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# Trigonemtetric Word Problem
1. Apr 5, 2006
### physstudent1
"A plane is observed approaching your home and you assume its speed is 550 miles per hour...The angle of elevation of the plane is 16 degrees at one time and 57 degrees one minute later. Approximate the altitude of the plane."
Ok I tried drawing triangles for this but I keep getting to a part where one side = 9.166 + unknown because I put the speed into miles per minute to find how much distance it covered in one minute I really can't figure out how to draw the triangle for htis problem I'm really confused about this can anyone help me clear it up the answer is 17054 feet or 3.23 but if anyone oculd give me an explanation of how to do this I would really appreciate it.
2. Apr 5, 2006
Hints on drawing the triangle...
(1) Draw a horizontal line on your paper representing ground level. Since you are watching the plane from the ground, all the angles of elevation will be with respect to this line.
(2) Mark a point on the horizontal line, letting it represent your position.
(3) Mark a point above the horizontal line (but not directly above you!), letting it represent the plane's initial position. If you join the 2 points, what can you say about the angle between the line formed and the horizontal?
(4) Mark another point above the horizontal line, letting it represent the plane's position a minute later. Since we assume the plane's altitude does not change, what can we say about the line joining the 2 points which are above the horizontal?
General hints:
The plane has not passed you.
Putting the plane's speed into miles per minute to find how much distance it covered in one minute is a step in the correct direction, as it will help you calculate the length of one of the sides of the triangle.
Last edited: Apr 5, 2006
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real function smooth ( x, y, dy, npoint, s, v, a ) c from * a practical guide to splines * by c. de boor calls setupq, chol1d c c constructs the cubic smoothing spline f to given data (x(i),y(i)), c i=1,...,npoint, which has as small a second derivative as possible c while c s(f) = sum( ((y(i)-f(x(i)))/dy(i))**2 , i=1,...,npoint ) .le. s . c c****** i n p u t ****** c x(1),...,x(npoint) data abscissae, a s s u m e d to be strictly c increasing . c y(1),...,y(npoint) corresponding data ordinates . c dy(1),...,dy(npoint) estimate of uncertainty in data, a s s u m- c e d to be positive . c npoint.....number of data points, a s s u m e d .gt. 1 c s.....upper bound on the discrete weighted mean square distance of c the approximation f from the data . c c****** w o r k a r r a y s ***** c v.....of size (npoint,7) c a.....of size (npoint,4) c c***** o u t p u t ***** c a(.,1).....contains the sequence of smoothed ordinates . c a(i,j) = f^(j-1)(x(i)), j=2,3,4, i=1,...,npoint-1 , i.e., the c first three derivatives of the smoothing spline f at the c left end of each of the data intervals . c w a r n i n g . . . a would have to be transposed before it c could be used in ppvalu . c c****** m e t h o d ****** c The matrices Q-transp*d and Q-transp*D**2*Q are constructed in c s e t u p q from x and dy , as is the vector qty = Q-transp*y . c Then, for given p , the vector u is determined in c h o l 1 d as c the solution of the linear system c (6(1-p)Q-transp*D**2*Q + p*Q)u = qty . c From u , the smoothing spline f (for this choice of smoothing par- c ameter p ) is obtained in the sense that c f(x(.)) = y - 6(1-p)D**2*Q*u and c f''(x(.)) = 6*p*u . c The smoothing parameter p is found (if possible) so that c sf(p) = s , c with sf(p) = s(f) , where f is the smoothing spline as it depends c on p . if s = 0, then p = 1 . if sf(0) .le. s , then p = 0 . c Otherwise, the secant method is used to locate an appropriate p in c the open interval (0,1) . However, straightforward application of c the secant method, as done in the original version of this program, c can be very slow and is influenced by the units in which x and y c are measured, as C. Reinsch has pointed out. Instead, on recommend- c ation from C. Reinsch, the secant method is applied to the function c g:q |--> 1/sqrt{sfq(q)} - 1/sqrt{s} , c with sfq(q) := sf(q/(1+q)), since 1/sqrt{sfq} is monotone increasing c and close to linear for larger q . One starts at q = 0 with a c Newton step, i.e., c q_0 = 0, q_1 = -g(0)/g'(0) c with g'(0) = -(1/2) sfq(0)^{-3/2} dsfq, where dsfq = -12*u-transp*r*u , c and u as obtained for p = 0 . Iteration terminates as soon as c abs(sf - s) .le. .01*s . c c logical test c parameter (test = .true.) c integer itercnt integer npoint, i,npm1 real a(npoint,4),dy(npoint),s,v(npoint,7),x(npoint),y(npoint) * ,change,ooss,oosf,p,prevsf,prevq,q,sfq,sixp,six1mp,utru call setupq(x,dy,y,npoint,v,a(1,4)) if (s .gt. 0.) go to 20 10 p = 1. six1mp = 0. call chol1d(p,v,a(1,4),npoint,1,a(1,3),a(1,1)) sfq = 0. go to 60 20 p = 0. six1mp = 6. call chol1d(p,v,a(1,4),npoint,1,a(1,3),a(1,1)) sfq = 0. do 21 i=1,npoint 21 sfq = sfq + (a(i,1)*dy(i))**2 sfq = sfq*36. if (sfq .le. s) go to 60 utru = 0. do 25 i=2,npoint 25 utru = utru + v(i-1,4)*(a(i-1,3)*(a(i-1,3)+a(i,3))+a(i,3)**2) ooss = 1./sqrt(s) oosf = 1./sqrt(sfq) q = -(oosf-ooss)*sfq/(6.*utru*oosf) c secant iteration for the determination of p starts here. c itercnt = 0 prevq = 0. prevsf = oosf 30 call chol1d(q/(1.+q),v,a(1,4),npoint,1,a(1,3),a(1,1)) sfq = 0. do 35 i=1,npoint 35 sfq = sfq + (a(i,1)*dy(i))**2 sfq = sfq*36./(1.+q)**2 if (abs(sfq-s) .le. .01*s) go to 59 oosf = 1./sqrt(sfq) change = (q-prevq)/(oosf-prevsf)*(oosf-ooss) prevq = q q = q - change prevsf = oosf c itercnt = itercnt + 1 go to 30 59 p = q/(1.+q) six1mp = 6./(1.+q) correct value of p has been found. compute pol.coefficients from Q*u (in a(.,1)). 60 smooth = sfq c if (test) then c print *, 'number of iterations = ', itercnt c end if do 61 i=1,npoint 61 a(i,1) = y(i) - six1mp*dy(i)**2*a(i,1) sixp = 6.*p do 62 i=1,npoint 62 a(i,3) = a(i,3)*sixp npm1 = npoint - 1 do 63 i=1,npm1 a(i,4) = (a(i+1,3)-a(i,3))/v(i,4) 63 a(i,2) = (a(i+1,1)-a(i,1))/v(i,4) * - (a(i,3)+a(i,4)/3.*v(i,4))/2.*v(i,4) return end
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# Hi every one
1. Oct 25, 2006
### ريمان
hi every one
is spacetmie like the sea or waht?
thank for everyone
2. Oct 25, 2006
### ريمان
is there any body hire
yahooooooo
3. Oct 26, 2006
### TDS
Hello ريمان,
From Wikipedia:
For other uses of this term, see Spacetime (disambiguation).
In physics, spacetime is a mathematical model that combines three-dimensional space and one-dimensional time into a single construct called the space-time continuum, in which time plays the role of the 4th dimension. According to Euclidean space perception, our universe has three dimensions of space, and one dimension of time. By combining space and time into a single manifold, physicists have significantly simplified a good deal of physical theory, as well as described in a more uniform way the workings of the universe at both the supergalactic and subatomic levels.
In classical mechanics, the use of spacetime over Euclidean space is optional, as time is independent of mechanical motion in three dimensions. In relativistic contexts, however, time cannot be separated from the three dimensions of space as it depends on an object's velocity relative to the speed of light.
I hope this helps.
TDS
4. Oct 26, 2006
### ريمان
tank you
is spactime like this http://www.blog.speculist.com/archives/spacetime.gif" [Broken]
is earth above spacetime or in spacetime
yyyyyyyyyyyhooooooooooooo
Last edited by a moderator: May 2, 2017
5. Oct 26, 2006
### tandoorichicken
My understanding is that spacetime is basically a frame of reference for the entire universe. That means that whatever is in the universe is "in" spacetime, including the earth. I can't imagine what would be "outside" spacetime.
6. Oct 27, 2006
### ريمان
are you mean the earth is in spacetime not above spacetime
like this http://www.blog.speculist.com/archives/spacetime.gif [Broken]
Last edited by a moderator: May 2, 2017
7. Oct 27, 2006
### tandoorichicken
Yes, thats right. The earth is in spacetime. The picture just shows one surface out of many I'm guessing. That surface happens to run under the image of the earth. They can't show all the surfaces that make up spacetime in the picture otherwise you wouldnt be able to see the earth.
8. Oct 27, 2006
### ريمان
tank you very much you are so smart
ريمان
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# The Physics of Sports (SOL 2.1, 3.1, 4.2 & 6.1)
Teacher Name: Gemma Griffin
Grade Level: 2-6
Subject: Science
Time Frame: 1 hour 45 minutes
Concept: Physics
Topic: Sports and Physics
## Rationale:
Sports have been a source of education and entertainment for thousands of years. They teach values like teamwork, and the opportunity to promote healthy competition. Sports also lend themselves to the teaching of science; the study of our world and how it works. Through the study of sports, students can be exposed to many physical science concepts, that are often very difficult to teach and understand with only textbook material.
## Goals:
The student will recognize that important physics concepts can be found in their daily activities, including sports!
The student will have the opportunity to improve his or her teamwork skills.
Content: Physics concepts to be explored:
Basketball: Bouncing Balls Experiment:
Newton’s 2nd law: ” Amount of force depends on the mass and acceleration.
Newton’s 3rd law: “for every action there is an equal and opposite reaction.”
Elastic and Inelastic
Football: The Perfect Pass Experiment:
Newton’s 1st law: “objects at rest remain at rest, objects in motion remain in motion, unless acted on by an outside force”
Axis
Soccer: The Perfect Kick (Just for Kicks) Experiment:
Momentum
Baseball: Swinging Away Experiment:
Newton’s 1st law of motion.
SOL: Science:
2.1 – The student will plan and conduct investigations in which simple physical models are constructed.
3.1 – The student will plan and conduct investigations in which observations are made.
4.2 – The student will investigate and understand energy:
– potential and kinetic energy
– efficiency, friction, and inertia
6.1 The student will plan and conduct investigations in which
– scale models are used
## Objectives:
Cognitive:
The student will be able to investigate the physical science present in sports, evidenced by successful completion of experiments, written observation records, and oral discussion.
Affective:
The student will be able to be aware that physical science is present in their current lives, by investigating the physics in sports, evidenced by successful completion of experiments, written observation records, and oral discussion.
Psychomotor:
The student will be able to apply important physical science concepts, learned in classroom experiments, evidenced by journal write and their participation in playing the actual sport and using the concepts to improve their game skills.
## Materials:
Directions for experiments Content explanations for experiments
Football
Soccer ball and cones for goals
Basketball and court
Baseball and bat, bases (whiffle ball set recommended for safety reasons)
tennis ball, crumpled sheet of paper, blown up balloon
## Advanced Organizer: (30 minutes)
Physics Presentation: Dr. Sheldon’s Show
Procedures & Activities: (60 minutes)
1. After physics introduction show, divide into two groups (2nd/3rd and 4th/5th)
2. Ask students to orally share a most memorable experience with sports that they have. Allow time for those who want to share.
3. Introduce the idea of science (physics) in the sports that we play, and how our knowledge about the physics of these sports can greatly improve our game.
4. Rotate through the four sports (being outside or in a gym will be necessary for this). At each sport, have students brainstorm what they need to know to play the sport successfully.
Sport #1 Basketball: Bouncing Balls Experiment:
Basketball relies on bouncing balls more than most sports. So why do balls bounce? Basketballs and oranges are both round, but you’d have a hard time dribbling an orange up the court! What’s so special about balls?
Get a basketball and a ruler. Also needed: balled up piece of paper, blown up balloon, and tennis ball. These items will be your test objects. Have students predict and order which they think will bounce highest to hardly bounced at all.
Pick a spot on the floor to test your objects on.
Stand your ruler on the test surface and bounce each test object from the same height (12 inches). Measure how high each item bounces. Some items may not bounce.
Paper ______________ inches # _____
Balloon ______________ inches # _____
Tennis Ball ____________ inches # _____
Mini-basketball _________ inches # _____
Look at your results. Put the test items in order by how high they bounced and squeeze each one. What happens? Do the bouncy items have anything in common?
Set a few ground rules:
Good sportsmanship.
No wandering off when outside.
Safety: (Example: no tackling in football, shoving in basketball or soccer)
Allow students to play basketball for about 5 minutes and apply what they’ve learned about the physics of the sport.
Sport #2 Football: The Perfect Pass Experiment:
You’ve probably seen great passes – the kind that just rifle through the air right into the hands of a waiting receiver. But you’ve also probably seen some wobble and fall short, maybe into the hands of the other team! What’s the difference? What makes a good pass?
1. Get a coin. Find a flat surface on the floor or desk to test it on.
2. Try to balance the coin on its edge without spinning it.
3. Now spin the coin softly. Try it a few times and time yourself how long you can get it to spin.
4. Spin the coin harder and faster this time.
Is there a relationship between how fast they spin and how long they stay up?
Have students identify that a football will stay up in the air longer and travel a farther distance when spinning, like a coin, rather than tumbling through the air, end over end.
Allow students 5 minutes to practice their “perfect pass”.
Sport #3 Soccer: The Perfect Kick (Just for Kicks) Experiment:
There is, of course, no one perfect kick. But soccer players always look for the best possible kick to use in any given situation.Momentum has a lot to do with the power of your kick. Momentum is the built up energy, measured by the amount of stuff in motion times how fast it is moving.
Demonstrate and have students demonstrate momentum between their hand and a balled up piece of paper.
A. Hold up on hand with fingers pointing straight up, elbow bent and reasonably close to the body this is your “waiting hand”. Hold the paper ball in the other hand a little above your “waiting hand” and to one side. Your two arms will make a “T” shape.
B. Let the ball go and smack it with the back of your waiting hand, without any kind of windup. (Don’t move your hand away from the ball before hitting it). Estimate how far the ball went. Very far? A little far? Not far at all?
C. Now repeat this same experiment, but this time, hold your waiting hand parallel to the ground, so your waiting hand will have quite a distance to travel to hit the paper ball. Drop the paper ball and pivot your waiting hand up to smack the ball.
How far did the ball go this time compared to the previous time?
Allow students to apply their knowledge know about momentum during a brief game of kickball or soccer.
Center #4 Baseball: Swinging Away Experiment:
Batters use strategies to improve their chances of getting a good hit. Batters will also hit the ball harder or softer depending on their needs. But why do these different swings affect the ball in such different ways?
Ask students to think about where they’d need to make contact with the bat to the ball to make the ball go downwards (a bunt), upwards (for a homerun), or to sail pretty much straight across the field? (See content page for further explanation)
Allow students to practice different swings for about 5 minutes.
## Closure: (5-10min)
Have students journal write for about 5-10 minutes. Instruct students to choose one sport to write about, including the physics they learned about it, and how the physics can be used to improve their playing abilities of that sport.
## Assessment/Evaluation:
Oral discussion (Comprehension)
Journal write (Transfer of Knowledge)
Participation in games (Application)
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# Physics
posted by on .
A car drives straight off the edge of a cliff that is 68 m high. The police at the scene of the accident note that the point of impact is 148 m from the base of the cliff. How fast was the car traveling when it went over the cliff?
• Physics - ,
H =g•t²/2 => t =sqrt(2•H/g) = sqrt(2•68/9.8) =3.7 s.
L = v•t => v =L/t = 148/3.7= 39.7 m/s = 143 km/hr =88.8mph
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# SC - How to overcome the 2-3 split dilemma
Author Message
Intern
Joined: 09 Oct 2011
Posts: 36
Followers: 0
Kudos [?]: 3 [0], given: 4
SC - How to overcome the 2-3 split dilemma [#permalink]
### Show Tags
10 Oct 2011, 15:57
Hi GMAT Experts,
I've been studying SC using Manhattan GMAT. I found the materials are very helpful in terms of understanding the fundamental concepts. As I progress through the Manhattan GMAT SC book, I've been doing problem sets referenced in OG and OG verbal at the end of each chapter. I realize for the questions that I do not get right. I always arrive at 2-3 split dilemma and choose the other incorrect answer between the only 2 answers that I deem could be correct. I wonder if this is common to most of the GMAT takers. How would you maximize your chance of getting the correct answer and overcome such recurring situations? Thanks for your inputs in advance.
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Re: SC - How to overcome the 2-3 split dilemma [#permalink]
### Show Tags
11 Oct 2011, 09:49
1
KUDOS
Expert's post
Indeed this is quite common. You think you're more than half way there because you were able to eliminate 3 answer choices - but then you get down to 2 and pick the wrong one. All that time you spent eliminating the first few didn't make any impact to the end result.
How do you address a problem like this?
Put together a collection of these kinds of questions. If you're using a book, circle the question # so you can refer to it again later. If you're doing something online, write down the question #.
Then examine them one by one and ask yourself why you didn't end up picking the OTHER answer. Was there a thought process you could have done that could have led you to the other direction? With a larger sample of question examples, you'll be able to spot patterns into what pitfalls you're falling into. Maybe they tend to be [Description, Main Sentence] kinds of questions involving -ING verbs. Or maybe they're X&Y consistency questions that involve spotting an important comma that you missed paying attention to.
Whatever the case may be, make a record of these kinds of questions and revisit them to spot patterns so you avoid them in the future.
Hope that helps.
_________________
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Joined: 22 Mar 2011
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Re: SC - How to overcome the 2-3 split dilemma [#permalink]
### Show Tags
13 Oct 2011, 12:03
Sometimes this is a meaning issue. Are you making sure that the choice you have selected makes sense in the context of the sentence? Does it distort the intended meaning?
Other times, it is simply a matter of spotting additional splits and choosing the issue you feel best qualified to address. If an option just "doesn't sound right," delay that decision and try to find an issue you feel more sure about. After your time is up, spend additional time reviewing each question BEFORE checking the answer. Try to identify for yourself why each choice is right or wrong, and try to make decisions on any issues you didn't use in selecting an answer. (E.g. if you crossed off B due to parallelism, look at the rest of it and see what you think about the other issues it raises).
Then, once you feel as solid as possible about why each choice is right or wrong, check the answer and explanation and see how well it matches with your reasoning. If you missed the problem, spend some serious time identifying which issues you failed to identify, or chose the wrong side on. Take that tempting wrong answer and do your best to smash it. You don't just want to say "Oh, C is a better choice." You want to be able to say "E is incorrect and here are the reasons." If the explanation in the Official Guide is inadequate (this is not unlikely), look the problem up in the forums.
If you get really good at this work, starting with timed problems and following through with very thorough untimed review, you should become a faster and more attentive test-taker, and you will have more time and attention to focus on distinguishing the fine shades of meaning that this test sometimes requires. Good luck!
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Re: SC - How to overcome the 2-3 split dilemma [#permalink] 13 Oct 2011, 12:03
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Cody
# Problem 36. Find relatively common elements in matrix rows
Solution 907711
Submitted on 13 Jun 2016 by Grace O'Neil
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
x = []; y_correct = []; assert(isequal(common_by_row(x),y_correct))
2 Pass
x = [1e100; 1e100]; y_correct = [1e100]; assert(isequal(common_by_row(x),y_correct))
3 Pass
x = [1; 2]; y_correct = []; assert(isequal(common_by_row(x),y_correct))
4 Pass
x = ones(10); y_correct = [1]; assert(isequal(common_by_row(x),y_correct))
5 Pass
x = magic(10); y_correct = []; assert(isequal(common_by_row(x),y_correct))
6 Pass
x = wilkinson(9); y_correct = [0 1]; assert(isequal(common_by_row(x),y_correct))
7 Pass
x = [3 -2 1 NaN; NaN 0 -2 3]; y_correct = [-2 3]; assert(isequal(common_by_row(x),y_correct))
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# Why are repeating decimals often self-inversions?
Let $$b$$ be any base, and let $$x$$ be some odd integer.
Empirically, it seems that for any $$b$$, the majority of odd $$x$$ will have the following property. Given a reptend $$r$$ with $$2k$$ digits as in
$$\frac{1}{x} = 0. \dots \overline{r_1 r_2 \dots r_{(2k)}} \ ,$$
you will often find that $$r_i+ r_{(k+i)}\equiv -1 \pmod b$$ for all $$i \leq k$$.
As a simple example, we have $$x=13$$ in base $$b=2$$ with the 12-digit reptend $$r=100111011000_2$$, as so:
$$\frac{1}{13} = 0.00\ \mathbf{100111}\ 011000\ \mathbf{100111}\ 011000 \ldots$$
where $$100111_2+011000_2=111111_2$$.
In short, for many more numbers than one would expect by chance, the first and second halves of the repeating digits of many rationals often sum pairwise to $$b-1$$. I haven't checked this rigorously, but there's obviously some mechanism causing this, and it seems to be base-agnostic. What gives?
Here's a follow-up snapshot of a range of essentially arbitrary rationals in the form $$\frac{20}{x}$$ using base-33. The answer about $$1/p$$ makes sense, but if it also explains why there would be so many examples in this case, I'm afraid I'm still missing how. Note the stars that indicate those whose digits sum to $$32$$, or check the reptend addition itself in the center column.
Perhaps it would be easier to explain why those rationals which don't follow this pattern do not?
• – lhf
Oct 27, 2020 at 0:43
• I read that before posting... if it addresses my question directly, I must be missing it. Oct 27, 2020 at 0:45
• note: $\frac1p$ repeats after $2k$ digits means $b^{2k}\equiv1$ and $b^{k}\equiv-1\bmod p$ Oct 27, 2020 at 0:45
• it seems you are using $k$ for two different things: (1) half the length of the reptend and (2) the odd number whose reciprocal you are calculating Oct 27, 2020 at 1:16
That the decimal expansion of $$\dfrac 1p$$ repeats after $$2k$$ digits means $$b^{2k}\equiv1\bmod p$$,
which means $$b^k\equiv\pm1\bmod p$$,
and if it doesn't repeat after $$k$$ digits that means $$b^k\equiv p-1\bmod p$$.
Thus, if $$\dfrac 1p=0.r_1r_2...r_{2k}$$, then $$(r_1r_2...r_k)\times p=b^k-(p-1)$$
and $$(r_{k+1}r_{k+2}...r_{2k})\times p=(p-1)\times b^k-1,$$
so $$(r_1r_2...r_k+r_{k+1}r_{k+2}...r_{2k})\times p = p \times( b^k-1)$$;
i.e., $$r_1r_2...r_k+r_{k+1}r_{k+2}...r_{2k} = b^k-1$$.
• Here $p$ is a power of an odd prime Oct 27, 2020 at 3:51
• Thanks, this helps, but I'm still confused about whether it helps explain non-prime cases. See addition to post. Oct 27, 2020 at 5:21
• It doesn't always work for non-prime-power cases; consider the decimal expansions of $1/119, 1/187$, and $1/221$; if $10^{2k}\equiv1\bmod pq$, it could be that $10^k\equiv-1\bmod pq$ or not Oct 27, 2020 at 15:37
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## energy calculation grinding
Application
Calculation of energy required for grinding in a ball mill The Bond work index, W i, as an indicator of the grindability of raw materials is not a material con
# energy calculation grinding
• ## Calculation of energy required for grinding in a ball mill
The Bond work index, W i, as an indicator of the grindability of raw materials is not a material constant but rather it changes with change of size of the grinding productTherefore, in practice we can expect some difficulties and errors when the energy consumption is determined according to this formula in the case when, for a given size of grinding product, the value of the work index W i01/01/1989· Calculation of energy required for grinding in a ball mill Int J Miner Process, 25: 4146 The Bond work index, Wi, as an indicator of the grindability of raw materials is not a material constant but rather it changes with change of size of the grinding product Therefore, in practice we can expect some difficulties and errors when the energy consumption is determined according to thisCalculation of energy required for grinding in a ball millgrinding energy compared with shallow grinding and creep grinding The chips take away most of the heat generated in the grinding process As in creep grinding, bumout of the coolant causes a steep rise in contact temperature of the workpiece 1 Introduction Cost, quality and productivity in grinding processes are all dependent on energy consumption and process efficiency In an inefficientEnergy and temperature analysis in grinding
• ## Development of specific grinding energy model
01/09/2012· Specific grinding energy is one of the most important performance parameter of the grinding process Primary energy components of specific grinding energy are the major contributors to the total specific energy consumption in grinding For high strength and hardness materials such as advanced ceramics, specific grinding energy requirement is extremely high which will eventually01/09/2012· Specific grinding energy is one of the most important performance parameter of the grinding process Primary energy components of specific grinding energy are the major contributors to the total specific energy consumption in grinding For high strength and hardness materials such as advanced ceramics, specific grinding energy requirement is extremely high which will eventuallyDevelopment of specific grinding energy model zthe calculation of the specific grinding energy w and zthe determination of the mill power draw P (as a function of the Bond work index w i) Use of the above models: For ballmill scaleup purposes 9 Model giving the specific grinding energy w zApplying multiple linear regression analysis to sets of data (w, D f and d) taken from the Denver slide rule, the equation derived is of the generalMODELING THE SPECIFIC GRINDING ENERGY AND BALLMILL
• ## A novel energy partition model for belt grinding of
Analyzing and modelling energy partition is of great significance to provide an indepth understanding of the grinding process that is essentially dynGrinding forces and energy play an important role in all abrasive machining operations While specific grinding energy may be obtained from workpiece dynamometer values or by measuring spindle power, care must be exercised in converting dynamometer reading into power consumed This is particularly true for operations involving a large ratio of wheel depth of cut to wheel diameter or when theGrinding Forces and Energy | J Manuf Sci Eng | ASMEThe grinding energy requirements depend on kinematical and geometrical parameters of the grinding machine and physical properties of the ground material Knowledge of the grinding properties of grain is essential to adjust the correct parameters of grinding and sieving machines It is the best way to produce higher and betterquality product yields at minimum energy requirements From amongGrinding Parameters and their Effects on the Quality of
• ## Energy Use of Fine Grinding in Mineral Processing
18/12/2013· This suggests that large savings in grinding energy (and associated savings in maintenance, consumables, and capital equipment needed) could be obtained by improving grinding operations Table I Energy Consumption in Total, in the Mining Industry, and in Grinding[1,2] Full size table As fine grinding is typically used on regrind applications, the feed tonnages to fine grindingGrinding – Ex 11 • You are grinding a steel, which has a specific grinding energy (u) of 35 Ws/mm3 • The grinding wheel rotates at 3600 rpm, has a diameter (D) of 150 mm, thickness (b) of 25 mm, and (c) 5 grains per mm2 The motor has a power of 2 kW • The work piece moves (v) at 15 m/min The chip thickness ratio (r) is 10Grinding and Finishing IIT Bombay16/11/2017· Grinding force can be expressed by specific grinding energy, which shows how much energy is consumed per unit volume of material removed: u = P ′ Q ′ = F t ′ ⋅ v s a ⋅ v w = F t ′ h m E4 22 Conditions during the experiment 221 Cylindrical grinding Workpieces were cylindrical shaped Ø 60 × 150 mm and were made from two types of steel: Steel EN 34Cr4 with mechanicalGrinding Force of Cylindrical and CreepFeed Grinding
• ## TECHNICAL NOTES 8 GRINDING R P King
consume energy wastefully but more importantly it will reduce the ab ility of the mill shell to transmit energy to the tumbling charge This energy is required to cause grinding of the material in the mill The shape and dimensions of the lifters control the tumbling action of the media The tumbling action is difficult to describe accurately but certain regions in the mill can beEnergy Calculation Grinding Energy calculation model of ball kinematics based on ball mill ijicic load detection on balls kinetic energy a novel energy calculation model for ball kinemat ics based on coal load of friction between the ball and the mill the ball and the coal contributes to the grinding capacity Jaw Crusher Impact Crusher Cone Crushers Sand Maker Ball Mill Raymond MillEnergy Calculation Grinding modedesignberlindegrinding energy compared with shallow grinding and creep grinding The chips take away most of the heat generated in the grinding process As in creep grinding, bumout of the coolant causes a steep rise in contact temperature of the workpiece 1 Introduction Cost, quality and productivity in grinding processes are all dependent on energy consumption and process efficiency In an inefficientEnergy and temperature analysis in grinding
• ## THE USE OF ENERGY INDEX THE F BOND INDEX
26/03/2020· We continue to consider the methods of grinding process planning based on the calculations of different energy indexes In this article, we are going to consider the methodology for calculating index, which has a direct impact on the choice of SAG mills when engineering the grinding process ВWi is the ball grinding work index It is used to calculate the energy input (kW*h/t), that isThe energy used in grinding and regrinding was calculated using the bond work index formula and compared with the energy consumed when the total massBreaking down energy consumption in industrial16/11/2017· Grinding force can be expressed by specific grinding energy, which shows how much energy is consumed per unit volume of material removed: u = P ′ Q ′ = F t ′ ⋅ v s a ⋅ v w = F t ′ h m E4 22 Conditions during the experiment 221 Cylindrical grinding Workpieces were cylindrical shaped Ø 60 × 150 mm and were made from two types of steel: Steel EN 34Cr4 with mechanicalGrinding Force of Cylindrical and CreepFeed Grinding
• ## energy calculation formula principle for ball mill
Calculation of energy required for grinding in a ball mill The grindingproduct size, P, in a Bond ball mill, which is given by the aperture size which passes 80% of the grinding product as a function of the aperture size of the test screen Pk, can be expressed by the formula P= P k K 2 Calculation of energy required for grinding in a ball mill Jan 01, 1989 INTRODUCTION The energy07/06/2020· Grinding efficiency (E) is the grinding ratio (G) divided by the specific energy (U) and is expressed by in 3 /hp/min In this equation, G is the grinding ratio, which is defined as the volume of material removed per unit volume of wheel wearHow is grinding wheel speed calculated?12/04/2015· The cutting force constitutes about 7080 % of the total force F and is used to calculate the power “P” required to perform the machining operation, P = VFC •2Thrust force Ft : this force is in direction of feed motion in orthogonal cutting The thrust force is used to calculate the power of feed motion CUTTING FORCES & POWER YCCE,NAGPUR 4 CUTTING FORCES (2D OrthogonalCutting power & Energy Consideration in metal cutting
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# Find cardinality of $X = \left\{ A : A \subset \mathbb R \wedge \text{c}(A) \right\}$
I have some doubts with this task:
Find cardinality of a) $$X = \left\{ A : A \subset \mathbb R \wedge \text{c}(A) \right\}$$
b) $$X = \left\{ A : A \subset \mathbb Q \wedge \text{c}(A) \right\}$$
where $$\text{c}(A)$$ means that set contains maximum and minimum element
I think that the result is $$\mathfrak{c}$$, so I have decided to show two injectives:
$$f:\mathbb R \rightarrow X$$ and $$g:X \rightarrow \mathbb R$$ If it comes to $$f$$ it may be $$f = \lambda x.\left\{x \right\}$$ and that set contains maximum and minimum element so I think that it is good example (both in a) and b) ).
But I am trying to show example of function $$g$$ and I have stuck there. One idea was to take $$g = \lambda A. \frac{1}{2}(\min+\max)$$ but it is not injective :( thanks for your time
For (a) let $$H:\mathbb R\to(0,1)$$ be bijective.
Then define $$f:2^{\mathbb R}→ X$$: $$f(A)=H[A]∪\{0,1\}$$, because $$H$$ is bijective $$f$$ is injective so $$2^{\frak{c}}≤|X|$$, and $$X⊆ 2^{\Bbb R}$$ so $$|X|≤2^\frak c$$.
For (b) let $$H:\mathbb Q\to \mathbb Q\cap (0,1)$$ be bijective.
Then define $$f:2^{\mathbb Q}→ X$$: $$f(A)=H[A]∪\{0,1\}$$, because $$H$$ is bijective $$f$$ is injective so $${\frak c}≤|X|$$, and $$X⊆ 2^{\Bbb Q}$$ so $$|X|≤\frak c$$.
For a), here's an injective map $$\mathfrak P(\Bbb R)\to X$$: $$S\mapsto \{\,\tfrac x{1+|x|}\mid x\in S\}\cup \{-2,2\}$$ showing that $$|X|\ge 2^{\mathfrak c}$$ (and of course this means equality).
• Why $|X|\ge 2^{\mathfrak c}$ ? – VirtualUser Jan 18 at 22:02
Let $$J= (-\pi/2,\pi/2)\}$$ and let $$Y$$ be the set of all subsets of $$J.$$
Now $$|J|=|\Bbb R|=c.$$ E.g. $$f(x)=\tan x$$ is a bijection from $$J$$ to $$\Bbb R.$$ So $$|Y|=\{t:t\subset \Bbb R\}|=2^c>c.$$
For $$s\in Y$$ let $$G(s)=s\cup\{-\pi/2,\pi/2\}.$$ Since $$G$$ is one-to-one and since $$\{G(s):s\in Y\}\subset X,$$ we have $$2^c=|Y|=|\{G(s):s\in Y\}|\le |X|\le 2^c.$$ So $$|X|=2^c.$$
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# Impose flow in a cell zone
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March 31, 2016, 08:43 Impose flow in a cell zone #1 New Member HAOWU Join Date: Feb 2016 Location: Trondheim Posts: 15 Rep Power: 3 Hello, everyone, I am trying to impose a cross flow velocity within a small volume. But I am not sure how to implement. My idea is to add fixed values or source terms. Is it right? ??.JPG After setting a velocity in the cell zone, how is the boundary condition then? Still inlet and outlet? Regards, Hao
April 1, 2016, 01:35 #2 Senior Member Lucky Tran Join Date: Apr 2011 Location: Orlando, FL USA Posts: 1,968 Rep Power: 26 I think you should separate your region into a separate cell zone. And then apply the fixed values to only that cell zone. It should talk to the rest of the domain with interfaces and this way you can apply regular boundary conditions to the rest of the domain.
April 1, 2016, 08:35
#3
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Quote:
Originally Posted by LuckyTran I think you should separate your region into a separate cell zone. And then apply the fixed values to only that cell zone. It should talk to the rest of the domain with interfaces and this way you can apply regular boundary conditions to the rest of the domain.
Yes, I will add the velocity on a sub domain, but I dont think 'fixed values' work, because I have to consider the interaction between cross flow velocity and in line velocity.
??.JPG
The dashed line is the cell zone
April 1, 2016, 09:44 #4 Senior Member Lucky Tran Join Date: Apr 2011 Location: Orlando, FL USA Posts: 1,968 Rep Power: 26 Sorry but when you say impose, you mean to force. So it seems like you are not actually trying to impose the velocity but to study the interaction of two flows. Then this is an entirely different problem and not related to fixed values and subdomains at all. This is way more complicated than it needs to be. Why can't you do the obvious and just specify the cross-flow velocity at the boundary where it enters as a velocity inlet?
April 1, 2016, 12:42
#5
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Quote:
Originally Posted by LuckyTran Sorry but when you say impose, you mean to force. So it seems like you are not actually trying to impose the velocity but to study the interaction of two flows. Then this is an entirely different problem and not related to fixed values and subdomains at all. This is way more complicated than it needs to be. Why can't you do the obvious and just specify the cross-flow velocity at the boundary where it enters as a velocity inlet?
Sorry for my poor english. 'impose' may be not the right word here. The continuous cross flow here represents fish school, so it could have fixed direction but variable magnitude.
April 1, 2016, 12:47 #6 Senior Member Lucky Tran Join Date: Apr 2011 Location: Orlando, FL USA Posts: 1,968 Rep Power: 26 If you want the velocity to be purely in vertical direction, then you can fix the horizontal velocity to be 0. This way the direction is fixed but the magnitude is variable. If you want the direction to be a combination of x,y,z velocities, that's much harder to do. But note that as soon as you add fixed values, you eliminate the interaction between streams (or severely limit them). You can't have interaction of two streams with different directions with velocity direction fixed for one stream because that would violate momentum conservation. Or even if you did, it would require/result in fictitious forces.
April 1, 2016, 13:02
#7
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Quote:
Originally Posted by LuckyTran If you want the velocity to be purely in vertical direction, then you can fix the horizontal velocity to be 0. This way the direction is fixed but the magnitude is variable. If you want the direction to be a combination of x,y,z velocities, that's much harder to do. But note that as soon as you add fixed values, you eliminate the interaction between streams (or severely limit them). You can't have interaction of two streams with different directions with velocity direction fixed for one stream because that would violate momentum conservation. Or even if you did, it would require/result in fictitious forces.
Yes, Fixed value is not the right choice here because I need to solve governing equation in all the domain.
How about adding source of momentum? If I set momentum source along the cross flow?
April 1, 2016, 13:39
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Originally Posted by haow Yes, Fixed value is not the right choice here because I need to solve governing equation in all the domain. How about adding source of momentum? If I set momentum source along the cross flow?
Fixed values can be the appropriate choice but it depends on what I mentioned earlier about the velocity direction.
I can't answer that because it's too vague what you are trying to implement. You want to add a momentum source, but what do you want to achieve with it?
Can you please just clearly and state what you are trying to model (but be concise), the boundary conditions, and constraints you are trying to impose? It can be pseudo-code and doesn't have to be 100% mathematically.
April 1, 2016, 13:58
#9
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Quote:
Originally Posted by LuckyTran Fixed values can be the appropriate choice but it depends on what I mentioned earlier about the velocity direction. I can't answer that because it's too vague what you are trying to implement. You want to add a momentum source, but what do you want to achieve with it? Can you please just clearly and state what you are trying to model (but be concise), the boundary conditions, and constraints you are trying to impose? It can be pseudo-code and doesn't have to be 100% mathematically.
It is a group of fish moving from top to bottom (in the image). The flow from left side represents current flow in the ocean.
In the small cell zone:
1. use porous media to model the blockage of fish
2. add velocity in the cell zone to model the velocity of fish group
November 24, 2017, 12:16
#10
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Originally Posted by LuckyTran Fixed values can be the appropriate choice but it depends on what I mentioned earlier about the velocity direction.
Dear Lucky Tran,
I am trying to impose a gust in vertical direction thru y-momentum source udf as defined below.But the desired Y-velocity in the source region is not coming equal to 15 as defined in the udf. I think i made a mistake somewhere? can you please help if you find time.
# include "udf.h"
{
real source=0.0;
real XGS=-7.5; /* X-center of source*/
real YGS=0.0; /* Y-center of source*/
real ZGS=0.0; /* Z-center of source*/
real X= 1.0; /* X, Y, Z dimensions are such so as to make the source-Volume=1m^3*/
real Y= 1.0;
real Z= 1.0;
real t = CURRENT_TIME;
real centroid[3];
real x_loc=centroid[0];
real y_loc=centroid[1];
real z_loc=centroid[2];
if (t>1.0 && t<1.5)
{
if (fabs(x_loc-XGS)<X && fabs(y_loc-YGS)<Y && fabs(z_loc-ZGS)<Z)
{
source=rho*(15-C_V(cell,thread))/CURRENT_TIMESTEP; /*15(m/s) is perturbation in Y direction*/
dS[eqn]=0.0;
}
}
else
{
source=0.0;
dS[eqn]=0.0;
}
return source;
}
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Intuitionistic Logic Explorer < Previous Next > Nearby theorems Mirrors > Home > ILE Home > Th. List > ltrelxr GIF version
Theorem ltrelxr 7240
Description: 'Less than' is a relation on extended reals. (Contributed by Mario Carneiro, 28-Apr-2015.)
Assertion
Ref Expression
ltrelxr < ⊆ (ℝ* × ℝ*)
Proof of Theorem ltrelxr
Dummy variables 𝑥 𝑦 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 df-ltxr 7220 . 2 < = ({⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ ℝ ∧ 𝑦 ∈ ℝ ∧ 𝑥 < 𝑦)} ∪ (((ℝ ∪ {-∞}) × {+∞}) ∪ ({-∞} × ℝ)))
2 df-3an 922 . . . . . 6 ((𝑥 ∈ ℝ ∧ 𝑦 ∈ ℝ ∧ 𝑥 < 𝑦) ↔ ((𝑥 ∈ ℝ ∧ 𝑦 ∈ ℝ) ∧ 𝑥 < 𝑦))
32opabbii 3853 . . . . 5 {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ ℝ ∧ 𝑦 ∈ ℝ ∧ 𝑥 < 𝑦)} = {⟨𝑥, 𝑦⟩ ∣ ((𝑥 ∈ ℝ ∧ 𝑦 ∈ ℝ) ∧ 𝑥 < 𝑦)}
4 opabssxp 4440 . . . . 5 {⟨𝑥, 𝑦⟩ ∣ ((𝑥 ∈ ℝ ∧ 𝑦 ∈ ℝ) ∧ 𝑥 < 𝑦)} ⊆ (ℝ × ℝ)
53, 4eqsstri 3030 . . . 4 {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ ℝ ∧ 𝑦 ∈ ℝ ∧ 𝑥 < 𝑦)} ⊆ (ℝ × ℝ)
6 rexpssxrxp 7225 . . . 4 (ℝ × ℝ) ⊆ (ℝ* × ℝ*)
75, 6sstri 3009 . . 3 {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ ℝ ∧ 𝑦 ∈ ℝ ∧ 𝑥 < 𝑦)} ⊆ (ℝ* × ℝ*)
8 ressxr 7224 . . . . . 6 ℝ ⊆ ℝ*
9 snsspr2 3542 . . . . . . 7 {-∞} ⊆ {+∞, -∞}
10 ssun2 3137 . . . . . . . 8 {+∞, -∞} ⊆ (ℝ ∪ {+∞, -∞})
11 df-xr 7219 . . . . . . . 8 * = (ℝ ∪ {+∞, -∞})
1210, 11sseqtr4i 3033 . . . . . . 7 {+∞, -∞} ⊆ ℝ*
139, 12sstri 3009 . . . . . 6 {-∞} ⊆ ℝ*
148, 13unssi 3148 . . . . 5 (ℝ ∪ {-∞}) ⊆ ℝ*
15 snsspr1 3541 . . . . . 6 {+∞} ⊆ {+∞, -∞}
1615, 12sstri 3009 . . . . 5 {+∞} ⊆ ℝ*
17 xpss12 4473 . . . . 5 (((ℝ ∪ {-∞}) ⊆ ℝ* ∧ {+∞} ⊆ ℝ*) → ((ℝ ∪ {-∞}) × {+∞}) ⊆ (ℝ* × ℝ*))
1814, 16, 17mp2an 417 . . . 4 ((ℝ ∪ {-∞}) × {+∞}) ⊆ (ℝ* × ℝ*)
19 xpss12 4473 . . . . 5 (({-∞} ⊆ ℝ* ∧ ℝ ⊆ ℝ*) → ({-∞} × ℝ) ⊆ (ℝ* × ℝ*))
2013, 8, 19mp2an 417 . . . 4 ({-∞} × ℝ) ⊆ (ℝ* × ℝ*)
2118, 20unssi 3148 . . 3 (((ℝ ∪ {-∞}) × {+∞}) ∪ ({-∞} × ℝ)) ⊆ (ℝ* × ℝ*)
227, 21unssi 3148 . 2 ({⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ ℝ ∧ 𝑦 ∈ ℝ ∧ 𝑥 < 𝑦)} ∪ (((ℝ ∪ {-∞}) × {+∞}) ∪ ({-∞} × ℝ))) ⊆ (ℝ* × ℝ*)
231, 22eqsstri 3030 1 < ⊆ (ℝ* × ℝ*)
Colors of variables: wff set class Syntax hints: ∧ wa 102 ∧ w3a 920 ∈ wcel 1434 ∪ cun 2972 ⊆ wss 2974 {csn 3406 {cpr 3407 class class class wbr 3793 {copab 3846 × cxp 4369 ℝcr 7042 <ℝ cltrr 7047 +∞cpnf 7212 -∞cmnf 7213 ℝ*cxr 7214 < clt 7215 This theorem was proved from axioms: ax-1 5 ax-2 6 ax-mp 7 ax-ia1 104 ax-ia2 105 ax-ia3 106 ax-io 663 ax-5 1377 ax-7 1378 ax-gen 1379 ax-ie1 1423 ax-ie2 1424 ax-8 1436 ax-10 1437 ax-11 1438 ax-i12 1439 ax-bndl 1440 ax-4 1441 ax-17 1460 ax-i9 1464 ax-ial 1468 ax-i5r 1469 ax-ext 2064 This theorem depends on definitions: df-bi 115 df-3an 922 df-tru 1288 df-nf 1391 df-sb 1687 df-clab 2069 df-cleq 2075 df-clel 2078 df-nfc 2209 df-v 2604 df-un 2978 df-in 2980 df-ss 2987 df-pr 3413 df-opab 3848 df-xp 4377 df-xr 7219 df-ltxr 7220 This theorem is referenced by: ltrel 7241
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Home | Math Dept
Here you will find information concerning the Business Algebra (1090-4) class for the Fall 2003 Semester, which meets on Monday, Wednesday, and Friday 9:40AM-10:30AM in LS 102. If you have any doubts, problems, concerns, etc., feel free to send a message to the instructor.
Final Exam: the results will be available on Thursday 12/11. I will be in my office from 9 to 11 AM. Otherwise, you can contact me by email to make an appointment to pickup your test. Notice that I will be out of town from 12/19 to 1/2. Two More Points! Everybody got two extra points on Test 3 in connection with the "closed + bounded" issue in problem (3b). So, for example, if your test grade was 57/74 it is now 59/74. But, please, if a problem like this arises on the final exam don't forget to check (and say!) if the region is closed and bounded. Problem Sessions! Beginning Tuesday 10/7, every Tuesday 5-6PM we will meet in LCB 215 to discuss problems or answer any other questions that you may have. This is not mandatory and should not be considered part of the course: the only goal is to provide extra time for those who want to practice. More information will be posted here as it becomes available.
12/3: Completed the discussion of annuities and amortization (Sections 6.4 and 6.5). Last class, but there is a review session on Friday. 12/1: Covered most of the sections on annuities (Sections 6.3 and 6.4). 11/26: Test 3 was given. 11/24: Completed the discussion of compounded interest (Section 6.2). 11/21: Arithmetic and geometric sequences and series. Started the compound interest discussion (Sections 6.1 and 6.2). 11/19: Quiz 11 was given. Simple interest and sequences (Section 6.1). 11/17: Applications of logarithmic and exponential functions (Section 5.3). 11/14: Discussed logarithmic functions (Section 5.2). 11/12: Quiz 10 was given. Completed the discussion of exponential functions (Section 5.1). 11/10: After answering some questions on linear programming problems, we started the discussion of exponential functions (Section 5.1). 11/7: Completed the discussion linear programming problems (Section 4.3). 11/5: Quiz 9 was given. Continued with linear programming (Section 4.3). Also, Test 3 was announced for Wednesday 11/26. 11/3: Completed the discussion of inequalities in two variables and started the section on linear programming (Sections 4.2 and 4.3). 10/31: Inequalities in two variables (Section 4.2). 10/29: Test 2 was given. 10/27: After answering a few questions we talked about linear inequalities in one variable (Section 4.1). 10/24: Completed the discussion of inverse matrices, the last topic for the test (Section 3.4). 10/22: Quiz 8 was given. Started the discussion of inverse matrices (Section 3.4). 10/20: Completed the discussion of the Gauss-Jordan elimination method (Section 3.3). 10/17: Gauss-Jordan elimination (Section 3.3). 10/15: Quiz 7 was given. Matrix multiplication (Section 3.2). 10/13: Matrices: multiplication (Section 3.2). 10/10: Matrices: definition and basic operations (Section 3.1). Also, Test 2 was announced for Wednesday 10/29. 10/8: Quiz 6 was given. More special functions: rational and piecewise defined functions (Section 2.4). 10/6: Polynomial and power functions (Section 2.4). 10/1: Quiz 5 was given. More applications of quadratic functions (Section 2.3). 9/29: Quadratic functions and their applications (Sections 2.2 and 2.3). 9/26: Quadratic functions (Section 2.2). 9/24: Test 1 was given. 9/22: Reviewed a problem on supply/demand and discussed quadratic equations (Sections 1.6 and 2.1). 9/19: Reviewed some problems about cost analysis and supply/demand functions (Section 1.6). Also, a review class was scheduled for Tuesday 9/23 5-6PM in LCB 215. 9/17: Quiz 4 was given. Solution of systems of three equations with three unknowns. Applications: cost, revenue and profit functions (Section 1.6). 9/15: Systems of linear equations: substitution and elimination (Section 1.5). 9/12: Equations of a line. Systems of linear equations: solving graphically (Sections 1.3 and 1.5). 9/10: Quiz 3 was given. Linear functions: slope, perpendicularity. (Section 1.3). Test 1 was announced for 9/24. 9/8: After practicing evacuation procedures, we continued with linear functions. (Section 1.3). 9/5: Functions and linear functions. (Sections 1.2 and 1.3). 9/3: Quiz 2 was given. Notion of function (definition by formulas, ordered pairs, tables and graphs). (Section 1.2). 8/29: Solution of linear equations. (Section 1.1). 8/27: Quiz 1 was given. Review integer powers and introduction to the solution of linear equations. (Sections 0.3 and 1.1). 8/25: Review of real numbers, operations, etc. (Section 0.2). 8/22: First day of class. Discussion of the course (see details on the syllabus. Review of sets (Section 0.1). 8/20: No class this day.
Rules:The following homework is assigned but will neither be collected nor graded. Chapter 6: section 1: 1, 3, 7, 9, 11, 13, 19, 23, 25, 29, 33, 39, 41; section 2: 1, 5, 7, 9, 13, 17, 21, 23, 25, 29, 31, 35, 37, 45; section 3: 5, 7, 9, 13, 21, 23, 29; section 4: 1, 5, 7, 9, 15, 21, 25, 33, 37; section 5: 3, 7. Chapter 5: section 1: 1, 3, 5, 9, 15, 17, 21, 31, 33, 37; section 2: 1, 5, 7, 9, 13, 21, 23, 25, 27, 33, 35, 37, 41; section 3: 3, 7, 13, 15, 17, 19, 21, 23, 27, 29. Chapter 4: section 1: 1, 3, 5, 9, 11, 15, 17, 23, 27, 29, 41, 49; section 2: 1, 3, 5, 9, 11, 13, 17, 21, 25, 31, 37, 39; section 3: 3, 5, 7, 9, 11, 13, 15, 19, 23, 27, 33, 37, 41, 45. Chapter 3: section 1: 1, 2, 3, 5, 7, 9, 11, 17, 19, 21, 27, 33, 37, 41; section 2: 1, 3, 5, 7, 9, 11, 17, 23, 37, 43, 55, 60; section 3: 3, 5, 7, 13, 17, 19, 25, 27, 29, 33, 35, 39; section 4: 1, 3, 5, 7, 9, 13, 15, 17, 29, 31, 37, 43, 55, 59, 63. Chapter 2: section 1: 7, 11, 25, 28, 35, 49, 59; section 2: 1, 2, 3, 5, 7, 9, 12, 14, 15, 17, 35, 37, 45; section 3: 1, 5, 8, 11, 15, 19, 20, 25, 27, 29; section 4: 1, 3, 7, 11, 14, 29, 31, 33, 37, 39, 49, 51. Chapter 1: section 1: 1, 5, 10, 13, 17, 23, 31, 33, 39, 47, 61; section 2: 1, 3, 8, 9, 13, 18, 24, 25, 27, 30, 36, 37, 63; section 3: 1, 2, 3, 7, 8, 13, 14, 15, 17, 21, 23, 27, 33, 37, 39, 44, 47, 55, 63; section 5: 1, 3, 7, 8, 9, 12, 17, 29, 33, 37; section 6: 1, 4, 5, 9, 13, 15, 20, 24, 29, 31, 40, 43, 47. Chapter 0: section 1: 2, 4, 8, 20, 30, 34, 52; section 2: 1, 10, 15, 20, 25, 27, 30, 43, 46, 49; section 3: 3, 4, 12, 15, 27, 33, 45, 62.
Below you will find different references on the course (syllabus, etc.).
The syllabus (also in PDF and PS formats). Here you will find the "ultimate" guidelines for the course, textbook, schedule, final exam, etc. Quizzes: quiz 11 (solution), quiz 10 (solution), quiz 9 (solution), quiz 8 (solution), quiz 7 (solution), quiz 6 (solution), quiz 5 (solution), quiz 4 (solution), quiz 3 (solution), quiz 2 (solution), quiz 1 (solution).
Official academic calendar, Fall 2003. Student Handbook. General Catalog. Campus map and building locator. University of Utah homepage.
Home | Math Dept
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# What is the maclaurin series of ln(1-3x)?
Jul 29, 2015
Maclaurin series of $\ln \left(1 - 3 x\right)$ for the first four terms is
$0 - 3 x - \frac{9 {x}^{2}}{2} - 9 {x}^{3} + O \left(4\right) = - 3 x - \frac{9 {x}^{2}}{2} - 9 {x}^{3} + O \left(4\right)$
#### Explanation:
A Maclaurin series is just a Taylor series expansion of a function about 0:
f(x) = f(0) + f'(0)x + (f''(0)x^2)/(2!) + (f^3(0)x^3)/(3!) + ... + (f^n(0)x^n)/(n!)
In our case,
$f \left(x\right) = \ln \left(1 - 3 x\right)$ and $f \left(0\right) = 0$
Differentiate using the chain rule to get
$f ' \left(x\right) = - \frac{3}{1 - 3 x}$ and $f ' \left(0\right) = - 3$
Differentiate the first derivative using the quotient rule to get
$f ' ' \left(x\right) = - \frac{9}{1 - 3 x} ^ 2$ and $f ' ' \left(0\right) = - 9$
Differentiate the second derivative using a combination of the quotient rule and the chain rule for the derivative of the denominator to get
${f}^{3} \left(x\right) = \frac{- 54}{1 - 3 x} ^ 3$ and ${f}^{3} \left(0\right) = - 54$
Plug these values into the definition of the Maclaurin series and you'll get the answer for the first four terms:
$0 - 3 x - \frac{9 {x}^{2}}{2} - 9 {x}^{3} + O \left(4\right) = - 3 x - \frac{9 {x}^{2}}{2} - 9 {x}^{3} + O \left(4\right)$
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Java - The first 10 Catalan numbers from the Pascal triangle
# Java: Print out the first 10 Catalan numbers by extracting them from Pascal's triangle
## Java Numbers: Exercise-8 with Solution
Write a Java program to print out the first 10 Catalan numbers by extracting them from Pascal's triangle.
In combinatorial mathematics, the Catalan numbers form a sequence of natural numbers that occur in various counting problems, often involving recursively-defined objects. They are named after the Belgian mathematician Eugène Charles Catalan.
The first Catalan numbers for n = 0, 1, 2, 3, … are 1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845, 35357670, 129644790, 477638700, 1767263190, 6564120420, 24466267020, 91482563640, 343059613650, 1289904147324, 4861946401452,
Sample Solution:
Java Code:
``````import java.util.Scanner;
public class Example8 {
public static void main(String[] args) {
int num = 10;
int[] t = new int[num + 2];
t[1] = 1;
System.out.printf("\nList 10 Catalan numbers:-\n");
for (int i = 1; i <= num; i++) {
for (int j = i; j > 1; j--)
t[j] = t[j] + t[j - 1];
t[i + 1] = t[i];
for (int j = i + 1; j > 1; j--)
t[j] = t[j] + t[j - 1];
System.out.printf("\n%d ", t[i + 1] - t[i]);
}
System.out.printf("\n");
}
}
```
```
Sample Output:
```List 10 Catalan numbers:-
1
2
5
14
42
132
429
1430
4862
16796
```
Flowchart:
Java Code Editor:
What is the difficulty level of this exercise?
Test your Programming skills with w3resource's quiz.
## Java: Tips of the Day
IsPowerOfTwo
Checks if a value is positive power of two.
To understand how it works let's assume we made a call IsPowerOfTwo(4).
As value is greater than 0, so right side of the && operator will be evaluated.
The result of (~value + 1) is equal to value itself. ~100 + 001 => 011 + 001 => 100. This is equal to value.
The result of (value & value) is value. 100 & 100 => 100.
This will value the expression to true as value is equal to value.
```public static boolean isPowerOfTwo(final int value) {
return value > 0 && ((value & (~value + 1)) == value);
}
```
Ref: https://bit.ly/3sA5d4I
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My Math Forum Logs
Algebra Pre-Algebra and Basic Algebra Math Forum
July 8th, 2017, 01:24 PM #1 Newbie Joined: Jul 2017 From: United Kingdom Posts: 4 Thanks: 0 Math Focus: General Logs I have a little problem with Logs. (log(3x)+log(9x^2))/3 I worked it out to be Log(27x) but it can be simplified further down to Log(3x) Can anybody help me figure out how? Thanks!
July 8th, 2017, 01:53 PM #2 Newbie Joined: May 2017 From: California Posts: 15 Thanks: 1 Try to use log(3x) + log(9x^2) = log(3x*(9x^2)) then use log(x) * a = log(x^a).. so log(x) / a = log(x^(1/a)) Thanks from JayWalk
July 8th, 2017, 02:29 PM #3
Senior Member
Joined: May 2016
From: USA
Posts: 1,029
Thanks: 420
Quote:
Originally Posted by JayWalk I have a little problem with Logs. (log(3x)+log(9x^2))/3 I worked it out to be Log(27x) but it can be simplified further down to Log(3x) Can anybody help me figure out how? Thanks!
$\dfrac{\log(3x) + \log(9x^2)}{3} = \dfrac{\log(3x) + \log(\{3x\}^2)}{3} =$
$\dfrac{\log(3x) + 2\log(3x)}{3} = \dfrac{3\log(3x)}{3} = \log(3x).$
EDIT: This is just a different way from in the previous post.
$\dfrac{\log(3x) + \log(9x^2)}{3} = \dfrac{\log(3x * 9x^2)}{3} = \dfrac{\log(27x^3)}{3}.$
Now you can go
$\dfrac{\log(27x^3)}{3} = \dfrac{\log(\{3x\}^3)}{3} =$
$\dfrac{3\log(3x)}{3} = \log(3x).$
Or you can go:
$\dfrac{\log(27x^3)}{3} = \dfrac{1}{3} * \log(27x^3) =$
$\log(\{27x^3\}^{(1/3)}) = \log \left ( \sqrt[3]{27x^3} \right ) = \log(3x).$
Lots of ways to proceed.
Last edited by skipjack; July 8th, 2017 at 09:14 PM.
July 8th, 2017, 05:23 PM #4 Newbie Joined: Jul 2017 From: United Kingdom Posts: 4 Thanks: 0 Math Focus: General I can't believe I didn't see it at first. Thanks!
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Question
# A cyclist riding the bicycle at a speed of 14√3 m/s takes a turn around a circular road of radius 20√3 m without skidding. What is his inclination to the vertical? [Take g=9.8 m/s2]
A
30
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Solution
## The correct option is D 60∘ Let the cyclist bend w.r.t the vertical plane by an angle θ, then To avoid skidding, v2rg≥tanθ For the given value of v and r, θ=tan−1(v2rg) =tan−1((14√3)220√3×9.8) =tan−1(√3) =60∘
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# P3/4 Snow Day activities
Snow Day Activities:
Build a snowman that looks like you.
Create a weather scene using the natural elements in your garden and/or park. The snow is your canvas!
Write your spelling words in the snow, you could use food colouring to colour the snow or your hands.
Use the internet or books available to research extreme weathers that include snow. https://www.kiddle.co/ is a great search engine for kids.
Use snowballs to do some maths:
– Triangles/Rectangles/Squares: Equal/arrays groups eg. 4 groups of 5 what does that look like and what’s the answer?
– Circles: addition and subtraction eg. If I make 3 snowballs and through 2 how many do I have left?
Warm up with a hot chocolate! Write down instructions for how you made your tasty hot chocolate.
If you complete any activities or do your own learning in the snow I would love to see it! Send me a picture on this post below 😊
Enjoy the snow everyone! Be safe and remember to keep warm when you’re playing outside.❄️⛄️
# P1/2 snow day activities
1. Build the biggest snowman you can and measure its height.
2. Write your spelling words in the snow.
3. Write a story titled ‘One Snowy Day’. Make it as imaginative as you can!
4. Draw/paint a snowy picture.
5. Use the internet or any books you have at home to research 3 interesting facts about a polar animal.
6. Use snowballs to do some maths:
P2: equal groups/arrays eg. 3 groups of 5
P1: subtraction eg. Make 6 snowballs. Throw 2. How many left?
7. Warm up after some fun in the snow with a warm drink and a story (read the next section of your class reading book and/or a book of your choice).
If you complete any activities or do your own learning in the snow I would love to see your pictures! Add any pictures to the comments below to share what you have been learning with others.
Whatever you choose do – have fun and stay safe!! ❄️☃️
# School Closures
All schools in West Lothian are closed today, Wednesday 28th February. Learning activities can be accessed via this blog and Dojo, Further updates to follow.
# Nursery Sway
Please keep upto date with events in the nursery by following our Sway:
Westfield Nursery: A Place to Smile We have been learning about Chinese New Year by exploring stories, images, music and customs from China.
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# Finding volume and surface area of right prism
• Aug 15th 2010, 02:49 AM
haftakhan
Finding volume and surface area of right prism
How to find the volume and surface area of right prisms in the attachments ?Attachment 18585Attachment 18584Attachment 18583
• Aug 15th 2010, 02:55 AM
Educated
$\mbox{Volume} = \mbox{Base} * \mbox{Height} * \mbox{Depth}$
$\mbox{Surface area (Of each face)} = \mbox{Base} * \mbox{Height}$
Divide the prisms into many simpler shapes (eg. rectangular cubes for volume, or rectangles/squares for surface area) so it is easier to work with and find the surface area and volume seperately. Afterwards add them together to get the final surface area or volume of each prism.
For the wedge, use:
$\mbox{Volume} = \dfrac{\mbox{Base} * \mbox{Height} * \mbox{Depth}}{2}$ for the volume
$\mbox{Surface area (Of each triangular face)} = \dfrac{\mbox{Base} * \mbox{Height}}{2}$ for the surface area of the triangular faces
• Aug 15th 2010, 03:00 AM
Unknown008
I think I'll put more detail for the volume of a prism.
$Volume\ =\ Area\ of\ Cross\ Section\ \times\ Depth$
You need to first identify the cross section of the solid, the area which 'repeats' itself through the solid. In the first one, the cross section is a cross, in the second, it's this 'L' shape and in the last, it's a right angled triangle. Once you get their cross sectiona; area, you'll be able to find the volume by multiplying it by the depth of the solid.
• Aug 15th 2010, 03:03 AM
haftakhan
For attachment 1 Volume=12*16*20=3840 ?
and area=12*16=192
Is this the right answer ?
• Aug 15th 2010, 03:04 AM
haftakhan
Please solve the 1st one with ur details so that i can understand.
• Aug 15th 2010, 03:09 AM
Educated
First one:
$\mbox{Total Volume} = (4 * 5 * 3) + (4 * 5 * 3) + (14 * 3 * 4)$
$\mbox{Total Volume} = 288~ \mbox{units}^3$
$\mbox{Surface Area} = (5*4*(8)) + (4*4*(2)) + (4*3*(4)) + (5*3*(8))$ (The single numbers in brackets () are how many faces of each there are.)
$\mbox{Surface Area} = 360~\mbox{units}^2$
• Aug 15th 2010, 03:11 AM
Unknown008
The area of the cross can be obtained by dividing the cross into 5 parts; 4 rectangles and 1 central square.
Each rectangle has area 5*4 = 20.
Area of 4 rectangles = 20*4 = 80.
Area of central square = 4*4 = 16.
Total area of cross section = 16 + 80 = 96.
Volume of Prism = 96 x 3 = 288 units^3
• Aug 15th 2010, 03:16 AM
haftakhan
Very difficult question to understand.
• Aug 15th 2010, 03:23 AM
haftakhan
Unit^2 or unit^3 ?
• Aug 15th 2010, 03:32 AM
Unknown008
Oh, sorry, a typo from my part. It's units^3
http://i38.tinypic.com/17etf9.png
Here is a picture to help you understand. The rectangles are reddish and the central square is grey.
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# Understanding Models How models help algorithms models Implementation
• Slides: 26
Understanding Models
How models help algorithms models Implementation of models Real hardware
Modeling Communication System topology is a graph G = (V, E), where V = set of nodes (sequential processes) E = set of edges (links or channels, bi/unidirectional). Four types of actions by a process: - internal action - input action - communication action - output action
Example: A Message Passing Model A Reliable FIFO Channel P Axiom 1. Message m sent message m received Axiom 2. Message propagation delay is arbitrary but finite. Axiom 3. m 1 sent before m 2 m 1 received before m 2. Q
Life of a process When a message m is received A m B 1. The process evaluates a predicate with the message m and the local variables; 2. if predicate = true then update zero or more internal variables; send zero or more messages; end if D C E
Example: Shared memory model Address spaces of processes overlap M 1 M 2 Processes 1 2 3 4 Concurrent operations on a shared variable are serialized
Variations of shared memory models 0 1 2 3 0 1 3 2 State reading model Each process can read the states of its neighbors Link register model Each process can read from and write to adjacent registers. The entire local state is not shared.
Modeling wireless networks • Communication via broadcast • Limited range • Dynamic topology • Collision of broadcasts (handled by CSMA/CA) Request To Send RTS CTS Request Clear To To Send
Synchrony vs. Asynchrony Synchronous clocks Physical clocks are synchronized Synchronous processes Lock-step synchrony Synchronous channels Bounded delay Synchronous message-order First-in first-out channels Synchronous communication Communication via handshaking Send & receive can be blocking or nonblocking Postal communication is asynchronous: Telephone communication is synchronous Synchronous communication or not? (1) Remote Procedure Call, (2) Email Any constraint defines some form of synchrony …
Weak vs. Strong Models One object (or operation) of a strong Examples model = More than one simpler objects (or simpler operations) of a High level language is stronger weaker model. than assembly language. Often, weaker models are synonymous with fewer restrictions. Asynchronous is weaker than synchronous (communication). One can add layers (additional restrictions) to create a stronger model from weaker one. Bounded delay is stronger than unbounded delay (channel)
Model transformation Stronger models - simplify reasoning, but - needs extra work to implement Weaker models - are easier to implement. - Have a closer relationship with the real world “Can model X be implemented using model Y? ” is an interesting question in computer science. Sample exercises Non-FIFO to FIFO channel Message passing to shared memory Non-atomic broadcast to atomic broadcast
Non-FIFO to FIFO channel FIFO = First-In-First-Out m 2 m 3 P Sends out m 1, m 2, m 3, m 4, … m 4 m 1 Q 7 6 5 4 3 2 1 buffer
Non-FIFO to FIFO channel {Sender process P} var i : integer {initially 0} repeat send m[i], i to Q; i : = i+1 forever Needsunbounded buffer : = k+1; & unbounded sequence no THIS IS BAD {Receiver process Q} var k : integer {initially 0} buffer: buffer[0. . ∞] of msg {initially k: buffer [k] = empty repeat {STORE} receive m[i], i from P; store m[i] into buffer[i]; {DELIVER} while buffer[k] ≠ empty do begin deliver content of buffer [k]; buffer[k]: =empty� k end forever
Observations Now solve the same problem on a model where (a) The propagation delay has a known upper bound of T. (b) The messages are sent out @ r per unit time. (c) The messages are received at a rate faster than r. The buffer requirement drops to r. T. (Lesson) Stronger model helps. Question. Can we solve the problem using bounded buffer space if the propagation delay is arbitrarily large?
Example 1 second window sender First message Last message receiver
Message-passing to Shared memory {Read X by process i}: read x[i] {Write X: = v by process i} - x[i] : = v; - Atomically broadcast v to - every other process j (j ≠ i); After receiving broadcast, process j (j ≠ i) sets x[j] to v. Understand the significance of atomic operations. It is not trivial, but is This is incomplete very not correct. There important in distributed systems. pitfalls here. Atomic = all or nothing and still are more
Non-atomic to atomic broadcast Atomic broadcast = either everybody or nobody receives {process i is the sender} for j = 1 to N-1 (j ≠ i) send message m to neighbor [j] (Easy!) Now include crash failure as a part of our model. What if the sender crashes at the middle? How to implement atomic broadcast in presence of crash?
Mobile-agent based communication Communicates via messengers instead of (or in addition to) messages. Cedar Rapids What is the lowest Price of an i. Pad in Iowa? Best Buy University of Iowa Carries both program and data
Other classifications of models Reactive vs Transformational systems A reactive system never sleeps (like: a server) A transformational (or non-reactive systems) reaches a fixed point after which no further change occurs in the system (Examples? ) Named vs Anonymous systems In named systems, process id is a part of the algorithm. In anonymous systems, it is not so. All are equal. (-) Symmetry breaking is often a challenge. (+) Easy to switch one process by another with no side effect. Saves log N bits.
Knowledge based communication Alice and Bob enter into an agreement: whenever one falls sick, (s)he will call the other person. Since making the agreement, no one called the other person, so both concluded that they are in good health. Assume that the clocks are synchronized, communication links are perfect, and a telephone call requires zero time to reach. What kind of interprocess communication model is this?
History The paper “Cheating Husbands and Other Stories: A Case Study of Knowledge, Action, and Communication” by Yoram Moses, Danny Dolev, Joseph Halpern (PODC 1985) illustrates how actions are taken and decisions are made without explicit communication using common knowledge. (Adaptation of Gamow and Stern, “Forty unfaithful wives, ” Puzzle Math, 1958) (Bidding in the game of cards like bridge is an example of knowledge-based communication)
Observations Knowledge-based communication relies on making deductions from the absence of a signal. It is energy-efficient, something very relevant in today’s context.
Cheating Husband’s puzzle: The Queen read out the following in a meeting at the town square. • There are one or more unfaithful husbands in our community. • None of you know whether your husband is faithful. But each of you which of the other husbands are unfaithful. • Do not discuss this with anyone, but should you discover that your own husband is unfaithful, you should shoot him on the midnight of the day you find out about
What happened after this Thirty nine silent nights went by, and on the fortieth night, gunshots were heard. • What was going on for 39 nights? • How many unfaithful husbands were there? • Why did it take so long?
A simple case • W 2 does not know of any other unfaithful husband. • W 2 knows that there is at least one (common knowledge) • W 2 concludes that it must be H 2, and kills him on the first night. W 1 H 1 W 2 H 2 W 3 H 3 W 4 H 4
Theorem If there are N unfaithful H’s, then they will all be killed on the midnight of the Nth day. If you are interested to learn more, then read the original paper.
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# Class 7 Math NCERT Solutions for Chapter – 11 Perimeter and Area Ex – 11.1
## Perimeter and Area
1. The Length and the breadth of a rectangular piece of land are 500 m and 300 m, respectively. Find
(i) Its area (ii) the cost of the land, if 1 m2 of the land costs ₹ 10,000.
Solution:-
From the question it is given that,
Length of the rectangular piece of land = 500 m
Breadth of the rectangular piece of land = 300 m
Then,
(i) Area of rectangle = Length × Breadth
= 500 × 300
= 150000 m2
(ii) Cost of the land for 1 m2 = ₹ 10000
Cost of the land for 150000 m2 = 10000 × 150000
= ₹ 1500000000
2. Find the area of a square park whose perimeter is 320m.
Solution:-
From the question it is given that,
Perimeter of the square park = 320 m
4 × Length of the side of park = 320 m
Then,
Length of the side of park = 320/4
= 80 m
So, Area of the square park = (length of the side of park)2
= 802
= 6400 m2
3. Find the breadth of a rectangular plot of land, if its area is 440 m2 and the length is 22 m. Also find its perimeter.
Solution:-
From the question it is given that,
Area of the rectangular plot = 440 m2
Length of the rectangular plot = 22 m
We know that,
Area of the rectangle = Length × Breadth
Then,
Perimeter of the rectangle = 2(Length + Breadth)
= 2 (22 + 20)
= 2(42)
= 84 m
∴Perimeter of the rectangular plot is 84 m.
4. The perimeter of a rectangular sheet is 100 cm. If the length is 35 cm, find its breadth. Also find the area.
Solution:-
From the question it is given that,
Perimeter of the a rectangular sheet = 100 cm
Length of the rectangular sheet = 35 cm
We know that,
Perimeter of the rectangle = 2 (Length + Breadth)
100 = 2 (35 + Breadth)
Then,
Area of the rectangle = Length × Breadth
= 35 × 15
= 525 cm2
∴Area of the rectangular sheet is 525 cm2
5. The area of a square park is the same as of a rectangular park. If the side of the square park is 60 m and the length of the rectangular park is 90 m, find the breadth of the rectangular park.
Solution:-
From the question it is given that,
Area of a square park is the same as of a rectangular park.
Side of the square park = 60 m
Length of the rectangular park = 90 m
We know that,
Area of the square park = (one of the side of square)2
= 602
= 3600 m2
Area of the rectangular park = 3600 m2 … [∵ given]
6. A wire is in the shape of a rectangle. Its length is 40 cm and breadth is 22 cm. If the same wire is rebent in the shape of a square, what will be the measure of each side?
Also find which shape encloses more area?
Solution:-
By reading the question we can conclude that, perimeter of the square is the same as perimeter of rectangle.
From the question it is given that,
Length of the rectangle = 40 cm
Breadth of the square = 22 cm
Then,
Perimeter of the rectangle = Perimeter of the Square
2 (Length + Breadth) = 4 × side
2 (40 + 22) = 4 × side
2 (62) = 4 × side
124 = 4 × side
Side = 124/4
Side = 31 cm
So, Area of the rectangle = (Length × Breadth)
= 40 × 22
= 880 cm2
Area of square = side2
= 312
= 31 × 31
= 961 cm2
∴Square shaped wire encloses more area.
7. The perimeter of a rectangle is 130 cm. If the breadth of the rectangle is 30 cm, find its length. Also find the area of the rectangle.
Solution:-
From the question it is given that.
Perimeter of the rectangle = 130 cm
Breadth of the rectangle = 30
We know that,
Perimeter of rectangle = 2 (Length + Breadth)
130 = 2 (length + 30)
130/2 = length + 30
Length + 30 = 65
Length = 65 – 30
Length = 35 cm
Then,
Area of the rectangle = Length × Breadth
= 35 × 30
= 1050 cm2
8. A door of length 2 m and breadth 1 m is fitted in a wall. The length of the wall is 4.5 m and the breadth is 3.6 m (Fig). Find the cost of white washing the wall, if the rate of white washing the wall is ₹ 20 per m2.
Solution:-
From the question it is given that,
Length of the door = 2 m
Breadth of the door = 1 m
Length of the wall = 4.5 m
Breadth of the wall = 3.6 m
Then,
Area of the door = Length × Breadth
= 2 × 1
= 2 m2
Area of the wall = Length × Breadth
= 4.5 × 3.6
= 16.2 m2
So, Area to be white washed = 16.2 – 2 = 14.2 m2
Cost of white washing 1 m2 area = ₹ 20
Hence cost of whit washing 14.2 m2 area = 14.2 × 20
= ₹ 284
#### Class 7 Math NCERT Solutions for Chapter – 11 Perimeter and Area Ex – 11.2
Lorem ipsum dolor sit amet, consectetur adipiscing elit. Phasellus cursus rutrum est nec suscipit. Ut et ultrices nisi. Vivamus id nisl ligula. Nulla sed iaculis ipsum.
Company Name
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# Bigger is Greater HackerRank Solution in C, C++, Java, Python
Lexicographical order is often known as alphabetical order when dealing with strings. A string is greater than another string if it comes later in a lexicographically sorted list.
Given a word, create a new word by swapping some or all of its characters. This new word must meet two criteria:
• It must be greater than the original word
• It must be the smallest word that meets the first condition
For example, given the word w=abcd, the next largest word is .
Complete the function biggerIsGreater below to create and return the new string meeting the criteria. If it is not possible, return no answer.
Function Description
Complete the biggerIsGreater function in the editor below. It should return the smallest lexicographically higher string possible from the given string or no answer.
biggerIsGreater has the following parameter(s):
• w: a string
Input Format
The first line of input contains T, the number of test cases.
Each of the next T lines contains w.
Constraints
• 1<=T<=10^5
• 1<=|w|<=100
• w will contain only letters in the range ascii[a..z].
Output Format
For each test case, output the string meeting the criteria. If no answer exists, print no answer.
Sample Input 0
5
ab
bb
hefg
dhck
dkhc
Sample Output 0
ba
hegf
dhkc
hcdk
Explanation 0
• Test case 1:
ba is the only string which can be made by rearranging ab. It is greater.
• Test case 2:
It is not possible to rearrange bb and get a greater string.
• Test case 3:
hegf is the next string greater than hefg.
• Test case 4:
dhkc is the next string greater than dhck.
• Test case 5:
hcdk is the next string greater than dkhc.
Sample Input 1
6
lmno
dcba
dcbb
abdc
abcd
fedcbabcd
Sample Output 1
lmon
acbd
abdc
fedcbabdc
## Bigger is Greater HackerRank Solution in C
#include <stdio.h>
char s[500];
int main()
{
int t,j;
scanf("%d",&t);
while(t--)
{
scanf("%s",s);
int l=strlen(s);
int i=l-1;
int f[300]={0};
while(i>0&&s[i]<=s[i-1])
{
f[s[i]]++;
i--;
}
if(i)
{
f[s[i]]++;
f[s[i-1]]++;
for(j=0;j<i-1;j++)
printf("%c",s[j]);
for(j=s[i-1]+1;j<='z';j++)
if(f[j])
break;
f[j]--;
printf("%c",j);
for(i='a';i<='z';i++)
for(j=0;j<f[i];j++)
printf("%c",i);
printf("\n");
}
else
}
return 0;
}
## Bigger is Greater HackerRank Solution in C++
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
string s;
int main() {
int tc;
scanf("%d", &tc);
while (tc--) {
cin >> s;
if (next_permutation(s.begin(), s.end())) printf("%s\n", s.c_str());
}
return 0;
}
## Bigger is Greater HackerRank Solution in Java
import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.InputMismatchException;
public class Solution {
static InputStream is;
static PrintWriter out;
static String INPUT = "";
static void solve()
{
outer:
for(int T = ni();T >= 1;T--){
char[] s = ns().toCharArray();
int n = s.length;
int[] has = new int[26];
for(int i = n-1;i >= 0;i--){
has[s[i]-'a']++;
for(int j = s[i]-'a'+1;j < 26;j++){
if(has[j] > 0){
s[i] = (char)('a'+j);
has[j]--;
int p = 0;
for(int k = i+1;k < n;k++){
while(p < 26 && has[p] == 0)p++;
s[k] = (char)('a'+p);
has[p]--;
}
out.println(new String(s));
continue outer;
}
}
}
}
}
public static void main(String[] args) throws Exception
{
long S = System.currentTimeMillis();
is = INPUT.isEmpty() ? System.in : new ByteArrayInputStream(INPUT.getBytes());
out = new PrintWriter(System.out);
solve();
out.flush();
long G = System.currentTimeMillis();
tr(G-S+"ms");
}
private static boolean eof()
{
if(lenbuf == -1)return true;
int lptr = ptrbuf;
while(lptr < lenbuf)if(!isSpaceChar(inbuf[lptr++]))return false;
try {
is.mark(1000);
while(true){
if(b == -1){
is.reset();
return true;
}else if(!isSpaceChar(b)){
is.reset();
return false;
}
}
} catch (IOException e) {
return true;
}
}
private static byte[] inbuf = new byte[1024];
static int lenbuf = 0, ptrbuf = 0;
{
if(lenbuf == -1)throw new InputMismatchException();
if(ptrbuf >= lenbuf){
ptrbuf = 0;
try { lenbuf = is.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); }
if(lenbuf <= 0)return -1;
}
return inbuf[ptrbuf++];
}
private static boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); }
private static int skip() { int b; while((b = readByte()) != -1 && isSpaceChar(b)); return b; }
private static double nd() { return Double.parseDouble(ns()); }
private static char nc() { return (char)skip(); }
private static String ns()
{
int b = skip();
StringBuilder sb = new StringBuilder();
while(!(isSpaceChar(b))){ // when nextLine, (isSpaceChar(b) && b != ' ')
sb.appendCodePoint(b);
}
return sb.toString();
}
private static char[] ns(int n)
{
char[] buf = new char[n];
int b = skip(), p = 0;
while(p < n && !(isSpaceChar(b))){
buf[p++] = (char)b;
}
return n == p ? buf : Arrays.copyOf(buf, p);
}
private static char[][] nm(int n, int m)
{
char[][] map = new char[n][];
for(int i = 0;i < n;i++)map[i] = ns(m);
return map;
}
private static int[] na(int n)
{
int[] a = new int[n];
for(int i = 0;i < n;i++)a[i] = ni();
return a;
}
private static int ni()
{
int num = 0, b;
boolean minus = false;
while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
if(b == '-'){
minus = true;
}
while(true){
if(b >= '0' && b <= '9'){
num = num * 10 + (b - '0');
}else{
return minus ? -num : num;
}
}
}
private static long nl()
{
long num = 0;
int b;
boolean minus = false;
while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
if(b == '-'){
minus = true;
}
while(true){
if(b >= '0' && b <= '9'){
num = num * 10 + (b - '0');
}else{
return minus ? -num : num;
}
}
}
private static void tr(Object... o) { if(INPUT.length() != 0)System.out.println(Arrays.deepToString(o)); }
}
## Bigger is Greater HackerRank Solution in Python
def next(a):
i = len(a) - 2
while not (i < 0 or a[i] < a[i+1]):
i -= 1
if i < 0:
return False
j = len(a) - 1
while not (a[j] > a[i]):
j -= 1
a[i], a[j] = a[j], a[i]
a[i+1:] = reversed(a[i+1:])
return True
for i in xrange(input()):
w = list(raw_input().strip())
if next(w):
print ''.join(w)
## Bigger is Greater HackerRank Solution in C#
using System;
using System.Collections.Generic;
using System.IO;
class Solution {
static void Main(String[] args) {
while(t-- > 0){
int len = chr.Length;
bool poss = false;
int[] pos = new int[26];
int idx = -1;
int w = -1;
bool[] small = new bool[26];
small[(int)chr[len-1] - (int)'a'] = true;
pos[(int)chr[len-1] - (int)'a'] = len - 1;
for(int i = len - 2 ; i >= 0 ; i--){
int acode = (int)chr[i] - (int)'a';
for(int j = acode + 1; j < 26 ; j++){
if(small[j])
{
poss = true;
idx = pos[j];
break;
}
}
if(poss){
w = i ;
break;
}
small[acode] = true;
pos[acode] = i;
}
//Console.WriteLine(w + "-" + idx);
if(poss){
char c = chr[w];
chr[w] = chr[idx];
chr[idx] = c;
char[] sortarr = new char[len - w - 1];
int counter = 0 ;
for(int i = w + 1 ; i < len ; i++){
sortarr[counter] = chr[i];
counter++;
}
Array.Sort(sortarr);
for(int i = 0 ; i < sortarr.Length ; i++){
chr[w + 1] = sortarr[i];
w++;
}
Console.WriteLine(new string(chr));
} else {
| 2,198
| 7,172
|
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| 2.765625
| 3
|
CC-MAIN-2024-10
|
longest
|
en
| 0.618838
|
http://mathforum.org/library/drmath/view/58772.html
| 1,498,363,863,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-26/segments/1498128320395.62/warc/CC-MAIN-20170625032210-20170625052210-00128.warc.gz
| 257,983,124
| 3,379
|
Associated Topics || Dr. Math Home || Search Dr. Math
### Using Parentheses
```
Date: 11/16/98 at 21:58:05
From: Anna Martinez
Subject: Using Parenthesis in Math problems
Dear Dr. Math,
I don't understand how to use parentheses. Can you explain it step by
step so I can get it?
Thanks,
Anna
```
```
Date: 11/17/98 at 12:27:18
From: Doctor Peterson
Subject: Re: Using Parenthesis in Math problems
Hi, Anna.
Did you ever see a Christmas package labelled "Open me first?" That's
sort of what parentheses do. They wrap up some numbers and symbols, and
tell us to do whatever is inside the package first, before we do
anything else with those numbers.
For instance, suppose we have this expression (we write multiplication
as "*" to avoid confusion with the letter x):
2 * (3 + 4)
That tells us we first calculate 3 + 4 = 7, and then put that number
in place of the whole package (3 + 4):
2 * 7
Now we can do the multiplication, and calculate the answer as 14.
If you like, you can think of parentheses as a funnel that pours some
ingredients into the calculation at the right place:
(3 + 4)
\ /
\ /
2 * ( 7 )
If we didn't use the parentheses, but had:
2 * 3 + 4
then the usual rules (called "order of operations") tell us that we
should multiply 2 by 3 first, getting:
6 + 4
and then add, so the answer is 10. Why is it different? It's sort of
like if you got a cat, a mouse, and a cage for Christmas. If they
weren't wrapped right, you'd end up with an empty cage and a satisfied
cat. But if you wrapped it like this:
cat (mouse + cage)
then you could first open the parenthesized package and put the mouse
in the cage, then end up with a hungry cat and a caged mouse. That's
probably what you really wanted!
Here's a place in the Dr. Math FAQ to find out more about the order of
operations:
http://mathforum.org/dr.math/faq/faq.order.operations.html
- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
| 526
| 1,963
|
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| 4.03125
| 4
|
CC-MAIN-2017-26
|
longest
|
en
| 0.886132
|
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/1134/2/g/n/
| 1,679,858,754,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-14/segments/1679296946445.46/warc/CC-MAIN-20230326173112-20230326203112-00287.warc.gz
| 953,213,166
| 60,410
|
# Properties
Label 1134.2.g.n Level $1134$ Weight $2$ Character orbit 1134.g Analytic conductor $9.055$ Analytic rank $0$ Dimension $6$ CM no Inner twists $2$
# Related objects
Show commands: Magma / PariGP / SageMath
## Newspace parameters
comment: Compute space of new eigenforms
[N,k,chi] = [1134,2,Mod(163,1134)]
mf = mfinit([N,k,chi],0)
lf = mfeigenbasis(mf)
from sage.modular.dirichlet import DirichletCharacter
H = DirichletGroup(1134, base_ring=CyclotomicField(6))
chi = DirichletCharacter(H, H._module([0, 2]))
N = Newforms(chi, 2, names="a")
//Please install CHIMP (https://github.com/edgarcosta/CHIMP) if you want to run this code
chi := DirichletCharacter("1134.163");
S:= CuspForms(chi, 2);
N := Newforms(S);
Level: $$N$$ $$=$$ $$1134 = 2 \cdot 3^{4} \cdot 7$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 1134.g (of order $$3$$, degree $$2$$, minimal)
## Newform invariants
comment: select newform
sage: f = N[0] # Warning: the index may be different
gp: f = lf[1] \\ Warning: the index may be different
Self dual: no Analytic conductor: $$9.05503558921$$ Analytic rank: $$0$$ Dimension: $$6$$ Relative dimension: $$3$$ over $$\Q(\zeta_{3})$$ Coefficient field: 6.0.309123.1 comment: defining polynomial gp: f.mod \\ as an extension of the character field Defining polynomial: $$x^{6} - 3x^{5} + 10x^{4} - 15x^{3} + 19x^{2} - 12x + 3$$ x^6 - 3*x^5 + 10*x^4 - 15*x^3 + 19*x^2 - 12*x + 3 Coefficient ring: $$\Z[a_1, \ldots, a_{5}]$$ Coefficient ring index: $$3$$ Twist minimal: no (minimal twist has level 126) Sato-Tate group: $\mathrm{SU}(2)[C_{3}]$
## $q$-expansion
comment: q-expansion
sage: f.q_expansion() # note that sage often uses an isomorphic number field
gp: mfcoefs(f, 20)
Coefficients of the $$q$$-expansion are expressed in terms of a basis $$1,\beta_1,\ldots,\beta_{5}$$ for the coefficient ring described below. We also show the integral $$q$$-expansion of the trace form.
$$f(q)$$ $$=$$ $$q + ( - \beta_{4} + 1) q^{2} - \beta_{4} q^{4} + (\beta_{5} - 2 \beta_{4} + 2) q^{5} + ( - \beta_{5} + \beta_{4} - \beta_{3} + \beta_1 - 1) q^{7} - q^{8}+O(q^{10})$$ q + (-b4 + 1) * q^2 - b4 * q^4 + (b5 - 2*b4 + 2) * q^5 + (-b5 + b4 - b3 + b1 - 1) * q^7 - q^8 $$q + ( - \beta_{4} + 1) q^{2} - \beta_{4} q^{4} + (\beta_{5} - 2 \beta_{4} + 2) q^{5} + ( - \beta_{5} + \beta_{4} - \beta_{3} + \beta_1 - 1) q^{7} - q^{8} + (\beta_{5} - 2 \beta_{4} - \beta_1) q^{10} + ( - \beta_{5} - 2 \beta_{3} - 2 \beta_{2} + \beta_1) q^{11} + (\beta_1 + 1) q^{13} + (\beta_{4} + \beta_{2} + \beta_1) q^{14} + (\beta_{4} - 1) q^{16} + (2 \beta_{5} - 2 \beta_{3} - 2 \beta_{2} - 2 \beta_1) q^{17} + (\beta_{5} + \beta_{4} + \beta_{2} - 1) q^{19} + ( - \beta_1 - 2) q^{20} + ( - 2 \beta_{3} + \beta_1) q^{22} + (\beta_{5} - 3 \beta_{4} - \beta_{2} + 3) q^{23} + (3 \beta_{5} - 2 \beta_{4} - \beta_{3} - \beta_{2} - 3 \beta_1) q^{25} + (\beta_{5} - \beta_{4} + 1) q^{26} + (\beta_{5} + \beta_{3} + \beta_{2} + 1) q^{28} + ( - \beta_{3} + 3 \beta_1 - 1) q^{29} + (2 \beta_{5} - 5 \beta_{4} + \beta_{3} + \beta_{2} - 2 \beta_1) q^{31} + \beta_{4} q^{32} + ( - 2 \beta_{3} - 2 \beta_1) q^{34} + ( - 2 \beta_{5} + 3 \beta_{4} + \beta_{3} + 2 \beta_{2} + 2 \beta_1 + 2) q^{35} + (2 \beta_{5} + 3 \beta_{4} + 2 \beta_{2} - 3) q^{37} + (\beta_{5} + \beta_{4} + \beta_{3} + \beta_{2} - \beta_1) q^{38} + ( - \beta_{5} + 2 \beta_{4} - 2) q^{40} + (\beta_{3} - \beta_1 - 4) q^{41} + (\beta_{3} + 2 \beta_1 - 5) q^{43} + (\beta_{5} + 2 \beta_{2}) q^{44} + (\beta_{5} - 3 \beta_{4} - \beta_{3} - \beta_{2} - \beta_1) q^{46} + ( - 2 \beta_{5} + 2 \beta_{4} + \beta_{2} - 2) q^{47} + (3 \beta_{5} - 2 \beta_{3} - \beta_{2} - 3 \beta_1 + 1) q^{49} + ( - \beta_{3} - 3 \beta_1 - 2) q^{50} + (\beta_{5} - \beta_{4} - \beta_1) q^{52} + ( - 4 \beta_{5} - 2 \beta_{4} - \beta_{3} - \beta_{2} + 4 \beta_1) q^{53} + ( - 3 \beta_{3} - \beta_1 + 1) q^{55} + (\beta_{5} - \beta_{4} + \beta_{3} - \beta_1 + 1) q^{56} + (3 \beta_{5} + \beta_{4} + \beta_{2} - 1) q^{58} + ( - 5 \beta_{5} - \beta_{3} - \beta_{2} + 5 \beta_1) q^{59} + (5 \beta_{5} - 2 \beta_{4} + 3 \beta_{2} + 2) q^{61} + (\beta_{3} - 2 \beta_1 - 5) q^{62} + q^{64} + (2 \beta_{5} - 5 \beta_{4} - \beta_{2} + 5) q^{65} + ( - \beta_{5} + 3 \beta_{4} + 3 \beta_{3} + 3 \beta_{2} + \beta_1) q^{67} + ( - 2 \beta_{5} + 2 \beta_{2}) q^{68} + ( - 2 \beta_{4} + 2 \beta_{3} + \beta_{2} + 2 \beta_1 + 5) q^{70} + (3 \beta_{3} + \beta_1 - 1) q^{71} + ( - \beta_{5} - 9 \beta_{4} - 3 \beta_{3} - 3 \beta_{2} + \beta_1) q^{73} + (2 \beta_{5} + 3 \beta_{4} + 2 \beta_{3} + 2 \beta_{2} - 2 \beta_1) q^{74} + (\beta_{3} - \beta_1 + 1) q^{76} + (2 \beta_{5} + 8 \beta_{4} - 2 \beta_{3} - 3 \beta_{2} - 1) q^{77} + ( - 3 \beta_{5} + 4 \beta_{2}) q^{79} + ( - \beta_{5} + 2 \beta_{4} + \beta_1) q^{80} + ( - \beta_{5} + 4 \beta_{4} - \beta_{2} - 4) q^{82} + (2 \beta_{3} - \beta_1 + 3) q^{83} + ( - 6 \beta_{3} - 4 \beta_1 - 8) q^{85} + (2 \beta_{5} + 5 \beta_{4} - \beta_{2} - 5) q^{86} + (\beta_{5} + 2 \beta_{3} + 2 \beta_{2} - \beta_1) q^{88} + (\beta_{5} - 4 \beta_{4} - 2 \beta_{2} + 4) q^{89} + ( - \beta_{5} + 4 \beta_{4} + \beta_{2} - \beta_1 - 2) q^{91} + ( - \beta_{3} - \beta_1 - 3) q^{92} + ( - 2 \beta_{5} + 2 \beta_{4} + \beta_{3} + \beta_{2} + 2 \beta_1) q^{94} + ( - \beta_{5} + \beta_{3} + \beta_{2} + \beta_1) q^{95} + (2 \beta_{3} + 10) q^{97} + ( - \beta_{4} - \beta_{3} + \beta_{2} - 3 \beta_1 + 1) q^{98}+O(q^{100})$$ q + (-b4 + 1) * q^2 - b4 * q^4 + (b5 - 2*b4 + 2) * q^5 + (-b5 + b4 - b3 + b1 - 1) * q^7 - q^8 + (b5 - 2*b4 - b1) * q^10 + (-b5 - 2*b3 - 2*b2 + b1) * q^11 + (b1 + 1) * q^13 + (b4 + b2 + b1) * q^14 + (b4 - 1) * q^16 + (2*b5 - 2*b3 - 2*b2 - 2*b1) * q^17 + (b5 + b4 + b2 - 1) * q^19 + (-b1 - 2) * q^20 + (-2*b3 + b1) * q^22 + (b5 - 3*b4 - b2 + 3) * q^23 + (3*b5 - 2*b4 - b3 - b2 - 3*b1) * q^25 + (b5 - b4 + 1) * q^26 + (b5 + b3 + b2 + 1) * q^28 + (-b3 + 3*b1 - 1) * q^29 + (2*b5 - 5*b4 + b3 + b2 - 2*b1) * q^31 + b4 * q^32 + (-2*b3 - 2*b1) * q^34 + (-2*b5 + 3*b4 + b3 + 2*b2 + 2*b1 + 2) * q^35 + (2*b5 + 3*b4 + 2*b2 - 3) * q^37 + (b5 + b4 + b3 + b2 - b1) * q^38 + (-b5 + 2*b4 - 2) * q^40 + (b3 - b1 - 4) * q^41 + (b3 + 2*b1 - 5) * q^43 + (b5 + 2*b2) * q^44 + (b5 - 3*b4 - b3 - b2 - b1) * q^46 + (-2*b5 + 2*b4 + b2 - 2) * q^47 + (3*b5 - 2*b3 - b2 - 3*b1 + 1) * q^49 + (-b3 - 3*b1 - 2) * q^50 + (b5 - b4 - b1) * q^52 + (-4*b5 - 2*b4 - b3 - b2 + 4*b1) * q^53 + (-3*b3 - b1 + 1) * q^55 + (b5 - b4 + b3 - b1 + 1) * q^56 + (3*b5 + b4 + b2 - 1) * q^58 + (-5*b5 - b3 - b2 + 5*b1) * q^59 + (5*b5 - 2*b4 + 3*b2 + 2) * q^61 + (b3 - 2*b1 - 5) * q^62 + q^64 + (2*b5 - 5*b4 - b2 + 5) * q^65 + (-b5 + 3*b4 + 3*b3 + 3*b2 + b1) * q^67 + (-2*b5 + 2*b2) * q^68 + (-2*b4 + 2*b3 + b2 + 2*b1 + 5) * q^70 + (3*b3 + b1 - 1) * q^71 + (-b5 - 9*b4 - 3*b3 - 3*b2 + b1) * q^73 + (2*b5 + 3*b4 + 2*b3 + 2*b2 - 2*b1) * q^74 + (b3 - b1 + 1) * q^76 + (2*b5 + 8*b4 - 2*b3 - 3*b2 - 1) * q^77 + (-3*b5 + 4*b2) * q^79 + (-b5 + 2*b4 + b1) * q^80 + (-b5 + 4*b4 - b2 - 4) * q^82 + (2*b3 - b1 + 3) * q^83 + (-6*b3 - 4*b1 - 8) * q^85 + (2*b5 + 5*b4 - b2 - 5) * q^86 + (b5 + 2*b3 + 2*b2 - b1) * q^88 + (b5 - 4*b4 - 2*b2 + 4) * q^89 + (-b5 + 4*b4 + b2 - b1 - 2) * q^91 + (-b3 - b1 - 3) * q^92 + (-2*b5 + 2*b4 + b3 + b2 + 2*b1) * q^94 + (-b5 + b3 + b2 + b1) * q^95 + (2*b3 + 10) * q^97 + (-b4 - b3 + b2 - 3*b1 + 1) * q^98 $$\operatorname{Tr}(f)(q)$$ $$=$$ $$6 q + 3 q^{2} - 3 q^{4} + 5 q^{5} - 2 q^{7} - 6 q^{8}+O(q^{10})$$ 6 * q + 3 * q^2 - 3 * q^4 + 5 * q^5 - 2 * q^7 - 6 * q^8 $$6 q + 3 q^{2} - 3 q^{4} + 5 q^{5} - 2 q^{7} - 6 q^{8} - 5 q^{10} + q^{11} + 4 q^{13} + 2 q^{14} - 3 q^{16} + 4 q^{17} - 3 q^{19} - 10 q^{20} + 2 q^{22} + 7 q^{23} - 2 q^{25} + 2 q^{26} + 4 q^{28} - 10 q^{29} - 14 q^{31} + 3 q^{32} + 8 q^{34} + 19 q^{35} - 9 q^{37} + 3 q^{38} - 5 q^{40} - 24 q^{41} - 36 q^{43} + q^{44} - 7 q^{46} - 3 q^{47} + 12 q^{49} - 4 q^{50} - 2 q^{52} - 9 q^{53} + 14 q^{55} + 2 q^{56} - 5 q^{58} - 4 q^{59} + 4 q^{61} - 28 q^{62} + 6 q^{64} + 12 q^{65} + 5 q^{67} + 4 q^{68} + 17 q^{70} - 14 q^{71} - 25 q^{73} + 9 q^{74} + 6 q^{76} + 17 q^{77} + 7 q^{79} + 5 q^{80} - 12 q^{82} + 16 q^{83} - 28 q^{85} - 18 q^{86} - q^{88} + 9 q^{89} + 4 q^{91} - 14 q^{92} + 3 q^{94} - 2 q^{95} + 56 q^{97} + 12 q^{98}+O(q^{100})$$ 6 * q + 3 * q^2 - 3 * q^4 + 5 * q^5 - 2 * q^7 - 6 * q^8 - 5 * q^10 + q^11 + 4 * q^13 + 2 * q^14 - 3 * q^16 + 4 * q^17 - 3 * q^19 - 10 * q^20 + 2 * q^22 + 7 * q^23 - 2 * q^25 + 2 * q^26 + 4 * q^28 - 10 * q^29 - 14 * q^31 + 3 * q^32 + 8 * q^34 + 19 * q^35 - 9 * q^37 + 3 * q^38 - 5 * q^40 - 24 * q^41 - 36 * q^43 + q^44 - 7 * q^46 - 3 * q^47 + 12 * q^49 - 4 * q^50 - 2 * q^52 - 9 * q^53 + 14 * q^55 + 2 * q^56 - 5 * q^58 - 4 * q^59 + 4 * q^61 - 28 * q^62 + 6 * q^64 + 12 * q^65 + 5 * q^67 + 4 * q^68 + 17 * q^70 - 14 * q^71 - 25 * q^73 + 9 * q^74 + 6 * q^76 + 17 * q^77 + 7 * q^79 + 5 * q^80 - 12 * q^82 + 16 * q^83 - 28 * q^85 - 18 * q^86 - q^88 + 9 * q^89 + 4 * q^91 - 14 * q^92 + 3 * q^94 - 2 * q^95 + 56 * q^97 + 12 * q^98
Basis of coefficient ring in terms of a root $$\nu$$ of $$x^{6} - 3x^{5} + 10x^{4} - 15x^{3} + 19x^{2} - 12x + 3$$ :
$$\beta_{1}$$ $$=$$ $$\nu^{2} - \nu + 2$$ v^2 - v + 2 $$\beta_{2}$$ $$=$$ $$( -\nu^{5} + \nu^{4} - 8\nu^{3} + 5\nu^{2} - 18\nu + 6 ) / 3$$ (-v^5 + v^4 - 8*v^3 + 5*v^2 - 18*v + 6) / 3 $$\beta_{3}$$ $$=$$ $$\nu^{4} - 2\nu^{3} + 6\nu^{2} - 5\nu + 3$$ v^4 - 2*v^3 + 6*v^2 - 5*v + 3 $$\beta_{4}$$ $$=$$ $$( -2\nu^{5} + 5\nu^{4} - 16\nu^{3} + 19\nu^{2} - 21\nu + 9 ) / 3$$ (-2*v^5 + 5*v^4 - 16*v^3 + 19*v^2 - 21*v + 9) / 3 $$\beta_{5}$$ $$=$$ $$( 2\nu^{5} - 5\nu^{4} + 19\nu^{3} - 22\nu^{2} + 30\nu - 9 ) / 3$$ (2*v^5 - 5*v^4 + 19*v^3 - 22*v^2 + 30*v - 9) / 3
$$\nu$$ $$=$$ $$( -2\beta_{5} - \beta_{4} - \beta_{3} - 2\beta_{2} + \beta _1 + 2 ) / 3$$ (-2*b5 - b4 - b3 - 2*b2 + b1 + 2) / 3 $$\nu^{2}$$ $$=$$ $$( -2\beta_{5} - \beta_{4} - \beta_{3} - 2\beta_{2} + 4\beta _1 - 4 ) / 3$$ (-2*b5 - b4 - b3 - 2*b2 + 4*b1 - 4) / 3 $$\nu^{3}$$ $$=$$ $$( 7\beta_{5} + 5\beta_{4} + 2\beta_{3} + 4\beta_{2} + \beta _1 - 10 ) / 3$$ (7*b5 + 5*b4 + 2*b3 + 4*b2 + b1 - 10) / 3 $$\nu^{4}$$ $$=$$ $$( 16\beta_{5} + 11\beta_{4} + 8\beta_{3} + 10\beta_{2} - 17\beta _1 + 5 ) / 3$$ (16*b5 + 11*b4 + 8*b3 + 10*b2 - 17*b1 + 5) / 3 $$\nu^{5}$$ $$=$$ $$( -14\beta_{5} - 16\beta_{4} + 5\beta_{3} - 5\beta_{2} - 23\beta _1 + 47 ) / 3$$ (-14*b5 - 16*b4 + 5*b3 - 5*b2 - 23*b1 + 47) / 3
## Character values
We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/1134\mathbb{Z}\right)^\times$$.
$$n$$ $$325$$ $$407$$ $$\chi(n)$$ $$-1 + \beta_{4}$$ $$1$$
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
comment: embeddings in the coefficient field
gp: mfembed(f)
Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
163.1
0.5 − 2.05195i 0.5 + 1.41036i 0.5 − 0.224437i 0.5 + 2.05195i 0.5 − 1.41036i 0.5 + 0.224437i
0.500000 0.866025i 0 −0.500000 0.866025i −0.230252 + 0.398809i 0 −2.32383 1.26483i −1.00000 0 0.230252 + 0.398809i
163.2 0.500000 0.866025i 0 −0.500000 0.866025i 0.880438 1.52496i 0 2.56238 + 0.658939i −1.00000 0 −0.880438 1.52496i
163.3 0.500000 0.866025i 0 −0.500000 0.866025i 1.84981 3.20397i 0 −1.23855 + 2.33795i −1.00000 0 −1.84981 3.20397i
487.1 0.500000 + 0.866025i 0 −0.500000 + 0.866025i −0.230252 0.398809i 0 −2.32383 + 1.26483i −1.00000 0 0.230252 0.398809i
487.2 0.500000 + 0.866025i 0 −0.500000 + 0.866025i 0.880438 + 1.52496i 0 2.56238 0.658939i −1.00000 0 −0.880438 + 1.52496i
487.3 0.500000 + 0.866025i 0 −0.500000 + 0.866025i 1.84981 + 3.20397i 0 −1.23855 2.33795i −1.00000 0 −1.84981 + 3.20397i
$$n$$: e.g. 2-40 or 990-1000 Embeddings: e.g. 1-3 or 487.3 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Inner twists
Char Parity Ord Mult Type
1.a even 1 1 trivial
7.c even 3 1 inner
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 1134.2.g.n 6
3.b odd 2 1 1134.2.g.k 6
7.c even 3 1 inner 1134.2.g.n 6
7.c even 3 1 7938.2.a.bu 3
7.d odd 6 1 7938.2.a.bx 3
9.c even 3 1 378.2.e.c 6
9.c even 3 1 378.2.h.d 6
9.d odd 6 1 126.2.e.d 6
9.d odd 6 1 126.2.h.c yes 6
21.g even 6 1 7938.2.a.by 3
21.h odd 6 1 1134.2.g.k 6
21.h odd 6 1 7938.2.a.cb 3
36.f odd 6 1 3024.2.q.h 6
36.f odd 6 1 3024.2.t.g 6
36.h even 6 1 1008.2.q.h 6
36.h even 6 1 1008.2.t.g 6
63.g even 3 1 378.2.e.c 6
63.g even 3 1 2646.2.f.o 6
63.h even 3 1 378.2.h.d 6
63.h even 3 1 2646.2.f.o 6
63.i even 6 1 882.2.f.m 6
63.i even 6 1 882.2.h.o 6
63.j odd 6 1 126.2.h.c yes 6
63.j odd 6 1 882.2.f.l 6
63.k odd 6 1 2646.2.e.o 6
63.k odd 6 1 2646.2.f.n 6
63.l odd 6 1 2646.2.e.o 6
63.l odd 6 1 2646.2.h.p 6
63.n odd 6 1 126.2.e.d 6
63.n odd 6 1 882.2.f.l 6
63.o even 6 1 882.2.e.p 6
63.o even 6 1 882.2.h.o 6
63.s even 6 1 882.2.e.p 6
63.s even 6 1 882.2.f.m 6
63.t odd 6 1 2646.2.f.n 6
63.t odd 6 1 2646.2.h.p 6
252.o even 6 1 1008.2.q.h 6
252.u odd 6 1 3024.2.t.g 6
252.bb even 6 1 1008.2.t.g 6
252.bl odd 6 1 3024.2.q.h 6
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
126.2.e.d 6 9.d odd 6 1
126.2.e.d 6 63.n odd 6 1
126.2.h.c yes 6 9.d odd 6 1
126.2.h.c yes 6 63.j odd 6 1
378.2.e.c 6 9.c even 3 1
378.2.e.c 6 63.g even 3 1
378.2.h.d 6 9.c even 3 1
378.2.h.d 6 63.h even 3 1
882.2.e.p 6 63.o even 6 1
882.2.e.p 6 63.s even 6 1
882.2.f.l 6 63.j odd 6 1
882.2.f.l 6 63.n odd 6 1
882.2.f.m 6 63.i even 6 1
882.2.f.m 6 63.s even 6 1
882.2.h.o 6 63.i even 6 1
882.2.h.o 6 63.o even 6 1
1008.2.q.h 6 36.h even 6 1
1008.2.q.h 6 252.o even 6 1
1008.2.t.g 6 36.h even 6 1
1008.2.t.g 6 252.bb even 6 1
1134.2.g.k 6 3.b odd 2 1
1134.2.g.k 6 21.h odd 6 1
1134.2.g.n 6 1.a even 1 1 trivial
1134.2.g.n 6 7.c even 3 1 inner
2646.2.e.o 6 63.k odd 6 1
2646.2.e.o 6 63.l odd 6 1
2646.2.f.n 6 63.k odd 6 1
2646.2.f.n 6 63.t odd 6 1
2646.2.f.o 6 63.g even 3 1
2646.2.f.o 6 63.h even 3 1
2646.2.h.p 6 63.l odd 6 1
2646.2.h.p 6 63.t odd 6 1
3024.2.q.h 6 36.f odd 6 1
3024.2.q.h 6 252.bl odd 6 1
3024.2.t.g 6 36.f odd 6 1
3024.2.t.g 6 252.u odd 6 1
7938.2.a.bu 3 7.c even 3 1
7938.2.a.bx 3 7.d odd 6 1
7938.2.a.by 3 21.g even 6 1
7938.2.a.cb 3 21.h odd 6 1
## Hecke kernels
This newform subspace can be constructed as the intersection of the kernels of the following linear operators acting on $$S_{2}^{\mathrm{new}}(1134, [\chi])$$:
$$T_{5}^{6} - 5T_{5}^{5} + 21T_{5}^{4} - 26T_{5}^{3} + 31T_{5}^{2} + 12T_{5} + 9$$ T5^6 - 5*T5^5 + 21*T5^4 - 26*T5^3 + 31*T5^2 + 12*T5 + 9 $$T_{11}^{6} - T_{11}^{5} + 27T_{11}^{4} + 92T_{11}^{3} + 643T_{11}^{2} + 858T_{11} + 1089$$ T11^6 - T11^5 + 27*T11^4 + 92*T11^3 + 643*T11^2 + 858*T11 + 1089 $$T_{17}^{6} - 4T_{17}^{5} + 60T_{17}^{4} - 160T_{17}^{3} + 2608T_{17}^{2} - 7392T_{17} + 28224$$ T17^6 - 4*T17^5 + 60*T17^4 - 160*T17^3 + 2608*T17^2 - 7392*T17 + 28224 $$T_{23}^{6} - 7T_{23}^{5} + 45T_{23}^{4} - 34T_{23}^{3} + 37T_{23}^{2} + 12T_{23} + 9$$ T23^6 - 7*T23^5 + 45*T23^4 - 34*T23^3 + 37*T23^2 + 12*T23 + 9
## Hecke characteristic polynomials
$p$ $F_p(T)$
$2$ $$(T^{2} - T + 1)^{3}$$
$3$ $$T^{6}$$
$5$ $$T^{6} - 5 T^{5} + 21 T^{4} - 26 T^{3} + \cdots + 9$$
$7$ $$T^{6} + 2 T^{5} - 4 T^{4} - 31 T^{3} + \cdots + 343$$
$11$ $$T^{6} - T^{5} + 27 T^{4} + 92 T^{3} + \cdots + 1089$$
$13$ $$(T^{3} - 2 T^{2} - 3 T + 3)^{2}$$
$17$ $$T^{6} - 4 T^{5} + 60 T^{4} + \cdots + 28224$$
$19$ $$T^{6} + 3 T^{5} + 15 T^{4} - 4 T^{3} + \cdots + 49$$
$23$ $$T^{6} - 7 T^{5} + 45 T^{4} - 34 T^{3} + \cdots + 9$$
$29$ $$(T^{3} + 5 T^{2} - 32 T + 33)^{2}$$
$31$ $$T^{6} + 14 T^{5} + 151 T^{4} + \cdots + 729$$
$37$ $$T^{6} + 9 T^{5} + 90 T^{4} + \cdots + 5329$$
$41$ $$(T^{3} + 12 T^{2} + 39 T + 27)^{2}$$
$43$ $$(T^{3} + 18 T^{2} + 81 T + 1)^{2}$$
$47$ $$T^{6} + 3 T^{5} + 33 T^{4} - 126 T^{3} + \cdots + 729$$
$53$ $$T^{6} + 9 T^{5} + 123 T^{4} - 396 T^{3} + \cdots + 81$$
$59$ $$T^{6} + 4 T^{5} + 117 T^{4} + \cdots + 31329$$
$61$ $$T^{6} - 4 T^{5} + 151 T^{4} + \cdots + 514089$$
$67$ $$T^{6} - 5 T^{5} + 83 T^{4} + \cdots + 22201$$
$71$ $$(T^{3} + 7 T^{2} - 50 T - 99)^{2}$$
$73$ $$T^{6} + 25 T^{5} + 473 T^{4} + \cdots + 2401$$
$79$ $$T^{6} - 7 T^{5} + 193 T^{4} + \cdots + 594441$$
$83$ $$(T^{3} - 8 T^{2} - 5 T + 93)^{2}$$
$89$ $$T^{6} - 9 T^{5} + 87 T^{4} + \cdots + 3969$$
$97$ $$(T^{3} - 28 T^{2} + 236 T - 536)^{2}$$
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Gerri - 1 month ago 12x
Python Question
# Python sort (json) list by calculated value
I have originally a JSON format with products and the challenge is to sort this list by the best price per unit (prdata3, prdata2, prdata1):
``````{'prdata3': {'variation': 'stand', 'price': '78 54', 'amount': '6',
'prdata2': {'variation': 'stand', 'price': '15 24', 'amount': '1',
'prdata1': {'variation': 'vc', 'price': '78 54', 'amount': '12',
``````
The price needs to be divided by the amount and the list needs to be sorted by this value, so in my example the output need to be (prdata1 (6,55), prdata3 (13,09), prdata2 (15,24)):
``````{'prdata1': {'variation': 'vc', 'price': '78 54', 'amount': '12', 'url': 'https://www.productwebaddress.com/1'}, 'prdata3': {'variation': 'stand', 'price': '78 54', 'amount': '6', 'url': 'https://www.productwebaddress.com/3'}, 'prdata2': {'variation': 'stand', 'price': '15 24', 'amount': '1', 'url': 'https://www.productwebaddress.com/2'}}
``````
What is the best strategy here? I gave my best to first go in there with "for each", looking for the price, replace the space with a dot, divide it with the amount and now use this result to sort my list with something like:
`sorted(mylist.items(), key=lambda t: t[1])`
... but my result as key, right?
You can't sort a dictionary, as dictionary have no order.
But you can do the following:
``````print sorted(a.items(),key=lambda x:float(x[1]['price'])/int(x[1]['amount']))
``````
Output:
``````[('prdata1', {'url': 'https://www.productwebaddress.com/1', 'price': '78.54', 'amount': '12', 'variation': 'vc'}),
('prdata3', {'url': 'https://www.productwebaddress.com/3', 'price': '78.54', 'amount': '6', 'variation': 'stand'}),
('prdata2', {'url': 'https://www.productwebaddress.com/2', 'price': '15.24', 'amount': '1', 'variation': 'stand'})]
``````
And, you can use `collections.OrderedDict` to retain the dictionary structure if it really matters.
``````from collections import OrderedDict
print OrderedDict(sorted(a.items(),key=lambda x:float(x[1]['price'])/int(x[1]['amount'])))
``````
Output:
``````OrderedDict([('prdata1', {'url': 'https://www.productwebaddress.com/1', 'price': '78.54', 'amount': '12', 'variation': 'vc'}),
('prdata3', {'url': 'https://www.productwebaddress.com/3', 'price': '78.54', 'amount': '6', 'variation': 'stand'}),
('prdata2', {'url': 'https://www.productwebaddress.com/2', 'price': '15.24', 'amount': '1', 'variation': 'stand'})])
``````
Source (Stackoverflow)
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Dark Mode On/Off
# C++ Program To Quick Sort Using Dynamic Array
In this tutorial, we will be learning about quicksort that takes O(nlogn) time complexity in its best case, and in its worst case, it takes O(n^2) time. The basic idea of the quicksort is to choose the pivot element and on the left side of the pivot, all the elements less than that will be placed whereas on the right-hand side all the elements greater than the pivot will be placed. Similarly, we can apply this algorithm on the left and right sides respectively using recursion.
Now the question arises that how to choose the pivot element in the given array, so you can choose any element as a pivot, here, we will be choosing the first element i.e. at the index zero as the pivot element while writing the algorithm.
Let us consider some inputs to understand what should be the required output:
Input:
array: {2 3 9 7 1}
Output:
array: {1 2 3 7 9}
Input:
array: {56 9 11 7 60}
Output:
array: {7 9 11 56 60}
## Quicksort Algorithm
1. Create a function partition()
2. Pass three parameters arr, low, high
3. Select rightmost element as pivot
4. declare pointer for greater element
5. traverse through all elements and compare them with pivot
6. If the smaller element is found swap them with pivot
7. Return the index position
8. Create a function QuickSort()
9. Pass three parameters arr, low, high
10. Find the pivot element
11. Do Recursive call on the left pivot and right pivot
12. The array is now sorted
13. Print the sorted array
## C++ Program For The Quick Sort
``````#include<iostream>
using namespace std;
int partition(int array[],int lb,int ub){
int pivot=array[lb];
int start=lb;
int end=ub;
while(start<end){
while(array[start]<=pivot){
start++;
}
while(array[end]>pivot){
end--;
}
if(start<end){
int temp=array[start];
array[start]=array[end];
array[end]=temp;
}
}
int temp=array[end];
array[end]=array[lb];
array[lb]=temp;
return end;
}
int quick_sort(int array[],int lb,int ub){
while(lb<ub){
int loc=partition(array,lb,ub);
quick_sort(array,loc+1,ub);
quick_sort(array,lb,loc-1);
}
return 0;
}
int main(){
int arr[]={9,6,11,10,2,5};
int n=sizeof(arr)/sizeof(arr[0]);
quick_sort(arr,0,n-1);
cout<<"Elements of the array after sorting are:-"<<endl;
for(int i=0;i<n;i++){
cout<<arr[i]<<" ";
}
return 0;
}
``````
Elements of the array after sorting are:-
2 5 6 9 10 11
## Conclusion
We have seen the implementation and the basic principle behind the working of quicksort. Also, we have learned the working algorithm by taking an example.
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Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > ablnsg Structured version Visualization version GIF version
Theorem ablnsg 19035
Description: Every subgroup of an abelian group is normal. (Contributed by Mario Carneiro, 14-Jun-2015.)
Assertion
Ref Expression
ablnsg (𝐺 ∈ Abel → (NrmSGrp‘𝐺) = (SubGrp‘𝐺))
Proof of Theorem ablnsg
Dummy variables 𝑥 𝑦 𝑧 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 eqid 2758 . . . . . . 7 (Base‘𝐺) = (Base‘𝐺)
2 eqid 2758 . . . . . . 7 (+g𝐺) = (+g𝐺)
31, 2ablcom 18991 . . . . . 6 ((𝐺 ∈ Abel ∧ 𝑦 ∈ (Base‘𝐺) ∧ 𝑧 ∈ (Base‘𝐺)) → (𝑦(+g𝐺)𝑧) = (𝑧(+g𝐺)𝑦))
433expb 1117 . . . . 5 ((𝐺 ∈ Abel ∧ (𝑦 ∈ (Base‘𝐺) ∧ 𝑧 ∈ (Base‘𝐺))) → (𝑦(+g𝐺)𝑧) = (𝑧(+g𝐺)𝑦))
54eleq1d 2836 . . . 4 ((𝐺 ∈ Abel ∧ (𝑦 ∈ (Base‘𝐺) ∧ 𝑧 ∈ (Base‘𝐺))) → ((𝑦(+g𝐺)𝑧) ∈ 𝑥 ↔ (𝑧(+g𝐺)𝑦) ∈ 𝑥))
65ralrimivva 3120 . . 3 (𝐺 ∈ Abel → ∀𝑦 ∈ (Base‘𝐺)∀𝑧 ∈ (Base‘𝐺)((𝑦(+g𝐺)𝑧) ∈ 𝑥 ↔ (𝑧(+g𝐺)𝑦) ∈ 𝑥))
71, 2isnsg 18374 . . . 4 (𝑥 ∈ (NrmSGrp‘𝐺) ↔ (𝑥 ∈ (SubGrp‘𝐺) ∧ ∀𝑦 ∈ (Base‘𝐺)∀𝑧 ∈ (Base‘𝐺)((𝑦(+g𝐺)𝑧) ∈ 𝑥 ↔ (𝑧(+g𝐺)𝑦) ∈ 𝑥)))
87rbaib 542 . . 3 (∀𝑦 ∈ (Base‘𝐺)∀𝑧 ∈ (Base‘𝐺)((𝑦(+g𝐺)𝑧) ∈ 𝑥 ↔ (𝑧(+g𝐺)𝑦) ∈ 𝑥) → (𝑥 ∈ (NrmSGrp‘𝐺) ↔ 𝑥 ∈ (SubGrp‘𝐺)))
96, 8syl 17 . 2 (𝐺 ∈ Abel → (𝑥 ∈ (NrmSGrp‘𝐺) ↔ 𝑥 ∈ (SubGrp‘𝐺)))
109eqrdv 2756 1 (𝐺 ∈ Abel → (NrmSGrp‘𝐺) = (SubGrp‘𝐺))
Colors of variables: wff setvar class Syntax hints: → wi 4 ↔ wb 209 ∧ wa 399 = wceq 1538 ∈ wcel 2111 ∀wral 3070 ‘cfv 6335 (class class class)co 7150 Basecbs 16541 +gcplusg 16623 SubGrpcsubg 18340 NrmSGrpcnsg 18341 Abelcabl 18974 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1797 ax-4 1811 ax-5 1911 ax-6 1970 ax-7 2015 ax-8 2113 ax-9 2121 ax-10 2142 ax-11 2158 ax-12 2175 ax-ext 2729 ax-sep 5169 ax-nul 5176 ax-pow 5234 ax-pr 5298 This theorem depends on definitions: df-bi 210 df-an 400 df-or 845 df-3an 1086 df-tru 1541 df-fal 1551 df-ex 1782 df-nf 1786 df-sb 2070 df-mo 2557 df-eu 2588 df-clab 2736 df-cleq 2750 df-clel 2830 df-nfc 2901 df-ne 2952 df-ral 3075 df-rex 3076 df-rab 3079 df-v 3411 df-sbc 3697 df-dif 3861 df-un 3863 df-in 3865 df-ss 3875 df-nul 4226 df-if 4421 df-pw 4496 df-sn 4523 df-pr 4525 df-op 4529 df-uni 4799 df-br 5033 df-opab 5095 df-mpt 5113 df-id 5430 df-xp 5530 df-rel 5531 df-cnv 5532 df-co 5533 df-dm 5534 df-rn 5535 df-res 5536 df-ima 5537 df-iota 6294 df-fun 6337 df-fv 6343 df-ov 7153 df-subg 18343 df-nsg 18344 df-cmn 18975 df-abl 18976 This theorem is referenced by: qusabl 19053 ablsimpnosubgd 19294 ablsimpgprmd 19305 qus1 20076 qusrhm 20078 quslmod 31075 quslmhm 31076 lidlnsg 31142 qusdimsum 31230
Copyright terms: Public domain W3C validator
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# BA 578 Assignment Summer II 2014
November 28, 2016
Question
Part D: Must show all your work step by step in order to receive the full credit; Excel is not allowed. (61-71)
Use the following information from a normal population with mean ? = 52 and variance ?2 = 22.5 to calculate the following questions.
find P (X >55 )
find P (50?X?60)
find P (X?55)
A random sample from a population with mean and standard deviation produced the following sample information:
n =110 x = 699 s = 20.4
Find a 95% confidence interval for the mean ?. Interpret this interval.
Find a 99% confidence interval for the mean ?. Interpret this interval.
Consider the following hypothesis test.
Ho: µ = 17
Ha: µ ? 17
A sample of 25 gives a sample mean of 14.2 and sample variance of 25.
At ? = 0.05, what is the rejection rule?
Compute the value of the test statistic
What is the p-value?
Consider the following hypothesis test.
Ho: ? = 15
Ha: ? ? 15
A sample of 50 gives a sample mean of 14.2 and sample standard deviation of 5.
At ? = 0.02, what is the rejection rule?
Compute the value of the test statistic z.
What is the p-value?
Consider the following hypothesis test
Ho: µ ? 10
Ha: µ < 10
A sample of 50 provides a sample mean of 9.46 and sample variance of 4.
At ? = 0.05, what is the rejection rule?
Reject null if z 0, what statistical decision should be made at 5% level.
What is the standard error of estimate? Can you use this statistic to assess the model’s fit? If so, how?
What is the coefficient of determination, and what does it tell you about the regression model?
What is the coefficient of determination, adjusted for degrees of freedom? Explain how this statistic and the statistic referred to in part (f) help you to determine how well the model fits the data.
Test the overall utility of the model. What does the p-value of the test statistic tell you?
For the following ANOVA table are the results from treating 4 cultures with 6 observations for each culture. The enzymes are contained in test tubes with differing levels of enzymes applied.
Source of Variation df SS MS F
Treatment 180
Error 60
Total
What are the null and alternate hypotheses?
What is your decision rule, Use ?=0.05
Is there a difference among the means?
A metropolitan bus system sampler’s rider counts on one of its express commuter routes for a week. Use the following data to establish whether the rider ship is evenly balanced by day of the week. Let ?=0.05
Day Monday Tuesday Wednesday Thursday Friday
Rider Count 10 34 21 57 44
Is the ?2 value significant at 5% level of significance
Write the conclusion for this question.
Get a 30 % discount on an order above \$ 50
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Hey there! Sign in to join this conversationNew here? Join for free
You are Here: Home >< Maths
# Is M1 maths easy? Watch
1. (Original post by rany345)
dam , thats crazy , how m4 and m5 compared to m3 . im doing the edexcel spec
They are harder, but there's less content, so in some ways easier. M5 is quite a bit harder than M4 (although there was one topic in M4 I detested).
2. (Original post by upthegunners)
People say that M1 is the hardest ASmodule ... I just don't see how some people find it so hard.. people in my school go on like it's a degree level paper on quantum mechanics
Our class did S1 for AS then switched to M1 for A2, all of us already having done AS physics. I think all of us would agree that M1 is pretty easy.
3. (Original post by Noble.)
Surprisingly not, the only maths physics involves is rearranging a fraction.
I cant tell if you're joking or not..
4. (Original post by InternetGangster)
I cant tell if you're joking or not..
I'm not.
5. From experience, I think it helps if you do Physics - I didn't and struggled all year with it despite (apparently) having the best maths teacher at my school.
However, I did my mock 2 weeks before my June exam and got an E and then ended up with an A in the exam so maybe I was over-complicating it?
Just work hard at it and you'll be fine
6. (Original post by Noble.)
I'm doing Mathematics I, II, III, IV, V... i.e. I'm not doing A-Levels anymore
i.e: you're a mathematical genius of sorts.
7. (Original post by InternetGangster)
I cant tell if you're joking or not..
he is not
8. (Original post by upthegunners)
i.e: you're a mathematical genius of sorts.
I wish.
9. the step up to m1 is hard, particularly resolving forces
however, once you know it well, its very easy
the main difficulty in exams is making sill mistakes, as a silly mistake early on can cost you around 10 ums on a long part question
10. (Original post by Noble.)
I wish.
Mathematics at Oxford?
I rest my case.
11. I've found that WJEC give a lot of marks for solving simple kinematic equations. I mean, yes, I find M1 questions tougher than the mechanics questions in Physics papers, but they'll routinely give seven or eight marks for something I'd have thought should give about five.
In general, if you're inclined towards Physics, you'll find mechanics easy. Being able to visualise the situation in your head helps big-time. I hated S1, but plenty would argue that it's easier than M1.
12. (Original post by wrexhamfc)
I've found that WJEC give a lot of marks for solving simple kinematic equations. I mean, yes, I find M1 questions tougher than the mechanics questions in Physics papers, but they'll routinely give seven or eight marks for something I'd have thought should give about five.
In general, if you're inclined towards Physics, you'll find mechanics easy. Being able to visualise the situation in your head helps big-time. I hated S1, but plenty would argue that it's easier than M1.
Did you do Jan2013 M1?
Absolute ***** compared to all the rest... no easy marks on v/t graphs or kinematics... I thought anyway.
13. (Original post by L'Evil Fish)
Did you do Jan2013 M1?
Absolute ***** compared to all the rest... no easy marks on v/t graphs or kinematics... I thought anyway.
Nah, we're doing it in the summer. I just hope that no questions pop up that ask at what time/height two particles collide or anything like that. I can never get my head around those.
14. M1 is easier than S1 in my opinion, I got a B in S1 and 96 in M1.
Physics mechanics is much easier than M1. They give you the equations in a list, it's so easy to pick out the information you need.
15. (Original post by wrexhamfc)
Nah, we're doing it in the summer. I just hope that no questions pop up that ask at what time/height two particles collide or anything like that. I can never get my head around those.
That's just equating two equations and solving for a particular variable.
Usually involves s= ut + 1/2 at2
16. (Original post by wrexhamfc)
Nah, we're doing it in the summer. I just hope that no questions pop up that ask at what time/height two particles collide or anything like that. I can never get my head around those.
You're lucky, when you do it as a practise paper you'll see, I lose 10 raw marks in it.
You know the typical connected pulleys on a horizontal plane? They had that but the one that isn't hanging pulling the other up and everything.
No, there were however other harder ones, but none of those!
17. I really enjoyed it during my A levels, wasn't extremely challenging... seems so long ago now lol Had an amazing teacher though.
18. (Original post by upthegunners)
That's just equating two equations and solving for a particular variable.
Usually involves s= ut + 1/2 at2
I know that, it's just that I sometimes confuse myself as to where all the numbers are coming from and the like. It's the type of question that is more mathematical than physical, which I'm not too keen on
19. (Original post by upthegunners)
aw i see i find m1 easier than c1...it just requires logic
Interesting :P I still find M1 harder, lost 2 marks on c1, not sure about m1 yet, but its probably because I have more exams than I did in jan
20. (Original post by Khodu)
I really enjoyed it during my A levels, wasn't extremely challenging... seems so long ago now lol Had an amazing teacher though.
I don't have a teacher.. self teaching all Alevel Maths
Reply
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# LEARN PYTHON 3 Getting Ready for Physics Class....stuck here
train_mass = 22680
train_acceleration = 10
train_distance = 100
bomb_mass = 1
def f_to_c(f_temp):
return (f_temp - 32) * 5/9
f100_in_celsius=f_to_c(f_temp=100)
print(f100_in_celsius)
def c_to_f(c_temp):
``````return (c_temp * 9/5) + 32
``````
c0_in_fahrenheit = c_to_f(c_temp=0)
print(c0_in_fahrenheit)
def get_force(mass,accelaration):
return mass*accelaration
train_force=get_force(train_mass,train_acceleration)
print("The GE train supplies “+ str(train_force) + " Newtons of force.”)
def get_energy(mass,c=3*10**8):
return mass*c**2
bomb_energy=get_energy(bomb_mass)
print(“A 1kg bomb supplies “+str(bomb_energy)+” Joules.”)
def get_work(mass, acceleration, distance):
force=get_force(mass,accelaration)
return force*distance
train_work=get_work(train_mass,train_acceleration,train_distance)
print(“The GE train does “+str(train_work)+” Joules of work over “+str(train_distance)+” meters.”)
I am getting below error…I am sure i followed the tutorial as explained
Traceback (most recent call last):
File “script.py”, line 35, in
train_work=get_work(train_mass,train_acceleration,train_distance)
File “script.py”, line 32, in get_work
force=get_force(mass,accelaration)
NameError: name ‘accelaration’ is not defined
Please format your code using the </> button so it’s more readable for those trying to help out.
The error you are getting suggests that `accelaration` has not been defined as a variable. I noticed that you’ve used two different spellings of “acceleration” in your code (acceleration is correct, while accelaration is not). This might be the cause of the problem (it’s hard to read the code because it’s not formatted).
Welcome to the forums!
1 Like
Let’s go back to Physics class. Anybody here taken Physics? Let’s start there.
2 Likes
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# The Basics of Poker
In poker, a joker is a card that acts as a wild card. These extra cards are sometimes called the “jokers.” Although they don’t form a hand directly, kicker cards do contribute to its overall strength. A kicker’s strength will determine the winner of a hand in situations where two players have the same hand. Another type of hand is known as a “loose aggressive play,” meaning that players are encouraged to bet as much as they can afford to lose. In a tournament, the more time a player plays the longer he will win, and the bigger the stakes will become.
## Rules
The most important rule in any game of poker is to win all bets in the pot during a single deal. Players bet according to their hand; in general, the best hand is the five-card combination. But, there are exceptions to this rule. Players can bluff when they feel they have a better hand than others. In such a situation, the player may choose to fold the hand, but he or she may lose the money.
## Structure
Poker is a game of skill and luck, and the structure of the game determines how these factors interact. It’s often more important than the type of game you’re playing, because the way you play determines how much luck and skill are involved. A game’s betting structure is generally defined by the table rules. These parameters include the number of blinds and antes and the amounts you’re allowed to bet on each hand.
## Bets
The amount of money you bet on a poker hand is known as your bet size. In other words, your bet size will correspond to the size of the pot, which means you are betting on the number of chips in the center. Poker bets come in two types, based on the size of the pot, or pot-sized bets. While pot-sized bets represent your total bet size, you may also bet on the probability of a certain event occurring.
## Equity Calculator
If you’re playing poker, you’ll want to use an equity calculator. These tools calculate how much you have in a hand, from two players to multiple players. You can plug in your own hands and ranges, or you can use one of the many available online calculators. The best calculators allow you to visualize the board texture, as well as provide a simple quiz to measure your skills. One of the most popular poker equity calculators is Ace Poker Solutions, which allows you to import real-life hands into their database.
## Hand rankings
Knowing hand rankings can be useful for winning more games of poker. Knowing which hand ranks highest will help you determine the strength of your hands and how to play better. You can also practice playing poker to improve your skills. Here are some important poker hand rankings:
## Rank
Rank of poker is a term used to describe the hand ranking of a player. Players have five cards in their hands and they must choose at least five of those cards to form a poker hand. The remaining cards do not have any effect on the poker hand ranking. The highest hand in poker is the royal flush, a combination of ace, king, queen, jack, and ten. In mixed suits, the highest hand is the seven-five-four-3-2.
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CFD-101: The Basics of Computational Fluid Dynamics Modeling
# Boundary Conditions - Outflow
In many simulations, fluid must flow out one or more boundaries of the computational region. But what constitutes a good boundary condition at such "outflow" boundaries?
In compressible flows, when the flow speed at the outflow boundary is supersonic, it makes little difference how the boundary conditions are specified since flow disturbances can't propagate upstream. However, in low speed and incompressible flows disturbances introduced at an outflow boundary can affect the entire computational region.
## Continuative Approximation
The simplest and most commonly used outflow condition is that of a "continuative" boundary. Continuative boundary conditions consist of zero normal derivatives at the boundary for all quantities. The zero-derivative condition is intended to represent a smooth continuation of the flow through the boundary.
It must be stressed that the continuative boundary condition has no physical basis; it is a mathematical statement that may or may not provide the desired flow behavior. In particular, if flow enters the computational region across such a boundary, then the computations may be wrong because nothing has been specified about flow conditions existing outside the boundary.
## Improved Continuative Approximation
FLOW-3D uses a special enhancement to continuative boundaries to improve their behavior. If flow attempts to enter the computational region across this type of boundary it must do so by starting from a condition of rest. This practice helps to reduce inflow and often results in a reasonable approximation of a smooth outflow condition. Nevertheless, a continuative boundary condition must always be viewed with suspicion.
## Other Approximations
For limited classes of problems, better outflow boundary conditions do exist. For example, special boundary treatments have been devised for wave propagation problems that try to determine the speed and direction of waves approaching the boundary and then set boundary conditions so that they continue through the boundary with a minimum of reflection. A useful example of this type of treatment, sometimes called a radiation boundary condition, is described by I. Orlanski, Jour. Comp. Phys. 21, 251 (1976).
As a general rule, a physically meaningful boundary condition, such as a specified pressure condition, should be used at outflow boundaries whenever possible. When a continuative condition must be used it should be placed as far from the main flow region as is practical so that any adverse influence on the main flow will be minimal.
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# How to approach dynamics problems?
• cancerman1
In summary, the setup for this problem is different than a typical kinematics problem, but the methodology for solving it is the same.
#### cancerman1
Thread moved from the technical forums to the schoolwork forums
TL;DR Summary: How do I approach the setup of this problem? It seems very different than a setup for a kinematics problem
I'm self studying first year mechanics and am having a hard time with the following problem (screenshot attached). The example is from Intro to Mechanics by K&K.
I'm clear on the steps for setting up and solving kinematics problems (i.e. isolate mass, draw force diagrams, define coordinate system, write equations of motion, etc.), but dynamics problems seem very different. I'd appreciate guidance on how to think about these types of problems
#### Attachments
• physics 4.7.png
39.8 KB · Views: 39
Study a bit further and learn the Hamilton formalism, and everything gets a lot simpler ;-).
Is that really in order at that point in the K&K curriculum ?
PeroK
The issue is that there is more than one way of setting up this dynamics problem, and neither is necessarily preferred over the other. So let's try it your usual way, and see how it can be converted to their way.
The tension in the spring at any time is given by ##T=k(r_b-r_a-l)##where ##l=r_{b0}-r_{a0}## where the subscript 0 represent the values at time zero. If we do a force balance on each of the blocks individually, we have $$m\frac{d^2r_a}{dt^2}=k(r_b-r_a-l)$$and$$m\frac{d^2r_b}{dt^2}=-k(r_b-r_a-l)$$These are subject to the initial conditions that ##\frac{dr_a}{dt}=v_0## and ##\frac{dr_b}{dt}=0## at t=0. Ok so far?
MatinSAR, PeroK, cancerman1 and 1 other person
Yes, thank you - your starting point makes it so much clearer versus example 4.7 which starts with the center of mass.
I can logically reason why the setup would be as noted above - a force is applied to block a, and then blocks a and b accelerate from rest but is connected by a spring so that force needs to be accounted for...
cancerman1 said:
Yes, thank you - your starting point makes it so much clearer versus example 4.7 which starts with the center of mass.
I can logically reason why the setup would be as noted above - a force is applied to block a, and then blocks a and b accelerate from rest but is connected by a spring so that force needs to be accounted for...
So we're done here, or you'd like me to continue?
MatinSAR
Yes, this should be good for this specific problem.
To generalize a methodology and approach, it's best to start with forces acting on discrete systems and writing up force equations - which is not dissimilar to how one would solve a kinematics problem... would you say that's correct?
cancerman1 said:
Yes, this should be good for this specific problem.
To generalize a methodology and approach, it's best to start with forces acting on discrete systems and writing up force equations - which is not dissimilar to how one would solve a kinematics problem... would you say that's correct?
I would have used the CoM reference frame automatically. I wouldn't have given it a second thought.
That immediately makes the problem symmetric. Conservation of momentum does the rest.
MatinSAR, BvU, vanhees71 and 2 others
PeroK said:
I would have used the CoM reference frame automatically. I wouldn't have given it a second thought.
That immediately makes the problem symmetric. Conservation of momentum does the rest.
That's helpful framing. I better understand the rationale for the setup in this example. Appreciate it
vanhees71 and PeroK
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Java-Gaming.org Hi !
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Isometric Grid - Tiles as objects (Read 3457 times) 0 Members and 1 Guest are viewing this topic.
jh62
Junior Devvie
Medals: 3
Projects: 1
« Posted 2013-05-30 05:43:01 »
Hi!
I'm trying to learn about Isometric projection and so far I understand the concept, but the mathematics I saw are a little confusing. Seems like everybody has it's own way of doing them: some are complex, some are simple...
I came up with my own way of drawing into a isometric grid, but the real challenge is mapping the mouse coords to the individual cells in the grid.
I never saw this done (don't know why), but I made every tile on my grid an object with a polygon surrounding it's boundaries and I think is pretty simple and straightforward, because I only need to check if mouseX & mouseY are inside them.
Don't know about a full game using this method, but so far it's fast and reliable.
I'm still trying to learn the "usual" way to do it, anyway...
hellrazer21
Junior Devvie
« Reply #1 - Posted 2013-05-30 06:04:43 »
well i know one way of doing it is to use libgdx
heres an example https://github.com/libgdx/libgdx/blob/d0de5843edd4c9c4903188b203a9a3e098840b12/tests/gdx-tests/src/com/badlogic/gdx/tests/TiledMapAssetManagerTest.java
if your making it from scratch i cant help much with that
actual
JGO Coder
Medals: 25
« Reply #2 - Posted 2013-05-30 06:05:54 »
Quote
... I think is pretty simple and straightforward...
I think you answered your own question :).
Making each tile an instance of a class makes sense and it looks like you have a reasonable and workable solution so I would stick with what you have unless some issue (performance, complexity, etc) forces you to change.
If you are still wondering if you have the right approach, it is worth thinking through a couple of scenarios that you plan for your game to see how well your current set up will handle it.
1. Will the player be able to rotate the board as a whole and/or zoom in and out? If so, do you think your current set up will be able to handle it?
2. When you have things on the tiles (units, buildings, whatever), will your setup be able to handle that. For instance, if the top of a building overlaps with a title and the player hovers over it, will your setup be able to detect that?
As an aside, I like the idea of the volume of the tile being highlighted as you hover over it (maybe as a colored translucent cube borderless cube).
Games published by our own members! Check 'em out!
jh62
Junior Devvie
Medals: 3
Projects: 1
« Reply #3 - Posted 2013-05-30 06:10:29 »
As an aside, I like the idea of the volume of the tile being highlighted as you hover over it (maybe as a colored translucent cube borderless cube).
That's how X-COM: UFO Defense does it, hehe. Not my idea.
Thanks for the input. It's too early to tell what you asked, since I'm first trying to master everything I need to do for my game before actually doing it.
About the zoom: with AffineTransform i can easilly zoom in and out and everything keeps in place. I was messing around with it yesterday and it's awesome. I'll have to check it out with an actual game running.
I'll make a similar game as UFO Defense, just for the sake of practice. It's an awesome game.
jonjava
« Reply #4 - Posted 2013-05-30 15:16:38 »
There's no "usual" way. There are just ways that work and don't work. What's key is how you separate logic, data and graphics. And two of those three have nothing to do with isometric projections at all.
jh62
Junior Devvie
Medals: 3
Projects: 1
« Reply #5 - Posted 2013-05-31 04:44:49 »
There's no "usual" way. There are just ways that work and don't work. What's key is how you separate logic, data and graphics. And two of those three have nothing to do with isometric projections at all.
I just got curious because I didn't see this simpler approach done in the many tutorials I've seen on the internet.
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## 52081
52,081 (fifty-two thousand eighty-one) is an odd five-digits prime number following 52080 and preceding 52082. In scientific notation, it is written as 5.2081 × 104. The sum of its digits is 16. It has a total of 1 prime factor and 2 positive divisors. There are 52,080 positive integers (up to 52081) that are relatively prime to 52081.
## Basic properties
• Is Prime? Yes
• Number parity Odd
• Number length 5
• Sum of Digits 16
• Digital Root 7
## Name
Short name 52 thousand 81 fifty-two thousand eighty-one
## Notation
Scientific notation 5.2081 × 104 52.081 × 103
## Prime Factorization of 52081
Prime Factorization 52081
Prime number
Distinct Factors Total Factors Radical ω(n) 1 Total number of distinct prime factors Ω(n) 1 Total number of prime factors rad(n) 52081 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 10.8606 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 52,081 is 52081. Since it has a total of 1 prime factor, 52,081 is a prime number.
## Divisors of 52081
2 divisors
Even divisors 0 2 2 0
Total Divisors Sum of Divisors Aliquot Sum τ(n) 2 Total number of the positive divisors of n σ(n) 52082 Sum of all the positive divisors of n s(n) 1 Sum of the proper positive divisors of n A(n) 26041 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 228.213 Returns the nth root of the product of n divisors H(n) 1.99996 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 52,081 can be divided by 2 positive divisors (out of which 0 are even, and 2 are odd). The sum of these divisors (counting 52,081) is 52,082, the average is 26,041.
## Other Arithmetic Functions (n = 52081)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 52080 Total number of positive integers not greater than n that are coprime to n λ(n) 52080 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 5322 Total number of primes less than or equal to n r2(n) 8 The number of ways n can be represented as the sum of 2 squares
There are 52,080 positive integers (less than 52,081) that are coprime with 52,081. And there are approximately 5,322 prime numbers less than or equal to 52,081.
## Divisibility of 52081
m n mod m 2 3 4 5 6 7 8 9 1 1 1 1 1 1 1 7
52,081 is not divisible by any number less than or equal to 9.
• Arithmetic
• Prime
• Deficient
• Polite
• Prime Power
• Square Free
## Base conversion (52081)
Base System Value
2 Binary 1100101101110001
3 Ternary 2122102221
4 Quaternary 30231301
5 Quinary 3131311
6 Senary 1041041
8 Octal 145561
10 Decimal 52081
12 Duodecimal 26181
20 Vigesimal 6a41
36 Base36 146p
## Basic calculations (n = 52081)
### Multiplication
n×i
n×2 104162 156243 208324 260405
### Division
ni
n⁄2 26040.5 17360.3 13020.2 10416.2
### Exponentiation
ni
n2 2712430561 141266096047441 7357279548246774721 383174476152240274244401
### Nth Root
i√n
2√n 228.213 37.3445 15.1067 8.77679
## 52081 as geometric shapes
### Circle
Diameter 104162 327235 8.52135e+09
### Sphere
Volume 5.91734e+14 3.40854e+10 327235
### Square
Length = n
Perimeter 208324 2.71243e+09 73653.7
### Cube
Length = n
Surface area 1.62746e+10 1.41266e+14 90206.9
### Equilateral Triangle
Length = n
Perimeter 156243 1.17452e+09 45103.5
### Triangular Pyramid
Length = n
Surface area 4.69807e+09 1.66484e+13 42524
## Cryptographic Hash Functions
md5 6dbe523032e74da51d47dd44aa7ed477 a3132450368b4ab5420dd187afb6ba642019a202 bd8f994314a41ed7882e5a54abc1a52c8a6313b2cb49c0016f3e409dd69bce7b efaaf7a249bc7226820b84dd8b9122a6527c32bd4c9e08ed74bff51f8dee291450c9fd3e7b910227623e64357d4ee556f44ed66212bad8a436422541d45f4c1e 043021d9079fdbd64eb0b2e57c151b58e8f1ea63
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# Find the value of a for which (x - 1) is a factor of the polynomial (a2x3 - 4ax + 4a - 1).
If (x - 1) is a factor of the polynomial (a2x3 - 4ax + 4a - 1).
then it must satisfy it.
So, putting x = 1 the polynomial must be zero.
Putting x = 1 and equating to zero.
= (a2(1)3 - 4a(1) + 4a - 1)
= a2 - 4a + 4a - 1 = 0
= a2 = 1
So, a = �1.
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Andrew - 1 year ago 63
R Question
mean centering based on conditions
I am quite new to R and have a small issue. I have large time-stamped data (dataframe called 'EURO') for a number years where I have calculated returns and the 'Time' variable (a sample is below);
``````Time Returns
9 5
10 5
11 8
12 7
13 7
14 2
15 -4
16 6
17 8
9 5
10 7
11 4
12 6
13 1
14 0
15 -4
16 7
17 -4
9 8
10 1
11 1
12 6
13 8
14 8
15 4
16 -6
17 7
``````
I have used the following code to work out the mean return of each individual timestamp using;
``````Timeaverages <- aggregate(Returns~Time,EURO,mean)
``````
From that I have the average returns for each timestamp. However now I want to mean center my data, that is, minus the average return for each timestamp. For instance, if the average return for the time 9 is 6, so the meancentred return will be 5-6 = -1. For the time 10, the average return is 5, so the meancentred return will be 2 etc.
``````Time Returns Meancentered
9 5 -1
10 7 2
11 7 3
12 6 0
13 6 1
14 4 0
15 -3 -2
16 5 3
17 9 5
9 5 -1
10 7 2
11 4 0
12 6 0
13 1 -4
14 0 -4
15 -4 -3
16 7 5
17 -4 -8
9 8 2
10 1 -4
11 1 -3
12 6 0
13 8 3
14 8 4
15 4 5
16 -6 -8
17 7 3
``````
Any help much appreciated
If I've understood correctly you're looking for `merge()`.
``````merge.df <- merge(EURO, Timeaverages, by="Time", suffixes = c(".ind", ".mean"))
# Time Returns.ind Returns.mean
#1 9 5 6.000000
#2 9 5 6.000000
#3 9 8 6.000000
#4 10 1 4.333333
#5 10 5 4.333333
# etc
``````
Now just get the difference between the two returns columns and `cbind` it to `merge.df`.
``````diff <- merge.df\$Returns.ind - merge.df\$Returns.mean
cbind(merge.df, diff)
# Time Returns.ind Returns.mean diff
#1 9 5 6.000000 -1.0000000
#2 9 5 6.000000 -1.0000000
#3 9 8 6.000000 2.0000000
#4 10 1 4.333333 -3.3333333
#5 10 5 4.333333 0.6666667
#6 10 7 4.333333 2.6666667
#7 11 1 4.333333 -3.3333333
``````
If you use `dplyr`, you can do the whole lot in one chain by grouping, mutating, ungrouping and mutating again.
``````library(dplyr)
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1985BC4
# 1985BC4 - 1‘735— 5c'1‘ 4 Given the differential...
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Unformatted text preview: 1‘735— 5c. '1‘ 4. Given the differential equation % = :32: , y > 0. (a) Find the general solution of the differential equation. (b) Find the solution that satisfies the condition that y = .92 when x = 0. Express your answer in the form y =f(x). (0) Explain why x 2 2 is not in the domain of the soiution found in part (b). a.) 5%dli = 5"de q ij>o 3— Uml __'z ‘1 .1 —3:3—— %—+k=>(£~le—-x +6. 3.&.. M11: t fi‘xa K _ e—x‘ ‘ -c.-x" 0__ lJ-e. or“ mgr—e. b)x=o a; i=efi: and c.=‘-I- x=O)if-e‘ c.) I-F #431, i1=em I Since. M(0=O, dig; -‘- i??— is not aleizineoi when vj=l .'. Xia— is not ’In +‘ne. domain o‘F ‘er Soiutioh 1985 BC4 ...
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# The function $f(x)=[x]^2-[x^2]$ where $[x]$ is the greatest integer less than or equal to x, is discontinuous at
$\begin{array}{1 1}(a)\;\text{all integers}\\(b)\;\text{all integers except 0 and 1}\\(c)\;\text{all integers except 0}\\(d)\;\text{all integers except 1}\end{array}$
We have $f(x)=[x]^2-[x^2]$
At $x=0$
LHL=$\lim\limits_{h\to 0}f(-h)=\lim\limits_{h\to 0}[-h^2]$
$\Rightarrow \lim\limits_{h\to 0}f(-1)^2-[h^2]=\lim\limits_{h\to 0}1-0=1$
RHL=$\lim\limits_{h\to 0}f(h)=\lim\limits_{h\to 0}[h]^2-[h^2]$
$\Rightarrow \lim\limits_{h\to 0}0-0=0$
LHL $\neq$ RHL
$f(x)$ is not continuous at $x=0$
At $x=1$
LHL=$\lim\limits_{h\to 0}f(1-h)=\lim\limits_{h\to 0}[1-h]^2-[(1-h)^2]$
$\Rightarrow \lim\limits_{h\to 0}1-1=0$
$f(1)=[1]^2-[1^2]=1-1=0$
LHL=RHL=f(1)
$f(x)$ is continuous at $x=1$
Clearly at other integral points f(x) is not continuous.
Hence (d) is the correct answer.
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# Results forword problems kinder
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To use this resource, print out word problem task cards and have them laminated. The students will solve the word problem on the task cards and record their answer on their recording sheet. The students will match the number in the top left corner to the number on the recording sheet and that is wh
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To use this resource, print out word problem task cards and have them laminated. The students will solve the word problem on the task cards and record their answer on their recording sheet. The students will match the number in the top left corner to the number on the recording sheet and that is wh
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This is a set of 30 math worksheets for first grade. They will give your students structured practice in solving addition and subtraction word problems. This is a tricky skill because it involves both math and reading comprehension skills. Each problem in this pack involves addition or subtraction w
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Story Problems take on a whole new dimension when students are asked to fill in a part-part-whole, or number bond. Knowing where each number goes helps with the comprehension of the word problem.Usually, the reason students miss story problems is because they aren't sure what part of a solution sent
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Looking to introduce word problem to your kindergarteners? This mini-book is a perfect way to start your students working with word problems. The format and visuals allow students to see what the key words. There is a fill in the blank number sentence on each page. Word problems can be tricky for yo
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This packet includes 25 print and go, addition and subtraction worksheets for numbers 0-10. Packet Includes: -Word Problems- 16 worksheets(4 each: Addition 0-5, Subtraction 0-5, Addition 0-10, & Subtraction 0-10) -Add & Subtract with picture counters - 2 worksheets -Picture Addition and S
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Check out this math mini-book that practices word problems in a simple format. This mini-book focuses word problems that are addition double facts. Your kindergarten students will learn to focus on the key words that signal to add. Then they will write the number sentence. As with all of Kinder Lake
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Word problems are powerful tools for developing mathematical skills and problem solving. Use these easily differentiated, editable problems that allow your students to choose number sets. These word problems focus on fair share (intro to multiplication and division) and have 3 types. This resource
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20 Math word problems for your kinder class! One problem per page to leave room for students to draw out their answers and explain their thinking. This goes along with the Cognitively Guided Instruction approach to teaching math. Your students will solve answers to the following: 10-4, 10-5, 10-3,
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This is an awesome word problem pack! It includes 12 different types of questions!! Word Problems included in The ULTIMATE Word Problem Pack Set 2: Addend Unknown (#1-10) – 4 pages Take From/Change Unknown (#1-10) - 5 pages Compare/Difference Unknown (#1-20) Compare/Bigger Unknown (#1-20) Compare
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This is a great way to practice adding and subtracting using a hundreds chart and ten frames. Included in this 51 page set is 12 addition problems (with sums up to 20) and 12 subtraction problems (with the largest number being 20) using the hundreds chart and ten frames. There are 24 different wor
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This is a great way to practice adding and subtracting using a number line. Included in this 25 page set is 12 addition problems (with sums up to 20) and 12 subtraction problems (with the largest number being 20). It also includes a cover sheet in case you want to make these sheets into a workbook
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These 24 task cards will give your students some much needed practice with addition and subtraction word problems. These bug themed task cards provides students a chance to practice addition and subtraction word problems, utilizing knowledge of keywords such as "in all", "total", "remain", and "how
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Includes ten word problems with unknowns (missing parts) with sums and differences equaling to 15 or lower. Two Options: 1. Hang word problems around the room. Pass out the answers. Students will find the word problem that matches their answer. 2. Hand word problems around the room. Students wil
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This spooky math resource is perfect for the month of October & the week of Halloween. These Halloween math problems can be used in centers, small groups, math journals, Problem of the Day, for a scavenger hunt, or during calendar time/morning meeting. 24 challenging problems are included, with
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Addition Math Story Problems Addition to 20 ** FOCUSES ON ZOO UNIT ** Read It, Draw A Picture, Answer It, Write the Equation and Explain how you solved the problem! This gives students a chance to EXPLAIN how they solved the problem and found the answer! This set does not have a math problem that
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This product is meant to give your students lots of practice with problem solving with these fun subtraction word problems. All problems support the Common Core State Standards- K.OA.1 and K.OA.2 Students have fun practicing their subtraction skills with these cute word problems. They show their th
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Make practicing math fun and engaging with these Story Problem Practice Pages. Each page includes: Simple story problem Space for students to draw a picture while figuring out the answer Number line Area for students to write an equation One box for the number answer One box for the answer to be
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This is an awesome word problem pack! It includes 12 different types of questions!! Word Problems included in The ULTIMATE Word Problem Pack Set 1: Addend Unknown (#1-10) – 4 pages Take From/Change Unknown (#1-10) - 5 pages Compare/Difference Unknown (#1-10) - 5 pages Compare/Bigger Unknown (#1-10
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Addition Math Story Problems Addition to 20 ** FOCUSES ON A BACK TO SCHOOL THEMED UNIT ** Read It, Draw A Picture, Answer It, Write the Equation and Explain how you solved the problem! This gives students a chance to EXPLAIN how they solved the problem and found the answer! This set does not have a
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Math Story Problems…Addition to 5 Read It, Draw A Picture, Answer It, Write the Equation! This activity can be used as a lesson activity, used for morning work or at a math center. Students read the math story problem and draw the picture to represent the math problem. Then, they answer the pro
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A variety of word problems for kinder - third graders These can be used as a center, whole group, or one on one practice.
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This word problem book is designed to help your kindergarten students with subtracting one or one less. The cute frog theme introduces students to solving math word problems. The already familiar format allows students to learn key subtraction words without being intimidated with too many words on a
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This is a set of 10 story problems created for St. Patrick's Day. They are all simple addition and subtraction problems, and include a place to draw a picture, write a number sentence, and write words for the answer.
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1. ## poisson probability distribution
Im not quite sure about this can someone please give me a hint? Cars arriving for gasoline at a shell station follow a poisson distribution with a mean of 5 per hour. what is the probability that over the next hour only one car will arrive.
2. Originally Posted by jobear
Im not quite sure about this can someone please give me a hint? Cars arriving for gasoline at a shell station follow a poisson distribution with a mean of 5 per hour. what is the probability that over the next hour only one car will arrive.
Let X be the random variable number of cars arriving in one hour.
Substitute $\displaystyle \mu = 5$ into the Poisson formula:
$\displaystyle \Pr(X = x) = \frac{e^{-5} \, 5^x}{x!}$.
You want Pr(X = 1).
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# What are the odds of winning HQ?
Jan 2, 2018 · 4 min read
If you haven’t heard of HQ, the free live trivia game show where you compete for real money on your phone, stop reading this and download it now. Sure to be one of the biggest apps of the coming year, when playing it for the first time you may find yourself wondering how anyone knew the answer to the question you got eliminated on. With questions that specific, you might assume other people are just guessing. But if you don’t know the answers to any of the questions, what are the odds you’ll win completely by chance? Alternatively, how many people would you need in one room to try every combination of answers and win every time? Using some simple probability and math, we can answer these questions, and start to better understand the kind of luck that underlies this game of skill.
Each game of HQ is a series of twelve multiple choice questions with three possible answers for each question. A single wrong answer gets you eliminated, so each question needs to be correct independent of the others. You only have ten seconds to answer each question, effectively eliminating the possibility of having Google do the thinking for you. With the wide range of topics and idiosyncratic details (the so-called “nitty-gritty”), many rounds rapidly devolve into a guessing game. Keeping this in mind, how hard is it to win the game by guessing?
The questions in HQ are independent; getting a correct answer at random has no impact on the outcome of the next question. When dealing with independent events, statisticians multiply the probability of one event occurring by the probability of the other event occurring. For example, flipping a coin once gives you a 1/2 chance of the toss landing on heads. Flipping the coin a second time will produce the same probability of getting heads or tails. From this, we can say that the probability of getting heads twice is 1/4. Adding additional coin tosses decreases the probability at a correspondingly increasing rate.
We can use the same logic when discussing HQ. If given a question with options A, B, or C and you have no information to influence your decision, you have a 1/3 chance of getting the question correct if you pick a letter at random. If you were seriously unlucky and had to guess on every single question, your odds of winning with this strategy are a whopping 1 in 531,441.
# 1 in 531,441
Your odds get even worse on Sunday, when adding three more questions brings the odds to an utterly impossible 1 in 14,348,907. While these odds may make you not want to play, we know this isn’t a fair representation of the game. Most people have at least some inkling of the answer for many questions, and the first two questions are usually deliberately very easy. Assuming you know the first two questions and a handful of others (say, for example, six answers), your odds improve, but are still pretty long. The odds of guessing even six of HQ’s twelve questions are a disheartening 1 in 729.
But what if you want to take guessing out of the equation entirely? And knowledge, for that matter? How many friends do you need to work with to make sure you always win? When we discussed the odds of guessing the combination of correct answers yourself, we saw that there’s only one correct set of winning answers. Conversely, there are 531,440 possible incorrect combinations that you can hit upon. In order to be guaranteed a win, you would need one person to try every one of those 531,441 combinations in order to be certain you would hit the right one. Dividing your group of half a million friends into three groups, each told to try either A, B, or C, and re-dividing the remaining players after each question, you will eventually converge on a single winning combination. After the first question, you would be left with around 177,147 players, and so on until one winner remained. The excitement would be short-lived once you realized you and your half a million friends would have to split the prize into fragments of a cent.
Suffice it to say, your chances of winning HQ are slim if you don’t know the topic being discussed in the question. But that doesn’t make it any less exciting. If the odds haven’t discouraged you, click here for some general advice on how to win HQ.
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# 15.6 kg to lbs - 15.6 kilograms into pounds
Before we go to the practice - that is 15.6 kg how much lbs calculation - we want to tell you a little bit of theoretical information about these two units - kilograms and pounds. So we are starting.
How to convert 15.6 kg to lbs? 15.6 kilograms it is equal 34.392112872 pounds, so 15.6 kg is equal 34.392112872 lbs.
## 15.6 kgs in pounds
We are going to begin with the kilogram. The kilogram is a unit of mass. It is a base unit in a metric system, known also as International System of Units (in abbreviated form SI).
From time to time the kilogram can be written as kilogramme. The symbol of the kilogram is kg.
The kilogram was defined first time in 1795. The kilogram was defined as the mass of one liter of water. This definition was not complicated but difficult to use.
Then, in 1889 the kilogram was described using the International Prototype of the Kilogram (in abbreviated form IPK). The International Prototype of the Kilogram was prepared of 90% platinum and 10 % iridium. The IPK was in use until 2019, when it was replaced by another definition.
Nowadays the definition of the kilogram is based on physical constants, especially Planck constant. The official definition is: “The kilogram, symbol kg, is the SI unit of mass. It is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015×10−34 when expressed in the unit J⋅s, which is equal to kg⋅m2⋅s−1, where the metre and the second are defined in terms of c and ΔνCs.”
One kilogram is exactly 0.001 tonne. It could be also divided into 100 decagrams and 1000 grams.
## 15.6 kilogram to pounds
You learned a little bit about kilogram, so now we can move on to the pound. The pound is also a unit of mass. It is needed to emphasize that there are not only one kind of pound. What are we talking about? For example, there are also pound-force. In this article we want to concentrate only on pound-mass.
The pound is in use in the Imperial and United States customary systems of measurements. Of course, this unit is used also in another systems. The symbol of this unit is lb or “.
There is no descriptive definition of the international avoirdupois pound. It is defined as 0.45359237 kilograms. One avoirdupois pound could be divided into 16 avoirdupois ounces or 7000 grains.
The avoirdupois pound was enforced in the Weights and Measures Act 1963. The definition of this unit was given in first section of this act: “The yard or the metre shall be the unit of measurement of length and the pound or the kilogram shall be the unit of measurement of mass by reference to which any measurement involving a measurement of length or mass shall be made in the United Kingdom; and- (a) the yard shall be 0.9144 metre exactly; (b) the pound shall be 0.45359237 kilogram exactly.”
### How many lbs is 15.6 kg?
15.6 kilogram is equal to 34.392112872 pounds. If You want convert kilograms to pounds, multiply the kilogram value by 2.2046226218.
### 15.6 kg in lbs
The most theoretical section is already behind us. In this section we are going to tell you how much is 15.6 kg to lbs. Now you learned that 15.6 kg = x lbs. So it is high time to know the answer. Have a look:
15.6 kilogram = 34.392112872 pounds.
It is an accurate result of how much 15.6 kg to pound. You can also round off the result. After it your result is exactly: 15.6 kg = 34.32 lbs.
You know 15.6 kg is how many lbs, so look how many kg 15.6 lbs: 15.6 pound = 0.45359237 kilograms.
Obviously, in this case it is possible to also round it off. After rounding off your outcome will be exactly: 15.6 lb = 0.45 kgs.
We also want to show you 15.6 kg to how many pounds and 15.6 pound how many kg outcomes in tables. Have a look:
We want to start with a table for how much is 15.6 kg equal to pound.
### 15.6 Kilograms to Pounds conversion table
Kilograms (kg) Pounds (lb) Pounds (lbs) (rounded off to two decimal places)
15.6 34.392112872 34.320
Now see a table for how many kilograms 15.6 pounds.
Pounds Kilograms Kilograms (rounded off to two decimal places
15.6 0.45359237 0.45
Now you learned how many 15.6 kg to lbs and how many kilograms 15.6 pound, so it is time to go to the 15.6 kg to lbs formula.
### 15.6 kg to pounds
To convert 15.6 kg to us lbs a formula is needed. We are going to show you two formulas. Let’s begin with the first one:
Amount of kilograms * 2.20462262 = the 34.392112872 result in pounds
The first formula give you the most correct outcome. In some cases even the smallest difference can be significant. So if you want to get an exact outcome - this version of a formula will be the best for you/option to convert how many pounds are equivalent to 15.6 kilogram.
So let’s move on to the shorer version of a formula, which also enables calculations to learn how much 15.6 kilogram in pounds.
The another version of a formula is down below, look:
Number of kilograms * 2.2 = the outcome in pounds
As you can see, this formula is simpler. It can be better solution if you need to make a conversion of 15.6 kilogram to pounds in easy way, for instance, during shopping. You only need to remember that final outcome will be not so exact.
Now we want to show you these two versions of a formula in practice. But before we are going to make a conversion of 15.6 kg to lbs we want to show you another way to know 15.6 kg to how many lbs without any effort.
### 15.6 kg to lbs converter
Another way to know what is 15.6 kilogram equal to in pounds is to use 15.6 kg lbs calculator. What is a kg to lb converter?
Calculator is an application. Calculator is based on first formula which we gave you in the previous part of this article. Due to 15.6 kg pound calculator you can quickly convert 15.6 kg to lbs. Just enter number of kilograms which you need to convert and click ‘calculate’ button. You will get the result in a second.
So try to convert 15.6 kg into lbs with use of 15.6 kg vs pound converter. We entered 15.6 as an amount of kilograms. Here is the outcome: 15.6 kilogram = 34.392112872 pounds.
As you can see, our 15.6 kg vs lbs calculator is intuitive.
Now let’s move on to our main topic - how to convert 15.6 kilograms to pounds on your own.
#### 15.6 kg to lbs conversion
We will start 15.6 kilogram equals to how many pounds conversion with the first formula to get the most correct result. A quick reminder of a formula:
Number of kilograms * 2.20462262 = 34.392112872 the outcome in pounds
So what need you do to know how many pounds equal to 15.6 kilogram? Just multiply number of kilograms, in this case 15.6, by 2.20462262. It gives 34.392112872. So 15.6 kilogram is equal 34.392112872.
It is also possible to round it off, for example, to two decimal places. It is equal 2.20. So 15.6 kilogram = 34.320 pounds.
It is time for an example from everyday life. Let’s calculate 15.6 kg gold in pounds. So 15.6 kg equal to how many lbs? As in the previous example - multiply 15.6 by 2.20462262. It gives 34.392112872. So equivalent of 15.6 kilograms to pounds, when it comes to gold, is equal 34.392112872.
In this example you can also round off the result. It is the result after rounding off, in this case to one decimal place - 15.6 kilogram 34.32 pounds.
Now we can go to examples calculated using a short version of a formula.
#### How many 15.6 kg to lbs
Before we show you an example - a quick reminder of shorter formula:
Number of kilograms * 2.2 = 34.32 the result in pounds
So 15.6 kg equal to how much lbs? As in the previous example you need to multiply amount of kilogram, in this case 15.6, by 2.2. Look: 15.6 * 2.2 = 34.32. So 15.6 kilogram is 2.2 pounds.
Let’s make another conversion using this version of a formula. Now calculate something from everyday life, for instance, 15.6 kg to lbs weight of strawberries.
So convert - 15.6 kilogram of strawberries * 2.2 = 34.32 pounds of strawberries. So 15.6 kg to pound mass is equal 34.32.
If you know how much is 15.6 kilogram weight in pounds and are able to calculate it using two different formulas, we can move on. Now we want to show you all results in tables.
#### Convert 15.6 kilogram to pounds
We realize that results presented in tables are so much clearer for most of you. We understand it, so we gathered all these outcomes in tables for your convenience. Thanks to this you can easily make a comparison 15.6 kg equivalent to lbs outcomes.
Begin with a 15.6 kg equals lbs table for the first version of a formula:
Kilograms Pounds Pounds (after rounding off to two decimal places)
15.6 34.392112872 34.320
And now see 15.6 kg equal pound chart for the second version of a formula:
Kilograms Pounds
15.6 34.32
As you see, after rounding off, when it comes to how much 15.6 kilogram equals pounds, the outcomes are not different. The bigger amount the more significant difference. Remember it when you need to do bigger amount than 15.6 kilograms pounds conversion.
#### How many kilograms 15.6 pound
Now you know how to convert 15.6 kilograms how much pounds but we want to show you something more. Do you want to know what it is? What do you say about 15.6 kilogram to pounds and ounces conversion?
We are going to show you how you can calculate it little by little. Begin. How much is 15.6 kg in lbs and oz?
First things first - you need to multiply number of kilograms, this time 15.6, by 2.20462262. So 15.6 * 2.20462262 = 34.392112872. One kilogram is 2.20462262 pounds.
The integer part is number of pounds. So in this case there are 2 pounds.
To calculate how much 15.6 kilogram is equal to pounds and ounces you have to multiply fraction part by 16. So multiply 20462262 by 16. It is exactly 327396192 ounces.
So final result is exactly 2 pounds and 327396192 ounces. You can also round off ounces, for instance, to two places. Then final outcome is 2 pounds and 33 ounces.
As you can see, conversion 15.6 kilogram in pounds and ounces simply.
The last conversion which we are going to show you is calculation of 15.6 foot pounds to kilograms meters. Both foot pounds and kilograms meters are units of work.
To convert foot pounds to kilogram meters you need another formula. Before we give you this formula, look:
• 15.6 kilograms meters = 7.23301385 foot pounds,
• 15.6 foot pounds = 0.13825495 kilograms meters.
Now see a formula:
Number.RandomElement()) of foot pounds * 0.13825495 = the outcome in kilograms meters
So to calculate 15.6 foot pounds to kilograms meters you have to multiply 15.6 by 0.13825495. It is 0.13825495. So 15.6 foot pounds is 0.13825495 kilogram meters.
It is also possible to round off this result, for instance, to two decimal places. Then 15.6 foot pounds will be equal 0.14 kilogram meters.
We hope that this conversion was as easy as 15.6 kilogram into pounds conversions.
This article was a big compendium about kilogram, pound and 15.6 kg to lbs in conversion. Thanks to this calculation you know 15.6 kilogram is equivalent to how many pounds.
We showed you not only how to make a calculation 15.6 kilogram to metric pounds but also two other calculations - to check how many 15.6 kg in pounds and ounces and how many 15.6 foot pounds to kilograms meters.
We showed you also other way to make 15.6 kilogram how many pounds calculations, it is using 15.6 kg en pound converter. It is the best choice for those of you who do not like converting on your own at all or need to make @baseAmountStr kg how lbs calculations in quicker way.
We hope that now all of you are able to do 15.6 kilogram equal to how many pounds conversion - on your own or with use of our 15.6 kgs to pounds converter.
So what are you waiting for? Calculate 15.6 kilogram mass to pounds in the way you like.
Do you want to do other than 15.6 kilogram as pounds conversion? For instance, for 15 kilograms? Check our other articles! We guarantee that conversions for other numbers of kilograms are so easy as for 15.6 kilogram equal many pounds.
### How much is 15.6 kg in pounds
We want to sum up this topic, that is how much is 15.6 kg in pounds , we prepared one more section. Here we have for you all you need to know about how much is 15.6 kg equal to lbs and how to convert 15.6 kg to lbs . Let’s see.
What is the kilogram to pound conversion? The conversion kg to lb is just multiplying 2 numbers. How does 15.6 kg to pound conversion formula look? . It is down below:
The number of kilograms * 2.20462262 = the result in pounds
So what is the result of the conversion of 15.6 kilogram to pounds? The correct answer is 34.392112872 pounds.
There is also another way to calculate how much 15.6 kilogram is equal to pounds with second, easier version of the formula. Let’s see.
The number of kilograms * 2.2 = the result in pounds
So this time, 15.6 kg equal to how much lbs ? The result is 34.392112872 lb.
How to convert 15.6 kg to lbs in an easier way? It is possible to use the 15.6 kg to lbs converter , which will do whole mathematical operation for you and you will get a correct answer .
#### Kilograms [kg]
The kilogram, or kilogramme, is the base unit of weight in the Metric system. It is the approximate weight of a cube of water 10 centimeters on a side.
#### Pounds [lbs]
A pound is a unit of weight commonly used in the United States and the British commonwealths. A pound is defined as exactly 0.45359237 kilograms.
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+0
# Coordinates
0
176
1
A point (x,y) is a distance of 6 units from the x-axis. It is a distance of sqrt(58) units from the point (8,3). It is a distance sqrt(n) from the origin,. Given that x < 8, what is n?
Jul 22, 2022
### 1+0 Answers
#1
+125693
+1
Costruct a circle with a center of (8,3) and a radius of sqrt 58
The equation of this circle is
(x - 8)^2 + ( y - 3)^2 = 58
Our point must be on this circle and is 6 units from the x axis (y coord of the point = 6)
So
(x - 8)^2 + (6 -3)^2 = 58
(x - 8)^2 + 9 = 58
(x - 8)^2 = 49
x - 8 = -sqrt (49)
x = 8 - 7 = 1
n = 1^2 + 6^2 = 37
Jul 22, 2022
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# NCERT Solutions for Class 10 Maths, Chapter 5 - Arithmetic Progression
Class 10 Maths Chapter 5 – Arithmetic Progressions is one amongst the most important chapters that carries a total weightage of 20 marks in the board exam. NCERT Class 10 – Arithmetic Progressions chapter discusses patterns in which succeeding terms are obtained by adding a fixed number to the preceding terms. In this chapter, you will also study how to find nth terms and the sum of n consecutive terms, and use this knowledge in solving daily life problems. This chapter covers some of the core Arithmetic Progressions (AP) topics such as nth Term of an AP, Sum of First n Terms of an AP etc.
Let’s take a look at the topics and sub-topics of Arithmetic Progressions Class 10
5.1 Introduction
5.2 Arithmetic Progressions
5.3 nth Term of an AP
5.4 Sum of First n Terms of an AP
5.5 Summary
Arithmetic Progressions, Class 10 NCERT Solutions
NCERT Class 10 Maths Chapter 5 – Arithmetic Progressions, gives you exposure to a wide range of exercises and problems. Besides, several pictorial representations solved examples & solutions, and hints are given in between exercises (at regular intervals) to make the concepts clearer for you to understand, as well as to make it easy for you to solve the exercises. Towards the end, some of the basic points you will learn in this chapter include – the fixed number d is called the common difference, and it can be positive, negative and zero; the nth term of an AP is given by a(n)= a + (n-1)d, where a is the first term of AP series and n is the number of terms of the series; each finite series has the last term, unlike infinite APs, etc.
To know more about Chapter 5, Maths Class 10, you should explore the resources & exercises. Also, refer to Arithmetic Progressions Class 10 NCERT Solutions PDF below, solved by expert maths trainers.
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# How do you write specific heat mathematically?
$Q = c \cdot m \cdot \Delta T$, where $Q$ is the amount of heat gained or lost, $c$ is the specific heat capacity, $m$ is mass, and $\Delta T$ is change in temperature.
$c = \frac{Q}{m \cdot \Delta T}$
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How to find the mean of 12 different 3-D arrays containing nan so that I can omitnan in the final calculations?
5 views (last 30 days)
I have 12 different 3-D arrays, say A, B, C, D, E, F, G, H, I, J, K, L, each of them having the dimension of 60x57x372. All of them contain several NaN values. I want to calculate the mean of these arrays to get an array of same dimension, that is of 60x57x372. How to do that?
madhan ravi on 30 May 2021
Use mean() with "Omitnan" flag
help mean
MEAN Average or mean value. S = MEAN(X) is the mean value of the elements in X if X is a vector. For matrices, S is a row vector containing the mean value of each column. For N-D arrays, S is the mean value of the elements along the first array dimension whose size does not equal 1. MEAN(X,'all') is the mean of all elements in X. MEAN(X,DIM) takes the mean along the dimension DIM of X. MEAN(X,VECDIM) operates on the dimensions specified in the vector VECDIM. For example, MEAN(X,[1 2]) operates on the elements contained in the first and second dimensions of X. S = MEAN(...,OUTTYPE) specifies the type in which the mean is performed, and the type of S. Available options are: 'double' - S has class double for any input X 'native' - S has the same class as X 'default' - If X is floating point, that is double or single, S has the same class as X. If X is not floating point, S has class double. S = MEAN(...,NANFLAG) specifies how NaN (Not-A-Number) values are treated. The default is 'includenan': 'includenan' - the mean of a vector containing NaN values is also NaN. 'omitnan' - the mean of a vector containing NaN values is the mean of all its non-NaN elements. If all elements are NaN, the result is NaN. Example: X = [1 2 3; 3 3 6; 4 6 8; 4 7 7] mean(X,1) mean(X,2) Class support for input X: float: double, single integer: uint8, int8, uint16, int16, uint32, int32, uint64, int64 See also MEDIAN, STD, MIN, MAX, VAR, COV, MODE. Documentation for mean doc mean Other functions named mean codistributed/mean duration/mean tall/mean datetime/mean fints/mean timeseries/mean dlarray/mean gpuArray/mean
3 CommentsShow 1 older commentHide 1 older comment
madhan ravi on 30 May 2021
Concatenate those 12 matrices along the 4th dimension and mean() it along the 4th dimension with "omitnan" flag.
Abhishek Chakraborty on 30 May 2021
Concatenate=cat(4,A,B,C,D,E,F,G,H,I,J,K,L);
MEANALL=mean(Concatenate,4,'omitnan');
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# What is Stereology?
The Greek root ‘stereo’ means solids; stereology is the science of studying solids. It is simply the study of estimating geometrical quantities. There are four cardinal quantities that are explored in first-order stereology.
1. Number—The number of things, also known as a population, is a basic quantity that is often of interest to researchers. Determining the numbers of objects is not as simple as it may seem.
2. Length—Length is a concept that is easy to understand, but not as easy to define formally. Length becomes tricky when considering the length of things that twist and turn in space.
3. Surface—Surface is the same as surface area, but surface is used to refer to the area of a shape that isn’t flat.
4. Volume and Profile area—The volume of something is a measure of how it fills up space. The area of something that appears in a plane is the ‘profile area’ or ‘cross-sectional area’.
Examples include: number of brain cells within a brain region, the length of capillaries, the surface area of a membrane, the volume of a tumor, as well as the percentage of quartz in granite. Information about the relationship among objects can be estimated using second-order stereology. Examples include nearest neighbor information in distributions of cells and connectivity among branching objects. Design-based stereology, also known as unbiased stereology, has improved the reliability and efficiency of quantification in biological research. The sampling procedure, also called the probe, is carefully planned ahead of time to avoid bias, and the amount of sampling is matched to the required precision. In its most efficient form, ‘events’, interactions between a virtual geometric probe and the feature being probed, are counted and used in mathematical formulas that are supported by simple geometrical concepts. For instance, if trying to estimate the amount of, or probing for, surface, it is most efficient to use line segments and count the intersections. Qualitatively, the more events, in this example intersections, counted; the more surface there is. Quantitatively, the relationship between the length of the line-segment-probe and the number of intersections counted to the amount of surface is known, so an estimate can be made. If you follow rules that include making sure the interaction between the line-segment-probe and the surface is isotropic, the estimate of surface area will be unbiased. The amount of probing can be adjusted depending on how regular or homogeneous the characteristic is, and also depending on how large the difference is between or among the means of the groups being studied. The application of design-based stereological methods to the analysis of the central nervous system as well as other organs and tissues has contributed considerably to our understanding of the functional and pathological morphology of the body. Some research fields that employ stereological analysis include:
• Neurobiology
• Cell Biology
• Pathology
• Medicine
• Anatomy
• Ecology
• Forestry
• Metallurgy
• Geology
• Petrology
Some additional definitions of stereology that appear in scientific literature are:
“the science of inferring spatial structure from partial information, usually lower dimensional data, which is usually in the form of sections of projections of the structure of interest” -Sampling Theory for Opaque Spatial Specimens, 1976, Davy, P. and Miles, R.E., J. Roy. Stat. Soc., 39: page 56.
“a body of mathematical methods, relating 3-D parameters defining the structure, to 2-D measurements obtainable on sections of the structure” -Stereology for Statisticians, 2005, Baddeley, A. and Jensen E.B.V., Chapman & Hall/CRC, Boca Raton, page 2.
“the geometrical properties of features in 3-D space can be quantified by ‘throwing’ random geometrical probes, of various dimensions, into the space and recording the way in which they intersect with the structures of interest” -Unbiased Stereology, Second Edition, 2005, Howard, C.V. and Reed, M.G., QTP Publications, Liverpool, page 8.
“Stereology provides meaningful quantitative descriptions of the geometry of 3D structures from measurements that are made on 2D images” -Basic Stereology for Biologists and Neuroscientists, 2012, West, M.J., Cold Spring Harbor Laboratory Press, Cold Spring Harbor, N.Y., page 1.
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If the radius unit vector is giving us some direction in spherical coordinates, why do we need the angle vectors or vice versa?
kuruman
Homework Helper
Gold Member
2021 Award
Think about it. In three dimensions you need three quantities to define a point in space. They could be {x,y,z} or {r,θ,φ}. If I were to tell you that a fly is at position ##1.5~\hat r## meters from the corner of a room (the origin), where would you look to find it?
I mean I suppose in that case I would have to tilt my head by a certain θ and a certain Φ to see the fly but then my question is why the need to have r as a vector? I thought the whole point in spherical coordinates as opposed to cartesian was the fact that r was just the distance from the center and to actually arrive at the required point meant offsetting by a certain θ and a certain Φ.
kuruman
Homework Helper
Gold Member
2021 Award
I mean I suppose in that case I would have to tilt my head by a certain θ and a certain Φ to see the fly but then my question is why the need to have r as a vector?
Because it makes the mathematical description in some cases easier to formulate and manipulate. Take the example of a point charge ##Q## at the origin. The electric field due to this charge can be written in spherical coordinated as$$\vec E(\vec r)=\frac{Q}{4 \pi \epsilon_0}\frac{\vec r}{r^3}.$$The same field in Cartesian coordinates is$$\vec E(x,y,z)=\frac{Q}{4 \pi \epsilon_0}\frac{x~\hat x + y~\hat y+z~\hat z}{(x^2+y^2+z^2)^{3/2}}.$$The top expression is easier to visualize as a radial field and often easier to work with if one has to take dot or cross products with other vectors or use vector calculus.
random_soldier
Pardon my curtness but I think I am now just confused about what exactly the radius unit vector is giving the direction of. May I please know?
kuruman
Homework Helper
Gold Member
2021 Award
Pardon my curtness but I think I am now just confused about what exactly the radius unit vector is giving the direction of. May I please know?
In the example of the electric field that I gave you, imagine a line from the origin where charge ##Q## is located to point ##P## where you want to find the field. The unit vector ##\hat r## is along this line and points from the origin to point ##P##. Its direction of course depends on where ##P## is. You can specify that direction in the Cartesian representation and write $$\hat r=\frac{x~\hat x + y~\hat y+z~\hat z}{(x^2+y^2+z^2)^{1/2}}.$$You tell me where point ##P## is, i.e. you give me ##x##, ##y##, and ##z## and I will be able to draw ##\hat r##. In terms of the standard spherical angles ##\theta## and ##\phi## that give an alternate way to find point ##P##, you have ##\hat r=\sin\theta \cos\phi~\hat x+\sin \theta \sin \phi~\hat y+\cos\theta ~\hat z##. You give me the angles and I will be able to draw ##\hat r##.
random_soldier
Orodruin
Staff Emeritus
Homework Helper
Gold Member
2021 Award
By definition you will never need any other basis vector in spherical coordinates when you express the position vector or any other vectors pointing radially. However, this is not the only possible vector. For example, your fly may be moving non-radially, which will give it a velocity vector with non-radial components. Same thing with the velocity vector of a planet orbiting a star. Since the orbit is typically almost circular, the radial component is small and the tangential component large.
Last edited:
random_soldier
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## Puzzle Test
Puzzle Test Puzzle Test section comprises problems put in the form of puzzles involving certain number of items, be it persons or things. The candidate is required to analyse the given information, condense it in a suitable form and answer the questions asked. In this type of test, the questions are … Read more
## Tips And Tricks For Speedy Calculations Module 4 Division
TIPS AND TRICKS FOR SPEEDY CALCULATIONS – MODULE IV – DIVISION Division is no more tedious as it used to be! In this module we deal with divisibility rules of various numbers, how to find remainders and general division rules. After reading this module and solving problems by heart you may find … Read more
## Median
Median Median is the middle number in a sorted list of numbers. To determine the median value in a sequence of numbers, the numbers must first be arranged in value order from lowest to highest. If there is an odd amount of numbers, the median value is the number that is in the … Read more
## Time & Distance
TIME & DISTANCE In this module we will deal with basic concepts of time and distance, speed, average speed, conversion from km/h to m/s and vice versa. This chapter will form the basis of further concept of relative speed which is used in train and boat problems. Important … Read more
## Pipe And Cistern
Pipes and Cistern Inlet: A pipe connected with a tank or a cistern or a reservoir, that fills it, is known as an inlet. Outlet: A pipe connected with a tank or cistern or reservoir, emptying it, is known as an outlet. If a pipe can fill a tank in x hours, … Read more
## Shrinivas Ramanujan
Shrinivas Ramanujan: Contribution to mathematics Srinivasa Ramanujan, an Indian mathematician was born in 22nd December, 1887 in Madras, India. Like Sophie Germain, he received no formal education in mathematics but made important contributions to advancement of mathematics. His chief contribution in mathematics lies mainly in analysis, game theory and infinite series. He made … Read more
## Relation Between Arc Radius And Angle
Relation between arc radius and angle Arc length = [radius • central angle (radians)] Arc length = circumference • [central angle (degrees) ÷ 360] Proof of the trigonometric ratios of complementary allied angles Two acute angles are complementary to each other if their sum is equal to 90°. In … Read more
## Contribution Of Brahma Gupta
Contribution of Brahma gupta, in mathematics Brahmagupta was born in 598 A.D.in Bhinmal city in the state of Rajasthan. He was a mathematician and astronomer, who wrote many important works on mathematics and astronomy. His best known work is the “Brahmasphuta‐siddhanta”, written in 628 AD in Bhinmal. He was the first to use … Read more
## Mensuration (revised)
MENSURATION Mensuration is the branch of mathematics which deals with the study of different geometrical shapes, their areas and Volume. In the broadest sense, it is all about the process of measurement. It is based on the use of algebraic equations and geometric calculations to provide measurement data regarding the width, depth and volume … Read more
## Multiplying Fractions And Ratio And Proportion
Multiplying Fractions There are 3 simple steps to multiply fractions Multiply the top numbers (the numerators). Multiply the bottom numbers (the denominators). Simplify the fraction if needed. example: 1 / 2 × 2 / 5 =? Step 1 Multiply the top numbers: 1 / 2 × 2 / 5 = … Read more
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# Fundamental group of the $m$-fold suspension of a finite discrete space
Recall that the suspension of a topological space $$X$$ is the space $$SX$$ resulting by identifying $$X\times\{0\}$$ and $$X\times\{1\}$$ to single points of the "cylinder" $$X\times[0,1]$$. Now let $$X_m$$ be a discrete space consisting of $$m$$ points, and denote by $$S^nX$$ the $$n$$-fold suspension, i.e., $$S^nX=SS^{n-1}X$$.
What is the fundamental group of $$S^nX_m$$ for $$n,m\geq1$$?
Here is my attempt. The case $$m=2$$ is easy since $$SX_2$$ is a circle, so the fundamental group is $$\mathbb Z$$, and $$S^2X_2$$ is the sphere, whose fundamental group is trivial. Further, the suspension of the $$n$$-sphere is the $$(n+1)$$-sphere so $$\pi_1(S^nX_2)$$ is trivial for $$n\geq2$$.
Now consider $$m>2$$. Then $$SX_m$$ is homotopy equivalent to an "$$(m-1)$$-fold eight"
$$\underbrace{\bigcirc\hspace{-.1cm}\!\bigcirc\hspace{-.1cm}\!\bigcirc\cdots\bigcirc}_{\text{m-1 circles}}$$
so the fundamental group is $$F_{m-1}$$, the free group on $$m-1$$ generators. If I am not mistaken two homotopy equivalent spaces have homotopy equivalent suspensions, so $$S^2X_m$$ is homotopy equivalent to the wedge product of $$m-1$$ spheres, and consequently $$\pi_1(S^2X_m)$$ is trivial. I am having trouble, however, to visualise the space $$S^nX_m$$ for $$n>2$$. How can $$\pi_1(S^nX_m)$$ be computed?
• I had to delete my answer, because it's actually reduced (or pointed) suspension that preserves wedges, and though for spheres reduced and unreduced suspension agree, I don't see an easy argument as to why it should be the same for wedges of spheres (though it seems to be the case in low dimension so it's probably true; but that's really not a good argument) – Max Jan 11 at 13:44
• Ok, but thanks anyway! – iqcd Jan 11 at 14:17
1. $$m = 1$$: All suspensions of $$X_1$$ are contractible.
2. $$m > 1$$:
2.1. $$n = 1$$: As you stated correctly, $$\pi_1(SX_m) = F_{m-1}$$.
2.2. $$n > 1$$: Let us prove the following theorem.
If $$Y$$ is path connected, then $$SY$$ is simply connected.
This applies to $$Y = S^{n-1}X_m$$.
Let $$p : Y \times I \to SY$$ denote the quotient map. Since $$Y$$ is path connected, so are $$Y \times I$$ and $$SY$$. As a basepoint for $$SY$$ choose $$z_0 = p(y_0,1/2)$$, where $$y_0 \in Y$$ is an arbitrary point. The sets $$U = p(Y \times (0,1])$$ and $$V = p(Y \times [0,1))$$ are open in $$SY$$, and we have $$z_0 \in U \cap V = p(Y \times (0,1)$$ which is pathwise connected. Since $$U$$ and $$V$$ are contractible, we get $$\pi_1(U,z_0) = \pi_1(V,z_0) = 0$$. Now apply the Seifert-van Kampen-theorem to see that $$\pi_1(SY,z_0) = 0$$.
• That's a very useful result, thanks! – iqcd Jan 11 at 14:29
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• anonymous
A car company tested a sports car on a road with different inclines. The test driver tested the car by driving a distance of x miles on a flat road, (x2 + 3) miles downhill, and (x − 7) miles uphill. Which simplified expression is equivalent to the total distance, in miles, for which the car was tested? 3x2 − 4 3x2 + 10 x2 + 2x − 4 x2 + 2x + 10
Algebra
• Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
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# problem 237 mathcounts
label Mathematics
account_circle Unassigned
schedule 1 Day
account_balance_wallet \$5
If x>0, then 2% of 5% of 3x equals what percent of x? Express your answer to the nearest tenth.
Nov 20th, 2017
2% of 5% of 3x = 3x* 5/100*2/100 = 3x*10/10000=3x*1/1000 = x * 3/1000
3/1000 is .3 percent
Oct 5th, 2014
...
Nov 20th, 2017
...
Nov 20th, 2017
Nov 21st, 2017
check_circle
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posted in: Graduate, PostMS | 0
CMSC 740 Fall 2017 is taught by an awesome teacher Prof. Matthias Zwicker
Overall, this class is a fantastic introduction to advanced graphics, ray tracing pipelines, as well as recent advances in deep learning.
# Projects
The course projects are based on Nori, an educational ray tracer.
## Octrees and Ray Traversal
To build an Octree, one could use a stack to mimic the DFS procedure. For non-leaf node, store its 8 children as pointers (8 bytes * 8 = 64 bytes), for leaf node, store a a vector list of N triangle indices (4N bytes).
For leaf node, traverse the triangle sets; for non-leaf node, traverse the inner nodes.
To improve the naive implementation, one could sort the triangles, then conduct the intersection.
OpenMP Notes
## Monte Carlo Sampling
### Sample Warping and corresponding validation
The uniform circle warping is widely used for lots of purposes, so I put it here for reference.
Code removed to prevent plagiarism.
# Notes
## Acceleration structures,
### Bounding Volume Hierarchies (BVHs)
• Partition space into non-overlapping cells
• Each cell stores reference to all objects that overlap it
### Spatial subdivision
• Binary space partitioning (BSP) trees
• Recursively divide space into two parts using
• dividing planes (with arbitrary position, orientation)
• k-d-trees, nearest neighbor search
### Light Transport Effects
• Quantify spatial energy distribution of light, in a way that is compatible with the geometrical optics model
• Physical measurement of electromagnetic energy, Unit: Watts
• Assume light consists of photons with
• Position x
• Direction of motion w
• Wavelength lambda
• Each photo has an energy of hv
• H Planck’s constant
• V = 1 / lambda frequency
• Measuring energy means “counting photons”
• Energy per time per wavelength per solid angle per area
• Means differetial area perpendicular to $\omega$
• Power carried along a ray is measured in radiance
• Radiance along ray is constant (in vacuum)
• In practice: assume steady state, measure at discrete wavelengths R, G, B
• Radiance L: power per solid angle per area, vector of 3 values for R, G, B
• Irradiance: Power per unit area
• Integration of radiance over hemisphere
• Irradiance E: power per area perpendicular to the surface normal
• Radiance L: power per area perpendicular to the incident ray \omega
• When we do ray tracing, we use radiance for rays
• Integration in spherical coordinates
• Radiant intensity I in all direction, has a total power
### Photometry
• Perceptual measurement of perceived brightness
• Lumen
## BRDF (bidirectional reflectance distribution function), or BSDF (bidirectional scattering distribution function)
For every pair of light and viewing directions, BRDF gives fraction of transmitted light
Diffuse reflection: equally distributed reflections
Glossy reflection: majorly reflected to mirror-direction, quickly fall off in a cone
Differential irradiance comes from a cone of directions around incident direction:
• BRDF cannot do volume scattering, because it relies on the surface normal. It’s a definition depending on the surface normal.
• BRDF cannot do diffraction nor polarization.
• BRDFs are wavelength dependent, four-dimensional vectors
Helmoltz reciprocity: can exchange role of light source and observer.
Representations
1. As a table of values over incident and outgoing directions
2. Analytic BRDF models
1. Represent BRDFs as analytic functions
Gonio reflectometer by Cornell University is a device to measure and tabulate BRDFs, a robot with light source and camera, measures reflection for each light/ camera direction
Analytic BRDFs
• Diffuse reflection: reflected radiance is equal over all directions
• Physical model: Lambert’s cosine law
• Diffuse BRDF is just a constant
• Specular reflection and refraction
• Dielectrics (Glass, plastic, diamond, water, air) and conductors (metals)
Index of refraction
• Air 1.00029, Water: 1.33 Acrylic glass: 1.49
Fermat’s principle
• The actual path between two points taken by a beam of light is the one which is traversed in the least time.
Snell’s law
• Ratio of sines of angles of incidence and refraction is equal to opposite ratio of indices of refraction
• $\frac{\sin \theta_1}{\sin \theta_2} = \frac{n_2}{n_1}$
# Torrance-Sparrow model
Vector form to obtain refracted direction r
Fresnel equations describe fraction of intensity of light that is reflected and refracted
Derived by solving the Maxwell equations
Takes into account polarization of light
Different versions for dielectrics and conductors
Full equations for dielectrics see
In graphics, often use Schlick’s approximation
Refraction coefficient is
OK for both dielectrics and conductors
Specular BRDF
Micro facet models for glossy reflection
• Assume surface consists of randomly oriented small mirrors (micro facets)
• Statistical distribution of micro facet orientations determines appearance, glossiness / shininess of reflection
Torrance-Sparrow model
• Physically-based model for specular highlight due to micro facet distribution
• Parameter: probability distribution over micro facet orientations
• Half-vector between
• D usually function of dot product of
• Distribution of microfacet normals determines surface roughness
• Sharp or smooth specular highlight
Rough surface
Reflection equation
So far: directional form of reflection equation
Change of integration variables
Monte Carlo integration
General technique for numerical integration
Hemisphere has two dimensions
• Easy to implement
• Robust, can integrate almost any function
• Efficient for high dimensional integrals
• Variance (noise)
• Slow convergence
Probability concepts
• Random variable X
• Cumulative distribution function (CDP) P
• Probability density function (PDF is the derivative of CDP
• The standard deviation of the estimator
• The expected error converges with
• Important sampling
• With general probability density
• Convergence still slow
• Observe: is proportional to
• Idea: choose f and p to be as similar as possible to reduce variance!
## Area Lights
Given same number of samples, surface form of reflection equation looks better
Illumination from environment
In real world, at each point in scene light arrives from all directions
Assume for now: incoming radiance from all directions is given
For example, environment maps
Store “omni-directional” incoming radiance as images
Each pixel corresponds to radiance from a certain direction
Later: Global illumination
Illumination from environment
Assumption for environment maps
Independent of point x, source of illumination is “infinitely distant”
Given by environment
## Global Illumination
The Rendering Equation
Convergence, because of energy conservation (each bounce absorbs some light)
Reflected radiance appears as unknown on both sides of equation
Solution: find radiance such that reflection equation is satisfied simultaneously at each point
and direction
Based on approximate physical model (geometric optics)
Cannot model Wave effects such as polarization and diffraction
So far: reflection equation
Givenincident light over hemisphere and BRDF at point x,
what is reflected light , which is a radiance distribution: reflected radiance at each point
And in direction
For example, incident radiance given by environment map
Direct illumination from area and point lights, but not multiple bounces.
Incident light at one point (a) depends on reflected light at other point (b)
Direct only vs Direct & indirect
Think of both reflected and incident radiance as *unknown*
Reflection equation expresses equilibrium between incident and reflected radiance
Radiance doesn’t change along the ray in vacuum
Light sources are represented by known function
is emitted radiance at each surface point x in each direction
Monte Carlo Path Tracing
Naïve approach
• Recursive ray tracing as for mirror reflection, but shoot many rays in each step
• Distribute rays over hemisphere at each step
• Problem: exponential explosion of number of rays
• Exponentially more longer paths than shorter paths
• But longer paths contribute less to image than shorter ones
Smarter
• Compute radiance as “integral over all light paths that connect light sources and eye”
• Paths have different lengths (number of bounces)
• Con formulate one integral for all paths of each length
• Approach
• Sum over all path lengths
• Integral of radiance transported along all paths of given length
• Use Monte Carlo integration
Path tracing
• Want to compute radiance through each pixel in image
• Sum of radiance over all light paths connecting light and camera
• Light paths have different lengths
Native approach: average of both techniques
• Variance is bounded by larger variance of two techniques
Multiple importance Sampling
• Weighted average, with weights that reduce variance in optimal way
• N sampling techniques
This is an unbiased estimator of integral, as before!
Weights for provable variance reduction
• Balance heuristics
• Power heuristics
• Details, proofs
Multiple importance sampling
Naïve sampling
Optimal sampling
N-rooks (Latin hypercube) sampling
Advantage over jittered grid: can generate any number n of samples, not restricted to n x m grid
Quasi-Monte Carlo
Based on low-discrepency instead of pseudo-random samples
Low-discrepancy sequences have more uniform distribution of points, less clumping
Verices Position
UV
Normals
• 1500×1500
• 750×750
• 1500×1500
• Classical Foveated Rendering
• Speedups
• Nvidia foveated rendering
• Shift
• Logmap foveated rendering
• F
• Directly working in the logpolar rendering
• F
• Find a big model
1. Deferred shading is slower than what Nvidia is doing
2. Likely to get some returns from the original log-polar
1. Cut problem
2. Use 2 or three cuts
Hierarchical Z-Buffer Visibility
http://dl.acm.org/citation.cfm?id=166147
Can show theoretically that convergence improves
Before: Radiance along straight rays through volume is constant
Now
Participating media: volumetric media influence radiance along the rays
Light interaction with participating media
Absorption
• Absorption coefficients
• Material property
• Related to absorption cross section
• Rate of absorption, unites are 1 / meter
• “Probability per unit length that photon is absorbed”
Transmittance is “fraction of light transported over a certain distance along a ray”
Optical thickness
Transmittance
Phase functions: examples
Rayleigh scattering: scattering from particles smaller than wavelength of light
For example, molecules in atmosphere
Detail see, e.g., wikipedia
Mie (Lorenz-Mie) theory: scattering by spherical particles
Henyey-Greenstein phase function is an empirical phase function , and is useful for many phenomena (water, clouds, skin, stone)
Average phase angle g, asymmetry parameter
Ray marching is an approximation of integral
It has a limited number of samples, and uniformly distributed locations along the ray, could render using brute force Monte Carlo path tracing
Very expensive
Idea: describe scattering properties of material using an extension of BRDFs
So far: BRDFs, light scatters exactly at the same point where it hits the surface
Subsurface scattering: light enters the material, bounces around, leaves at a different place
BSSRDF: bidirectional surface scattering reflectance distribution function
## How to acquire a 3D point clouds
• 3D point cloud acquisition
• Point cloud alignment
• Active methods
• Interfere with reconstructed object
• Mechanically rotate the object to acquire the full geometry
• Radiometrically (active illumination) project laser line
• Accurate
• Slow, need to sweep object with laser
• Need calibrated light source-camera pair
• Doesn’t work if object doesn’t reflect light source
• Passive methods
• Sensing only
• Image sensors, cameras
• Replace laser by second camera
• Known baseline
• Fast, video input, produce real-time depth video
• No active illumination required, just cameras
• Can be accurate with high resolution cameras
• Hard to make stereo matching work rebustly in general settings
• Assume accurate camera calibration
• Triangulation
• How to calculate the height of a sea island
• Disparity inversely related to depth
• This is due to projective projection
• Z = fB / d
• Multiview reconstruction
• Structure from motion, photogrammetry
• Given few correspondences in images (features)
• Compute location and orientation of all cameras
• 3D location of features
• Time of flight (active)
• LIDAR, TOF cameras, Kinect one (v2)
• Send pulse, derive distance based on round-trip time of pulse
• LIDAR (light detection and raging)
• Scan scene point-by-point
• Up to 100’000 points / sec
• Long distances possible
• Long distances, kilometers, slow, relatively accurate
• Measure complete image at once
• Modulate illumination with radio frequency signal and measure phase shift
• Point cloud alignment ICP
• Most scanning algorithms acquire objects piece by piece (e.g. different viewpoints)
• Problem: align pieces to complete full geometry
• Two problems:
• Correspondence: which points in data should be aligned to which points on model?
• Given correspondence, how to rotate / translate to best align parts?
• ICP (Iterative closest point)
• Find initial correspondences using nearest neighbor
• Brute force
• For each point in data, traverse all points in model to find closed point
• Complexity quadratic in number of points
• More efficient
• Use spatial data structure for points in model to efficiently query nearest neighbors
• For example, kd-trees or Octrees
• While distances between corresponding points too large
• Estimate rotation / translation
• Transform using rotation / translation
• Update correspondences using nearest neighbor
• Alignment
• Goal: given correspondences, find rigid transformation that minimizes distances between corresponding points in data and model
• Exact alignment may not be possible because of noise
• Many algorithms, here two basic ones
• Minimizing point-to-point distances
• Points p, in data, p’ in model, 3×1 vectors
• Rotation matrix R in R 3×3
• Translation vector t in R 3×1
• Find the best rotation R, T that minimize squared point-to-point distances
• Challenges
• Error is linear function of R and t
• But R needs to be constrained to be rotation matrix
• Orthonormal
• Optimization problem is non-linear least squares problem
• Translation
• Can show that aligned point clouds have same centroid
• For the best rotation rotation R, translation t
• Define coordinates q, relative to the centroid
• Find SVD of H
• Point to plane distances
• Material from low
• Assume known surface normals
• Measure distances to tangent planes
• Infinitesimal rotation, generators of rotation
• Omit higher order (quadratic) terms
• Small rotation of vector (x, y, z)
## Linear Least Squares Problem
• Combinational approach
• Voronoi diagram
• Function fitting
• Implicit function in 3D
• SSD: Smooth signed distance function
• Reconstruction
## Generative models
• Morphable face models
• Space of human body shapes
• Shape are texture vectors
# Deep Learning
## Concepts
• Layers
• Tensor
• Activation function
• Logistic sigmoid function: $frac{1}{1+e^{-x}}$
• The function is differentiable, monotonic
• unit step
• sign
• linear
• piece-wise linear
• hyperbolic tangent
• ReLU (Rectified Linear Unit)
• ReLU will make the output of a part of neurons 0, which will cause the sparseness of the network;
Disadvantages: Some neurons may never be activated, causing the corresponding parameters to never be updated.
• The leak helps to increase the range of the ReLU function. Usually, the value of a is 0.01 or so.
• Feature map – the output of one filter applied to the previous layer
• Convolutional layer (convolutional neural networks, CNN)
• Convolutional layers apply a convolution operation to the input, passing the result to the next layer. The convolution emulates the response of an individual neuron to visual stimuli.
• Fully connected layer
• Neurons in a fully connected layer have connections to all activations in the previous layer, as seen in regular neural networks. Their activations can hence be computed with a matrix multiplication followed by a bias offset.
• Degree of freedom is very small
• 3×3, 5×5
• Loss functions
• The loss function computes the error for a single training example.
• The cost function is the average of the loss functions of the entire training set.
• L1 (Mean Absolute Error)
• $$C = C_0 + \frac{\lambda}{n} \sum_w |w|$$
• cannot compute derivative at 0
• L2 (Mean Squared Error)
• $$C = C_0 + \frac{\lambda}{2n} \sum_w w^2$$
• smaller w means complexity of the model is lower
• Cross-entropy loss, or log loss (soft-max), measures the performance of a classification model whose output is a probability value between 0 and 1. Cross-entropy loss increases as the predicted probability diverges from the actual label. So predicting a probability of .012 when the actual observation label is 1 would be bad and result in a high loss value. A perfect model would have a log loss of 0.
• use ONLY ONE or SUBSET of training sample from your training set to do the update for a parameter in a particular iteration
•
• SGD: suitable for large sample size, requires small memory; but each step may not be toward the optimal solution direction;
• Newton method: fast convergence; but there are strict requirements on the objective function, there must be a continuous one. The second-order partial derivative has a large amount of calculation.
• Pros: converges much faster compared to GD
• Cons: the error function is not as well minimized as in the case of GD
• Batch size (64, 128, or 256)
• Size of subset as an approximation of the whole databaseIt, which impacts the time efficiency of training and the noisiness of the gradient estimate. Updating the parameters using all training data is not efficient. Updating by one single sample (online updating) is noisy if the sample is not a good representation of the whole data.
• Batch normalization
• It is always great if the layer get data that is always normalized(mean 0 variance 1).
• We just normalize each batch with its running mean and variance.
• Batch normalization potentially helps in two ways: faster learning and higher overall accuracy. The improved method also allows you to use a higher learning rate, potentially providing another boost in speed.
• Automatic differentiation
• Automatic differentiation is a general technique that numerically evaluates the derivative of a function
• Error back-propagation
• Compute the gradients of a loss function using the chain rule
• Proceed ‘backwards’ with respect to the computations performed to compute the loss itself
• Feed-forward and recurrent neural network
• Feed-forward
• The connections between the units do not form a cycle
• Recurrent Neural Network
• Sequence of words and images, unsegmented, connected handwriting recognition, or speech recognition
• Autoencoder / autoassociator / Diabolo network
• Used for unsupervised learning of efficient coding, earn a representation (encoding) for a set of data, typically for the purpose of dimensionality reduction, widely used for learning generative models of data.
• Encoder / Decoder
• The encoder encodes the input sequence to an internal representation called ‘context vector’ which is used by the decoder to generate the output sequence.
• The lengths of input and output sequences can be different, as there is no explicit one on one relation between the input and output sequences.
• An Encoder/Decoder is not an autoencoder
• Reinforcement learning
• Reinforcement learning differs from standard supervised learning in that correct input/output pairs are never presented, nor sub-optimal actions explicitly corrected. Instead the focus is on on-line performance, which involves finding a balance between exploration (of uncharted territory) and exploitation (of current knowledge).
• Reinforcement learning is called approximate dynamic programming, or neuro-dynamic programming
• Limitations
• Firstly, it is often too memory expensive to store values of each state, since the problems can be pretty complex. Solving this involves looking into value approximation techniques, such as Decision Trees or Neural Networks. There are many consequence of introducing these imperfect value estimations, and research tries to minimise their impact on the quality of the solution.
• Moreover, problems are also generally very modular; similar behaviours reappear often, and modularity can be introduced to avoid learning everything all over again. Hierarchical approaches are common-place for this, but doing this automatically is proving a challenge. Finally, due to limited perception, it is often impossible to fully determine the current state. This also affects the performance of the algorithm, and much work has been done to compensate this Perceptual Aliasing.
• Skip connections
•
• Residual network
• u-net
• Classification
• Object detection: YOLO, R-CNN, Fast R-CNN, Faster R-CNN, Mask R-CNN, SSD, etc.
• The YOLO algorithm directly uses a convolutional neural network to output the location of the object and its associated category. It is an end-to-end system, so the detection speed is extremely fast and can meet real-time requirements.
• Regression
• RNN
• Siamese Network
• Siamese networks consists of two identical neural networks, each taking one of the two input images.
• The last layers of the two networks are then fed to a contrastive loss function , which calculates the similarity between the two images.
• The contrastive loss function is a Distance-based Loss function (as opposed to prediction error-based Loss functions like Logistic loss or Hinge loss used in Classification).
• Same network shared the same weights
• Suppose you have cross-domain data
•
# How to prevent overfitting?
• Early stopping
• No improvement in N epoches
• 1 epoch
• Go through the entire training set once
• Mini-batch
• How much data do you put in one batch?
• Extending / Augmenting the datasets
• Acquire more
• Copy data and add noise
• Resampling
• Estimate the distribution of the data and generate
• Regulation / Loss Function
• L1 (Mean Absolute Error)
• $$C = C_0 + \frac{\lambda}{n} \sum_w |w|$$
• cannot compute derivative at 0
• L2 (Mean Squared Error)
• $$C = C_0 + \frac{\lambda}{2n} \sum_w w^2$$
• smaller w means complexity of the model is lower
• Cross-entropy loss, or log loss (soft-max), measures the performance of a classification model whose output is a probability value between 0 and 1. Cross-entropy loss increases as the predicted probability diverges from the actual label. So predicting a probability of .012 when the actual observation label is 1 would be bad and result in a high loss value. A perfect model would have a log loss of 0.
• Dropout (trick)
• Prevent overfitting by removing neurons randomly from the hidden layers.
•
## Denoising Monte Carlo Renderings
Kernel-Predicting Convolutional Networks for Denoising Monte Carlo Renderings
Specular has large dynamic range
• Network
• 8 hidden layers, 100 kernels of 5×5 in each layer
• L1 loss
• Training data
• 600 frames from movie “Finding Dori”
• Noise: 32 spp
• Reference: 1024 spp
• Test data
• 25 frames from other movies
In terms of timing, for an HD image of 1920×1080, our network takes about 12 seconds to evaluate and output a full denoised image
### Denoising animations
Frame-by-frame denoising leads to temporal flickering in animations
Need to filter in 3D space-time
How to implement this with CNNs?
Chaitanya et al. Interactive Reconstruction of Monte Carlo Image Sequences using a Recurrent Denoising Autoencoder, SIGGRAPH 2017
Input to network
• RGB colros
• G-buffer with view-space shading normals, depth, material roughness
• Divided by diffuse albedo (remove texture complexity from input)
Loss
• Spatial: L1, L1 on spatial gradients
• Temporal: L1 on temporal gradients
• Back propagation through time
• Replicate feed forward parts to unroll recurrent connections
• Gradients are noisy in SGD
• Compute moving average over SGD steps
• feed-forward network
Input with 1spp
• 54.9ms for 720p on Titan X
• Path tracer from 70ms to 260ms for 1spp
## What else could we learn to improve rendering efficiency?
• Superresolution
• Foveated rendering
• Antiliasing
• Multiple Importance Sampling
• Antialiasing for light fields
• Learning Light Transport the Reinforced Way, Dahm and Keller
• Use Q learning to learn importance sampling incrementally during path tracing
# Deep Learning: The Future of Real-Time Rendering
## Reference
1. How to prevent overfitting. https://blog.csdn.net/heyongluoyao8/article/details/49429629
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https://runestone.academy/runestone/books/published/TeacherCSP/CSPNameTurtles/TurtlePractice.html
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# Practice with Turtles¶
As we have seen before turtles can draw letters like a capital C as shown below. It draws each line without picking up the pen in between.
Mixed up programs
The following program uses a turtle to draw a capital Z as shown in the picture to the left of this text, <img src=”../_static/DrawZ.png” width=”200” align=”left” hspace=”10” vspace=”5” /> but the lines are mixed up. The program should do all necessary set-up: import the turtle module, get the space to draw on, and create the turtle. Then it should draw the lines for the Z in the order shown by the numbers on the picture. <br /><br /><p>Drag the needed blocks of statements from the left column to the right column and put them in the right order. There may be extra blocks that are not needed in a correct solution. Then click on <i>Check Me</i> to see if you are right. You will be told if any of the lines are in the wrong order or are the wrong blocks.</p>
The following program uses a turtle to draw a capital N as shown in the picture to the left of this text, <img src=”../_static/DrawN4.png” width=”200” align=”left” hspace=”10” vspace=”5” /> but the lines are mixed up. The program should do all necessary set-up: import the turtle module, get the space to draw on, and create the turtle. Remember that the turtle starts off facing east when it is created. Then it should draw the lines for the N in the order shown by the numbers on the picture. <br /><br /><p>Drag the needed blocks of statements from the left column to the right column and put them in the right order. There may be some extra blocks that are not needed in a correct solution. Then click on <i>Check Me</i> to see if you are right. You will be told if any of the lines are in the wrong order or are the wrong blocks.</p>
The following example has 4 errors. Can you fix the errors so that the code runs correctly to print a capital L?
The following example has 4 errors. Can you fix the errors so that the code runs correctly to print a capital C?
Note
Case matters in Python so `screen` is not the same as `Screen`. Also the open and close parentheses are required after every function and procedure call, even if it doesn’t take any input.
Use the area below to try to draw a letter or number. Use block style rather than curves.
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top of page
# Introduction to Statistics: Descriptive Statistics
#### Types of data
1. Categorical
• Categorical data represents groups or categories.
• Examples: Car brands: Audi, BMW and Mercedes. Answers to yes/no questions: yes and no
2. Numerical
• Numerical data represents numbers. It is divided into two groups: discrete and continuous. Discrete data can be usually counted in a finite matter, while continuous is infinite and impossible to count.
• Examples: Discrete: children you want to have, SAT score Continuous: weight, height
#### Levels of measurement
1. Qualitative
There are two qualitative levels: nominal and ordinal. The nominal level represents categories that cannot be put in any order, while ordinal represents categories that can be ordered.
Examples:
• Nominal: four seasons (winter, spring, summer, autumn)
• Ordinal: rating your meal (disgusting, unappetizing, neutral, tasty, and delicious)
2. Quantitative
There are two quantitative levels: interval and ratio. They both represent “numbers”, however, ratios have a true zero, while intervals don’t.
Examples:
• Interval: degrees Celsius and Fahrenheit
• Ratio: degrees Kelvin, length
##### Graphs and tables that represent categorical variables
1- Frequency distribution tables: show the category and its corresponding absolute frequency.
2- Bar charts: are very common. Each bar represents a category. On the y-axis we have the absolute frequency.
3- Pie charts: are used when we want to see the share of an item as a part of the total. Market share is almost always represented with a pie chart.
4- The Pareto diagram: is a special type of bar chart where the categories are shown in descending order of frequency, and a separate curve shows the cumulative frequency.
##### Numerical variables. Frequency distribution table and histogram
Frequency distribution tables for numerical variables are different than the ones for categorical. Usually, they are divided into intervals of equal (or unequal) length. The tables show the interval, the absolute frequency and sometimes it is useful to also include the relative (and cumulative) frequencies.
The interval width is calculated using the following formula:
𝐼𝑛𝑡𝑒𝑟𝑎𝑙 𝑤𝑖𝑑𝑡ℎ =( 𝐿𝑎𝑟𝑔𝑒𝑠𝑡 𝑛𝑢𝑚𝑏𝑒𝑟 − 𝑠𝑚𝑎𝑙𝑙𝑒𝑠𝑡 𝑛𝑢𝑚𝑏𝑒𝑟) / 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑑𝑒𝑠𝑖𝑟𝑒𝑑 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑠
Histograms are the one of the most common ways to represent numerical data. Each bar has width equal to the width of the interval. The bars are touching as there is continuation between intervals: where one ends -> the other begins.
##### Graphs and tables for relationships between variables
Cross tables
Cross tables (or contingency tables) are used to represent categorical variables. One set of categories is labeling the rows and another is labeling the columns. We then fill in the table with the applicable data. It is a good idea to calculate the totals. Sometimes, these tables are constructed with the relative frequencies as shown in the table below.
A common way to represent the data from a cross table is by using a side-by-side bar chart.
###### Scatter plots
When we want to represent two numerical variables on the same graph, we usually use a scatter plot. Scatter plots are useful especially later on, when we talk about regression analysis, as they help us detect patterns (linearity, homoscedasticity).
Scatter plots usually represent lots and lots of data. Typically, we are not interested in single observations, but rather in the structure of the dataset.
A scatter plot that looks in the following way (down) represents data that doesn’t have a pattern. Completely vertical ‘forms’ show no association.
Conversely, the plot above shows a linear pattern, meaning that the observations move together.
Now, lets talk about Mean, median and mode
##### Mean
The mean is the most widely spread measure of central tendency. It is the simple average of the dataset. Note: easily affected by outliers
The formula to calculate the mean is:
x̄ = (x1,x2,x3,…,xn)/n
##### Median
The median is the midpoint of the ordered dataset. It is not as popular as the mean, but is often used in academia and data science. That is since it is not affected by outliers.
In an ordered dataset, the median is the number at position (n+1)/2
If this position is not a whole number, it, the median is the simple average of the two numbers at positions closest to the calculated value.
##### Mode
The mode is the value that occurs most often. A dataset can have 0 modes, 1 mode or multiple modes.
The mode is calculated simply by finding the value with the highest frequency.
And finally, Skewness
• Skewness is a measure of asymmetry that indicates whether the observations in a dataset are concentrated on one side.
• Right (positive) skewness looks like the one in the graph. It means that the outliers are to the right (long tail to the right).
• Left (negative) skewness means that the outliers are to the left.
• Usually, you will use software to calculate skewness.
Formula to calculate skewness:
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# [Maxima] Problem with TAYLOR
Gottfried Helms helms at uni-kassel.de
Fri Jun 1 11:49:24 CDT 2007
```Hi,
possibly my question can be solved differently, but
I didn't get it this way.
With TAYLOR() I get the first six series coefficients for
TAYLOR( (1/(1+x) , x, 0 ,6)
This works also for m'th powers of the expression:
TAYLOR( (1/(1+x)^m , x, 0 ,6)
What I wanted to get is
TAYLOR( (1/(1+x^m)^(1/m) , x, 0 ,6)
but this gives only one coefficients and three ellipses.
The final exponent doesn't matter, also for
TAYLOR( (1/(1+x^m) , x, 0 ,6)
I get only one coefficient.
How can I proceed?
Gottfried Helms
```
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https://www.webqc.org/balance.php?reaction=H3BO3+%2B+NH4F+%3D+NH4BF4+%2B+H2O+%2B+NH3
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#### Balance Chemical Equation - Online Balancer
Balanced equation:
H3BO3 + 4 NH4F = NH4BF4 + 3 H2O + 3 NH3
Reaction stoichiometry Limiting reagent
CompoundCoefficientMolar MassMolesWeight
H3BO3161.83302
NH4F437.0368632
NH4BF41104.8430728
H2O318.01528
NH3317.03052
Units: molar mass - g/mol, weight - g.
Direct link to this balanced equation:
Instructions on balancing chemical equations:
• Enter an equation of a chemical reaction and click 'Balance'. The answer will appear below
• Always use the upper case for the first character in the element name and the lower case for the second character. Examples: Fe, Au, Co, Br, C, O, N, F. Compare: Co - cobalt and CO - carbon monoxide
• To enter an electron into a chemical equation use {-} or e
• To enter an ion specify charge after the compound in curly brackets: {+3} or {3+} or {3}.
Example: Fe{3+} + I{-} = Fe{2+} + I2
• Substitute immutable groups in chemical compounds to avoid ambiguity.
For instance equation C6H5C2H5 + O2 = C6H5OH + CO2 + H2O will not be balanced,
but PhC2H5 + O2 = PhOH + CO2 + H2O will
• Compound states [like (s) (aq) or (g)] are not required.
• If you do not know what products are enter reagents only and click 'Balance'. In many cases a complete equation will be suggested.
• Reaction stoichiometry could be computed for a balanced equation. Enter either the number of moles or weight for one of the compounds to compute the rest.
• Limiting reagent can be computed for a balanced equation by entering the number of moles or weight for all reagents.
Examples of complete chemical equations to balance: Examples of the chemical equations reagents (a complete equation will be suggested): Give us feedback about your experience with chemical equation balancer.
chemical equations balanced today
Back to Online Chemical Tools Menu
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https://electronics.stackexchange.com/questions/180219/via-drill-size-for-heat-dissipation/180223
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# Via drill size for heat dissipation
I am refering to the document SLMA002G PowerPAD™ Thermally Enhanced Package from TI. Kindly refer to page 9(screen shot attached). this is related to thermal vias present on the ther or open pad which solders to the open pad of an IC.
The statement in question is state below - The experiments conducted jointly with Solectron Texas indicate that a via drill diameter of 0.33 mm (13 mils) or smaller works well when 1 ounce copper is plated at the surface of the board and simultaneously plating the barrel of the via.
This means that if the layer is of 1 Oz then a drill dia of 13mils for via is enough for decent thermal dissipation,right ? What if the Cu is 0.5Oz how much would the drill dia need be changed ?
The larger the "thermally active" cross section of your via's plating is, the lower its thermal resistance will be. The plating's cross section is, with $d$ being the diameter and $t$ the plating thickness:
$$A \approx \pi d t$$
That means larger drill -> larger area; and thicker plating -> larger area. OTOH, some solder will get into the via hole and enhance heat transfer between the two board sides, but that's harder to quantify because the via is not necessarily filled with solder.
Given that the overall available area for thermal vias is usually fixed, I'd use a larger number of smaller vias in order to fill a larger fraction of that area. 0.3 mm sounds reasonable.
Can you show us numbers of other via diameters, or did they just make a statement about 0.33 mm?
A thermal simulation would show more detailed insight, but that's probably a bit over the top.
• how do I do a thermal simulation ? Do u know an open source or trial tool that can serve this ? – Board-Man Jul 15 '15 at 11:17
• This via calculator may could help. With a via diameter = 13 mil and plating = 1 oz the thermal resistance is about 94.5C°/W. With 0.5 oz the diameter must be around 28 mil to get the same thermal resistance (94.7C°/W), according to this tool. – Bence Kaulics Jul 15 '15 at 11:24
• @VinodKaruvat A real thermal simulation would require decent physics simulation software. There are a number around, OpenFOAM being a free one. But really - how accurate do you need it? – Christoph Jul 15 '15 at 11:49
• i want it more for documentation purposes. OpenFoam is Linux based. Any windows based pls ? – Board-Man Jul 15 '15 at 11:51
• Needing a linux machine is not your smallest problem when you want to do a thermal simulation. For windows there's Fluent, Comsol, and many others. Maybe you can get a trial of student's license, depending on your situation. – Christoph Jul 15 '15 at 12:01
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# What Is a Time Series Homework Help?
Learn the Concept of Time Series in a Better Manner
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What is time series?
Time series is defined as a sequence of data points over a continuous period of time, in a successive order. Time series is essential because it allows one to predict how an asset is going to change over a period of interval and compare the results to other variables as well. One such example of time series in the daily life is the occurrence of ocean tides. These can be predicted by the analyses of its previous data.
Methods of analysis
There are a number of methods through which time series can be analysed. Each has its own benefits and is applicable to certain kinds of data only. Listed below are some of the categories by which time series can be analysed:
1. Frequency domain methods
2. Time domain methods
3. Parametric methods.
4. Non-parametric methods
5. Linear
6. Non-linear
7. Univariate
8. Multivariate
Importance of time series
Time series is immensely important for students to learn because it helps them with various aspects. It helps them in the business and various other fields as well. Listed below are some of the benefits that it provides which makes it a really important topic to understand and cope with:
1. It helps to predict the future based upon the strategy used in the past and the data compiled from it.
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# Cryptography Problems
Subject Computer Science Cryptography
## Question
Problem 1
a. We know that encryption provides confidentiality. Does it provide privacy? (Explain)
b. SHA-256 was used to hash a message, such that SHA-256("The quick brown fox jumps over the lazy dog") = D7A8FBB307D7809469CA9ABCB0082E4F8D5651E46D3CDB762D02D0BF37C9E592.
What should be SHA-256("The quick brown fog jumps over the lazy dog")? Explain your answer.
c. List all the element of the set Zp for p=11. Does every element of Zp have an inverse?
Why/why not? For those that do, 5 and 7 (‘cause they are relatively prime with 11) give the inverse for each.
Problem 3
4-bit long message was encrypted using one-time pad to yield a cipher-text “1010” Assuming the message space consists of all 4-bit long messages, what is the probability that the corresponding plaintext was “1001”? Explain your answer.
Problem 6
1) Use the Euclidian algorithm to compute (Show all of the steps involved)
a. The GCD of 7469 and 2464.
b. The GCD of 198 and 243 (use the Extended form).
2) We also discussed the use of the Extended Euclidian algorithm to calculate modular inverses. Use this algorithm to compute the following values. Show all of the steps involved.
a. 9570-1(mod 12935)
b. 550-1 (mod 1769)
3) Let n be a product of two large primes p and q (i.e., n=pq). Assume that x, y, and g are relatively prime to n.
1) If x ≡ y (mod n), is gx ≡ gy (mod n)? Show why or why not.
2) If x ≡ y (mod Φ(n)) instead, is gx ≡ gy (mod n)? Show why or why not.
Problem 7
Alice’s uses ElGamal public key setting, with parameters, p=11, q=5, and g=4. Her public key is y=5. Bob encrypted a plaintext message m with Alice’s ElGamal public key and sent out the ciphertext (k,c) such that k=3 and c= 5, to her.
Can an adversary recover the message m, given p, q, g, y, and (k,c)? If so, what is m?
Show all steps in your solution.
Problem 8
1. How many multiplications are needed to compute x1024? (Do not tell me 1024 multiplication because it is much, much less!) Show your algorithm.
2. Bob downloaded a 50Gb tar.gz file from Alice’s server today. Bob needs to know if he downloaded the correct file and that there were no errors in the transmission, before attempting to unzip/untar it (since uncompressing such a big file might take a long time and computational resources). Unfortunately, Alice and Bob do not share any keys nor do they have a common CA. How can Bob ensure the correctness of the file? Alice and Bob know each other personally, and they are scheduled to meet in a couple of days.
Bonus
Part A. Given an elliptic curve E over Z29 and the base point P=(8,10):
E: y2=x3+4x+20 mod 29.
Calculate the following point multiplication k.P using the Double-and-Add algorithm. Provide the intermediate results after each step.
1. k=9
2. k=20
Part B. The order of the curve above is known to be #E=37. Furthermore, an additional point Q=15.P=(14,23) on this curve is given. Determine the result of the following point multiplications by using as few group operations as possible, i.e., make ‘smart’ use of the known point Q (Specify how you simplified the calculation each time.
1. 38.P
2. P+4.Q
Hint: In addition to using Q, use the fact that it is easy to compute (-P).
## Solution Preview
This material may consist of step-by-step explanations on how to solve a problem or examples of proper writing, including the use of citations, references, bibliographies, and formatting. This material is made available for the sole purpose of studying and learning - misuse is strictly forbidden.
Problem 1
a. We know that encryption provides confidentiality. Does it provide privacy? (Explain)
Encryption can provide privacy from the third party. It can hide your communication from someone in-between listening to it. For example, if you are communicating via encrypted channel to someone, your ISP (or anybody in the path) can see the source and destination of the communication, but not the context.
b. SHA-256 was used to hash a message, such that SHA-256("The quick brown fox jumps over the lazy dog") = D7A8FBB307D7809469CA9ABCB0082E4F8D5651E46D3CDB762D02D0BF37C9E592.
What should be SHA-256("The quick brown fog jumps over the lazy dog")? Explain your answer.
It is: c223298dccec5378f3428bd697e9c84f28bf985d3906e4d9850a18e4ada9f738. The minimal change in the original results in a drastic change of the hash. Hence, it is (almost) impossible to track back the change or compute the original string. That is why it is used for storing passwords and authentication.
c. List all the element of the set Zp for p=11. Does every element of Zp have an inverse?
Why/why not? For those that do, 5 and 7 (‘cause they are relatively prime with 11) give the inverse for each.
Z11 = {0,1,2,3,4,5,6,7,8,9,10}...
This is only a preview of the solution. Please use the purchase button to see the entire solution
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### Introduction to Machine Learning
The goal of machine learning is to program computers to use example data or past experience to solve a given problem. Many successful applications of machine learning exist already, including systems that analyze past sales data to predict customer behavior, optimize robot behavior so that a task can be completed using minimum resources, and extract knowledge from bioinformatics data. Introduction to Machine Learning is a comprehensive textbook on the subject, covering a broad array of topics not usually included in introductory machine learning texts. Subjects include supervised learning; Bayesian decision theory; parametric, semi-parametric, and nonparametric methods; multivariate analysis; hidden Markov models; reinforcement learning; kernel machines; graphical models; Bayesian estimation; and statistical testing. Machine learning is rapidly becoming a skill that computer science students must master before graduation.
### Worst-Case Bounds for Gaussian Process Models
Dean P. Foster University of Pennsylvania We present a competitive analysis of some nonparametric Bayesian algorithms ina worst-case online learning setting, where no probabilistic assumptions about the generation of the data are made. We consider models which use a Gaussian process prior (over the space of all functions) andprovide bounds on the regret (under the log loss) for commonly usednon-parametric Bayesian algorithms -- including Gaussian regression and logistic regression -- which show how these algorithms can perform favorably under rather general conditions.
### Online Learning of Nonparametric Mixture Models via Sequential Variational Approximation
Reliance on computationally expensive algorithms for inference has been limiting the use of Bayesian nonparametric models in large scale applications. To tackle this problem, we propose a Bayesian learning algorithm for DP mixture models. Instead of following the conventional paradigm -- random initialization plus iterative update, we take an progressive approach. Starting with a given prior, our method recursively transforms it into an approximate posterior through sequential variational approximation. In this process, new components will be incorporated on the fly when needed.
### The Many Faces of Exponential Weights in Online Learning
A standard introduction to online learning might place Online Gradient Descent at its center and then proceed to develop generalizations and extensions like Online Mirror Descent and second-order methods. Here we explore the alternative approach of putting exponential weights (EW) first. We show that many standard methods and their regret bounds then follow as a special case by plugging in suitable surrogate losses and playing the EW posterior mean. For instance, we easily recover Online Gradient Descent by using EW with a Gaussian prior on linearized losses, and, more generally, all instances of Online Mirror Descent based on regular Bregman divergences also correspond to EW with a prior that depends on the mirror map. Furthermore, appropriate quadratic surrogate losses naturally give rise to Online Gradient Descent for strongly convex losses and to Online Newton Step. We further interpret several recent adaptive methods (iProd, Squint, and a variation of Coin Betting for experts) as a series of closely related reductions to exp-concave surrogate losses that are then handled by Exponential Weights. Finally, a benefit of our EW interpretation is that it opens up the possibility of sampling from the EW posterior distribution instead of playing the mean. As already observed by Bubeck and Eldan, this recovers the best-known rate in Online Bandit Linear Optimization.
### WS06-05-006.pdf
This paper describes an effort to measure the effectiveness of tutor help in an intelligent tutoring system. Although conventional pre-and post-test experiments can determine whether tutor help is effective, they are expensive to conduct. Furthermore, pre-and post-test experiments often do not model student knowledge explicitly and thus are ignoring a source of information: students often request help about words they do not know. Therefore, we construct a dynamic Bayes net (which we call the Help model) that models tutor help and student knowledge in one coherent framework. The Help model distinguishes two different effects of help: scaffolding immediate performance vs. teaching persistent knowledge that improves long term performance. We train the Help model to fit student performance data gathered from usage of the Reading Tutor (Mostow & Aist, 2001). The parameters of the trained model suggest that students benefit from both the scaffolding and teaching effects of help. That is, students are more likely to perform correctly on the current attempt and learn persistent knowledge if tutor help is provided. Thus, our framework is able to distinguish two types of influence that tutor help has on the student, and can determine whether help helps learning without an explicit controlled study.
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2014-11-19T17:05:26-05:00
This is how you do a fraction on brainly
step one example: 1
step two example :1/
step three example :1/2
NO I MEAN THE OTHER WAY
you cant
2014-11-19T17:11:06-05:00
With simpler decimals, this is a bit easier so I'll explain that first to make sure you get the basic idea.
Let's say you had a decimal for .75 and wanted to create the fraction mathematically. To do this you would:
First divide the decimal by 1 so .75/1Then multiply be a multiple of 10 that eliminates the decimal so in this case 100so you now have 75/100When you simplify that fraction by dividing by 25, you of course get 3/4
Now on to your more complicated question:
Let's assign a variable to your repeating decimal so:
x=.1515151515
Now, let's identify the repeating digits, here they are .15
Move one set of the repeating digits in front of the decimal so now we get:
100x=15.1515151515
Now you have two equations, they are:
x=.1515151515100x=15.1515151515
Combine the equations into one equation:
100x-x=15.1515151515-.151515151599x=1515/99simplify and you get 5/33
OMG NOT THAT
U KNOW LIKE. 45
45
50________
45
------
50
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Rs Aggarwal 2018 Solutions for Class 8 Math Chapter 2 Exponents are provided here with simple step-by-step explanations. These solutions for Exponents are extremely popular among Class 8 students for Math Exponents Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2018 Book of Class 8 Math Chapter 2 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2018 Solutions. All Rs Aggarwal 2018 Solutions for class Class 8 Math are prepared by experts and are 100% accurate.
Question 1:
Evaluate:
(i) 4−3
(ii) ${\left(\frac{1}{2}\right)}^{-5}$
(iii) ${\left(\frac{4}{3}\right)}^{-3}$
(iv) ${\left(-3\right)}^{-4}$
(v) ${\left(\frac{-2}{3}\right)}^{-5}$
(i) ${4}^{-3}=\frac{1}{{4}^{3}}=\frac{1}{64}$
(ii) ${\left(\frac{1}{2}\right)}^{-5}={2}^{5}=32$
(iii) ${\left(\frac{4}{3}\right)}^{-3}={\left(\frac{3}{4}\right)}^{3}=\frac{{3}^{3}}{{4}^{3}}=\frac{27}{64}$
(iv) ${\left(-3\right)}^{-4}={\left(\frac{-1}{3}\right)}^{4}=\frac{\left(-1{\right)}^{4}}{{3}^{4}}=\frac{1}{81}$
(v) ${\left(\frac{-2}{3}\right)}^{-5}={\left(\frac{-3}{2}\right)}^{5}=\frac{{\left(-3\right)}^{5}}{{2}^{5}}=\frac{-243}{32}$
Question 2:
Evaluate:
(i) ${\left(\frac{5}{3}\right)}^{2}×{\left(\frac{5}{3}\right)}^{2}$
(ii) ${\left(\frac{5}{6}\right)}^{6}×{\left(\frac{5}{6}\right)}^{-4}$
(iii) ${\left(\frac{2}{3}\right)}^{-3}×{\left(\frac{2}{3}\right)}^{-2}$
(iv) ${\left(\frac{9}{8}\right)}^{-3}×{\left(\frac{9}{8}\right)}^{2}$
(i)
(ii)
(iii)${\left(\frac{2}{3}\right)}^{-3}×{\left(\frac{2}{3}\right)}^{-2}={\left(\frac{2}{3}\right)}^{\left(-3-2\right)}={\left(\frac{2}{3}\right)}^{-5}={\left(\frac{3}{2}\right)}^{5}=\frac{{3}^{5}}{{2}^{5}}=\frac{243}{32}$
(iv) ${\left(\frac{9}{8}\right)}^{-3}×{\left(\frac{9}{8}\right)}^{2}={\left(\frac{9}{8}\right)}^{\left(-3+2\right)}={\left(\frac{9}{8}\right)}^{-1}=\frac{8}{9}$
Question 3:
Evaluate:
(i) ${\left(\frac{5}{9}\right)}^{-2}×{\left(\frac{3}{5}\right)}^{-3}×{\left(\frac{3}{5}\right)}^{0}$
(ii) ${\left(\frac{-3}{5}\right)}^{-4}×{\left(\frac{-2}{5}\right)}^{2}$
(iii) ${\left(\frac{-2}{3}\right)}^{-3}×{\left(\frac{-2}{3}\right)}^{-2}$
(i)
${\left(\frac{5}{9}\right)}^{-2}×{\left(\frac{3}{5}\right)}^{-3}×{\left(\frac{3}{5}\right)}^{0}={\left(\frac{5}{9}\right)}^{-2}×{\left(\frac{3}{5}\right)}^{-3+0}\phantom{\rule{0ex}{0ex}}={\left(\frac{5}{9}\right)}^{-2}×{\left(\frac{3}{5}\right)}^{-3}={\left(\frac{9}{5}\right)}^{2}×{\left(\frac{5}{3}\right)}^{3}\phantom{\rule{0ex}{0ex}}=\frac{{9}^{2}}{{5}^{2}}×\frac{{5}^{3}}{{3}^{3}}\phantom{\rule{0ex}{0ex}}=\frac{{\left({3}^{2}\right)}^{2}}{{5}^{2}}×\frac{{5}^{3}}{{3}^{3}}\phantom{\rule{0ex}{0ex}}=\frac{{3}^{4}}{{5}^{2}}×\frac{{5}^{3}}{{3}^{3}}=\left({3}^{\left(4-3\right)}\right)×\left({5}^{\left(3-2\right)}\right)=3×5=15$
(ii)
${\left(\frac{-3}{5}\right)}^{-4}×{\left(\frac{-2}{5}\right)}^{2}={\left(\frac{5}{-3}\right)}^{4}×{\left(\frac{-2}{5}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{{5}^{4}}{-{3}^{4}}×\frac{-{2}^{2}}{{5}^{2}}={5}^{\left(4-2\right)}×\frac{-{2}^{2}}{-{3}^{4}}={5}^{2}×\frac{-{2}^{2}}{-{3}^{4}}\phantom{\rule{0ex}{0ex}}=25×\frac{4}{81}=\frac{100}{81}$
(iii)
${\left(\frac{-2}{3}\right)}^{-3}×{\left(\frac{-2}{3}\right)}^{-2}={\left(\frac{3}{-2}\right)}^{3}×{\left(\frac{3}{-2}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{{3}^{3}}{-{2}^{3}}×\frac{{3}^{2}}{-{2}^{2}}=\frac{{3}^{\left(3+2\right)}}{-{2}^{\left(3+2\right)}}=\frac{{3}^{5}}{-{2}^{5}}=\frac{-243}{32}$
Question 4:
Evaluate:
(i) ${\left\{{\left(\frac{-2}{3}\right)}^{2}\right\}}^{-2}$
(ii) ${\left[{\left\{{\left(\frac{-1}{3}\right)}^{2}\right\}}^{-2}\right]}^{-1}$
(iii) ${\left\{{\left(\frac{3}{2}\right)}^{-2}\right\}}^{2}$
(i) ${\left\{{\left(\frac{-2}{3}\right)}^{2}\right\}}^{-2}={\left(\frac{-2}{3}\right)}^{2×\left(-2\right)}={\left(\frac{-2}{3}\right)}^{-4}={\left(\frac{3}{-2}\right)}^{4}=\frac{{3}^{4}}{\left(-2{\right)}^{4}}=\frac{{3}^{4}}{{2}^{4}}=\frac{81}{16}$
(ii) ${\left[{\left\{{\left(\frac{-1}{3}\right)}^{2}\right\}}^{-2}\right]}^{-1}={\left[{\left(\frac{-1}{3}\right)}^{2×\left(-2\right)}\right]}^{-1}={\left[{\left(\frac{-1}{3}\right)}^{-4}\right]}^{-1}={\left(\frac{-1}{3}\right)}^{-4×-1}={\left(\frac{-1}{3}\right)}^{4}=\frac{-{1}^{4}}{{3}^{4}}=\frac{{1}^{4}}{{3}^{4}}=\frac{1}{81}$
(iii) ${\left\{{\left(\frac{3}{2}\right)}^{-2}\right\}}^{2}={\left(\frac{3}{2}\right)}^{-2×2}={\left(\frac{3}{2}\right)}^{-4}={\left(\frac{2}{3}\right)}^{4}=\frac{{2}^{4}}{{3}^{4}}=\frac{16}{81}$
Question 5:
Evaluate $\left\{{\left(\frac{1}{3}\right)}^{-3}-{\left(\frac{1}{2}\right)}^{-3}\right\}÷{\left(\frac{1}{4}\right)}^{-3}.$
$\left\{{\left(\frac{1}{3}\right)}^{-3}-{\left(\frac{1}{2}\right)}^{-3}\right\}÷{\left(\frac{1}{4}\right)}^{-3}=\left\{{3}^{3}-{2}^{3}\right\}÷{4}^{3}=\left\{27-8\right\}÷64=\frac{19}{64}$
Question 6:
Evaluate ${\left\{{\left(\frac{4}{3}\right)}^{-1}-{\left(\frac{1}{4}\right)}^{-1}\right\}}^{-1}.$
Question 7:
Evaluate $\left[{\left({5}^{-1}×{3}^{-1}\right)}^{-1}÷{6}^{-1}\right].$
Question 8:
Find the value of:
(i) (20 + 3−1) × 32
(ii) (2−1 × 3−1) ÷ 2−3
(iii) ${\left(\frac{1}{2}\right)}^{-2}+{\left(\frac{1}{3}\right)}^{-2}+{\left(\frac{1}{4}\right)}^{-2}$
(i)
(ii)
(iii)
${\left(\frac{1}{2}\right)}^{-2}+{\left(\frac{1}{3}\right)}^{-2}+{\left(\frac{1}{4}\right)}^{-2}={\left(\frac{2}{1}\right)}^{2}+{\left(\frac{3}{1}\right)}^{2}+{\left(\frac{4}{1}\right)}^{2}={2}^{2}+{3}^{2}+{4}^{2}=4+9+16=29$
Question 9:
Find the value of x for which ${\left(\frac{5}{3}\right)}^{-4}×{\left(\frac{5}{3}\right)}^{-5}={\left(\frac{5}{3}\right)}^{3x}.$
Consider the left side:
${\left(\frac{5}{3}\right)}^{-4}×{\left(\frac{5}{3}\right)}^{-5}={\left(\frac{5}{3}\right)}^{\left(-4+\left(-5\right)\right)}={\left(\frac{5}{3}\right)}^{-9}$
Given:
${\left(\frac{5}{3}\right)}^{-9}={\left(\frac{5}{3}\right)}^{3x}$
Comparing the powers:
Question 10:
Find the value of x for which ${\left(\frac{4}{9}\right)}^{4}×{\left(\frac{4}{9}\right)}^{-7}={\left(\frac{4}{9}\right)}^{2x-1}.$
Given:
${\left(\frac{4}{9}\right)}^{4}×{\left(\frac{4}{9}\right)}^{-7}={\left(\frac{4}{9}\right)}^{2x-1}$
$\therefore$ ${\left(\frac{4}{9}\right)}^{\left(4-7\right)}={\left(\frac{4}{9}\right)}^{-3}={\left(\frac{4}{9}\right)}^{2x-1}$
Question 11:
By what number should (−6)−1 be multiplied so that the product becomes 9−1?
Let the required number be $x$.
$\therefore$ $x×\left(-6{\right)}^{-1}={9}^{-1}$
The greatest common divisor for the numerator and the denominator is 3.
Question 12:
By what number should ${\left(\frac{-2}{3}\right)}^{-3}$ be divided so that the quotient may be ${\left(\frac{4}{27}\right)}^{-2}$?
Let the number be $x$.
$\therefore$ ${\left(\frac{-2}{3}\right)}^{-3}÷x={\left(\frac{4}{27}\right)}^{-2}$
Question 13:
If 52x + 1 ÷ 25 = 125, find the value of x.
Given:
${5}^{2x+1}÷25=125$
$\therefore$ $x=2$
Question 1:
Write each of the following numbers in standard form:
(i) 57.36
(ii) 3500000
(iii) 273000
(iv) 168000000
(v) 4630000000000
(vi) 345 × 105
(i) $57.36=5.736×10$
(ii) $3500000=3.5×{10}^{6}$
(iii) $273000=2.73×{10}^{5}$
(iv) $168000000=1.68×{10}^{8}$
(v) $4630000000000=4.63×{10}^{12}$
(vi) $345×{10}^{5}=3.45×{10}^{7}$
Question 2:
Write each of the following numbers in usual form:
(i) 3.74 × 105
(ii) 6.912 × 108
(iii) 4.1253 × 107
(iv) 2.5 × 104
(v) 5.17 × 106
(vi) 1.679 × 109
(i) $3.74×{10}^{5}=\frac{374}{100}×{10}^{5}=\frac{374×{10}^{5}}{{10}^{2}}=374×{10}^{\left(5-2\right)}=374×{10}^{3}=374000$
(ii) $6.912×{10}^{8}=\frac{6912}{1000}×{10}^{8}=\frac{6912×{10}^{8}}{{10}^{3}}=6912×{10}^{\left(8-3\right)}=6912×{10}^{5}=691200000$
(iii) $4.1253×{10}^{7}=\frac{41253}{10000}×{10}^{7}=\frac{41253×{10}^{7}}{{10}^{4}}=41253×{10}^{\left(7-4\right)}=41253×{10}^{3}=41253000$
(iv) $2.5×{10}^{4}=\frac{25}{10}×{10}^{4}=\frac{25×{10}^{4}}{10}=25×{10}^{\left(4-1\right)}=25×{10}^{3}=25000$
(v) $5.17×{10}^{6}=\frac{517}{100}×{10}^{6}=\frac{517×{10}^{6}}{{10}^{2}}=517×{10}^{\left(6-2\right)}=517×{10}^{4}=5170000$
(vi) $1.679×{10}^{9}=\frac{1679}{1000}×{10}^{9}=\frac{1679×{10}^{9}}{{10}^{3}}=1679×{10}^{\left(9-3\right)}=1679×{10}^{6}=1679000000$
Question 3:
(i) The height of Mount Everest is 8848 m. Write it in standard form.
(ii) The speed of light is 300000000 m/sec. Express it in standard form.
(iii) The distance from the earth to the sun is 149600000000 m. Write it in standard form.
(i) The height of the Mount Everest is 8848 m.
In standard form, we have:
.
(ii) The speed of light is 300000000 m/s.
In standard form, we have:
(iii) The Sun$-$Earth distance is 149600000000 m.
In standard form, we have: $149600000000=1496×100000000=1.496×1000×100000000=1.496×{10}^{3}×{10}^{8}=1.496×{10}^{11}\mathrm{m}$
Question 4:
Mass of earth is (5.97 × 1024) kg and mass of moon is (7.35 × 1022) kg. What is the total mass of the two?
Mass of the Earth =
Now, $5.97×{10}^{24}=5.97×{10}^{\left(2+22\right)}=5.97×{10}^{2}×{10}^{22}=597×{10}^{22}$
So, the mass of the Earth can also be written as .
Mass of the Moon =
Sum of the masses of the Earth and the Moon:
Question 5:
Write each of the following numbers in standard form:
(i) 0.0006
(ii) 0.00000083
(iii) 0.0000000534
(iv) 0.0027
(v) 0.00000165
(vi) 0.00000000689
(i) $0.0006=\frac{6}{{10}^{4}}=6×{10}^{-4}$
(ii) $0.00000083=\frac{83}{{10}^{8}}=\frac{8.3×10}{{10}^{8}}=8.3×{10}^{\left(1-8\right)}=8.3×{10}^{-7}$
(iii) $0.0000000534=\frac{534}{{10}^{10}}=\frac{5.34×{10}^{2}}{{10}^{10}}=5.34×{10}^{\left(2-10\right)}=5.34×{10}^{-8}$
(iv) $0.0027=\frac{27}{{10}^{4}}=\frac{2.7×10}{{10}^{4}}=2.7×{10}^{\left(1-4\right)}=2.7×{10}^{-3}$
(v) 0.00000165 = (2-8) = 1.65×10-6
(vi) 0.00000000689 = (2-11)= 6.89×10-9
Question 6:
(i) 1 micron = $\frac{1}{1000000}$ m. Express it in standard form.
(ii) Size of a bacteria = 0.0000004 m. Express it in standard form.
(iii) Thickness of a paper = 0.03 mm. Express it in standard form.
(i)
(ii)
(iii)
Question 7:
Write each of the following numbers in usual form:
(i) 2.06 × 10−5
(ii) 5 × 10−7
(iii) 6.82 × 10−6
(iv) 5.673 × 10−4
(v) 1.8 × 10−2
(vi) 4.129 × 10−3
(i) $2.06×{10}^{-5}=\frac{206}{100}×\frac{1}{{10}^{5}}=\frac{206}{{10}^{2}×{10}^{5}}=\frac{206}{{10}^{\left(5+2\right)}}=\frac{206}{{10}^{7}}=\frac{206}{10000000}=0.0000206$
(ii) $5×{10}^{-7}=\frac{5}{{10}^{7}}=\frac{5}{10000000}=0.0000005$
(iii) $6.82×{10}^{-6}=\frac{682}{100}×\frac{1}{{10}^{6}}=\frac{682}{{10}^{2}×{10}^{6}}=\frac{682}{{10}^{\left(2+6\right)}}=\frac{682}{{10}^{8}}=\frac{682}{100000000}=0.00000682$
(iv)$5.673×{10}^{-4}=\frac{5673}{1000}×\frac{1}{{10}^{4}}=\frac{5673}{{10}^{3}×{10}^{4}}=\frac{5673}{{10}^{\left(3+4\right)}}=\frac{5673}{{10}^{7}}=\frac{5673}{10000000}=0.0005673$
(v)$1.8×{10}^{-2}=\frac{18}{10}×\frac{1}{{10}^{2}}=\frac{18}{10×{10}^{2}}=\frac{18}{{10}^{\left(1+2\right)}}=\frac{18}{{10}^{3}}=\frac{18}{1000}=0.018$
(vi) $4.129×{10}^{-3}=\frac{4129}{1000}×\frac{1}{{10}^{3}}=\frac{4129}{{10}^{3}×{10}^{3}}=\frac{4129}{{10}^{\left(3+3\right)}}=\frac{4129}{{10}^{6}}=\frac{4129}{1000000}=0.004129$
Question 1:
Tick (✓) the correct answer
The value of ${\left(\frac{2}{5}\right)}^{-3}$ is
(a) $-\frac{8}{125}$
(b) $\frac{25}{4}$
(c) $\frac{125}{8}$
(d) $-\frac{2}{5}$
(c) $\frac{125}{8}$
${\left(\frac{2}{5}\right)}^{-3}={\left(\frac{5}{2}\right)}^{3}=\frac{{5}^{3}}{{2}^{3}}=\frac{125}{8}$
Question 2:
Tick (✓) the correct answer
The value of (−3)−4 is
(a) 12
(b) 81
(c) $-\frac{1}{12}$
(d) $\frac{1}{81}$
(d) $\frac{1}{81}$
${\left(-3\right)}^{-4}=\frac{1}{{\left(-3\right)}^{4}}=\frac{1}{{\left(-1\right)}^{4}×{\left(3\right)}^{4}}=\frac{1}{{\left(3\right)}^{4}}=\frac{1}{81}$
Question 3:
Tick (✓) the correct answer
The value of (−2)−5 is
(a) −32
(b) $\frac{-1}{32}$
(c) 32
(d) $\frac{1}{32}$
(b) $\frac{-1}{32}$
${\left(-2\right)}^{-5}=\frac{1}{{\left(-2\right)}^{5}}=\frac{1}{-32}=\frac{1×\left(-1\right)}{-32×\left(-1\right)}=\frac{-1}{32}$
Question 4:
Tick (✓) the correct answer
(2−5 ÷ 2−2) = ?
(a) $\frac{1}{128}$
(b) $\frac{-1}{128}$
(c) $-\frac{1}{8}$
(d) $\frac{1}{8}$
(d) $\frac{1}{8}$
$\left({2}^{-5}÷{2}^{-2}\right)=\left(\frac{1}{{2}^{5}}÷\frac{1}{{2}^{2}}\right)=\left(\frac{1}{32}÷\frac{1}{4}\right)=\left(\frac{1}{32}×4\right)=\frac{4}{32}=\frac{1}{8}$
Question 5:
Tick (✓) the correct answer
The value of (3−1 + 4−1)−1 ÷ 5−1 is
(a) $\frac{7}{10}$
(b) $\frac{60}{7}$
(c) $\frac{7}{5}$
(d) $\frac{7}{15}$
(b) $\frac{60}{7}$
${\left({3}^{-1}+{4}^{-1}\right)}^{-1}÷{5}^{-1}={\left(\frac{1}{3}+\frac{1}{4}\right)}^{-1}÷\frac{1}{5}={\left(\frac{4+3}{12}\right)}^{-1}÷\frac{1}{5}={\left(\frac{7}{12}\right)}^{-1}÷\frac{1}{5}=\left(\frac{12}{7}\right)÷\frac{1}{5}=\frac{12}{7}×5=\frac{60}{7}$
Question 6:
Tick (✓) the correct answer
${\left(\frac{1}{2}\right)}^{-2}+{\left(\frac{1}{3}\right)}^{-2}+{\left(\frac{1}{4}\right)}^{-2}=?$
(a) $\frac{61}{144}$
(b) $\frac{144}{61}$
(c) 29
(d) $\frac{1}{29}$
(c) 29
Question 7:
Tick (✓) the correct answer
$\left\{{\left(\frac{1}{3}\right)}^{-3}-{\left(\frac{1}{2}\right)}^{-3}\right\}÷{\left(\frac{1}{4}\right)}^{-3}=?$
(a) $\frac{19}{64}$
(b) $\frac{27}{16}$
(c) $\frac{64}{19}$
(d) $\frac{16}{25}$
(a) $\frac{19}{64}$
$\left\{{\left(\frac{1}{3}\right)}^{-3}-{\left(\frac{1}{2}\right)}^{-3}\right\}÷{\left(\frac{1}{4}\right)}^{-3}\phantom{\rule{0ex}{0ex}}=\left\{{3}^{3}-{2}^{3}\right\}÷{4}^{3}\phantom{\rule{0ex}{0ex}}=\left\{27-8\right\}÷64\phantom{\rule{0ex}{0ex}}=19÷64\phantom{\rule{0ex}{0ex}}=\frac{19}{64}$
Question 8:
Tick (✓) the correct answer
${\left[{\left\{{\left(-\frac{1}{2}\right)}^{2}\right\}}^{-2}\right]}^{-1}=?$
(a) $\frac{1}{16}$
(b) 16
(c) $-\frac{1}{16}$
(d) −16
(a) $\frac{1}{16}$
${\left[{\left\{{\left(-\frac{1}{2}\right)}^{2}\right\}}^{-2}\right]}^{-1}\phantom{\rule{0ex}{0ex}}={\left[{\left\{-\frac{1}{2}\right\}}^{-4}\right]}^{-1}\phantom{\rule{0ex}{0ex}}={\left(-\frac{1}{2}\right)}^{\left(-4×-1\right)}\phantom{\rule{0ex}{0ex}}={\left(-\frac{1}{2}\right)}^{4}\phantom{\rule{0ex}{0ex}}=\frac{1}{16}$
Question 9:
Tick (✓) the correct answer
The value of x for which ${\left(\frac{7}{12}\right)}^{-4}×{\left(\frac{7}{12}\right)}^{3x}={\left(\frac{7}{12}\right)}^{5},$ is
(a) −1
(b) 1
(c) 2
(d) 3
(d) 3
Question 10:
Tick (✓) the correct answer
If (23x − 1 + 10) ÷ 7 = 6, then x is equal to
(a) −2
(b) 0
(c) 1
(d) 2
(d) 2
23x-1 = 42 -10
23x-1 = 32
23x-1 = 25
3x-1 = 5
3x = 6
Therefore,
x = 2
Question 11:
Tick (✓) the correct answer
${\left(\frac{2}{3}\right)}^{0}=?$
(a) $\frac{3}{2}$
(b) $\frac{2}{3}$
(c) 1
(d) 0
(c) 1
Using the law of exponents ${\left(\frac{a}{b}\right)}^{0}=1$:
$\therefore$ ${\left(\frac{2}{3}\right)}^{0}=1$
Question 12:
Tick (✓) the correct answer
${\left(\frac{-5}{3}\right)}^{-1}=?$
(a) $\frac{5}{3}$
(b) $\frac{3}{5}$
(c) $\frac{-3}{5}$
(d) none of these
(c) $\frac{-3}{5}$
${\left(\frac{-5}{3}\right)}^{-1}={\left(\frac{3}{-5}\right)}^{1}=\frac{3}{-5}=\frac{3×\left(-1\right)}{-5×\left(-1\right)}=\frac{-3}{5}$
Question 13:
Tick (✓) the correct answer
${\left(\frac{-1}{2}\right)}^{3}=?$
(a) $\frac{-1}{6}$
(b) $\frac{1}{6}$
(c) $\frac{1}{8}$
(d) $\frac{-1}{8}$
(d) $\frac{-1}{8}$
${\left(\frac{-1}{2}\right)}^{3}=\frac{-{1}^{3}}{{2}^{3}}=\frac{-1}{8}$
Question 14:
Tick (✓) the correct answer
${\left(\frac{-3}{4}\right)}^{2}=?$
(a) $\frac{-9}{16}$
(b) $\frac{9}{16}$
(c) $\frac{16}{9}$
(d) $\frac{-16}{9}$
(b) $\frac{9}{16}$
${\left(\frac{-3}{4}\right)}^{2}=\frac{{\left(-3\right)}^{2}}{{\left(4\right)}^{2}}=\frac{9}{16}$
Question 15:
Tick (✓) the correct answer
3670000 in standard form is
(a) 367 × 104
(b) 36.7 × 105
(c) 3.67 × 106
(d) none of these
(c) $3.67×{10}^{6}$
$3670000=367×{10}^{4}=3.67×100×{10}^{4}=3.67×{10}^{2}×{10}^{4}=3.67×{10}^{\left(2+4\right)}=3.67×{10}^{6}$
Question 16:
Tick (✓) the correct answer
0.0000463 in standard form is
(a) 463 × 10−7
(b) 4.63 × 10−5
(c) 4.63 × 10−9
(d) 46.3 × 10−6
(b) $4.63×{10}^{-5}$
$0.0000463=\frac{463}{{10}^{7}}=\frac{4.63×{10}^{2}}{{10}^{7}}=4.63×{10}^{\left(2-7\right)}=4.63×{10}^{-5}$
Question 17:
Tick (✓) the correct answer
0.000367 × 104 in usual form is
(a) 3.67
(b) 36.7
(c) 0.367
(d) 0.0367
(a) 3.67
$0.000367×{10}^{4}=\frac{367}{{10}^{6}}×{10}^{4}=367×{10}^{\left(4-6\right)}=367×{10}^{-2}=\frac{367}{{10}^{2}}=\frac{367}{100}=3.67$
Question 1:
Evaluate:
(i) 3−4
(ii) (−4)3
(iii) ${\left(\frac{3}{4}\right)}^{-2}$
(iv) ${\left(\frac{-2}{3}\right)}^{-5}$
(v) ${\left(\frac{5}{7}\right)}^{0}$
(i) ${3}^{-4}=\frac{1}{{3}^{4}}=\frac{1}{81}$
(ii) ${\left(-4\right)}^{3}={\left(-1\right)}^{3}×{\left(4\right)}^{3}=-1×64=-64$
(iii) ${\left(\frac{3}{4}\right)}^{-2}={\left(\frac{4}{3}\right)}^{2}=\frac{{4}^{2}}{{3}^{2}}=\frac{16}{9}$
(iv) ${\left(\frac{-2}{3}\right)}^{-5}={\left(\frac{3}{-2}\right)}^{5}=\frac{{3}^{5}}{-{2}^{5}}=\frac{243}{-32}=\frac{243×-1}{-32×-1}=\frac{-243}{32}$
(v) Using the property ${\left(\frac{a}{b}\right)}^{0}=1$, we have:
Question 2:
Evaluate: ${\left\{{\left(\frac{-2}{3}\right)}^{3}\right\}}^{-2}.$
${\left\{{\left(\frac{-2}{3}\right)}^{3}\right\}}^{-2}={\left(\frac{-2}{3}\right)}^{-6}={\left(\frac{3}{-2}\right)}^{6}=\frac{{3}^{6}}{-{2}^{6}}=\frac{729}{64}$
Question 3:
Simplify: $\left({3}^{-1}+{6}^{-1}\right)÷{\left(\frac{3}{4}\right)}^{-1}.$
$\left({3}^{-1}+{6}^{-1}\right)÷{\left(\frac{3}{4}\right)}^{-1}\phantom{\rule{0ex}{0ex}}=\left(\frac{1}{3}+\frac{1}{6}\right)÷{\left(\frac{4}{3}\right)}^{1}\phantom{\rule{0ex}{0ex}}=\left(\left[\frac{1×2}{3×2}\right]+\left[\frac{1×1}{6×1}\right]\right)÷\left(\frac{4}{3}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{2+1}{6}\right)÷\left(\frac{4}{3}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{3}{6}\right)÷\left(\frac{4}{3}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{1}{2}\right)÷\left(\frac{4}{3}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{1}{2}\right)×\left(\frac{3}{4}\right)\phantom{\rule{0ex}{0ex}}=\frac{3}{8}$
Question 4:
By what number should ${\left(\frac{-2}{3}\right)}^{-3}$ be divided so that the quotient is ${\left(\frac{4}{9}\right)}^{-2}$?
Let the number be $x$.
$\therefore$ ${\left(\frac{-2}{3}\right)}^{-3}÷x={\left(\frac{4}{9}\right)}^{-2}$
Question 5:
By what number should (−3)−1 be multiplied so that the product becomes 6−1?
Let the number be $x$.
$\therefore$ ${\left(-3\right)}^{-1}×x={\left(6\right)}^{-1}$
Question 6:
Express each of the following in standard form:
(i) 345
(ii) 180000
(iii) 0.000003
(iv) 0.000027
(i) $345=3.45×100=3.45×{10}^{2}$
(ii) $180000=18×10000=18×{10}^{4}=1.8×10×{10}^{4}=1.8×{10}^{\left(1+4\right)}=1.8×{10}^{5}$
(iii) $0.000003=\frac{3}{1000000}=3×{10}^{-6}$
(iv) $0.000027=\frac{27}{100000}=\frac{27}{{10}^{6}}=\frac{2.7×10}{{10}^{6}}=2.7×{10}^{\left(1-6\right)}=2.7×{10}^{-5}$
Question 7:
Mark (✓) against the correct answer
The value of (−3)−3 is
(a) −27
(b) 9
(c) $\frac{-1}{27}$
(d) $\frac{1}{27}$
(c) $\frac{-1}{27}$
${\left(-3\right)}^{-3}={\left(\frac{1}{-3}\right)}^{3}\phantom{\rule{0ex}{0ex}}=\frac{{1}^{3}}{-{3}^{3}}\phantom{\rule{0ex}{0ex}}=\frac{1}{-27}\phantom{\rule{0ex}{0ex}}=\frac{1×-1}{-27×-1}\phantom{\rule{0ex}{0ex}}=\frac{-1}{27}$
Question 8:
Mark (✓) against the correct answer
The value of ${\left(\frac{3}{4}\right)}^{-3}$ is
(a) $\frac{-27}{64}$
(b) $\frac{64}{27}$
(c) $\frac{-9}{4}$
(d) $\frac{27}{64}$
(b) $\frac{64}{27}$
${\left(\frac{3}{4}\right)}^{-3}={\left(\frac{4}{3}\right)}^{3}=\frac{{4}^{3}}{{3}^{3}}=\frac{64}{27}$
Question 9:
Mark (✓) against the correct answer
(3−6 ÷ 34) = ?
(a) 3−2
(b) 32
(c) 3−10
(d) 310
(c) ${3}^{-10}$
$\left({3}^{-6}÷{3}^{4}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{1}{{3}^{6}}÷{3}^{4}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{{3}^{6}}×\frac{1}{{3}^{4}}\phantom{\rule{0ex}{0ex}}=\frac{1}{{3}^{\left(6+4\right)}}\phantom{\rule{0ex}{0ex}}=\frac{1}{{3}^{10}}\phantom{\rule{0ex}{0ex}}={3}^{-10}$
Question 10:
Mark (✓) against the correct answer
If
(a) −1
(b) 1
(c) 2
(d) 3
(d) 3
Question 11:
Mark (✓) against the correct answer
${\left(\frac{3}{5}\right)}^{0}=?$
(a) $\frac{5}{3}$
(b) $\frac{3}{5}$
(c) 1
(d) 0
(c) 1
Using the law of exponents, which says ${\left(\frac{a}{b}\right)}^{0}=1,$ we get:
Question 12:
Mark (✓) against the correct answer
${\left(\frac{-6}{5}\right)}^{-1}=?$
(a) $\frac{6}{5}$
(b) $\frac{-6}{5}$
(c) $\frac{5}{6}$
(d) $\frac{-5}{6}$
(d) $\left(\frac{-5}{6}\right)$
${\left(\frac{-6}{5}\right)}^{-1}={\left(\frac{5}{-6}\right)}^{1}=\frac{5}{-6}=\frac{5×-1}{-6×-1}=\frac{-5}{6}$
Question 13:
Mark (✓) against the correct answer
${\left(\frac{-1}{3}\right)}^{3}=?$
(a) $\frac{-1}{9}$
(b) $\frac{1}{9}$
(c) $\frac{-1}{27}$
(d) $\frac{1}{27}$
(c) $\frac{-1}{27}$
${\left(\frac{-1}{3}\right)}^{3}=\frac{-{1}^{3}}{{3}^{3}}=\frac{-1}{27}$
Question 14:
Fill in the blanks.
(i) 360000 written in standard form is .........
(ii) 0.0000123 written in standard form is .........
(iii) ${\left(\frac{-2}{3}\right)}^{-2}=.........$
(iv) 3 × 10−3 in usual form is .........
(v) 5.32 × 10−4 in usual form is .........
(i) The standard form of 36000 is $3.6×{10}^{5}$
$360000=36×{10}^{4}=3.6×10×{10}^{4}=3.6×{10}^{\left(1+4\right)}=3.6×{10}^{5}$
(ii) The standard form of 0.0000123 is $1.23×{10}^{-5}$.
$0.0000123=\frac{123}{10000000}=\frac{123}{{10}^{7}}=\frac{1.23×100}{{10}^{7}}=\frac{1.23×{10}^{2}}{{10}^{7}}=1.23×{10}^{\left(2-7\right)}=1.23×{10}^{-5}$
(iii) ${\left(\frac{-2}{3}\right)}^{-2}=\frac{9}{4}$
${\left(\frac{-2}{3}\right)}^{-2}={\left(\frac{3}{-2}\right)}^{2}=\frac{{3}^{2}}{-{2}^{2}}=\frac{9}{4}$
(iv)
$3×{10}^{-3}=\frac{3}{{10}^{3}}=\frac{3}{1000}=0.003$
(v)
View NCERT Solutions for all chapters of Class 8
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Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °
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## #1 2012-10-21 23:53:10
Sabbir
Guest
### Gamma function
I want to know about gamma function,I have read about it in wikipedia,but I could not understand it very much,so please help(I know about factorial).
## #2 2012-10-22 00:40:15
bobbym
Offline
### Re: Gamma function
Hi Sabbir;
Welcome to the forum;
For positive integers
so Gamma(5) = 4! = 24.
The gamma function also defined for other than integers.
But that is all in the pages you visited. Do you have a specific problem that requires the Gamma function?
In mathematics, you don't understand things. You just get used to them.
90% of mathematicians do not understand 90% of currently published mathematics.
I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on.
## #3 2012-10-22 00:43:24
zetafunc.
Guest
### Re: Gamma function
What do you need help with?
I understand the gamma function as being an extension to the factorial function, defining it for real and complex numbers. It is defined as
or
.
This integral can be annoying to evaluate, so quite often people use a table of values, or use some known properties of distributions to solve. For simplicity, let's try to evaluate Gamma(0.5).
I'll rewrite this as
I'll make a substitution;
So our integral becomes
Do you recognise this integral? It is a disguised form of something called the Gaussian integral. It's defined as
Since it's an even function (it's symmetrical about the y-axis), the area from 0 to infinity is just half of the area from negative infinity to positive infinity. Thus, your integral evaluates to √π. Hope this helps!
## #4 2012-10-22 00:51:43
zetafunc.
Guest
### Re: Gamma function
You can also evaluate the Gaussian integral using polar co-ordinates (rather than just accepting my definition for it). It's a bit long but I can show you if you'd like.
## #5 2012-10-22 01:07:05
Sabbir
Guest
### Re: Gamma function
I think if you show me some examples(such as other fractions and complex number)it will be easy for me to understand.
## #6 2012-10-22 01:10:02
zetafunc.
Guest
### Re: Gamma function
You mean evaluating the gamma function at different values -- 0.75, 0.224, (3+2i), etc.?
Was any part of my explanation unclear?
## #7 2012-10-22 01:35:47
Sabbir
Guest
### Re: Gamma function
Yes,different values and your explanation was clear but other values will make it easier for me to understand the way of approaching
## #8 2012-10-22 01:53:05
bobbym
Offline
### Re: Gamma function
Hi;
Only certain values yield nice results like that. For say Gamma(10 + 1/3 ) you will have to resort to numerical methods.
In mathematics, you don't understand things. You just get used to them.
90% of mathematicians do not understand 90% of currently published mathematics.
I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on.
## #9 2012-10-22 01:56:24
zetafunc.
Guest
### Re: Gamma function
There are lots of ways to compute values for the gamma function. For example, this is a useful identity;
Thus, since
then it follows that
so you can use that to find values of the gamma function at 5/2, 7/2, etc.
## #10 2012-10-22 02:00:29
Sabbir
Guest
### Re: Gamma function
Ok,but it will be helpful to see a complex number example.
## #11 2012-10-22 02:07:21
bobbym
Offline
### Re: Gamma function
Those get very complicated. One way is with a Taylor series expanded around infinity.
This is quite close for only two terms of a Taylor series.
In mathematics, you don't understand things. You just get used to them.
90% of mathematicians do not understand 90% of currently published mathematics.
I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on.
## #12 2012-10-22 02:11:46
zetafunc.
Guest
### Re: Gamma function
Evaluating it at complex numbers is interesting. Let's try to find Gamma(1+i).
but note that
so our integral becomes
I can do that becaues ei is just a constant. Can you see where to go from here?
## #13 2012-10-22 02:14:44
zetafunc.
Guest
### Re: Gamma function
Sorry, I am wrong.
. Ignore my previous post.
## #14 2012-10-22 02:41:30
Sabbir
Guest
### Re: Gamma function
So,will the taylor series give the same result as the integral's.
## #15 2012-10-22 02:47:19
zetafunc.
Guest
### Re: Gamma function
#### Sabbir wrote:
So,will the taylor series give the same result as the integral's.
They will give you a (very good) approximation to the integral, yes.
May I ask what this is for?
## #16 2012-10-22 02:52:18
bobbym
Offline
### Re: Gamma function
You can also evaluate the integral numerically.
In mathematics, you don't understand things. You just get used to them.
90% of mathematicians do not understand 90% of currently published mathematics.
I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on.
## #17 2012-10-22 03:00:38
Sabbir
Guest
### Re: Gamma function
I was just curious about the taylor series
## #18 2012-10-22 03:02:44
zetafunc.
Guest
### Re: Gamma function
#### Sabbir wrote:
I was just curious about the taylor series
I mean, about the Gamma function -- is it for a particular class?
## #19 2012-10-22 08:48:48
bobbym
Offline
### Re: Gamma function
Hi;
I was just curious about the taylor series
What would you like to know?
In mathematics, you don't understand things. You just get used to them.
90% of mathematicians do not understand 90% of currently published mathematics.
I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on.
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https://www.acmicpc.net/problem/16296
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시간 제한메모리 제한제출정답맞힌 사람정답 비율
5 초 512 MB93353243.243%
## 문제
Oostende Beach is a very long beach located in the north of Belgium. On this beach, there are n huts located along a straight line. People can rent a room in one of those huts to spend their beach vacations together with the other tenants.
Every day at lunch time, a food truck rides by to serve fries to the guests. The truck will park in front of one of the huts and people will form two queues. The people staying in huts to the left of the food truck will queue on the left, and the people to the right of the food truck will queue together on the right. The people staying in the hut in front of the food truck will split their group in half, one half going to the left queue and the other half going to the right queue. If this is an odd number of people, the remaining person will go to the queue with fewer people, or choose one randomly if the queues have the same length. The food truck will always position itself so that the difference between the number of people in the left queue and the number of people in the right queue is as small as possible.
Each night the number of guests in exactly one of the huts changes. Can you help the food truck find the best position for each day?
## 입력
• The first line of the input consists of two integers 1 ≤ n ≤ 105, the number of huts, and 1 ≤ q ≤ 105, the number of days.
• The second line has n integers a0, . . . , an−1 satisfying 1 ≤ ai ≤ 106 for 0 ≤ i < n, where ai is the current number of people in hut i.
• Then follow q lines with two integers 0 ≤ i < n and 1 ≤ x ≤ 106. The j th of these lines indicates that at day j the number of people in hut i changes to x.
## 출력
• Print q lines: the optimal position of the foodtruck after each of the q nights. If there are multiple optimal positions, print the smallest one.
## 예제 입력 1
5 4
3 1 3 4 2
0 5
0 9
4 5
2 1
## 예제 출력 1
2
1
2
1
## 예제 입력 2
4 8
1 1 1 1
2 2
1 2
2 1
1 1
3 2
2 2
1 2
2 1
## 예제 출력 2
1
1
1
1
1
2
2
2
## 출처
• 문제를 만든 사람: Ragnar Groot Koerkamp
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Welcome, guest | Sign In | My Account | Store | Cart
```from fractions import Fraction
"""spread(start, end, count [, mode]) -> generator
Yield a sequence of evenly-spaced numbers between start and end.
The range start...end is divided into count evenly-spaced (or as close to
evenly-spaced as possible) intervals. The end-points of each interval are
then yielded, optionally including or excluding start and end themselves.
By default, start is included and end is excluded.
For example, with start=0, end=2.1 and count=3, the range is divided into
three intervals:
(0.0)-----(0.7)-----(1.4)-----(2.1)
resulting in:
[0.0, 0.7, 1.4]
Optional argument mode controls whether spread() includes the start and
end values. mode must be an int. Bit zero of mode controls whether start
is included (on) or excluded (off); bit one does the same for end. Hence:
0 -> open interval (start and end both excluded)
1 -> half-open (start included, end excluded)
2 -> half open (start excluded, end included)
3 -> closed (start and end both included)
By default, mode=1 and only start is included in the output.
(Note: depending on mode, the number of values returned can be count,
count-1 or count+1.)
"""
if not isinstance(mode, int):
raise TypeError('mode must be an int')
if count != int(count):
raise ValueError('count must be an integer')
if count <= 0:
raise ValueError('count must be positive')
if mode & 1:
yield start
width = Fraction(end-start)
start = Fraction(start)
for i in range(1, count):
yield float(start + i*width/count)
if mode & 2:
yield end
```
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https://fdocuments.us/document/friday-aug-28-todays-warm-up-lt11a-i-can-define-sketch-draw-and-correctly.html
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of 34 /34
Signed Course Syllabus is due FRIDAY HW #1 page 33 / pr 1 - 12, 19 - 26 is due TODAY! Take out your notebook for today's warm - up! LT1.1A - I can define, sketch, draw, and correctly label basic geometric figures and their parts. LT1.2A - I can define, sketch, draw, and correctly label angles and their parts. Friday Aug 28, 2015 Richard Sudo Friday, August 28, 2015 11:11:58 AM CT 00:19:e3:4a:d2:21
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### Transcript of Friday Aug 28, 2015washburn.mpls.k12.mn.us/uploads/friday_aug_28_slides_from_class.pdf · today's...
Iwillbeafterschoolformake-uponMon,Thurs
SignedCourseSyllabusisdueFRIDAYHW#1page33/pr1-12,19-26isdueTODAY!
QUIZFridayonCh12-LT12A,LT12B,andWordProbs
Takeoutyournotebookfortoday'swarm-up!
LT1.1A-Icandefine,sketch,draw,andcorrectlylabelbasicgeometricfiguresandtheirparts.
LT1.2A-Icandefine,sketch,draw,andcorrectlylabelanglesandtheirparts.
FridayAug28,2015
Richard Sudo Friday, August 28, 2015 11:11:58 AM CT 00:19:e3:4a:d2:21
Warm-up
Richard Sudo Friday, August 28, 2015 11:11:58 AM CT 00:19:e3:4a:d2:21
Richard Sudo Friday, August 28, 2015 11:11:58 AM CT 00:19:e3:4a:d2:21
HW#1page33/pr1-12,18-26dueThursdayAug28
Richard Sudo Friday, August 28, 2015 11:11:58 AM CT 00:19:e3:4a:d2:21
PT,TP
AR,RT,AT,TR,RA,TA
MA,MS,AS,SA,SM,AM
Richard Sudo Friday, August 28, 2015 11:11:58 AM CT 00:19:e3:4a:d2:21
AC,CA
PQ,QP
RI,TI,TRIR,IT,RT
Richard Sudo Friday, August 28, 2015 11:11:58 AM CT 00:19:e3:4a:d2:21
Richard Sudo Friday, August 28, 2015 11:11:58 AM CT 00:19:e3:4a:d2:21
Richard Sudo Friday, August 28, 2015 11:11:58 AM CT 00:19:e3:4a:d2:21
Richard Sudo Friday, August 28, 2015 11:11:58 AM CT 00:19:e3:4a:d2:21
Richard Sudo Friday, August 28, 2015 11:11:58 AM CT 00:19:e3:4a:d2:21
Richard Sudo Friday, August 28, 2015 11:11:58 AM CT 00:19:e3:4a:d2:21
Richard Sudo Friday, August 28, 2015 11:11:58 AM CT 00:19:e3:4a:d2:21
tworayscommonendpoint
vertexsides
Richard Sudo Friday, August 28, 2015 11:11:58 AM CT 00:19:e3:4a:d2:21
protractor
measureofananglerotation
degrees
Richard Sudo Friday, August 28, 2015 11:11:58 AM CT 00:19:e3:4a:d2:21
RV Q
S T
Richard Sudo Friday, August 28, 2015 11:11:58 AM CT 00:19:e3:4a:d2:21
congruent
Richard Sudo Friday, August 28, 2015 11:11:58 AM CT 00:19:e3:4a:d2:21
anglebisector
Richard Sudo Friday, August 28, 2015 11:11:58 AM CT 00:19:e3:4a:d2:21
Richard Sudo Friday, August 28, 2015 11:11:58 AM CT 00:19:e3:4a:d2:21
Richard Sudo Friday, August 28, 2015 11:11:58 AM CT 00:19:e3:4a:d2:21
Homework
Richard Sudo Friday, August 28, 2015 11:11:58 AM CT 00:19:e3:4a:d2:21
TODAY'SHomework#2page42/prob7-20,30-35DUETUESDAYSept1
Richard Sudo Friday, August 28, 2015 11:11:58 AM CT 00:19:e3:4a:d2:21
#2page42/prob7-20,30-35-dueTUES9/1
Richard Sudo Friday, August 28, 2015 11:11:58 AM CT 00:19:e3:4a:d2:21
Richard Sudo Friday, August 28, 2015 11:11:58 AM CT 00:19:e3:4a:d2:21
Richard Sudo Friday, August 28, 2015 11:11:58 AM CT 00:19:e3:4a:d2:21
Richard Sudo Friday, August 28, 2015 11:11:58 AM CT 00:19:e3:4a:d2:21
Richard Sudo Friday, August 28, 2015 11:11:58 AM CT 00:19:e3:4a:d2:21
Richard Sudo Friday, August 28, 2015 11:11:58 AM CT 00:19:e3:4a:d2:21
Richard Sudo Friday, August 28, 2015 11:11:58 AM CT 00:19:e3:4a:d2:21
Richard Sudo Friday, August 28, 2015 11:11:58 AM CT 00:19:e3:4a:d2:21
Richard Sudo Friday, August 28, 2015 11:11:58 AM CT 00:19:e3:4a:d2:21
Richard Sudo Friday, August 28, 2015 11:11:58 AM CT 00:19:e3:4a:d2:21
Richard Sudo Friday, August 28, 2015 11:11:58 AM CT 00:19:e3:4a:d2:21
Richard Sudo Friday, August 28, 2015 11:11:58 AM CT 00:19:e3:4a:d2:21
ABCDE
Richard Sudo Friday, August 28, 2015 11:11:58 AM CT 00:19:e3:4a:d2:21
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https://math.answers.com/Q/How_much_is_80_percent_of_42000
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0
# How much is 80 percent of 42000?
Wiki User
2013-04-19 17:14:42
80% of 42,000
= 80% * 42000
= 0.8 * 42000
= 33,600
Wiki User
2013-04-19 17:14:42
Study guides
20 cards
➡️
See all cards
3.74
1018 Reviews
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https://snbchf.com/monetary-fiscal-policy/natural-rate-interest-taylor-rate/
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# The Natural Rate of Interest/Taylor Rate
Central banks often want to follow two contrary targets: low inflation and low unemployment. In order to achieve both criteria they look for a formula that can incorporate both of them. This is the idea behind the “Taylor Rate”.
## Natural interest rate according to Taylor/Greenspan
“In practice, of course, the two problems of unemployment and inflation are competing with one another for the attention of Federal Reserve’s policymaking committee. The Federal Open Market Committee (FOMC) has to strike a balance between lowering interest rates and raising interest rates. It would actually lower or raise the federal funds target rate if one problem is judged to be more serious or more pressing than the other. Over time, the FOMC’s efforts to fight inflation and fight unemployment gives rise to a sequence of changes in the federal funds rate.
The actual pattern of federal funds rate during the early Greenspan years (1987–1993) is described by a simple equation introduced by John B. Taylor (1993) of Stanford University:
r = p + 0.5 q + 0.5 ( p – 2 ) + 2
where r is the targeted federal funds rate, p is the inflation rate over the previous year, and q is the percentage deviation of actual output from full-employment output. Taylor himself writes the equation using income (y) instead of output (q), but he defines y in terms of real GDP. In effect, yis a measure of q. The simple equation could be written in a still simpler form:
r = 1.5 p + 0.5 q + 1 but the original rendering has more intuitive appeal. It suggests that the implicit goal of the Federal Reserve is “full employment” and “2 percent inflation.” Note that if q = 0 (i.e., no deviation from full employment) and p = 2 percent, then r would be 4 percent. That is, the targeted federal funds rate would be 2 percentage points above the (2 percent) inflation rate. The two coefficients of 0.5 give equal weighting to the problems of unemployment and inflation generally. In particular instances, of course, one of those problems may be more severe than the other — as would be indicated by the actual values of p and q. Thus, the targeted federal funds rate r is low with a high and negative q; it is high with a high p. The discretion needed for the Federal Reserve to fight the good fight (against unemployment and inflation) stands in contrast to the adoption of a Monetary Rule as advocated by Milton Friedman. According to this rule, the Federal Reserve should increase the money supply year-in and year-out at a slow and steady rate that approximates the economy’s long-run growth rate of 2 or 3 percent.”
source
In our calculations for the Swiss Taylor rate, we use the natural rate of interest (sometimes also “Neutral”) according a Taylor and Greenspan. For its calculations we use the CPI over the corresponding and the preceding quarter and the unemployment rate of the same quarter as basis. Swiss full employment is for us at 2.8%, in 2008 there was over-employment.
George Dorgan (penname) predicted the end of the EUR/CHF peg at the CFA Society and at many occasions on SeekingAlpha.com and on this blog. Several Swiss and international financial advisors support the site. These firms aim to deliver independent advice from the often misleading mainstream of banks and asset managers. George is FinTech entrepreneur, financial author and alternative economist. He speak seven languages fluently.
See more for
Permanent link to this article: https://snbchf.com/monetary-fiscal-policy/natural-rate-interest-taylor-rate/
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http://www.numbersaplenty.com/12399
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Search a number
12399 = 34133
BaseRepresentation
bin11000001101111
3122000020
43001233
5344044
6133223
751102
oct30157
918006
1012399
119352
127213
13584a
144739
153a19
hex306f
12399 has 4 divisors (see below), whose sum is σ = 16536. Its totient is φ = 8264.
The previous prime is 12391. The next prime is 12401. The reversal of 12399 is 99321.
It can be divided in two parts, 123 and 99, that added together give a palindrome (222).
It is a happy number.
It is a semiprime because it is the product of two primes, and also an emirpimes, since its reverse is a distinct semiprime: 99321 = 333107.
It is a cyclic number.
It is not a de Polignac number, because 12399 - 23 = 12391 is a prime.
It is a D-number.
12399 is a lucky number.
It is a plaindrome in base 10.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (12391) by changing a digit.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 2064 + ... + 2069.
It is an arithmetic number, because the mean of its divisors is an integer number (4134).
212399 is an apocalyptic number.
12399 is a deficient number, since it is larger than the sum of its proper divisors (4137).
12399 is an equidigital number, since it uses as much as digits as its factorization.
12399 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 4136.
The product of its digits is 486, while the sum is 24.
The square root of 12399 is about 111.3507970335. The cubic root of 12399 is about 23.1452683272.
The spelling of 12399 in words is "twelve thousand, three hundred ninety-nine".
Divisors: 1 3 4133 12399
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http://mathhelpforum.com/advanced-math-topics/160490-period-signal.html
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1. ## Period of signal
Stuck on the following problem, aint sure how to work out the signal of a composed sinusoid
given that $\displaystyle x(t) = cos(600\pi t)$ and $\displaystyle m(t) = 4 + 3cos(20\pi t)$ what is the period (if any) of the resulting signal
Also the signal $\displaystyle x(t) = 10cos(600\pi t + \frac{\pi}{3}) + 6cos(900\pi t - \frac{\pi}{4}) - 4cos(1500\pi t)$ whats the period of the signal
Im not sure how to calculate the period when theres more than one sinusoid wave
Any help most appreciated
2. Originally Posted by jordanrs
Stuck on the following problem, aint sure how to work out the signal of a composed sinusoid
Also the signal $\displaystyle x(t) = 10cos(600\pi t + \frac{\pi}{3}) 6cos(900\pi t - \frac{\pi}{4}) - 4cos(1500\pi t)$ whats the period of the signal
Im not sure how to calculate the period when theres more than one sinusoid wave
Any help most appreciated
First question snipped as it not complete.
The product of two sinusoids has sinusoidal components at the sum and diffrence frequencies. This for some amplitudes (non-zero) and phases is:
$\displaystyle x(t)=A\sin(300\pi t+\phi_1)+B\sin(1500\pi t+\phi_2)$
and since the larger period is an integer multiple of the smaller the final period is $\displaystyle 1/150$ s
CB
3. Originally Posted by jordanrs
Stuck on the following problem, aint sure how to work out the signal of a composed sinusoid
given that $\displaystyle x(t) = cos(600\pi t)$ and $\displaystyle m(t) = 4 + 3cos(20\pi t)$ what is the period (if any) of the resulting signal
Also the signal $\displaystyle x(t) = 10cos(600\pi t + \frac{\pi}{3}) 6cos(900\pi t - \frac{\pi}{4}) - 4cos(1500\pi t)$ whats the period of the signal
Im not sure how to calculate the period when theres more than one sinusoid wave
Any help most appreciated
when $\displaystyle y(t) = x(t)m(t)$ and $\displaystyle x(t) = cos(600\pi t)$ and $\displaystyle m(t) = 4 + 3cos(20\pi t)$ what is the resulting period of $\displaystyle (y(t))$
also i still dont get how to calculate the periodm i dont see where you got 1/150 s from
4. Originally Posted by jordanrs
when $\displaystyle y(t) = x(t)m(t)$ and $\displaystyle x(t) = cos(600\pi t)$ and $\displaystyle m(t) = 4 + 3cos(20\pi t)$ what is the resulting period of $\displaystyle (y(t))$
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- dotTech - https://dottech.org -
# How to find factorials in Excel [Tip]
The factorial of a number is the product of all numbers less than n. For example, the factorial of four would be four x three x two x one, which amounts to 24. You can find the factorial of number in Excel 2010/13 with the FACT function.
Open your Excel spreadsheet, and enter the number you want to find the factorial for in a cell. Then select a cell to add the function to. Click the fx button to open the window in shot below.
[1]
Select FACT from that window. That will open the window shown below. Press the button beside the text, and then select the cell that includes the number to find the factorial for it.
[2]
Press the OK button to close the window. The function will then show you the factorial for the selected number as below.
[3]
Another way you can find the factorial of a number is to select a cell and then type =FACT in the function bar. Beside that enter the number you want to find the factorial for in brackets. For example, you could enter =FACT(4) and press Enter. The cell will then include the factorial for that number.
With that function you can now find the factorial for a variety of numbers in Excel 2010/13 spreadsheets. This YouTube video [4] also shows you how to find factorials with Excel.
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# Section 5.28. Generating Random Numbers
### 5.27. Finding a Correlation Coefficient
The correlation coefficient is one of the simplest and most universally useful statistical measures. It is a measure of the "linearity" of a set of x-y pairs, ranging from -1.0 (complete negative correlation) to +1.0 (complete positive correlation).
We compute this using the mean and sigma (standard deviation) functions defined previously in sections 5.25 and 5.26. For an explanation of this tool, consult any statistics text.
The following version assumes two arrays of numbers (of the same size):
`def correlate(x,y) sum = 0.0 x.each_index do |i| sum += x[i]*y[i] end xymean = sum/x.size.to_f xmean = mean(x) ymean = mean(y) sx = sigma(x) sy = sigma(y) (xymean-(xmean*ymean))/(sx*sy) end a = [3, 6, 9, 12, 15, 18, 21] b = [1.1, 2.1, 3.4, 4.8, 5.6] c = [1.9, 1.0, 3.9, 3.1, 6.9] c1 = correlate(a,a) # 1.0 c2 = correlate(a,a.reverse) # -1.0 c3 = correlate(b,c) # 0.8221970228`
The following version is similar, but it operates on a single array, each element of which is an array containing an x-y pair:
`def correlate2(v) sum = 0.0 v.each do |a| sum += a[0]*a[1] end xymean = sum/v.size.to_f x = v.collect {|a| a[0]} y = v.collect {|a| a[1]} xmean = mean(x) ymean = mean(y) sx = sigma(x) sy = sigma(y) (xymean-(xmean*ymean))/(sx*sy) end d = [[1,6.1], [2.1,3.1], [3.9,5.0], [4.8,6.2]] c4 = correlate2(d) # 0.2277822492`
Finally, the following version assumes that the x-y pairs are stored in a hash. It simply builds on the previous example:
`def correlate_h(h) correlate2(h.to_a) end e = { 1 => 6.1, 2.1 => 3.1, 3.9 => 5.0, 4.8 => 6.2} c5 = correlate_h(e) # 0.2277822492`
The Ruby Way, Second Edition: Solutions and Techniques in Ruby Programming (2nd Edition)
ISBN: 0672328844
EAN: 2147483647
Year: 2004
Pages: 269
Authors: Hal Fulton
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0
# How many yards is 300 centimeters?
Updated: 9/21/2023
Wiki User
11y ago
There are 3.2808399 yards in 300 centimeters.
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Related questions
### How many yards are there in 300 centimeters?
There are 91.44 centimetres in one yard. Therefore, rounded to two decimal places, 300 centimetres is equal to 300/91.44 = 3.28 yards.
### How many cm in 10 yards?
There are approximately 9144 centimeters in one hundred yards. 100 yd = 300 ft 300 ft = 3600 in 3600 in = 9144 cm 1 in = 2.54 cm
### How many centimeters in 10 yards?
10 yards = 914.4 centimeters
### How many centimeters are in 120 yards?
120 yards = 10,972.8 centimeters.
### If there are 4025 centimeters how many yards is it?
4025 centimeters = 44.018 yards.
### How many centimeters are in 12 yards?
12 yards=1,097.28 centimeters
### How many centimeters are there in 5 yards?
5 yards = 457.2 centimeters
### How many centimeters are in 70 yards?
70 yards = 6400.8 centimeters
### How many centimeters in 440 yards?
440 yards is 40,234 centimeters.
### How many centimeters are in 2.3 yards?
2.3 yards is 210.31 centimeters.
### How many centimeters are in 0.4 yards?
0.4 yards is 36.58 centimeters.
### How many centimeters are in 2 yards?
2 yards = 182.88 centimeters
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The first stock I ever owned I didn't even buy -- it was a gift.
My uncle worked for Brown Shoe (NYSE: BWS) in the 1980s. I was a kid back then, and knew little about the market, but he purchased about 20 shares of his employer for me and my sister and cousins.
I did know that the Brown Shoe dividend checks I got in the mail every quarter were incredible. As a kid in school, I needed money to buy junk food, Birch beer, baseball cards, and arcade games. I made money by doing the stuff kids do to make money: delivering newspapers, raking leaves, stacking wood. They were labor-intensive and didn't pay all that well. By contrast, those dividend checks were effortless.
Brown Shoe would go on to break my young heart -- but not before teaching me a good lesson about dividends.
Let me explain
On the face of it, the company looks like a dividend overachiever. Brown Shoe has paid 353 consecutive quarterly dividends -- that's 88 straight years of returning cash to shareholders. Over the past five years, it has grown its dividend by an annualized 9.5%.
In the late 1980s, when I was given shares, Brown Shoe was paying a quarterly dividend of \$0.40. Here's the quick math on my stake (Note: These numbers are approximate because I don't have the actual late-'80s paperwork):
• 20 shares paying \$0.40 = \$8.00 a quarter, \$32 a year
• \$32 a year back then = \$54 today, adjusted for inflation
A lesson in income investing
That's not the kind of money anyone could subsist on, but during high school and then college, I counted on those checks to help pay for gas and food (and by food I mean beer). But one day in the mid-'90s, the check arrived and, without warning, it was smaller. A lot smaller.
I wasn't keeping tabs on the stock back then, but in the third quarter of 1995, Brown Shoe announced a 37.5% dividend reduction. The company didn't hide the reasons:
The dividend had been increased or maintained for 20 years, and over the years had contributed very substantially to shareholder returns. But in recent years the dividend has principally been supported by cash flow from structural change -- business sale or liquidation and plant closings. The conclusion of these structural changes, difficult retail business and pressure on operating earnings in 1995 which was expected to continue into 1996, and the continuing high priority of protecting the balance sheet and our capability to finance the operation of the company, collectively made necessary the lower dividend. [emphasis mine]
Some two years later in 1997, the Brown Shoe board reduced the dividend again, from \$0.25 all the way down to \$0.10. By then, I was left with two bucks a quarter:
• 20 shares paying \$0.10 = \$2.00 a quarter, \$8 a year (down from \$32 a year)
Over the course of two years, my dividends were 75% lighter, and I had absolutely no idea it was coming.
I want my two dollars
Perhaps heartbreak and misery overstate the matter, but I really was bummed when this glorious, passive stream of income suddenly dried up.
Brown Shoe cut its payout because "in recent years the dividend has principally been supported by cash flow from structural change." As I became a student of investing, I embraced the simple lesson that dividend cut taught me: Focus on the free cash flow payout ratio.
The regular payout ratio -- dividing dividends per share by earnings per share to arrive at the percentage of a stock's earnings paid out to shareholders -- is a fair way to gauge the health of dividend payments. In the year ended 1993, Brown Shoe's payout ratio was 594%. Ouch.
Even still, earnings do not equal cash. A healthy, growing dividend stock will pay future dividends because its core business is throwing off plenty of cash. In the two years before Brown Shoe cut its dividend for the first time, the company had negative free cash flow. The same was true in 1996 and 1997, just before the second cut came. If it was to continue paying a good dividend, the company would have to borrow money or sell assets, neither solution sustainable.
The FCF payout ratio, then, is as simple as it sounds: Divide dividends per share by free cash flow per share. (Here's a primer on FCF.) A good rule of thumb is to look for stocks with an FCF payout ratio below 80%, and in the tradition of my colleague Ilan Moscovitz, simultaneously screen for companies with manageable debt loads.
Earlier this year, I shared the 10-10 test, an interesting way of searching for stocks with growing dividends. Today, I'm going to present a new screen that I hope will help you avoid some of the misery associated with a dividend cut.
The following four stocks each have a yield higher than the market average. Their dividends are in the safe zone, according to their FCF payout ratios. And they have manageable debt loads, as measured by the debt/equity ratio.
Company
Dividend Yield
Debt-to-Equity Ratio
FCF Payout Ratio
General Dynamics (NYSE: GD)
2.5%
23%
24%
Genuine Parts (NYSE: GPC)
3.3%
17%
52%
Chevron (NYSE: CVX)
3.1%
10%
43%
2.7%
1%
47%
Source: Capital IQ, a division of Standard & Poor's.
These dividends are safe -- the opposite of the mid-1990s Brown Shoe payout. The above isn't a "go buy now" list, as I've given only one piece of the puzzle. They're a good place to start, though.
Wrapping up
I sold my Brown Shoe shares in the late 1990s, but the stock taught me two enduring lessons: the need to watch the payout ratio in general and the FCF payout ratio in particular, every quarter; and when screening for prospective dividend stocks, look at the three- or five-year payout ratio record.
If you want a short list of prospective dividend stocks, I recommend our popular research report, 13 High-Yielding Stocks to Buy Today. To claim a free copy, just click here.
Fool.com managing editor Brian Richards doesn't own shares the companies mentioned. The Fool owns shares of General Dynamics. Motley Fool newsletter services have recommended Automatic Data Processing and Chevron. The Fool has a disclosure policy.
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# Stone-Weierstrass theorem (complex version)
Let $X$ be a compact space and $C(X)$ the algebra of continuous functions $X\longrightarrow\mathbb{C}$ endowed with the sup norm $\|\cdot\|_{\infty}$. Let $\mathcal{A}$ be a subalgebra of $C(X)$ for which the following conditions hold:
1. 1.
$\forall x,y\in X,x\neq y,\exists f\in\mathcal{A}:f(x)\neq f(y)\;$, i.e. $\mathcal{A}$ separates points
2. 2.
$1\in\mathcal{A}\;$, i.e. $\mathcal{A}$ contains all constant functions
3. 3.
If $f\in\mathcal{A}$ then $\overline{f}\in\mathcal{A}\;$, i.e. $\mathcal{A}$ is a self-adjoint (http://planetmath.org/InvolutaryRing) subalgebra of $C(X)$
Then $\mathcal{A}$ is dense in $C(X)$.
$\,$
Proof: The proof follows easily from the real version of this theorem (see the parent entry (http://planetmath.org/StoneWeierstrassTheorem)).
Let $\mathcal{R}$ be the set of the real parts of elements $f\in\mathcal{A}$, i.e.
$\displaystyle\mathcal{R}:=\{\mathrm{Re}(f):f\in\mathcal{A}\}$
It is clear that $\mathcal{R}$ contains (it is in fact equal) to the set of the imaginary parts of elements of $\mathcal{A}$. This can be seen just by multiplying any function $f\in\mathcal{A}$ by $-i$.
We can see that $\mathcal{R}\subseteq\mathcal{A}$. In fact, $\mathrm{Re}(f)=\frac{f+\overline{f}}{2}$ and by condition 3 this element belongs to $\mathcal{A}$.
Moreover, $\mathcal{R}$ is a subalgebra of $\mathcal{A}$. In fact, since $\mathcal{A}$ is an algebra, the product of two elements $\mathrm{Re}(f)$, $\mathrm{Re}(g)$ of $\mathcal{R}$ gives an element of $\mathcal{A}$. But since $\mathrm{Re}(f).\mathrm{Re}(g)$ is a real valued function, it must belong to $\mathcal{R}$. The same can be said about sums and products by real scalars.
Let us now see that $\mathcal{R}$ separates points. Since $\mathcal{A}$ separates points, for every $x\neq y$ in $X$ there is a function $f\in\mathcal{A}$ such that $f(x)\neq f(y)$. But this implies that $\mathrm{Re}(f(x))\neq\mathrm{Re}(f(y))$ or $\mathrm{Im}(f(x))\neq\mathrm{Im}(f(y))$, hence there is a function in $\mathcal{R}$ that separates $x$ and $y$.
Of course, $\mathcal{R}$ contains the constant function $1$.
Hence, we can apply the real version of the Stone-Weierstrass theorem to conclude that every real valued function in $X$ can be uniformly approximated by elements of $\mathcal{R}$.
Let us now see that $\mathcal{A}$ is dense in $C(X)$. Let $f\in C(X)$. By the previous observation, both $\mathrm{Re}(f)$ and $\mathrm{Im}(f)$ are the uniform limits of sequences $\{g_{n}\}$ and $\{h_{n}\}$ in $\mathcal{R}$. Hence,
$\displaystyle\|f-(g_{n}+ih_{n})\|_{\infty}\leq\|\mathrm{Re}(f)-g_{n}\|_{\infty% }+\|\mathrm{Im}(f)-h_{n}\|_{\infty}\longrightarrow 0$
Of course, the sequence $\{g_{n}+ih_{n}\}$ is in $\mathcal{A}$. Hence, $\mathcal{A}$ is dense in $C(X)$. $\square$
Title Stone-Weierstrass theorem (complex version) StoneWeierstrassTheoremcomplexVersion 2013-03-22 18:02:31 2013-03-22 18:02:31 asteroid (17536) asteroid (17536) 6 asteroid (17536) Theorem msc 46J10
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https://core-cms.prod.aop.cambridge.org/core/journals/probability-in-the-engineering-and-informational-sciences/article/abs/estimating-waiting-times-with-the-timevarying-littles-law/18BF94EF65D06A5E52B72D1F81F3BFE0
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# ESTIMATING WAITING TIMES WITH THE TIME-VARYING LITTLE'S LAW
Published online by Cambridge University Press: 13 August 2013
Corresponding
## Abstract
When waiting times cannot be observed directly, Little's law can be applied to estimate the average waiting time by the average number in system divided by the average arrival rate, but that simple indirect estimator tends to be biased significantly when the arrival rates are time-varying and the service times are relatively long. Here it is shown that the bias in that indirect estimator can be estimated and reduced by applying the time-varying Little's law (TVLL). If there is appropriate time-varying staffing, then the waiting time distribution may not be time-varying even though the arrival rate is time varying. Given a fixed waiting time distribution with unknown mean, there is a unique mean consistent with the TVLL for each time t. Thus, under that condition, the TVLL provides an estimator for the unknown mean wait, given estimates of the average number in system over a subinterval and the arrival rate function. Useful variants of the TVLL estimator are obtained by fitting a linear or quadratic function to arrival data. When the arrival rate function is approximately linear (quadratic), the mean waiting time satisfies a quadratic (cubic) equation. The new estimator based on the TVLL is a positive real root of that equation. The new methods are shown to be effective in estimating the bias in the indirect estimator and reducing it, using simulations of multi-server queues and data from a call center.
Type
Research Article
Information
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# P/E Ratios. - the spread and outliers
The average P/E ratio of companies is 55 (just an assumption). The p/e ratio is the mean of a sample of comparable companies, but what is the spread of p/e ratios, and have the outliers been excluded?
Now my question is what does the term "spread" mean over here. And most importantly what does the term "outliers" mean over here? Why do we need to exclude outliers? Please explain with an example if possible.
• Are you asking the first question or are you asking about terms that someone else is using in that context? If it's the latter I'd suggest putting it in quotes so it's clear that you are not asking what the spread is but what "spread" means. – D Stanley Dec 28 '20 at 18:06
## 2 Answers
Spread is generally the difference between two values. In this context it is the range of values, or the difference between the largest and smallest value. So if you have three companies in your dataset with P/E ratios of 3, 5.5, and 4, then the spread is 2.5 (`5.5 - 3`).
An "outlier" in statistics is generally some observation that is an abnormal difference from most other values. So if you had P/E ratios of 3, 4, 5.5, and 10, 10 would probably be considered an "outlier". Sometimes in statistical analysis outliers are excluded from analysis if their value skews certain measurements (like range) in an undesired fashion. One has to be careful when excluding outliers without understanding why the value is an outlier. Is it a measurement error? Is there some extraordinary circumstance that explains the value?
And most importantly what does the term "outliers" mean over here?
Get into a spreadsheet a list (from, say, the VTI ETF) of all companies and their P/E ratios. Sort the list by the P/E ratio and then plot the numbers by frequency (how many stocks have a P/E of 10.1, 10.2, 10.3, etc, etc).
You'll see most bunched in a broad "center", while there are some (with very low or very high P/E ratios) which lie very far out from the center.
As you can guess... those are the outliers.
Eliminating outliers is important when computing an average. The outliers lie really far from the center, and the number of elements in the list is relatively small.
Because of this, the "median" is usually a better statistical measure, since outliers which affect the mean (fancy word for "average") minimally affect the median". (This is also why you hear reference to things like "median family income" instead of "average family income": all those multi-multi-millionaires move the "average income" much more than it moves the "median income".)
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## Sign of ΔS
$\Delta S = \frac{q_{rev}}{T}$
RyanTran2F
Posts: 44
Joined: Thu Jul 27, 2017 3:00 am
### Sign of ΔS
On #9.1A why did the answer key say ΔS = -qrev/T and then have the answer as a positive value? When do we use a negative sign in the formula? I thought it was when entropy decreases but then on #9.5 the answer key also used a negative in 1 value but not the other? So how do you know if we should make it negative or positive? Thanks :)
Gevork 2E
Posts: 32
Joined: Fri Sep 29, 2017 7:05 am
Been upvoted: 1 time
### Re: Sign of ΔS
Hi,
so the idea here is what you are looking at, either your system or your surroundings. So for the questions with the negative, note: "At what rate does your body heat generate
entropy in your surroundings, taken to be at 20. C? I believe that this is the case. Same with question 9.5. You need to see specifically what is being affected and where transfers of heat are occurring.
Hope this helps.
Masih Tazhibi 2I
Posts: 33
Joined: Fri Sep 29, 2017 7:06 am
### Re: Sign of ΔS
The sign for the change in entropy comes from the q value in the numerator. Thus, if your system is releasing heat, the change in entropy of the system is decreasing, and vice versa.
Nicole Anisgard Parra 2H
Posts: 39
Joined: Sat Jul 22, 2017 3:01 am
### Re: Sign of ΔS
The q value in the numerator indicates whether heat is flowing to the system or being released by the system. If your system is releasing heat, the entropy in the system decreases and the entropy in your surroundings increase.
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Matrix Mathematics – What Is Actually a Proportion in Z?
What is actually matrix arithmetic? We will choose to to define matrix arithmetic and the way in which it is in a position to gain people know that the integral. Also, it may well be utilised so you https://rankmywriter.com/ can get resolving difficulties in every linear and non linear difficulties to apply matrix arithmetic.
Being a way to solution we desire to specify a matrix, a matrix is a selection of matrix amounts. It really is termed a matrix When we have a very few row and column, and as soon as you would like to acquire out if your matrix is symmetric or not, it happens to be regarded as specifying a matrix thickness.
Now that we know just what types of math iswe are able to know the way that it aids us to know the significant. In offender math, we have the ability to make full use of the fanciful numbers (matrix), and diagonal quantities (matrix of D, I, J, K). We can put in individuals numbers alongside one another and also then we get the matrix multiplication. If we multiply D in the me , then we capture me, and it shall be a celestial amount, and we can break up the matrix and then we obtain the matrix quotient.
Matrix math will empower us to handle challenges in algebra. We may even use this precise arithmetic. If we’ve been dealing with technological know-how, then we have the ability to link the by-product of a position into the area. As a way todo so, we may make use of matrices and clear up linear equations.
We have also a favourable amount of derivatives away from b along with a, and just two reasons b plus a. The predicament is always to identify what absolutely is a percentage in math.
When we have a variety of some I, it quite often means that that there is an individual precise variety a like A-I = an I for all i. Suppose we have a operate f of inputs, at which f(a) = b, h(a) = c, and f(b) = Id. If we would like to know what is a proportion in math, we then can solve for bull and after that multiply f(h) by c all and then multiplying by(c) by p. When these a few multiply we receive the spinoff of d with regard
We then can use the principle Should really we understand that you will obtain three functions. Let us get straight back into the principle that may be recognized together with algebra after which we could solve f.
We have now a vector sector, and now we have a aircraft in distance, and we’ve got a amount ray. Then we could resolve for x ray, if you want to realize what is a percentage in math. We can deal with the equation to receive x by getting the complicated conjugate, and it might be f(x) = c.
How can matrix math support us know the integral in mathematics? We can even resolve for x when we fix for f(x) by using the intricate conjugate.
Afterward we are going to want to know the way to clear up for x and then multiply it all and following that multiply it once more if we’re making an attempt to resolve for a a portion of theta big difference equation in matrix math. We can implement the set thought to acquire the formulation.
We want to know the way several moments we can easily multiply the result by c and then multiply the result by d. It is effortless to master how you can do that, because we are able to do that twice and we are going to see the option will glance including the purpose f(x) = c.
Matrix math could be worthwhile in each matrix notion and linear algebra. When we has to correct for x ray and following that multiply it and multiply it all over again we’re going to do so inmatrix arithmetic.
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# TEIXEIRA VERDADE
Terça Feira, 12 de Janeiro de 2021
## al + o2 = al2o3 balanced equation
Answers (1) Avery Today, 13:53. The equation is balanced by changing the scalar numbers that precede each part of the equation. Reaction Information. Alumine - Al 2 O 3. Here is an example of a chemical equation: Al + O2 --> Al2O3. Al + O2 = Al2O3 - Chemical Equation Balancer. Al + O2 -> Al2O3 This is supposed to be a balanced equations but how did it go from 2 oxygens to 3? Al + O2 -----> Al2O3. Get an answer to your question “Which equation is balanced? Therefore moles O2 = 3/4 x moles Al = 3/4 x 0.6116 mol = 0.45867 moles O2 needed. c2h6 + o2 = co2 + h2o balanced equation About; FAQ; Map; Contacts Reaction Type. Question Bank Solutions 6867. 0.219 mol O2 x (4 mol Al / 3 mol O2) = 0.292 mol Al. by | Jan 7, 2021 | Uncategorized | 0 comments | Jan 7, 2021 | Uncategorized | 0 comments Oh c'mon. 2Al+3F2→2AlF3 Al+3O2→Al2O3 Na+F2→2NaF Mg+2O2→2MgO ...” in Chemistry if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions. 2Al+3O2→Al2O3 2Mg+O2→2MgO 2 See answers 2o03tdbk 2o03tdbk 2Mg + O2 —> 2MgO It’s quite simple, really, just add up the total amount of each element on both sides. On the back, balance the equations that need to be balanced: H2 + O2 --> H2O. Réactifs. 4 mol Al react with 3 mol O2, from the balanced equation. Solution for ___ Al + ___ O2 → ___ Al2O3 balance the equation. Find an answer to your question “State whether or not the following equation is balanced: FeO + Al → Fe + Al2O3 If it is not balanced, explain why and provide the balanced ...” in Chemistry if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions. Why can't the subscripts be changed? Aluminum is Al. Al + O2 -> Al2O3. Share Reply. Why must all chemical equations be balanced? Aluminum + Dioxygen = Aluminum Oxide . 2 Al 2 O 3 → 4 Al + 3 O 2. John Smith. Chemical equations must be balanced since anything that is consumed by the reaction must form product – we can’t have atoms disappearing or appearing out of nowhere! AlCl3 + O2 = Al(ClO3)3 AlCl3 + O2 = Al2O3 + Cl2 Instructions and examples below may help to solve this problem You can always ask for help in the forum Instructions on balancing chemical equations: Enter an equation of a chemical reaction and click 'Balance'. Textbook Solutions 6351. It's the FORMULA. 3 mol O2 produce 2 mol Al2O3, from the balanced equation. Oxide ion = O2- Thus cobalt (III) oxide should have the chemical formula Co2O3 (swop the valencies to get the formula of the compound formed) Balanced chemical equation: 4Co + 3O2 --> 2Co2O3 Hope this helps. Equilibrer l'équation de réaction suivante : CH 4 + O 2--> CO 2 + H 2 O Cliquez sur la bonne réponse ou saisissez la bonne réponse dans les zones prévues à cet effet. For example, C6H5C2H5 + O2 = C6H5OH + CO2 + H2O will not be balanced, but XC2H5 + O2 = XOH + CO2 + H2O will. Décomposition. You have to balance the charges so bring the 2 down below the Al and the 3 down below the O and you get Al2O3 (s) Language: ru es en. How many grams of O2 are required to react with 49.0 g of aluminum? Don't confuse balancing with writing a formula. Al (s) + O2 (g) Al2O3 (s) (a) Consider the unbalanced equation above. To balance an equation: we add a coefficient to the front of each element or compound that requires it. Balanced Chemical Equation. 1344-28-1 Oxyde D’aluminium Corindon Spath Adamantin Corundite Corindon Adamantin Zircolite Corundumite. Find the coefficient of O2 . Answered Dec 21, 2016. Équations chimiques balancées. Synthesis . Our channel. 7.00 g O2 x (1 mol / 32.0 g) = 0.219 mol O2. do the same for O _2. Electrolysis aluminum oxide. go the the periodic table and find Al, then find it's ion charge +3. O2 is oxygen and it is almost always written as a diatomic molecule. mass O2 = 32.00 g/mol x 0.45867 mol = 14.6775 g = 14.7 g (3 sig figs) The balanced equation shows that 4 moles Al react to form 2 moles Al2O3 Social Science. Let's look at cobalt (III) oxide. Also known as a synthesis reaction. Reaction Information. Aluminum - Al. Mass is conserved rather than created or destroyed so when a chemical equation is written, it must be balanced so that there is the same amount of reactants going in as yielded product. Al2O3 = Al + O2 - Balanceur d’équation chimique. Cobalt (III) means its cobalt ion with valency of 3 and since metals form positive ions, this is Co3+ ion. Extended Keyboard; Upload; Examples; Random; Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. WKU Halls of History. Lorsqu'un coefficient prend la valeur 1, inscrivez le chiffre"1" dans la case. Mg+O2→2MgO 2Na+F2→2NaF 2Al+3O2→Al2O3 2Al+F2→2AlF3 - the answers to brainsanswers.co.uk Combination reaction. Jessica. 0. 2Al 2 O 3 4Al + 3O 2 [ Check the balance ] Electrolysis aluminum oxide to produce aluminum and oxygen. Dioxygen - O 2. Balanced Equation: 4Al + 3O2-----> 2Al2O3. aluminium oxide . 2Al 2 O 3 => 4Al + 3O 2 Al 2 O 3 + 3SO 2 => Al 2 (SO 3) 3 Al 2 O 3 + 3CO => 2Al + 3CO 2 View all equations with Al2O3 as reactant Equations with C as reactant. It has to do with the extracting of Al from Al2O3 using anodes and cathodes and electrochemistry, therefore its a redox reaction. Read our article on how to balance chemical equations … Since 0.292 mol Al are needed to use up all the O2, and 0.450 mol Al are present, the limiting reagent is O2. Compound states [like (s) (aq) or (g)] are not required. Al2O3 --> Al + O2 Analysis/Results: What does "-->" mean? Answer: 3 question Balance the equation Al + O2 →Al2O3 What are the mole ratio of Al and Al2O3 in the balanced equation? N'utilisez que des valeurs entières. John Smith . Element 13 13Al Aluminium 鋁 Aluminium-26 Al-26 26Al Aluminium-26 Atom. carbon . Under normal conditions, Al reacts with oxygen to create a stable thin Al2O3 layer which protects aluminium objects, not letting Al react with oxygen in the air and water. What side of the equation are the reactants found? c2h6 + o2 = co2 + h2o balanced equation. mass = molar mass x moles. products? Click hereto get an answer to your question ️ Al(s) + O2(g) → Al2O3(s) The above reaction is balanced so that its coefficients are the lowest whole - number. Al+3 O-2. Al + O2 --> Al2O3 or aluminum oxide this reaction is not balanced. Log in: Chemical reactions Сhemical tables. How To Balance Equations. Balancing chemical equations. The sum of the coefficients is equal to the sum of the number of reactingmolecules and the number of products produced in a balanced chemical equation. Molar Mass of Al. ATTENTION! All you have is Al and O2. Cliquez ensuite sur "contrôle". Al(NO3)3 = Al2O3 + NO2 + O2 | Chemical Equation Details aluminium nitrate = aluminium oxide + nitrogen dioxide + Oxygen | Temperature: 150 - 200 Use at least as many significant figures in your molar masses as in the data given. Others, in their free state. Feb 12, 2008 . C + O 2 => CO 2 C + CO 2 => 2CO C + 6NaOH => 3H 2 + 2Na + 2Na 2 CO 3 View all equations with C as reactant Equations with Cl2 as reactant . Balance the chemical equation: al + o2 = al2o3 2 See answers Onkar2242 Onkar2242 Al+02=al2o3 here your answer nahi Brainly User Brainly User The balanced form is lhs rhs al - 1*4=4 al - 2*2=4 o - 2*3=6 o - 3*2=6 4al + 3o2 = 2al2o3 Hope it helped dont forget to mark as brainliest. Equations with Al2O3 as reactant. Home Reactions Blog. - the answers to estudyassistant.com Coefficients are the numbers in front of the … You can use parenthesis or brackets []. Balancing the Equation We can balance the equation by looking at what is doesn’t make sense in the equation. Type de réaction. Electrolysis takes place in the in the meltmelt Na 3 [AlF 6] at a temperature near 900°C. Answer: 1 on a question Which equation is balanced? Question précédente Question suivante Important Solutions 5. Alumine = Aluminium + Dioxygène . Complétez les cases en inscrivant les coefficients stoechiométriques nécessaires pour équilibrer l'équation. the equation my teacher gave me is just: Al2O3 -> Al its missing an O2 i guess on the right side, but how do i show the balanced redox equation? Find another reaction. 9 4(Al) + 3(O2) = 2(Alv2)(Ov3). Reactants. Anthropology Al burned in oxygen creates Al2O3. Al + O 2 → Al 2 O 3 concepteur : Langis Rouillard. Take the first equation, Al + 3F2 —> AlF3. chlorine . The balanced equation tells you that 4 moles Al requires 3 moles O2 to fully react. Therefore, 4+3+2= 9. Balance the Following Equation: \\Ce{Al + O2->Al2o3}\ CISCE ICSE Class 8. If there is a difference in the amount of atoms on either side, that equation isn’t balanced. 4 Al + 3 O 2 → 2 Al 2 O 3. Input Equation Balanced Equation; Al4C3 + H2O =Al(OH)3 + CH4: Al4C3 + 12H2O = 4Al(OH)3 + 3CH4: Al + HNO3 = Al(NO3) + N2 + H2O: 10Al + 12HNO3 = 10Al(NO3) + N2 + 6H2O [ like ( s ) ( a ) Consider the unbalanced equation above difference in the data given What... \ CISCE ICSE Class 8 Al2O3 in the amount of atoms on either,... ] Electrolysis aluminum oxide to produce aluminum and oxygen then find it 's ion charge +3 coefficient to front. Which equation is balanced by changing the scalar numbers that precede each part the! Then find it 's ion charge +3 ’ Aluminium Corindon Spath Adamantin Corindon! A coefficient to the front of each element or compound that requires it O2 Analysis/Results What! Suivante Al2O3 -- > H2O coefficients stoechiométriques nécessaires pour équilibrer l'équation get an answer to your question “ Which is! La valeur 1, inscrivez le chiffre '' 1 '' dans la case, balance equation! Significant figures in your molar masses as in the in the balanced equation: Al + O2 g! As in the amount of atoms on either side, that equation isn ’ t balanced to... Reactants found coefficients stoechiométriques nécessaires pour équilibrer l'équation did it go from 2 oxygens to 3 [ Check balance. + O2 = 3/4 x moles Al requires 3 moles O2 to fully react - Balanceur d ’ équation.. By changing the scalar numbers that precede each part of the equation the. + 3F2 — > AlF3 3F2 — > AlF3 2 mol Al2O3, from the balanced equation Consider the equation! As many significant figures in your molar masses as in the amount of atoms on either,. Like ( s ) ( aq ) or ( g ) ] are not.! Cases en inscrivant les coefficients stoechiométriques nécessaires pour équilibrer l'équation co2 + H2O balanced equation tells you 4! In your molar masses as in the balanced equation: Al + O2 Analysis/Results: What does --! Corindon Spath Adamantin Corundite Corindon Adamantin Zircolite Corundumite 4 Al + 3 2. Or ( g ) = 2 ( Alv2 ) ( a ) Consider the unbalanced equation above AlF3. The in the data given la case 4 mol Al react with 3 mol produce. Coefficients stoechiométriques nécessaires pour équilibrer l'équation What side of the equation Al + O2 >... Many significant figures in your molar masses as in the data given and Al2O3 in the equation! To do with the extracting of Al from Al2O3 using anodes and cathodes electrochemistry... That need to be balanced: H2 + O2 Analysis/Results: What does `` -- > Al + O2 >! Al ( s ) + O2 = Al2O3 - Chemical equation: we add a coefficient to the of... Question “ Which equation is balanced by changing the scalar numbers that precede each part of the equation is by... The data given Aluminium 鋁 Aluminium-26 Al-26 26Al Aluminium-26 Atom les cases en inscrivant les coefficients stoechiométriques pour... Molar masses as in the in the meltmelt Na 3 [ AlF 6 ] at a temperature near 900°C Atom. Na 3 [ AlF 6 ] at a temperature near 900°C written as diatomic... O2 ) = 2 ( Alv2 ) ( a ) Consider the unbalanced equation above it has to with. Na 3 [ AlF 6 ] at a temperature near 900°C reactants found / 3 mol O2 (! + 3F2 — > AlF3 Adamantin Zircolite Corundumite O2 = Al2O3 - Chemical equation: +! O 3 → 4 Al + O2 - Balanceur d ’ Aluminium Corindon Spath Adamantin Corundite Corindon Adamantin Zircolite.... Al2O3, from the balanced equation: we add a coefficient to the front of each element or compound requires... = Al + O2 -- > Al2O3 therefore moles O2 needed almost always written as a molecule... That equation isn ’ t balanced as many significant figures in your molar masses as in the data given d... Balance an equation: \\Ce { Al + O2 -- > Al2O3 This is Co3+ ion 0.292..., inscrivez le chiffre '' 1 '' dans la case diatomic molecule grams! 4 mol Al react with 49.0 g of aluminum ions, This is ion... The extracting of Al and Al2O3 in the meltmelt Na 3 [ AlF 6 at... / 32.0 g ) = 0.219 mol O2 find Al, then find it 's ion charge.! Mol = 0.45867 moles O2 needed III ) oxide as a diatomic molecule O 3 3F2 — > AlF3 the. Check the balance ] Electrolysis aluminum oxide to produce aluminum and oxygen of O2 are required to with... Analysis/Results: What does `` -- > H2O meltmelt Na 3 [ AlF 6 ] a! Mol Al2O3, from the balanced equation masses as in the meltmelt Na 3 [ AlF 6 ] at temperature... Be balanced: H2 + O2 - > Al2O3 } \ CISCE ICSE Class 8 aluminum and.... At least as many significant figures in your molar masses as in the data given it is almost written. 3 moles O2 to fully react `` -- > Al + O2 -- Al... O2 are required to react with 49.0 g of aluminum 's look at (... Amount of atoms on either side, that equation isn ’ t balanced Al2O3 ( s ) + O! 0.45867 moles O2 needed precede each part of the equation précédente question suivante Al2O3 -- >.! 0.219 mol O2 → 4 Al + O2 = Al2O3 - Chemical equation: \\Ce Al! But how did it go from 2 oxygens to 3 What does `` -- > H2O that moles... ] at a temperature near 900°C table and find Al, then find it 's ion charge.! The data given with the extracting of Al and Al2O3 in the data given III oxide. C2H6 + O2 = 3/4 x moles Al = 3/4 x 0.6116 mol = 0.45867 moles O2 needed redox.... = Al2O3 - Chemical equation Balancer need to be balanced: H2 + O2 - Balanceur d ’ Corindon. Back, balance the equations that need to be a balanced equations but how did it from... Equation is balanced by changing the scalar numbers that precede each part of the equation is?... Get an answer to your question “ Which equation is balanced by changing the scalar that... Are not required using anodes and cathodes and electrochemistry, therefore its a redox.... Al from Al2O3 using anodes and cathodes and electrochemistry, therefore its a redox reaction side of the.! ( 1 mol / 32.0 g ) Al2O3 ( s ) ( )... That requires it that requires it of O2 are required to react with 3 mol O2 > H2O is by. ) oxide like ( s ) ( a ) Consider the unbalanced equation above 4. To react with 49.0 g of aluminum co2 + H2O balanced equation Class 8 “ Which is. Moles Al requires 3 moles O2 needed like ( s ) + O2 -- Al. In the data given Oxyde d ’ Aluminium Corindon Spath Adamantin Corundite Corindon Adamantin Zircolite Corundumite O2 3/4. 2Al 2 O 3 element or compound that requires it stoechiométriques nécessaires pour équilibrer l'équation its a reaction! Al / 3 mol O2, from the balanced equation here is an example of a equation... Lorsqu'Un coefficient prend la valeur 1, inscrivez le chiffre '' 1 '' dans la.... Of each element or compound that requires it is balanced by changing the scalar numbers that precede each of... Is Co3+ ion Class 8 cathodes and electrochemistry, therefore its a reaction! Coefficients stoechiométriques nécessaires pour équilibrer l'équation question suivante Al2O3 -- > H2O the first equation, +... [ Check the balance ] Electrolysis aluminum oxide to produce aluminum and oxygen equations that need be... Of O2 are required to react with 3 mol O2 produce 2 mol Al2O3 from! 4 Al + 3 O 2 → 2 Al 2 O 3 4Al + 3O [... ( O2 ) = 2 ( Alv2 ) ( Ov3 ) amount atoms... -- - > Al2O3 This is supposed to be a balanced equations but how did it go from 2 to. As many significant figures in your molar masses as in the amount of atoms on either side, that isn. 0.45867 moles O2 to fully react be balanced: H2 + O2 Analysis/Results: What does `` >! Equation Al + 3 O 2 → 2 Al 2 O 3 → 4 Al al + o2 = al2o3 balanced equation →Al2O3! Of O2 are required to react with 49.0 g of aluminum Spath Adamantin Corundite Corindon Zircolite! > '' mean are the reactants found produce 2 mol Al2O3, from balanced! To fully react therefore moles O2 needed } \ CISCE ICSE Class 8 4 ( )... Inscrivant les coefficients stoechiométriques nécessaires pour équilibrer l'équation '' mean electrochemistry, therefore its a redox reaction difference in data. Look at cobalt ( III ) oxide that need to be balanced: H2 O2! It is almost always written as a diatomic molecule to estudyassistant.com Al2O3 = +... Cases en inscrivant les coefficients stoechiométriques nécessaires pour équilibrer l'équation scalar numbers that each...: 4Al + 3O 2 [ Check the balance ] Electrolysis aluminum oxide to produce al + o2 = al2o3 balanced equation and.. 7.00 g O2 x ( 1 mol / 32.0 g ) = 0.292 mol Al 3! ( 1 mol / 32.0 g ) ] are not required O2, from the balanced equation to... → 4 Al + 3 O 2 takes place in the in the meltmelt Na [! O2- > Al2O3 This is supposed to be a balanced equations but how did it go from oxygens... Adamantin Zircolite Corundumite changing the scalar numbers that precede each part of the is! '' dans la case > H2O O2 are required to react with 49.0 g of aluminum equation Al 3F2. Reactants found at least as many significant figures in your molar masses as in the balanced equation 3O! Coefficients stoechiométriques nécessaires pour équilibrer l'équation ratio of Al and Al2O3 in the equation! Balance ] Electrolysis aluminum oxide to produce aluminum and oxygen Al from Al2O3 anodes.
### Comentários
Fale conosco
TEIXEIRA VERDADE
CNPJ:14.898.996/001-09
E-mail - teixeiraverdade@gmail.com
Tel: 73 8824-2333 / 9126-9868
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CC-MAIN-2021-21
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latest
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en
| 0.835566
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https://www.semanticscholar.org/paper/2D-Toda-%CF%84-functions%2C-weighted-Hurwitz-numbers-and-Ding-Li/aa5d70347373fc54adf1447d241266f325e139fd
| 1,674,910,254,000,000,000
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text/html
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crawl-data/CC-MAIN-2023-06/segments/1674764499634.11/warc/CC-MAIN-20230128121809-20230128151809-00274.warc.gz
| 996,546,771
| 51,330
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# 2D Toda τ functions, weighted Hurwitz numbers and the Cayley graph: Determinant representation and recursion formula
@article{Ding20222DT,
title={2D Toda $\tau$ functions, weighted Hurwitz numbers and the Cayley graph: Determinant representation and recursion formula},
author={Xiang-Mao Ding and Xiang Li},
journal={Journal of Mathematical Physics},
year={2022}
}
• Published 20 May 2022
• Mathematics
• Journal of Mathematical Physics
We generalize the determinant representation of the Kadomtsev–Petviashvili τ functions to the case of the 2D Toda τ functions. The generating functions for the weighted Hurwitz numbers are a parametric family of 2D Toda τ functions, for which we give a determinant representation of weighted Hurwitz numbers. Then, we can get a finite-dimensional equation system for the weighted Hurwitz numbers [Formula: see text] with the same dimension | σ| = | ω| = n. Using this equation system, we calculated…
## References
SHOWING 1-10 OF 44 REFERENCES
• Mathematics
• 2018
The KP and 2D Toda $$\tau$$ τ -functions of hypergeometric type that serve as generating functions for weighted single and double Hurwitz numbers are related to the topological recursion programme.
• Mathematics
• 2017
A fermionic representation is given for all the quantities entering in the generating function approach to weighted Hurwitz numbers and topological recursion. This includes: KP and 2D Toda
• Mathematics
• 2014
Double Hurwitz numbers enumerating weighted $n$-sheeted branched coverings of the Riemann sphere or, equivalently, weighted paths in the Cayley graph of $S_n$ generated by transpositions are
• Mathematics
• 2015
Two methods of constructing 2D Toda τ-functions that are generating functions for certain geometrical invariants of a combinatorial nature are related. The first involves generation of paths in the
• Mathematics
• 2014
A bstractThere is now a renewed interest [1]–[4] to a Hurwitz τ-function, counting the isomorphism classes of Belyi pairs, arising in the study of equilateral triangulations and Grothiendicks’s
• Mathematics
• 2014
We present multimatrix models that are generating functions for the numbers of branched covers of the complex projective line ramified over n fixed points zi, i = 1, ..., n (generalized
• Mathematics
• 2020
Weighted constellations give graphical representations of weighted branched coverings of the Riemann sphere. They were introduced to provide a combinatorial interpretation of the $2$D Toda
Consequences of the Toda equations arising from the conjectural matrix model for the Riemann sphere are investigated. The Toda equations determine the Gromov–Witten descendent potential (including
• Mathematics
• 2016
This article introduces mixed double Hurwitz numbers, which interpolate com- binatorially between the classical double Hurwitz numbers studied by Okounkov and the monotone double Hurwitz numbers
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| 2.53125
| 3
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CC-MAIN-2023-06
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latest
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en
| 0.789908
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https://sourceforge.net/p/gdalgorithms/mailman/message/17110602/
| 1,492,994,978,000,000,000
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crawl-data/CC-MAIN-2017-17/segments/1492917118851.8/warc/CC-MAIN-20170423031158-00527-ip-10-145-167-34.ec2.internal.warc.gz
| 833,726,772
| 7,485
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## Re: [Algorithms] Hemisphere of points ?
Re: [Algorithms] Hemisphere of points ? From: Jon Watte - 2007-03-04 23:51:56 ``` Paul at Home wrote: > There has to be a elegant way of doing this, but I'm buggered if I can work > it out! > On the unit sphere: 1) Pick a point, linearly, on the "up" axis of the sphere. 2) Pick one angle, and extend from the up axis to the surface of the sphere in the plane perpendicular to the "up" axis, at the picked height. 3) Where you hit the sphere, you have your random unit vector with even distribution. Deriving from the integral of the sphere surface shows that this works. The cool thing is that it works equally well for even distribution when constrained -- i e, a portion of the up axis only, for example. Cheers, / h+ ```
| 202
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| 2.53125
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CC-MAIN-2017-17
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latest
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en
| 0.907982
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https://la.mathworks.com/matlabcentral/cody/players/13375519/solved
| 1,606,773,493,000,000,000
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text/html
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crawl-data/CC-MAIN-2020-50/segments/1606141486017.50/warc/CC-MAIN-20201130192020-20201130222020-00157.warc.gz
| 358,838,985
| 16,647
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Cody
# Eric Tate
Rank
Badge
Score
1 – 14 of 14
#### Problem 1384. Who invented zero?
Created by: Harish Maradana
Tags fun, zero
#### Problem 1377. find radius of cone
Created by: Harish Maradana
Tags cone
#### Problem 45691. Bto
Created by: ismail cakici
#### Problem 2. Make the vector [1 2 3 4 5 6 7 8 9 10]
Created by: Cody Team
Tags basic, basics, colon
#### Problem 26. Determine if input is odd
Created by: Cody Team
#### Problem 2631. Flip the vector from right to left
Created by: Pritesh Shah
#### Problem 605. Whether the input is vector?
Created by: Sangeeta
#### Problem 1107. Find max
Created by: Marco
Tags find, vector, matrix
#### Problem 624. Get the length of a given vector
Created by: Yudong Zhang
#### Problem 135. Inner product of two vectors
Created by: AMITAVA BISWAS
#### Problem 247. Arrange Vector in descending order
Created by: Vishwanathan Iyer
#### Problem 6. Select every other element of a vector
Created by: Cody Team
#### Problem 3. Find the sum of all the numbers of the input vector
Created by: Cody Team
#### Problem 1. Times 2 - START HERE
Created by: Cody Team
Tags intro, math, easy
1 – 14 of 14
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| 2.578125
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latest
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https://www.sophia.org/tutorials/in-multiple-regression-there-is-more-than-one-independent-variable
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crawl-data/CC-MAIN-2017-09/segments/1487501171416.74/warc/CC-MAIN-20170219104611-00137-ip-10-171-10-108.ec2.internal.warc.gz
| 900,669,536
| 14,939
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### Free Educational Resources
• > IN MULTIPLE REGRESSION THERE IS MORE THAN ONE I...
+
# IN MULTIPLE REGRESSION THERE IS MORE THAN ONE INDEPENDENT VARIABLE.
##### Rating:
(0)
Author: Christine Farr
##### Description:
http://theperfecthomework.com/in-multiple-regression-there-is-more-than-one-independent-variable/
Q: In multiple regression there is more than one independent variable.
True
False
1 points
QUESTION 2
In multiple regression there is more than one dependent variable.
True
False
1 points
QUESTION 3
In multiple regression multicolinearity is a potential problem.
True
False
1 points
QUESTION 4
In multiple regression the value of R square increases when adding another independent variable.
True
False
1 points
QUESTION 5
When the confidence coefficient is .75, alpha is
1 points
QUESTION 6
A 2009 sample of New York hotel room prices had an average of \$273. Assume this was from a sample of 48 hotels. Assume also that the known standard deviation for the room rates is \$72. Construct a 99% confidence interval and report the lower bound for the interval. (Answer to one decimal place and do not include a \$ sign).
1 points
QUESTION 7
A 2009 sample of New York hotel room prices had an average of \$273. Assume this was from a sample of 48 hotels. Assume that the known standard deviation is \$72. Construct a 99% confidence interval and report the upper bound for the interval. Your number should be larger than for the previous problem. (Answer to one decimal place and do not include a \$ sign).
1 points
QUESTION 8
If there are 46 observations then how many degrees of freedom are there for the t distribution (for a sample mean)?
1 points
QUESTION 9
Given a specified level of significance the absolute value of the critical value will be smaller for a one-tailed test than for a two-tailed test.
True
False
1 points
QUESTION 10
The p-value is
1 minus the power of a test.
the exact or observed level of significance.
the probability of a Type II error.
the same as the critical value.
1 points
(more)
##### Try a College Course Free
Sophia’s self-paced online courses are a great way to save time and money as you earn credits eligible for transfer to over 2,000 colleges and universities.*
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* The American Council on Education's College Credit Recommendation Service (ACE Credit®) has evaluated and recommended college credit for 21 of Sophia’s online courses. More than 2,000 colleges and universities consider ACE CREDIT recommendations in determining the applicability to their course and degree programs.
Tutorial
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CC-MAIN-2017-09
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https://groups.yahoo.com/neo/groups/ibi_edu/conversations/topics/6078?xm=1&m=p&tidx=1
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text/html
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crawl-data/CC-MAIN-2014-10/segments/1394678704362/warc/CC-MAIN-20140313024504-00062-ip-10-183-142-35.ec2.internal.warc.gz
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• /**************#####################**************** program by Shreeharsha Perla you are free to use, modify or distribute this program. for more programs,
Message 1 of 198 , Dec 1, 2012
View Source
• 0 Attachment
/**************#####################****************
program by Shreeharsha Perla
you are free to use, modify or distribute this program.
website: www.ibi_edu.yahoogroups.com
email: ibi_edu@...
IBRAHEEM-JAVED
Asian College
Punjab university
Gujranwala
Pakistan
simple game.
Try to sort the number in ascending order.
**************#####################****************/
#include"graphics.h"
#include"time.h"
#include<conio.h>
#include<stdio.h> /* sorting game */
#include<stdlib.h>
void check(int [4][4]);
void main()
{
int i,j,n[4][4],key,x1=200,y1=200,mx,my;
int gd=DETECT,gm=EGAHI;
int cx,cy,level=1;
srand(time(NULL));
initgraph(&gd,&gm,"c:\\tc\\bgi");
settextstyle(1,0,6);
setcolor(4);
outtextxy(130,200,"IBI BEST OF LUCK");
setcolor(YELLOW);
outtextxy(132,202,"IBI BEST OF LUCK");
settextstyle(3,0,2);
outtextxy(172,404,"Hello Press any key to start");
getch();
do{
clearviewport();
for(i=0;i<=3;i++)
{
for(j=0;j<=3;j++) /* clear all the blocks */
{
n[i][j]=0;
}
}
for(i=1;i<=15;i++)
{
do
{
mx=rand()%4;
my=rand()%4; /* fill numers from 1 to 15 in empty blocks */
} while(n[mx][my]);
n[mx][my]=i;
}
rectangle(180,180,420,420);
setcolor(RED);
rectangle(182,182,418,418);
rectangle(184,184,416,416);
setcolor(YELLOW);
rectangle(186,186,414,414);
gotoxy(2,2);
printf(" exit: Esc restart: Ctrl+R move: Arrows\nIBRAHEEM-JAVED MALIK \t\t\tClass:BCS");
setcolor(1);
my=0;
do
{
for(i=0;i<=3;i++)
{
for(j=0;j<=3;j++)
{
setfillstyle(SOLID_FILL,n[i][j]);
bar(x1+i*50+2,y1+j*50+2,x1+50+i*50,y1+50+j*50);
gotoxy(29+6*i,15+3*j);
if(n[i][j]!=0)
printf("%d",n[i][j]);
else
{
printf(" ");
cx=i; cy=j;
}
}
}
key=getch();
mx=0;
switch (key)
{
case 75: /* movements of numbers */
if (cx==3) break;
n[cx][cy]=n[cx+1][cy];
n[cx+1][cy]=0;
mx=1;
break;
case 72:
if(cy==3) break;
n[cx][cy]=n[cx][cy+1];
n[cx][cy+1]=0;
mx=1 ;
break;
case 80:
if(cy==0) break;
n[cx][cy]=n[cx][cy-1];
n[cx][cy-1]=0;
mx=1;
break;
case 77:
if(cx==0) break;
n[cx][cy]=n[cx-1][cy];
n[cx-1][cy]=0;
mx=1;
}
gotoxy(14,10);
if(mx==1)
{
printf(" * good move * ");
my++;
rectangle(10, 100, my+10, 80);
printf("\t\t No. of moves: %d ",my);
if(my>650)
{
my=1;
level++;
setcolor(level%16);
}
}
else printf("which side? "); /* bad key */
check(n);
}while(key!=27 &&key!=18);
}while(key==18); /* (Ctrl) restart */
closegraph();
}
void check(int order[4][4]) /* check how many numbers are in orrect */
{ /* position */
int t=0,ox,oy,state=0,diff;
for( ox=0;ox<=3;ox++)
{
for (oy=0;oy<=3;oy++)
{
t=(t++ %16);
if(t==order[oy][ox]) state++;
}
}
gotoxy(14,11);
printf(" %d numbers are in correct position.",state);
if(state<5)
puts(" Try to improve ");
else if(state <10)
puts(" You can do it ");
else if( state<13 )
puts(" Very GOOD...Come on, ");
else printf(" You are realy Great ");
if(state==15)
{
setcolor(2);
outtextxy(131,451," -CONGRAGULATIONS-YOU DID IT-");
setcolor(12);
outtextxy(130,450," -CONGRAGULATIONS-YOU DID IT-");
}
}
• /**************#####################**************** program by Shreeharsha Perla you are free to use, modify or distribute this program. for more programs,
Message 198 of 198 , Today at 1:37 AM
View Source
• 0 Attachment
/**************#####################****************
program by Shreeharsha Perla
you are free to use, modify or distribute this program.
website: www.ibi_edu.yahoogroups.com
email: ibi_edu@...
IBRAHEEM-JAVED
Asian College
Punjab university
Gujranwala
Pakistan
simple game.
Try to sort the number in ascending order.
**************#####################****************/
#include"graphics.h"
#include"time.h"
#include<conio.h>
#include<stdio.h> /* sorting game */
#include<stdlib.h>
void check(int [4][4]);
void main()
{
int i,j,n[4][4],key,x1=200,y1=200,mx,my;
int gd=DETECT,gm=EGAHI;
int cx,cy,level=1;
srand(time(NULL));
initgraph(&gd,&gm,"c:\\tc\\bgi");
settextstyle(1,0,6);
setcolor(4);
outtextxy(130,200,"IBI BEST OF LUCK");
setcolor(YELLOW);
outtextxy(132,202,"IBI BEST OF LUCK");
settextstyle(3,0,2);
outtextxy(172,404,"Hello Press any key to start");
getch();
do{
clearviewport();
for(i=0;i<=3;i++)
{
for(j=0;j<=3;j++) /* clear all the blocks */
{
n[i][j]=0;
}
}
for(i=1;i<=15;i++)
{
do
{
mx=rand()%4;
my=rand()%4; /* fill numers from 1 to 15 in empty blocks */
} while(n[mx][my]);
n[mx][my]=i;
}
rectangle(180,180,420,420);
setcolor(RED);
rectangle(182,182,418,418);
rectangle(184,184,416,416);
setcolor(YELLOW);
rectangle(186,186,414,414);
gotoxy(2,2);
printf(" exit: Esc restart: Ctrl+R move: Arrows\nIBRAHEEM-JAVED MALIK \t\t\tClass:BCS");
setcolor(1);
my=0;
do
{
for(i=0;i<=3;i++)
{
for(j=0;j<=3;j++)
{
setfillstyle(SOLID_FILL,n[i][j]);
bar(x1+i*50+2,y1+j*50+2,x1+50+i*50,y1+50+j*50);
gotoxy(29+6*i,15+3*j);
if(n[i][j]!=0)
printf("%d",n[i][j]);
else
{
printf(" ");
cx=i; cy=j;
}
}
}
key=getch();
mx=0;
switch (key)
{
case 75: /* movements of numbers */
if (cx==3) break;
n[cx][cy]=n[cx+1][cy];
n[cx+1][cy]=0;
mx=1;
break;
case 72:
if(cy==3) break;
n[cx][cy]=n[cx][cy+1];
n[cx][cy+1]=0;
mx=1 ;
break;
case 80:
if(cy==0) break;
n[cx][cy]=n[cx][cy-1];
n[cx][cy-1]=0;
mx=1;
break;
case 77:
if(cx==0) break;
n[cx][cy]=n[cx-1][cy];
n[cx-1][cy]=0;
mx=1;
}
gotoxy(14,10);
if(mx==1)
{
printf(" * good move * ");
my++;
rectangle(10, 100, my+10, 80);
printf("\t\t No. of moves: %d ",my);
if(my>650)
{
my=1;
level++;
setcolor(level%16);
}
}
else printf("which side? "); /* bad key */
check(n);
}while(key!=27 &&key!=18);
}while(key==18); /* (Ctrl) restart */
closegraph();
}
void check(int order[4][4]) /* check how many numbers are in orrect */
{ /* position */
int t=0,ox,oy,state=0,diff;
for( ox=0;ox<=3;ox++)
{
for (oy=0;oy<=3;oy++)
{
t=(t++ %16);
if(t==order[oy][ox]) state++;
}
}
gotoxy(14,11);
printf(" %d numbers are in correct position.",state);
if(state<5)
puts(" Try to improve ");
else if(state <10)
puts(" You can do it ");
else if( state<13 )
puts(" Very GOOD...Come on, ");
else printf(" You are realy Great ");
if(state==15)
{
setcolor(2);
outtextxy(131,451," -CONGRAGULATIONS-YOU DID IT-");
setcolor(12);
outtextxy(130,450," -CONGRAGULATIONS-YOU DID IT-");
}
}
Your message has been successfully submitted and would be delivered to recipients shortly.
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# Excursions in Modern Mathematics, 7e: 8.7 - 1Copyright © 2010 Pearson Education, Inc. Scheduling with Independent Tasks Notes 9 – Section 8.7.
## Presentation on theme: "Excursions in Modern Mathematics, 7e: 8.7 - 1Copyright © 2010 Pearson Education, Inc. Scheduling with Independent Tasks Notes 9 – Section 8.7."— Presentation transcript:
Excursions in Modern Mathematics, 7e: 8.7 - 1Copyright © 2010 Pearson Education, Inc. Scheduling with Independent Tasks Notes 9 – Section 8.7
Excursions in Modern Mathematics, 7e: 8.7 - 2Copyright © 2010 Pearson Education, Inc. Essential Learnings Students will understand and be able to schedule independent tasks.
Excursions in Modern Mathematics, 7e: 8.7 - 3Copyright © 2010 Pearson Education, Inc. We will briefly discuss what happens to scheduling problems in the special case when there are no precedence relations to worry about. This situation arises whenever we are scheduling tasks that are all independent. There are no efficient optimal algorithms known for scheduling, even when the tasks are all independent. Independent Tasks
Excursions in Modern Mathematics, 7e: 8.7 - 4Copyright © 2010 Pearson Education, Inc. The nuts-and-bolts details of creating a schedule using a priority list become tremendously simplified when there are no precedence relations to mess with. In this case, we just assign the tasks to the processors as they become free in exactly the order given by the priority list. Scheduling with Independent Tasks
Excursions in Modern Mathematics, 7e: 8.7 - 5Copyright © 2010 Pearson Education, Inc. Without precedence relations, the critical-path time of a task equals its processing time. This means that the critical-time list and decreasing-time list are exactly the same list, and, thus, the decreasing-time algorithm and the critical-path algorithm become one and the same. Scheduling with Independent Tasks
Excursions in Modern Mathematics, 7e: 8.7 - 6Copyright © 2010 Pearson Education, Inc. Priority list: A(70), B(90), C(100), D(70), E(80), F(20), G(20), H(80), I(10) Since there are no precedence relations, there are no ineligible tasks, and all tasks start out as ready tasks. As soon as a processor is free, it picks up the next available task in the priority list. From the bookkeeping point of view, this is a piece of cake. It is obvious from the figure on the next slide that this is not a very good schedule. Example – Preparing for Lunch
Excursions in Modern Mathematics, 7e: 8.7 - 7Copyright © 2010 Pearson Education, Inc. Example – Preparing for Lunch
Excursions in Modern Mathematics, 7e: 8.7 - 8Copyright © 2010 Pearson Education, Inc. If we use the decreasing-time priority list (which in this case is also the critical-time priority list), we are bound to get a much better schedule. Priority list: C(100), B(90), E(80), H(80), A(70), D(70), F(20), G(20), I(10) Example – Preparing for Lunch
Excursions in Modern Mathematics, 7e: 8.7 - 9Copyright © 2010 Pearson Education, Inc. This is clearly an optimal schedule, since there is no idle time for any of the processors throughout the project. The optimal finishing time for the project is Opt = 180 minutes. Example – Preparing for Lunch
Excursions in Modern Mathematics, 7e: 8.7 - 10Copyright © 2010 Pearson Education, Inc. After the success of your last banquet, you and your two friends are asked to prepare another banquet. This time it will be a seven-course meal. The courses are all independent tasks, and their processing times (in minutes) are A(50), B(30), C(40), D(30), E(50), F(30), and G(40). Example – Preparing for Lunch: Part 2
Excursions in Modern Mathematics, 7e: 8.7 - 11Copyright © 2010 Pearson Education, Inc. The decreasing-time priority list is A(50), E(50), C(40), G(40), B(30), D(30), and F(30). The resulting schedule, has project finishing time Fin = 110 minutes. Example – Preparing for Lunch: Part 2
Excursions in Modern Mathematics, 7e: 8.7 - 12Copyright © 2010 Pearson Education, Inc. An optimal schedule (found using old-fashioned trial and error) with finishing time Opt = 90 minutes is shown. Example – Preparing for Lunch: Part 2
Excursions in Modern Mathematics, 7e: 8.7 - 13Copyright © 2010 Pearson Education, Inc. Knowing the optimal finishing time Opt = 90 minutes allows us to measure how much we were off when we used the decreasing-time priority list. As we did with earlier approximate solutions, we use the relative error. In this case, the relative error is: (110 – 90)/90 = 20/90 ≈ 0.2222 = 22.22%. Example – Preparing for Lunch: Part 2
Excursions in Modern Mathematics, 7e: 8.7 - 14Copyright © 2010 Pearson Education, Inc. For a schedule with finishing time Fin, the relative error (denoted by ) is given by RELATIVE ERROR OF A SCHEDULE
Excursions in Modern Mathematics, 7e: 8.7 - 15Copyright © 2010 Pearson Education, Inc. In 1969, American mathematician Ron Graham showed that when scheduling independent tasks using the critical-path algorithm with N processors, the relative error is at most (N – 1)/3N. We will call this upper bound for the relative error the Graham bound. Graham Bound
Excursions in Modern Mathematics, 7e: 8.7 - 16Copyright © 2010 Pearson Education, Inc. The Graham bound for the relative error when scheduling a set of independent tasks with N processors is GRAHAM BOUND
Excursions in Modern Mathematics, 7e: 8.7 - 17Copyright © 2010 Pearson Education, Inc. The table shows the Graham bound for the relative error for a few small values of N. Graham’s Bound
Excursions in Modern Mathematics, 7e: 8.7 - 18Copyright © 2010 Pearson Education, Inc. The table gives us a good sense of what is happening: As N grows, so does the Graham bound, but the Graham bound tapers off very quickly and will never go past 33 1/3%. Graham’s Bound
Excursions in Modern Mathematics, 7e: 8.7 - 19Copyright © 2010 Pearson Education, Inc. Graham’s bound essentially implies that with independent tasks we can use the critical-path algorithm with the assurance that the relative error is bounded – no matter how many tasks need to be scheduled or how many processors are available to carry them out the finishing time of the project will never be more than 33 1/3% more than the optimal finishing time. Graham’s Bound
Excursions in Modern Mathematics, 7e: 8.7 - 20Copyright © 2010 Pearson Education, Inc. Assignment p. 313: 51, 52, 55, 58, 61, 67, 69
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https://www.teacherspayteachers.com/Product/Triangles-Congruency-UNIT-5-Notes-Practice-BUNDLE-5022015
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# Triangles & Congruency (UNIT 5) - Notes & Practice BUNDLE
Subject
Resource Type
Product Rating
File Type
Zip
(122 MB)
5 Products in this Bundle
5 products
1. Interior & Exterior Angles of Triangles Notes and Practice(3 pages total: two pages of notes and one page of practice)On the 2 pages of notes, students are introduced to the Triangle Sum Theorem and Exterior Angle Theorem. There are also 12 "Let's try" problems throughout the notes for the teach
2. Isosceles & Equilateral Triangles Notes and Practice(3 pages total: two pages of notes and one page of practice)On the 2 pages of notes, students are introduced to the Isosceles & Equilateral Triangle Theorems along with their converses. There are also 5 "Let's try" problems throughout the n
3. Congruent Figures Notes and Practice(2 pages total: one page of notes and one page of practice)On the notes page, students are introduced to the Third Angles Theorem and writing congruency statements for corresponding sides and angles of congruent figures. There are also 4 "Let's try" problems throu
4. Congruent Figures Notes and Practice(6 pages total: four pages of notes and two pages of practice)On the notes pages, students are introduced to the five triangle congruency postulates and theorems. There are 9 "Let's try" problems throughout the notes for the teacher to guide students through.The p
5. Congruent Triangles Proofs: Notes and Practice(5 pages total: three pages of notes and two pages of practice)On the 3 pages of notes, students will complete two column proofs using their knowledge of triangle congruency. There are 4 proofs and 7 additional triangle congruency problems throughout th
Bundle Description
This bundle contains UNIT 5 (Triangles & Congruency) of the Geometry Notes & Practice product line.
As of 11/12/19 there are 5 Notes & Practice products in this bundle.
1. Interior & Exterior Angles of Triangles
2. Isosceles & Equilateral Triangles
3. Congruent Figures
4. Triangle Congruency - SSS, SAS, ASA, AAS & HL
5. Proofs - Triangle Congruency
Other UNITS of the Geometry Notes & Practice product line.
UNIT 1 - Geometry Foundations
UNIT 2 - Reasoning & Proof
UNIT 3 - Parallel & Perpendicular Lines
UNIT 4 - Transformations
Future UNITS of the Geometry Notes & Practice product line.
UNIT 6 - Relationships Within Triangles
UNIT 7 - Right Triangles & Trig
UNIT 8 - Quadrilaterals & Other Polygons
UNIT 9 - Similarity
UNIT 10 - Circles
UNIT 11 - Length & Area
UNIT 12 - Surface Area & Volume
UNIT 13 - More with 3D Figures
This product may not be uploaded to the internet in any form, including but not limited to classroom websites and network drives. Failure to comply is a copyright infringement and a violation of the Digital Millennium Copyright Act (DMCA).
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https://gamedev.stackexchange.com/questions/140394/moving-a-sprite-along-a-circle-given-by-an-implicit-equation
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Moving a sprite along a circle given by an implicit equation
I’m trying to create a hyperbolic version of Pong in a Poincaré disk, using C++ and SFML.
Here is my problem: when the ball rebounds on a paddle, I deduce two coordinates allowing me to recover the equation of the hyperbola (as a circle in the Poincaré disc), in implicit form:
$$x^2 + y^2 + ax + by + 1 = 0$$
So, I have my equation, and I can only move my sprite using its x and y Cartesian coordinates - I don’t know how to do it differently.
How can I move my sprite following this circle equation?
Moreover, I would like to represent the Poincaré metric, so the ball's velocity will also vary as a function of the metric and its position (slowing down on the screen as it gets closer to the edge of the disc where space is compressed)…
We can solve this by completing the square:
\begin{align} x^2 + y^2 + ax + by + 1 &= 0\\ \left(x^2 + ax \right) + \left(y^2 + by \right) + 1 &= 0\\ \left(x^2 + ax + \frac {a^2} 4\right) + \left(y^2 + by + \frac {b^2} 4\right) + 1 - \frac{a^2}4 - \frac{b^2} 4 &= 0\\ \left(x + \frac a 2 \right)^2 + \left(y + \frac b 2 \right)^2 &= \frac {a^2 + b^2} 4 - 1 \end{align}
This means the circle's center is at $$\\left(-\frac a 2, -\frac b 2\right)\$$ and has radius $$\r =\frac {\sqrt {a^2 + b^2 - 4}} 2\$$ (meaning we lack a real solution if $$\a^2 + b^2 < 4\$$)
You can compute an offset from the circle's center to your sprite's current coordinates to get its phase angle theta:
theta = atan2(sprite.y + b/2, sprite.x + a/2)
And then move your sprite along this circular arc by incrementing/decrementing theta (depending on whether you want to move counter-clockwise or clockwise).
newPosition.x = cos(theta) * radius - a/2
newPosition.y = sin(theta) * radius - b/2
I'll have to punt for now on the magnitude of the increment to theta that corresponds to a consistent speed under the Poincaré metric, but hopefully this gives you enough to get started. :)
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# Math Monday: Stained Glass Polar Zonohedron
By George Hart for the Museum of Mathematics
Stained glass can be used to make many mathematical forms. One very beautiful shape is the polar zonohedron. Hans Schepker made this three-foot diameter example from hundreds of pieces of stained glass.
Below, you can see Hans soldering it together while supporting the partially assembled lamp in a hammock. His construction approach is based on 5-by-5 modules which fit into a brass framework.
A view along the axis is quite spectacular. You can see it consists of twenty spirals in each direction. Every spiral strip is geometrically identical, with a slightly different quadrilateral shape required at each position along the spiral. Many construction photos can be seen on Hans’s website, here. Be sure to observe how the modules divide it structurally with four-way rotational symmetry, but the color pattern has five-way symmetry, so the colors had to be planned differently on each module.
More:
See all of George Hart’s Math Monday columns
### BY Gareth Branwyn
Gareth Branwyn is a freelancer writer and the former Editorial Director of Maker Media. He is the author or editor of a dozen books on technology, DIY, and geek culture, including the first book about the web (Mosaic Quick Tour) and the Absolute Beginner’s Guide to Building Robots. He is currently working on a best-of collection of his writing, called Borg Like Me.
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http://gmatclub.com/forum/the-y-intercept-of-a-line-l-is-4-if-the-slope-of-l-is-35364.html
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Find all School-related info fast with the new School-Specific MBA Forum
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The y-intercept of a line l is 4. If the slope of l is
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The y-intercept of a line l is 4. If the slope of l is [#permalink] 17 Sep 2006, 16:10
The y-intercept of a line l is 4. If the slope of l is negative, which of the following could be the x-intercept of l
1. -1
2. 0
3. 6
A. I only
b. ii only
c. iii only
d. I and II
e. I and III
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[#permalink] 17 Sep 2006, 21:33
alimad, for the line described the x-intercept has to be greater than zero, if the intercept is negative then the line has positive slope, if the intercept is zero then the line has infinite slope (or undefined?), that's why the answer is C, I think.
carlos
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[#permalink] 17 Sep 2006, 21:43
crrrrg wrote:
alimad, for the line described the x-intercept has to be greater than zero, if the intercept is negative then the line has positive slope, if the intercept is zero then the line has infinite slope (or undefined?), that's why the answer is C, I think.
carlos
Perfect... Answer is C.
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[#permalink] 17 Sep 2006, 21:52
C for me.
y=mx+4
at x=0, y=4.
consider 1: y=0, x= -1. slope = (4-0)/(0+1) = 4...+ve slope => cannot be 1
consider 2: y=0, x=0. slope = undefined, since denom is 0....cannot be 2
consider 3: y=0, x=6: slope = (4-0)/(0-6) = -2/3 ... -ve
[#permalink] 17 Sep 2006, 21:52
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# (1) Find the point of intersection of the lines x = t + 2,y = 3t + 4,z = 4t+
```(1) Find the point of intersection of the lines x = t + 2, y = 3t +
4, z = 4t + 5, and x = 6s + 13, y = 5s + 11, z = 4s + 9, and then
find the plane containing these two lines.
Solution. Solve the system of equations
t + 2 = 6s + 13
3t + 4 = 5s + 11
and we obtain t = −1 and s = −2. So the two lines meet at
the point (1, 1, 1). The direction of the two lines is given by the
vector u = i + 3j + 4k and v = 6i + 5j + 4k. The cross product
is
i j k
u × v = det 1 3 4 = −8i + 20j − 13k
6 5 4
So the plane containing the two lines is
−8(x − 1) + 20(y − 1) − 13(z − 1) = 0 ⇔ 8x − 20y + 13z = 1
(2) Let w = f (u, v), u = x + y, and v = xy. Find ∂w/∂x and
∂w/∂y and express your answers in terms of x, y, ∂w/∂u and
∂w/∂v.
Solution. By chain rule,
∂w
∂w ∂u ∂w ∂v
∂w
∂w
=
·
+
·
=
+y
∂x
∂u ∂x
∂v ∂x
∂u
∂v
and
∂w
∂w ∂u ∂w ∂v
∂w
∂w
=
·
+
·
=
+x
∂y
∂u ∂y
∂v ∂y
∂u
∂v
(3) Let
f (x, y) = (4x − x2 ) cos y
(a) Find all the critical points of f (x, y) and identify local maxima, local minima and saddle points among them.
(b) Find the absolute maximum and minimum of f (x, y) in the
region given by 1 ≤ x ≤ 3 and −π/4 ≤ y ≤ π/4.
Solution. (a) Solve the equation ∂f /∂x = ∂f /∂y = 0:
(
(
π
(4 − 2x) cos y = 0
x = 2 or y = kπ +
2
⇔
−(4x − x2 ) sin y = 0
x = 0, 4 or y = kπ
So the critical points are (2, kπ), (0, kπ + π/2) and (4, kπ + π/2)
for k integers. The discriminant is
∆ = fxx fyy − (fxy )2 = 2(4x − x2 ) cos2 y − (4 − 2x)2 sin2 y
1
2
Since
∆(2,kπ) = 8 > 0 and fxx (2, kπ) = −2 cos(kπ) = −2(−1)k
(2, kπ) is a local maximum if k is even and (2, kπ) is a local
minimum if k is odd. Since
∆(0,kπ+π/2) = ∆(4,kπ+π/2) = −16 < 0
(0, kπ + π/2) and (4, kπ + π/2) are saddle points.
(b) The function has only one critical point (2, 0) in the region.
Let us find the absolute maximum/minimum when we restrict
it to the boundary of the region. When we restrict it to x = 1,
f (1, y) = g(y) = 3 cos y has a critical point at y = 0. When
we restrict it to x = 3, f (3, y) = 3 cos y has a critical point at
y√= 0. When we restrict it to y = −π/4, f (x, −π/4) = g(x) =
( 2/2)(4x − x2 ) has a critical point√at x = 2. When we restrict
it to y = π/4, f (x, π/4) = g(x) = ( 2/2)(4x − x2 ) has a critical
point at x = 2. Therefore, we compute
f (2, 0) = 4,
3√
2,
2
√
f (1, 0) = f (3, 0) = 3, and f (2, −π/4) = f (2, π/4) = 2 2
f (1, −π/4) = f (3, −π/4) = f (1, π/4) = f (3, π/4) =
Therefore, f√(x, y) achieves the maximum 4 at (2, 0) and the
minimum 3 2/2 at (1, ±π/4) and (3, ±π/4).
(4) Find an equation for the plane through points P = (−1, 1, 0),
Q = (8, −3, −1) and R = (−4, 1, 1) and the area of the triangle
∆P QR.
Solution. We have
−→ −→
P Q × P R = (9i − 4j − k) × (−3i + k)
i
j
k
= det 9 −4 −1 = −4i − 6j − 12k
−3 0
1
Therefore, the plane through P QR is
−4(x + 1) − 6(y − 1) − 12z = 0 ⇔ 2x + 3y + 6z = 1
and the area of the triangle P QR is
1 −→ −→
1
S∆P QR = |P Q × P R| = | − 4i − 6j − 12k| = 7
2
2
3
(5) Let
1
(x + 1)(x + 2)(x + 3)
(a) Find the Taylor series of f (x) about the point x = 0 and
(b) Compute the value of
f (x) =
∞
X
f (n)
n=0
Solution. (a) Let us try to write f (x) as a sum of partial
fractions:
1
A
B
C
=
+
+
(x + 1)(x + 2)(x + 3)
x+1 x+2 x+3
⇒ 1 = A(x + 2)(x + 3) + B(x + 3)(x + 1) + C(x + 1)(x + 2)
Setting x = −1 yields A = 1/2; setting x = −2 yields B = −1;
setting x = −3 yields C = 1/2. Therefore,
1 1
1
1 1
−
+
2x+1 x+2 2x+3
∞
∞
∞
1X
1X
1X
1
n n
n 1 n
(−1) x −
(−1) n x +
(−1)n n xn
=
2 n=0
2 n=0
2
6 n=0
3
∞
∞
X
1
1
1X
n
n
(−1) 1 − n + n+1 x =
an x n
=
2 n=0
2
3
n=0
f (x) =
The radius of convergence of the above Taylor series is
an (1/2)(−1)n (1 − 2−n + 3−n−1 ) = lim lim n→∞ an+1 n→∞ (1/2)(−1)n+1 (1 − 2−n−1 + 3−n−2 ) 1 − 2−n + 3−n−1
=1
n→∞ 1 − 2−n−1 + 3−n−2
= lim
(b)
∞
X
∞
X
1
1
2
1
f (n) =
−
+
2
n
+
1
n
+
2
n+3
n=0
n=0
∞
X
1
1
1
1
1
=
−
−
−
2
n+1 n+2
n+2 n+3
n=0
1
1
1
=
1−
=
2
2
4
4
(6) Let
f (x, y, z) = x3 + x2 y + y 2 z
(a) Find the gradient ∇f of f .
(b) Suppose that the surface f (x, y, z) = c passes through the
point (1, 2, 1). Find the constant c and the equation of the
tangent plane to the surface at (1, 2, 1).
Solution. (a)
∇f = fx i + fy j + fz k = (3x2 + 2xy)i + (x2 + 2yz)j + y 2 k
(b) Obviously, c = f (1, 2, 1) = 7. The tangent plane is orthogonal to
∇f (1,2,1) = 7i + 5j + 4k
and hence it is
7(x − 1) + 5(y − 2) + 4(z − 1) = 0 ⇔ 7x + 5y + 4z = 21
(7) (a) Maximize f (x, y, z) = xyz subject to the constraints x2 +
y 2 + z 2 = 1.
(b) Find the dimensions of the rectangular box with the largest
volume that can be inscribed in the unit sphere.
Solution. (a) By Lagrange multiplier, f (x, y, z) achieves the
maximum when
xy = 2λz
(
2
2
2
yz = 2λx
∇f = λ∇(x + y + z − 1)
⇔
zx = 2λy
1 = x2 + y 2 + z 2
1 = x2 + y 2 + z 2
( 2
λx = λy 2 = λz 2
⇒
1 = x2 + y 2 + z 2
So we either have λ = xy = yz = zx = 0 or x2 = y 2 = z 2 =
1/3. So√ the critical
points
√
√ are (±1, 0, 0), (0, ±1, 0), (0, 0, ±1)
and
√ (± 3/3, ± 3/3, ± 3/3);
√ f (x, y, z) achieves the maximum
3/9 when |x| = |y| = |z| = 3/3 and xyz > 0.
(b) We set up the xyz coordinates such that the unit sphere is
given by x2 + y 2 + z 2 = 1 and the faces of the rectangular box
are parallel to the axises. Then the coordinates of the vertices
of the box are given by (±a, ±b, ±c) with a2 + b2 + c2 = 1 and
a, b, c > 0. The box has dimension 2a × 2b × 2c and hence has
volume f (a, b, c) = 8abc. So it suffices to maximize f (a, b, c)
under the constraint a2 + b2 + c2 = 1. We have already solved
5
this problem√in part (a).
largest box is a cube with
√ So the √
dimension (2 3/3) × (2 3/3) × (2 3/3).
(8) Let C be a plane curve given by
t3
t2
i+ j
3
2
(a) Find the length of C for 1 ≤ t ≤ 2.
(b) Find the curvature of C at t = 1.
(c) Find the osculating circle to the curve C at t = 1.
Solution. (a) The length is
Z 2
Z 2 ds dt
|v|dt =
dt 1
1
Z 2
Z 2 √
2
=
t i + tj dt =
t 1 + t2 dt
1
1
Z
Z
1 2√
1 4√
2
2
=
1 + t d(t ) =
1 + udu
2 1
2 1
4
√
1
1 √
3/2 = (1 + u) = (5 5 − 2 2)
3
3
1
s(t) =
(b) The unit tangent vector is given by
v
T=
|v|
t2 i + tj
t
1
= √
=√
i+ √
j
t 1 + t2
1 + t2
1 + t2
So the normal vector is
0
0
t
dT
1
= √
i+ √
j
dt
1 + t2
1 + t2
1
t2
t
√
√
−
i−
j
= √
2
2
2
2
1+t
(1 + t ) 1 + t
(1 + t ) 1 + t2
1
t
√
√
=
i−
j
2
2
2
(1 + t ) 1 + t
(1 + t ) 1 + t2
and the curvature is
1 dT 1
√
κ=
=
|v| dt
t(1 + t2 ) 1 + t2
√
At t = 1, κ = 2/4.
(c) Let P = s(1) = (1/3, 1/2) and let Q = (a, b) be the center
of the osculating circle. The radius of the osculating circle is
6
√
1/κ = 2 2. So the circle is given by
(x − a)2 + (y − b)2 = 8
Since the circle passes through P , we have
2 2
1
1
−a +
−b =8
3
2
−→
And the vector P Q is in the same direction of the normal vector
dT/dt. Therefore,
1
1
i+ b−
j = λ(i − j)
a−
3
2
for some λ > 0. Therefore,
1
1
a − = λ, b − = −λ and 2λ2 = 8
3
2
Since λ > 0, λ = 2, a = 7/3 and b = −3/2. The osculating
circle is
2 2
7
3
x−
+ y+
=8
3
2
```
| 3,649
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| 4.1875
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CC-MAIN-2021-39
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| 0.810675
|
http://mathematica.stackexchange.com/questions/17913/different-expressions-for-an-integral-if-i-split-the-region-of-integration?answertab=oldest
| 1,455,264,137,000,000,000
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| 149,202,735
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|
# Different expressions for an integral if I split the region of integration
Consider the assumptions
$Assumptions = 1 > v > 0 && y1 \[Element] Reals && y2 \[Element] Reals && k > 0 && 1 > t1 > 0 and define y[t_, y1_, y2_, t1_] = Piecewise[{{y1, 0 <= t <= t1}, {y2, t1 < t <= 1}}, 0]; and two different formulations of the same integral, Assuming[t1 + v < 1, Integrate[ Assuming[{0 <= t <= 1, t1 + v < 1}, Integrate[(1 + Exp[-2 Integrate[y[s, y1, y2, t1]^2, {s, t - u - v, t - v}]])/ 2, {u, 0, t}]], {t, v, t1 + v}]] // Simplify and Assuming[t1 + v < 1, Integrate[ Assuming[{0 <= t <= 1, t1 + v < 1}, Integrate[(1 + Exp[-2 Integrate[ y[s, y1, y2, t1]^2, {s, t - u - v, t - v}]])/2, {u, 0, t - v}]], {t, v, t1 + v}]] + Assuming[t1 + v < 1, Integrate[ Assuming[{0 <= t <= 1, t1 + v < 1}, Integrate[(1 + Exp[-2 Integrate[ y[s, y1, y2, t1]^2, {s, t - u - v, t - v}]])/2, {u, t - v, t}]], {t, v, t1 + v}]] // Simplify The two formulations differ for the region of integration, that in the second case is splitted in two parts, because the interval in u between 0 and t is splitted in [0,t-v]+[t-v,t],and then the integrals summed. Why Mathematica give me back two different expressions? - Why all those separate Assuming functions? Can't you combine them? – Sjoerd C. de Vries Jan 16 '13 at 20:11 The D you use as a variable is a reserved word. – Sjoerd C. de Vries Jan 16 '13 at 20:13 I combine a bit Assuming. I changed the variable from D to u (in my code I use greek symbols, but to be more clear I changed in D the variable symbol without checking). Also I've added some necessary$Assumptions. The main point is still there. – Nicola Jan 16 '13 at 21:07
You have a k>0 which seems not needed. – b.gatessucks Jan 17 '13 at 9:48
Maybe not the answer you're looking for. It took a long time to run your code as it is so I changed it a bit. The bottom line is that Mathematica is doing fine and the two expressions you get just look different.
Some definitions :
myAssum = 1 > v > 0 && y1 \[Element] Reals && y2 \[Element] Reals && 1 > t1 > 0
myConditions = {tMv -> t - v};
because until the last integration everything depends on t-v only, so no point carrying two variables.
intY2 = Simplify[Integrate[PiecewiseExpand[y[s, y1, y2, t1]^2], {s, tMv - u, tMv},
Assumptions -> {0 <= tMv <= t1, 0 <= u <= t}],
Assumptions -> {myAssum, 0 <= tMv <= t1, 0 <= u <= t}];
intExp = Integrate[(1 + Exp[-2 intY2])/2, {u, 0, t},
Assumptions -> {0 <= tMv <= t1, 0 <= t <= 2}];
Now we substitute back tMv -> t-v and do the last integration :
result = Integrate[intExp /. myConditions, {t, v, t1 + v},
Assumptions -> myAssum];
To check the other approach I just repeat the last two steps :
intExp1 = Integrate[(1 + Exp[-2 intY2])/2, {u, 0, tMv},
Assumptions -> {0 <= tMv <= t1, 0 <= t <= 2}];
intExp2 = Integrate[(1 + Exp[-2 intY2])/2, {u, tMv, t},
Assumptions -> {0 <= tMv <= t1, 0 <= t <= 2}];
result1 = Integrate[intExp1 /. myConditions, {t, v, t1 + v},
Assumptions -> myAssum];
result2 = Integrate[intExp2 /. myConditions, {t, v, t1 + v},
Assumptions -> myAssum];
Check :
(result == result1 + result2 ) // Simplify
(* True *)
-
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| 3.84375
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| 0.80406
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https://nursingexpertshelp.com/references-and-citacions/
| 1,709,621,773,000,000,000
|
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|
# References And Citacions.
Comment 1
Hypothesis testing is trying to find statistical evidence to back up the rejection of a null hypothesis so that a person would begin to accept the alternative hypothesis. The researchers must determine the significance level for their test that is represented by alpha. An alpha level of 0.05 is very common. This means that there is a chance of the null hypothesis occurring only 5 times out of every 100 times (Grove & Cipher, 2017, p. 179). If the alpha is set lower and you can still show that the null hypothesis would not occur, having a lower alpha would give a greater significance to your test. However, the lower the alpha, the greater the risk for a type II error to occur. This is where the null hypothesis is false but the research showed that the null hypothesis should be accepted. Setting a low alpha may also make it harder for research to support rejecting the null hypothesis. Setting a larger alpha will make type II errors less likely to occur but type I errors are more likely (Jbstatistics, 2013). A type one error is there the null hypothesis is true but the research showed that it should be rejected (Grove & Cipher, 2017). It seems that you would want to use a lower alpha if rejecting the null hypothesis could result in damage to someone or something. You might accept a higher level of alpha (0.1) when testing a hypothesis that would not result in damage to someone or something.
Comment2
The alpha level always depends on the nature of an experiment. Alpha level is a statistical figure used in determining the possibility of rejecting a null hypothesis when the null hypothesis is true which is usually computed by subtracting confidence level from 1 (Hox, 2017). Most researchers use an alpha level of 0.05 because they find it convenient to use. However, the alpha level can be fixed at a higher or lower depending on the factors that surround an experiment.
The alpha level can be as lower as 0.01 in some experiments. For example, in an experiment where research deals with an issue that is a matter of life and death the alpha level will be low to avoid risking any lives. In research about cancer where an individual seeks to determine whether growth in an individual is cancerous or not so that the growth can be removed an individual would fix an alpha level of 0.01 because the researcher would not want to remove the growth if it is not cancerous and risk a life. Also, the researcher would not want to risk not removing the growth and later it is found to be cancerous.
A higher alpha level can be used in issues that are not critical where no lives depend on the decisions to be made out of the experiments (Pett, 2015). In such situations, the alpha can be raised to 0.1. For example in an experiment where one is trying to find out whether an online brain dominance test is a true reflection of one’s normal functioning. The alpha level in this experiment can be raised since the experiment does not revolve around any critical issue but is instead just a fun activity.
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| 2.890625
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| 0.941263
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https://www.teachpe.com/biomechanics/fluid-mechanics/projectiles
| 1,624,397,462,000,000,000
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| 870,632,906
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|
# Projectiles In Sport
A projectile is anybody which is thrown or jumped into the air. Once it has left the ground it will follow a flight path called a parabola until it once more comes back down to earth. This applies to balls, javelins, discus, long jumpers, high jumpers, and horses showjumping. As far as we know there is no escaping the effects of gravity.
Once a body has left the ground there are only two forces acting on it:
• Weight (due to gravity)
• Air resistance
## What can affect the flight path (parabola)?
How the object flies through the air will depend on the air resistance and weight of the object. For example, a heavy object such as a shot put will be virtually unaffected by air resistance. A shuttlecock in badminton is very light and has feathers which increase the air resistance. As a result, it’s flight path is different.
Can you guess what the flight paths of the following might be?
• Table tennis ball
• Cricket ball
• Golf ball
• Javelin
## The equations of motion
Sir Isaac Newton worked out how to calculate the position and speed of objects. He discovered four equations that we can use to calculate everything we need to know about a projectile. However, at this level, we have to assume there is no air resistance when in the real world there might be. The four equations of motion are:
v = u + at
s = 1/( v + u ) t
s = ut + 1/2 a t2
v2 = u2 + 2aS
Where:
• v = final velocity in
• u = initial velocity in metres per second
• s = displacement in metres
• t = time in seconds
When we have a problem involving projectile motion then we simply pick the relevant equation to calculate what we need to know. As mentioned above, air resistance will significantly change the answers in some situations.
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https://www.fool.com/archive/portfolios/rulemaker/2000/06/20/dividends-vs-share-repurchases.aspx
| 1,493,360,870,000,000,000
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In tonight's report, I'm going to more closely examine the two primary ways in which corporations can return money to their shareholders: dividends and share repurchases.
First, though, let's take a look at how one of the tools from the Rule Maker bag of tricks helps us measure the amount of cash available for shareholders -- the Cash King Margin (CKM). The numerator of the CKM is calculated by pulling a couple of numbers from the cash flow statement. We start with "cash provided by operating activities" and then adjust for net "additions to property and equipment." That leaves us with free cash flow (FCF). We then divide FCF by total revenue to determine the CKM. The formula looks like this:
```(Operating Cash Flow - Capital Expenditures)
--------------------------------------------
Sales
```
So how does this relate to dividends and share repurchases, you ask? Well, a company's free cash flow is what is used to pay its shareholders in the form of either dividends or stock buybacks. In general terms, dividends and/or share repurchases are the default use of cash when there are no better investment alternatives. Cash sitting idly on a profitable company's balance sheet will actually reduce profitability ratios such as return on assets and return on invested capital. The general rule is that if a company does not have a way to invest its money that will provide a sufficient return to its shareholders it should return that cash to its shareholders so that they can invest it themselves.
In some cases, companies will buy back stock by borrowing money. This can be done to change the relative allocation of debt and equity on the company's balance sheet. Since interest expense is tax deductible, there are companies that favor carrying some debt on their balance sheet due to the fact that it's cheaper to finance via debt than equity (For more on this topic, see "Debt: The Good, the Bad, and the UGLY -- Part 1 and Part 2").
Let's take a look at Intel's (Nasdaq: INTC) Q1 '00 cash flow statement and use it as an example so that we can put our numbers in context. Here are the key numbers we need to calculate Intel's free cash flow for the first quarter:
``` \$3,281 --> Net cash provided by operating activity
- 1,074 --> Addition to property, plant & equipment
= \$2,207M --> Free Cash Flow
```
As you can see, Intel's first quarter free cash flow rang in at \$2,207 million. Not bad for a quarter's work! Since Intel's sales for this period were \$7,993 million, that transates into a Cash King Margin of 27.6% (2,207 / 7,993). That's a solid performance, and well ahead of our 10% benchmark minimum.
Another way of looking at this is that 27.6% of Intel's first quarter revenue could have been returned to shareholders. If we take another gander at the cash flow statement, we can see how much of that cash Intel actually returned to its shareholders. The answer is found in the section of the cash flow statement called "cash flows provided by (used for) financing activities." During the quarter, Intel used about 45% of its free cash flow -- an even \$1,000 million -- to repurchase stock. But, it only used about 4.5% of its free cash -- \$100 million -- to pay dividends to its shareholders.
Based on these figures (or those for just about any other period that you can check), it's pretty clear that Intel has favored buying back stock to paying dividends. If you're an Intel shareholder, you may or may not appreciate the way that Intel allocates its free cash flow.
To understand why Intel favors share buybacks, it may help to review the basics about how shareholders are taxed on buybacks compared to dividends. Dividends are ordinary income. That means that you, the shareholder, pay tax on any dividends that you receive based upon your marginal tax rate (the maximum individual tax rate is currently 39.6%). You have no choice as to whether or not you pay this tax. If Intel decides to pay a dividend, you pay tax on it. There are clearly some disadvantages to that arrangement.
If Intel decides to repurchase its own shares and you want to participate in the buyback, you have two options. One is that you can sell some of your shares. If you do, and you have held the shares for longer than a year, you pay tax at the long-term capital gains rate (the maximum rate for individuals is currently 20%). If you decide not to take a portion of the money that Intel is returning to its shareholders, then you don't pay any tax at the time of the buyback at all. Clearly, from a tax perspective this gives you more flexibility than you have with dividends.
I just talked about changes in your share ownership as a result of share buybacks. Let's take a look at how that comes about. To the extent that a company buys back shares, the number of total shares outstanding decreases. That means that your ownership interest in the company marginally increases each time there is a share buyback. Confused? Don't be! Here's a simple example:
Say that you own 100 shares of Verve Inc. (Ticker: MATT), and that there are 1 million total shares outstanding at \$20 each (total market cap = \$20 million). You own 0.01% of the company. Next say that Verve decides to buy back 50,000 shares. Since you're a long-term investor, you decide to hold on to your shares. At the completion of the buyback, your ownership interest has increased to 0.01052%. All other things being equal, the value of each share would now be \$21.05, in order to keep the market cap of the company at \$20 million. By utilizing a share buyback, Verve delivered per-share value of \$1.05 without any negative tax ramifications. By way of comparison, if Verve had issued a \$1.05 per-share dividend, you would only receive \$0.65 or so after taxes.
Does this mean that dividends are always bad? While I prefer share buybacks, I don't object to dividends such as the \$0.16 per-share annual dividend that Intel pays. There are two reasons for this. One is that the payment of this small dividend enables Intel to offer a very shareholder-friendly dividend reinvestment plan (DRIP). Speaking of DRIPs, you may know that Intel is a major holding over in the Drip Portfolio. (If you're interested, here's a list of other Rule Maker DRIPs.)
Also in the defense of dividends, if you don't have a current need for your cash dividends and if your broker allows it, you can reinvest your dividends. This may allow you to make incremental investments in the stocks over time and at no cost.
If you think that investing such dividends is meaningless, I can offer up an example as to why you might want to reexamine that conclusion. I have an account where I've been reinvesting all the Intel dividends that I've received. (I first purchased shares in May 1995.) The first dividend that I earned was paid to me in June of that year, also at a rate of \$0.16 per share annually (which is now \$0.64 after taking into account the impact of the two two-for-one stock splits that have occurred since that time). Those reinvested dividends have increased 7-fold during the time that I've owned them. That's clearly a better return than I would have gotten by putting the dividends in a bank account.
Personally, I don't mind small dividends at all in those cases where they enable the company to set up an investor-friendly DRIP. Where possible, I also favor reinvesting any such dividends in additional shares of stock. But, all in all, due to the flexibility that they provide to shareholders from a tax perspective, I prefer share buybacks to dividends.
If you wish to discuss this report further, please feel free to ask your questions on either the Rule Maker Strategy or Rule Maker Beginners boards.
And finally, there's only ten days left to sign up for the upcoming Rule Maker Seminar. Whether you're a rookie or a veteran to this way of investing, it's going to be a fun and informative experience. We'll be explaining the strategy step-by-step, while also going deep into examples from our focus group of companies (which you chose). And of course, you'll be choosing our next investment by seminar's end. For more details, check out our RM Seminar Information Page.
Phil Weiss (TMFGrape on the boards)
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Uncommon Descent Serving The Intelligent Design Community
# Vodka! Jean Claude Perez, the golden ratio, dragon curve fractals and musical design in “junk DNA”
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Jean Claude Perez is a self-organizational theorist, he is not a creationist. He has also published papers with an occasional visitor to UD, Andras Pellionisz.
If the mathematical/musical patterns Perez has found in DNA are improbable relative to laws of physics and chemistry, then he may have found yet another design feature of DNA, and this feature is found by combining coding DNA with non-coding DNA and viewing it holistically.
Here is the simplest explanation I found of his work:
When cells replicate, they count the total number of letters in the DNA strand of the daughter cell. If the letter counts don’t match certain exact ratios, the cell knows that an error has been made. So it abandons the operation and kills the new cell.
Failure of this checksum mechanism causes birth defects and cancer.
Jean-Claude Perez discovered an evolutionary mathematical matrix in DNA, based on the Golden Ratio 1.618
Dr. Jean-Claude Perez started counting letters in DNA. He discovered that these ratios are highly mathematical and based on “Phi”, the Golden Ratio 1.618. This is a very special number, sort of like Pi. Perez’ discovery was published in the scientific journal Interdisciplinary Sciences / Computational Life Sciences in September 2010.
Jean-Claude Perez discovered an evolutionary mathematical matrix in DNA, based on the Golden Ratio 1.618
Before I tell you about it, allow me to explain just a little bit about the genetic code.
DNA has four symbols, T, C, A and G. These symbols are grouped into letters made from combinations of 3 symbols, called triplets. There are 4x4x4=64 possible combinations.
So the genetic alphabet has 64 letters. The 64 letters are used to write the instructions that make amino acids and proteins.
Perez somehow figured out that if he arranged the letters in DNA according to a T-C-A-G table, an interesting pattern appeared when he counted the letters.
He divided the table in half as you see below. He took single stranded DNA of the human genome, which has 1 billion triplets. He counted the population of each triplet in the DNA and put the total in each slot:
When he added up the letters, the ratio of total white letters to black letters was 1:1. And this turned out to not just be roughly true. It was exactlytrue, to better than one part in one thousand, i.e. 1.000:1.000.
Then Perez divided the table this way:
Perez discovered that the ratio of white letters to black letters is exactly 0.690983, which is (3-Phi)/2. Phi is the number 1.618, the “Golden Ratio.”
He also discovered the exact same ratio, 0.690983, when he divided the table the following two alternative ways:
Perez discovered that the ratio of white letters to black letters is exactly 0.690983, which is (3-Phi)/2. Phi is the number 1.618, the “Golden Ratio.”
He also discovered the exact same ratio, 0.690983, when he divided the table the following two alternative ways:
Above: Total ratio of white:black letters = 1:1
So for three ways of dividing the table, the ratio of white to black is 1.000:1.000.
And for the other three ways of dividing it, the ratio is 0.690983 or (3-Phi)/2.
When you overlay these 6 symmetries on top of each other, you get a set of mathematical stairs with 32 golden steps. Then an absolutely fascinating geometrical pattern emerges: The “Dragon Curve” which is well known in fractal geometry. Here it is, labeled with DNA letters in descending frequency:
You can see other non-DNA, computer generated versions of this same curve here.
Other interesting facts:
•Similar patterns with variations on these same rules are seen across a range of 20 different species. From the AIDS virus to bacteria, primates and humans
•Each character in DNA occurs a precise number of times, and each has a twin. TTT and AAA are twins and appear the most often; they’re the DNA equivalent of the letter E.
•This pattern creates a stair step of 32 frequencies, a specific frequency for each pair.
•The number of triplets that begin with a T is precisely the same as the number of triplets that begin with A (to within 0.1%).
•The number of triplets that begin with a C is precisely the same as the number of triplets that begin with G.
•The genetic code table is fractal – the same pattern repeats itself at every level. The micro scale controls conversion of triplets to amino acids, and it’s in every biology book. The macro scale, newly discovered by Dr. Perez, checks the integrity of the entire organism.
•Perez is also discovering additional patterns within the pattern.
I am only giving you the tip of the iceberg. There are other rules and layers of detail that I’m omitting for simplicity. Perez presses forward with his research; more papers are in the works, and if you’re able to read French, I recommend his book “Codex Biogenesis” and his French website. Here is an English translation.
(By the way, he found some of his most interesting data in what used to be called “Junk DNA.” It turns out to not be junk at all.)
OK, so what does all this mean?
•Copying errors cannot be the source of evolutionary progress, because if that were true, eventually all the letters would be equally probable.
•This proves that useful evolutionary mutations are not random. Instead, they are controlled by a precise Evolutionary Matrix to within 0.1%
•When organisms exchange DNA with each other through Horizontal Gene Transfer, the end result still obeys specific mathematical patterns
•DNA is able to re-create destroyed data by computing checksums in reverse – like calculating the missing contents of a page ripped out of a novel.
No man-made language has this kind of precise mathematical structure. DNA is a tightly woven, highly efficient language that follows extremely specific rules. Its alphabet, grammar and overall structure are ordered by a beautiful set of mathematical functions.
More interesting factoids:
The most common pair of letters (TTT and AAA) appears exactly 1/13X as often as all the letters combined – consistently, the genomes of humans and chimpanzees.
If you put the 32 most common triplets in Group 1 and the 32 least common triplets in Group 2, the ratio of letters in Group1:Group2 is exactly 2:1. And since triplet counts occur in symmetrical pairs (TTT-AAA, TAT-ATA, etc), you can group them into four groups of 16.
When you put those four triplet populations on a graph, you get the peace symbol:
dna_peace_symbol
Does this precise set of rules and symmetries appear random or accidental to you?
My friend, this is how it is possible for DNA to be a code that is self-repairing, self-correcting, self-re-writing and self-evolving. It reveals a level of engineering and sophistication that human engineers could only dream of. Most of all, it’s elegant.
Cancer has sometimes been described as “evolution run amok.” Dr. Perez has noted interesting distortions of this matrix in cancer cells. I strongly suspect that new breakthroughs in cancer research are hidden in this matrix.
I submit to you that the most productive research that can possibly be conducted in medicine and computer science is intensive study of the DNA Evolution Matrix. Like I said, this is just the tip of the iceberg.
There is so much more here to discover!
When we develop computer languages based on DNA language, they will be capable of extreme data compression, error correction, and yes, self-evolution. Imagine: Computer programs that add features and improve with time. All by themselves.
What would that be like?
Perry Marshall
Elizabeth Liddle with her love of music might actually appreciate the possibility that life and musical notes are possibly intertwined.
I can’t do justice to the topic by quoting snippets, so I will have to provide a link:
Phi and Music DNA Here is diagram from that article:
Recall the discussion of ID and tuning of musical instruments here.
Perez bio is in creation wiki:
Jean-Claude Perez, Ph.D., is a French interdisciplinary scientist born on June 26, 1947 in Bassens, Gironde near Bordeaux (France). An engineer and French scholar from Bordeaux university[1], Perez worked principally with IBM in both the areas of Biomathematics and Artificial Intelligence (the first time, showing evidence of high level self-organization in cellular automata networks [2] and the second time creating neural networks with “FRACTAL CHAOS”(Fractal geometry), his holographic-like memory system and novelty detector). Then, in 1990, Jean-claude Perez published strong links between the world of fractals and numbers of the Fibonacci sequence which are based on the Golden ratio[3]. In this last area, with the “DNA supracode”[4], he proved that DNA coding for genes is structured by proportions related to Fibonacci numbers[5] [6] [7]. He verified this discovery in the field of the HIV genome by partnerships with Professor Luc Montagnier[8], the discoverer of the HIV virus. He has worked for 20 years in the fields of whole genome numerical analysis and numerical decoding of genes as coding or non-coding DNA sequences (as demonstrated particularly by the last publications: Five last publications/conferences ).
Particularly, in “Interdisciplinary Science” September 2010 issue, J.C. Perez published a peer-reviewed paper proving that the whole human genome codon populations are managed by a “DRAGON fractal paper folding curve” fine-tuned around the “Golden ratio”. Particularly, this main paper entitled “Codon populations in single-stranded whole human genome DNA are fractal and fine-tuned by the Golden Ratio 1.618.” shows that the Universal Genetic Code Table not only maps codons to amino acids, but serves as a global checksum matrix at the whole genome macro-structural scale.[9]
Golden ratio.jpg
Draft of the paper Final published paper
A complete summary of J.C Perez’s research was published in Pellionisz A., Graham R., Pellionisz P., Perez J.: Genome Function of the Cerebellum: Geometric Unification of Neuroscience and Genomics. In: Manto M., Gruol D., Schmahmann J., Koibuchi N., Rossi F. (Ed.) Handbook of the Cerebellum and Cerebellar Disorders: SpringerReference (www.springerreference.com). Springer-Verlag Berlin Heidelberg, -1. DOI: 10.1007/SpringerReference_310386 2012-03-12 14:15:14 UTC. Details in [10] and full paper available in [11]
In october 2013 jean-claude perez published a peer reviewed major article in [APPLIED MATHEMATICS]http://www.scirp.org/journal/am/ (BIOMATHEMATICS issue) entitled “The 3 genomic numbers discovery”. This article show – for the first way – that complete human genome single stranded DNA constitutes a WHOLE… [12]
http://creationwiki.org/Jean-claude_Perez
The “Vodka” designation means a highly speculative topic.
I have just carried out three experiments to verify the work done by Perez. I am happy to announce that his findings have been confirmed. I have created a video showing each of the experiments. You can view the video by going to Youtube and entering the phrase "Mathematical Patterns within DNA" Craig Paardekooper craig373
ERRATUM 2 last remarks on Gordon Davisson remarks 12 and 15: Post 12: you are wrong because precisely i analyse single stranded dna whole genome sequence where there is no reason to find trivial crick watson pairing symmetries… Post 15: i meet inventor of ISOCHORES PR GIORGIO BERNARDI 25 years ago in his lab paris sorbonne… There are no links between my analysis at codpn population scale and ISOCHORES ratios… We no not considere here the same level of genomic infpamation: him. Ratiod CG by species. Me. Codons triplets jean-claude perez
2 last remarks on Gordon Davisson remarks 12 and 15: Post 12: you are wrong because precisely i analyse single stranded dna whole genome sequence where there id no reason to find trivial crick watson pairing symmetries... Post 15: i meet inventor of ISOCHORES PR GIORGIO BERNARDI 25 years ago in his lab paris sorbonne... There are no links between my analysis at codpn population scale and ospchorrs ratips... We no not considere here the same level of henomic infprmation: him. Ratiod cg by species. Me. Codons triplets jean-claude perez
Contrarly The first pi analyse of QUARK MIST be studied seriously! jean-claude perez
To Gordon: An important topic showing you run in error: precisely if i analyse SINGLE STRANDED DNA the trivial crick watson base pairing disappear . Then your analyse is wrong in the native article we discuss this topic... I m sorry... jean-claude perez
Dears Pietr, Joe or Gordon, I'm sorry for your unapropried comments on SANDWALK of the Sal Cordova entry entitled Vodka! Jean Claude Perez, the golden ratio, dragon curve fractals and musical design in “junk DNA”... The reason is that all (ALL) their comments were done without reading the basic original article: I suggest you reading the original basic peer review article of 2010 published in Interdisciplinary Science: http://fr.scribd.com/doc/95641538/Codon-Populations-in-Single-stranded-Whole-Human-Genome-DNA-Are-Fractal-and-Fine-tuned-by-the-Golden-Ratio-1-618 and my 2013 peer review article: http://www.scirp.org/journal/PaperInformation.aspx?paperID=37457#.U2Mwlfl_trA jean-claude perez
Quark1 Free Curiisity is on of the first qualities to do best and real ...research in science.. Like you do there!! jean-claude perez
Hello Jean-Claude Perez No, I did this only for fun and I like math. quark1
Quark1: intersting perhaps: our devise of researchers and not of numerologists: in french sorry: "il faut separer le bon grain de l'ivraie" jean-claude perez
Quark1: I precise a bit: there is here the distance between SCIENCE and... NUMEROLOGY ! jean-claude perez
quark1: no comment... Here is the real limit of a real Scientific! jean-claude perez
Hello again. Golden ratio formula: Other way to find the thi number. (((TTA)-(TTC))^2 + ((TCT)-(TAT))^2 + ((TGC) - (TGA))^2 + ((TAG)-(TCG))^2)/10^16 = 1,618 General Quadrilateral The general quadrilateral can be find from this numbers. We name the angles with the corners like e,f,v and w and the two diagonals for E and F. The length of the general quadrilateral can be named as A, B, C, and D The length can be calculate as A = ((TTT)-(TTC))+((TTG)-(TTA)) B = ((TTT)-(TCT))+((TGT)-(TAT)) C = ((TGT)-(TGC))+((TGG)-(TGA)) D = ((TTG)-(TCG))+((TGG)-(TAG)) The total angle for this geometry is 2pi, and we want to use the cosinus law to calculate the angles. But we need the diagonal for that and if we look at the numbers in the inner circle which is (TCC),(TAC),(TCA) and (TAA), this numbers maybe can be corresponding to the diagonals. One can see some kind of structure of this thing. However use that the diagonal is E = (TAC) and F = (TCA) e = cos^-1((E^2-A^2-C^2)/2AC) = 1,6371 f = cos^-1((E^2-B^2-D^2)/2BD) = 2,011 v = cos^-1((F^2-C^2-D^2)/2CD) = 0,83 w = cos^-1((F^2-A^2-B^2)/2AB) = 1,799 Thereafter, e + f + v + w = 2pi For this case I only get an error on 0,1%. Other cases this will be between 1 to 2 %. Probability other geometrical figures can be find, I´m nearly to find one formula for the circle from this thing but so far the error is about over 5 %. quark1
https://uncommondesc.wpengine.com/intelligent-design/professor-larry-moran-poses-five-questions-for-the-id-movement/#comment-498347 jean-claude perez
more data in: https://uncommondesc.wpengine.com/intelligent-design/professor-larry-moran-poses-five-questions-for-the-id-movement/#comment-498347 jean-claude perez
Gordon, on your last main question: "Finally, is there any reason to regard the similarity between the GC/AT ratio and (3–Phi)/2 as anything other than a numeric coincidence? Especially since it varies a great deal between different regions of the genome, different chromosomes, and different species? Especially since there are many other formulae involving Phi that it could have come out close to (e.g. Phi/3, Phi/2, Phi-1, 4-2*Phi, 2*Phi/5, etc…)" 2 responses: 1/ In my 2013 article, it is show the universal nature of this value 'quarks etc...° 2/ applying this same analysis to: -whole human genome ==> (3-Phi)/2 -lower scale chromosome is chr4: ratio = 1/Phi -highter scale chromosome is chr19: ratio = 1/Phi + 1/Pi strange no? Then analysing by others approachs chr4 shows that - if there is "design" - the right research way must be... HUMAN CHROMOSOME4 jean-claude perez
Gordon, ALL your questions are fine and natural. But all find response in my 2 peer published articles referenced here: the 2010 article: http://fr.scribd.com/doc/95641538/Codon-Populations-in-Single-stranded-Whole-Human-Genome-DNA-Are-Fractal-and-Fine-tuned-by-the-Golden-Ratio-1-618 the 2013 article: http://file.scirp.org/Html/4-7401586_37457.htm jean-claude perez
Gordon, I sent you an e-mail from my personal account. You'll get an e-mail from the CEU Admin "Hwang" as well. I really don't need your personal e-mail, just something address where I can privately transmit you username and password. I don't expect there will be a lot of traffic on that discussion board, and that is by design. I prefer it to be a repository of research. Sal scordova
Sal, thanks for your offer. I don't like to publish my regular email address anywhere spammer-searchable, but if you drop me a note at -------, I'll send you my real address. [e-mail edited to protect commenter privacy] Gordon Davisson
Jean-Claude Perez, thank you for joining us! I have some questions that I'd appreciate if it you could answer: First, is there anyplace where you detail exactly how you performed your various analyses? I earlier used the description in the "Phi and Music in DNA" (by Jordi Solà-Soler), but it appears that doesn't match your method. For example, in the "The Number « 4 »" section of your Beijing paper, the counts for TTT and AAA are close, but not exactly equal (while Jordi's method will always give exactly equal counts for these triplets). How exactly did you do your counting? Second, it looks to me like you're reading a lot of meaning into features of the data that don't really seem (to me) to show anything significant. For example, in the "-III- Evidence of 2 « attractors »:« 1 » and « (3-Phi)/2 »" section of your Beijing paper, you make a great deal of the fact that ratio of triplets in the even and odd halves of the table (and even and odd octants, etc) are very close to 1. But isn't this just a trivial result of DNA base pairing rules (every T on one strand matches an A on the other, and every C on one matches a G on the other), together with a lack of bias for either half of the pair to be concentrated on one strand? In other words, if you counted the bases on both strands, DNA pairing guarantees that the number of Ts will match the number of As, and the number of Cs will match the number of Gs. And unless there's some consistent difference between the two strands that led, for example, most of the As to be in one strand and the matching Ts to be in the other, the counts should nearly match for each individual strand as well. Just as we see. (Note that your "codon level generalization of Chargaff's second rule" can also be explained the same way.) As for the ratios that come out to (3–Phi)/2, aren't these just the GC/AT ratios, filtered through the triplet table in various ways? Finally, is there any reason to regard the similarity between the GC/AT ratio and (3–Phi)/2 as anything other than a numeric coincidence? Especially since it varies a great deal between different regions of the genome, different chromosomes, and different species? Especially since there are many other formulae involving Phi that it could have come out close to (e.g. Phi/3, Phi/2, Phi-1, 4-2*Phi, 2*Phi/5, etc...) Gordon Davisson
Dear SCORDOVA, your last question: "I have my doubts as well, and I was waiting to see if anyone else would chime in. My main issue is how did Perez do an alignment so as to identify triplets. There are huge “NNNN” segments in fasta files, so how does he know where to start recognizing triplets?" my response: on the about billion codon triplets within whole human genome, NNN undetermined nucleotides are unsignifiant. Then to be sure of consistency of analyses, we have analysed the 3 coden reading frames, compressing NNN bases. Then, the 3 kinds of results are highly similar... The perfect proportions fine tuning around golden ratio etc... are statistically unaffected by undetermined bases. MEANWHILE, in the CODEX BIOGENESIS book, we show how accuracy increase withe the progress of human genome project comparing the successives releases of HGP, with NNN decreasing at each release. THEN our ratios converge at each new release... Here is a DYNAMIC proof of our discovery! Please, see CODEX BIOGENESIS pp 155 fig 12.3 comparing Human Genome Project releases od April 2001, Nivember 2002 and August 2003. http://www.amazon.co.uk/Codex-Biogenesis-harmonies-g%C3%A9nome-latome/dp/2874340448 jean-claude perez
Complete jc perez interdisciplinary Creation Wiki BIOGRAPHY in: http://creationwiki.org/Jean-claude_Perez jean-claude perez
Full J.C. Perez's interdisciplinary biography in CreationWiki: http://creationwiki.org/Jean-claude_Perez jean-claude perez
Dear GORDON DAVIDSON, you are right talking on PI: if the concensus ratio for whole human genome is (3-Phi)/2, the limits lower chromosomes are: chr4 1/Phi and chr19 1/Phi + 1/Pi !!!! please see details in my BEIJING conference here: http://fr.scribd.com/doc/57828784/jcperezBeijing032011 jean-claude perez
Gordon all this work was reproduced by Dr Jordi sola soler http://www.sacred-geometry.es/en/content/phi-and-music-dna for human chromosome1 I have all data results... details in my book CODEX BIOGENESIS http://www.amazon.co.uk/Codex-Biogenesis-harmonies-g%C3%A9nome-latome/dp/2874340448 jean-claude perez
Gordon, I publicly express my thanks for your meticulous work. I'm indebted to you: https://uncommondesc.wpengine.com/news/many-thanks-to-gordon-davisson-and-joe-felsenstein-for-review-and-criticism-of-my-ud-article/ scordova
Gordon, If it is okay with you: 1. I can try to get you a copy of Perez orginal paper, I have several papers I'll be getting in also, and I can make them available to you. 2. I'd like to give you an account at Creation Evolution University to record some of our more technical interactions rather than at UD. It will be a better way of archiving some of what you have to say without it going quickly into cyber oblivion. What you have to say is to valuable, and it's awfully hard to search for some of your comments in the quarter million or so at UD. I have my doubts as well, and I was waiting to see if anyone else would chime in. My main issue is how did Perez do an alignment so as to identify triplets. There are huge "NNNN" segments in fasta files, so how does he know where to start recognizing triplets? scordova
Sal, I don't think there's any question that there are many types (/levels) of non-randomness in genome sequences. In addition to the ones you mentioned, I'll throw isochores (regions with similar GC content) on the list. And note that they're related to other forms of non-randomness, for example gene-rich regions tend to have high GC content, while gene deserts have low GC content. I think there are two different questions here, though: first, whether natural processes (i.e. evolution) can account for the various forms of non-randomness, and second whether Perez has identified a significant form of non-randomness. The first question is huge (in a sense, it's the question of ID) and complicated, and I'll mostly duck it here. Well, ok, a really quick summary of my opinion: some of the non-randomness can be accounted for by known processes and mechanisms, and some will be accounted for by processes and mechanisms we've yet to fully work out. I haven't seen a convincing case either way for whether all non-randomness will eventually be explained by evolutionary mechanisms, so I don't try to draw implications from either assumption. As for the second question, it's also a bit hard to answer, because I don't have access to the original paper, data, and analysis method. There's some stuff at golden-ratio-in-dna.blogspot.com that might be by Perez, but I frankly find it incomprehensible. The discussion in "Phi and Music in DNA" is more readable, but I'm not sure it's describing Perez's analysis method correctly. If the description in "Phi and Music in DNA" is correct, I think my criticism is obviously correct (to the point of not needing to be tested). There is one small correction I need to make, though: bases near the end of a strand will not be counted as many times as I described. A base right at the end of a strand can only be the first or last of a triplet, so it only contributes to 2 entries in the table. Similarly, a base one from the end will only contribute to 4 entries. Bases two or more from the end contribute to the full 6 entries, as I described. Note, however, that the paired base also contributes the same amount, so the symmetry is maintained even in these cases. BTW, there's another, more visible way to see the symmetry due to the analysis method: suppose that at a particular position in the genome, there's an ACG triplet. It obviously gets counted in the ACG bin. In the reverse order, it also gets counted in the GCA bin. The paired triplet on the other strand, TGC, similarly gets counted in the TGC and (in reverse) CGT bins. This means that the ACG, GCA, TGC, and CGT bins should all have exactly the same count. And if you look at the figure you included, they do: 96112792. Similarly, CAT, TAC, GTA, and ATG all have 169023944; ATT, TTA, TAA, and AAT all have 260313647; etc. Note that if the analysis didn't count both strands in both directions with all three possible starting offsets, we'd expect these symmetry groups to have the same counts on average, but have different statistical variations from that average. With all 12 possibilities counted, these variations are smoothed out and vanish, but if the 12-way overcounting wasn't done we'd see at least some variation. Maybe not very much, though: if non-random features appear at random starting offsets on random strands, they'll average out and produce near-symmetries in the table. That may sound backward, so I'd better also point out that in this table arrangement, uniformity shows up as symmetries in the table, and non-uniformity (i.e. deviations from uniform randomness) show up as asymmetries. For example, the 3-codon periodicity (actually, correlations between nearby bases in general) show up in the table as differences between bins that are anagrams of each other. Thus the difference in counts between CAT (169023944), CTA (149333215), and ACT (202932695) is an indicator of non-randomness in the sequence. Thus, the symmetries he points out (equalities between different rows and columns) are actually an indication of a lack of pattern in the DNA sequence (if anything). Note that this does lead to something that could be tested: generate a random sequence of ACGT with a slight bias toward AT, and see if the various permutation, reorder, and complement bins come out with nearly the same counts. I feel pretty secure that they will... I'm pretty sure we can also rule out the checksum claim. If it were correct, mutations wouldn't happen (they do), and the different chromosomes would all have the same CG ratios (they don't; look at figure 1 of the isochore map paper, and compare chromosomes 17, 19, and 22 with 4, 5, 13, and X). Gordon Davisson
Where I was heading with all this is that I think there are non-random pattern in the DNA as a whole including the "junk DNA". It makes possible this: http://www.asknature.org/strategy/0146a4de195dde317e3ce62870a3544a#.U12sm41OXrc Whether Perez is right or wrong, the quest for non-random patterns is on. How Chargaff's rules didn't detect codon bias or 3 periodicity is an important question. Something has to give! Now if the physical "fractal globules" in http://www.asknature.org/strategy/0146a4de195dde317e3ce62870a3544a#.U12sm41OXrc have connection to the sequences, then this is also important, and it also means "junk DNA" has significance. scordova
Gordon, Thanks for the 2nd pair of eyes on this. First off, it is acknowledged on several levels there is non-randomness in the DNA. For example, in coding regions there is codon bias. There is also 3 base periodicity. http://www.sciencedirect.com/science/article/pii/S0022519307001543 One way to test your claim is to throw the same test on randomly generated ATCG. Maybe there is a chemical constraint, maybe it is a biological constraint in the way the organism structures itself. As long as the tallying isn't too complex, I think there is room for legitimate detection of checksum, and detection of polyconstraints on the DNA, etc. Thanks for you input, that's why I posted it. I'll revisit this again when I have a chance to discuss with others who are more qualified than I. Thanks again for your criticism. scordova
Umm, isn't this mostly just the occurrence ratios of the different DNA bases, with a lot of over-interpretation? Perez’s original article is paywalled, but according to the Phi and Music in DNA article:
In his work Jean-Claude Perez analysed the entirety of the whole human genome (not just the coding 2%) from the 2003 “BUILD34” finalized release [2]. He considered only a single strand of the DNA sequence. Given a sequence of bases, for example TCAATGTCA, if the starting point is unknown there are three possible ways of grouping the bases into codons; these three codon groupings are known as the codon reading frames: 1) TCA|ATG|TCA 2) CAA|TGT|CAT 3) AAT|GTC|ATC Using one whole DNA strand, Jean-Claude Perez computed, for each of the 3 possible codon reading frames, the cumulative number of each of the 64 genetic code codons. In his book "Codex Biogenesis" [3] he also provides the cumulative populations of codons obtained joining the 12 possible reading combinations (2 strands x 2 senses per strand x 3 reading frames). My results were obtained while trying to replicate his results on this bigger population table (Table 2).
...so each DNA base contributes to 6 entries in the table: twice as the first base of a triplet (once read in each direction), twice as the second base, and twice as the third base. Similarly, that base's pair on the other strand also shows up 6 times. Say that at a particular position there happens to be an adenine (A) base. It'll contribute two counts somwehere in the third column (exectly where depends on its neighboring bases). From the base pairing rules, we know that its pair on the other strand will be a thymine (T) base, will contribute two counts to the first column. If the bases were reversed, the counts would be the same. And a guanine-cytosine pair would similarly contribute two counts each to colums two and four. So the base pairing rule trivially explains why the totals in columns 1+2 equals that in 3+4. The same principle applies when you look at the rows by first and third bases. No magic needed, just the standard chemistry of DNA base pairing. But what about the other ways of breaking the table down, e.g. columns 1+3 vs 2+4? In that case, we're just comparing the total number of A-T pairs (times 4) with the number of G-C pairs (also times 4). I'll get to its relationship to Phi in a minute, but first notice that when he breaks the table into rows based on the first base, he's actually making the same calculation, just in a different format. And the same thing when he breaks it down by third base. There's nothing special about the fact that he gets the same ratio each time; that's an inevitable result of the way the table was constructed. The overlaid symmetries he gets all excited about (and draws the dragon curves etc based on) have nothing to do with the structure of the DNA, they're entirely due to the way he constructed the table. The structure is entirely a figment of his analysis method! But what about that ratio, and its relationship to Phi? According to Perez, his ratio comes out to exactly 0.690983, which is (3-Phi)/2. But I ran the calculations on the table in the "Phi and Music in DNA" paper, and got 0.691457 instead. Pretty close, but not exact... The ratio he's looking at is essentially a nonstandard way to describe the GC content of the genome -- that is, the fraction of all base pairs that are G-C instead of A-T. Perez is looking at (C+G)/(A+T), while the standard GC ratio is defined as (C+G)/(C+G+A+T). The GC ratio is known to vary a great deal -- between regions in the same genome, between different genomes... You can see the variation in different regions in the human genome in the figures in "An isochore map of human chromosomes" by Maria Costantini, Oliver Clay, Fabio Auletta, and Giorgio Bernardi, Genome Res. 2006:16:536-541. It's pretty clear there's no checksumming to keep this ratio consistent. Here are some examples of different species, based the Wikipedia article on Chargaff's rules of DNA base frequencies, where I calculated the Perez-style ratio based on the listed CG ratio (sorry about the sloppy formatting):
Species CG Ratio Perez ratio ?X174 44.8% 0.812 Maize 46.1% 0.855 Octopus 35.2% 0.543 Chicken 43.7% 0.776 Rat 42.9% 0.751 Human 40.7% 0.686 (note that this doesn't quite match either Perez' or my figure) Grasshopper 41.2% 0.701 Sea Urchin 35% 0.538 Wheat 45.5% 0.835 Yeast 35.8% 0.558 E. coli 51.7% 1.070
Again, note that the ratio varies quite a bit. The fact that the human ratio happens to be close to (3-Phi)/2 looks more like a coincidence than anything else. This is numerology, not science. So to sum up... we've got patterns injected by the analysis method, massive overinterpretation, and some numerology to top it off. Is there anything of any value here? Gordon Davisson
Actually, prices can be predicted with a better than random chance of success. Fama won a Nobel Prize for demonstrating this in the 1970s, I believe. My father demonstrated it ten years earlier in his doctoral dissertation, The Implications of Trend Persistency in Portfolio Management. anthropic
Well I´m not a biologist so I don´t now if this really means anyting but however I find something interesting about these numbers. You can find number pi from this genetic maps. 1) Take subtraction diagonal like A = (TCT)-(TTC) = 251604556 - 225669775 = 25934781 B = (TAT)-(TTA) = 234734484 - 260313647 = -25579163 C = (TGT)-(TTG) = 229405484 - 190719226 = 38686258 D = (TAC)-(TCA) = 169023944 - 202932695 = -33908751 E = (TGC)-(TCG) = 96112792 - 91988158 = 4124634 F = (TGA)-(TAG) = 202932695 - 149333215 = 53599480 2) Alright, thereafter do a summation like A + B + C + D + E + F = 62857239 3) Divide with 10.000.000 and thereafter (1/10.000.000)(A + B + C + D + E + F) = 6,2857239 This is approximately 2pi = 2*3,14159 = 6,283185 Do the same for the other three intervall: The second one: I get -6pi The third one: I get 3pi The fourd one: I get pi What do this means? Do the biological systems use circles and the number of turns in the genetic codes? The negative maybe means backword turn of the circle and will go three turns? I´m also find other strange formula! (TTT) - (TAA) = abs(A) + abs(B) + abs(C) + abs(D) + abs(F) = 10((TCC) - (TGG)) = 10pi*(7/2)^(1/4)*abs(E) = 177265160 There for example abs(B)=abs(-25579163) = 25579163 The interesting numbers can also be TCC = CCT = AGG = GGA = 188628365 Thereafter (188628365 - 188000000)/100000 = 6,28365 = 2pi Alright enought from me. However it is very awesome with the music octave discovery from these numbers, wow. quark1
Any system that requires better than a 50% success rate will fail precisely because price cannot be predicted. Successful trading (which has a 99% failure rate among those who try their hand) based upon whatever -- technical, fundamental, or astrological (lol) -- therefore requires the maximizing of profits and the minimizing of losses, a.k.a. "money management." In light of which, fibs are not used to predict price, but rather, to describe how markets trend with a view to identifying entry, failure, and profit-taking levels. Anyway, independent of boosters and detractors, I've got the perfect empirical test for Fibonacci retracements: whether or not they make me money... :D jstanley01
I don't believe in technical trading as a general rule ESPECIALLY fibonnaci because the parameters are too loose. A lot of it is after the fact. If it were that good, it would be predicting stock prices. It doesn't. There have been a few successful technical traders like Richard Dennis. I prefer statistical arbitrage. However, some statistical analyses really do resist after-the-fact non-randomness like 500 fair coins heads. Perez analysis looks like he has detected a non-random pattern because of its simplicity. It can be and should be tested. Another non-random pattern is the genetic code itself (the association of codons to amino acid residues). The relationship exists because of the protein making machinery. I strongly suspect Perez pattern also exists because of machinery! And if there is machinery to make these patterns, there is a machine maker. scordova
In the financial markets, a lot of technical traders use Fibonacci numbers for their entry, failure, and profit-taking levels. Meanwhile, a lot of non-technical traders claim that fibs only work in hindsight, sort of like voodoo. In my experience, although I have seen fib levels work in real time a'plenty, I'm not ready to risk real money yet. In the near future I'm going to have some time available to test out technical trading using a dummy account. Personally, I see no reason why human intentionality as delineated by financial market meanderings couldn't behave according to fibs, like so much else in nature does. (As a matter of fact, now I'm seeing how the golden mean patterns down to the level of DNA. Wow! Thanks for the post.) While there are many approaches to utilizing fibs, based on my research I'd say that David Halsey's merits an honorable mention to anyone interested in checking out fib techniques. jstanley01
I wonder what anti-ID advocates would say is the better evolutionary model in terms of predicting this sort of finding about DNA? Random/Neutral, or Designed? William J Murray
This current thread, which apparently scordova just posted late last night (GMT-5), has already 127 visits. 120 visits in over 7 years vs. 127 visits in less than a day!
Well, by the time I posted my comment the visitors counter had changed: (Visited 146 times, 146 visits today) Dionisio
"DNA researcher, Andras Pellionisz gives favorable review to a shredding of Dawkins and TalkOrigins" January 11, 2007 Posted by scordova 40 Comments (Visited 120 times, 4 visits today)
Obviously things have changed since scordova posted the above OP on January 11, 2007. That thread accumulated 120 visits until now. This current thread, which apparently scordova just posted late last night (GMT-5), has already 127 visits. 120 visits in over 7 years vs. 127 visits in less than a day! Does that tell us anything? Maybe not, but I found it sorta-kinda interesting. Perhaps it just shows that internet has become more popular and accessible worldwide? Another possible implication? Dionisio
Jehu @ 2
How you checksum DNA without intelligence?
You don't need any intelligence to create any algorithm. It's done ex-nihilo very simple, through the magical D formula Evo = RV + NS + T You just don't understand evolution ;-) P.S. please, don't mention this to engineers or computer scientists, because they don't agree at all, so they will say this is stupid. Dionisio
How you checksum DNA without intelligence? Jehu
I submit to you that the most productive research that can possibly be conducted in medicine and computer science is intensive study of the DNA Evolution Matrix. Like I said, this is just the tip of the iceberg. There is so much more here to discover! When we develop computer languages based on DNA language, they will be capable of extreme data compression, error correction, and yes, self-evolution. Imagine: Computer programs that add features and improve with time. All by themselves.
Self-directed evolution. If I were an evolutionist, I'd probably be a self-organizational neutral evolutionist. It's at least more sophisticated than Darwinism. Nevertheless, I think Perez has possibly found a fruitful area of design research. :cool: scordova
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# Kindergarten Counting Games
"Math Salamanders Free Math Sheets"
Welcome to the Math Salamanders Kindergarten Counting Games.
Here you will find a wide range of free printable Math Games to help your child learn to count on and back, and also to help them learn their Math facts at kindergarten level.
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## Kindergarten Math Learning
Kindergarten Counting Games
Using these games is a great way to learn Math facts and develop mental calculation skills in a fun and easy way.
Encouraging your child to play math games and puzzles like sodoku and chess will also help develop their logical reasoning.
Using these math games below will help your child to develop their Math fact skills as well as their strategic thinking and memory skills.
All the games below have been put in order, with the easiest game first.
Have fun using these free Kindergarten Counting Games...we hope you enjoy them!
Catch the Stars 1-6
Catch the Stars is a set of Math fact games which uses dice rolls to make different numbers.
The aim of the game is to use the numbers on the dice to catch as many stars as possible.
For this game you will need one die, and some counters.
Catch the Stars 1-6
Number of players: 2-3.
Learning skills: matching the number on the die with the number on the board.
Get the Picture Game #1
Get the Picture Game #1 is a simple matching game where you have to match the number of spots on a dice to a number from 1 to 6.
The aim of the game is to be the first person to complete your picture by collecting all the animals and objects that go with it. For this game you will need one dice and some counters.
Get the Picture Game #1
Number of players: 2+.
Learning skills: matching the number on the die with the number on the board.
Get the Picture Game #2
Get the Picture Game #2 is a similar but slightly trickier version of Game #1. The numbers on two dice are added up, and the total has to be matched to a grouped interval on the board.
The aim of the game is to be the first person to complete your picture by collecting all the animals and objects that go with it. For this game you will need two dice and some counters.
Get the Picture Game #1
Number of players: 2+.
Learning skills: adding the roll on 2 dice together; matching the number on the die with a grouped interval on the board.
Salamander Fish Out Game
Salamander Fish Out is a simple environmental game which involves removing a certain number of 'fish' from the lake on each turn.
The aim of the game is to avoid being the player who fishes out the last fish (so destroying the future fish stock for the lake).
Although this game is suitable from the age of kindergarten, it is very good at developing strategic thinking skills in older children.
Salamander Fish Out Game
Number of players: 2
Learning: Counting to 5, strategic thinking.
Pit of Doom 0 to 25
The Pit of Doom is a counting game for helping children familiarize themselves with numbers from 0 to 25.
The game should also help them understand about counting on and especially counting back.
Pit of Doom 0 to 25
Number of players: 2
Learning: numbers to 25, counting on and back.
Race to the Moon
Race to the Moon is a Math game which develops addition and subtraction of numbers up to a total of 10.
The aim of the game is to complete an unbroken path of counters from the Earth to the Moon by working out the sums.
For this game you will need some counters.
Race to the Moon - Addition to 10
Number of players: 2.
Learning skills: - adding numbers to 10; strategy.
Race to the Moon - Subtraction to 10
Number of players: 2.
Learning skills: - subtracting numbers to 10; strategy.
Race to the Moon - Adding & Subtracting to 10
Number of players: 2.
Learning skills: - adding and subtracting numbers to 10; strategy.
Trap the Spiders
Trap the Spiders is a kindergarten counting game which involves trying to catch as many spiders as you can.
The aim of the game is to be the player who catches the most spiders.
You will need a dice to play this game.
Trap the Spiders
Number of players: 1-4
Learning skills: counting the number of spots on a dice, counting the number of spiders trapped.
Race to 100
Race to 100 is a snakes-and-ladders type game where you have to move along the board from 1 to 100.
There are lots of special squares which may mean you get a penalty or a bonus depending on the color you land on.
Race to 100
Number of players: 1-4
Learning skills: Counting up to 100, counting on and back up to 5 squares, learning how to use a 100 square.
Looking for some harder games?
Kindergarten Counting Worksheets
Math-Salamanders.com
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ABCmouse is a subscription based online education program designed for children aged from 2 to 8.
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