the hand, we have an acceleration > g?
The hand ITSELF is accelerated with respect to the Earth. This is your accelerated reference frame. You should treat that as a non-inertial frame, with fictitious forces, or you should FIRST find the ball's velocity and the hand's velocity with respect to the Earth, and THEN subtract.
Gold Member
The hand ITSELF is accelerated with respect to the Earth. This is your accelerated reference frame. You should treat that as a non-inertial frame, with fictitious forces, or you should FIRST find the ball's velocity and the hand's velocity with respect to the Earth, and THEN subtract.
I see. I think I have it now: Let ##v_{XY}## = vel of X relative to Y So,
##v_{BH} = u - g(t - t_1) - at##.
Which means, distance of ball from hand relative to hand: ##\delta(s) = ut - \frac{1}{2}gt^2 - \frac{1}{2}at^2 + gt_1t##
Rather, ##\delta(s) = ut - ut_1 - \frac{1}{2}gt^2 - \frac{1}{2}gt_1^2 - \frac{1}{2}at^2 - \frac{1}{2}at_1^2 + gt_1t - gt_1^2## and then I can collect terms. This is nearly the answer in the other solution, but they don't have the last 2 terms.
Last edited:
I see. I think I have it now: Let ##v_{XY}## = vel of X relative to Y So,
##v_{BH} = u - g(t - t_1) - at##.
Let ## t = t1 ##. Then ##v_{BH} ## must be ## u ##. But it is not.
Which means, distance of ball from hand relative to hand: ##\delta(s) ... ##
That should be ## v_{BH} ##. What is ##\delta(s) ## and what is ## s ## to begin with?
Gold Member
Let ## t = t1 ##. Then ##v_{BH} ## must be ## u ##. But it is not.
Stupid mistake. When I set up the expression I neglected that t1 wasn't 0. So it should read vBH = u -g(t-t1) -a(t-t1)
That should be ## v_{BH} ##. What is ##\delta(s) ## and what is ## s ## to begin with?
I am finding the displacement of the ball from the hand. We know the velocity so I integrated. In the end, I get $$x(t) - x(t1) = ut - ut_1 - 1/2 gt^2 + 1/2 gt_1^2 + gt_1t -gt_1^2 - 1/2 at^2 - 1/2 at_1^2 + at_1t -at_1^2$$
Frankly, I am lost in all those terms. I would let vBH = u - (a + g)T, where T = t - t1, and integrate that. Then I would convert back to t and t1 if required.
What is the value of x at t1, by the way? | 2,546 | 9,430 | {"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} | 4.125 | 4 | CC-MAIN-2021-25 | latest | en | 0.941128 |
http://math.stackexchange.com/questions/54735/restricted-three-body-problem?answertab=active | 1,455,408,760,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701168065.93/warc/CC-MAIN-20160205193928-00139-ip-10-236-182-209.ec2.internal.warc.gz | 144,657,945 | 22,975 | # Restricted Three-Body Problem
The movement of a spacecraft between Earth and the Moon is an example of the infamous Three Body Problem. It is said that a general analytical solution for TBP is not known because of the complexity of solving the effect of three bodies which all pull on each other while moving, a total of six interactions. Mathematician Richard Arenstorf while at NASA solved a special case of this problem, by simplifying the interactions to four, because, the effect of the spacecraft's gravity upon the motion of the vastly more massive Earth and Moon is practically non-existent. Arenstorf found a stable orbit for a spacecraft orbiting between the Earth and Moon, shaped like an '8'
http://en.wikipedia.org/wiki/Richard_Arenstorf
Was Arenstorf's solution purely analytical, or did he use numerical mechanisms? Is the '8' shape an optimal path, meaning the route on which the spacecraft would expand the least amount of energy? If yes, how was this requirement included in the derivation in mathematical form?
If anyone has a clean derivation for this problem, that would be great, or any links to books, other papers, etc.
Note: Apparently there was an earlier related mathoverflow question on this as well:
http://mathoverflow.net/questions/52489/on-the-non-rigorous-calculations-of-the-trajectories-in-the-moon-landings
Arenstorf's technical report is here
http://hdl.handle.net/2060/19630005545
The most clean, succint description of the equations I could find were here
http://farside.ph.utexas.edu/teaching/celestial/Celestial.pdf
http://www.uibk.ac.at/mathematik/personal/csomos/arenstorf_en.pdf
Restricted 3 Body Problem mentioned in this book (chapter 8)
Some vpython (visual python) code can simulate Apollo 13 trip,
the result is the 8-orbit as seen here
Some explanation on the math, and superposition of gravities is also mentioned in this link.
http://maths.anu.edu.au/comptlsci/Tutorial-Gravity/tutorial_apollo.html
I have no idea yet how this would relate to Arenstorf's method, I believe in this code Earth and Moon are not moving, but I thought it'd be nice to share.
Regards,
-
Arenstorf actually found his orbits through numerics. There's a short discussion in Hairer/Norsett/Wanner with references. I'll post it if nobody beats me to it... – J. M. Jul 31 '11 at 11:53
he used numerics.. interesting. thanks. Hairer/Norsett/Wanner discussion would be much appreciated. – BB_ML Jul 31 '11 at 14:22
I'm far away from my notes, so it might take me a while to write something. :( I was about to suggest looking at Fehlberg's "Runge-kutta type formulas of high-order accuracy and their application to the numerical integration of the restricted problem of three bodies" but there seems to be no digital copy of that available. Additionally, there's related work (in German) by Filippi that should interest you as well. – J. M. Jul 31 '11 at 14:41
This was crossposted to MO. In the future, please wait some time before posting your question in multiple fora, and when you do, provide links to the other posts - as you can imagine, it would be frustrating for someone to put time into answering your question here, only to see hear from you that you'd already gotten the solution elsewhere. – Zev Chonoles Jul 31 '11 at 15:45
See this as well... – J. M. Jul 31 '11 at 15:49
In the comments, you were already informed that numerics had to be used so I will explain from that point on.
The figure 8 design isn't because it is an optimal path. It occurs due to the gravity of moon and the Earth. When the spacecraft comes within the sphere of influence (SOI) of the moon, the spacecraft is pulled towards it. If the spacecraft is moving at escape velocity, the moon will perturb the flight but the spacecraft won't do a fly by. With the current speed, the moon's gravity is enough to cause an orbital fly by. Upon exiting the moon's SOI, the spacecraft is being pulled in by the Earth. Since the trajectory of the fly by was throwing the spacecraft away from the moon, it crosses its original path but this is short lived since the Earth then pulls it back in. If the spacecraft would have picked up enough velocity from the orbital maneuver to be on a parabolic or hyperbolic trajectory, it could have escaped the pull of Earth and been sent out into space.
One way to determine test speeds in designing a flight is to find the Jacobi constant, $C$. For a given $C$, the zero velocity curves are determined. Since we wanted to reach the moon, $C\geq -1.6649$ which corresponds to an initial velocity of at least $10.85762$ but a velocity of $11.01$ is the required escape velocity from the Earth so the initial speed has to be less than $v_{esc}$.
We can derive the equations of motion only up to a point. I am skipping the basic derivation of the two body problem so we can move on to the crux of the matter. \begin{alignat*}{4} r_{12} & = \sqrt{(x_1 - x_2)^2} & \qquad & x_1 &&{}= \text{Is the } x \text{ location of } m_1\\ x_2 & = x_1 + r_{12} & & &&{}\phantom{=} \text{relative to the center of gravity.}\\ x_1 & = \frac{-m_2}{m_1 + m_2}r_{12} & & \pi_1 &&{}= \frac{m_1}{m_1 + m_2}\\ x_2 & = \frac{m_1}{m_1 + m_2}r_{12} & & \pi_2 &&{}= \frac{m_2}{m_1 + m_2}\\ 0 & = m_1x_1 + m_2x_2 \end{alignat*} We can describe the position of $m$ as $\mathbf{r} = x\hat{\mathbf{i}} + y\hat{\mathbf{j}} + z\hat{\mathbf{k}}$ in relation to the center of gravity, i.e., the origin. \begin{align} \mathbf{r}_1 & = (x - x_1)\hat{\mathbf{i}} + y\hat{\mathbf{j}} + z\hat{\mathbf{k}}\\ & = (x + \pi_2r_{12})\hat{\mathbf{i}} + y\hat{\mathbf{j}} + z\hat{\mathbf{k}}\tag{1}\\ \mathbf{r}_2 & = (x - x_2)\hat{\mathbf{i}} + y\hat{\mathbf{j}} + z\hat{\mathbf{k}}\\ & = (x - \pi_1r_{12})\hat{\mathbf{i}} + y\hat{\mathbf{j}} + z\hat{\mathbf{k}}\tag{2} \end{align} Let's define the absolute acceleration where $\omega$ is the initial angular velocity which is constant. Then $\omega = \frac{2\pi}{T}$. $$\ddot{\mathbf{r}}_{\text{abs}} = \mathbf{a}_{\text{rel}} + \mathbf{a}_{\text{CG}} + \mathbf{\varOmega}\times(\mathbf{\varOmega}\times\mathbf{r}) + \dot{\mathbf{\varOmega}}\times\mathbf{r} + 2\mathbf{\varOmega}\times\mathbf{v}_{\text{rel}} \tag{3}$$ where \begin{alignat*}{4} \mathbf{a}_{\text{rel}} & = \text{Rectilinear acceleration relative to the frame} & \quad & \mathbf{\varOmega}\times(\mathbf{\varOmega}\times\mathbf{r}) &&{}= \text{Centripetal acceleration}\\ 2\mathbf{\varOmega}\times\mathbf{v}_{\text{rel}} & = \text{Coriolis acceleration} \end{alignat*} Since the velocity of the center of gravity is constant, $\mathbf{a}_{\text{CG}} = 0$, and $\dot{\mathbf{\varOmega}} = 0$ since the angular velocity of a circular orbit is constant. Therefore, $(3)$ becomes: $$\ddot{\mathbf{r}} = \mathbf{a}_{\text{rel}} + \mathbf{\varOmega}\times(\mathbf{\varOmega}\times\mathbf{r}) + 2\mathbf{\varOmega}\times\mathbf{v}_{\text{rel}} \tag{4}$$ where \begin{align} \mathbf{\varOmega} & = \varOmega\hat{\mathbf{k}}\tag{5}\\ \mathbf{r} & = x\hat{\mathbf{i}} + y\hat{\mathbf{j}} + z\hat{\mathbf{k}}\tag{6}\\ \dot{\mathbf{r}} & = \mathbf{v}_{\text{CG}} + \mathbf{\varOmega}\times\mathbf{r} + \mathbf{v}_{\text{rel}}\tag{7}\\ \mathbf{v}_{\text{rel}} & = \dot{x}\hat{\mathbf{i}} + \dot{y}\hat{\mathbf{j}} + \dot{z}\hat{\mathbf{k}}\tag{8}\\ \mathbf{a}_{\text{rel}} & = \ddot{x}\hat{\mathbf{i}} + \ddot{y}\hat{\mathbf{j}} + \ddot{z}\hat{\mathbf{k}}\tag{9} \end{align} After substituting $(5)$, $(6)$, $(8)$, and $(9)$ into $(3)$, we obtain $$\ddot{\mathbf{r}} = \left(\ddot{x} - 2\varOmega\dot{y} - \varOmega^2x\right)\hat{\mathbf{i}} + \left(\ddot{y} + 2\varOmega\dot{x} - \varOmega^2y\right)\hat{\mathbf{j}} + \ddot{z}\hat{\mathbf{k}}. \tag{10}$$ Newton's $2^{\text{nd}}$ Law of Motion is $m\mathbf{a} = \mathbf{F}_1 + \mathbf{F}_2$ where $\mathbf{F}_1 = -\frac{Gm_1m}{r_1^3}\mathbf{r}_1$ and $\mathbf{F}_2 = -\frac{Gm_2m}{r_2^3}\mathbf{r}_2$. Let $\mu_1 = Gm_1$ and $\mu_2 = Gm_2$. \begin{align} m\mathbf{a} & = \mathbf{F}_1 + \mathbf{F}_2\\ m\mathbf{a} & = -\frac{m\mu_1}{r_1^3}\mathbf{r}_1 - \frac{m\mu_2}{r_2^3}\mathbf{r}_2\\ \mathbf{a} & = -\frac{\mu_1}{r_1^3}\mathbf{r}_1 - \frac{\mu_2}{r_2^3}\mathbf{r}_2\\ \bigl(\ddot{x} - 2\varOmega\dot{y} - \varOmega^2x\bigr)\hat{\mathbf{i}} + \bigl(\ddot{y} + 2\varOmega\dot{x} - \varOmega^2y\bigr)\hat{\mathbf{j}} + \ddot{z}\hat{\mathbf{k}} & = -\frac{\mu_1}{r_1^3}\mathbf{r}_1 - \frac{\mu_2}{r_2^3}\mathbf{r}_2\\ \bigl(\ddot{x} - 2\varOmega\dot{y} - \varOmega^2x\bigr)\hat{\mathbf{i}} + \bigl(\ddot{y} + 2\varOmega\dot{x} - \varOmega^2y\bigr)\hat{\mathbf{j}} + \ddot{z}\hat{\mathbf{k}} & = \begin{aligned} - & \frac{\mu_1}{r_1^3}\Bigl[(x + \pi_2r_{12})\hat{\mathbf{i}} + \hat{\mathbf{j}} + \hat{\mathbf{k}}\Bigr]\\ - & \frac{\mu_2}{r_2^3}\Bigl[(x - \pi_1r_{12})\hat{\mathbf{i}} + \hat{\mathbf{j}} + \hat{\mathbf{k}}\Bigr] \end{aligned}\tag{12} \end{align} Now all we have to do is equate the coefficients. \begin{align} \ddot{x} - 2\varOmega\dot{y} - \varOmega^2x & = -\frac{\mu_1}{r_1^3}(x + \pi_2r_{12}) - \frac{\mu_2}{r_2^3}(x - \pi_1r_{12})\\ \ddot{y} + 2\varOmega\dot{x} - \varOmega^2y & = -\frac{\mu_1}{r_1^3}y - \frac{\mu_2}{r_2^3}y\\ \ddot{z} & = -\frac{\mu_1}{r_1^3}z - \frac{\mu_2}{r_2^3}z \end{align} We now have system of nonlinear ODEs. We can assume the trajectory is in plane and we do that by letting $z = 0$ so we only have two equations remaining: \begin{align} \ddot{x} - 2\varOmega\dot{y} - \varOmega^2x & = -\frac{\mu_1}{r_1^3}(x + \pi_2r_{12}) - \frac{\mu_2}{r_2^3}(x - \pi_1r_{12})\\ \ddot{y} + 2\varOmega\dot{x} - \varOmega^2y & = -\frac{\mu_1}{r_1^3}y - \frac{\mu_2}{r_2^3}y \end{align} Now to achieve the require trajectory we can use the true anomaly of $-90^{\circ}$, a flight path angle of $20^{\circ}$, and an initial velocity of $v_0 = 10.9148$ km/s. At this point, we have no other choice but to numerically integrate. Here is a code that will achieve the desired plot. However, in the actually apollo flight, they had mid course corrections so they arrived exactly on the other side of Earth.
## Python
#!/usr/bin/env ipython
import numpy as np
import scipy
from scipy.integrate import odeint
from mpl_toolkits.mplot3d import Axes3D
from scipy.optimize import brentq
from scipy.optimize import fsolve
import pylab
me = 5.974 * 10 ** 24 # mass of the earth
mm = 7.348 * 10 ** 22 # mass of the moon
G = 6.67259 * 10 ** -20 # gravitational parameter
re = 6378.0 # radius of the earth in km
rm = 1737.0 # radius of the moon in km
r12 = 384400.0 # distance between the CoM of the earth and moon
M = me + mm
d = 200 # distance the spacecraft is above the Earth
pi2 = mm / M
mue = 398600.0 # gravitational parameter of earth km^3/sec^2
mum = G * mm # grav param of the moon
omega = np.sqrt(mu / r12 ** 3)
vbo = 10.9148
nu = -np.pi*0.5
gamma = 20*np.pi/180.0 # angle in radians of the flight path
vx = vbo * (np.sin(gamma) * np.cos(nu) - np.cos(gamma) * np.sin(nu))
# velocity of the bo in the x direction
vy = vbo * (np.sin(gamma) * np.sin(nu) + np.cos(gamma) * np.cos(nu))
# velocity of the bo in the y direction
xrel = (re + d)*np.cos(nu)-pi2*r12
# spacecraft x location relative to the earth
yrel = (re + 200.0) * np.sin(nu)
u0 = [xrel, yrel, 0, vx, vy, 0]
def deriv(u, dt):
return [u[3], # dotu[0] = u[3]
u[4], # dotu[1] = u[4]
u[5], # dotu[2] = u[5]
(2 * omega * u[4] + omega ** 2 * u[0] - mue * (u[0] + pi2 * r12) /
np.sqrt(((u[0] + pi2 * r12) ** 2 + u[1] ** 2) ** 3) - mum *
(u[0] - pi1 * r12) /
np.sqrt(((u[0] - pi1 * r12) ** 2 + u[1] ** 2) ** 3)),
# dotu[3] = that
(-2 * omega * u[3] + omega ** 2 * u[1] - mue * u[1] /
np.sqrt(((u[0] + pi2 * r12) ** 2 + u[1] ** 2) ** 3) - mum * u[1] /
np.sqrt(((u[0] - pi1 * r12) ** 2 + u[1] ** 2) ** 3)),
# dotu[4] = that
0] # dotu[5] = 0
dt = np.linspace(0.0, 535000.0, 535000.0) # secs to run the simulation
u = odeint(deriv, u0, dt)
x, y, z, x2, y2, z2 = u.T
# 3d plot
fig = pylab.figure()
ax.plot(x, y, z, color = 'r')
# 2d plot if you want
#fig2 = pylab.figure(2)
#ax2.plot(x, y, color = 'r')
phi = np.linspace(0, 2 * np.pi, 100)
theta = np.linspace(0, np.pi, 100)
xm = rm * np.outer(np.cos(phi), np.sin(theta)) + r12 - pi2 * r12
ym = rm * np.outer(np.sin(phi), np.sin(theta))
zm = rm * np.outer(np.ones(np.size(phi)), np.cos(theta))
ax.plot_surface(xm, ym, zm, color = '#696969', linewidth = 0)
ax.auto_scale_xyz([-8000, 385000], [-8000, 385000], [-8000, 385000])
xe = re * np.outer(np.cos(phi), np.sin(theta)) - pi2 * r12
ye = re * np.outer(np.sin(phi), np.sin(theta))
ze = re * np.outer(np.ones(np.size(phi)), np.cos(theta))
ax.plot_surface(xe, ye, ze, color = '#4169E1', linewidth = 0)
ax.auto_scale_xyz([-8000, 385000], [-8000, 385000], [-8000, 385000])
ax.set_xlim3d(-20000, 385000)
ax.set_ylim3d(-20000, 80000)
ax.set_zlim3d(-50000, 50000)
pylab.show()
## Matlab
script:
days = 24*3600;
G = 6.6742e-20;
rmoon = 1737;
rearth = 6378;
r12 = 384400;
m1 = 5974e21;
m2 = 7348e19;
M = m1 + m2;
pi_1 = m1/M;
pi_2 = m2/M;
mu1 = 398600;
mu2 = 4903.02;
mu = mu1 + mu2;
W = sqrt(mu/r12^3);
x1 = -pi_2*r12;
x2 = pi_1*r12;
d0 = 200;
phi = -90;
v0 = 10.9148;
gamma = 20;
t0 = 0;
tf = 6.2*days;
r0 = rearth + d0;
x = r0*cosd(phi) + x1;
y = r0*sind(phi);
vx = v0*(sind(gamma)*cosd(phi) - cosd(gamma)*sind(phi));
vy = v0*(sind(gamma)*sind(phi) + cosd(gamma)*cosd(phi));
f0 = [x; y; vx; vy];
[t,f] = rkf45(@rates, [t0 tf], f0);
x = f(:,1);
y = f(:,2);
vx = f(:,3);
vy = f(:,4);
xf = x(end);
yf = y(end);
vxf = vx(end);
vyf = vy(end);
df = norm([xf - x2, yf - 0]) - rmoon;
vf = norm([vxf, vyf]);
plot(x,y)
functions: 2 separate ones
function [tout, yout] = rkf45(ode_function, tspan, y0, tolerance)
a = [0 1/4 3/8 12/13 1 1/2];
b = [0 0 0 0 0
1/4 0 0 0 0
3/32 9/32 0 0 0
1932/2197 -7200/2197 7296/2197 0 0
439/216 -8 3680/513 -845/4104 0
-8/27 2 -3544/2565 1859/4104 -11/40];
c4 = [25/216 0 1408/2565 2197/4104 -1/5 0];
c5 = [16/135 0 6656/12825 28561/56430 -9/50 2/55];
if nargin < 4
tol = 1.e-8;
else
tol = tolerance;
end
t0 = tspan(1);
tf = tspan(2);
t = t0;
y = y0;
tout = t;
yout = y';
h = (tf - t0)/100; % Assumed initial time step.
while t < tf
hmin = 16*eps(t);
ti = t;
yi = y;
for i = 1:6
t_inner = ti + a(i)*h;
y_inner = yi;
for j = 1:i-1
y_inner = y_inner + h*b(i,j)*f(:,j);
end
f(:,i) = feval(ode_function, t_inner, y_inner);
end
te = h*f*(c4' - c5');
te_max = max(abs(te));
ymax = max(abs(y));
te_allowed = tol*max(ymax,1.0);
delta = (te_allowed/(te_max + eps))^(1/5);
if te_max <= te_allowed
h = min(h, tf-t);
t = t + h;
y = yi + h*f*c5';
tout = [tout;t];
yout = [yout;y'];
end
h = min(delta*h, 4*h);
if h < hmin
fprintf(['\n\n Warning: Step size fell below its minimum\n'...
' allowable value (%g) at time %g.\n\n'], hmin, t)
return
end
end
second function:
function dfdt = rates(t,f)
r12 = 384400;
m1 = 5974e21;
m2 = 7348e19;
M = m1 + m2;
pi_1 = m1/M;
pi_2 = m2/M;
mu1 = 398600;
mu2 = 4903.02;
mu = mu1 + mu2;
W = sqrt(mu/r12^3);
x1 = -pi_2*r12;
x2 = pi_1*r12;
x = f(1);
y = f(2);
vx = f(3);
vy = f(4);
r1 = norm([x + pi_2*r12, y]);
r2 = norm([x - pi_1*r12, y]);
ax = 2*W*vy + W^2*x - mu1*(x - x1)/r1^3 - mu2*(x - x2)/r2^3;
ay = -2*W*vx + W^2*y - (mu1/r1^3 + mu2/r2^3)*y;
dfdt = [vx; vy; ax; ay];
end
- | 5,680 | 15,121 | {"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": 5, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2016-07 | longest | en | 0.909946 |
https://chercher.tech/python-certification/python-certification-exam-preparation-set-27 | 1,621,360,325,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991288.0/warc/CC-MAIN-20210518160705-20210518190705-00375.warc.gz | 190,154,263 | 8,132 | ## Python Certification Exam Preparation Set 27
##### What is the output when we execute list("hello")?
Options are :
• ['h', 'e', 'l', 'l', 'o'] (Correct)
• ['hello']
• ['llo']
• ['olleh']
Answer :['h', 'e', 'l', 'l', 'o']
Options are :
• 5 (Correct)
• 4
• None
• Error
##### What will be the output of the following Python code?`x = ['ab', 'cd']for i in x: x.append(i.upper())print(x)`
Options are :
• none of the mentioned (Correct)
• ['AB', 'CD']
• ['ab', 'cd']
• ['ab', 'cd', 'AB', 'CD']
##### What will be the output of the following Python code snippet?`['hello', 'morning'][bool('')]`
Options are :
• no output
• error
• morning
• hello (Correct)
##### Which of these about a set is not true?
Options are :
• Data type with unordered values
• Mutable data type
• Allows duplicate values
• Immutable data type (Correct)
##### Which of the following Boolean expressions is not logically equivalent to the other three?
Options are :
• not(-6<10 or-6==10)
• -6>=0 and -6<=10
• not(-6<0 or-6>10)
• not(-6>10 or-6==10) (Correct)
##### What will be the output of the following Python code?`>>>t1 = (1, 2, 4, 3)``>>>t2 = (1, 2, 3, 4)``>>>t1 < t2`
Options are :
• None
• Error
• False (Correct)
• True
##### What will be the output of the following Python code?`>>>t = (1, 2, 4, 3, 8, 9)``>>>[t[i] for i in range(0, len(t), 2)]`
Options are :
• [1, 4, 8] (Correct)
• (1, 4, 8)
• [1, 2, 4, 3, 8, 9]
• [2, 3, 9]
##### What will be the output of the following Python code?`>>>my_tuple = (1, 2, 3, 4)``>>>my_tuple.append( (5, 6, 7) )``>>>print len(my_tuple)`
Options are :
• Error (Correct)
• 2
• 1
• 5
##### What will be the output of the following Python code?`>>>t=(1,2,4,3)``>>>t[1:-1]`
Options are :
• (2, 4, 3)
• (1, 2, 4)
• (2, 4) (Correct)
• (1, 2)
##### What will be the output of the following Python code snippet?`X="get-job"print("%56s",X)`
Options are :
• 56 blank spaces after get-job
• 56 blank spaces before get-job (Correct)
• 56 blank spaces before get-job
• no change
Answer :56 blank spaces before get-job
##### What will be the output of the following Python code?`import math[str(round(math.pi)) for i in range (1, 6)]`
Options are :
• ['3.1', '3.14', '3.142', '3.1416', '3.14159', '3.141582']
• ['3', '3', '3', '3', '3'] (Correct)
• ['3', '3', '3', '3', '3', '3']
• ['3.1', '3.14', '3.142', '3.1416', '3.14159']
Answer :['3', '3', '3', '3', '3']
##### What will be the output of the following Python code?`l1=[1,2,3]l2=[4,5,6]l3=[7,8,9]for x, y, z in zip(l1, l2, l3): print(x, y, z)`
Options are :
• Error
• 1 4 7 2 5 8 3 6 9 (Correct)
• [(1, 4, 7), (2, 5, 8), (3, 6, 9)]
• (1 4 7) (2 5 8) (3 6 9)
Answer : 1 4 7 2 5 8 3 6 9
##### What will be the output of the following Python code?`x = ['ab', 'cd']for i in x: i.upper()print(x)`
Options are :
• none of the mentioned
• [None, None]
• ['AB', 'CD']
• ['ab', 'cd'] (Correct)
##### Which of the following statements is used to create an empty set?
Options are :
• [ ]
• ( )
• set() (Correct)
• { }
Options are :
• 45
• "john"
• "peter"
• 40 (Correct)
##### What will be the output of the following Python code?`>>>t=(1,2,4,3)``>>>t[1:3]`
Options are :
• (1, 2)
• (2, 4, 3)
• (2, 4) (Correct)
• (1, 2, 4)
Options are :
• f
• Error
• t (Correct)
• No output
##### What will be the output of the following Python code?`l=[1, 0, 2, 0, 'hello', '', []]list(filter(bool, l))`
Options are :
• [1, 0, 2, 0, 'hello', ", []]
• Error
• [1, 0, 2, 'hello', ", []]
• [1, 2, 'hello'] (Correct)
##### Which of the following is not the correct syntax for creating a set?
Options are :
• set((1,2,3,4))
• set([1,2,2,3,4])
• set([[1,2],[3,4]]) (Correct)
• {1,2,3,4}
##### Which of the following is a Python tuple?
Options are :
• {1, 2, 3}
• (1, 2, 3) (Correct)
• [1, 2, 3]
• {}
##### What will be the output of the following Python code?`l=[[1 ,2, 3], [4, 5, 6], [7, 8, 9]][[row[i] for row in l] for i in range(3)]`
Options are :
• Error
• (1 4 7) (2 5 8) (3 6 9)
• [[1, 4, 7], [2, 5, 8], [3, 6, 9]] (Correct)
• 1 4 7 2 5 8 3 6 9
Answer :[[1, 4, 7], [2, 5, 8], [3, 6, 9]]
Options are :
• 33 (Correct)
• 24
• 30
• 12
##### What will be the output of the following Python code?`class Truth: passx=Truth()bool(x)`
Options are :
• pass
• true (Correct)
• false
• error
##### What will be the output of the following Python code?`>>>t = (1, 2)``>>>2 * t`
Options are :
• [1, 1, 2, 2]
• (1, 2, 1, 2) (Correct)
• (1, 1, 2, 2)
• [1, 2, 1, 2]
##### What will be the output of the following Python code?`>>> a={5,4}>>> b={1,2,4,5}>>> a<b`
Options are :
• False
• True (Correct)
• {1,2}
• Invalid operation
##### What will be the output of the following Python code?`nums = set([1,1,2,3,3,3,4,4])print(len(nums))`
Options are :
• Error, invalid syntax for formation of set
• 8
• 7
• 4 (Correct)
##### What will be the output of the following Python code?`if (9 < 0) and (0 < -9): print("hello")elif (9 > 0) or False: print("good")else: print("bad")`
Options are :
• error
• hello
• good (Correct)
##### What will be the output of the following Python code snippet?`X="hi"print("05d"%X)`
Options are :
• hi000
• 00000hi
• error (Correct)
• 000hi
##### What will be the output of the following Python code?`a = [5,5,6,7,7,7]b = set(a)def test(lst): if lst in b: return 1 else: return 0for i in filter(test, a): print(i,end=" ")`
Options are :
• 5 5 6 7 7 7 (Correct)
• 5 6 7
• 5 5 6
• 5 6 7 7 7
Answer :5 5 6 7 7 7
Comment / Suggestion Section
Point our Mistakes and Post Your Suggestions | 2,137 | 5,517 | {"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} | 2.765625 | 3 | CC-MAIN-2021-21 | latest | en | 0.580438 |
http://www.jiskha.com/display.cgi?id=1266767442 | 1,498,158,500,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128319688.9/warc/CC-MAIN-20170622181155-20170622201155-00051.warc.gz | 564,835,228 | 3,776 | # MATH156
posted by on .
made 135 free throws in 180 attempts. Of the next 50 free
throws attempted, how many would she have to make to
raise her percent of free throws made by 5%?
• MATH156 - ,
Her current percentage of successes is 135/180 = 75%.
She now makes 50 more free throws, for a total of 180+50 = 230. She wants to get 75% + 5% = 80%. How many of those 230 does she need to make?
Subtract 135 to find the number of free throws out of the extra 50. | 137 | 464 | {"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-2017-26 | latest | en | 0.970522 |
https://kotlin.link/articles/Feedback-on-the-Josephus-problem.html | 1,723,547,771,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641076695.81/warc/CC-MAIN-20240813110333-20240813140333-00058.warc.gz | 257,395,194 | 3,919 | Feedback on the Josephus problem
Nicolas Frankel
My last week article was about the solving the Josephus problem in Kotlin. For ease of comparison, here’s the version I wrote originally:
``````class Soldier(val position: Int) {
var state = State.Living
lateinit var next: Soldier
fun suicide() {
}
}
enum class State {
}
class Circle(private val size: Int, private val step: Int) {
private val first = Soldier(0)
init {
var person = first
while (person.position < size - 1) {
person = createNext(person)
}
val last = person
last.next = first
}
private fun createNext(soldier: Soldier): Soldier {
val new = Soldier(soldier.position + 1)
soldier.next = new
return new
}
fun findSurvivor(): Soldier {
var soldier: Soldier = first
while (numberOfDead < size - 1) {
var count: Int = 0
while (count < step) {
soldier = nextLivingSoldier(soldier)
count++
}
soldier.suicide()
}
return nextLivingSoldier(soldier)
}
private fun nextLivingSoldier(soldier: Soldier): Soldier {
var currentSoldier = soldier.next
currentSoldier = currentSoldier.next
}
return currentSoldier
}
}
``````
The post ended with an open question: was the code the right way to do it? In particular:
• Is the code idiomatic Kotlin?
• Lack of `for` means using `while` with `var`s
• Too many mutability (`var`) for my own liking
I’ve received great feedback from the community on many different channels, including Ilya Ryzhenkov from JetBrains, Cédric Beust, Peter Somerhoff and Gaëtan Zoritchak. Thanks guys!
I think the most interesting is this one, very slightly modified from the original Gist by Gaëtan:
``````class Soldier(val position: Int, var state:State = State.Living) {
fun suicide() {
}
fun isAlive() = state == State.Living
}
enum class State {
}
class Circle(val size: Int, val step: Int) {
val soldiers = Array( size, {Soldier(it)}).toList()
fun findSurvivor(): Soldier {
var soldier = soldiers.first()
(2..size).forEach {
(1..step).forEach {
soldier = soldier.nextLivingSoldier()
}
soldier.suicide()
}
return soldier.nextLivingSoldier()
}
tailrec private fun Soldier.nextLivingSoldier():Soldier =
if (next().isAlive())
next()
else
next().nextLivingSoldier()
private fun Soldier.next() = soldiers.get(
if (position == size - 1)
0
else
position + 1
)
}
``````
I like this code a lot because it feels more Kotlin-esque. Improvements are many.
• The code is shorter without any loss of readability
• It’s entirely functional, there’s only a single `var` involved:
• the 2 `while` loops with their associated `var` counter have been replaced by simple `forEach` on `Int` ranges.
• Chaining `Soldier` instances is not handled in the `Soldier` class itself through the `next()` method but by the containing `Circle`. Thus, a simple backing array can store them and there’s no need for custom code with mutable variables.
• The recursive `nextLivingSoldier()` function has been “annotated” with `tailrec` in order for the compiler to run its optimization magic.
• The `Soldier` class doesn’t know about its container `Circle`‘s `size`, so functions using it have been moved inside the `Circle` class as extension functions to the `Soldier` class. This is a great usage of Kotlin’s extension functions.
This experience reinforced my belief that learning a language by just reading about it is not enough. To truly make it yours, steps should be those:
1. obviously, learn the syntax of the language – and its API,
2. code a solution or an app with this language, | 866 | 3,456 | {"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} | 2.765625 | 3 | CC-MAIN-2024-33 | latest | en | 0.725195 |
https://www.numbersaplenty.com/4745331 | 1,725,765,958,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650958.30/warc/CC-MAIN-20240908020844-20240908050844-00357.warc.gz | 882,416,336 | 3,410 | Search a number
4745331 = 33175753
BaseRepresentation
bin10010000110100001110011
322221002101000
4102012201303
52203322311
6245413043
755222533
oct22064163
98832330
104745331
112751268
12170a183
13ca1bb6
148b74c3
1563b056
hex486873
4745331 has 8 divisors (see below), whose sum is σ = 7030160. Its totient is φ = 3163536.
The previous prime is 4745317. The next prime is 4745339. The reversal of 4745331 is 1335474.
4745331 = T301 + T302 + ... + T381.
It is not a de Polignac number, because 4745331 - 26 = 4745267 is a prime.
It is a super-2 number, since 2×47453312 = 45036332599122, which contains 22 as substring.
It is a Harshad number since it is a multiple of its sum of digits (27), and also a Moran number because the ratio is a prime number: 175753 = 4745331 / (4 + 7 + 4 + 5 + 3 + 3 + 1).
It is a d-powerful number, because it can be written as 47 + 72 + 411 + 55 + 33 + 312 + 1 .
It is a Duffinian number.
It is a junction number, because it is equal to n+sod(n) for n = 4745295 and 4745304.
It is not an unprimeable number, because it can be changed into a prime (4745339) by changing a digit.
It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 87850 + ... + 87903.
It is an arithmetic number, because the mean of its divisors is an integer number (878770).
Almost surely, 24745331 is an apocalyptic number.
4745331 is a deficient number, since it is larger than the sum of its proper divisors (2284829).
4745331 is a wasteful number, since it uses less digits than its factorization.
4745331 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 175762 (or 175756 counting only the distinct ones).
The product of its digits is 5040, while the sum is 27.
The square root of 4745331 is about 2178.3780663604. The cubic root of 4745331 is about 168.0436748694.
The spelling of 4745331 in words is "four million, seven hundred forty-five thousand, three hundred thirty-one".
Divisors: 1 3 9 27 175753 527259 1581777 4745331 | 652 | 2,050 | {"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} | 3.203125 | 3 | CC-MAIN-2024-38 | latest | en | 0.861243 |
http://www.algebra.com/cgi-bin/show-question-source.mpl?solution=35548 | 1,369,270,116,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368702652631/warc/CC-MAIN-20130516111052-00037-ip-10-60-113-184.ec2.internal.warc.gz | 320,027,048 | 791 | Question 53307
You're right you must be checking it wrong:
5(2)-3(-1)=13
10+3=13
13=13
The first equation checks, now check the second equation:
4(2)-3(-1)=11
8+3=11
11=11
The second equation checks.
Congratulations, you were right!!!! | 78 | 235 | {"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} | 3.1875 | 3 | CC-MAIN-2013-20 | latest | en | 0.834463 |
https://www.physicsforums.com/threads/position-velocity-and-acceleration-yes-or-no-question.374158/ | 1,544,929,689,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376827252.87/warc/CC-MAIN-20181216025802-20181216051802-00390.warc.gz | 975,725,919 | 14,304 | # Homework Help: Position, velocity, and acceleration yes or no question
1. Jan 31, 2010
### mybrohshi5
1. The problem statement, all variables and given/known data
A ball is lodged in a hole in the floor near the outside edge of a merry-go-round that is turning at constant speed. assuming that the position is measured from an origin at the center of the merry-go-round do the magnitudes of the position, velocity, and acceleration vectors change with time?
2. Relevant equations
N/A
3. The attempt at a solution
I believe the answer is yes because at any given time the position will be different and at any given time the velocity and acceleration will be different because they change with direction since they are vectors.
Am i correct about this?
Thank you
2. Jan 31, 2010
### mybrohshi5
now i am rethinking this because it says magnitude so is it maybe no because the magnitude of the position will be the same?
3. Jan 31, 2010
### PhanthomJay
Yes, No is correct, now you got it right.
4. Jan 31, 2010
### mybrohshi5
ok thank you :)
5. Jan 31, 2010
### mybrohshi5
just to make sure i am understanding this correctly
would none of the magnitudes change with time because the magnitude of a vector changes with time only if the length changes with time and the length of this never changes correct?
6. Jan 31, 2010
### mybrohshi5
bump^^^^
anyone know if i have the right theory behind this?
would none of the magnitudes change with time because the magnitude of a vector changes with time only if the length changes with time and the length of this never changes correct?
thanks :)
7. Feb 1, 2010
### PhanthomJay
The position, velocity, and acceleration vectors all change with time, because their direction changes with time. Their magnitudes , however, stay constant; as you noted, the length of the position vector (the distance between the ball and the center of the carousel) does not change with time (it stays the same no matter where the ball is as it rotates along with the carousel). The magnitude of the velocity vector and the magnitude of the acceleration vector also do not change with time.......why?
8. Feb 1, 2010
### mybrohshi5
Does the magnitude of velocity vector and the magnitude of the acceleration vector not change with time because the speed of the carousel is constant?
9. Feb 1, 2010
### PhanthomJay
That is correct.
10. Feb 1, 2010
### mybrohshi5
Thank you for all the help and great explanations and other non-math related information.
I really appreciate the help
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook | 624 | 2,625 | {"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-2018-51 | latest | en | 0.914281 |
https://www.tetsa.com.tr/2021/06/08/dropping-odds/ | 1,721,372,594,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514866.83/warc/CC-MAIN-20240719043706-20240719073706-00299.warc.gz | 872,486,837 | 22,392 | Home Uncategorized Dropping Odds
# Dropping Odds
A point spread evens things out and gives both sides of the bet some wagering appeal. American odds, also known as moneyline odds, are primarily used by sites that cater to US sports bettors. Having knowledge about the raw math behind betting odds explained is useful, but of course you want to know about your potential payout too. Each different format for betting odds will have a different formula for calculating your winnings. Just like in all the betting odds explained in this guide, the higher your risk, the higher your potential payout.
## Betting On Horse Racing, Explained
Implied probability is useful because if your estimate of the probability of an event occurring is different than a How To Create A Waterfall Chart In Excel sportsbook’s you can and should adjust your bet accordingly. Therefore, if you wager \$40 on Pittsburgh, you will win \$30.77, and your payout would be \$70.77 . In order to win \$100 on the Penguins , you’ll need to wager \$130. If you wagered \$100 on the Senators , you’d be set to win \$110. On the flip side, the plus-minus juice can also encourage gamblers to wager on a longshot.
## Baseball Totals Explained
In other words, you are 4 times more likely to lose this pot than you are to win it. Your opponent is holding two cards, but we ignore those as our calculations in online Texas Hold’em poker are only based on the cards you can see and what could be left in the deck. Here’s our at-a-glance poker chart guide to pot odds and which hands to play. You can download and print out this Texas Hold’em poker cheat sheet to have next to you when you play. Click the image below to enlarge the poker odds chart or download the pdf here. We recommend checking our starting hands page for more information.
## Nba Betting Odds
Betting odds are traditionally formulated by the odds maker at the specific sports book. These lines and odds are formulated using sophisticated mathematical models that help the sports books predict the outcome of the game. The chance of a particular outcome occurring that a sportsbook has calculated is called the implied probability.
## Betting Odds Explained: How Are Football Odds Calculated
If Jason believes the coin flip will be heads, he must wager \$100 to win \$100. If Jamal believes Bonds will have more runs, RBI and home runs than Thomas, he must wager \$125 to win \$100. After all, sports handicappers are regular people who use their judgment and knowledge to assess probability; they’re subject to the same biases and errors as everyone else. In order to understand how American odds work, let’s use an MLB example from the 2018 World Series.
Bookmakers usually hire specialists, like traders and odds compilers to compile all the data possible and make sense of it. They have the best tools possible and work with the best software to ensure that they get near-perfect results and objective statistical evaluation of each game and the possibilities. How likely is it that an event will occur during the match? That’s why these people answer every day by providing odds for hundreds of thousands of games in different sports. These days there is just too much information for an average person to take in, so bookies employ these specialists.
## Low Risk Football Betting Strategy
Much like other major sports, total betting is an option on most matches. When betting the total, you are betting whether the total number of games played will exceed or fall below the total posted. It does not matter which player wins the match, or even how lopsided the score is. All that matters is the total numbers of games played when the match is completed.
If you are calling heads, then the favourable outcome will be heads. Therefore to get your probability value, you divide the favourable outcome by the number of possible outcomes . In the simplest of terms, probability is a scale running from 0 to 1 .
We are solely dedicated to providing helpful advice and information on everything related to online betting and sports betting online. Find everything you need from bookmaker comparison in the UK, Africa, India, Australia, New Zealand and Canada, to general betting advice and even free, expert betting tips. When you’ve found a match you want to bet on it’s always worth comparing odds from different betting sites to find the best available odds. You’ll find that if you’re willing to spend a little time shopping around you’ll find better odds than the odds you got from the first betting site. However, don’t expect to find a huge amount of difference as betting companies often use the same statistics to calculate their odds.
The variations can take some getting used to, but we’ll give you a breakdown on each format of betting odds explained. Converting implied probability into moneyline is a bit more complicated than converting it into decimal or fractional odds. The calculation depends on whether the implied probability is above 50% or below. They all reflect the same thing – the return you will receive as a ratio of the sum of money placed on a bet. To convert your odds to implied probabilities or an implied probability to odds you can use an odds conversion calculator.
0 comment | 1,063 | 5,258 | {"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} | 2.578125 | 3 | CC-MAIN-2024-30 | latest | en | 0.927241 |
https://www.coursehero.com/file/6514381/inversefunctions/ | 1,516,250,878,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887065.16/warc/CC-MAIN-20180118032119-20180118052119-00268.warc.gz | 877,887,215 | 112,774 | inversefunctions
# inversefunctions - n the last example from the previous...
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n the last example from the previous section we looked at the two functions and and saw that and as noted in that section this means that there is a nice relationship between these two functions. Let’s see just what that relationship is. Consider the following evaluations. In the first case we plugged into and got a value of -5. We then turned around and plugged into and got a value of -1, the number that we started off with.
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In the second case we did something similar. Here we plugged into and got a value of , we turned around and plugged this into and got a value of 2, which is again the number that we started with. Note that we really are doing some function composition here. The first case is really, and the second case is really, Note as well that these both agree with the formula for the compositions that we found in the previous section. We get back out of the function evaluation the number that we originally plugged into the composition. So, just what is going on here? In some way we can think of these two functions as undoing what the other did to a number. In the first case we plugged into and then plugged the result from this function evaluation back into and in some way undid what had done to and gave us back the original x that we started with. Function pairs that exhibit this behavior are called inverse functions . Before formally defining inverse functions and the notation that we’re going to use for them we need to get a definition out of the way. A function is called one-to-one if no two values of x produce the same y . Mathematically this is the same as saying,
So, a function is one-to-one if whenever we plug different values into the function we get different function values. Sometimes it is easier to understand this definition if we see a function that isn’t one-
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inversefunctions - n the last example from the previous...
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Ask a homework question - tutors are online | 518 | 2,482 | {"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-2018-05 | latest | en | 0.961443 |
https://richardwiseman.wordpress.com/2012/07/23/answer-to-the-friday-puzzle-164/ | 1,529,611,795,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864257.17/warc/CC-MAIN-20180621192119-20180621212119-00116.warc.gz | 707,277,542 | 34,051 | On Friday I posted this puzzle…..
Here are 10 numbered statements. How many of them are true?
1) Exactly one of these statements is true.
2) Exactly two of these statements are true.
3) Exactly three of these statements are true.
4) Exactly four of these statements are true.
5) Exactly five of these statements are true.
6) Exactly six of these statements are true.
7) Exactly seven of these statements are true.
8) Exactly eight of these statements are true.
9) Exactly nine of these statements are true.
10) Exactly ten of these statements are true.
If you haven’t tried to solve it, have a go now. For everyone else the answer is after the break.
All of the statements contradict one another, and therefore at most one of them can be true, in which case the other nine statements will all be false, which is what statement 1 asserts. Therefore statement 1 is the only true statement.
Note: I originally posted the wrong answer, saying it was statement 9. This is correct if you substitute the word ‘false’ for ‘true’ in the puzzle. Apologies for mix up.
Did you solve it?
I have produced an ebook containing 101 of the previous Friday Puzzles! It is called PUZZLED and is available for the Kindle (UK here and USA here) and on the iBookstore (UK here in the USA here). You can try 101 of the puzzles for free here.
1. Joy says:
But statement 9 is saying “Exactly nine of these statements are true.” rather than “Exactly nine of these statements are false”. Maybe you mean statement 1 is true?
1. SofARMaths says:
I agree. 1 is true…
2. -M- says:
?
No…
If they contradict they are not true. Only one statement can be true and that is statement 1. An alternative is that none of the statements is true.
1. -M- says:
I think the puzzle was wrong and should have been “exactly -n- of these statements are false”. Pity. Would have been a nice puzzle.
2. ivan says:
You are correct, all of them being false is also a consistent interpretation. Therefore the correct answer to the puzzle as (mis)printed is “we cannot say”, since there are two consistent interpretations.
3. Slightly Less Wise One says:
M
Any chance of you explaining your “none of the statements is true” answer Oh Wise One?
4. -M- says:
Answer 1: if 1 statement is true, it is statement 1 that is true.
Alternative: If none of the statements is true, they are all false. That is a correct possibility. They CAN all be false.
Try any other alternatives:
“3 statements are true”. Then statement 3 is true, but no others. That is incorrect.
Wise enough?
5. Slightly Less Wise One says:
Not really
3. Ian says:
Isn’t this tactic of posting the wrong answer to provoke debate getting a bit old?
1. Steve says:
It seems to work for FtB…
2. I came to the comment section to make the same point 🙂
Just to follow up on “Slightly Less Wise One”‘s comment:
The question is “How many of these statements are true”. You can answer 1, which is true, because then statement one is true. You can also answer 0, because if 0 is the answer, then none of the statements are true, so 0 is a correct answer.
3. Slightly Less Wise One says:
Sorry
Still don’t understand
You seem to be saying that your answers themselves determine the truth of the statements in some wierd kind of Heisenberg scenario.
You will have to spell it out in a more simple way for a dunce like me to understand.
4. 9 says “exactly nine of these statements are true”, not “exactly nine of these statements are false. 1 is the only one that is true, as far as I can tell.
5. Bazooka says:
Statement 9 asserts that exactly 9 statements are true. If 9 are false, then statement 1 is the only possible correct answer.
6. Anders says:
I wonder if it was the question that was wrong, and that all the statements should really have said …are false. As it was phrased it was trivial, the other way would I think have been more confusing.
1. extremepoker says:
I suspect this is the case, as you say trivial puzzle in current form. So where are all the ‘there are two correct answers’ brigade now? Those comments had me fairly puzzled over the weekend.
2. -M- says:
I said there where 2 correct answers: Statement 1 is true or none of them.
3. Photon Boy says:
I like it -M- .
4. Martha says:
How can there be 2 correct answers if they contradict each other? There are 2 possible answers, but they can’t both be correct!
5. Davis says:
@Martha — two correct answers is okay here because one of the correct answers is not one of “these statements.”
7. Surely it depends on whether the tree is on this side of the island, or whether the river flows faster than the fire can spread?
Yes, but only of you are being epistemic
2. Wigan FC says:
Gonna explain where trees and rivers come into it ?
3. -M- says:
Guess he found them between two flag poles
4. Anders says:
How could he? The dog ate all the marbles
8. Mervulon says:
Yes. Only one can be true, and 1) is the only one that so states. Richard, do you not read these prior to posting the solutions?
9. Niva says:
Exactly one of the Richard’s solution to this puzzler is not true.
10. Anataboga says:
Statement 1!
11. Michael Sternberg says:
I agree with all above and had pegged 1, for exactly the same reasons stated.
Perhaps, it’s “opposite day” today.
12. Roland says:
What I said last week. Statement 9 3/4 is true 😉
14. edgar1975 says:
I think the answer and puzzle are both wrong.
NB THIS IS A COMMENT ON RW’s ORIGINAL ANSWER
RW’s answer works if you substitute “true” with “false”.
I think he has done an incomplete cut and paste from a similar question on the Internet
16. This doesn’t make sense, surely?
Statement 1 in this case is true.
If it were “exactly ____ of these statements are FALSE…”, then 9 would be correct.
Am I right? I feel like this might be a trick in itself…
17. Furie says:
Oh thank god for that. I thought I was being really thick then not understanding that. I had 1 as the only possible true one too.
18. Klaas De Smedt says:
Did I solved it? 5 minutes ago I thought so, then I saw your answer and your logic that I can’t follow. The read some of the comments.
The real question is: Did Richard Wiseman solved it 😉
19. Roland says:
The answer to Friday’s question: it took Richard the full weekend and a number of our comments to solve it 🙂
20. Martha says:
That one was easy!!!
21. Lazy T says:
I still have problems…
exactly doesn’t mean exclusivley, so 1, 0, any of them, all of them, could be true.
The statements only contradict each other if you interpret ‘exactly’ to mean more than it does.
“at most one of them can be true, ….Therefore statement 1 is the only true statement.”
‘can be true’ doesn’t mean ‘is true’
I have other problems too but they’re for HungoverPedant.nit
1. Anders says:
What are you talking about? “Exactly one” can never mean 0 or two. It can only mean “one”. It is exactly the meaning of “exactly”
When someone says to you “be here at exactly 2pm” and you show up at 5am and wait, you won’t get far by claiming that “exactly” doesn’t mean exclusively so you decided to include some more hours.
If you want to be pedantic, wait for something that actually has multiple meanings
You are right that “can be” isn’t the same as “is”, but in this puzzle it’s kinda moot
I do wish there had been one more statement “exactly 0 of these statements are true” to remove the second solution that they’re all false, as M pointed out
2. Photon Boy says:
Exactly DOES imply exclusively, Lazy T. Thats kind of what it means !
3. Lazy T says:
I’m not looking for mutiple meanings for ‘exactly’ I just think it was superfluous, it’s use actually adds to my confusion as to which solution was being asked for.
‘Only seven of these…’, ‘Just four of……’, or even ‘Three of..’ would have been clearer.
and
they didn’t say I couldn’t turn up at 5am, but it was a mistake to sleep in their herbacious border.
22. jumbleguitar says:
1 or 0, surely?
23. Duncan says:
Shame the wording was wrong, it made the puzzle too easy.
24. Anonymous says:
What is true about, “This statement is true.”? It implies neither truth nor false.
25. Eddie says:
Can I recommend listening to the ‘Logic’ episode of ‘In Our Time’ with Melvyn Bragg. It’s right up this street.
26. Christian says:
“here are 10 numbered statements” is also a true statement, so there are exactly 2 true statements.
1. Anonymous says:
…except it asks how many of the numbered statements are true, and “here are 10 numbered statements” is not numbered.
I’m tempted to say “Read the question” but if Richard Wiseman doesn’t RTQ, why should we expect Christian to?
27. Too many book tours, Richard. How many puzzles have required a correction after posting this year? Half?
28. Christian says:
I did read the question, and my statement still holds. I am not disputing the answer, but had the question been phrased just slightly different it could contain the twist I mentioned, thus making it a more interesting puzzle.
29. I said that there are two solutions.
One of them is that only one is true. The other solution is that none of them are true.
1. Just cause everyone disagrees, doesn’t mean one of them has to be right.
I am irresistibly prompted to mention religion at this point.
30. And “none are true” is a better solution (or, at least it’s more fun). A string of words does not have to be either true or false, consider: “are of true these Exactly five statements”, “are exactly five of these true?” or “flibble”.
One solution to the liar paradox uses this; “this sentence is false” as it refers to itself is taken to be just “flibble”, meaningless nonesense. All of the above numbered “statements” refer to themselves, are thus flibble and not statments at all. It is the assertion in the question, “here are 10 numbered statements”, that is false!
1. Photon Boy says:
Statements can be false (ask any politician). Therefore there is nothing false about “here are 10 numbered statements”.
31. Bob says:
Actually, according to widely accepted rules of logic (in mathematics) a statement cannot determine whether it’s true, nor can it determine the validity of any other stamement referring to the validity of the first statement. Therefore, none of these statements are either true or false. This rule was established in order to avoid paradoxes such as: 1. Statement number 2 is true. 2. statement numer 1 is false.
32. There is a subtlety here that Richard overlooked, and a quite interesting one: Statement number 1 is the only one that *can* be true, but it doesn’t therefore follow that it *is* true. In essence, the statement says “this sentence is true”, and there is no way to resolve whether or not it in fact is. This is kind of a converse to the Liar’s Paradox; the sentence “this sentence is false” has indeterminate truth value, because assuming it’s true implies it is false and vice versa. The given sentence, “this sentence is true”, confirms any assumption we make of it, but we can’t get anything else out of it, so it too has indeterminate truth value. So the correct answer is either zero or “at most one”, depending on your preferences.
33. joshthelegend says:
I’m not really sure what there is to argue about on this one. It’s clearly either just No. 1 or none at all. I’m 14, and I got that in about ten seconds. | 2,745 | 11,345 | {"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-2018-26 | latest | en | 0.956643 |
https://www.physicsforums.com/threads/clep-test-question.351833/ | 1,508,748,803,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825812.89/warc/CC-MAIN-20171023073607-20171023093607-00136.warc.gz | 961,526,281 | 16,095 | # CLEP Test Question
1. Nov 4, 2009
### 3.141592654
1. The problem statement, all variables and given/known data
I was looking into taking the Calculus CLEP Exam and came across this practice problem:
Derive the general solution to the following differential equation: y' = sin x.
A. sin y + cos x = C
B. tan y = C
C. sin y - cos x = C
D. cos x = C
E. cos x - sin y = C
3. The attempt at a solution
The first thing I thought was that the question was asking for me to integrate y'=sin X. If y' = sin X, then y = -cos x + C. I didn't know what they wanted me to do so I simply set this equal to zero, so -cos x + C = 0 and therefore cos X=C. However, D is not the correct answer. So, my question is: what is it that the question is asking me to do when it says it wants a "general solution to ...[a] differential equation"?
2. Nov 4, 2009
### 3.141592654
It might be relevant to state that choices A-E are answers for a multiple choice problem and not five parts to the problem.
3. Nov 4, 2009
### Staff: Mentor
Your first inclination was correct. The solution to this differential equation is y = -cos x + C. That is the general solution. I have no idea why this is not listed as an option. Is what you showed the exact wording of the problem?
4. Nov 4, 2009
### 3.141592654
Yea, the text I provided is the exact wording. The answer, which was listed at the bottom of the page, was listed as (A) and states:
A. In order to be correct, a general solution to a differential equation must satisfy both the homogenous and non-homogenous equations.
5. Nov 4, 2009
### Staff: Mentor
Our solution satisfies the DE. If y = -cos x + C, then dy/dx = sin x.
The homogeneous problem would be y' = 0, for which the solution is y = C. The nonhomogeneous problem is y' = sin x, for which the general solution is y = -cos x + C.
I think there might be a typo in the CLEP answer. They might have meant to say y + cos x = C, which would be equivalent to our answer, instead of sin y + cos x = C.
6. Nov 5, 2009
### 3.141592654
Can you explain what the homogeneous and nonhomogeneous problems are? I haven't come across those terms before.
7. Nov 5, 2009
### Staff: Mentor
A differential equation can be represented as f(y, y', y'', ..., y(n)) = g(x). On the left side you have some function of y and its derivatives. On the right side there is some other function of the independent variable.
The homogeneous equation is f(y, y', y'', ..., y(n)) = 0. The one above is the nonhomogeneous equation. For your problem, you have y' = sin x. (There is no y term.) The homogeneous equation is y' = 0, and the nonhomogeneous equation is y' = sin x.
Another example is y'' + y = x, a nonhomogeneous DE. The related homogeneous equation is y'' + y = 0, whose solution is y = Asin x + Bcos x. The general solution of the nonhomogeneous equation turns out to be y = Asin x + Bcos x + x, and is made up of the solution to the homogeneous equation plus what is called a particular solution to the nonhomogeneous problem.
The term "homogeneous" is also used in a different context in differential equations, which can lead to some confusion. In that sense of the word, a differential equation in the form y' = f(y/x) is said to be homogeneous. My sense is that in the world of differential equations, the description I gave before is much more commonly used. | 899 | 3,357 | {"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-2017-43 | longest | en | 0.947568 |
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View Full Version : Anyone ever play the game 31?
09-04-2005, 05:30 AM
Rules:
Each player is dealt 3 cards face down, discard/draw piles in middle. Each player has option to take face up card or draw and must discard after hand, like gin. Goal is to get highest total; cards are counted like blackjack but must be suited, with the exception that a set counts 30.5. AsTsTs = 31 wins instantly; all other hands must knock. Knocker lays 2:1 that he can beat at least one other player; each other player immediately following knock has one final turn to improve. Low hand loses.
Questions:
I'm struggling with the exact probabilities to find a) at what total one should knock first hand, dependent upon position [note that the 'UTG' player, upon knocking, would limit opps to one additional card each, while the 'button' player would face two chances - albeit only one where the opps knew the finite horizon] and b) what a reasonable rate of improvement would be over the course of several rounds of a hand [how quickly does a particular holding devalue?]
Any help, or just random thoughts, would be highly appreciated.
Cheers!
MickeyHoldem
09-04-2005, 07:59 AM
I calculated the average starting hand to be ~13.0
Given the chance to pick 1 additional card the average hand will improve to ~15.7
This was calculated through a sim.
Edit: Just noticed I didn't take into account the fact a set was 30.5... but I doubt that will change the numbers more than .1
magiluke
09-04-2005, 12:21 PM
I played it once. And that was years ago. I barely remember what the game is about even after reading you post.
09-04-2005, 03:29 PM
Sim, eh? Where could I find one of those? Are they available in such a flavor as to input any opening number of cards and draws?
Also, note that the average hand might not be the important one as the knocker has to lay 2:1.
magiluke
09-04-2005, 06:08 PM
Usually, especially with off the path games like this, you have to write it yourself.
Maybe this tutorial (http://www.cplusplus.com/doc/tutorial/) will help.
Man, that was sarcastic... I'm sorry =D
Piz0wn0reD!!!!!!
09-04-2005, 09:06 PM
i play a game called 13, or tien len. its the best.
09-05-2005, 06:26 AM
har har.
No worries, I figured as much. Unfortunately, I haven't done any programming for a while, nor do I know any C++. Of course, the saddest thing of all is that I just finished a degree in economics from a top-10 program, with a year's worth of statistics and a healthy helping of game theory, and I still get stuck doing what should be relatively basic [censored] like this.
And I'm unemployed.
Che sera, eh? Time for law school.
Tater10
09-05-2005, 01:23 PM
Here is the "random thoughts" you were asking for...
I played a game called schwimmen (sp). Very similar
http://www.pagat.com/commerce/schwim.html | 738 | 2,818 | {"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} | 3.296875 | 3 | CC-MAIN-2021-17 | latest | en | 0.959079 |
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Is n prime?
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Please note that nowhere we are given in the question that 'n' is an integer.
Statement 1. n is between 1 and 4.
Now if n is an integer, then there are only 2 possible values: 2 and 3 - both of which are prime.
But if n is not an integer, then there are infinite values possible: 2.01, 2.456, 3.97 etc- which are obviously not prime.
So Insufficient.
Statement 2. n^2 - 5n + 6 = 0
We can factorise this quadratic equation:
n^2 - 3n - 2n + 6=0
(n-3)(n-2) = 0
So either n=3 or n=2, Both of which are prime .
Sufficient - this gives YES as an answer to the question asked.
Hence B is the anwer
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Stmnt 1: 1<n<4 . Can be 1.5, 1.2,...3.9 : Not Suff
Stmnt 2: n^2 − 5n + 6 = 0 : => (n-3)(n-2)=0
So either n = 3 or n=2 , and since both 3 and 2 are prime numbers therefore B is sufficient.
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http://au.metamath.org/mpegif/reueq1.html | 1,531,985,598,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590711.36/warc/CC-MAIN-20180719070814-20180719090814-00620.warc.gz | 31,602,904 | 3,407 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > reueq1 Structured version Unicode version
Theorem reueq1 2906
Description: Equality theorem for restricted uniqueness quantifier. (Contributed by NM, 5-Apr-2004.)
Assertion
Ref Expression
reueq1
Distinct variable groups: , ,
Allowed substitution hint: ()
Proof of Theorem reueq1
StepHypRef Expression
1 nfcv 2572 . 2
2 nfcv 2572 . 2
31, 2reueq1f 2902 1
Colors of variables: wff set class Syntax hints: wi 4 wb 177 wceq 1652 wreu 2707 This theorem is referenced by: reueqd 2914 isplig 21765 isfrgra 28380 frgra3v 28392 1vwmgra 28393 3vfriswmgra 28395 hdmap14lem4a 32672 hdmap14lem15 32683 This theorem was proved from axioms: ax-1 5 ax-2 6 ax-3 7 ax-mp 8 ax-gen 1555 ax-5 1566 ax-17 1626 ax-9 1666 ax-8 1687 ax-6 1744 ax-7 1749 ax-11 1761 ax-12 1950 ax-ext 2417 This theorem depends on definitions: df-bi 178 df-or 360 df-an 361 df-tru 1328 df-ex 1551 df-nf 1554 df-sb 1659 df-eu 2285 df-cleq 2429 df-clel 2432 df-nfc 2561 df-reu 2712
Copyright terms: Public domain W3C validator | 502 | 1,147 | {"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} | 2.578125 | 3 | CC-MAIN-2018-30 | latest | en | 0.230856 |
http://mathhelpforum.com/calculus/8171-help-please-optimization-equation.html | 1,480,965,736,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541783.81/warc/CC-MAIN-20161202170901-00462-ip-10-31-129-80.ec2.internal.warc.gz | 172,848,514 | 10,403 | I can't get the equation for this question, I spent almost 2 hours!! This is all new to me so sorry for any inconveniance.
Here's the question:
A Car rental agency has 24 identical cars. The owner of the agency finds that a price of $10 per day, all the cars can be rented. However for each$1 increase in rental, on of the cars is not rented. What should he carge to maximize income?
2. If your price is $10, you rent all 24 cars; yielding: 10*24 =$240.
If your price is $11, you rent only 23 cars; yielding: 11*23 =$253.
If your price is $12, you're down to 22 cars; yielding: 12*22 =$264.
You see the pattern? As you can see, the profit is rising.
However, you can't keep doing this since the number of cars will go to zero.
In general, if the price is (10+n), you'll be renting (24-n) cars.
The income I(n) is still the product of the price with the number of cars.
3. Thank you so much....we did an example similar to this in class, just a different variation which confused me a bit, hence the long hours of thinking. Anyway thanks a lot.
4. You're welcome. Did you manage to solve it, what do you get as solution?
5. He needs to charge \$17 to maximize income, and I also checked the second order condition to make sure its correct. Thanks again!
6. I found the same answer | 334 | 1,288 | {"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-2016-50 | longest | en | 0.963857 |
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# angles are formed by adjacent sides of the polygon at the vertices, whereas angles are formed by the sides of the polygon when an adjacent side is extended.
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## Angles formed by adjacent sides of the polygon at the vertices are called interior angles. Examples of interior angles: ∠CBA,∠BAE Angles formed by the sides of the polygon when an adjacent side is extended are called exterior angles. Here, ∠B/BC is an exterior angle.
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https://www.foxinfotech.in/2022/12/create-vector-in-julia.html | 1,675,477,085,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500080.82/warc/CC-MAIN-20230204012622-20230204042622-00026.warc.gz | 777,212,321 | 66,522 | Home » Julia » How to Create Vector in Julia?
# How to Create Vector in Julia?
In Julia, a vector is a one-dimensional array of elements that can be accessed by their index. Vectors are commonly used to store and manipulate data in scientific and mathematical computing. There are several ways to create a vector in Julia. The most common way is to use the collect function, which takes an iterable object (such as an array or a range) as an argument and returns a vector with the same elements. Below are some examples:
## Create Vector in Julia Examples
```julia> collect(1:5)
5-element Array{Int64,1}:
1
2
3
4
5```
In this example, the `collect` function is used to create a vector with the elements `1`, `2`, `3`, `4`, and `5`. The `1:5` syntax is used to create a range of numbers from `1` to `5`, which is then passed as an argument to the `collect` function.
## Using vec Function in Julia
Another way to create a vector in Julia is to use the `vec` function, which is similar to `collect` but returns a vector of type `Vector` rather than a regular array. For example:
```julia> vec(1:5)
5-element Vector{Int64}:
1
2
3
4
5```
In this example, the `vec` function is used to create a vector with the elements `1`, `2`, `3`, `4`, and `5`. As you can see, the `vec` function produces the same result as the `collect` function, but the resulting vector is of type `Vector` rather than a regular array.
## Using the Vector Function in Julia
You can also use the `Vector` function to create a vector in Julia. This function takes a list of elements as an argument and returns a vector with the same elements. For example:
```julia> Vector([1, 2, 3, 4, 5])
5-element Vector{Int64}:
1
2
3
4
5```
In this example, the `Vector` function is used to create a vector with the elements `1`, `2`, `3`, `4`, and `5`. The list of elements is passed as an argument to the `Vector` function, which returns a vector with the same elements.
## Using the fill Function in Julia
You can also use the `fill` function to create a vector of a specific size with the same element repeated. For example:
```julia> fill(1, 5)
5-element Array{Int64,1}:
1
1
1
1
1
julia> fill(1, 5, vec)
5-element Vector{Int64}:
1
1
1
1
1
julia> fill(1, 5, Vector)
5-element Vector{Int64}:
1
1
1
1
1
julia> fill(1, 5, Vector{Int64})
5-element Vector{Int64}:
1
1
1
1
1```
In these examples, the `fill` function is used to create vectors of size 5 with the element `1` repeated. The `fill` function takes the element to repeat as the first argument and the size of the vector as the second argument. You can also specify a vector constructor as the third argument to control the type of the resulting vector.
Here are a few more examples of creating vectors with random elements in Julia:
```julia> rand(5)
5-element Array{Float64,1}:
0.572428
0.691394
0.591458
0.826224
0.209213
julia> rand(5, vec)
5-element Vector{Float64}:
0.687733
0.749905
0.413824
0.914159
0.250619
julia> rand(5, Vector)
5-element Vector{Float64}:
0.768625
0.454217
0.586898
0.0247973
0.998432
julia> rand(5, Vector{Int64})
5-element Vector{Int64}:
0
0
0
0
0```
In these examples, the `rand` function is used to create vectors with five random elements. The `rand` function takes the size of the vector as an argument and returns a vector with random elements of type `Float64`. You can also specify a vector constructor as the second argument to control the type of the resulting vector.
For example, in the fourth example above, the `rand` function is used to create a vector of type `Vector{Int64}` with five random elements. Since the elements of the vector are of type `Int64`, the random elements are rounded to the nearest integer. | 1,039 | 3,698 | {"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} | 3.0625 | 3 | CC-MAIN-2023-06 | latest | en | 0.817539 |
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# SUBJECT : Calculating net positive suction head (NPSH) in non-metric units 11-12.
The definition of NPSHA is simple: Static head + surface pressure head - the vapor
pressure of your product - the friction losses in the piping, valves and fittings.
But to really understand it, you first have to understand a couple of other concepts:
Cavitation is what net positive suction head (NPSH) is all about, so you need to
Vapor Pressure is another term we will be using. The product's vapor pressure
varies with the fluid's temperature.
Specific gravity play an important part in all calculations involving liquid. You
have to be familiar with the term.
You have to be able to read a pump curve to learn the N.P.S.H. required for your
pump.
You need to understand how the liquid's velocity affects its pressure or head.
It is important to understand why we use the term Head instead of Pressure when
we make our calculations.
Head loss is an awkward term, but you will need to understand it.
o You will have to be able to calculate the head loss through piping, valves
and fittings.
You must know the difference between gage pressure and absolute pressure.
Vacuum is often a part of the calculations, so you are going to have to be familiar
with the terms we use to describe vacuum.
## Lets look at each of these concepts in a little more detail :
Cavitation means cavities or holes in liquid. Another name for a hole in a liquid is
a bubble, so cavitation is all about bubbles forming and collapsing.
o Bubbles take up space so the capacity of our pump drops.
o Collapsing bubbles can damage the impeller and volute. This makes
cavitation a problem for both the pump and the mechanical seal.
Vapor pressure is about liquids boiling. If I asked you, "at what temperature does
water boil ?" You could say 212 F. or 100 C., but that is only true at
atmospheric pressure. Every product will boil (make bubbles) at some
combination of pressure and temperature. If you know the temperature of your
product you need to know its vapor pressure to prevent boiling and the formation
of bubbles. In the charts section of this web site you will find a vapor pressure
chart for several common liquids.
Specific gravity is about the weight of the fluid. Using 4C (39 F) as our
temperature standard we assign fresh water a value of one. If the fluid floats on
this fresh water it has a specific gravity is less than one. If the fluid sinks in this
water the specific gravity of the fluid is greater than one.
Look at any pump curve and make sure you can locate the values for head,
capacity, best efficiency point (B.E.P.), efficiency, net positive suction head
(NPSH), and horse power required. If you cannot do this, have someone show you
where they are located.
Liquid velocity is another important concept. As a liquid's velocity increases, its
pressure (90 to the flow) decreases. If the velocity decreases the pressure
increases. The rule is : velocity times pressure must remain a constant.
"Head" is the term we use instead of pressure. The pump will pump any liquid to
a given height or head depending upon the diameter and speed of the impeller.
The amount of pressure you get depends upon the weight (specific gravity) of the
liquid. The pump manufacturer does not know what liquid the pump will be
pumping so he gives you only the head that the pump will generate. You have to
figure out the pressure using a formula described later on in this paper.
Head (feet) is a convenient term because when combined with capacity (gallons
or pounds per minute) you come up with the conversion for horsepower (foot
pounds per minute).
"Head loss through the piping, valves and fittings" is another term we will be
using. Pressure drop is a more comfortable term for most people, but the term
"pressure" is not used in most pump calculations so you could substitute the term
"head drop" or "loss of head" in the system. To calculate this loss you will need to
be able to read charts like those you will find in the "charts you can use" section
in the home page of this web site. They are labeled Friction loss for water and
Resistance coefficients for valves and fittings.
Gage and absolute pressure. Add atmospheric pressure to the gage pressure and
you get absolute pressure.
Vacuum is a pressure less than atmospheric. At sea level atmospheric pressure is
14.7 psi. (760 mm of Mercury). Vacuum gages are normally calibrated in inches
or millimeters of mercury.
To calculate the net positive suction head (NPSH) of your pump and determine if you are
information:
The curve for your pump. This pump curve is supplied by the pump manufacturer.
Someone in your plant should have a copy. The curve is going to show you the
Net Positive Suction Head (NPSH) required for your pump at a given capacity.
Each pump is different so make sure you have the correct pump curve and use the
numbers for the impeller diameter on your pump. Keep in mind that this NPSH
required was for cold, fresh water.
A chart or some type of publication that will give you the vapor pressure of the
fluid you are pumping. You can find a typical vapor pressure chart in the "charts
you can use" section in the home page of this web site
If you would like to be a little more exact, you can use a chart to show the
possible reduction in NPSH required if you are pumping hot water or light
hydrocarbons. I will cover this subject in great detail in another paper.
You need to know the specific gravity of your fluid. Keep in mind that the
number is temperature sensitive. You can get this number from a published chart,
using a hydrometer.
Charts showing the head loss through the size of piping you are using between the
source and the suction eye of your pump. You will also need charts to calculate
the loss in any fittings, valves, or other hardware that might have been installed in
the suction piping. You can find these charts in the "charts you can use" section in
Is the tank you are pumping from at atmospheric pressure or is it pressurized in
some manner? Maybe it is under a vacuum ?
You need to know the atmospheric pressure at the time you are making your
calculation. We all know atmospheric pressure changes through out the day, but
you have to start somewhere.
The formulas for converting pressure to head and head back to pressure in the
imperial system are as follows:
## o sg. = specific gravity
o pressure = pounds per square inch
You also need to know the formulas that show you how to convert vacuum
To convert surface pressure to feet of liquid; use one of the following formulas:
## Inches of mercury x 1.133 / specific gravity = feet of liquid
Pounds per square inch x 2.31 / specific gravity = feet of liquid
Millimeters of mercury / (22.4 x specific gravity) = feet of liquid
There are different ways to think about net positive suction head (NPSH) but they
all have two terms in common.
## NPSHA (net positive suction head available)
NPSHR (net positive suction head required)
NPSHR (net positive suction head required) is defined as the NPSH at which the pump
total head (first stage head in multi stage pumps) has decreased by three percent (3%) due
to low suction head and resultant cavitation within the pump. This number is shown on
your pump curve, but it is going to be too low if you are pumping hydrocarbon liquids or
hot water.
Cavitation begins as small harmless bubbles before you get any indication of loss of head
or capacity. This is called the point of incipient cavitation. Testing has shown that it takes
from two to twenty times the NPSHR (net positive suction head required) to fully
suppress incipient cavitation, depending on the impeller shape (specific speed number)
and operating conditions.
To stop a product from vaporizing or boiling at the low pressure side of the pump the
NPSHA (net positive suction head available) must be equal to or greater than the NPSHR
## As I mentioned at the beginning, NPSHA is defined as static head + surface pressure
head - the vapor pressure of your product - loss in the piping, valves and fittings .
In the following paragraphs you will be using the above formulas to determine if
you have a problem with NPSHA. Here is where you locate the numbers to put into
the formula:
Static head. Measure it from the centerline of the pump suction to the top of the
liquid level. If the level is below the centerline of the pump it will be a negative or
minus number.
Surface pressure head. Convert the gage absolute pressure to feet of liquid using
the formula:
o Pressure = head x specific gravity / 2.31
Vapor pressure of your product . Look at the vapor pressure chart in the "charts
you can use" section in the home page of this web site. You will have to convert
the pressure to head. If you use the absolute pressure shown on the left side of the
chart, you can use the above formula
Specific gravity of your product. You can measure it with a hydrometer if no one
in your facility has the correct chart or knows the number.
Loss of pressure in the piping, fittings and valves. Use the three charts in the
"charts you can use" section in the home page of this web site
o Find the chart for the proper pipe size, go down to the gpm and read across
to the loss through one hundred feet of pipe directly from the last column
in the chart. As an example: two inch pipe, 65 gpm = 7.69 feet of loss for
each 100 feet of pipe.
o For valves and fittings look up the resistance coefficient numbers (K
numbers) for all the valves and fittings, add them together and multiply
the total by the V2/2g number shown in the fourth column of the friction
loss piping chart. Example: A 2 inch long radius screwed elbow has a K
number of 0.4 and a 2 inch globe valve has a K number of 8. Adding them
together (8 + 0.4) = 8.4 x 0.6 (for 65 gpm) = 5 feet of loss.
In the following examples we will be looking only at the suction side of the pump. If we
were calculating the pump's total head we would look at both the suction and discharge
sides.
Let's go through the first example and see if our pump is going to cavitate:
Given:
## Atmospheric pressure = 14.7 psi
Gage pressure =The tank is at sea level and open to atmospheric pressure.
Liquid level above pump centerline = 5 feet
Piping = a total of 10 feet of 2 inch pipe plus one 90 long radius screwed elbow.
Pumping =100 gpm. 68F. fresh water with a specific gravity of one (1).
Vapor pressure of 68F. Water = 0.27 psia from the vapor chart.
Specific gravity = 1
NPSHR (net positive suction head required, from the pump curve) = 9 feet
## Now for the calculations:
- vapor pressure of your product - loss in the piping, valves and fittings
## Static head = 5 feet
Atmospheric pressure = pressure x 2.31/sg. = 14.7 x 2.31/1 = 34 feet absolute
Gage pressure = 0
Vapor pressure of 68F. water converted to head = pressure x 2.31/sg = 0.27 x
2.31/1 = 0.62 feet
Looking at the friction charts:
o 100 gpm flowing through 2 inch pipe shows a loss of 17.4 feet for each
100 feet of pipe or 17.4/10 = 1.74 feet of head loss in the piping
o The K factor for one 2 inch elbow is 0.4 x 1.42 = 0.6 feet
Adding these numbers together, 1.74 + 0.6 = a total of 2.34 feet friction loss in the
pipe and fitting.
NPSHA (net positive suction head available) = 34 + 5 + 0 - 0.62 - 2.34 = 36.04 feet
The pump required 9 feet of head at 100 gpm. And we have 36.04 feet so we have plenty
to spare.
Example number 2 . This time we are going to be pumping from a tank under
vacuum.
Given:
## Gage pressure = - 20 inches of vacuum
Atmospheic pressure = 14.7 psi
Liquid level above pump centerline = 5 feet
Piping = a total of 10 feet of 2 inch pipe plus one 90 long radius screwed elbow.
Pumping = 100 gpm. 68F fresh water with a specific gravity of one (1).
Vapor pressure of 68F water = 0.27 psia from the vapor chart.
NPSHR (net positive suction head required) = 9 feet
## Now for the calculations:
- vapor pressure of your product - loss in the piping, valves and fittings
## Atmospheric pressure = 14.7 psi x 2.31/sg. =34 feet
Gage pessure pressure = 20 inches of vacuum converted to head
o inches of mercury x 1.133 / specific gravity = feet of liquid
o -20 x 1.133 /1 = -22.7 feet of pressure head absolute
Vapor pressure of 68F water = pressure x 2.31/sg. = 0.27 x 2.31/1 = 0.62 feet
Looking at the friction charts:
o 100 gpm flowing through 2.5 inch pipe shows a loss of 17.4 feet or each
100 feet of pipe or 17.4/10 = 1.74 feet loss in the piping
o The K factor for one 2 inch elbow is 0.4 x 1.42 = 0.6 feet
Adding these two numbers together: (1.74 + 0.6) = a total of 2.34 feet friction loss
in the pipe and fitting.
NPSHA (net positive suction head available) = 34 + 5 - 22.7 - 0.62 - 2.34 = 13.34 feet.
This is enough to stop cavitation also.
For the third example we will keep everything the same except that we will be
pumping 180 F. hot condensate from the vacuum tank.
The vapor pressure of 180F condensate is 7 psi according to the chart. We get the
specific gravity from another chart and find that it is 0.97 sg. for 180 F. Fresh water.
## pressure x 2.31/sg. = 7 x 2.31 / 0.97 = 16.7 feet absolute
- vapor pressure of your product - loss in the piping, valves and fittings
NPSHA (net positive suction head available) = 34 + 5 - 22.7 - 16.7 - 2.34 = -2.74 feet.
## A negative NPSHA is physically impossible because it implies that the friction
losses exceed the available head and that cannot happen. The rule when pumping
a boiling fluid is: The NPSHA equals the Static Suction Head minus the Suction
friction head because the suction surface pressure and the vapor pressure equalize
one another. The absolute pressure in the tank is 34 -22.7 = 11.3 ft. The vapor
pressure of the condensate in the tank converts to 16.7 ft of head (see above) so
the condensate is boiling /flashing and reaching a state of equilibrium.
When pumping a boiling liquid, the Static Head must exceed the Suction Friction
Head (2.34 feet) by the amount of NPSH Required (9 feet) or: (9 ft. + 2.34 feet =
11.34 feet.) We can do this by raising the level in the suction tank an additional
6.34 feet to get the 11.34 feet required (6.34 feet + 5 feet existing = 11.34 feet)
In some instances you could reduce the Suction Friction Head to get the same
result, but in this example there is not enough friction head available to reduce.
This example also allows you to shortcut NPSHA calculations any time you are
pumping from a tank where the liquid is at its vapor pressure. Oil refineries are
full of these applications.
If you are given the absolute and vapor pressures in psia, and you forgot how to convet to
feet of head; you can use the following formula, providing you know the specific weight
of the liquid you are pumping :
Pp = Absolute pressure expressed in psia. In an open system, Pp equals
atmospheric pressure, Pa, expressed in psia.
Pvpa = Vapor pressure expressed in psia.
W = Specific weight of liquid at the pumping temperature in pounds per cubic
foot.
Search
## Web The Engineering ToolBox
Resources, Tools and Basic Information for Engineering and Design of Technical Applications!
## NPSH - Net Positive Suction Head
A definition and an introduction to Net
## Low pressure at the suction side of a pump can
encounter the fluid to start boiling with
reduced efficiency
cavitation
damage
## of the pump as a result. Boiling starts when the
pressure in the liquid is reduced to the vapor pressure
of the fluid at the actual temperature.
To characterize the potential for boiling and
cavitation, the difference between the total head on
the suction side of the pump - close to the impeller,
and the liquid vapor pressure at the actual
temperature, can be used.
## Based on the Energy Equation - the suction head in
the fluid close to the impeller can be expressed as the
sum of the static and the velocity head:
hs = ps / + vs2 / 2 g (1)
where
to the impeller
## = specific weight of the fluid
vs = velocity of fluid
g = acceleration of gravity
## The liquids vapor head at the actual temperature
can be expressed as:
hv = pv / (2)
where
pv = vapor pressure
## Note! The vapor pressure in a fluid depends on
temperature. Water, our most common fluid, starts
boiling at 20 oC if the absolute pressure in the fluid is
2,3 kN/m2. For an absolute pressure of 47,5 N/m2,
the water starts boiling at 80 oC. At an absolute
pressure of 101.3 kN/m2 (normal atmosphere), the
boiling starts at 100 oC.
## The Net Positive Suction Head - NPSH - can be
expressed as the difference between the Suction
NPSH = hs - hv (3)
## The required NPSH for a particular pump, often
named NPSHr, is in general determined
experimentally by the pump manufacturer and a
part of the documentation of the pump.
## The pumps required NPSHr should always exceeded
the systems available NPSHa to avoid the
vaporization and cavitation of the impellers eye. The
required NPSHr should in general be significant
higher than the available NPSHa to avoid that head
loss in the suction pipe and in the pump casing, local
velocity accelerations and pressure decreases, start
boiling the fluid on the impeller surface.
square capacity.
## The available NPSH for an actual system is often
named NPSHa. The NPSHa can be determined during
design and construction, or determined
experimentally from the actual physical system.
The available NPSHa can be calculated with the
Energy Equation. For a common application - where
the pump lifts a fluid from an open tank at one level
to an other, the energy or head at the surface of the
tank is the same as the energy or head before the
pump impeller and can be expressed as:
h0 = hs + hl (4)
where
## hl = head loss from the surface to
impeller - major and minor loss in the
suction pipe
## In an open tank the head at surface can be expressed
as:
h0 = p0 / = patm / (4b)
## For a closed pressurized tank the absolute static
pressure inside the tank must be used.
## The head before the impeller can be expressed as:
hs = ps / + vs2 / 2 g + he (4c)
where
he = elevation from surface to pump -
positive if pump is above the tank,
negative if the pump is below the tank
## Transforming (4) with (4b) and (4c):
patm / = ps / + vs2 / 2 g + he + hl
(4d)
expressed as:
## If the pump is positioned above the tank, the
elevation - he - is positive and the NPSHa decreases
when the elevation of the pump increases.
## At some level the NPSHa will be reduced to zero and
the fluid starts to evaporate.
## If the pump is positioned below the tank, the
elevation - he - is negative and the NPSHa increases
when the elevation of the pump decreases (lowering
the pump).
## It's always possible to increase the NPSHa by
lowering the pump (as long as the major and minor
head loss due to a longer pipe don't increase it more).
This is important and it is common to lower the
pump when pumping fluids close to evaporation
temperature.
## Example - Pumping Water from an Open
Tank
When increasing the the elevation for a pump located
above a tank, the fluid will start to evaporate at a
maximum level for the actual temperature.
## At the maximum elevation NPSHa is zero. The
maximum elevation can therefore be expressed by
(4f):
NPSHa = patm / - he - hl - pv / = 0
## For optimal theoretical conditions we neglect the
then be expressed as:
he = patm / - pv / (5)
## The maximum elevation or suction head for an open
tank depends on the atmospheric pressure - which in
general can be regarded as constant, and the vapor
pressure of the fluid - which in general vary with
temperature, especially for water.
## The absolute vapor pressure of water at temperature
20 oC is 2.3 kN/m2. The maximum theoretical
elevation height is therefore:
## he = (101.33 kN/m2) / (9.80 kN/m3) -
(2.3 kN/m2) / (9.80 kN/m3)
= 10.1 m
Due to the head loss in the suction pipe and the local
conditions inside the pump - the theoretical
maximum elevation is significantly decreased.
## The maximum theoretical elevation of a pump above
an open water tank at different temperatures can be
found from the table below:
(oC) (kN/m2) (m)
0 0.6 10.3
5 0.9 10.2
10 1.2 10.2
15 1.7 10.2
20 2.3 10.1
25 3.2 10.0
30 4.3 9.9
35 5.6 9.8
40 7.7 9.5
45 9.6 9.4
50 12.5 9.1
55 15.7 8.7
60 20 8.3
65 25 7.8
70 32.1 7.1
75 38.6 6.4
80 47.5 5.5
85 57.8 4.4
90 70 3.2
95 84.5 1.7
100 101.33 0.0
## Net Positive Suction Head Required (NPSHR)
The net positive suction head required is a function of the pump design at the operating point on the pump
performance curve. In our example on the Centrifugal Pump Performance Curve -->page, the NPSHR by the
pump at the operating point is 5 ft.
NPSHA
Net Positive Suction Head Available (NPSHA)
The net positive suction head available is a function of the pump suction system.
The Net Positive Suction Head is the absolute total suction head in feet.
## The NPSH available in a flooded suction system is:
Atmospheric Pressure (- ) Vapor Pressure (+) Liquid Height (-) Friction in the Suction Line.
The NPSH available in a suction lift system is:
Atmospheric Pressure (-) Vapor Pressure (-) Liquid Ht. (-) Friction in the Suction Line.
## Calculating TDH-->page we would calculate the NPSHA as follows:
NPSHA = Atmospheric Pressure (-) Elevation Correction (-) Vapor Pressure (+) Suction
## Atmospheric Pressure = 33.96 ft.
Elevation Correction for 2000 ft. = 33.96 ft. (-) 31.58 ft. = 2.38 ft. | 5,484 | 21,408 | {"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} | 3.734375 | 4 | CC-MAIN-2019-39 | latest | en | 0.927396 |
https://math.stackexchange.com/questions/2380103/a-simple-concrete-example-of-the-product-topology | 1,563,844,601,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195528635.94/warc/CC-MAIN-20190723002417-20190723024417-00486.warc.gz | 462,870,957 | 37,133 | A simple, concrete example of the product topology
I am having some hard time understanding the concept of product topology.
Could somebody give me a simple, concrete example of a product topology? For example, what is the product topology of $A\times B$ when both $A$ and $B$ are equipped with the discrete topology?
Better, let's say that $A$ and $B$ each consist of two points $a_1,a_2$ and $b_1,b_2$ respectively. What is the product topology of $A$ and $B$?
• For a finite product, product topology coincides with box topology. – Sahiba Arora Aug 2 '17 at 14:45
• Do you know that the collection of sets $U\times V$ with $U$ open in $A$ and $V$ open in $B$ is a base of the product topology? So a set is open in the product topology iff you can write the set as a union of this sort of sets. Now find these bases in the cases you mention and have a look at the unions of their elements. – drhab Aug 2 '17 at 14:46
• For better understanding, compare the box and product topology on $\mathbb R^{\omega}.$ – Sahiba Arora Aug 2 '17 at 14:47
• @SahibaArora Regarding the nature of the question it might be that the OP is not familiar yet with box topology. – drhab Aug 2 '17 at 14:49
• The plane, $\Bbb R^2$, with the usual topology, is the product $\Bbb R\times \Bbb R$ of the real line with itself. – MJD Aug 2 '17 at 15:11
If both $A$ and $B$ are equipped with the discrete topology, the the product topology on $A\times B$ is again the discrete topology. This is so because if $a\in A$ and $b\in B$, then $\{(a,b)\}=\{a\}\times\{b\}$ and therefore $\{(a,b)\}$ is the cartesian product of two open sets. So, $\{(a,b)\}$ is an open set. Since all singletons are open sets, the topology is the discrete topology.
• Thanks Jose for the reply. If $A$ and $B$ are both discrete spaces, they each have 4 elements. How many elements are there in the product topology? Is it 16, the simple Cartesian product of both topologies? – user60264 Aug 2 '17 at 14:54
• @student Why do you say that if $A$ is a discrete space, then it has $4$ elements? It can have any number of elements. – José Carlos Santos Aug 2 '17 at 14:56
• because each space has two points – user60264 Aug 2 '17 at 14:57
• @student What I wrote was about any two discrete spaces $A$ and $B$. – José Carlos Santos Aug 2 '17 at 14:58
Let $T_A$ be the topology on $A$ and let $T_B$ be the topology on $B.$ Consider any topology $T$ on $A\times B$ such that the projections $p_A: (A\times B)\to A$ and $p_B:(A\times B)\to B$ are continuous, where $p_A((a,b))=a$ and $p_B((a,b))=b.$
If $A'\in T_A$ and $B'\in T_B$ we must have $A'\times B=p_A^{-1}A'\in T$ and $A\times B'=p_B^{-1}B'\in T,$ so $$T \supset U=\{A'\times B:A'\in T_A\}\cup \{A\times B':B'\in T_B\}.$$ Now $U$ is a sub-base for a topology $T^*$ on $A\times B,$ so $T\supset T^*.$
So any $T$ for which $p_A$ and $p_B$ are continuous is stronger than $T^*.$ And we can confirm that $p_A$ and $p_B$ are continuous when $A\times B$ is given the topology $T^*.$ So $T^*$ is the weakest topology on $A\times B$ such that $p_A$ and $p_B$ are continuous.
$T^*$ is called the product topology.
Observe that when $A'\in T_A$ and $B'\in T_B$ we have $A'\times B'=(A'\times B)\cap (A\times B')\in T^*.$ And the set $$\{A'\times B': A'\in T_A\land B'\in T_B\}$$ is a base for $T^*.$
If $\beta_A$ is a base for $T_A$ and $\beta_B$ is a base for $T_B$ then $\{A'\times B': A'\in \beta_A \land B'\in \beta_B\}$ is also a base for $T^*$.
Examples: 1. $A=B=\mathbb R$ with the usual topology on $\mathbb R.$ A base for the product topology $T^*$ on $\mathbb R^2$ is $\{I\times J: I,J\in \beta_{\mathbb R}\}$ where $\beta_{\mathbb R}$ is the set of bounded open real intervals. The topology $T^*$ on $\mathbb R^2$ is equal to the topology generated by the base of "open disks": If $C\subset \mathbb R^2$ then $C\in T^*$ iff for every $(x,y)\in C$ there exists $r>0$ such that $(-r+x,r+x)\times (-r+y,r+y)\subset C$ iff there exists $r'>0$ such that $C\supset \{(x',y'): {r'}^2>(x'-x)^2+(y'-y)^2\}.$
1. Suppose $B=\{b\}.$ Then $A\times B,$ with the product topology, is homeomorphic to $A.$ The projection $p_A$ is not only continuous, it is a homeomophism.
2. Let $A=B=S^1$...... $S^1$ is standard notation for $\{(x,y)\in \mathbb R^2: x^2+y^2=1\},$ with the usual topology on $S^1$ as a subspace of $\mathbb R^2,$ where $\mathbb R^2$ has the product topology of Example 1..... With the product topology, $S^1\times S^1$ is homeomorphic to the surface of a donut. | 1,490 | 4,478 | {"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} | 3.84375 | 4 | CC-MAIN-2019-30 | latest | en | 0.907821 |
https://reproducibility.org/RSF/book/rsf/school/marm.html | 1,643,073,579,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304749.63/warc/CC-MAIN-20220125005757-20220125035757-00438.warc.gz | 532,742,013 | 2,987 | ```from rsf.proj import * eta=0.1 n1=751 n2=2301 d1=4 d2=4 Fetch('marmvel.hh','marm') Flow('marm','marmvel.hh','dd form=native | window j1=10 j2=10') Plot('marm',''' window j1=1 j2=1 | grey allpos=y screenratio=.327 screenht=4.7 wanttitle=n wantaxis=y label1='Depth (m)' label2='Lateral (m)' bias=1500 polarity=y ''') Flow('marm2','marmvel.hh','dd form=native') Plot('marm2',''' window j1=1 j2=1 | grey allpos=y screenratio=.327 screenht=4.7 wanttitle=n wantaxis=y label1='Depth (m)' label2='Lateral (m)' bias=1500 polarity=y ''') Flow('eta',None,'makevel v000=%g n1=%d n2=%d d1=4 d2=4 | window j1=10 j2=10' % (eta,n1,n2)) Plot('eta','window j1=1 j2=1 | grey allpos=y screenratio=.327 screenht=4.7 wanttitle=n') Flow('velx','marm','math y=%s output="input*sqrt(1+2*y)" | window j1=10 j2=10' % ('eta.rsf')) Plot('velx','window j1=1 j2=1 | grey allpos=y screenratio=.327 screenht=4.7 wanttitle=n') Result('medium','marm','SideBySideIso') Flow('marms','marm','smooth rect1=2 rect2=2') Flow('etas','eta','smooth rect1=1 rect2=1') Flow('velxs','velx','smooth rect1=2 rect2=2') eiko = 'eiko1' Flow(eiko,'marm','eikEta sorder=%d yshot=6700 zshot=2400 vb1=5 b2=5 eta=%s vx=%s eta=0.1 btime=n' % (3,'eta.rsf','velx.rsf')) Plot(eiko, ''' window j1=1 j2=1 | contour screenratio=.327 screenht=4.7 wanttitle=n wantaxis=n dash=%d nc=50 ''' % (0,3)[1]) Flow('eiko','marm','eikonal order=%d yshot=6700 zshot=2400 b1=5 b2=5 eta=%s vx=%s btime=n' % (2,'eta.rsf','velx.rsf')) Plot('eiko', ''' window j1=1 j2=1 | contour screenratio=.327 screenht=4.7 wanttitle=n wantaxis=y dash=%d nc=50 label1='Depth (m)' label2='Lateral (m)' ''' % (0,5)[1]) Flow('eiko2','marms etas velxs','wkbjTI yshot=6700 zshot=2400 b1=5 b2=5 eta=\${SOURCES[1]} vx=\${SOURCES[1]} btime=n') Plot('eiko2', ''' window j1=1 j2=1 | contour screenratio=.327 screenht=4.7 wanttitle=n wantaxis=y nc=50 label1='Depth (m)' label2='Lateral (m)' ''') Result('marmousi','marm2 eiko eiko1 eiko2','Overlay') Flow('marmfoc','marm','window min2=4200 max2=9200') Plot('marmfoc','window j1=2 j2=2 | grey allpos=y screenratio=.327 screenht=4.7 wanttitle=n') Flow('t0','marmfoc','ScanCoef yshot=6700 zshot=2400 vb1=5 b2=5 t1=%s t2=%s btime=n' % ('t1.rsf','t2.rsf')) Flow('tt','t0 t1 t2','math x=\${SOURCES[1]} y=\${SOURCES[2]} output="input+x*0.1+y*0.1*0.1" ') Plot('tt', ''' window j1=2 j2=2 | contour screenratio=.327 screenht=4.7 wanttitle=n wantaxis=n nc=50 ''') Result('marmousi2','marmfoc tt','Overlay') Flow('t0s','t0','window max1=0') Flow('t1s','t1','window max1=0') Flow('t2s','t2','window max1=0') eta=0.0 file=0 while eta<0.5: tt = 'tt%d' % file ttt = 'ttt%d' % file ttr = 'ttr%d' % file Flow(tt,'t0s t1s t2s','math x=\${SOURCES[1]} y=\${SOURCES[2]} output="input+x*x*%g/(x-y*%g)" ' % (eta,eta)) Flow(ttt,tt,'rotate ') # Flow(ttt,tt,'transp | put n1=209 o1=4200 n2=1 o2=0') Flow(ttr,[tt,ttt],'math x=\${SOURCES[1]} output="input+x" ') Plot(tt, ''' graph screenratio=.55 screenht=9.7 wanttitle=n wantaxis=y max2=1.3 min2=0.9 label1='Lateral (m)' label2='Time (s)' labelsz=10 ''') eta += 0.05 file += 1 Result('moveout','tt0 tt1 tt2 tt3 tt4 tt5 tt6 tt7 tt8 tt9', 'Overlay') for i in range(3): Flow('t%de' % i,'t%ds' % i,'spray axis=2 n=81 d=0.01 o=0 label=eta') Flow('tall','t0e t1e t2e','math x=\${SOURCES[1]} y=\${SOURCES[2]} output="input+x*x*x2/(x-y*x2)" ') Result('tall', ''' grey wanttitle=n wantaxis=y bias=0.9 allpos=y transp=n color=j label1=Lateral unit1=m label2="\F10 h" maxval=2 minval=0.9 barreverse=n scalebar=y barlabel="Time (s)" ''') End()```
sfdd sfwindow sfgrey sfmakevel sfmath sfsmooth sfeikEta sfcontour sfeikonal sfwkbjTI sfScanCoef sfrotate sfgraph sfspray | 1,484 | 3,625 | {"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} | 2.703125 | 3 | CC-MAIN-2022-05 | latest | en | 0.215587 |
https://premiumqualityessay.com/1solve-x-7rootx-10-0-2-fourth-grade-class-decides-enclose-rectangular-garden/ | 1,656,152,865,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103034930.3/warc/CC-MAIN-20220625095705-20220625125705-00578.warc.gz | 516,028,683 | 11,544 | # 1solve x 7rootx 10 0 2 fourth grade class decides enclose rectangular garden
1.solve : x – 7rootx + 10 = 0 2. A fourth grade class decides to enclose a rectangular garden, using the side of the school as one side of the rectangle. suppose they have 80 feet to enclose the garden. A.Express the area of the garden as a function of x. B.What should the dimensions of the garden be in order to maximize the area ? 3.x^3 is greater than or less than 4x. solve, express solutions in interval form. 4. use any method to solve the system. x-2y+3z=4 2x+y-4z=3 -3x+4y-z=-2 | 171 | 566 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2022-27 | latest | en | 0.827952 |
http://physics.stackexchange.com/questions/123411/traveling-between-two-planets-at-rest-to-one-another | 1,469,405,761,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824201.28/warc/CC-MAIN-20160723071024-00029-ip-10-185-27-174.ec2.internal.warc.gz | 200,711,571 | 24,127 | Traveling between two planets at rest to one another [duplicate]
If I travel at relativistic speed from planet A to planet B which are at rest relative to one another, I will be younger than people on A or B when I arrive. However how does this mesh with the fact that the change in proper time should be symmetrical, i.e. I should observe events on A as well as B as moving at a slower rate while they observe events for me to be moving at a slower rate, so when I arrive at B why would I be younger? I understand this is similar to the twin paradox and other questions I have asked but I still don't understand how you can resolve the discrepancy since you remain in one inertial frame for the entire journey. Is it because I have to de-accelerate and thus change reference frames to arrive at B, and if so would the effect be the same if I never accelerated or de-accelerated from A to B I just merely flew past them with some velocity set for me at lets say the big bang?
-
marked as duplicate by Brandon Enright, Colin McFaul, Qmechanic♦Jul 6 '14 at 23:29
possible duplicate of How can time dilation be symmetrical? – ACuriousMind Jul 6 '14 at 18:38
I asked that question, the reason I'm asking this one is that I did not find the answer on the last one satisfactory so figured I'd ask a better version of the question. Linking it as a duplicate does not help me much. – Krel Jul 6 '14 at 18:40
And, as to all your other questions, the answer is: Proper time is a Lorentz invariant. I really don't know where you get the weird idea from that it should be "symmetrical". – ACuriousMind Jul 6 '14 at 18:40
Because it has been taught that if you are flying in a ship at relativistic speeds and pass another ship you will see time running slower for them while they will see time running slower for you. Thus it seems symmetrical. – Krel Jul 6 '14 at 18:43
@bright magus: SR speaks clearly about Lorentz invariance. The whole talk of frames and who sees what is an unfortunate pedagogical failure, just as dmckee says. Proper time is Lorentz invariant no matter whether anything is accelerating or not. If you do not understand the power of Lorentz invariant quantities, you have not understood SR. dmckee is absolutely right. – ACuriousMind Jul 6 '14 at 21:27
so when I arrive at B why would I be younger?
I believe I addressed this in another question of yours.
Once again, assume that when you pass planet A, your clock and planet A's clock both read $t = t_A =0$.
Now, according to the inhabitants of planet A, planet B's clock is synchronized with their clock.
However, in your inertial frame of reference, planet B's clock is ahead of planet A's clock.
Assume for simplicity that planet B is 1 light-second from planet A, in the rest frame of the planets, and that your are travelling at $0.5c$ towards planet B.
Then, when you pass planet A, you observe planet A's clock to read $t_A = 0$ and you observe planet B's clock to read $t_B = 0.5s$
When you fly past planet B, you observe your clock to read $t = 1.732s$ and you observe planet B's clock to read $t_B = 2s$.
So, in fact, you observe that planet B's clock runs slow; your elapsed time is $\tau = 1.732s$ while planet B's elapsed time is $\Delta t_B = 2s - 0.5s = 1.5s$
Moreover, the inhabitants of planet B observe your clock to run slow. They observe that you passed planet A when their clock read $t_B =t_A = 0$ so, according their clock, you took $2s$ to make the trip while your clock only showed $1.732s$.
Thus, the time dilation is symmetrical - you observe planet B's clock to run slow and they observe your clock to run slow.
Note that this is not a contradiction and is made possible by the fact that the two planetary clocks are not synchronized in your frame of reference.
These are the calculations for the above numbers...
When your clock reads $t=0$, planet B's clock reads
$$t_B = \frac{0.5c \cdot 1ls}{c^2} = 0.5s$$
Since you cover 1 light-second at a speed of $0.5c$ in the rest frame of the planets, your elapsed time is
$$\Delta t = 2s \cdot\sqrt{1 - 0.5^2} = 1.732s = \tau$$
Since, according to you, planet B's clock is moving, you should calculate that
$$\Delta t_B = 1.732s \cdot \sqrt{1 - 0.5^2} = 1.5s$$
which agrees with what you observe
$$\Delta t_B = 2s - 0.5s = 1.5s$$
Still what would happen if the traveler decided to drastically decelerate as they passed B? How would their clock go from reading 2 seconds to 1.5 seconds for me?
As long you stay inertial, the time dilation is symmetric.
However, if you suddenly decelerated to zero speed (relative to the planets) upon arriving at planet B, you would now find that your clock runs at the same rate as the planetary clocks, which you now observe to be synchronized, and that you are behind them by $2s - 1.732s = .268s$.
Since you know that your clock read $t=0$ when $t_A=0$ you know you aged less than the inhabitants on planet A. Essentially, you would 'see' that the inhabitants of planet A 'jumped' in age by 0.5s during the deceleration
Just before the deceleration, you would observe planet A's clock to read $t_A = 1.5s$.
Just after the deceleration, you would observe planet A's clock to read $t_A = 2s$.
Since you are co-located with planet B just before and after the deceleration, you would not observe planet B's clock to change.
-
Ah right simultaneity is broken between A & B if your in a different inertial frame. I have continued to make that mistake in the scenarios I've asked about. Still what would happen if the traveler decided to drastically decelerate as they passed B? How would their clock go from reading 2 seconds to 1.5 seconds for me? Also if B was to drastically accelerate to match the travelers velocity as they passed, I'm assuming the inhabitants of planet B would then be the younger ones? – Krel Jul 6 '14 at 22:05
What I've been trying to ask is how the separate inertial frames come to agree on their previous observations when they align themselves in the same frame. If I've been orbiting a planet at .5 c for some time and I would observe everyone on the planet as experiencing less time until I decelerate, at which point I'd realize they all experienced more time. That process is confusing to me. Likewise if someone from the planet were to match my speed while I orbit, suddenly they become the younger ones and I the older. Again confusing. What am I missing here? – Krel Jul 6 '14 at 22:11
Alfred Centauri: "as long you stay inertial, the time dilation is symmetric. However, if you suddenly decelerated to zero speed..." - this means that the source of time dilatation here is claimed to be deceleration only, and not the difference in uniform speed before. Otherwise, the total difference after landing on the planet would also depend on the time of travel at uniform speed. But then time dilatation due to uniform speed would not be symmetric. – bright magus Jul 6 '14 at 22:31
@krel, see updates to my answer – Alfred Centauri Jul 6 '14 at 22:31
@krel, and have you drawn the spacetime diagrams of your various scenarios yet? You really should because, once you start thinking in terms of spacetime diagrams and events, you can 'picture' the solution in your mind's eye. – Alfred Centauri Jul 6 '14 at 23:09
While travelling in an inertial reference frame, you perceive the time of objects moving relative to you to go slower than your time. For such situations, you may apply the naive notion of time dilation. As soon as you accelerate anywhere, you should forget about time dilation as a way to gain what you will read on any clock. Time dilation is not the only concept in SR. The right concept for elapsed times is:
Regardless of acceleration, for any path $\gamma$ in spacetime travelled, the elapsed time on a clock on the end of that path is the proper time $\tau = \int_\gamma \sqrt{\mathrm{d}x^\mu\mathrm{d}x_\mu}$.
In order for you to meaningfully say that you are "younger" or "older" than anyone else, you must both be in the same inertial frame.
If anyone is travelling from anywhere to anywhere else, they must always accelerate or decelerate in order to compare their ages to the people living at the end of such paths.
Therefore, time dilation will not yield any meaningful result about who is "older" or "younger" since it is only formulated for inertial frames.
There is no paradox because calculating the proper time for each travelled path through spacetime will yield unambiguous results about which clock reads what since proper time is a Lorentz invariant.
-
Many comments on this answer (by myself, and others) have apparently been removed recently. I hereby restate my objection. ACuriousMind: "for any path $\gamma$ in spacetime travelled, the elapsed time on a clock on the end of that path is the proper time $\tau = \int_{\gamma} \sqrt{dx^\mu dx_\mu}$. [...] calculating proper time for each travelled path [...] will yield unambiguous results about which clock reads what" -- What do you suppose do "readings" of a clock have to do with "proper time" values for any or all particular path segments of a given clock at all; or even unambiguously? – user12262 Jul 9 '14 at 16:21
@user12262: Why proper time is the reading of a clock at the end of a spacetime path is discussed here. This is unambiguous, as by Lorentz invariance all observes must agree on the proper time a path in spacetime has. – ACuriousMind Jul 9 '14 at 16:38
ACuriousMind: "Why proper time is the reading of a clock at the end of a spacetime path is discussed here [ Why do clocks measure arc-length? ]" -- There's no mentioning on that page of "clock reading", "a clock reads", or variations thereof. So ... what (which terminology) on that page do you call "reading of a clock", please? – user12262 Jul 9 '14 at 16:54
@user12262: The reading of a clock is, quite naturally in my opinion, the $I_{ba} = \int_{\lambda_a}^{\lambda_b} \mathrm{d}\lambda \dot{t(\lambda)}$ looked at in joshphysics' question, defined as simply the time $t$ that passes in the frame in which the clock is stationary at all times. It is then shown that this can be expressed as the invariant proper time. – ACuriousMind Jul 9 '14 at 17:05
ACuriousMind: "The reading of a clock is, quite naturally in my opinion, [...] simply the time $t$ that passes in the frame in which the clock is stationary at all times." -- Is the phrase "the time $t$ that passes" just another way of saying "the elapsed time on a clock", or "the proper time" (as in your answer above), or "the duration of that clock"? If so, then why the separate symbol: "$t$" instead "$\tau$"? (And for what it's worth, in my humble opinion any "reading of a clock" is quite simply some real or integer number read off an en.wikipedia.org/wiki/Clock#Indicator ) – user12262 Jul 9 '14 at 17:45
If I travel at relativistic speed
... let's say at constant speed $\beta ~ c$ ...
from planet A to planet B which are at rest relative to one another
then
(1) A and B succeed in determining which indication of A had been simultaneous to which indication of B (and vice versa); and
(2) A's duration from indicating your departure, until the (A's) indication simultaneous to B's indication of your arrival is equal to
B's duration from the (B's) indication simultaneous to A's indication of your departure until indicating your arrival; and
(3) your duration from indicating A's departure until indicating B's arrival is
$\sqrt{ 1 - \beta^2 }$ of
the duration(s) described above in (2).
I will be younger than people on A or B when I arrive.
That's not at all guaranteed, but it depends on
• whether you had been as young as the people of A at your departure,
• whether the people of B, at their indication simultaneous to A's indication of your departure, had been as young as you (and the people of A) at your departure,
• whether the people of B and the people of A had been aging equally (as determined by comparing their simultaneous indications), and
• whether the people of A or B had been aging equally as you did; in proportion to the duration ratio described above in (3).
In other words:
The duration ratio of (3) can be derived completely independent of any comparisons of "youthfulness of appearance".
However how does this mesh with the fact that the change in proper time should be symmetrical [...]
For a symmetric setup we should consider someone, say Q, who is and remains at rest to you, such that A travelled at speed $\beta~c$ from you to Q.
Then, symmetric to (3) above:
A's duration from indicating your departure until indicating Q's arrival is
$\sqrt{ 1 - \beta^2 }$ of
your duration from indicating A's departure until the (your) indication simultaneous to Q's indication of A's arrival.
-
You have got a synchronization problem because you are describing 2 events (departure planet $A$ and arrival planet $B$), but there are 4 events in spacetime to be taken into account (state of planet $B$ at the departure and state of planet $A$ of the arrival are missing).
The contradiction is relatively simple to track, only applying the time dilation formula $$T=\gamma\tau$$ and the proper time formula $$\tau = T/\gamma$$:
Lets say the travel from $A$ to $B$ $(v=0,6, \gamma=1,25)$ takes $10$ years, so the proper time of the spaceship is $\tau_1 = 10$ years. Now you will see that we will get two different values for the proper time $\tau_2$ of the planets:
1.The observed time of the travel, observed by the planets' reference frame $T_1 = 12,5$ years. Thus the proper time of the planets is $\tau_2=10$ years.
2.The observed time of the relative movement of the planets, observed by the spaceship $T_2 = \tau_1=10$ years.
Applying the proper time formula, the spaceship will calculate for the planets a proper time of $\tau_2=8$ years.
The difference between these two different values for $\tau_2$ is due to the fact that one starting time and one arrival time have not been defined for each planet. Thus, the planets consider a proper time of $10$ years, while the spaceship considers only a proper time for the planets of $8$ years.
-
It gets worse, because "the state of planet B at the departure" is not uniquely defined as departure occurs at A and they are spatially separated. Likewise for "state of planet A at arrival" with arrival occurring at B. – dmckee Jul 6 '14 at 20:31 | 3,573 | 14,307 | {"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} | 2.78125 | 3 | CC-MAIN-2016-30 | latest | en | 0.964833 |
http://brandweerheerdeinbeeld.nl/composite/8671.html | 1,563,456,893,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525634.13/warc/CC-MAIN-20190718125048-20190718151048-00157.warc.gz | 21,692,011 | 5,358 | 20 x 16 deck how many sq foot
# 20 x 16 deck how many sq foot
Decks.com. Decking CalculatorLocate Local deck Builders for a Quote. Go! Go! Menu. Search. Personal menu. decking Floor Calculator. Total square Feet. Enter Total square Feet. square Feet. Or, Enter Dimensions. Length. Feet * Required. Inches. Width. Feet * Required. Inches. Board Size. 2"x4" or 5/4"x4" 2"x6" or 5/4"x6". Joist Spacing. 12" o.c. 16" o.c. 19.4" o.c. 24" o.co. decking Angle. 90° to joist 45° to joist. Total Length Of Perimeter. Feet. Was this calculator helpful? Yes Some What No. Additional feedback?
How Much Does a Deck Cost? - CostHelper.comHiring someone to construct a wood deck can cost \$8-\$75 or more a square foot for labor and materials, depending on whether it is built with pressure-treated wood (\$8-\$50 or more a square foot, or about \$2,560-\$16,000 for 16'x20'), cedar ( \$20-\$75 or more a square foot, or \$6,400-\$24,000 for 16'x20'), redwood (\$30-\$75 or .... how much should I pay to construct and tear down the old deck for a 8 x 12 deck made of composite and vinyl railings with three steps. 4 feet off the ground ...
Cost to Build a Composite Deck: Deck Pricing | Planning your dream deck can be a lot of work. But with the deck Cost Calculator, we've made it easier to estimate how much it costs to build a deck. By factoring in the anticipated size of your project along with the estim
How to Calculate How Much Decking You Need | Hunker29 Mar 2010 ... The typical 5/4-inch-by-6-inch treated deck board, 8 feet long, will cover 4 square feet of deck. Divide 225 by 4, ... deck boards. Multiply the length of your deck boards, 8 feet, by the number of boards per row, 2, and subtract the length of the deck: 16 feet minus 15 feet equals one wasted foot of board per row. Divide the width of your deck by the width of the boards to find out how many wasted feet of decking board you will have: 15 feet divided by 6 inches equals 30.
How many 16"x16" pavers do I need for a 100 square ... | Zillow1 Jul 2013 ... You can't get it exactly that size with those pavers unless you can use 2 rows of half size pavers as well (8"x16"). If you want to use 16"x16" pavers without cutting any then you will need a patio that is either 9ft 4in by 9ft 4in or go slightly bigger at 10ft 8in by 10ft 8in. If you go with the smaller size you would need 49 pavers (7x7). If you go with the larger size you would need 64 pavers (8x8). If you wanted to be exactly at 10 feet by 10 feet you would need 49 full pavers, ...
Cost of a Pressure-Treated Wood Deck - Estimates and Prices Paid16 Apr 2015 ... A PT wood deck can cost \$8-\$50 a square foot (\$2,560-\$16,000 for a 16'x20' deck), but averages about \$25-\$35 a square foot (\$8,000-\$11,200 for 16'x20'), depending on the quality of materials, project complexity, and whether ... This Old House estimates that building a simple, ground-level 10'x16' deck of PT wood with no railings is of moderate to hard difficulty (it helps to have at least two workers) and takes 4-6 days of construction plus a week for concrete to cure.
2018 Decking Cost Calculators & Estimators | Average Deck PricesThe cost of your deck is broken down into many small parts, and if you can plan ahead, your budget will be better cushioned for any unexpected costs. ... These prices will vary by square footage and whether you have other features added to your deck that involve more materials. .... Total cost for 80 feet of facing, 20 feet of 2" x 10" facing, 80 feet of level railing and 30 feet of stair railing hand-built and installed to code, plus permits, a gate and a two-tread stair. . .just under \$5,000.00. | 1,010 | 3,660 | {"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} | 2.546875 | 3 | CC-MAIN-2019-30 | latest | en | 0.872455 |
https://www.physicsforums.com/threads/how-to-apply-the-wkb-approximation-in-this-case.728412/ | 1,532,106,417,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591718.31/warc/CC-MAIN-20180720154756-20180720174756-00181.warc.gz | 969,229,656 | 14,184 | # Homework Help: How to apply the WKB approximation in this case?
1. Dec 14, 2013
### a1111
1. The problem statement, all variables and given/known data
I'm trying to learn how to apply the WKB approximation. Given the following problem:
An electron, say, in the nuclear potential
$$U(r)=\begin{cases} & -U_{0} \;\;\;\;\;\;\text{ if } r < r_{0} \\ & k/r \;\;\;\;\;\;\;\;\text{ if } r > r_{0} \end{cases}$$
1. What is the radial Schrödinger equation for the $\ell=0$ state?
2. Assuming the energy of the barrier (i.e. $k/r_{0}$) to be high, how do you use the WKB approximation to estimate the bound state energies inside the well?
2. Relevant equations
For the first question, I thought the radial part of the equation of motion was the following
$$\left \{ - {\hbar^2 \over 2m r^2} {d\over dr}\left(r^2{d\over dr}\right) +{\hbar^2 \ell(\ell+1)\over 2mr^2}+U(r) \right \} R(r)=ER(r)$$
3. The attempt at a solution
For the first part, do I simply just let $\ell=0$ and obtain the following? Which of the two potentials do I use?
$$\left \{ - {\hbar^2 \over 2m r^2} {d\over dr}\left(r^2{d\over dr}\right) +U(r) \right \} R(r)=ER(r)$$
For the other question, do I use $\int \sqrt{2m(E-U(r))}=(n+1/2)\hbar π$, where $n=0,1,2,...$ ? If so, what are the turning points? And again, which of the two potentials do I use? | 450 | 1,327 | {"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} | 3.6875 | 4 | CC-MAIN-2018-30 | latest | en | 0.764626 |
https://www.allinterview.com/interview-questions/80-113/civil-engineering.html | 1,545,203,158,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376831715.98/warc/CC-MAIN-20181219065932-20181219091932-00430.warc.gz | 811,302,170 | 8,414 | Civil Engineering Interview Questions
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5. Two long boundary walls run parallel to each other at a center to center distance of 1 meter apart. The width and height of the first wall are 0.30 m and 2.5m respectively. While those of the second are respectively 0.18m and 3.5m. Plot the distribution of vertical stress intensity due to walls on a horizontal plane, 1m below ground level. The walls have negligible depth of foundation and are made of brick masonry (y = 19.2 kN/m cu)
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1.what is the mix proportion for cement soil compressed earth blocks? 2.how to calculate the mix proportion for cement stablised earth soil blocks?
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925 | 736 | 2,874 | {"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} | 3.125 | 3 | CC-MAIN-2018-51 | longest | en | 0.884198 |
https://adlmag.net/how-many-bags-of-soil-do-i-need-for-a-4x8-raised-bed/ | 1,716,330,473,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058522.2/warc/CC-MAIN-20240521214515-20240522004515-00082.warc.gz | 67,327,664 | 22,887 | For example, if you have a length of 5 feet, a width of 4 feet and a depth of 2 feet, the volume would be 40 cubic feet (5 x 4 x 2 = 40). This is the amount of dirt you will need to fill your raised planter box. This would equal 20 bags of soil sold in bags containing 2 cubic feet of dirt each.
Also, How much is a bag of soil?
If you buy topsoil by the bag instead of in bulk, expect to pay about \$100 per cubic yard. Bagged material from local home and garden or big box stores ranges from \$2 to \$5 per 40-pound bag, or \$35 to \$180 per cubic yard. Due to its higher price point, bagged topsoil should only be used for small areas.
In this way, What is in Miracle Gro potting soil? MiracleGro® Potting Mixes contain a blend of sphagnum peat moss, aged bark fines, perlite, plant food, and a wetting agent. MiracleGro® Moisture Control® Potting Mix also contains coir (coconut husks) to help protect against over and under watering.
Contents
## How many 40lb bags of topsoil are in a cubic yard?
For reference: One cubic yard equals 27 cubic feet. A 40 pound bag of topsoil usually contains about . 75 Cubic Feet of soil.
## How much soil does a plant need?
Nevertheless, vegetable plants concentrate their root systems in the top 1 to 3 feet of soil, and most crops can make adequate growth in only 12 inches of specially prepared soil.
## How many cubic feet are in a 5 gallon bucket of dirt?
668 = 2.004 (cubic feet). Therefore, 12 five-gallon buckets make 8 cubic feet (rounded off). To simplify mattes further, one needs 4 pails each of coarse vermiculite, peat moss, and at least 5 composts for each 4’x4′ x 6″ box.
## How many pounds is a cubic foot of potting soil?
Based on my research on potting soil, each quart weights approximately 0.875 pounds; thus, 10 pounds means roughly 11.43 quarts. 10 Quarts equals 9.4 Liters. There are about 25 and 3/4 dry quarts in a cubic foot. A 20 dry quarts package of potting soil is approximately 3/4 of a cubic foot.
## How many quarts are in a gallon of soil?
Three Dry Quarts
Filling a 1-gallon plant pot takes 3 dry quarts of potting soil. A typical 1-gallon plastic flower pot measures 6.5 inches in diameter and 6.5 inches high. Three dry quarts is equivalent to one-eighth of a cubic foot. So 1 cubic foot of potting soil will fill about eight 1-gallon pots.
## How much does .75 cubic feet of soil weigh?
One cubic yard of topsoil weighs approximately 1,080 pounds. The estimate is based on the cubic yard calculation. One cubic foot of topsoil weighs around 40 pounds. The exact weight will depend on various conditions.
## What is the difference between potting soil and potting mix?
When you get specific though, potting soil refers to any growth media which contains dirt, either partially or completely, and which is used to grow plants in a container. Potting mix, however, is any soil-less media which was specifically developed to produce better gardening better results inside containers.
2,000 pounds
## How much does 1 cubic yard of dirt weigh?
The average cubic yard of dry fill dirt will typically weigh as much as 2,000 pounds. If it is made up of a mixture of sand, stone and gravel, the weight can easily exceed 3,000 pounds per cubic yard. Depending on the composition and moisture content of the dirt, a cubic yard can weigh from 2,000 – 2,700 pounds.
## How much does Miracle Gro cost?
Miracle-Gro Water Soluble All Purpose Plant Food, 5 lbs.
Was: \$9.98
You Save: \$0.54 (5%)
## How heavy is a 5 gallon bucket of dirt?
A typical loose sample of fairly dry topsoil weighs around 1600 pounds per cubic yard. There are 27 cubic feet in a yard and 7.48 gallons in a cubic foot. 1600/27=59.26 pounds a cubic foot. 59.26/7.48=7.92 pounds per gallon.
## How long is Miracle Grow Potting soil good for?
Unopened. Unopened bags of Miracle Grow potting soil kept in proper storage conditions should keep for five years or more. As long as the bag stays dry, the fertilizer cannot release the nutrients.
## How many pounds is a cubic foot?
The pounds amount 150.23 lb converts into 1 cu ft – ft3, one cubic foot. It is the EQUAL concrete volume value of 1 cubic foot but in the pounds mass unit alternative.
## How many pounds are in a gallon of soil?
Soil taken from your yard or a garden bed is too dense to use in a pot or raised bed. Potting mix is too light for use in raised beds, while garden soil is too heavy. The “just right” solution is MiracleGro® Raised Bed Soil, a pre-mixed blend of the two.
## How many pounds are in a gallon of soil?
For the most part, no it does not! As long as your potting soil does not have a foul smell, a bad insect problem, or a disease issue, it is perfectly fine to use to grow your plants successfully! Even though potting soil may be old or used over and over, it can still be used again the next season!
## Does Miracle Gro potting soil have fertilizer?
MiracleGro potting mix contains nontoxic amounts of nitrogen, phosphorus and potassium for fertilizer and is recommended for container vegetables at the manufacturer’s own website. All plants need these three basic nutrients for healthy growth.
## How do you calculate the weight of soil?
To calculate the weight of a cubic yard of soil, you simply have to multiply the volume by its density. Just type the density of soil (you will probably find it on the packaging) into the topsoil calculator and this calculation will be performed effortlessly.
## How long does potting soil last?
Age and improper storage degrade potting soil. The useful life of potting soil depends on whether or not it is currently in use. Unused potting soil lasts roughly six months before it degrades in quality, while used potting soil should be replaced every year or two.
## Is Miracle Gro potting mix good?
The two best-looking soil mixes are also the most expensive – MiracleGro and the Square Foot Gardening Soil. They are also the smallest size bags. When adding in the positive benefits of an OMRI listing, the MiracleGro becomes the initial winner – something that surprised us.
12-13 lbs
## How many 40lb bags of topsoil are in a yard?
One cubic yard equals 27 cubic feet. A 40 pound bag of topsoil usually contains about . 75 Cubic Feet of soil.
## How much is a dump truck load of top soil?
The average cost of a dump truck load of topsoil is \$150 to \$500 for a typical 10 to 15-yard load delivered. Prices depend on the amount ordered, the local cost of the topsoil, and the hauling distance. Charges do not include spreading or installation.
## How much is a dump truck load of top soil?
Plain soil from a garden can weigh 12 pounds per 1 gallon. Add water, which weighs 8.3 pounds per gallon at room temperature, and a large container can become an immovable object.
## How many bags of dirt makes a yard?
Bagged Material Conversions
Cubic Yards 1 Cubic Foot Sized Bags 2 Cubic Foot Sized Bags
1 27 14
2 54 27
3 81 41
4 108 54
## How many bags of topsoil do I need?
Coverage. One yard of topsoil will cover 324 square feet of soil, spread at a depth of 1 inch, or 100 square feet of soil spread at a depth of 3 inches. To calculate how much topsoil you’ll actually need, multiply the number of square feet in the area by the number of inches of topsoil you want to install.
## Can I mix garden soil and potting mix?
Potting soil can be mixed with garden soil for particular cases such as raised beds, but it’s not a good mix for containers. Learn more about these different types of soil and how to use them in various types of gardens.
## Can you use potting soil in garden?
Potting soil is best used for when your plants are still in containers. When combined with soil outdoors it can cause the soil in your garden to dry out because it can often drain too well. Topsoil on the other hand is best combined with outdoor soil that already exists in your garden or flowerbed.
## How many pounds is 16 quarts of soil?
Coverage. One yard of topsoil will cover 324 square feet of soil, spread at a depth of 1 inch, or 100 square feet of soil spread at a depth of 3 inches. To calculate how much topsoil you’ll actually need, multiply the number of square feet in the area by the number of inches of topsoil you want to install.
## What is the white stuff in potting soil?
Perlite used in soil resembles tiny white plastic foam balls, but it’s actually a naturally occurring volcanic glass. When processed for use in potting soil, perlite is heated to 1,600 degrees Fahrenheit, so that it puffs like popcorn. When it puffs up, it expands to several times its original volume.
12-13 lbs | 2,104 | 8,607 | {"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} | 3.078125 | 3 | CC-MAIN-2024-22 | latest | en | 0.925607 |
https://www.audiosciencereview.com/forum/index.php?threads/vera-audio-class-d-amp-build-quality.9895/page-8#post-942699 | 1,719,222,181,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865348.3/warc/CC-MAIN-20240624084108-20240624114108-00874.warc.gz | 580,003,760 | 20,285 | # Vera Audio Class-D Amp Build Quality
#### Armand
##### Member
Audio Company
Is is a 20kHz 3,1V sine wave mixed with a 440kHz 0,3V sine wave.
There are small amounts of unavoiadable remaints of the switching frequency left on the output on class D amplifiers. It cannot be heard and it does not affect tweeters. Let me explain why.
The switching frequency current and power
A typical tweeter inductance is 30uH and that alone will represent an impedance of 80 Ohms. Adding the tweeter DC resistance of typical 6 ohms we end up with 86 ohms resistance to the 440kHz signal. The current in the tweeter will then be 0,3v/86= 0,0035A (3,5mA). Calculating the power we get 0.0035*0.0035*86=0,001W
On top of that I don't think it is neccassary to say that 440kHz is very very far from being audible either.
The high frequency audio signal
The 20kHz signal is 3,1V and the AC resistance (impedance) of the 30uH coil is 3.8 ohm. Adding the DC resistance og 6 ohms we get 9,8 ohms.
The current in the tweeter will be 3,1/9,8=0,316A. Calculating the power we get 0.316*0.316*9,8=1W
To sum up the math in words.
The signal we see in the oscilloscope does not look that nice. The 440kHz frequency is only about 10 times smaller than the 20kHz auido signal.
But because of the inductance in the tweeter the power delivery of the high frequency is 1000 times smaller. Even when a fairly small 3,1V 20kHz signal. At higher sound levels it will be many thousands time smaller. Its neglible.
0,001W power is way to little to matter at all. Also understand that this power is continous from the moment the amp is turned on. It is not something that changes. It is constant and independant of the music signal. It will heat up the tweeter by something like 0,01 degrees. I haven't seen anyone claim that a 0,01 degree constant temperature difference has a huge impact on the sound yet.
This supposedly damaging high frequency noise is the last grasping at straws to disqualify class D amplifiers.
I have here explained why it has no effect.
Why cannot you explain why it is so bad?
#### tmtomh
##### Major Contributor
Forum Donor
I am sure that John does not speak about audibility, he rather speaks about technical accuracy.
BTW, is this a sine wave or something else? And we are sure it does not matter?
View attachment 159494
The entire point of this part of the discussion - and the point @amirm and @Armand are trying to make - is precisely that one cannot answer the question "are we sure it does not matter" simply by looking at that waveform and saying, "oh, that looks ugly, therefore there must be a problem."
OP
#### Bjorn
##### Major Contributor
Audio Company
Forum Donor
The Vera Audio P400/1000 is back in stock now. Only a few for now, but numbers will gradually increase in the weeks to come.
#### Feyire
##### Active Member
The Vera Audio P400/1000 is back in stock now. Only a few for now, but numbers will gradually increase in the weeks to come.
Safe to assume that the latest NCOREx amplifier range from Hypex will be available in the future, right?
#### Bjorn
##### Major Contributor
Audio Company
Forum Donor
Safe to assume that the latest NCOREx amplifier range from Hypex will be available in the future, right?
The new NCx500 comes with an internal buffer, something Vera Audio doesn't need since VA has developed its own buffer with world class performance. The new module offers somewhat better cooling. However, great cooling no matter how the amplifier is being used is already take care of in the P400/1000 unit. So we're primarily left with sligthly more power with very low impedances. We need to weigh that against the cost and see what more is coming.
#### Bjorn
##### Major Contributor
Audio Company
Forum Donor
A teaser of our next amplifier.
#### Bjorn
##### Major Contributor
Audio Company
Forum Donor
Something we also spent a lot time on in the design, was the cooling solution. The product was built to have a long life span by never exceeding a certain temperature. That's also explained more in the detail in the document.
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2K | 1,070 | 4,181 | {"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} | 3.125 | 3 | CC-MAIN-2024-26 | latest | en | 0.900486 |
https://www.ragazzi-pizza.com/tips-for-making-pizza/quick-answer-how-many-inches-is-pizza-hut-large.html | 1,656,695,262,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103943339.53/warc/CC-MAIN-20220701155803-20220701185803-00337.warc.gz | 979,340,695 | 20,583 | # Quick Answer: How Many Inches Is Pizza Hut Large?
## How big is a Pizza Hut large?
Large pizzas are 14 inches in diameter and will offer approximately 10 slices. Extra- large pizzas come in between 16 and 18 inches in diameter and will provide at least 12 slices.
## Is a large pizza 14 inches?
Pizza size generally varies between: Small Pizza: 8-10 inches with 6 slices. Medium Pizza: 12 inches with 8 slices. Large Pizza: 14 inch with 10 slices.
## What size is a Panormous pizza?
The Panormous has 16 slices and is 40 percent larger than Pizza Hut’s regular old large pan pizza – that’s the selling point (or as we call it in Drive-Thru Land … the gimmick).
## How many inches is a large Domino’s pizza?
Domino’s large pizzas are the perfect answer to feeding a hungry group. These 14— inch pizzas feed approximately three to five people who are craving the Hand Tossed, Brooklyn Style, or Crunchy Thin Crust pizzas.
You might be interested: How Many Calories Does A Little Caesars Pizza Have?
## Are two medium Pizza Hut pizzas bigger than a large?
That large pizza has the same area as 2 mediumpizzas (14 inches) and 6.3 smalls (8 inches). To equalthe same amount of pizza as the large, it would setyou back an additional \$8.82 and \$30.79 respectively. How big is a medium Domino’s Pizza?
Square inches Total
2 medium pizzas 226.19 \$11.98
1 large pizza 153.94 \$11.99
## Is 2 Medium Pizzas bigger than a large?
If you’re planning on ordering a takeaway pizza this week, you might want to rethink your order. Mathematicians have revealed that getting one large pizza is a better idea than ordering two medium pizzas. Surprisingly, one 18-inch pizza actually has more ‘ pizza ‘ than two 12-inch pizzas.
## How much bigger is a 12 inch pizza?
Pizza Comparison (old version)
Diameter (in) Area (sq in) Times bigger than 10in
12 113 1.4
14 154 2.0
16 201 2.6
18 254 3.2
## How many large pizzas do I need for 10 adults?
10 People = 4 Pizzas. 15 People = 6 Pizzas. 20 People = 8 Pizzas. 30 People = 12 Pizzas.
## How much bigger is a 16-inch pizza?
A 16 – inch pie is 201 square inches — approximately 31 percent larger, in terms of area, than a 14- inch pie.
## What is the size of a personal size pizza?
By definition, a personal pizza should be an adequate serving for one person’s meal. While there is no definitive size, individual- sized pizzas usually range between 6 to 8 inches, says Tom Lehmann, director at the American Institute of Baking in Manhattan, Kansas.
## Is 12 Pizza a personal size?
A personal pizza could be anywhere between 6 to 8 inches, but it can’t be more than 12 inches. Restaurants usually consider at around 10 inches of pizza when you want a personal pan pizza size.
You might be interested: FAQ: What Time Does Marco's Pizza Close?
## What are Pizza Hut sizes?
Pizza Hut serves three sizes: personal pan, medium, and large.
## How many large pizzas do I need for 6 adults?
A Medium 12″ inch Pizza is normally cut into 8 slices and serves 3-4 people. A Large 14″ inch Pizza is normally cut into 8 or 10 slices and serves 3-5 people. An Extra Large 16″ inch Pizza is normally cut into 6 or 12 slices and serves 5- 6 people. An 18″ inch Pizza is normally cut into 6 or 12 slices and serves 6 -7
## Which pizza is bigger regular or medium?
Regular size with its 4 slices serves a single person, medium size with 6 slices serves two buddies or a couple. And our large pizza with 8 big slices serves a family of four. Look around the circle of your friends, contemplate their appetite and decide the size accordingly.
## How many slices are in a 20 inch pizza?
Extra-large pizzas have a 16 to 20 – inch diameter and yield 12 slices. Make sure you check with the restaurant before you order, though, as the sizes and slices might vary. | 961 | 3,801 | {"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} | 2.515625 | 3 | CC-MAIN-2022-27 | latest | en | 0.896752 |
https://www.solutioninn.com/box-and-liu-1999-describe-an-experiment-flying-paper-helicopters | 1,713,286,780,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817103.42/warc/CC-MAIN-20240416155952-20240416185952-00838.warc.gz | 900,605,007 | 17,442 | # Box and Liu (1999) describe an experiment flying paper helicopters where the objective is to maximize flight
## Question:
Box and Liu (1999) describe an experiment flying paper helicopters where the objective is to maximize flight time. They used the central composite design shown in Table 14E.9. Each run involved a single helicopter made to the following specifications: x1 = wing area (in2), ˆ’1 = 11.80 and +1 = 13.00; x2 = wing-length-to-width ratio,
ˆ’1 = 2.25 and +1 = 2.78; x3 = base width (in), ˆ’1 = 1.00 and +1 = 1.50; and x4 = base length (in), ˆ’1 = 1.50 and
+1 = 2.50. Each helicopter was flown four times, and the average flight time and the standard deviation of flight time were recorded.
The design and data are in the Minitab worksheet Ex14-16.MTW.
(a) Fit a second-order model to the average flight time response.
(b) Fit a second-order model to the standard deviation of flight-time response.
(c) Analyze the residuals for both models from parts (a) and (b). Are transformations on the response(s) necessary? If so, fit the appropriate models.
(d) What design would you recommend to maximize the flight time?
(e) What design would you recommend to maximize the flight time while simultaneously minimizing the standard deviation of flight time?
Fantastic news! We've Found the answer you've been seeking!
## Step by Step Answer:
Related Book For
Question Posted: | 348 | 1,391 | {"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} | 2.703125 | 3 | CC-MAIN-2024-18 | latest | en | 0.921744 |
https://community.alteryx.com/t5/Alteryx-Designer-Discussions/how-to-adding-rows-with-conditions/m-p/74398/highlight/true | 1,590,773,805,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347405558.19/warc/CC-MAIN-20200529152159-20200529182159-00339.warc.gz | 299,380,099 | 43,906 | # Alteryx designer Discussions
SOLVED
## how to adding rows with conditions
Highlighted
8 - Asteroid
This table includes the historical item cost with the week of when the cost has been updated.
For example, the cost in WK1 was \$10, and this cost remained the same until WK4 when it increased to \$15.
item Cost Week Apple \$ 10.0 1 Apple \$ 15.0 4 Apple \$ 12.0 7 Apple \$ 8.0 10
I want to make this table with full consecutive weeks as below, so the cost in WK 2 and WK3 should be 10.00 as well.
item Cost Week Apple \$ 10.0 1 Apple \$ 10.0 2 Apple \$ 10.0 3 Apple \$ 15.0 4 Apple \$ 15.0 5 Apple \$ 15.0 6 Apple \$ 12.0 7
I have created another table with full yearly weeks from WK1 to WK52... but I can't figure out a way to join these two tables or use a formula to make the cost table be come consecutively.
Also, some data do not star from wk1 or has an ending week as below
Item Cost Week Peach 25 4
Therefore, for the above case, since we are now in WK21, I wish the formula can generate the lines for cost from WK1 - WK21 at \$25
As the item cost table update by weekly, I want to make a query keep updating with consecutive week cost.
Thank you
Highlighted
12 - Quasar
Does this do what you want to accomplish?
1. Use the Generate Rows Tool to create all the weeks.
2. Use the Append Tool to create all combinations of items and weeks.
3. Join to assign the known values to items and weeks.
4. Sort, the forward fill missing cost values using the Multi-Row Tool.
5. Reverse sort, then forward fill the remaining missing cost values that occur before the first known value (your Peaches example).
Highlighted
Alteryx Certified Partner
Combination of Generate Rows, Append, Join, Sort, Multi-Row Formula would do the magic!
Highlighted
Alteryx Certified Partner
@Philip Just realized you posted a solution while I am working on it :). Hi5 to you. We both have a same thought process!
Highlighted
8 - Asteroid
Hi @Philip
Thank you very much for your helps!!
8 - Asteroid
Hi @Philip
Then I copied exact query by using Alteryx 10.4 version since this is the one my company is using now, I got the following error message in the Generate Rows tools
Parse error at char (0); unknown variable "allweek"
Could you please advise what's wrong with it? I just dragged the Generate Rows tools and typed exact what you typed in your query....... because of this error, I even can't run the query to pass the appendent tool.
Thank you.
Highlighted
7 - Meteor
But what if you do not want to start with week one and want to cross year boundaries? For example, a one year period beginning with the current week?
Labels | 685 | 2,719 | {"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} | 2.796875 | 3 | CC-MAIN-2020-24 | latest | en | 0.840634 |
https://help.qlik.com/en-US/sense/November2019/Subsystems/Hub/Content/Sense_Hub/Scripting/TrigonometricAndHyperbolicFunctions/trigonometric-hyperbolic-functions.htm?tr=zh-CN | 1,575,690,880,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540495263.57/warc/CC-MAIN-20191207032404-20191207060404-00344.warc.gz | 394,550,982 | 9,008 | # Trigonometric and hyperbolic functions
This section describes functions for performing trigonometric and hyperbolic operations. In all of the functions, the arguments are expressions resolving to angles measured in radians, where x should be interpreted as a real number.
All angles are measured in radians.
All functions can be used in both the data load script and in chart expressions.
Examples:
The following script code loads a sample table, and then loads a table containing the calculated trigonometric and hyperbolic operations on the values.
SampleData: LOAD * Inline [Value -1 0 1]; Results: Load *, cos(Value), acos(Value), sin(Value), asin(Value), tan(Value), atan(Value), atan2(Value, Value), cosh(Value), sinh(Value), tanh(Value) RESIDENT SampleData; Drop Table SampleData;
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Can you tell us why it did not help you and how we can improve it? | 214 | 982 | {"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} | 2.546875 | 3 | CC-MAIN-2019-51 | latest | en | 0.858607 |
https://myassignments-help.com/2022/11/03/stochastic-process-dai-xie-math3801-2/ | 1,695,447,165,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506479.32/warc/CC-MAIN-20230923030601-20230923060601-00514.warc.gz | 453,945,654 | 22,882 | ## 统计代写|随机过程代写stochastic process代考|Wald’s fundamental identit
Let $X_1, X_2, \ldots$ are i.i.d. r.v.s with $S_n=X_1+X_2+\ldots+X_n$ and $N$ is a stopping rule.
Let $F_n(x)=P\left[S_n \leq x\right], F_1(x)=F(x)=P\left[X_1 \leq x\right]$ and m.g.f. of $X_1$ is given by $\phi(\theta)=\int_{-\infty}^{\infty} e^{\theta x} d F(x)<\infty$ if $\phi(\sigma)<\infty$, where $\sigma=\operatorname{Re}(\theta)$. We also assume that $$\phi(\sigma)<\infty \text { for all } \sigma,-\beta<\sigma<\alpha<\infty, \alpha, \beta>0 .$$
Under these conditions, $P\left[e^X<1-\delta\right]>0$ and $P\left[e^X>1+\delta\right]>0, \delta>0$. $\phi(\theta)$ has a minimum at $\theta=\theta_0 \neq 0$, where $\theta_0$ is the root of the equation $\phi(\theta)=1$.
Wald’s Sequential Analysis presented the so-called Wald’s identify
$$E\left(e^{\theta S_N} /[\phi(\theta)]^N\right)=1 \text { for } \phi(\theta)<\infty \text { and }|\phi(\theta)| \geq 1 .$$
Actually we shall give the proof of a more general theorem in Random walk due to Miller and Kemperman (1961).
Define $F_n(x)=P\left[S_n \leq x ; N \geq n\right], N=\min \left{n \mid S_n \notin(-b, a), 0<a, b<\infty\right}$ and the series $F(z, \theta)=\sum_{n=0}^{\infty} z^n \int_{-b}^a e^{\theta x} d F_n(x)$
Then
$$E\left(e^{\theta S_N} z^N\right)=1+[z \phi(\theta)-1] F(z, \theta) \text { for all } \theta$$
which is known as Miller and Kemperman’s Identity.
## 统计代写|随机过程代写stochastic process代考|Fluctuation Theory
In this section $X_1, X_2, \ldots, X_n, \ldots$ are i.i.d. r.v.s.
Theorem $3.3$ If $E\left|X_i\right|<\infty$, then \begin{aligned} P[N(b)&<\infty]=1 \text { if } E X_i \leq 0 \ &<1 \text { if } E X_i>0 \end{aligned}
For Proof see Chung and Fuchs (1951) and Chung and Ornstein (1962), Memoirs of American Math. Society.
Definition $3.2$ If $S$ is uncountable, and $S_n=X_1+\ldots+X_n$ are Markov, $X_i$ ‘s being independent, then $x$ is called a possible value of the state space $S$ of the Markoy chain if there exits an $n$ such that
$P\left[\left|S_n-x\right|<\delta\right]>0$ for all $\delta>0$. A state $x$ is called recurrent if $P\left[\left|S_n-X\right|<\delta\right.$ i.o. $]=1$ i.e. $S_n \varepsilon(x-\delta, x+\delta)$ i.o. with probability one.
We shall conclude this section by stating two very important and famous theorems whose proofs are beyond the scope of this book.
Theorem $3.4$ (Chung and Fuchs)
Either every state is recurrent or no state is recurrent. (ref. Spitzer-Random Walk (1962)).
Theorem $3.5$ (Chung and Ornstein)
If $E\left|X_i\right|<\infty$, then recurrent values exist iff $E\left(X_i\right)=0$.
# 随机过程代考
## 统计代写|随机过程代写stochastic process代考|Wald’s fundamental identit
Wald’s Sequential Analysis 提出了所谓的 Wald 标识
$$E\left(e^{\theta S_N} /[\phi(\theta)]^N\right)=1 \text { for } \phi(\theta)<\infty \text { and }|\phi(\theta)| \geq 1 .$$
$$E\left(e^{\theta S_N} z^N\right)=1+[z \phi(\theta)-1] F(z, \theta) \text { for all } \theta$$
## 统计代写|随机过程代写stochastic process代考|Fluctuation Theory
$P\left[\left|S_n-x\right|<\delta\right]>0$ 对所有人 $\delta>0$. 一个状态 $x$ 称为循环如果 $P\left[\left|S_n-X\right|<\delta\right.$ io $]=1 \mathrm{E}$ $S_n \varepsilon(x-\delta, x+\delta)$ io 概率为 1 。
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Math作业代写、数学代写常见问题
myassignments-help擅长领域包含但不是全部: | 1,510 | 3,579 | {"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} | 3.140625 | 3 | CC-MAIN-2023-40 | latest | en | 0.578911 |
https://www.physicsforums.com/threads/why-modulo-m1-and-modulo-m2-implies-modulo-m1-m2.429764/ | 1,544,927,561,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376827175.38/warc/CC-MAIN-20181216003916-20181216025916-00194.warc.gz | 1,002,083,192 | 12,823 | # Why modulo m1 and modulo m2 implies modulo [m1, m2]
1. Sep 17, 2010
### crypto_rsa
Why "modulo m1" and "modulo m2" implies "modulo [m1, m2]"
If $$a \equiv r$$ (mod m1) and $$a \equiv r$$ (mod m2) then $$a \equiv r$$ (mod [m1, m2]), where [a, b] is the least common multiple of a and b.
I have tried to prove that.
Assume that
$$[m_{1}, m_{2}] = l_{1}m_{1} = l_{2}m_{2}$$
and
$$a = k_{1}m_{1} + r$$
$$a = k_{2}m_{2} + r$$
Then
$$al_{1} = k_{1}l_{1}m_{1} + rl_{1}$$
$$al_{2} = k_{2}l_{2}m_{2} + rl_{2}$$
thus
$$a(l_{1} - l_{2}) = [m_{1}, m_{2}](k_{1} - k_{2}) + r(l_{1} - l_{2})$$
and
$$a = [m_{1}, m_{2}] {(k_{1} - k_{2}) \over (l_{1} - l_{2})} + r$$
In order to
$$a \equiv r\ (mod [m_{1}, m_{2}])$$, or $$a = K[m_{1}, m_{2}] + r$$
we have to prove that
$$(l_{1} - l_{2})\; | \; (k_{1} - k_{2})$$
But how?
2. Sep 17, 2010
### Gokul43201
Staff Emeritus
Re: Why "modulo m1" and "modulo m2" implies "modulo [m1, m2]"
This doesn't directly answer your question, but wouldn't the proof be a lot easier if you started off with m1|(a-r) and m2|(a-r) ... ?
3. Jan 3, 2011
### crypto_rsa
Re: Why "modulo m1" and "modulo m2" implies "modulo [m1, m2]"
You are right, the proof is actually very simple:
If m1 | (a-r) and m2 | (a-r) then a-r is a common multiple of m1 and m2. Therefore it is divisible by the least common multiple, [m1, m2]. Thus [m1, m2] | (a-r), or $$a$$ $$\equiv r$$ (mod [m1, m2])
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook | 628 | 1,501 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2018-51 | latest | en | 0.682726 |
https://tex.stackexchange.com/questions/193402/picture-indicating-the-fifth-roots-of-unity | 1,653,676,676,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662675072.99/warc/CC-MAIN-20220527174336-20220527204336-00027.warc.gz | 634,328,073 | 69,533 | # Picture indicating the fifth roots of unity
I would like to have the unit circle, centered at the origin, on the Cartesian plane drawn. The origin is to be marked with a dot and labeled "O" and five dots are to be drawn on the circle, one on the x-axis, and the others at k(2\pi/5) radians from the positive x-axis for each integer 1 \leq k < 5, and they are to be labeled "w_{k}". Arrows are to be drawn from the origin to each w_{k}, too. One angle is to be drawn and labeled - the one from the positive x-axis to the ray through w_{1}.
I have included some of the code. It starts with the following commands.
\tikzset{mydot/.style={fill,circle,inner sep=1.5pt}}
Is there a manual that explains each of these commands? I guess that this is instructing LaTeX how to make the dots indicating the coordinates each time "\mydot" is in the code.
\documentclass{amsart}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage{newlfont}
\usepackage{mathtools}
\usepackage{tikz}
\tikzset{
mydot/.style={
fill,
circle,
inner sep=1.5pt
}
}
\begin{document}
\begin{tikzpicture}[>=latex]
% the coordinates of the vertices
\coordinate (O) at (0,0);
\coordinate (w_{1}) at (\cos(2\pi/5), \sin(2\pi/5));
% the axes
\draw[help lines,->] (-1.5,0) -- (1.5,0);
\draw[help lines,->] (0,-1.5) -- (0,1.5);
% labelling the vertices
\node[mydot,label={below:$O$}] at (O) {};
\node[mydot,label={right:$w_{1} = \bigl(\cos(2\pi/5), \sin(2\pi/5)\bigr)$}] at (w_{1}) {};
% the arcs for the angle
\begin{scope}[gray]
\draw[->]
(1,0) +(0:0.5cm) arc [radius=1cm,start angle=0,end angle=2\pi/5] node[midway,right] {$2\pi/5$};
\end{scope}
\end{tikzpicture}
\end{document}
To learn more about TikZ read pgfmanual.pdf. Via the macro \n you can adjust the order of the root to be taken.
\documentclass[tikz]{standalone}
\usepackage{amsmath}% for \Re and \Im
\def\n{5}
\begin{document}
\begin{tikzpicture}[
dot/.style={draw,fill,circle,inner sep=1pt}
]
\draw[->] (-2,0) -- (2,0) node[below] {$\Re$};
\draw[->] (0,-2) -- (0,2) node[left] {$\Im$};
\draw[help lines] (0,0) circle (1);
\node[dot,label={below right:$O$}] (O) at (0,0) {};
\foreach \i in {1,...,\n} {
\node[dot,label={\i*360/\n-(\i==\n)*45:$w_{\i}$}] (w\i) at (\i*360/\n:1) {};
\draw[->] (O) -- (w\i);
}
\draw[->] (0:.3) arc (0:360/\n:.3);
\node at (360/\n/2:.5) {$\alpha$};
\end{tikzpicture}
\end{document}
For 1 < n < 13:
• Shouldn't the labels w_2 and w_3 be placed symmetrically? They are conjugate, after all. Likewise for w_1 and w_4. Jul 26, 2014 at 17:51
• @egreg Yes, you're right. I also was unsatisfied and looked up in the documentation how to write a Kronecker delta. Now the angle is computed by \i*360/5-(\i==5)*45. Jul 26, 2014 at 17:54
• +1 Now you have just to generalize it to the n-th roots for any n > 2. ;-) Jul 26, 2014 at 17:57
• @egreg Done, but the label placement doesn't really work as expected, e.g. for the angle 270° the label isn't placed below the node. Jul 26, 2014 at 18:12
• @user143462 To get only fifth root change every occurrence of \n to 5 and remove \def\n{5}. In the argument of the tikzpicture I define a new style with the name dot. What it does is, it applies all options given in the braces to the path where I apply it. For a tutorial see section "2.8 Adding a Touch of Style", for the reference see section "82.4.4 Defining Styles" in the pgfmanual.pdf. Jul 27, 2014 at 17:50
Just for fun with PSTricks.
\documentclass[pstricks,border=24pt,12pt]{standalone}
\usepackage{pst-node,pst-plot}
\def\Object#1{%
\begin{pspicture}[saveNodeCoors,arrows=->](-2,-2)(2,2)
\pscircle{2}
\curvepnodes[plotpoints=\numexpr#1+1]{0}{360}{2 t PtoC}{w}
\multido{\i=0+1}{\wnodecount}{\psline(w\i)\uput[!N-w\i.y N-w\i.x atan](w\i){$w_{\i}$}}
\end{pspicture}}
\begin{document}
\foreach \n in{1,2,...,10}{\Object{\n}}
\end{document}
• More about angle (made from node expression) passed to \uput can be found here(click). Jul 26, 2014 at 18:39
With some additional notations
\documentclass[tikz]{standalone}
\usepackage{amsmath}% for \Re and \Im
\def\n{8}
\begin{document}
\begin{tikzpicture}[
dot/.style={draw,fill,circle,inner sep=1pt}
]
\draw[->] (-2,0) -- (2,0) node[below] {$\Re$};
\draw[->] (0,-2) -- (0,2) node[left] {$\Im$};
\draw[help lines] (0,0) circle (1);
\node[dot] (O) at (0,0) {};
\node[label={above right:$+i$}] (I1) at (0,1) {};
\node[label={below right:$-i$}] (I2) at (0,-1) {};
\node[label={above right:$1$}] (U1) at (1,0) {};
\node[dot,label={above right:$-1$}] (I2) at (-1,0) {};
\foreach \i in {1,...,\n} {
\node[dot,label={\i*360/\n-(\i==\n)*45:$w_{\i}$}] (w\i) at ( \i*360/\n:1) {};
\draw[->] (O) -- (w\i);
}
\draw[->] (0:.3) arc (0:360/\n:.3);
\node at (360/\n/2:.5) {$\alpha$};
\end{tikzpicture}
\end{document}
• I took the liberty of adding a picture. I'd also recommend not using \def\n{8}, but something like \newcommand{\lastnumber}{8} (or any other command name): with \def you risk redefining important commands, which could lead to puzzling errors or output. Just to be picky: \Re and \Im don't need amsmath, which is recommended anyway when math typesetting is involved. Sep 6, 2015 at 22:00 | 1,859 | 5,152 | {"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} | 2.90625 | 3 | CC-MAIN-2022-21 | latest | en | 0.690165 |
https://solvedlib.com/what-significance-did-the-grand-council-an,32923 | 1,685,745,050,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648858.14/warc/CC-MAIN-20230602204755-20230602234755-00672.warc.gz | 584,616,332 | 16,390 | # What significance did the Grand Council, an experiment in democracy, have for the Iroquois?
###### Question:
What significance did the Grand Council, an experiment in democracy, have for the Iroquois?
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How do you divide (2x^3 - 7x^2 - 17x - 3) / (2x+3)?... | 1,337 | 4,663 | {"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} | 2.96875 | 3 | CC-MAIN-2023-23 | latest | en | 0.81567 |
http://math.ubooks.pub/Books/ON/M1/1704/C23S7M004.html | 1,521,871,813,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257649931.17/warc/CC-MAIN-20180324054204-20180324074204-00185.warc.gz | 180,805,008 | 4,146 | 23-7. Back Emf
Learning Objectives
• Explain what back emf is and how it is induced.
It has been noted that motors and generators are very similar. Generators convert mechanical energy into electrical energy, whereas motors convert electrical energy into mechanical energy. Furthermore, motors and generators have the same construction. When the coil of a motor is turned, magnetic flux changes, and an emf (consistent with Faraday’s law of induction) is induced. The motor thus acts as a generator whenever its coil rotates. This will happen whether the shaft is turned by an external input, like a belt drive, or by the action of the motor itself. That is, when a motor is doing work and its shaft is turning, an emf is generated. Lenz’s law tells us the emf opposes any change, so that the input emf that powers the motor will be opposed by the motor’s self-generated emf, called the back emf of the motor. (See Figure 1.)
Figure 1: The coil of a DC motor is represented as a resistor in this schematic. The back emf is represented as a variable emf that opposes the one driving the motor. Back emf is zero when the motor is not turning, and it increases proportionally to the motor’s angular velocity.
Back emf is the generator output of a motor, and so it is proportional to the motor’s angular velocity $\omega$. It is zero when the motor is first turned on, meaning that the coil receives the full driving voltage and the motor draws maximum current when it is on but not turning. As the motor turns faster and faster, the back emf grows, always opposing the driving emf, and reduces the voltage across the coil and the amount of current it draws. This effect is noticeable in a number of situations. When a vacuum cleaner, refrigerator, or washing machine is first turned on, lights in the same circuit dim briefly due to the $IR$ drop produced in feeder lines by the large current drawn by the motor. When a motor first comes on, it draws more current than when it runs at its normal operating speed. When a mechanical load is placed on the motor, like an electric wheelchair going up a hill, the motor slows, the back emf drops, more current flows, and more work can be done. If the motor runs at too low a speed, the larger current can overheat it (via resistive power in the coil, $P={I}^{2}R$), perhaps even burning it out. On the other hand, if there is no mechanical load on the motor, it will increase its angular velocity $\omega$ until the back emf is nearly equal to the driving emf. Then the motor uses only enough energy to overcome friction.
Consider, for example, the motor coils represented in Figure 1. The coils have a $\text{0.400}\phantom{\rule{0.25em}{0ex}}\Omega$ equivalent resistance and are driven by a 48.0 V emf. Shortly after being turned on, they draw a current $I=\text{V/R}=\left(\text{48}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{V}\right)/\left(0\text{.}\text{400}\phantom{\rule{0.25em}{0ex}}\Omega \right)=\text{120}\phantom{\rule{0.25em}{0ex}}\text{A}$ and, thus, dissipate $P={I}^{2}R=5\text{.}\text{76}\phantom{\rule{0.25em}{0ex}}\text{kW}$ of energy as heat transfer. Under normal operating conditions for this motor, suppose the back emf is 40.0 V. Then at operating speed, the total voltage across the coils is 8.0 V (48.0 V minus the 40.0 V back emf), and the current drawn is $I=\text{V/R}=\left(8\text{.}0\phantom{\rule{0.25em}{0ex}}\text{V}\right)/\left(0\text{.}\text{400}\phantom{\rule{0.25em}{0ex}}\Omega \right)=\text{20}\phantom{\rule{0.25em}{0ex}}\text{A}$. Under normal load, then, the power dissipated is $P=\text{IV}=\left(\text{20}\phantom{\rule{0.25em}{0ex}}\text{A}\right)/\left(8\text{.}0\phantom{\rule{0.25em}{0ex}}\text{V}\right)=\text{160}\phantom{\rule{0.25em}{0ex}}\text{W}$. The latter will not cause a problem for this motor, whereas the former 5.76 kW would burn out the coils if sustained.
Section Summary
• Any rotating coil will have an induced emf—in motors, this is called back emf, since it opposes the emf input to the motor.
Conceptual Questions
Exercise 1
Suppose you find that the belt drive connecting a powerful motor to an air conditioning unit is broken and the motor is running freely. Should you be worried that the motor is consuming a great deal of energy for no useful purpose? Explain why or why not.
Problems & Exercises
Exercise 1
Suppose a motor connected to a 120 V source draws 10.0 A when it first starts. (a) What is its resistance? (b) What current does it draw at its normal operating speed when it develops a 100 V back emf?
Show/Hide Solution
Solution
(a) $12.00 \Omega$
(b) 1.67 A
Exercise 2
A motor operating on 240 V electricity has a 180 V back emf at operating speed and draws a 12.0 A current. (a) What is its resistance? (b) What current does it draw when it is first started?
Exercise 3
What is the back emf of a 120 V motor that draws 8.00 A at its normal speed and 20.0 A when first starting?
Show/Hide Solution
Solution
72.0 V
Exercise 4
The motor in a toy car operates on 6.00 V, developing a 4.50 V back emf at normal speed. If it draws 3.00 A at normal speed, what current does it draw when starting?
Exercise 5
Integrated Concepts
The motor in a toy car is powered by four batteries in series, which produce a total emf of 6.00 V. The motor draws 3.00 A and develops a 4.50 V back emf at normal speed. Each battery has a $\text{0.100 Ω}$ internal resistance. What is the resistance of the motor?
Show/Hide Solution
Solution
$\text{0.100 Ω}$ | 1,489 | 5,493 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 12, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2018-13 | latest | en | 0.947732 |
https://www.majortests.com/essay/Week-2-Team-d-Compilation-Of-611208.html | 1,719,201,624,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198864986.57/warc/CC-MAIN-20240624021134-20240624051134-00825.warc.gz | 752,607,215 | 5,094 | # Week 2 Team D compilation of successes and issues Essay examples
Submitted By shedragon
Words: 594
Pages: 3
Team D Week Two
Math Case Study Review from Chapter 5 and 6
LaChundra Bray, Silvia P. Reyes Hernandez, Thomas Laing,
Susan Rodgers, and Mona Roulaine
QRB/501 – Quantitative Reasoning
Walden Gemmil
Math Case Study Review from Chapter 5 and 6
This week Team D was tasked with reviewing and solving case study 5.2 and 6.2 from respective chapters 5 and 6 of the ninth edition of Business Math. Additionally the team members were required to submit a paragraph explaining any issues or successes we individually had in answering the case study questions. The following are a compilation of Team D’s thoughts on this assignment.
Susan Rodgers
I completed case study 6-2 question 2. I found 2A to be simple but effective in finding the answers I needed. However 2B was a problem for me. I understand that it is just finding the difference and that is just subtracting unfortunately I could not understand what equation I was supposed to use to find my answers. I had to ask help from my team mates to find the answers for 2B. Thankfully I think I figured it out. I hope that I have been able to use the equations correctly.
Mona Roulaine
This was a hard week for me. While I understand the concept of what we are doing, since I work in excel and build formulas every day. I struggled greatly with the format the assignment was in. Having to look in multiple places for the information was no fun either. Especially using this new format for textbooks where you cannot scroll the entire book per page or search within other chapters that are not open at that time.
LaChundra Bray
I have to admit that this week was a struggle for myself. This class has opened my eyes to different equations that I have not used in years and it may take some time to get adjusted to it all. I was able to look over both cases found it to be difficult to figure out which formulas went where and when to plug in different figures. Again this is my first online Math class and I can honestly say that it frustrating not knowing what is needed or expected of me at all times. I have looked at the equations and it seemed to come together after looking at the | 499 | 2,251 | {"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} | 2.953125 | 3 | CC-MAIN-2024-26 | latest | en | 0.976623 |
https://phantran.net/individual-demand/ | 1,718,471,251,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861605.77/warc/CC-MAIN-20240615155712-20240615185712-00788.warc.gz | 410,099,696 | 27,904 | # Individual Demand
This section shows how the demand curve of an individual consumer follows from the consumption choices that a person makes when faced with a budget constraint. To illustrate these concepts graphically, we will limit the avail- able goods to food and clothing, and we will rely on the utility-maximization approach described in Section 3.3 (page 86).
### 1. Price Changes
We begin by examining ways in which the consumption of food and cloth- ing changes when the price of food changes. Figure 4.1 shows the consump- tion choices that a person will make when allocating a fixed amount of income between the two goods.
Initially, the price of food is \$1, the price of clothing \$2, and the consum- er ’s income \$20. The utility-maximizing consumption choice is at point B in Figure 4.1 (a). Here, the consumer buys 12 units of food and 4 units of clothing, thus achieving the level of utility associated with indifference curve U2.
Now look at Figure 4.1 (b), which shows the relationship between the price of food and the quantity demanded. The horizontal axis measures the quantity of food consumed, as in Figure 4.1 (a), but the vertical axis now measures the price of food. Point G in Figure 4.1 (b) corresponds to point B in Figure 4.1 (a). At G, the price of food is \$1, and the consumer purchases 12 units of food.
Suppose the price of food increases to \$2. As we saw in Chapter 3, the budget line in Figure 4.1 (a) rotates inward about the vertical intercept, becoming twice as steep as before. The higher relative price of food has increased the magnitude of the slope of the budget line. The consumer now achieves maximum utilityat A, which is found on a lower indifference curve, U1. Because the price of food has risen, the consumer ’s purchasing power—and thus attainable utility—has fallen. At A, the consumer chooses 4 units of food and 6 units of clothing. In Figure 4.1 (b), this modified consumption choice is at E, which shows that at a price of \$2, 4 units of food are demanded.
Finally, what will happen if the price of food decreases to 50 cents? Because the budget line now rotates outward, the consumer can achieve the higher level of utility associated with indifference curve U3 in Figure 4.1 (a) by selecting D, with 20 units of food and 5 units of clothing. Point H in Figure 4.1 (b) shows the price of 50 cents and the quantity demanded of 20 units of food.
### 2. The Individual Demand Curve
We can go on to include all possible changes in the price of food. In Figure 4.1 (a), the price-consumption curve traces the utility-maximizing combinations of food and clothing associated with every possible price of food. Note that as the price of food falls, attainable utility increases and the consumer buys more food. This pattern of increasing consumption of a good in response to a decrease in price almost always holds. But what happens to the consumption of clothing as the price of food falls? As Figure 4.1 (a) shows, the consumption of clothing may either increase or decrease. The consumption of both food and clothing can increase because the decrease in the price of food has increased the consumer ’s ability to purchase both goods.
An individual demand curve relates the quantity of a good that a single consumer will buy to the price of that good. In Figure 4.1 (b), the individual demand curve relates the quantity of food that the consumer will buy to the price of food. This demand curve has two important properties:
1. The level of utility that can be attained changes as we move along the
The lower the price of the product, the higher the level of utility. Note from Figure 4.1 (a) that a higher indifference curve is reached as the price falls. Again, this result simply reflects the fact that as the price of a product falls, the consumer ’s purchasing power increases.
for clothing equals the ratio of the prices of food and clothing. As the price of food falls, the price ratio and the MRS also fall. In Figure 4.1 (b), the price ratio falls from 1 (\$2/\$2) at E (because the curve U1 is tangent to a budget line with a slope of -1 at A) to 1/2 (\$1/\$2) at G, to 1/4 (\$0.50/\$2) at H. Because the con- sumer is maximizing utility, the MRS of food for clothing decreases as we move down the demand curve. This phenomenon makes intuitive sense because it tells us that the relative value of food falls as the consumer buys more of it.
The fact that the MRS varies along the individual’s demand curve tells us something about how consumers value the consumption of a good or service. Suppose we were to ask a consumer how much she would be willing to pay for an additional unit of food when she is currently consuming 4 units. Point E on the demand curve in Figure 4.1 (b) provides the answer: \$2. Why? As we pointed out above, because the MRS of food for clothing is 1 at E, one additional unit of food is worth one additional unit of clothing. But a unit of clothing costs \$2, which is, therefore, the value (or marginal benefit) obtained by consuming an additional unit of food. Thus, as we move down the demand curve in Figure 4.1 (b), the MRS falls. Likewise, the value that the consumer places on an addi- tional unit of food falls from \$2 to \$1 to \$0.50.
### 3. Income Changes
We have seen what happens to the consumption of food and clothing when the price of food changes. Now let’s see what happens when income changes.
The effects of a change in income can be analyzed in much the same way as a price change. Figure 4.2 (a) shows the consumption choices that a consumer will make when allocating a fixed income to food and clothing when the price of food is \$1 and the price of clothing \$2. As in Figure 4.1 (a), the quantity of clothing is measured on the vertical axis and the quantity of food on the horizontal axis. Income changes appear as changes in the budget line in Figure 4.2 (a). Initially, the consumer ’s income is \$10. The utility-maximizing consumption choice is then at A, at which point she buys 4 units of food and 3 units of clothing.
This choice of 4 units of food is also shown in Figure 4.2 (b) as E on demand curve D1. Demand curve D1 is the curve that would be traced out if we held income fixed at \$10 but varied the price of food. Because we are holding the price of food constant, we will observe only a single point E on this demand curve.
What happens if the consumer ’s income is increased to \$20? Her budget line then shifts outward parallel to the original budget line, allowing her to attain the utility level associated with indifference curve U2. Her optimal consump- tion choice is now at B, where she buys 10 units of food and 5 units of clothing. In Figure 4.2 (b) her consumption of food is shown as G on demand curve D2. D2 is the demand curve that would be traced out if we held income fixed at \$20 but varied the price of food. Finally, note that if her income increases to \$30, she chooses D, with a market basket containing 16 units of food (and 7 units of clothing), represented by H in Figure 4.2 (b).
We could go on to include all possible changes in income. In Figure 4.2 (a), the income-consumption curve traces out the utility-maximizing combina- tions of food and clothing associated with every income level. The income- consumption curve in Figure 4.2 slopes upward because the consump- tion of both food and clothing increases as income increases. Previously, we saw that a change in the price of a good corresponds to a movement along a demand curve. Here, the situation is different. Because each demand curve is
measured for a particular level of income, any change in income must lead to a shift in the demand curve itself. Thus A on the income-consumption curve in Figure 4.2 (a) corresponds to E on demand curve D1 in Figure 4.2 (b); B corresponds to G on a different demand curve D2. The upward-sloping income-consumption curve implies that an increase in income causes a shift to the right in the demand curve—in this case from D1 to D2 to D3.
### 4. Normal versus Inferior Goods
When the income-consumption curve has a positive slope, the quantity demanded increases with income. As a result, the income elasticity of demand is positive. The greater the shifts to the right of the demand curve, the larger the income elasticity. In this case, the goods are described as normal: Consumers want to buy more of them as their incomes increase.
In some cases, the quantity demanded falls as income increases; the income elasticity of demand is negative. We then describe the good as inferior. The term inferior simply means that consumption falls when income rises. Hamburger, for example, is inferior for some people: As their income increases, they buy less hamburger and more steak.
Figure 4.3 shows the income-consumption curve for an inferior good. For relatively low levels of income, both hamburger and steak are normal goods. As income rises, however, the income-consumption curve bends backward (from point B to C). This shift occurs because hamburger has become an inferior good—its consumption has fallen as income has increased.
### 5. Engel Curves
Income-consumption curves can be used to construct Engel curves, which relate the quantity of a good consumed to an individual’s income. Figure 4.4 shows how such curves are constructed for two different goods. Figure 4.4 (a), which shows
an upward-sloping Engel curve, is derived directly from Figure 4.2 (a). In both figures, as the individual’s income increases from \$10 to \$20 to \$30, her consump- tion of food increases from 4 to 10 to 16 units. Recall that in Figure 4.2 (a) the verti- cal axis measured units of clothing consumed per month and the horizontal axis units of food per month; changes in income were reflected as shifts in the budget line. In Figures 4.4 (a) and (b), we have replotted the data to put income on the vertical axis, while keeping food and hamburger on the horizontal.
The upward-sloping Engel curve in Figure 4.4 (a)—like the upward-sloping income-consumption curve in Figure 4.2 (a)—applies to all normal goods. Note that an Engel curve for clothing would have a similar shape (clothing consump- tion increases from 3 to 5 to 7 units as income increases).
Figure 4.4 (b), derived from Figure 4.3, shows the Engel curve for hamburger. We see that hamburger consumption increases from 5 to 10 units as income increases from \$10 to \$20. As income increases further, from \$20 to \$30, con- sumption falls to 8 units. The portion of the Engel curve that slopes downward is the income range within which hamburger is an inferior good.
### 3 thoughts on “Individual Demand”
1. Jack says:
Quality posts is the important to invite the visitors to pay a quick visit the site,
that’s what this web site is providing.
2. Granville Wnuk says:
Hi my loved one! I want to say that this post is awesome, nice written and include almost all vital infos. I’d like to see extra posts like this. | 2,565 | 10,985 | {"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} | 3.03125 | 3 | CC-MAIN-2024-26 | longest | en | 0.926621 |
http://stuver.blogspot.com/2012/07/journey-of-gravitational-wave-i-gws.html | 1,503,119,885,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105304.35/warc/CC-MAIN-20170819051034-20170819071034-00655.warc.gz | 426,674,879 | 24,106 | ## Thursday, July 12, 2012
### The Journey of a Gravitational Wave I: GWs Cast No Shadows!
What happens to a gravitational wave between when it is produced and when LIGO can detect it? It turns out not much, which makes it a key new medium in which to observe the Universe!
In order to make this information more digestible, I will address one aspect of a gravitational wave's journey through space. Today's topic discusses how the Universe is essentially transparent to a gravitational wave. Future editions will discuss how matter can bend gravitational waves (gravitational lensing) and how the expanding Universe can stretch out (redshift) gravitational waves.
GRAVITATIONAL WAVES CAST NO SHADOWS:
First, let's think about what happens to light. As light travels through the Universe, any time that it encounters other matter, some of the light is absorbed by the matter or reflected away from its original path. The opposite happens for a gravitational wave; it can pass through matter and come out the other side unchanged (although there are some negligible effects)! That means that there is no such thing as a gravitational-wave shadow and nothing can obscure our detection of a gravitational wave!
The Spacetime Explanation:
But why is this? In a previous post, I described a gravitational wave as a change in the gravitational field moving out into the Universe. This change in gravitational field is often illustrated as a ripple, or wave, on spacetime (where the steepness of the curvature of spacetime represents the strength of the gravitational field, or the gravitational force a mass would feel, there). Let's look at what the Earth sitting on space-time looks like:
This picture isn't a perfect representation since the size of the Earth will affect the shape of the depression and this has no effect on real spacetime. Also, this is a simplified 2-dimenional representation of 3-dimensional (or a snapshot of the 4-dimensional spacetime) space. But if you were to imagine giving a corner of this flexible grid (spacetime) a swift shake, the Earth in the middle would be affected by it but it would not impede the wave. So, this is an example of how matter doesn't interact with gravitational waves, but I am still somewhat unsatisfied with this since you may think that the Earth will bounce after a wave passes creating more waves of its own (here is another aspect where this representation of spacetime is not perfect).
FYI: A better 3-dimensional representation of spacetime is shown in this clip from the American Museum of Natural History's short documentary called Gravity: Making Waves (which can be seen in its entirety on my Viewing Fun! page). This animation shows a grid-like scaffolding filling space in which there is a depression caused by mass. While it still isn't a perfect representation of spacetime, it is much better than the trampoline approximation above.
The Lunar Eclipse Explanation:
Recently, I thought of another more intimate example that most of us can identify with: a total lunar eclipse (which I have also blogged about). This is a situation where the Sun, Earth, and Moon line up in that order so that the Moon is completely in the Earth's shadow:
Image from Wikipedia
When the Moon is completely in the Earth's shadow (or the umbra in the diagram above), a viewer on the Moon would not be able to see any part of the Sun. If the gravitational field from the Sun were blocked by the Earth, then the Moon's orbit would appear to change. Since there is no change in the Moon's orbit during a total lunar eclipse, then the Earth does not block the Sun's gravitational field. By extension, the Earth also would not be able to block any changes in the gravitational field (which are gravitational waves). If you are familiar with physics, this is an application of the principle of superposition.
Great! But Why do We care?
Since mass does not absorb or reflect gravitational fields, the Universe is transparent to a gravitational wave. This is a huge advantage when using gravitational waves to make astronomical observations since nothing can block our view of a gravitational wave! If you have ever seen the our own Milky Way galaxy in the sky on a clear, dark night, you've seen the billions and billions of stars that live in our "backyard":
Fish-eye mosaic of the Milky Way galaxy as seen from Chile. [Image from Wikipedia]
While this is beautiful, it also it almost impossible to see past all the stars and dust there to observe what is behind our "backyard". We will be able to see right through the Milky Way with gravitational waves!
P.S. We can also detect gravitational waves from the other side of the Earth with LIGO since gravitational waves can travel through matter. Read more about this in this previous post (under the subheading of "Why are there 2 LIGOs?")!
1. If I take some plasticine or rubber and compress it with a quadrupole motion like a gravitational wave, it starts to get hot, because some of the applied force has been turned into heat. So would a really thick wall of a lossy material would absorb a gravitational wave?
Tried to look up stored energy in a gravity wave vs. losses in dampers just now, Google wasn't being terribly informative but one pair of equations suggested the absorption length has factor of c^3 in it, so it might be long!
2. I really don't know much about this other than what energy is absorbed from a gravitational wave is negligible compared to the total energy of the wave.
After a quick search, I found this old (1974) article dealing with bar detectors and energy absorption.
I do have a friend nearby at LSU who has worked on bar detectors until the one at LSU (Allegro) was decommissioned about 5 years ago. I am going to ask him the next time I see him and try to get you a better answer!
3. Thanks! No need to go out of your way, but it's one of those things that's kind of educational to know about (like working out the absorption length of neutrinos when I was working on neutrino beams).
Interesting about the bar detectors - if they're still vibrating after the wave has passed, they've extracted energy from the wave (so absorption doesn't even need a lossy material, just a bound system).
4. Thanks for the post! I'm thinking about the general rule from physics - if a system's inherent resonance matches the frequency of the applied wave, energy transfer will occur.
In this case if you had something that was already generating lower intensity gravity waves of the same frequency, couldn't it absorb some of the energy from incident, higher intensity gravity waves? | 1,396 | 6,646 | {"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} | 2.53125 | 3 | CC-MAIN-2017-34 | latest | en | 0.908631 |
http://betterlesson.com/lesson/resource/2086438/3-6-number-categories-docx | 1,477,379,920,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719960.60/warc/CC-MAIN-20161020183839-00117-ip-10-171-6-4.ec2.internal.warc.gz | 27,344,140 | 23,015 | ## 3.6 Number Categories.docx - Section 2: Number Categories
3.6 Number Categories.docx
3.6 Number Categories.docx
# Rational Numbers and Integer Practice
Unit 3: Integers and Rational Numbers
Lesson 7 of 17
## Big Idea: What is a rational number? Which is greater: -1.6 or -1.5? Students work to answer these questions while also practicing adding integers.
Print Lesson
16 teachers like this lesson
Standards:
Subject(s):
Math, Number Sense and Operations, adding integers, rational numbers, 6th grade, master teacher project
60 minutes
### Andrea Palmer
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Environment: Suburban | 334 | 1,462 | {"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-2016-44 | longest | en | 0.781269 |
http://lasp.colorado.edu/~bagenal/MATH/problems/problems5.html | 1,580,210,768,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251778168.77/warc/CC-MAIN-20200128091916-20200128121916-00314.warc.gz | 92,248,861 | 2,724 | ## Problems on Trigonometry
#### Section A: Angles (units and conversions)
1. Convert the following angles to radians.
a. 30º = d. 18º = g. 352º = b. 60º = e. 97º = h. 360º = c. 15º = f. 163º = i. 270º =
2. Convert the following angles to degrees (and to arcseconds/arcminutes if you wish to practice more!).
3. Convert the following angles to radians (you might want to recall the definition of arcminutes and arcseconds):
a. 28º 18'
b. 57º 25' 42"
c. 178º 52' 33"
4. Conver the following from radians to degree, arcminute, arecsecond notation:
#### Section B: Trigonometric Functions and Right Triangles
1. Evaluate the following quantities. This is for you to practice taking the sines and cosines and tangents of angles that are given to you in radians or in degrees.
a. sin (p/2) = ? d. cos (p/2) = ? g. tan (p) =? b. sin (42º) = ? e. cos (42º) = ? h. tan (45º) c. sin (162º) = ? f. cos (162º) = ?
2. Suppose that instead of being given the angle you encounter a situation - for instance in setting up a problem or solving for a quantitiy in a given situation - in which you know what the sine or cosine of some angle is equal to, but you don't know the angle! In these cases you take the inverse of the function to get the angle. Practice doing the following to see how this is done. On your calculator you simply do the "inverse" (often called <shift> or <2nd>). Sometimes - if the number is a nice number (such as b. or f. below) you ask yourself: "what angle would I have to take the sine/cosine of in order to get 0.0 or 1.0?".
a. sin (x) = 0.2, x = ? d. cos (x) = 1.0, x = ? b. sin (x) = 0.0, x = ? e. cos (x) = 0.34, x = ? c. sin (q) = 1.9, q = ? f. cos (q) = -1.0, q = ?
3. A right triangle has one leg equal to 3.0 cm. The angle opposite to this leg is measured to be 30º. Calculate the length of the other leg and the length of the hypotenuse. Refer to the figure below.
#### Section C: Small Angle Approximation, Arc Length ... and more
1. For which of the following situations drawn below is the small angle approximation - namely that sin(q)=q - valid?
2. A hill is at a distance of 10.0 km from you. You stand up and measure the angular size that it spans and you notice that is is about 2º. From this information, will you be able to determine the height of the hill? If so, what is the height? If not, explain why not?
#### Section D: Solving
1. Given "a" = (7/8)π radians, "b"= 70 degrees, and "d"= 80 degrees: Express angle "c" in:
a) degrees
c) arcminutes and arcseconds
2. find angle alpha in the following diagram: | 774 | 2,563 | {"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} | 4.5 | 4 | CC-MAIN-2020-05 | latest | en | 0.800343 |
http://docs.sunpy.org/en/stable/guide/units-coordinates.html | 1,555,671,300,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578527566.44/warc/CC-MAIN-20190419101239-20190419123239-00531.warc.gz | 51,470,273 | 7,718 | # Units and Coordinates in SunPy¶
This section of the guide will talk about representing physical units and physical coordinates in SunPy. SunPy makes use of Astropy for both these tasks.
## Units in SunPy¶
All functions in SunPy that accept or return numbers associated with physcial quantities accept and return Quantity objects. These objects represent a number (or an array of numbers) and a unit. This means SunPy is always explicit about the units associated with a value. Quantities and units are powerful tools for keeping track of variables with a physical meaning and make it straightforward to convert the same physical quantity into different units.
In this section of the guide we will give a quick introduction to astropy.units and then demostrate how to use units with SunPy.
To use units we must first import them from Astropy. To save on typing we usually import units as u:
>>> import astropy.units as u
Once we have imported units we can create a quantity by multiplying a number by a unit:
>>> length = 10 * u.meter
>>> length
<Quantity 10. m>
A Quantity has both a .unit and a .value attribute:
>>> length.value
10.0
>>> length.unit
Unit("m")
These Quantity objects can also be converted to other units, or unit systems:
>>> length.to(u.km)
<Quantity 0.01 km>
>>> length.cgs
<Quantity 1000. cm>
Probably most usefully, Quantity objects will propogate units through arithmetic operations when appropriate:
>>> distance_start = 10 * u.mm
>>> distance_end = 23 * u.km
>>> length = distance_end - distance_start
>>> length
<Quantity 22.99999 km>
>>> time = 15 * u.minute
>>> speed = length / time
>>> speed
<Quantity 1.53333267 km / min>
However, operations which do not make physical sense for the units specified will cause an error:
>>> length + time
Traceback (most recent call last):
...
astropy.units.core.UnitConversionError: Can only apply 'add' function to quantities with compatible dimensions
## Quantities as function arguments¶
An extremely useful addition to the base functionality of Quanitities is the @u.quantity_input decorator. This allows you to specify required units for function arguments to ensure that the calculation within that function always make physical sense. For instance, if we defined a function to calculate speed as above, we might want the distance and time as inputs:
>>> def speed(length, time):
... return length / time
However, this requires that length and time both have the appropriate units. We therefore want to use quantity_input to enforce this, here we use function annotations to specify the units (this is a Python 3.5+ feature, see the quantity_input documentation for more details and Python 2 instructions):
>>> @u.quantity_input
... def speed(length: u.m, time: u.s):
... return length / time
Now, when this function is called, if the units of length and time are not convertible to the units specified, an error will be raised stating that the units are incorrect or missing:
>>> speed(1*u.m, 10*u.m)
Traceback (most recent call last):
...
astropy.units.core.UnitsError: Argument 'time' to function 'speed' must be in units convertible to 's'.
>>> speed(1*u.m, 10)
Traceback (most recent call last):
...
TypeError: Argument 'time' to function 'speed' has no 'unit' attribute. You may want to pass in an astropy Quantity instead.
Note that the units of the inputs do not have to be exactly the same as those in the function definition, as long as they can be converted to those units. So for instance, passing in a time in minutes still works even though we specified time: u.s:
>>> speed(1*u.m, 1*u.minute)
<Quantity 1. m / min>
This may still not be quite as we want it, since we wanted the input time in seconds but the output is in m/min. We can correct this by defining the function with an additional annotation:
>>> @u.quantity_input
... def speed(length: u.m, time: u.s) -> u.m/u.s:
... return length / time
This will force the output of the function to be converted to m/s before returning, so that you will always have the same units on the output from this function:
>>> speed(1*u.m, 1*u.minute)
<Quantity 0.01666667 m / s>
## Physical Coordinates in SunPy¶
In much the same way as units are used for representing physical quantities, SunPy uses astropy.coordinates to represent points in physical space. This applies to both points in 3D space and projected coordinates in images.
The astropy coordinates module is primarily used through the SkyCoord class:
>>> from astropy.coordinates import SkyCoord
To enable the use of the solar physics specific frames defined in SunPy we also need to import them:
>>> from sunpy.coordinates import frames
A SkyCoord object to represent a point on the Sun can then be created:
>>> c = SkyCoord(70*u.deg, -30*u.deg, obstime="2017-08-01",
... frame=frames.HeliographicStonyhurst)
>>> c
<SkyCoord (HeliographicStonyhurst: obstime=2017-08-01 00:00:00): (lon, lat, radius) in (deg, deg, km)
(70., -30., 695508.)>
This SkyCoord object can then be transformed to any other coordinate frame defined either in Astropy or SunPy, for example:
>>> c.transform_to(frames.Helioprojective)
<SkyCoord (Helioprojective: obstime=2017-08-01 00:00:00, rsun=695508.0 km, observer=<HeliographicStonyhurst Coordinate (obstime=2017-08-01 00:00:00): (lon, lat, radius) in (deg, deg, AU)
(0., 5.78339799, 1.01496923)>): (Tx, Ty, distance) in (arcsec, arcsec, km)
(769.74997696, -498.75932128, 1.51668819e+08)>
It is also possible to convert three dimensional positions to astrophysical frames defined in Astropy, for example ICRS.
>>> c.transform_to('icrs')
<SkyCoord (ICRS): (ra, dec, distance) in (deg, deg, km)
(49.85053118, 0.05723938, 1417577.0297545...)>
### Observer Location¶
Both Helioprojective and Heliocentric frames are defined based on the position of the observer. Therefore to transform either of these frames to a different frame the location of the observer must be known. The default observer is the Earth. A different observer can be specified for a coordinate object using the observer argument to SkyCoord. For SunPy to calculate the location of the Earth, it must know the time for which the coordinate is valid; this is specified with the obstime argument.
Using the observer location it is possible to convert a coordinate as seen by one observer to a coordinate seen by another:
>>> hpc1 = SkyCoord(0*u.arcsec, 0*u.arcsec, observer="earth",
... obstime="2017-07-26",
... frame=frames.Helioprojective)
>>> hpc1.transform_to(frames.Helioprojective(observer="venus",
... obstime="2017-07-26"))
<SkyCoord (Helioprojective: obstime=2017-07-26 00:00:00, rsun=695508.0 km, observer=<HeliographicStonyhurst Coordinate (obstime=2017-07-26 00:00:00): (lon, lat, radius) in (deg, deg, AU)
(77.03547231, 3.17032536, 0.72510629)>): (Tx, Ty, distance) in (arcsec, arcsec, km)
(-1285.11970265, 106.17983302, 1.08317783e+08)>
### Using Coordinates with SunPy Map¶
SunPy Map uses coordinates to specify locations on the image, and to plot overlays on plots of maps. When a Map is created, a coordinate frame is constructed from the header information. This can be accessed using .coordinate_frame:
>>> import sunpy.map
>>> from sunpy.data.sample import AIA_171_IMAGE # doctest: +REMOTE_DATA
>>> m = sunpy.map.Map(AIA_171_IMAGE) # doctest: +REMOTE_DATA
>>> m.coordinate_frame # doctest: +REMOTE_DATA
<Helioprojective Frame (obstime=2011-06-07 06:33:02.770000, rsun=696000000.0 m, observer=<HeliographicStonyhurst Coordinate (obstime=2011-06-07 06:33:02.770000): (lon, lat, radius) in (deg, deg, m)
(0., 0.048591, 1.51846026e+11)>)>
This can be used when creating a SkyCoord object to set the coordinate system to that image:
>>> from astropy.coordinates import SkyCoord
>>> import astropy.units as u
>>> c = SkyCoord(100 * u.arcsec, 10*u.arcsec, frame=m.coordinate_frame) # doctest: +REMOTE_DATA
>>> c # doctest: +REMOTE_DATA
<SkyCoord (Helioprojective: obstime=2011-06-07 06:33:02.770000, rsun=696000000.0 m, observer=<HeliographicStonyhurst Coordinate (obstime=2011-06-07 06:33:02.770000): (lon, lat, radius) in (deg, deg, m)
(0., 0.048591, 1.51846026e+11)>): (Tx, Ty) in arcsec
(100., 10.)>
This SkyCoord object could then be used to plot a point on top of the map:
>>> import matplotlib.pyplot as plt
>>> ax = plt.subplot(projection=m) # doctest: +REMOTE_DATA
>>> m.plot() # doctest: +REMOTE_DATA
<matplotlib.image.AxesImage object at ...>
>>> ax.plot_coord(c, 'o') # doctest: +REMOTE_DATA
For more information on coordinates see SunPy coordinates section of the Code Reference. | 2,268 | 8,673 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2019-18 | longest | en | 0.819601 |
https://gatestack.in/t/problem-on-basic-theorm/735 | 1,586,300,539,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371806302.78/warc/CC-MAIN-20200407214925-20200408005425-00542.warc.gz | 471,480,965 | 2,825 | # Problem on basic theorm
#1
Determine if each of the following objects is a member of Z;
{5}, {3, −1}, 7.12,√5, a = the 2,00th decimal digit in the base-10 expression for π
#2
2000th decimal digit will be from 0 to 9 so it belongs to Z ,set of integers
{3,-1} is not a single integer so i think not belong to Z
7.12 also not a integer
√5
is an irrational number (how do u proove it) | 128 | 387 | {"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} | 3.203125 | 3 | CC-MAIN-2020-16 | latest | en | 0.879583 |
https://fddcn.cn/number-sequence-2.html | 1,558,442,492,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256381.7/warc/CC-MAIN-20190521122503-20190521144503-00518.warc.gz | 459,395,881 | 10,168 | # Number Sequence
### Problem Description
A number sequence is defined as follows:
f(1) = 1, f(2) = 1, f(n) = (A * f(n – 1) + B * f(n – 2)) mod 7.
Given A, B, and n, you are to calculate the value of f(n).
### Input
The input consists of multiple test cases. Each test case contains 3 integers A, B and n on a single line (1 <= A, B <= 1000, 1 <= n <= 100,000,000). Three zeros signal the end of input and this test case is not to be processed.
### Output
For each test case, print the value of f(n) on a single line. | 164 | 525 | {"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} | 2.828125 | 3 | CC-MAIN-2019-22 | latest | en | 0.809366 |
http://forums.wolfram.com/mathgroup/archive/2013/Feb/msg00121.html | 1,553,085,391,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202326.46/warc/CC-MAIN-20190320105319-20190320131319-00191.warc.gz | 77,806,233 | 7,743 | Re: Differencing two equations
• To: mathgroup at smc.vnet.net
• Subject: [mg129757] Re: Differencing two equations
• From: Murray Eisenberg <murrayeisenberg at gmail.com>
• Date: Mon, 11 Feb 2013 04:36:32 -0500 (EST)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com
• Delivered-to: l-mathgroup@wolfram.com
• Delivered-to: mathgroup-newout@smc.vnet.net
• Delivered-to: mathgroup-newsend@smc.vnet.net
• References: <20130210082414.C3A5F6937@smc.vnet.net>
```Yeah, Mathematica stubbornly refuses to thread Equal over the difference. I seem to recall that there are good design reasons for that.
The following works, but of course is more involved than one might like:
Equal @@ Subtract @@@ {a == r, b == s}
Which is an abbreviated form of:
Apply[Equal, Apply[Subtract, {a == r, b == s}, 1]]
Essentially, you're taking out the equalities and after subtractions putting an equality back.
On Feb 10, 2013, at 3:24 AM, g.c.b.at.home at gmail.com wrote:
> I'm brand new to Mathematica, so I apologize for the naive questions...
>
> I'm trying to figure out how to difference two equations. Basically if I have:
> a==r
> b==s
>
> I'd like to get:
> a-b == r-s
>
> What I'm getting is more like (a==r) - (b==s). I'm not sure how that's a useful result, but is there a function to do what I'm looking for?
>
> A quick search of the archives seem to bring up ways of doing this from using transformation rules to swap heads to unlocking the Equals operator and hacking its behavior. I'd like to avoid doing that kind of rewiring for a simple operation, and I'd like to keep the syntax clean.
>
> The Core Language documentation makes a big point of how everything is basically a list with different heads. In this case, what I'm trying to do would work if it were treated as a list ({a,b}-{r,s} returns {a-b,r-s}) but doesn't work under Equal.
>
> Thanks for any suggestions.
---
Murray Eisenberg murray at math.umass.edu
Mathematics & Statistics Dept.
Lederle Graduate Research Tower phone 413 549-1020 (H)
University of Massachusetts 413 545-2838 (W)
710 North Pleasant Street fax 413 545-1801
Amherst, MA 01003-9305
```
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At a Glance - Distributive Property
This video shows the distributive property in action on a slightly advanced problem.
Officially, we use the distributive property to multiply binomials. If you need a quick refresher (or introduction) to binomials, those are expressions that consist of two terms. You know, as in "a + b" or "Mike + Brittney 4EVA." We’ll go with the first example, since we found the second one carved into the side of a maple tree and we’re not actually sure if it’s public domain.
If you want to multiply a + b and c + d, the distributive property tells you how to do exactly that. All we need to do is multiply a by both letters in the second term, then multiply b by everything in the second term. That sounds kind of abstract, so here's what we mean:
(a + b)(c + d) =
a
(c + d) + b(c + d) =
ac +ad + bc + bd
That may not look too helpful right now, but this thing will be a lifesaver once you’ve got a problem with a few numbers to plug in. A cherry-flavored lifesaver. | 298 | 1,230 | {"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-2016-22 | longest | en | 0.930915 |
https://www.totaltypescript.com/workshops/type-transformations/conditional-types-and-infer/refine-conditional-logic-in-a-type-helper | 1,695,551,334,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506632.31/warc/CC-MAIN-20230924091344-20230924121344-00110.warc.gz | 1,161,273,067 | 27,350 | Conditional Types and Infer 10 exercises
Problem
# Refine Conditional Logic in a Type Helperself.__wrap_balancer=(t,e,n)=>{n=n||document.querySelector(`[data-br="\${t}"]`);let o=n.parentElement,r=E=>n.style.maxWidth=E+"px";n.style.maxWidth="";let i=o.clientWidth,s=o.clientHeight,c=i/2,u=i,d;if(i){for(;c+1<u;)d=~~((c+u)/2),r(d),o.clientHeight==s?u=d:c=d;r(u*e+i*(1-e))}};self.__wrap_balancer(":Rhd9j6:",1)
The starting point is our solution from the last exercise:
`type YouSayGoodbyeAndISayHello<T> = T extends "hello" ? "goodbye" : "hello";`
## Challenge
There's one more thing to add to this code. Recall that if we pass something in that is not "hello" that we will be returned "hello."
Y | 213 | 699 | {"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} | 2.546875 | 3 | CC-MAIN-2023-40 | longest | en | 0.502287 |
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