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https://www.physicsforums.com/threads/laplace-eq-in-cylindrical-coordinates.915420/ | 1,532,258,083,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676593208.44/warc/CC-MAIN-20180722100513-20180722120513-00438.warc.gz | 962,942,535 | 19,131 | # A Laplace Eq. in Cylindrical coordinates
1. May 22, 2017
### chimay
Hi,
I need to solve Laplace equation:$\nabla ^2 \Phi(x,r)=0$ in cylindrical domain $0<r<r_0$, $0<x<L$ and $0<\phi<2\pi$. The boundary conditions are the following ones:
\left\{ \begin{aligned} &C_{di}\Phi(x,r_0)=\epsilon \frac{\partial \Phi(x,r)}{\partial r}\rvert_{r=r_0} \\ &\Phi(0,r)=f_1(r) \\ &\Phi(L,r)=f_2(r) \end{aligned} \right.
$\epsilon$ being the dielectric constant of the medium and $C_{di}$ a constant capacitance; $f_1$ and $f_2$ are two known function.
Forcing no $\phi$ dependence, and solving the equation by separation of variable:
$\Phi(x,r)=(Ae^{\lambda x/r_0}+Be^{-\lambda x/r_0})J_0(\lambda r/r_0)$
$\lambda$ being the separation constant and $J_0$ the first type Bessel function of 0-order.
By applying the first boundary condition:
$\lambda \frac{J_1(\lambda)}{J_0(\lambda)}=C_r$
$C_r=r_0 C_{di}/\epsilon$
which can be solved to compute all the values of $\lambda$.
By applying the remaining boundary conditions, I obtain the following set of equations:
\left\{ \begin{aligned} &\sum_{m=1}^{\infty} (A_m+B_m)J_0(\lambda_m r/r_0)=f_1(r) \\ &\sum_{m=1}^{\infty} (A_m e^{\lambda_m L/r_0}+B_m e^{-\lambda_m L/r_0})J_0(\lambda_m r/r_0)=f_2(r) \\ \end{aligned} \right.
which allow me to compute the values of all the coefficients, by exploiting the orthogonality between differently scaled Bessel functions:
\left\{ \begin{aligned} &A_i=\frac{1}{(2\sinh{\lambda_i L/r_0})(\frac{1}{2}{J_1(\lambda_i)}^2)}(I_2-I_1 e^{-\lambda_i L/r_0}) \\ &B_i=\frac{1}{(2\sinh{\lambda_i L/r_0})(\frac{1}{2}{J_1(\lambda_i)}^2)}(I_1e^{\lambda_i L/r_0}-I_1) \\ \end{aligned} \right.
$I=\int_0^1 \frac{r}{r_0} f(r) J_0(\lambda_i r/r_0) d(\frac{r}{r_0})$
Now the point that is driving me crazy is the following: how do you expect $A_i$ and $B_i$ to change as a function of the order $i$? I expect them to decrease in modulus; indeed, this is the case for $A_i$, but $B_i$ presents a strange behaviour, oscillating in sign and increasing in modulus. Can you see any mistake? If you want, I can provide you with all the mathematical passages that I omitted for brevity, the trends of $A_i$ and $B_i$ or show you whatever plot you may need. Any suggestion is really appreciated here.
Thank you all.
2. May 23, 2017
I spotted one algebraic mistake I believe: On $B_i$, the last $-I_1$ should be $-I_2$. I didn't check every other detail, but the rest of it looks like it might be correct. $\\$ Editing: A little further reading on the Bessel functions shows that for the functions to be orthogonal, I believe it requires $J_o(\lambda_m)=0$. I'm not sure that your functions meet this condition. $\\$ Additional editing: From what I googled on the subject, you may need to also use $J_n$ of all orders in your series rather than limiting it to $J_o$.
Last edited: May 23, 2017
3. May 24, 2017
### chimay
The algebraic mistake s a transcription error; in my implementation $I_2$ correctly takes the place of $I_1$.
I am not very strong on Bessel equation and I completely forgot the condition you pointed me out!
Can you clarify your second comment? Do you mean:
\left\{ \begin{aligned} &\sum_{m=1}^{\infty} (A_m+B_m)J_m(\lambda_m r/r_0)=f_1(r) \\ &\sum_{m=1}^{\infty} (A_m e^{\lambda_m L/r_0}+B_m e^{-\lambda_m L/r_0})J_m(\lambda_m r/r_0)=f_2(r) \\ \end{aligned} \right.
?
Thank you a lot, your help is really really appreciated.
4. May 24, 2017
### Orodruin
Staff Emeritus
This is not correct. Any homogeneous boundary condition, be it of Dirichlet, Neumann, or Robin type, will do. Each of the choices results in a Sturm-Liouville problem.
This would be true if the problem was not rotationally symmetric. As it stands, the rotational symmetry directly implies that only the $n = 0$ terms are of relevance.
5. May 24, 2017
### chimay
6. May 24, 2017
### Orodruin
Staff Emeritus
7. May 24, 2017
### Orodruin
Staff Emeritus
I did not check your arithmetics, but let me point out that it is significantly simpler to use $C \cosh(\lambda x/r_0) + D \sinh(\lambda x/r_0)$ for the $x$-dependent part, since the boundary condition at $x = 0$ will give you one of the constants directly. You should find that both $C$ and $D$ are decrease sufficiently fast with $m$ as long as $I_1$ and $I_2$ do (which they should if $f_1$ and $f_2$ are in the correct Hilbert space).
8. May 24, 2017
### chimay
I do not get how I can compute both constants if I cannot exploit orthogonality. Could you help me with this too?
Thank you very much.
9. May 24, 2017
### Orodruin
Staff Emeritus
Also note that it is not necessarily a problem that the coefficients increase in modulus. The norm of the basis functions $J_0(\lambda_m r/r_0)$ is not constant.
What do you mean? I just told you you can exploit orthogonality - it is a Sturm-Liouville problem!
10. May 24, 2017
### chimay
As I mentioned before, I am very weak on this topic. Is this correct?
In general
$\int_0^1 x J_0(\lambda_m x)J_0(\lambda_i x)=\frac{{J^{'}_0(\lambda_m)}J_0(\lambda_i)-J_0(\lambda_m){J^{'}_0(\lambda_i)}}{{\lambda_i}^2-{\lambda_m}^2}$
In my case, since
$\lambda \frac{J_1(\lambda)}{J_0(\lambda)}=C_r$
I can write
$J^{'}_0(\lambda_m)=-J_1(\lambda_m)=-J_0(\lambda_m) \frac{C_r}{\lambda_m}$
and
$J^{'}_0(\lambda_i)=-J_1(\lambda_i)=-J_0(\lambda_i) \frac{C_r}{\lambda_m}$
and so
$\int_0^1 x J_0(\lambda_m x)J_0(\lambda_i x)=\frac{{J^{'}_0(\lambda_m)}J_0(\lambda_i)-J_0(\lambda_m){J^{'}_0(\lambda_i)}}{{\lambda_i}^2-{\lambda_m}^2}=0$
So I can exploit orthogonality even if $\lambda$ are not zeros of $J_0$.
If this is true, my calculations are correct, and I cannot explain the strange behaviour of the constants I depicted before...
Last edited: May 24, 2017
11. May 24, 2017
### chimay
I tried to sum numerically the first 16 values of $A_i+B_i$ and I am very far from fitting the boundary condition in $x=0$ and $x=L$.
Also, $\sum_i^n B_i$ seems to increase with $n$.
12. May 24, 2017
### Orodruin
Staff Emeritus
As I said, your basis functions do not have constant norm so there is no a priori reason to expect the coefficients to decrease. Without knowing more of the functions $f_i$, I don't know what more I can say about this.
Yes, this is correct. The orthogonality follows from the $\lambda$ being such that the Robin boundary condition is satisfied.
13. May 24, 2017
### chimay
These are the remaining functions:
$f_1=\Phi_1-\Phi_0(r)$ and $f_2=\Phi_2-\Phi_0(r)$,
$\Phi_0(r)=V-k_a \log(\frac{k_b}{{(4k_b-r^2)}^2}k_c)$
where $V$ and $k_{a,b,c}$ are some constants.
Can you guess any $A_i$ and $B_i$ behaviour?
I am starting to think there is some hidden mistake in my implementation.
Thank you a lot!
14. May 24, 2017
### Orodruin
Staff Emeritus
What are the possible values of $k_b$?
15. May 24, 2017
### chimay
$k_b$ is the solution of another equation.
Do you mean numerical values ?
16. May 24, 2017
### Orodruin
Staff Emeritus
I am just trying to figure out if that function is square integrable or not. Without looking at it in detail I am not sure I can help you. A suggestion would be to try to see what happens with your expressions in some cases (different fs) where you know the solutions to check your implementation.
17. May 24, 2017
### chimay
I will take another look tomorrow in the morning. Anyway, my approach should be correct, and the equations for $A_i$ and $B_i$ in the first post should be valid...right?
18. May 28, 2017
### chimay
Hi,
I may have found one tricky point. Can someone tell me the result of $\int_0^1 x J_0(\lambda_m x)J_0(\lambda_i x)$ when $\lambda_m=\lambda_i$ and $\lambda_i$ is not a zero of the Bessel function?
Edit:
I have finally found the mistake. I used the following relation:
$\int_0^1 x J_0^2(\lambda x) dx=\frac{1}{2}J_1^2(\lambda)$
but when $\lambda$ is not a zero of $J_0$, like in my case, this holds:
$\int_0^1 x J_0^2(\lambda x) dx=\frac{1}{2} (J_0^2(\lambda)+J_1^2(\lambda))$.
By using the correct relation I find $B_i$ to descrease...
Last edited: May 28, 2017
19. Jun 5, 2017
### chimay
Hi,
since you helped me so much in my previous doubt, I would like to make another question about Bessel functions.
Assume that my boudary condition is expressed like this:
$\sum_{m=1}^{\infty} A_m J_0(\lambda_m r/r_0) + B_m Y_0(\lambda_m r/r_0)=C$
$C$ being a certain constant and $Y$ second type Bessel function.
How can I proceed to obtain a relation between $A_m$ and $B_m$ now, following a similar path as before?
Thank you
20. Jun 5, 2017
### Orodruin
Staff Emeritus
The same way as before. First you need to find the appropriate boundary conditions so that you can find what combination of Bessel functions form a complete basis on the domain of interest.
21. Jun 5, 2017 | 2,823 | 8,717 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2018-30 | latest | en | 0.743613 |
https://www.itetgroup.com/dog-zodiac-traits/ | 1,623,915,397,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487629632.54/warc/CC-MAIN-20210617072023-20210617102023-00168.warc.gz | 713,924,678 | 10,160 | # Dog Zodiac Traits
The Life Course number is the most important due to the fact that it is your primary purpose in life and the main factor you were born; dog zodiac traits the expression number is exactly how you express yourself to the globe and exactly how others see you.
The heart’s urge number makes you happy in life and what brings you joy, and the birthday celebration number is all the skills and skills and capacities that you boil down into this lifetime with to support your life’s objective.
So I’m mosting likely to show you just how to compute the Life Course number first because that is one of the most critical number in your graph, and it is your key objective in life, so the Life Path number is figured out by your day of birth.
So all we’re doing is we are simply adding all of the numbers with each other in your date of birth till we break them down and get a single digit.
So in this instance, we have December 14, 1995.
So we require to get a solitary figure for each and every area.
First, we require to obtain a solitary figure for the month.
We need to get a solitary number for the day of the month, and we need to obtain a solitary number for the birth year, and afterwards when we get a solitary number for each one of these, we can add these three together.
So December is the 12th month of the year, so we require to include the one in the two to obtain a solitary digit.
One plus two equates to 3, and for the day of the month, we require to include the one in the 4 with each other, which will certainly offer us number 5 for the day of the month.
And after that we require to add 1995 with each other.
Which will offer us a single digit for the year, so one plus nine plus 9 plus 5 amounts to twenty-four, and after that we need to include the 2 in the four together due to the fact that we have to damage it down to one digit for the year.
So two plus four amounts to six.
So as soon as we have a single number for each one of these areas, a single number for the month, a solitary number for the day, and a single digit for the year, after that we’re mosting likely to include the three of these with each other, and in this instance.
It would certainly be three plus five plus 6 equals 14, and after that we need to add the one and the four with each other due to the fact that we’re attempting to break all of this down until we get a solitary figure.
So if you include the one in the four together in 14, you get a number 5.
So in this scenario, this person’s life course number is a number five.
He or she is a 14 five, and it’s critical to keep in mind of the two last numbers that we totaled to get the Life Path number five due to the fact that these 2 last digits are important to the person’s life path number.
So, in the instance of a Life Path, the number five could be a fourteen 5 because the four equals number five.
They could likewise be a 2 and a 3 a twenty-three number 5. Still, those two numbers that you included together to obtain that Life Course number are substantial because they inform you what energies you will require to utilize during your lifetime to attain your life purpose. In this situation, dog zodiac traits, this individual has a Life Path variety of number 5.
Still, to accomplish their life course, the number of a number 5, they are going to require to use the energy of the number one and the number Four to achieve their function, so remember of those 2 last numbers that you added together to get your last Life Course number due to the fact that those are very important.
A fourteen-five is going to be really various than a twenty-three-five.
If you have any concerns concerning these two last numbers made use of to acquire your final Life Path, number comment below, and I will try to address your concerns currently.
I wished to reveal you this instance because, in some situations, we do not break down every one of the numbers to acquire a solitary digit.
So in this instance, we have December 14, 1992, and when we added every one of the numbers together in the month, the day, and the year, we wound up with a number 11 and 11 in numerology is a master number and the master numbers.
We do not add the two numbers with each other, so there are three master numbers in numerology, and the three master numbers are 11, 22, and 33.
So after you have actually included all of the numbers together in your birth date and if you end up with either an 11, a 22, or a 33, you will certainly not add these 2 figures with each other due to the fact that you have a master number Life Course.
Number and the master numbers are various from the other numbers in numerology due to the fact that they hold the dual digits’ energy, and we do not include both digits with each other in these situations.
So if you have either an 11, a 22, or 33, you will certainly not include both numbers with each other.
You will keep it as is, and you have a master number as a life course.
In this circumstance, the number that we combined to obtain the 11 were 8 and 3.
Those are considerable numbers in this situation because this master number 11 will certainly need to utilize the eight and the 3.
To obtain their number 11 life objective, so in 83, 11 will be a lot different from on 92 11 since an 83 11 will certainly have to use the power of the 8 in the 3 to acquire their life function. The 9211 will certainly Have to use the nine and the 2 factors to obtain their 11 life function.
So the birthday celebration number is probably the most obtainable number to compute in your chart due to the fact that for this number, all you need to do is include the figures together of the day you were birthed on, so this individual was born upon December 14, 1995.
So we will include the one and the 4 together due to the fact that those are the digits of the day. dog zodiac traits
This person was born, so 1 plus 4 amounts to 5.
So he or she’s birthday celebration number is a number 5.
Currently, in this example, December 11, 1995.
He or she was born upon the 11th day of the month, and 11 is a master number, so we do not include the 2 ones with each other because 11 is a master number, and there are 3 master numbers in numerology, 11, 22, and 33.
So if you were born upon the 11th of a month or the 22nd of a month, you would not include the 2 figures together because your birthday celebration number is a master number.
So your birthday number is either a master number 11 or a master number 22. The master numbers are the only numbers in numerology that we do not include the two numbers with each other to get a last number.
So you will certainly maintain those 2 digits alone, and you will certainly not include them with each other.
For the last two numbers, you have actually utilized your date of birth to compute those two numbers, but also for the expression number and the soul’s impulse number, you will make use of the complete name on your birth certificate.
You’re going to use your initial, center, and surname on your birth certificate to compute your expression and your spirit’s urge number, therefore we’re now mosting likely to use the Pythagorean number system to compute these numbers.
And it’s called the Pythagorean system because Pythagoras, a Greek mathematician, developed it. He was the mathematician that formed the Pythagorean thesis. He is the dad of numerology, and he discovered that all numbers hold power. They all have a Details resonance, and so after figuring out that all numbers hold details power, Pythagoras developed the Pythagorean number system. From that, we have contemporary numerology today.
It is a chart with all of the letters of the alphabet and all letters representing a certain number.
So essentially, all letters in the alphabet have the energy of a number.
And if you take a look at this chart, you can determine what number each letter has the power of.
So a has the energy of a primary B has the power of a number.
Two C has the power of a number 3 and so on, and more.
So, for the expression number, all you need to do is add every one of the letters in your full birth name, so the initial, center, and last name on your birth certification and minimize them down to one number. dog zodiac traits
So in this instance, we have Elvis Presley.
So what I did was I checked out the number graph, and I discovered the corresponding number per letter in Elvis’s name.
So E is a 5, l is a three V is a four.
I am a 9, and I just found all of the numbers representing all of the letters in his name; and I included every one of those numbers with each other, and the final number I got was an 81.
After that I added the eight and the one together due to the fact that we need to keep damaging these down up until we obtain a single-digit, and when I claimed the eight and the one together, I obtained a nine, so Elvis’s expression number is a 9.
Currently the one circumstance where you would certainly not remain to add these numbers together till you obtained a solitary digit would certainly be if you obtained a master number, so the three master numbers are 11, 22, and 33.
If you got a master number, you would not remain to include these numbers with each other.
You will certainly maintain them as either 11, 22, or 33.
So it’s the same scenario as it was with the life half number and the birthday celebration number.
The master numbers are special, and we do not add the 2 numbers with each other, so going on to the soles prompt number so occasionally the soles motivate number can be called the single number, and it can also be called the heart’s desire number.
These words are utilized mutually however understand whenever you see the single number or heart’s desire number that they are essentially the same thing as a soles advise number.
For this number, we will add all of the vowels in your complete birth name.
All of the vowels in your very first, center, and surname on your birth certificate, and we’re going to lower them to one number.
So we’re going to use the same Pythagorean graph that we did previously, and we’re mosting likely to seek out every one of the numbers that represent the vowels in your first, middle, and last name.
So here we have Kate Middleton.
I recently simply did a video clip on her numerology, so I figured why not utilize her today.
So her very first center and last name are Catherine, Elizabeth Middleton, so I sought out the numbers matching to just the vowels in her name.
So, as you can see, an equates to 1 B amounts to 5, I equals nine, and E amounts to 5, and afterwards I did that for all of the vowels in her complete name.
And afterwards I just included all of those numbers with each other, which provided me 60, and then 60 minimizes to number 6 due to the fact that we remember we’re simply trying to damage these numbers down until we get a single digit.
So in this circumstance, Kate’s Seoul’s impulse number is a number 6.
Now you will certainly not continue to break down the numbers if you get an 11, a 22, or a 33, so, as I said with all the other numbers if you get one of these numbers, this is a master number, and we do not include the two figures with each other so 11, 22 and 33 are master numbers.
Dog Zodiac Traits
If you obtain among those as your spirits prompt numbers, you will certainly not remain to add the numbers with each other; you will maintain them as dual figures. | 2,591 | 11,420 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2021-25 | latest | en | 0.957948 |
https://atomparticles.com/how-do-you-solve-a-particle-in-a-box-problem/ | 1,725,973,257,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651255.81/warc/CC-MAIN-20240910125411-20240910155411-00528.warc.gz | 88,647,057 | 12,337 | # How Do You Solve A Particle In A Box Problem
## How do you solve a particle in a box problem?
1. Define the Potential Energy, V.
2. Solve the Schrödinger Equation.
3. Solve for the wavefunctions.
4. Solve for the allowed energies.
## What is the particle in a 1d box problem?
The particle in a box problem is a common application of a quantum mechanical model to a simplified system consisting of a particle moving horizontally within an infinitely deep well from which it cannot escape. The solutions to the problem give possible values of E and ψ that the particle can possess.
## What is the principle of particle in a box?
The particle in a box model (also known as the infinite potential well or the infinite square well) describes a particle free to move in a small space surrounded by impenetrable barriers. The model is mainly used as a hypothetical example to illustrate the differences between classical and quantum systems.
## What is a particle in a box in quantum physics?
Because of its mathematical simplicity, the particle in a box model is used to find approximate solutions for more complex physical systems in which a particle is trapped in a narrow region of low electric potential between two high potential barriers.
## What is the formula for solving a box?
You can calculate the volume of a box by multiplying length x width x height.
## What is the formula of the particle?
Quantity Defining equation
Number of atoms N0=N+ND
Mass number A=Z+N
Radioactive decay d N d t = − λ N N = N 0 e − λ t
## What is the direction of a particle?
If I understand your question right, the direction of the particle is the same as the direction of its velocity vector, which is dr(t)dt.
## What is the Hamiltonian of a particle in 1D?
Continuous single-particle Hamiltonian in 1D is a usual sum of the kinetic energy differential operator and potential energy (multiplicative) operator.
## What is the difference between a free particle and a particle in a box?
For the free particle, the absence of confinement allowed an energy continuum. Note that, in both cases, the number of energy levels is infinite-denumerably infinite for the particle in a box, but nondenumerably infinite for the free particle. The state of lowest energy for a quantum system is termed its ground state.
## What are the 5 rules of particle theory?
• All matter is made up of tiny particles known as atoms.
• Particles of matter are constantly in motion.
• Particles of matter attract each other.
• Particles of matter have spaces between them.
• As temperature increases, particles of matter move faster.
## What is particle law?
Laws of Conservation in Particle Physics. The conservation laws of classical physics, such as the conservation of energy, linear and angular momentum and electric charge all readily hold in particle physics; that is, when particles interact with one another, energy, charge and momentum are all conserved.
## How is the particle theory?
All matter is composed of tiny indivisible particles too small to see. These particles do not share the properties of the material they make up. There is nothing in the space between the particles that make up matter. The particles which make up matter are in constant motion in all physical states.
## What is the formula for the energy level of a particle in a box?
Thus the normalized wave function of a particle in a box is. The energy of the particle in a 1-d box can be mathematically expressed as, E n = n 2 π 2 ℏ 2 2 m L 2 where n is a principal quantum number (n=1,2,3,4,……), L is the length of the box.
## How do you solve for the number of particles?
Formula: number of particles = number of moles x 6 x 10 Since 1 mole of substance contains 6 x 1023 particles, 2 moles of substance contains 2 x 6 x 1023 particles. 0.5 moles of substance contains 0.5 x 6 x 1023 particles.
## How do you find the particle in a sentence?
A particle in a sentence is a word that is added to a verb to enhance it. A particle is typically a preposition, one that adds a colloquial meaning to the verb.
## How do you solve Schrodinger equations?
(x) = Bsin n x L , n =1,2,3.. This satisfies the uncertainty principle. If the minimum energy were 0, then the momentum would be precisely 0, and then the location of the particle would be unknown – it would not be confined to the box. A particle is in a box of length L in the ground state (lowest energy state). | 994 | 4,437 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2024-38 | latest | en | 0.917529 |
http://www.edweek.org/ew/articles/1992/12/09/14phelps.h12.html | 1,490,342,999,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218187744.59/warc/CC-MAIN-20170322212947-00359-ip-10-233-31-227.ec2.internal.warc.gz | 495,007,826 | 24,721 | Published Online:
# The Case for U.S. Metric Conversion Now
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The news by now is familiar and expected. U.S. elementary and high school students perform poorly in comparison with their foreign counterparts on international tests of mathematics and science achievement. Several explanations have been posed to explain the poor performance, such as a relatively shorter school year in the United States, fewer required math and science courses in U.S. high schools, sub-populations taking the exams that differ from country to country in the proportions of students at each grade level represented, and (need I mention it?) inferior schooling.
I'd like to pose another possible explanation. The United States, almost alone in the world, does not use the metric system of measurement in its daily life. But it does teach the metric system in the schools. It teaches two systems of measurement in the schools and, the confusion from learning two systems aside, there is a cost to the time spent in teaching two systems. A full year of mathematics instruction is lost to the duplication of effort.
Mostly in the elementary grades, our schools spend a few weeks a year teaching two measurement systems when teaching just metric could be done in one-third the time. Elementary school mathematics textbooks generally give equal weight to the two systems, as do the newly completed curriculum standards of the National Council of Teachers of Mathematics. Educators feel forced to teach both systems because, even though we Americans still use the inch-pound system in our daily lives, the metric system is used in many professions (medicine, science, and engineering, for example) and now in much of industry. High school science courses now use the metric system exclusively.
The year's worth of classroom time savings would accumulate from several sources, including: elimination of the duplication of teaching two systems; the conciseness of the metric system, which requires much less time to teach and in which calculations can be performed more than twice as quickly; and a reduced amount of time devoted to teaching fractions.
Teaching the metric system in the schools would be of little help to our students, however, if the society in which they live continues to use the inch-pound system. The solution is to implement a nationwide "soft'' conversion to the metric system. This would involve dual labelling of all measured items used publicly, such as many consumer products and highways. Dual labelling would enable all adults familiar with the inch-pound system to continue using the inch-pound system at the same time students who learn the metric system could fully function in society using their metric knowledge.
It should be emphasized that, beyond obliging firms and public authorities to provide dual labelling, no one would be forced to do anything. No adult would be forced to learn the metric system. No company would be forced to size its products in metric, or retrofit equipment to metric sizes. The conversion of products and equipment to metric sizes is happening anyway, albeit very slowly, and most quickly in those firms involved in international buying and selling. And, sensible exceptions should be made even to the dual-labelling requirement--on football fields and in hardware stores, for example.
Soft conversion to the metric system might cause some inconvenience to adults as they get used to dual labelling on highway signs and in public places. And perhaps careful effort should be made to clearly distinguish the metric numbers from the inch-pound numbers, by using different colors or different scripts, for example. But the inconvenience would be slight. Think of how effortlessly we got used to the labelling on the many consumer products that are already dual-labelled.
The benefits could be enormous. Though it would cost some money to redo signs and labels nationwide, the one-time-only cost amounts to only a few hundred million dollars. By contrast, we spend over one-and-a-half billion dollars per year as a nation for a year-long mathematics course in the public schools. With another year of mathematics, our students could either learn their math a lot better, or take a second year of algebra or a year of geometry, statistics, or economics. And, they should be able to perform much better on those international math tests.
One may vaguely recall that metric conversion in the United States was already tried once, and failed. There was a plan in the late 1970's to convert, at about the same time as Canada did. The U.S. plan met some opposition, the most colorful coming from some unapologetic luddites and xenophobes. (The xenophobes were somewhat misinformed. The inch-pound system is really of British, not American, derivation, and the British themselves abandoned it in favor of metric years ago.)
The few serious studies of metric conversion at the time did not attempt to measure the benefits and costs of conversion. They merely asked firms in various industries how they felt about conversion, and the responses were more unfavorable than favorable. The intervening years have shown, however, that metric conversion in industry, when it is done as part of the normal product-replacement cycle, is far less expensive and disruptive than had been feared.
The studies did not attempt, in particular, to estimate the savings in classroom time that a metric conversion would effect, nor did they anticipate that our schools would be stuck with both systems of measurement in their curricula. Neither did the studies consider the possibility of a soft conversion plan, such as I have described.
Most everyone favors technological advance. Metric conversion offers one that is simple and familiar. Just about every other country in the world has adopted it. Yet, despite its simplicity and familiarity, the benefit of this technological advance can be very large. To fully know an inch-pound system consisting of 12 measures requires memorization of over a thousand different names and conversion ratios. A metric system of equivalent size includes only 36 names and conversion ratios, and all calculations can be done more simply in decimal notation.
Progress involves not only adopting new, more efficient technologies, but an ability to recognize when and which new technologies are superior and a willingness to let go of old, less efficient ones. And, some progress inevitably requires collective effort and government action.
Richard Phelps works as an analyst on education issues in the U.S. General Accounting Office in Washington. He formerly taught middle and high school mathematics in the West African nation Burkina Faso. The views expressed here are his own and should not be construed to be the policy or position of the U.S. General Accounting Office.
Web Only | 1,301 | 6,888 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2017-13 | longest | en | 0.961282 |
https://us.metamath.org/mpeuni/leeq2d.html | 1,719,093,504,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862420.91/warc/CC-MAIN-20240622210521-20240623000521-00225.warc.gz | 518,342,059 | 3,525 | Mathbox for Stanislas Polu < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > leeq2d Structured version Visualization version GIF version
Theorem leeq2d 40794
Description: Specialization of breq2d 5054 to reals and less than. (Contributed by Stanislas Polu, 9-Mar-2020.)
Hypotheses
Ref Expression
leeq2d.1 (𝜑𝐴𝐶)
leeq2d.2 (𝜑𝐶 = 𝐷)
leeq2d.3 (𝜑𝐴 ∈ ℝ)
leeq2d.4 (𝜑𝐶 ∈ ℝ)
Assertion
Ref Expression
leeq2d (𝜑𝐴𝐷)
Proof of Theorem leeq2d
StepHypRef Expression
1 leeq2d.1 . 2 (𝜑𝐴𝐶)
2 leeq2d.2 . 2 (𝜑𝐶 = 𝐷)
31, 2breqtrd 5068 1 (𝜑𝐴𝐷)
Colors of variables: wff setvar class Syntax hints: → wi 4 = wceq 1538 ∈ wcel 2114 class class class wbr 5042 ℝcr 10525 ≤ cle 10665 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 2116 ax-9 2124 ax-ext 2794 This theorem depends on definitions: df-bi 210 df-an 400 df-or 845 df-3an 1086 df-ex 1782 df-sb 2070 df-clab 2801 df-cleq 2815 df-clel 2894 df-v 3471 df-un 3913 df-sn 4540 df-pr 4542 df-op 4546 df-br 5043 This theorem is referenced by: imo72b2lem0 40802
Copyright terms: Public domain W3C validator | 583 | 1,197 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-26 | latest | en | 0.26232 |
https://codegolf.stackexchange.com/questions/61371/energy-levels-of-electrons | 1,585,949,917,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370518622.65/warc/CC-MAIN-20200403190006-20200403220006-00435.warc.gz | 427,495,182 | 36,393 | # Energy levels of electrons [closed]
## Energy levels of electrons
Electrons are ordered into "shells". However, sometimes electrons get excited, for example by heat, and leave their shell to go to another! These electrons will go back at some point, but they cannot go back to their original shell without emitting energy in the form of radio waves.
For example, the energy levels for hydrogen look like this:
The base energy level of an electron in hydrogen is at n = 1. If it gets 'excited', it might go to n = 4. However, on the way back, the electron does not always simply go from n = 4 to n = 1, it has different options. See the example.
Write a program or function that takes an integer and outputs the amount of options the electron has to get back to its un-excited state. For this task, you can take granted that an electron can only emit energy from its excited state. It will not get more energy once it has been excited.
This is , the answer with the least amount of bytes wins.
If you are interested in these type of questions, check out this similar question. Also check out this for more info on these matters.
## Input
The input of your function/program is a whole integer. It is greater than 0 and less than 2^31.
## Output
An integer, the amount of options the electron has to return to its original state, which is n = 1 (for hydrogen). When the input is 1, return 0, as it is already in its original energy level.
## Examples
Input: 4
Output: 4
Because the options are:
4 -> 3 -> 2 -> 1
4 -> 3 -> 1
4 -> 2 -> 1
4 -> 1
Visualised:
• When I saw "the electron does not always simply go from n = 4 to n = 1" i.imgur.com/FNxsSSC.png – DanTheMan Oct 21 '15 at 16:00
• @DanTheMan I didn't get the joke... – user41805 Oct 21 '15 at 18:46
• The simple mathematical expression for the solution makes the whole problem rather silly. Before posting a challenge, you should write code to generate test cases, and here it would have let you catch the pattern. – xnor Oct 21 '15 at 19:34
• To add to what @xnor said, it's also a good idea to post challenges to the Sandbox. That way people can offer useful feedback before the challenge goes live. – Alex A. Oct 22 '15 at 5:26
• I'm voting to close this question as off-topic because it's too narrow - there's a simple mathematical formula to solve it, no ingenuity involved. – isaacg Oct 22 '15 at 8:11
# Pyth, 5 bytes
/^2Q4
I integer-divide by 4 instead of subtracting two from the power and flooring.
# CJam, 68 7 bytes
1li2-m<
Try it online
Unless I completely misunderstood the question, the result is simply 2^(n-2). For each level between n and 1, you can either skip or not skip it on the way back.
Using the left shift operator instead of the power operator is one byte more, but gives the specified result 0 for n=1, while the power operator gives 0.5 for that case.
Explanation:
1 Push 1 for later shift operator.
li Get input and convert to int.
2- Subtract 2.
m< Left-shift 1 by (n-2).
If I shamelessly stole Thomas Kwa's approach, this would be 6 bytes in CJam:
2li#4/
• Ouch. Did not realize the answer was that simple. Nice work though. – Thomas W Oct 21 '15 at 15:47
# Dyalog APL, 6 bytes
⌊2*-∘2
This is a monadic function train. It is equivalent to the following, trainless function.
{⌊2*⍵-2}
Try it online on TryAPL.
### How it works
-∘2 Take the right argument and subtract 2.
2* Elevate 2 to the difference.
⌊ Round down to the nearest integer.
• Note to self: If you beat Dennis to a CJam answer, he's just going to find something else that's shorter. ;) – Reto Koradi Oct 21 '15 at 16:02
• 6 bytes? What character encoding are you using? – ev3commander Oct 21 '15 at 19:10
• @ev3commander There are legacy APL code pages (that predate Unicode by a few decades) where one character is one byte. – Dennis Oct 21 '15 at 19:29
# JavaScript ES7, 12 11 bytes
i=>2**(i-2)
Defines an anonymous function.
• Anonymous functions are fine. I'm pretty sure that ^ is bitwise XOR though, not power. – Dennis Oct 21 '15 at 17:07
• @Dennis Frick, you're right! Gotta add another byte :( – DanTheMan Oct 21 '15 at 17:14
• Gives the wrong result for i=1 – Peter Taylor Oct 21 '15 at 17:30
• the ** Opererator is ES7 and presently supported in almost nothing – Shaun H Oct 21 '15 at 17:41
• i=>(1<<i)>>2 1 char more, works everywhere, correct for i=1 – edc65 Oct 22 '15 at 0:14
## Python, 2922 16 bytes
lambda n:2**n//4
Technique shamelessly stolen from Thomas's answer. Try it online
# Stuck, 6 bytes
2i2-^)
Hey look I almost outgolfed Pyth outgolfed CJam
• Does Stuck implicitly floor? Interesting. – lirtosiast Oct 21 '15 at 16:44
• I didn't execute this, but from the documentation, 2^ seems to square instead of calculating a power of 2 – Dennis Oct 21 '15 at 16:46
• ^ pops off two values x and y and calculates x^y. So this does do power of 2 not square. – a spaghetto Oct 21 '15 at 16:48
• I wish Stuck had an online interpreter... – a spaghetto Oct 21 '15 at 16:48
• @ThomasKwa Actually it appears to implicitly ceiling since an input of 1 outputs 1. I'll have to fix that... – a spaghetto Oct 21 '15 at 16:49
# dc, 6
2?^4/p
Algorithm expertly lifted from other correct answers.
### Testcases:
$for t in 1 4; do dc -e'2?^4/p' <<<$t; done
0
4
$ # Julia, 9 bytes n->2^n÷4 Like several other answers, this uses Thomas's approach. ÷ performs integer division in Julia (but unfortunately is 2 bytes in UTF-8). • Does Julia support the ISO 8859-1 encoding? – Dennis Oct 21 '15 at 18:12 • @Dennis I have no idea. – Alex A. Oct 21 '15 at 18:24 # PHP, 28 bytes I hope I got it right. floor handles the case for 1 where it should output 0 <?=floor(pow(2,$argv[1]-2));
Works from command line like:
electrons.php 4
+1 and thanks to Reto Koradi for providing the formula. ;)
• Note that you can left-shift 1 to simulate a power of 2 (as is the nature of binary). The result is necessarily an integer. For the n=1 case, this would imply a negative number - so you can simply compare with zero. Suggestion: <?=max(0,1<<\$argv[1]-2); (24 chars) – Jake Oct 21 '15 at 17:08
• Never mind. This doesn't work with the full range defined by the question. Neither does pow, as it assumes large numbers are infinity, but it works for a larger scope than the left shift. If PHP allowed one to specify a var as a long, that would be more useful. – Jake Oct 21 '15 at 17:15
• @JArkinstall Yeah, even with 64 bit int it will return INF. It's not made for this domain. – insertusernamehere Oct 21 '15 at 19:58
## Minkolang 0.9, 7 bytes
Copying Thomas like everyone else...
2n;4:N.
This is a full program, and the first five characters are equivalent to 2**n//4 in Python 3. Try it online. | 1,960 | 6,736 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2020-16 | latest | en | 0.923498 |
https://it.scribd.com/document/263619781/03-Main-and-Tail-Rotor-Theory | 1,603,909,785,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107900200.97/warc/CC-MAIN-20201028162226-20201028192226-00259.warc.gz | 379,995,895 | 106,517 | Sei sulla pagina 1di 46
# He Wharekura-tini
Kaihautu 0 Aotearoa
THE OPE N
P0l.YTE(HN|(
OF NEW ZEALAND
\\
R
\$_\\
~<
II
55533
CONTENTS
Basic Rotors
## Forces of the Rotor Head
11
Dissymmetry of Lift
13
Gyroscopic Effect
21
Coriolis Effect
23
25
Drag Effect
27
## The Tail Rotor
30
Principle of Operation
31
## Forces of the Tail Rotor
32
Dissymmetry of Lift
33
Drift
36
10/91
This rn'a'Yerial is for the sole use of enrolled students and may not be
reproduced without the written authority of the Principal, TOPNZ.
555/3/3
AIRCRAFT Er IHEERING
\$-
CI)
HELICOPTERS
ASSIGNMENT 3
s
BASIC ROTORS
## The main rotor of a helicopter converts the power supplied
to it from the engine into a lifting force. When tilted forwards,
backwards, or to either side, the lifting force propels the
helicopter. The mechanism within the rotor head that tilts the
lifting force is controlled by the pilot through the collective
pitch lever and the cyclic pitch~control column.
By tilting
the lifting force, the pilot controls the helicopter about the
lateral and longitudinal axes.
The tail rotor is the helicopter's rudder.
It consists of
a rotor mounted vertically and at 90 to the centre line of the
helicopter.
## train used to drive the main rotor and is arranged to turn
whenever the main rotor turns. The tail rotor provides an opposing
force to the torque reaction of the main rotor and controls the
helicopter about the vertical axis, especially when hovering.
is
It
## In this assignment, we will show you how the lifting force of
a main rotor is tilted and resolved into lift and thrust. We will
consider the tail rotor and discuss the aerodynamic and mechanical
forces acting on the helicopter.
THE MAIN ROTOR
WW
## As with an aircraft propeller, the thrust generated by a
helicopter rotor acts at right angles to the tip-path plane.
Opposing this force and exactly equalling it, when hovering, is
the weight of the helicopter. In this condition, iii; is equal
555/3/8
## to weight and, because the
helicopter is not moving, 532
Lift
TOTAL
Tta|
REACTION
To obtain
## helicopter, the tip-path plane
is tilted and the total reaction
resolves into lift and thrust,
with the lift supporting the
weight and the thrust being
equal to the drag for straight
## (El) Hovering Qrillair); Tip path Plane horizontal.
rout
'i
as/\c'r |ou
E vvmic-=1
>
:( lift J
Ifnrr|1rIMI'\f
_ _
'4Ferward (thrill!)
wmpumni
Div-cc! ion of
mot] In
## shows these forces acting on a
helicopter in hover and in
forward flight.
wucur
## unbalanced, the helicopter will
climb or descend. When thrust
and drag are unbalanced, the
LIFT
-rnausr
Figure l
e-
X
_ ;
i
slow down.
DRAG
Dlndinn J
Ivnhnn
For example, in
## the thrust exceeds the drag and
the lift exceeds the weight.
(c) Total reaction resolved into lift and thrust for
for-ward flight.
obtain thrust
l.
engine assembly;
2.
## By changing the centre of gravity of the helicopter
by, for example, moving the cabin assembly;
555/3/3
_ 3 _
3.
4.
## By using aerodynamic forces to lift and depress
rigidly mounted on its drive shaft; or
5.
## By using aerodynamic forces to lift and depress
rotor blades rigidly fixed to the rigidly mounted
root ends.
In practice, the last three methods are used, and the rotors
that use these methods are
1.
2.
3.
## Thehingeless and articulated rotor heads tilt the tip~path plane
by simply increasing the angle of attack of the retreating blade and
As a
flaps down, as shown in Fig. 2 (a) and(b).
555/3/3
_ u _
## The tilting is controlled
by the cyclic-pitch control column
usually called the cyclic, and
can be effected in any direction.
The cyclic column is moved
forward to tilt the tip-path
plane forward, which moves the
helicopter forward.
It is
moved to the right to tilt the
(a)
right.
___ ~
F;
-Et?fI:D
## of the cyclic column and, of
course, for any intermediate
position.
The tip~path plane of the
semi-rigid rotor is tilted in
the same way, with the rising
(b)
## equalled by the dropping of
both blades are mounted on a
___===r-
## rigid, centrally pivoted yoke,
or seesaws in the direction
chosen.
Figure 2 (c) shows
a tilted tip-path plane for
a semirigid rotor.
(c)
FIG. 2
Types of rotor
## For vertical flight, the
angle of attack of all the blades
is increased or decreased
simultaneously. This is
controlled by the collectivepitch control column, usually called the collective. The
pilot raises the collective for the helicopter to go up and
lowers it to go down.
555/3/3
_ 5 _
The control inputs to the rotor head from the cyclic and the
collective pitch controls are superimposed upon each other
mechanically so that, for example, a climbing, turning, forward
flight path is possible.
\$9
I7 7
7 7
' ""7"
77
SUMMARY
The main rotor converts engine power into a force that
## can both lift and propel the helicopter.
The total reaction of a rotor is at right angles to
the tip-path plane.
tip-path plane.
lift and thrust.
I
l.
2.
3.
## The tail rotor counteracts the torque of the main rotor
and gives directional control during hovering.
## Coning and Flapping
when the helicopter is hovering in still air, the airflow
comes from directly above and goes straight down through the rotor
The lift force generated by the turning rotor acts vertically
upward and is equal to the weight (mass) of the helicopter acting
vertically downward.
1.
2.
## The centrifugal force of each blade.
555/3/3
_ 5 _
The lift force acts to lift each blade tip upwards, and the
centrifugal force acts to keep each blade horizontal. The two
forces resolve into a single force, that results in a small upward
The angle formed between the blade
and a plane at right angles to the rotor shaft is called the
coning angle.
See Fig.
3.
## The articulated rotor has each blade mounted on a horizontal
or flapping hinge, which permits its blade to freely move up and
down or flap.
## The rigid rotor permits the blade tip to move up andcbwn by
the bending of the blade and the bending of the rotor head just
Z/B
kg
A7
T
l?
1%
%Z
/?\\
Coning angle
=>- w:>
## Hovering in sffll air
Comng angk dbphced.
FIG. 3
EIIJ
Coning angle
Neither the articulated nor the rigid rotor system can give
flight other than hovering unless a coning angle is generated.
The coning angle must be displaced about the main rotor shaft
by the flapping of the blades to give the tilted tip-path plane
necessary for horizontal movement. See Fig. 2 (a) and (b).
Because the semi-rigid rotor has its tip-path plane tilted by
point, both blades flap together but in opposite directions.
by an equal amount.
See Fig. 2 (c). Thus, the semi-rigid rotor
does not need to make a coning angle, and the lift force bends
the stiff, heavily built blades evenly but slightly along their
span.
555/3/3
_ 7 _
Coriolis Effect.
## In fact, the coning angle hardly varies during all stages of
flight because the rotor and engine rev/min are held in a narrow
operating range.
## winched aboard or released from a cargo hook.
Any increases/decreases
in load make only a small difference to the coning angle and soonly small changes in bladepitch angle are made to correct for
them.
The following simplified example shows the change in coning
angle caused by an increase in the weight of a helicopter with a
exact.
Helicopter AUW
8000 lbf
llO lbf
C of G location
13 ft
Rotor rev/min
210
gggg
2000 lbf
## alloz ><<1TXn2n2n0*x 210)
21 477 lbf
2
Centrifugal force
555/3/3
_ 8 _
**
4*
FIG. 4
.7]
.Wt
m+77mf
fjzmmur
## Generating a coning angle
2000
tans = are
6
tan 0.093l2
Coning angle or 9
5 19
## The helicopter is now loaded to ll 000 lb max. AUW.
Lift generated
Centrifugal force
iiggg
2750 lb
21477 lb
2750
## tan " nan"
9
tan 0.1280
Coning angle or G
7 18
## During flight, the rotor and engine rev/min can be considered
as constant, with the power being changed by alteration to the
engine induction manifold pressure or fuel flow and, at the
same time, collectively altering the main rotor blade angles.
555/3/3
i Q _
The power is changed because the inertia of the rotor head and
blades assembly makes an increase or decrease in rotor rev/min
impossible to get in a reasonable time, but a quick response
to a change in power may be obtained by changing the blade pitchangles with a change in engine fuel flow.
We cannot overemphasise the importance of keeping the rev/min
in the correct range.
If the recommended rev/min are exceeded,
result, the severity of the damage depending upon the amount and
duration of the overspeed.
If the rev/min fall below those recommended, the collective
pitch must be decreased or the engine power increased and the
rev/min allowed to increase to their normal value.
However,
## if no more engine power is available, the helicopter will have to
descend with the reduced collective pitch.
If the collective
pitch is increased to maintain altitude, the increase in blade
angle above normal will produce more drag for the lift generated,
the rev/min will decay (slow down) further, and the helicopter will
descend rapidly.
## the coning angle increases due to the reduced centrifugal force
until a position is reached where it takes a long time to bring
the blades down again by reducing collective pitch.
A very
## heavylandingp is then unavoidable.
Early helicopters with articulated rotors could get their
## This condition, known
as candling,
Modern articulated
## resulted in a crash landing.
prevent excessive coning of the blades.
## an angle well outside the normal coning angle of the blades
but small enough to allow a reasonably rapid increase in rev/min
when collective pitch is reduced.
555/3/3
_ 19 _
SUMMARY
In flight, the engine and rotor rev/min are kept in a
## narrow operating range .
The coning angle is caused by the balance of lift and
centrifugal forces.
The tip-path plane is tilted to give forward flight
by flapping the blades up at the rear and down at the
front of the helicopter.
3
## In a semi-rigid rotor, as one blade flaps up, the
other blade flaps down by an equal amount.
A very low rotor rev/min will produce a large coning
## angle, which could endanger flight.
PRACTICE EXERCISE A
or false:
l.
## The main rotor is controlled by the pilot through
the collective and cyclic pitch control columns.
2.
## The main purpose of the tail rotor is to control
the helicopter about the vertical axis when in
level flight.
3.
## The total reaction of the main rotor is resolved
into lift and drag when the tip-path plane is
tilted.
4.
## When the tip-path plane is tilted to the right,
the helicopter moves to the right.
5.
## During hovering, the lift of the main rotor must
slightly exceed the weight of the helicopter.
6.
## In tilting the tip~path plane of a semi-rigid
rotor, the complete rotor head assembly is tilted.
7.
8.
555/3/3
-11..
9.
10.
## The horizontal or flapping hinges of an articulated
rotor permit the blades to freely flap up and down.
A rigid rotor has its blades rigidly attached to
the rotor head, which is itself rigidly attached
to the rotor drive shaft.
## Forces of the Rotor Head
To achieve flight, the tip-path plane is tilted in the desired
assembly becomes resolved into lift to support the weight and
thrust to propel the helicopter. As the helicopter moves, the
airflow direction into the rotor head changes from directly
## rotor changes from straight down, forming a ground cushion, to aft
and down.
These changes in the airflow through the rotor disc, and the
fact that the tip-path plane has been tilted, create extra forces
and effects above those experienced when hovering.
555/3/3
_12_
7
1
If
/"" -\\
*_ i4._ _
is
r\
5.
lire-2
__&
.._/
(a)
7%%
When hovering
4*
\l_l
\l__h
'
///
so
if
_r
;_
it ,_%;; w |~
n
1.
Z--\\
.
\
Q;
_\
\\
i\
(la)
In eve| flight
(b)
In level flight
FIG. 5
## Airflow through the rotor
555/3/3
_ 13 _
Dissymmetry of Lift
The area within the tip~path plane of the main rotor is called
the disc area or, more generally, the rotor disc. when the
Q
## helicopter is hovering in still air, lift is generated by the
rotor blades equally at all positions around the rotor disc.
## As the helicopter moves, or as a wind is felt, the velocity of the
airflow over the rotor blades changes, with a higher velocity
in one half of the rotor disc and a lower velocity in the other
half.
## As a result of the different air velocities, different
lift forces will be generated from one side of the rotor disc to
the other.
## Unless this unequal distribution of lift is counteracted,
the helicopter would roll over in the direction of the side with
the least lift.
## dissgngetrg of 1ift,was a considerable problem to the designers
of early helicopters and autogiros.
Figure 6 (a) shows typical velocities at different positions
on a rotor blade when the helicopter is hovering in still air.
Because the air is still and the helicopter is hovering, the
rotor blade velocity is also the velocity of the air over the
The blade will thus experience the same air velocities
at all positions in the rotor disc.
Figure 6 (b) shows the same helicopter in forward flight
with an IAS of 100 kt.
## by the tip of the blade when it is advancing and at 90 to the
line of flight is the tip velocity pigs the 100 kt TAS, giving
a total of 500 kt. This increase in air velocity of 100 kt is
retreating and at 90 to the line of flight, the air velocity
felt at the tip is the tip velocity minus the lU0kt IAS, giving
2
## a total of 300 kt.
This decrease in air velocity of 100 kt is
felt along the span of the retreating blade.
For any given angle
of attack, the lift generated increases as the velocity of the
## airflow over the airfoil increases.
In fact, the lift increases
as the square of the air velocity. That is
If an air velocity of A m/s gives l unit of lift, then an air
velocity of ZXA m/s gives 4 units of lift, and an air velocity
555/3/3
.. 11; _
FORWARD
{
so
'b
\-
Iv
'5
8*
Drrecon
RETREATING
|4gq
\\
I lgw
/////~_\\\\
I00
i H
_0
\lO0
\3QD
HALF
HALF
/_
400
\
AFT
(a)
Hovering
A FORWARD
<\
-;~
63
'7
ho
r'|-EC
_
Q
RIETREAHNG
zoo
Y
1-=00
0
V V*T1gp_
zoo
Loo
W 500
F
HALF
HALF
AFT
(b)
FIG. 6
## Forward flight of 100 kt
Rotorblade velocities in hovr and in forward flight
555/3/3
_ 15 _
## Bearing this in mind, a study of Fig. 6 (b) will show that
more lift will be generated in the advancing half than in the
retreating half of the rotor disc unless some correction is used
an
We have seen that the articulated rotor and the rigid rotor
systems have blades that either flap or bend in the vertical
4%
plane.
## blade due to forward motion of the helicopter will cause the
blade to flap or bend up.
This upward movement will decrease
the angle of attack because the relative wind will change from
a horizontal direction to more of a downward direction while the
See Fig.
7.
AIRFLOW
FROM ABOVE
its ,d
Space
R55!-1:.rAN r
-_L>_AlRFLOw
-__ __ _ __ __
6:
FIG. 7
555/3/ 3
_ 15 _
## The decreased lift on the retreating blade will cause the
blade to flap or bend downward. This downward movement will
increase the angle of attack because the relative wind will
change from a horizontal direction to an upward direction while the
as
shown in Fig.
7.
## The combination of decreased angle of attack on the advancing
through blade flapping tends to equalise the lift over the two
halves of the rotor disc.
The position of the cyclic pitchcontrol
column in forward flight also causes a decrease in the angle
of attack on the advancing blade and an increase in the angle of
## column gives the major correction for dissymmetry of lift, with
the correction for blade flap being a minor but necessary contribution
The semi-rigid rotor behaves as a seesaw. As one blade flaps
change in angle of attack of each blade tends to equalise the
lift over the rotor disc.
## for dissymetry of lift is supplied by the forward movement of the
control column.
Another method that can be used on the articulated and rigid
rotor systems to decrease the angle of attack and the consequent
lift of an advancing blade flapping up and to increase the angle
of attack and the lift of a retreating blade flapping down is
to slightly offset the pitchchange horn on the blade in relation
to the flapping hinge.
exaggerated form.
## the input from the pilot to the rotor blade is attached to
face A on the control horn and the blade can rotate on the blade
spindle.
555/3/3
_ 17 l
\.
. _r.\" { ''a'5>n\_ \
/ -_
r"
i.
Q,
/.-X \
K.
_ .
*3
"
1'
Ftapping hinge
I
1
____|
I
.
91
_
Piich corliroi
horn
EII
gl
'
of-f5et
"
I Pikfn control
'
Angle of attack
mcP.ET.s_
d
J
A .
.-I
.-a:---
D5-"M55
(a) Normai
FIG. 8
horn
(5) Offs
Offset pitch~change horn
## In Fig. 8 (a), when the advancing blade flaps up because of
the increased airflow, there will be no mechanical change in the
angle of attack of the blade because the centre lines of the
flapping hinge and the contro1rod attachment to the control horn
coincide.
## up, the angle of attack of the blade is mechanically decreased
because the centre line of the control rod attachment to the
control horn is outboard of the centre line of the flapping hinge
As the blade flaps up, it also rotates on its blade spindle, with
## retreats and starts to flap down.
In flight, the blades are allowed to flap as they wish.
No damping devices or mechanicalrestraintsare used to inhibit or
prevent flapping other than the limits of movement imposed by the
design of the rotor head and,in some helicopters, an upper coning
stop.
For practical design and construction reasons, the flapping
hinges are offset.
## has a useful dynamic effect in the control of the helicopter.
555/3/3
_ 13 _
In Fig. 9, the strings represent the rotor blades, the arrows
show the centrifugal force, and the weight represents the fuselage
of the helicopter.
In Fig. 9 (a), the tip-path plane is tilted,
but because the blades are hinged in the centre of the rotor, the
fuselage hangs straight down and will be slow to adapt its
attitude to the tilt of the rotor.
## If its centre of gravity was
anywhere but in the same lateral plane as the lift vector of the
rotor, the helicopter would be unmanageable.
In Fig. 9 Cb), the
fuselage quickly follows the tilt of the tip-path plane, and
the position of the centre of gravity is now not so critical.
The
result is a helicopter that is sensitive to the control of the
pilot and has a useful working range of permissible centreofgravity movement.
Offset of hi--51.;
if}
,_ _.v__>__\3_H
__
ti
.___
-1
FIG. 9
D-
\\V
(b) ovrszr
## The offset distance of the flapping hinges determines the
size of this dynamic effect.
The blades in a rigid rotor are,
in effect, stiff continuous flapping hinges, and the overall
effect is similar to widely offset flapping hinges.
The semi-rigid rotor also uses a dynamic effect to give a
manageable and sensitive response to the pilots controls.
This
effect is obtained by having the rotor assembly underslung on
555/3/3
.-...]_Q..
its pivot.
## lies below its central pivot axis.
Figure l0 Ca) shows a semi-rigid rotor helicopter hovering,
with the lift vector acting vertically upward and the weight
vector acting vertically downward and in the same plane.
Figure l0
Cb) shows the tip-path plane tilted for forward flight, with the
lift vector moved aft because of the tilt of the assembly. As a
result of this movement, a couple is formed by the lift and weight
vectors, which lowers the nose of the helicopter.
The underslung
## mounting of the semi-rigid rotor assembly has another important
service to perform, which we shall discuss later on in this
assignment under Coriolis Effect.
LIFT
/'7
.. 5..
WEi6H'|'
(a)
Hovering
Tofal
reaciion
Rotor pivu+
L";-r
pn'|v1~|THRUST
Q--i
c J
tcf;Z'
WEIGHT
(b)
FIG. 10
-
## Semi-rigid rotor helicopter
555/3/3
/,.\
in
5///\}\\
_ 29 _
SUMMARY
Dissymmetry of lift is caused by horizontal flight or
by wind during hover.
Y
\
## Dissymmetry of lift is the difference in lift that occurs
half of the rotor disc area.
## Unless corrected, dissymetry of lift will roll the
Dissymmetry of lift is corrected by
l.
T
## An aerodynamic reduction in the angle of attack
as the retreating blade flaps down;
l
2.
## The blade's angle of attack being reduced as it
advances and increased as it retreats by the
position of the cyclic control column; and sometimes by
3.
## Mechanically reducing the angle of attack of the
respect to the flapping hinge.
PRACTICE EXERCISE B
l.
## Show, with the aid of a sketch, why the adyancing
blade of a helicopter in horizontal flight tends
to develop more lift than the retreating blade
unless corrected.
2.
## an airflw, and show the chord line and angle of
attack of the airfoil.
3.
## With the aid of a sketch, show that, when a rotor
blade flaps up, its angle of attack is aerodynamically reduced.
555/3/3
_ 21 _
Gyroscopic Effect
The turning main rotor assembly behaves as a large gyroscope
in that it tilts at right angles to the direction of a push that
## Figure ll shows how a gyroscope tilts or precesses in a
reaction to an applied force or push.
t
s
(Q) 11,, grm,c,,p_
## (5) The rg':eJegr?;;:>l;!;::I:Pi41Y~F"
13!!
(C) 5ubpose"t|;::'Inr;-is split into
Ii; ~23
::;:see%m&n3,ca;ifgidIy
11
## fhei, 5|-mp: 5, not man"
-;~t>
(gen tgwtlrrxovgggeztgd tltggnulilguoaie
2 /ct_j):{,
(i) Now suppose we apply a torque
to the axle an the horizontal plane,
lg
"lhis imparts a mation in the horizontal (lg) 7,4,", the ngmenls mm M" both
dlrecnnn to the segments, one to the right _ 0 ;,,,|z,,nm/ and G "mm, momm
and the other to the left.
L\
t -_.
._\_-.L2.|
## 1-My me";-are mu" dwgcnay
_.__t
eff!
..
<;
\ I
%%%g;%%%:%?%?%E
(M) This is the key diagram. Study it carefH"Y- The axle is rigidly connected In the
segments and must therefore tilt when the
segments move diagonally.
gggg
(H) All the other segment: must
n in we mm, wan
FIG. ll
C::(@;%:::D EiEi!!;g%iiifir
(0) T-h='=l're the whole wheel trlu.
555/3/3
## (P) Thu: when a gyroscope is given a push
il "'15 1" Pith! angle: to the directwn uf
the push.
._ 22 _
## Figure 12 shows the gyroscope
rotor without its gimbals, and
f~\
i?T
r/,/IF
\\\\'i._M/
A'/
Ll
## lying in a horizontal plane such
as a main rotor assembly.
If
we apply a push to the rotor
'
at A,it will move in the direction
marked tilt at B. Compare this
movement with that shown in
PUSH
## Fig. ll and you will find that
FIG. 12
it is the same.
Tilting of a rotor
## see that the tilting of the
o
rotor occurs 90 later in the direction of rotation from, and in
the same sense as, the applied force.
We have seen earlier that, for horizontal flight, the tip~
to flap up and the advancing blade being allowed to flap down, and
that a blade is made to flap up or down by its angle of attack,
and thus the lift-force generated being increased or decreased.
Because of the gyroscopic effect of the turning rotor, the change
in angle of attack must be made 90 before it is to take effect.
Thus, the desired change for forward flight is made at 90 to the
centre line on the lefthand side of the helicopter see Fig. 13.
LOW NTCH APPLIED
HIGH FLAP RESULT
W-I
F ""_\\,-__
\%f"ilktow rm aesutt
- FIG. 13
id
\|
"
## This applies to a rotor which is turning in the conventional
direction, which is counterclockwise when viewed from above.
555/3/3
_ 23 _
Bell helicopter if
direction and then
forwards, watching
## in the control can be seen very easily on a
you position the blades in a fore~andaft
move the cyclic control column backwar d s and
the blades at the same time.
No movement
takes place at all when this is done, but when the control column
is moved laterally, the blades increase and decrease their pitch
as the control is moved.
## The same effect can be seen, less easily,
in a fully articulated rotor if you position the control horn pushpull rod of one blade directly over a control rod into the swashp lat e and then move the cyclic control column.
Coriolis Effect
As a blade of a fully articulated or rigid rotor flaps up, the
centre of mass of the blade moves in towards the centre of the
rotor disc, and as the blade flaps down, it moves outward.
See Fig. lu.
Axis
of rotacn
UP
X3
X2
\\|
\
\
X:
DOWN
_
'
Xwn
\KC8ntre of mass
FIG. 14
## Remember that, because of coning, a blade of these two types
.
0 rotor will not flap down below a plane passing through the
ro t or hub and perpendicular to the axis of rotation.
f
## The product of mass and velocity yields momentum.
Mass X Velocity
Momentum
555/3/3
Thatis
_ gu _
Thus, when a rotor is turning, each blade has a certain
amount of momentum.
The law of Conservation of Momentum states that
"the momentum
acts on it".
## As a blade flaps up, its centre of mass moves towards
the axis of rotation, and so the length of its path around the
axis becomes shorter.
## which it must, its angular velocity must therefore increase.
reverse holds true as a blade flaps down.
The
## This law is well demonstrated by the exhibition ice skater.
When the skater pirouettes with her arms outstretched, her rate
of spin is not very great, but as she lowers her arms, the rate
increases markedly.
When the blade flaps up and increases its angular velocity,
it is said to 551, and as it flaps down and decreases its angular
velocity, it is said to 52.
--Dirch.~
.... --
LE
W
-:4.
ca
___.--
___.
I
;
'.
_
1_
.7.
,.
.
Vcrhcal hmge
_________ __
, '=,
/__
::'
--- 'I 1.
.
___
=I
-A . __ _ __
## ' ' '' '
I:-5
. ~' "
\.- _
.__
If
"--\ ":
3
:'
.'
"-~~~.':.AGGVG
FIG. l5
555/3/3
: I
'
-25..
The blade is not free to lead and lag without restraint, which
would cause mechanical damage to the rotor head and would also
create an unbalanced rotor. It is attached to a damping device,
which is adjusted to give a specified rate of movement under a
The semi-rigid rotor is underslung and has a small preset
coning angle.
effect.
## tend to increase their velocity to maintain their momentum but,
because the C of G of the head has moved away from the axis of
rotation, it will generate some added momentum of its own to the
system.
## long as the total momentum of the system stays the same.
Thus,
the blade flapping up will have little tendency to increase its
velocity to conserve momentum, and the blade flapping down will
have little tendency to decrease its velocity.
The small
## lead-lag forces that are generated by the remaining coriolis
braces that locate the blades in their grips.
## Ho0ke's Joint Effect
Horizontal flight is obtained by tilting the tip~path plane.
When the tip-path plane is tilted, its plane of rotation differs
from that of the rotor drive shaft.
to Hooke's
driving member.
## Figure l6 shows the effect on a fourbladed
articulated rotor.
## to the rotor drive-shaft
at 90 to each other.
## is no coriolis effect, and so the blades will not move about
their vertical hinges.
In horizontal flight with the tip~path
plane tilted(for the rotor shaft plane of rotation to maintain
a constant velocity),the two athwartships blades must move on
555/3/3
_ 25 _
## accelerate and decelerate twice in each revolution of the rotor
shaft.
__|_./L ..ml-n
,........-dd;-~
wat
.~<i"at
<* rgtajg
.,_
if
Qt:
_ /
-a
Til
,_
-_._~""
---...-x._.._....----
.-
"t"
/-'
(a) Hovering
gb) Horizontal
FIG. 16
ight
## The articulated rotor caters for this effect by allowing the
blades to move on their vertical hinges, the rigid rotor by
bending the blades at or near their attachments to the rotor head,
and the semi-rigid rotor by bending its stiff and heavy blades.
555/3/3
_ 27 _
Wmwwi
1HLJ%a
_ _
v
3.
1-
## velocity universal joint where
torque is transferred at constant
'_:>.:, .;;|
\
4
1|.-4mll-=w|-I-||f"" "'w'
'.g.-.-.fi
a/..q_ :3
'1
nfrm
I
l 1
"=1|li\'1
2-.:\\1_i,1
HE
n'
wmi
1.
g '55
## speed by the use of free-moving
.
.
steel balls between the driving
- -i;+%
Q
iv
\
/A J./'
in "K -_ agave"
1
vawng
,5
Each member
*hjf;&:,
fY /;
pix,/
.
~;
?QE
Z,_
in
## the sides of which are specially
shaped grooves. A steel ball forms
the driving connection in the
FIG. l7
Cqnstantvelocity universal
## two curved channels formed when
joint
the joint is assembled.
The
shape of these channels is such that, irrespective of the angle at
which the joint operates, the balls always lie in a plane that makes
equal angles with both driving and driven members of the joint.
This feature is common to all makes of constant~velocity universal
joints.
Rotor heads have been designed and built to behave as a
constant-velocity joint but, so far, this type of rotor head has
not been used in production-run helicopters.
Drag Effect
of the airflow over the blade varies as does the drag generated.
This changing value in drag causes the blade to move about its
vertical hinge or, in the case of the rigid and semirigid rotors,
for the blade itself to bend.
The five main effects discussed, that is
l.
2.
Dissymmetry of lift,
j
Gyroscopic effect,
3.
Coriolis effect,
4.
5.
Drag effect,
555/3/3
_28..
## all take place together when the helicopter is in horizontal
flight. However, for hovering in still air, only gyroscopic
and drag effects occur.
SUMMARY
In an articulated rotor
l.
## Dissymmetry of lift is corrected by the blade
flapping up and down about the horizontal (flapping)
hinge;
2.
## Gyroscopic effect is allowed for by the angle of
attack of the blade being changed approximately
90 before the result of the pitch change is
desired; and
3.
## Coriolis, Hookes joint, and drag effects are
absorbed by each blade being mounted on a vertical
(drag) hinge, with its variations in velocity
with respect to the rotor head being controlled
1.
2.
## Dissymmetry of lift is corrected by the blades
point;
Gyroscopic effect is allowed for as in the
articulated rotor;
3.
## Coriolis effect is absorbed by the rotor assembly
being underslung on the rotor drive shaft;
4.
and
In a rigid rotor
l.
## Gyroscopic effect is allowed for as in the
articulated rotor, and
2.
## All other effects are absorbed by the blades
bending at or very near their attachment to the
555/3/3
_ 29 _
PRACTICE EXERCISE C
State whether each of the following statements is true
or false:
l.
in still air.
2.
3.
4.
5.
6.
7.
## No allowance for the gyroscopic effect of the
rotating rotor is needed in its control.
8.
9.
## Because of gyroscopic effect, the angle of attack
rotation before the desired effect of the change
is to take place.
10.
in still air.
ll.
## An underslung semi-rigid rotor with between.2 to 6 of
preconing will experience little coriolis effect.
l2.
## The movement of rotor blades in the vertical plane
is called flagging.
13.
called flapping.
14.
15.
hinge.
16.
l7.
## A cyclic pitch change alters the pitch angles
of all the blades by the same amount at the
same time.
18.
vertical and horizontal hinges.
555/3/3
_ 30 _
19.
## A semi-rigid rotor seesaws spanwise about a central
point.
20.
a horizontal hinge but is damped in its dragging
on page
41)
## The Tail Rotor
A tail rotor is used to counteract the torque PacTiOn frcm
a single main rotor and, to a much lesser extent, to provide
directional control in flight.
## counteract their torque reaction by counterrotating the rotors.
Thus, they obtain directional control by mixing the cyclic
inputs to each rotor head and so don't need a tail rotor.
The tail rotor is mounted vertically, or nearly so, on one side
of the fuselage, with its centre line at right angles to the
direction of normal forward flight.
It is driven
## through shafting and gearboxes from the main rotor and is
connected mechanically with the main rotor so that, when the main
rotor turns, so must the tail rotor.
This mechanical connection
between the two rotors means that, in autorotation, the pilot
has normal behaviour from the tail rotor.
Ideally, in level
flight, the tail rotor uses little or no power, nearly all power
being available at the main rotor for lifting and propelling the
helicopter.
This is
## a climbing, left-hand turn if the main rotor turns in the
conventional direction of counter-clockwise when looked at from
above,
## with a maximum AUW the power used by the tail rotor
in such a turn can exceed 10% of the total power available from
the engine.
555/3/3
_ 31 i
/ .'.;.;.;~-~
nu
if
se\$5'
..
lllli?/_,=i\'!-'
i
.
-1-ulna-Q
..
an
F1
Q!
MI
//
%A:'|-
V,
7' 7
/"|
I/
Itli
L1 |"
g
// --. \
ii
y ,,~=7
'
'
//* \\
I 41 I
\
I
Rt-'te\
l________l
FIG
18
## We determine the direction of rotation by viewing the tail
rotor from the side that it is mounted on the helicopter. A
tail rotor may have between two and six blades and will turn
much faster than the main rotor, but usually slower than the
engine rev/min.
## differs from one type of helicopter to another.)
Principle of Operation
The tail rotor, which is a type of reversible pitch propeller,
is controlled by the pilot through conventional rudder pedals.
Movement of the rudder pedals increases or decreases the pitch
of all the blades by the same amount and in the same_direction,
Q!
## thereby increasing or decreasing the thrust generated and the
lateral force felt by the tail of the helicopter.
can be moved from a positive-pitch angle through 0 to a
negative-pitch angle so that a thrust to the right or left may be
obtained.
S55/3/3
._ 32 _
FORWARD
(1-innnnnu
'::T'i1>
AIR FLOW
FIG. 19
-: "'
-'_'jIi-'*"
,""_
1; <>
\_ +.
## Forces of a Tail Rotor
The tail rotor is a rotating airfoil sited in an airflow. As
with the main rotor, the airflow causes dissymmetry of lift to be
felt across the disc of the tail rotor.
In correcting for
## dissymmetry of lift, the effects of drag, Hooke's joint, and
coriolis are introduced and absorbed by the relatively stiff
blades and heavily built hub assembly. The presence of gyroscopic
effect or phase lag does not matter because the pitch of the
tail~rotor blades is always changed collectively.
As with the
## articulated and semi-rigid main rotors, dissymmetry of lift is
catered for by blade flapping or by the assembly seesawing in a
vertical plane, and by the geometry of the pitchchange mechanism
555/3/3
_ 33 _
Dissymmetry of Lift
From Fig. 18, you will see that the velocity of the airflow
over the top tailrotor blade is
## Thus, if both blades have the same angle of attack, then
much more thrust will be produced by the top half of the disc
than by the lower half.
## This uneven distribution of the thrust
will cause vibration and will unevenly load the tail rotor and
the tailrotor gearbox assembly.
## This problem is overcome by
the blades flapping in much the same way as the main rotor
## helicopter, its angle of attack becomes less, and less thrust is
produced. At the same time, the lower blade flaps inwards, its
angle of attack is increased, and more thrust is produced.
The
net result of the flapping action is an even distribution of
thrust over the disc area.
Figure 20 shows schematically a twobladed tail rotor with
both the blades mounted on a yoke freely pivoted in the centre
axis and is connected to the pitch-change mechanism by a push-pull
rod.
## The trunnion is mounted so that its axis lies at an angle
to the centre line of the yoke, which gives an angled hinge called
a delta three hinge, This hinge reduces the angle of attack of
as the tail rotor flaps.
## by the pitchmchange linkage, because each push-pull rod is
attack of an outward-flapping blade is reduced.
The result of
## tail~rotor flapping is that for level flight in calm air, the
assembly assumes a less than vertical angle. Note angle 6 in Fig.2O
555/3/3
_ 34 _
____,/
Fuih-Pu" Nd
./
"'
I
-iv
Fnfch-change
Yo kg
Yakg
G \$
..
Trunnion
"W \\
an
/3
|F@jrg|{@7
Fa
'
""'
,-
>
~
"Fail-rota:
dnv: shaft
"_'
/6
'
---_--4 "L
'
r''\""
Trunnion
Pushpunrod
:1
r
5+
## P=-+-.+.-.,.. .r-1., r=""'
L
in --vi FLAP -------->- Ouk
FIG. 20
Tail--rotor flapping
You can see the change in the angle of attack of the blades
due to tail~rotor flapping on the delta three hinge very easily
if you balance a 30cm rule on a pencil with the rule inclined
(offset) at a small angle to the pencil.
555/3/8
## See Fig. 21.
Seesaw
_ 35 _
the rule on the pencil.
As
## the 30 cm end of the rule lifts
up, the numbers between 15 and 30
incline down, and the opposite
happens with the other end of
the rule.
Increase the angle
'
'
edge
n5
Tra'
ragedge
## of the rule relative to the
pencil and see the effect.
Position the pencil almost
lengthwise under the rule and
QQR
## see the effect.
The delta-hinge-mounted
tail rotor can have only two
edge
ng
Tra
FIG; 21
ag:
## because more thrust is necessary
1-
or because a large~diameter
L.;d';g
## tail rotor cannot be used, a
different type of tail rotor
Deltathree-hinge effect
is called for.
One common
type of tail rotor that can have as few as two and as many as
six blades has a central hub rigidly fixed to the tailrotor
gearbox output shaft, with each blade attached to the hub by a
axis and is connected to the pitchchange head by a pushpull
rod. The geometry of the pitch~change head and the attachment
of the pushpull rod to the blade is arranged so that, as the
blade flaps outward, its angle of attack is reduced and vice
versa.
## Figure 22 shows schematically this angle change and the
tip-path plane of this tail rotor during level flight in calm
air.
555/3/3
'- as -
I
Ft? Our
I
ROTATION
-4----
I
4
-~_a_
HI
C)
Hing
e
l ., 7
H_'PLKE}
"
Pikh - change. hand
\
_Q
p/.
WW
\
\\l
QTH
TP
/
f
L?
.\'1-4';';;;n
FIG. 22
## Tail rotor with flapping hinge
Drift
The force or thrust from the tail
' rotor 15
' u sed to counteract
the torque reaction of the main rotor. A couple is a pair of
equal and opposite parallel forces that tend to produce rotation,
that is, a torque. The force produced by the tail rotor acts
perpendicularly to an arm.
## That is, the tail rotor produces
The main
' rotor torqu e is balanced by the moment, which
stops its rotational effect but results in a s mall translationa
l
force that drifts the helicop
'
ter sideways.
a moment.
555/3/3
- 37 _
Dlrechon of rotation
of mam rater
F
Q
\ I
RESULTANT
Reaction for ue
t: 1*;=. : a.*:.
\
,
_2_+_
- FIG. 23
Tail~rotor drift
## Drift is counteracted by tilting the main rotor to one side
The tilt can be achieved by the design of the mount supporting
the main transmission or by the cyclic controls being rigged so
that neutral on the cyclic control column results in the tipa
path plane of the main rotor being tilted.
A combination of
## both methods is often used.
The main rotor control system is often designed to give a
progressively increasing tilt of the tip-path plane as the
collective is raised.
## the collective and as more tailrotor thrust is applied by the
pilot, the resulting increase in drift is automatically opposed.
SUMMARY
## The tail rotor counteracts the torque of the main rotor.
The tail rotor's blade angles are changed collectively.
That is, all blades have their pitch angle changed by
the same amount and in the same direction at the
same time.
555/3/3
_33_
## The tail rotor, like the main rotor, experiences
dissymmetry of lift, which is corrected by blade
flapping.
Whenever the main rotor turns, so does the tail
rotor. They are mechanically connected.
The tail rotor is controlled by the pilot through
the tail rotor (rudder) pedals.
PRACTICE
EXERCISE D
false.
l.
the helicopter.
2.
3.
4.
5.
pitch angle.
6.
work.
7.
8.
main rotor.
9.
tail rotor.
10.
during hovering.
555/3/3
_ 39 i
EXERCISE A
## Statements 1, H, 6, 7, 9, and 10 are true.
Statement 2 is false. The main purpose of the tail rotor
is to counteract the torque reaction of the main rotor.
Statement 3 is false. The total reaction of the main
rotor is resolved into lift and thrust.
Statement 5 is false. when the lift exceeds the weight,
the helicopter climbs. To hover (neither gain nor
lose height), lift must exactly equal weight.
Statement 8 is false. The coning angle is the angle
formed between the blades and a plane at right angles to
the rotor shaft.
EXERCISE
1.
1
E-4
<
Q.
I./'\.S. 60 knbfs
->
'>
fer,/8
0/
1
W;
E
*,
s' O
REYREANNG
0
'7:1.' ' O.
HALF
r|Au=
80
I
a' Q
an
K. Q
4&0
AH
FIG. 24
555/3/3
...L[Q_
## If, instead of the helicopter flying forward at 60 kt IAS,
60 IAS. when the advancing blade is at 90 to the aircraft centre line, the velocity of the air over the blade
tip is now the stillair tip velocity plus the air
velocity of 60 kt, and the retreating blade tip experiences
the still-air tip velocity minus the air velocity of 60 kt.
lift by the retreating blade. As a result, the helicopter
will tend to roll to the side of the retreating blade.
2.
Relam Iir'_f|OW
FIG. 25
3.
chord ne
,._ L
-_->-
-v'- E
"
"7
## when a rotor blade is turning and flapping up, it
has two velocities.
One velocity is in the direction
of rotation and the other is upwards and at right
angles to the first.
If we hold the blade still and
apply to it the air velocities it felt when turning
and flapping up, we will have an airflow from ahead
and an airflow from above.
Figure 26 shows the
space diagram of the two velocities.
Air moon
'
7L"i'
t|n
lI"Qr\
irigle of aliids
FIG. 26
555/3/3
-141-
## These two velocities are combined to give a
triangle of velocities.
Its resultant gives us
the new velocity and direction.
See Fig.27.
/'
A"_f|w
mm abut
, 7,
-;:.:. .
"'
_ _ -
FIG. 27
~ ~ _ _ ___
* ~ -___ __ 5
## The change in direction gives a decrease in the
angle of attack of the blade and, it follows, a
decrease in the lift generated. The opposite
occurs as a blade flaps down.
EXERCISE C
are true.
## Statement l is false. Dissymmetry of lift is caused by
an airflow meeting the rotor disc and causing differing
air velocities over the advancing and retreating rotor
Thus, during hover in still air, there is no
dissymmetry of lift.
Statement 3 is false. A retreating blade experiences a
Statement 6 is false. An offset pitch change horn
changes the pitch angle of the blade as it flaps
up and down. The offset is arranged so that, as the
blade flaps up, the pitch angle is reduced.
Statement 7 is false. Because of the gyroscopic effect
the change in angle of attack must be made 90 of rotor
"rotation ahead of where the effect is to take place.
Statement 10 is false. Coriolis effect occurs because
of the flapping up and down of the blades needed to
tilt the rotor disc for flight other than hover. During
hover in still air, no tilting of the rotor disc is
needed, and so no Coriolis effect will be felt.
Statement l3 is false.
555/3/3
_ n2 _
## Statement 16 is false. A rotor blade damper controls
Statement 17 is false. A collective pitch change alters
the pitch angles of all the blades by the same amount
at the same time.
F
e
EXERCISE D
\$3
ii
## Statements 2, 3, 5, 6, 8 and 10 are true.
Statement l is false. The propulsive force for the
helicopter is supplied by the main rotor.
Statement H is false.
The tail rotor turns at all
times that the main ro tor turns.
Statement 7 is false.
## Tail rotor-blade angles are only
changed collectively.
Statement 9 is false.
The rev/min of the tail rotor
are higher than those of the main rotor.
TEST PAPER 3
l.
2.
(a)
(b)
A tail rotor.
(a)
(b)
## In each sketch, show the main rotor force resolved into
lift and thrust forces and also show the drag and weight
forces. Assign values to the lift, drag, and thrust
forces.
555/3/3
..L1_3
## Discuss briefly the main differences between semi-rigid,
articulated, and hingeless rotors.
## With the aid of a diagram, show how dissymmetry of lift
may be felt by a main rotor unless corrected. What
would happen to the helicopter if no correction were
%
%
e
(a)
## A vertical hinge, and
(b)
A horizontal hinge.
(c)
hinges?
## Explain why the angle of attack of a main rotor blade is
changed 90 of rotor~head rotation before the desired
effect of the change is to take place.
8-
(a)
(b)
## with the aid of a diagram, show how dissymmetry of lift
can be felt by a tail rotor unless corrected. What
would happen to the tail rotor if no correction were made?
## Briefly describe one method used to correct dissymmetry of
lift of a tail rotor.
9
-3
## Why is the tail rotor mechanically connected to the main
rotor so that it must turn when the main rotor turns?
555/3/3 | 12,096 | 44,904 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2020-45 | latest | en | 0.866623 |
https://engineering.stackexchange.com/questions/48871/what-are-disadvantages-if-increasing-the-gain-kd-for-a-pdproportional-plus-deri | 1,718,498,269,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861618.0/warc/CC-MAIN-20240615221637-20240616011637-00837.warc.gz | 211,067,067 | 39,597 | # What are disadvantages if increasing the gain Kd for a PD(Proportional Plus Derivative ) controller
What are disadvantages if I increase the gain Kd for a PD(Proportional Plus Derivative) controller, someone told me larger gain for derivative error for will reduce the stability margin, while the text books said increase the Kd could increase the damping ratio and high damping ratio means more stability margin.
• How did you conclude that high damping ratio means high stability margin? What is the system being controlled? Take an example (a fourth order system would be good) and increase KD; you will notice that the stability margins will reduce with increase in KD (after some point)
– AJN
Commented Dec 23, 2021 at 12:26
• at least for second order system, increasing kd does increase the damping. There is app at SimulationOfFeedbackControlSystemWithControllerAndSecondOrde which you can use to try and see. Commented Dec 23, 2021 at 14:06
• @Nasser - cool app! But wouldn't Kd would improve phase margin at first, then after some point, push the crossover frequency high enough that additional real-world poles would become significant, thus degrading it again? Commented Dec 23, 2021 at 14:30
Depending on the plant, a reduction in phase margin could happen. Here is a theoretical example. Consider a plant with a double integrator and an additional pole:
$$\frac{K_1}{s^2(1+s/\omega_P)}$$
Now let's have a PD controller, which could be written either $$[K_P(1+{K_D}s)]$$ , or as $$[K_P + {K_D}s]$$ . For convenience, both of these could be re-written in the form
$$K_0(1+s/\omega_Z)$$
The loop's stability can then be analyzed by looking at the product of the gains going around the loop -- i.e. controller and plant. We have two poles at the origin, a zero at $$\omega_Z$$ and a pole at $$\omega_P$$.
An hand-drawn Bode plot is below, followed by an Octave plot of essentially the same thing. Each graph shows two different values of $$\omega_Z$$, corresponding to two different values of $$K_D$$.
For completeness, here are the corresponding closed loop bode plots and closed loop step responses
Hopefully it's possible to see from first bode plots (the ones looking at around-the-loop), that reducing $$\omega_Z$$ (i.e. increased $$K_D$$), by 10x in this example, moves the first breakpoint left by 10x. This then lifts the section of the line with the -20dB/decade slope up by 20dB, causing it to intersect the horizontal axis about 10x further to the right, meaning the loop has about a 10x higher bandwidth.
The movement of the gain crossover frequency also results in a different phase margin. In the example shown, it reduces phase margin from something like 80 degrees for line (1), to 50 degrees for line (2).
However, if the overall gain were smaller, the opposite could also be the case -- if the gain crossover started out to the left of the "peak" in the phase plot, then the phase margin would increase at first, but then decrease again after the gaincrossover frequency passes the peak in the phase plot. | 726 | 3,047 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 10, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-26 | latest | en | 0.932474 |
https://byjus.com/question-answer/simplify-the-following-x-2y-y-3x-x-y-x-3y-y-3x-4x-5y-1/ | 1,723,759,861,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641316011.99/warc/CC-MAIN-20240815204329-20240815234329-00319.warc.gz | 118,606,752 | 31,286 | 1
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Question
# Simplify the following:(x−2y)(y−3x)+(x+y)(x−3y)−(y−3x)(4x−5y)
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Solution
## (x−2y)(y−3x)+(x+y)(x−3y)−(y−3x)(4x−5y)=(xy−3x2−2y2+6xy)+(x2−3xy+xy−3y2)−(4xy−5y2−12x2+15xy)=7xy−3x2−2y2+(x2−2xy−3y2)−(19xy−5y2−12x2)=7xy−3x2−2y2+x2−2xy−3y2−19xy+5y2+12x2=(−3+1+12)x2+(7−2−19)xy+(−2−3+5)y2=10x2−14xy=2x(5x−7y)
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Join BYJU'S Learning Program | 288 | 562 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2024-33 | latest | en | 0.654325 |
http://mathematica.stackexchange.com/questions/33552/long-waiting-time-for-computing-a-summation/33559 | 1,467,380,177,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783402746.23/warc/CC-MAIN-20160624155002-00162-ip-10-164-35-72.ec2.internal.warc.gz | 209,637,935 | 18,404 | # Long waiting time for computing a summation
It takes a long time to compute the summation below, and I'd like to know if there are alternative ways
to compute things faster. When replacing $15$ by $\infty$, then I should get $3^{1/3}$. I need Mathematica solution
not mathematical solutions (since here I can handle things on my own).
Clear[b, k]
b[0] := 1;
b[k_] := b[k - 1] + 1/(b[k - 1]^2 + b[k - 1] + 1)
N[1/(15)^(1/3) Sum[1/(b[k]^2 + b[k] + 1), {k, 0, 15}]]
-
I'd like to see what happens when I set $1000$ instead of $15$. – A_math_ninja Oct 6 '13 at 18:57
## 3 Answers
EDIT
Clear[b, k]
b[0] := 1;
b[k_] := ( b[k] = N[b[k - 1] + 1/(b[k - 1]^2 + b[k - 1] + 1), 100] )
1/(1000)^(1/3) Sum[1/(b[k]^2 + b[k] + 1), {k, 0, 1000}]
(*1.28966188938645818444514220128278717806515296115261423714827563492762\
4979228386408865750383532338923*)
very close to 3^(1/3) = 1.4422...
The convergence is slow. So you should do extrapolation.
-
Thank you for your answer (+1). It seems that $N$ in front caused some trouble. – A_math_ninja Oct 6 '13 at 19:19
yeah, the limit is precisely $3^{1/3}$. – A_math_ninja Oct 6 '13 at 19:32
I actually put N in the definition of b[k], since its exact values are huge rational numbers. – Vahagn Poghosyan Oct 6 '13 at 19:32
Note that here k is local variable, so it is not necessary to Clear it at the beginning of the code. – Vahagn Poghosyan Oct 6 '13 at 19:39
I got the point. Thanks! – A_math_ninja Oct 6 '13 at 19:46
b[0] := 1;
b[k_] := b[k] = N[b[k - 1] + 1/(b[k - 1]^2 + b[k - 1] + 1)]
f[x_] := f[x] = f[x - 1] + (1/(b[x]^2 + b[x] + 1))
f[0] := 0
Manipulate[
ListPlot[Table[1/x^(1/3) f[x], {x, 1, a}], MaxPlotPoints -> 100,
Epilog -> {Green, Line[{{1, 3^(1/3)}, {a, 3^(1/3)}}]},
PlotRange -> {0, 4}], {a, 10, 100000, 1}]
-
Here is an alternative,
Nest[# + 1/(1 + # + #^2) &, 1., n + 1] - 1
that will compute the sum over 0 <= k <= n. For 100-digit precision use
Nest[# + 1/(1 + # + #^2) &, 1.100, n + 1] - 1
Examples
With[{n = 1000},
1/n^(1/3) (Nest[# + 1/(1 + # + #^2) &, 1., n + 1] -
1)] // AbsoluteTiming
(* {0.000359, 1.28966} *)
With[{n = 1000},
1/n^(1/3) (Nest[# + 1/(1 + # + #^2) &, 1.100, n + 1] - 1)] // AbsoluteTiming
(* {0.005293,
1.28966188938645818444514220128278717806515296115261423714827563492762\
4979228386408865750383532338923} *)
With[{n = 1000000},
1/n^(1/3) (Nest[# + 1/(1 + # + #^2) &, 1.100, n + 1] - 1)] // AbsoluteTiming
(* {5.204264,
1.42720128317421809656439478563254431280673078216998116597726119064353\
220814168138689903801038204268} *)
It avoids the recursion and memory limits that the OP's b runs into. (With n = 1000000, my kernel crashed using b and Sum.) And you can use NestList to get all the sums up to n.
- | 1,095 | 2,709 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2016-26 | latest | en | 0.740401 |
https://micquality.com/six_sigma_glossary/f_test.htm | 1,701,805,692,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100555.27/warc/CC-MAIN-20231205172745-20231205202745-00864.warc.gz | 448,671,780 | 4,739 | SIX SIGMA GLOSSARY PROCESS IMPROVEMENT AND SIX SIGMA
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:: Home ::Six Sigma ::SIX SIGMA GLOSSARY: F Test :: CURRENT STUDENTS LOGIN
F Test
The F Test is used to compare the variances of two samples to test the hypothesis that the samples are drawn from populations with different variances:
H0 the samples are drawn from populations with equal variances
H1 the samples are drawn from populations with different variances
The F test is based on the ratio of the variances of the two samples:
1 2 3 4 5 Variance Sample A 8.7 9.8 12.2 12.9 9.5 3.327 Sample B 23.8 21.5 18.3 24.3 22.9 5.788
This gives an F statistic:
(Most tables give critical values greater than 1, so divide the larger by the smaller)
This can be compared to the critical F value from tables, or the p-value can be obtained using the Excel function:
=FDIST(Fo, DOFn, DOFd) = 0.302
DOFn and DOFd are the Degrees of Freedom of the numerator and denominator (both are 4 in this example)
The F Test is based on the F Distribution.
the F Test, and other types of hypothesis test, is covered in the MiC Quality online course Advanced Statistics. Try out our courses by taking the first module of the Primer in Statistics free of charge.
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:: Six Sigma Primer :: Primer in Statistics :: Advanced Statistics :: Statistical Process Control SPC :: Advanced SPC :: Design of Experiments :: Advanced DOE :: Measurement Systems Analysis MSA/ Gage R&R Primer in Statistics FREE First Module "Introduction to Statistics" Primer in Statistics FREE Reference Booklet FREE Excel Primer | 492 | 1,967 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2023-50 | latest | en | 0.787442 |
http://stackoverflow.com/questions/8154079/add-text-to-faces-of-polyhedron/8155008 | 1,369,402,145,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368704662229/warc/CC-MAIN-20130516114422-00077-ip-10-60-113-184.ec2.internal.warc.gz | 241,380,485 | 12,908 | # Add text to faces of polyhedron
Is it possible to automate the addition of any text to the faces of a polyhedron, like this manually-drawn graphic shows (the example's odd numbering scheme isn't relevant):
It was easy enough to label the vertices:
c = 1;
Show[{Graphics3D[
Text[c++, #] & /@ PolyhedronData["Dodecahedron", "VertexCoordinates"]],
PolyhedronData["Dodecahedron"]},
Boxed -> False]
(even though some of the text is placed in front of the shape for vertices that are hidden. That's probably soluble.)
But when I tried to do the same thing for faces, nothing worked. PolyhedronData["Dodecahedron", "Faces"] returns a GraphicsComplex, rather than coordinates.
Am I overlooking an easy solution/option?
Edit: thanks for these answers, they're all brilliant. If I could combine the text placing of szabolcs' answer with the text quality of belisarius', the perfect solution is in sight!
-
Please see my edit. Now labels should be oriented to point roughly "up", and also align with edges. – Szabolcs Nov 16 '11 at 18:07 That's great! Even better. – cormullion Nov 16 '11 at 18:15
Here's a funky approach:
(* this function just transforms the polygon onto the [0,1] 2D square *)
vtc[face_, up_:{0,0,1}] := Module[{pts, pts2, centre, r, r2, topmost},
pts = N@face;
centre = Mean[pts];
pts = (# - centre & /@ pts);
r = SingularValueDecomposition[pts][[3]];
(* these two lines ensure that the text on the outer face
of a convex polyhedron is not mirrored *)
If[Det[r] < 0, r = -r];
If[Last[centre.r] < 0, r = r.RotationMatrix[\[Pi], {1, 0, 0}]];
pts2 = Most /@ (pts.r);
topmost = Part[pts2, First@Ordering[up.# & /@ pts, -1]];
r2 = Transpose[{{#2, -#1} & @@ topmost, topmost}];
r2 /= Norm[r2];
Rescale[pts2.r2]
]
faces = First /@ First@Normal@PolyhedronData["Dodecahedron", "Faces"];
numbers =
Graphics[Text[
Style[#, Underlined, FontFamily -> "Georgia",
FontSize -> Scaled[.3]]]] & /@ Range@Length[faces];
Graphics3D[
Polygon[#2, VertexTextureCoordinates -> vtc[#2]]} &, {numbers,
faces}],
Boxed -> False
]
Demoing a "SmallRhombicosidodecahedron":
-
Congrats on your 5k! – belisarius Nov 16 '11 at 16:59
@belisarius Thank you! :-) – Szabolcs Nov 16 '11 at 17:05
Thanks, great idea. – cormullion Nov 16 '11 at 17:50
a = PolyhedronData["Dodecahedron", "Faces"] /. GraphicsComplex -> List;
c = 1;
Show[{Graphics3D[
Text[c++, #] & /@ (Mean /@ (a[[1, #]] & /@ a[[2, 1]]))],
PolyhedronData["Dodecahedron"]}, Boxed -> False]
Edit
Perhaps better:
Show[{Graphics3D[
MapIndexed[Text[#2, #1] &,
Mean /@ (PolyhedronData["Dodecahedron", "VertexCoordinates"][[#]] & /@
PolyhedronData["Dodecahedron", "FaceIndices"])]],
PolyhedronData["Dodecahedron"]}, Boxed -> False]
Edit
Or
text = Style[#, 128] & /@ Range[12]
Graphics3D@
Riffle[Texture /@ text,
(Append[#1, {VertexTextureCoordinates ->
With[{n = Length[First[#1]]}, Table[1/2 {Cos[2 Pi i/n], Sin[2 Pi i/n]}+
{1/2, 1/2}, {i, 0, n - 1}]]}] &) /@
Flatten[Normal[PolyhedronData["Dodecahedron", "Faces"]]]]
-
+1, but why not use a single Map operation instead of three? And toss in a little infix spice while you're at it. ;-) Graphics3D[c++ ~Text~ Mean@a[[1, #]] & /@ a[[2, 1]]] – Mr.Wizard Nov 16 '11 at 15:47
Regarding your second edit: I consider the challenge to be doing this for an arbitrary polyhedron. – Szabolcs Nov 16 '11 at 16:58
@Sza And I upvoted you for that :) – belisarius Nov 16 '11 at 17:05
Great answer, thanks. It will take me much longer for me to study it. I still haven't heard of half of these functions, let alone learnt how to use them! – cormullion Nov 16 '11 at 17:27
@cormullion Don't hesitate in asking for help here! – belisarius Nov 16 '11 at 17:56 | 1,189 | 3,650 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2013-20 | latest | en | 0.812974 |
https://statvizor.com/bits-to-mebibytes | 1,721,644,711,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517846.73/warc/CC-MAIN-20240722095039-20240722125039-00775.warc.gz | 464,395,182 | 5,378 | # Bits to Mebibytes Converter
Bits and mebibytes are units of digital information storage, but they differ in magnitude. One mebibyte (MiB) is equal to 8,388,608 bits (1024 kibibytes), while one bit is simply one binary digit (either 0 or 1). Therefore, to convert from bits to mebibytes, you need to divide the number of bits by 8,388,608.
For example:
To convert 8 bits to mebibytes, divide by 8,388,608: 8 / 8,388,608 = 0.00000095367 mebibytes
To convert 10,000 bits to mebibytes, divide by 8,388,608: 10,000 / 8,388,608 = 0.00000119209 mebibytes
To convert 1,000,000 bits to mebibytes, divide by 8,388,608: 1,000,000 / 8,388,608 = 0.11920929 mebibytes
So, to convert any number of bits to mebibytes, divide the number of bits by 8,388,608. | 267 | 749 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2024-30 | latest | en | 0.786686 |
https://answercult.com/question/what-is-the-actual-use-of-the-median-and-mode-in-statistics-compared-to-the-average-mean/ | 1,642,582,059,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301264.36/warc/CC-MAIN-20220119064554-20220119094554-00543.warc.gz | 154,045,440 | 21,326 | # What is the actual use of the median and mode in statistics compared to the average (mean)?
144 views
What is the actual use of the median and mode in statistics compared to the average (mean)?
In: Mathematics
The average is affected by extreme values. Like if in Fentucky everybody makes \$1 a year except for one person who makes \$1,000,000,000 a year, the average will not be useful for determining how a normal person lives. That’s an extreme example but for things where outlier values are an issue but can’t just be tossed out, median can be more useful.
Mode is usually used for categorical data. Like if Coke is comparing the sales of Coke Zero, Vault (RIP), and Mellow Yellow then they probably just care about how much of each one is sold instead of trying to construct an average.
I use median much more often than mean, just to say. If you don’t know the distribution, and most likely it’s gonna be skewed in one direction, then taking the mean is not too good to estimate the mode. Many processes in nature create log normal distributions, or many distributions are too close to an absolute limit, like 0 or so. In these cases i prefer taking the median, it ignores outliers much more easily if you got enough samples
One example is if you have a wide range with a few outliers that would fuck up your data point. For example for average income or net worth. Because of a few billionaires tossed in the average would be way higher so you would go with one of the other two for a more accurate representation of the data set. Can’t you there is ways to calculate if you should include an outlier or not.
Let’s say you have 99 randomly selected people in a room with Jeff Bezos. The average net worth of everybody in the room will be approximately \$2 billion, because Jeff Bezos is worth \$200 billion and there are 100 people in there, and the other people have effectively zero money (compared to \$200 billion). But saying “the average net worth in the room is \$2 billion” is utterly meaningless because it’s really just one person bringing up the average for everybody else. So a more meaningful picture might be the median or the mode… the median is the middle person’s net worth. The mode is the most common net worth. These give you a better picture of what the “average” person looks like in the room, than what the actual average tells you.
They each tell you different things about the data set, so the “actual use” would be dependent on what meaning you want to extract from the data set.
Let’s say you’re being graded on something. You do it four times perfectly, 10/10, but you are sick on day, or just have a run of bad luck, and mess it up, 1/10.
You’re “mean” rating would be 8.2. But does that accurately reflect your abilities? Does a single fluke really mean you are 20% less capable than all those other times?
Both the median and the mode would report a value of 10.
The thing to consider and remember is that they are summarizing some aspect of the data, necessarily focusing on some property at the expense of others. With the mean, you sacrifice any information about the distribution of data and are influence by extreme outliers (such as that 1/10 fluke above).
With the mode, you are explicitly being told *something* about the distribution of the data (namely the most common element in the set), but you don’t really know how it compares to the remaining elements of the set.
And, finally, with the median you are learning what the “middle” of the data set looks like while losing information about any extreme outliers.
Median is a really great measurement, and frankly, should be used for a lot of things that people traditionally use average for.
Median takes the “middle” value of a series. So: 1, 5, 10, 11, 87 The median is the middle number, 10.
Why is this interesting? Median is good because it operates somewhat similar to taking an average, but it generally gets rid of outliers, as above had 87 as one of the values. The average of the above is 22.8. But is that even interesting when all of your values except 1 outlier is way lower than that? Median helps deal with series that have outliers. It also tells you a middle point, for example, what if I add another number? Well, just for theory sake, you can say there’s a 50% chance it’ll be below the median of 10.
Median is good for say how long it takes people to pay bills.
If I said the average time it took to pay a bill is 22.8 days, but look above, thats a meaningless value, nothing took 22.8 days, in fact 4/5 of them took 11 or less. But if you say the median amount is 10 days, you can say, there’s probably about a 50% chance it gets paid in 10 days or less. That 87 out there is probably some special case, so we don’t want to use a method that adjusts for that
Mode is just the most common number in a series. 1, 2, 2, 3, 4, 5, 6 ,7
The mode is 2. It appears twice, all other numbers appear once. This is just telling me what valued occurred the most times. There’s lots of uses for this, especially in series with a lot of numbers
The median can be more useful than the mean in situations where there are large extremes at either end of the data set. It’s often helpful in money – let’s say that we’re looking at how much money people have saved up, and we pick 10 people at random for a simplified example. 3 of them have no money in savings, 3 have around \$20k saved up, two have \$50k, one has \$100k, and one has \$1million saved. The mean of that set is \$126k in savings, but that doesn’t really give us any useful information, because 9/10 of our group actually have less saved up than that, most of them *far* less.
The median of the set, on the other hand, is \$20k. Which doesn’t tell us anything about how far the extreme ends are, but it does mean that if we pick an average person, then 50% will fall at or below that amount, and 50% will fall at or above. So it actually tells us a slightly clearer story than the mean – we don’t know how many people have no savings or how many are millionaires, but we do know where the dividing line is.
The mode can also be handy in sets like this – there are two modes in this set, \$0, and \$20k. The problem with modes in small sets is that they tend to swing frequently, but a mode like \$0 definitely tells you a lot of the story in a data set, because it shows you that lots of those empty accounts exist, and that they’re dragging down your other measurements.
Median is a good simple way to remove outliers from a data set. For example I have a system that measures a distance. As with all real world systems it has a certain level of noise in the output, averaging the output is a good way to reduce this noise. Most of the time the distance is about correct, you could take the mean and get a very accurate result. But every now and then something goes wrong, it picks up a reflection or some other issue and the result will be completely wrong. E.g. it would measure 10 meters rather than 60.
If I took the mean and this was to happen then my result would be significantly wrong. But such large errors are rare, if instead I use the median these rare outliers are ignored and don’t impact the results.
As others have indicated median is often used when looking at wealth related things like income, it avoids a small number of very wealthy people skewing the results. Any time when you have a small number of significantly different outliers median will often give a better indication of the typical value than mean.
Mode is less commonly used. I can’t think of a time I’ve seen it used in the real world.
I heard this ELI5-ish analogy a while ago and it was worth remembering:
There are 10 people in a room. 9 of the people have no apples at all. The 10th person has ten apples. The *average* person has 1 apple…which is true mathematically, but utterly false realistically. *Averages* can be dragged far to one side or the other.
The *median* person has no apples, which is true both mathematically and realistically. The *median* is much less vulnerable to distortion from a single data point far outside all the rest.
The better question to ask is….why do you even use mean in the first place? If your answer is “well….I need an aggregate quantity and I pick the mean because I was told to do so….”, that’s where the confusion came about.
Mean has specific usage, it’s not something that you just throw out there because you want a way to measure central tendency. In fact, it might seems like no-brainer today, but mean used to be quite controversial, until Gauss showed that it is actually useful. The more intuitive, default, quantity to measure central tendency used to be the *median*.
The purposes of these numbers are simple: they are the minimizers of “surprise”. Let’s say you were to need to use a single number to represent the value of every elements in the sample sets, but then you are given a single element, and if the element deviate from the number, you get punished by an amount dependent on how they differed. Then what you want is a number that minimize the expected amount of punishment.
Which number you use depends on how much punishment scale up with the deviation. The most intuitive measure is for the punishment to equal the deviation: the minimizer is the median, which is what people used for centuries. At an extreme end, any deviations at all is bad, the minimizer of this is the mode. And if the square of the deviation is the punishment, the minimizer is the mean.
The reason why the mean see a lot of use in theoretical statistics and probability theory is because: (a) by framing everything into a game theory context, lots of questions can be converted into about the mean of *something* (not necessarily the original data but something computed from it), so you don’t really lose generality by just studying the mean; (b) it has mathematically nice properties, such as linearity and central limit theorem.
But that doesn’t mean you should blindly use the mean in every situation. Think about what is your punishment. How bad it would be for your number to deviate from each samples? And this depends on practical context. If you are reporting numbers generally with no specific purposes, it might be useful to just report a whole bunch of different central tendency so that people have a choice of different numbers to make use of. | 2,331 | 10,397 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2022-05 | latest | en | 0.928793 |
http://mathhelpforum.com/pre-calculus/25202-solved-help-3-assignments-late.html | 1,481,161,292,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542288.7/warc/CC-MAIN-20161202170902-00364-ip-10-31-129-80.ec2.internal.warc.gz | 187,395,948 | 11,552 | # Thread: [SOLVED] Help with 3 assignments that are late!
1. ## [SOLVED] Help with 3 assignments that are late!
Write the slope-intercept form of the equation passing through the points (-3, 4) and
(3, 6).
Show all your steps in a Word document, save the document and attach it in the dropbox.
Go left –3 , up4 .. put a dot
Go up 3 , go right 6 , put a dot
Slide them across equal to eachother.
----------------------------------------------------------------
Is the blue line part of the solution set?
Is the red line part of the solution set?
How would you describe the solution to the above system of linear inequalities?
• The overlapping area of red and blue shading
------------------------------------------------------------------------------------------------------
Given the following sets of systems of equations, choose the one that you would want to solve using substitution. Write one sentence explaining why you would use substitution to solve this problem.
A. 3x + 3y = 18
4x – 3y = 0
B. 5x + 3y = 3
X + 9y = 2
C. 3x -4y = -1
9x + 12y = 15
-----------------------------------------------------------------------
Name all pairs of vertical angles:
Name all pairs of corresponding angles:
Name all pairs of alternate interior angles:
2. since this is a graded assignment to be handed in, i will not do the problem for you. i will rather give you the method, or hints on how to solve the problems.
Originally Posted by JetPack360
Write the slope-intercept form of the equation passing through the points (-3, 4) and
(3, 6).
you may want to search the forum for something like "equation of a line," there are many problems of this nature that has been done here.
the slope intercept form of a line is: $y = mx + b$ where $m$ is the slope, and $b$ is the y-intercept
here's how to get that equation given two points on a line.
we first find the slope:
recall that the slope is rise over run, that is, change in y divided by the change in x. so given two points on a line, $(x_1,y_1)$ and $(x_2,y_2)$, we find the slope, m, as follows:
$m = \frac {y_2 - y_1}{x_2 - x_1}$
we then use the point-slope form:
use either of the points given and the slope you found, and plug them into the slope-intercept form shown below:
$y - y_1 = m(x - x_1)$
where $(x_1,y_1)$ is the point you used, and m is the slope
finally, solve for y:
solving for y in the above form and simplifying will give you the slope-intercept form of the equation of the line
3. Originally Posted by JetPack360
Is the blue line part of the solution set?
Is the red line part of the solution set?
we draw a line broken if it is not in the solution set of an inequality
How would you describe the solution to the above system of linear inequalities?
• The overlapping area of red and blue shading
the red area defines the solution set of one of the inequalities in the system. the blue area is the solution set for the other. where they overlap is the solution set to the system, that is, both inequalities are satisfied at the same time. anywhere that is not shaded is, of course, not in the solution set of either inequality. BEWARE OF THE BOUNDARIES. bold and broken lines make a difference
4. Originally Posted by JetPack360
Given the following sets of systems of equations, choose the one that you would want to solve using substitution. Write one sentence explaining why you would use substitution to solve this problem.
A. 3x + 3y = 18
4x – 3y = 0
B. 5x + 3y = 3
X + 9y = 2
C. 3x -4y = -1
9x + 12y = 15
you want to use the substitution method if it is easy to solve for one variable in terms of the other in one of the equations. which system above meets that requirement?
5. Originally Posted by JetPack360
Name all pairs of vertical angles:
Name all pairs of corresponding angles:
Name all pairs of alternate interior angles:
see here | 962 | 3,857 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 8, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2016-50 | longest | en | 0.89215 |
https://carlrocks.com/docs/QuestionBank/Leetcode/LC236 | 1,675,357,814,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500035.14/warc/CC-MAIN-20230202165041-20230202195041-00243.warc.gz | 177,124,793 | 7,004 | LC236. Lowest Common Ancestor of a Binary Tree
Problem Description
Given a binary tree, return the lowest common ancestor (LCA) of two given nodes in the tree.
According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”
For example: Given binary tree [3, 5, 1, 6, 2, 0, 8, null, null, 7, 4], where p = 5, and q = 1,
`` 3 / \ 5 1 / \ / \ 6 2 0 8 / \ 7 4``
return 3.
Solution
High level strategy
Our approach to this problem is to solve it recursively. In our base case, we return the current node if it is equal to 'p', 'q', or null. This way, we will receive falsy values as we recurse down a branch that does not contain either 'p' or 'q'. Therefore, if either the left or right side of the current node returns null, then we know that that side of the tree does not contain the target values, and we can simply check the opposite side. We repeat this process until we arrive at a subtree that contains both target values, at which point we return the root node of the subtree. The time complexity of this solution is O(n), where 'n' is equal to the number of nodes. The space complexity is O(logn), or O(h), where 'h' is equal to the height of the tree.
Code
``/** * Definition for a binary tree node. * function TreeNode(val, left, right) { * this.val = (val===undefined ? 0 : val) * this.left = (left===undefined ? null : left) * this.right = (right===undefined ? null : right) * } **/const lowestCommonAncestor = (root, p, q) => { if (root === null || root === p || root === q) return root; let left = lowestCommonAncestor(root.left, p, q); let right = lowestCommonAncestor(root.right, p, q); if (!left) return right; if (!right) return left; return root;};`` | 523 | 1,930 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2023-06 | latest | en | 0.802215 |
https://rdrr.io/cran/PST/man/predict.html | 1,548,248,048,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547584331733.89/warc/CC-MAIN-20190123105843-20190123131843-00041.warc.gz | 603,694,346 | 16,406 | # predict: Compute the probability of categorical sequences using a... In PST: Probabilistic Suffix Trees and Variable Length Markov Chains
## Description
Compute the probability (likelihood) of categorical sequences using a Probabilistic Suffix Tree
## Usage
1 2 ## S4 method for signature 'PSTf' predict(object, data, cdata, group, L=NULL, p1=NULL, output="prob", decomp=FALSE, base=2)
## Arguments
object a probabilistic suffix tree, i.e., an object of class "PSTf" as returned by the pstree, prune or tune function. data a sequence object, i.e., an object of class 'stslist' as created by TraMineR seqdef function, containing the sequences to predict. cdata not implemented yet. group if object is a segmented PST, providing a vector of group membership so that each sequence probability will be predicted with the conditional probability distributions for the group it belongs to. If object is a segmented PST and group is not provided, each sequence will be predicted by each of the submodel, and the output will be a matrix with nbgroup columns, where nbgroup is the number of segments in the PST. L integer. Maximal context length for sequence prediction. This is the same as pruning the PST by removing all nodes of depth
## Details
A probabilistic suffix tree (PST) allows to compute the likelihood of any sequence built on the alphabet of the learning sample. This feature is called sequence prediction. The likelihood of the sequence a-b-a-a-b given a PST S1 fitted to the example sequence s1 (see example) is
P^{S1}(abaab)= P^{S1}(a) \times P^{S1}(b|a) \times P^{S1}(a|ab) \times P^{S1}(a|aba) \times P^{S1}(b|abaa)
The probability of each of the state is retrieved from the PST. To get for example P(a|a-b-a), the tree is scanned for the node labelled with the string a-b-a, and if this node does not exist, it is scanned for the node labelled with the longest suffix of this string, that is b-a, and so on. The node a-b-a is not found in the tree (it has been removed during the pruning stage), and the longest suffix of a-b-a found is b-a. The probability P(a|b-a) is then used instead of P(a|a-b-a).
The sequence likelihood is returned by the predict function. By setting decomp=TRUE the output is a matrix containing the probability of each of the symbol composing the sequence. The score P^S(x) of a sequence x represents the probability that the VLMC model stored by the PST S generates x. It can be turned into a more readable prediction quality measure such as the average log-loss
logloss(S,x)=-\frac{1}{\ell} ∑_{i=1}^{\ell} \log_{2} P^{S}(x_{i}| x_{1}, …, x_{i-1})=-\frac{1}{\ell} \log_{2} P^{S}(x)
by using 'output=logloss'. The returned value is the average log-loss of each state in the sequence, which allows to compare the prediction for sequences of unequal lengths. The average log-loss can be interpreted as a residual, that is the distance between the prediction of a sequence by a PST S and the perfect prediction P(x)=1 yielding logloss(P^{S},x)=0. The lower the value of logloss(P^{S},s) the better the sequence is predicted. For more details, see Gabadinho 2016.
## Value
Either a vector of sequence probabilities (decomp=FALSE) or a matrix (if decomp=FALSE) containing for each sequence (row) the probability of each state in columns.
Alexis Gabadinho
## References
Gabadinho, A. & Ritschard, G. (2016) Analyzing State Sequences with Probabilistic Suffix Trees: The PST R Package. Journal of Statistical Software, 72(3), 1-39.
## Examples
1 2 3 4 5 6 7 8 data(s1) s1 <- seqdef(s1) S1 <- pstree(s1, L=3, nmin=2, ymin=0.001) S1 <- prune(S1, gain="G1", C=1.20, delete=FALSE) predict(S1, s1, decomp=TRUE) predict(S1, s1)
### Example output
Loading required package: TraMineR
TraMineR stable version 2.0-7 (Built: "Sat,)
Website: http://traminer.unige.ch
Please type 'citation("TraMineR")' for citation information.
Loading required package: RColorBrewer
PST version 0.94 (Built: 2017-09-22)
Website: http://r-forge.r-project.org/projects/pst
[>] 2 distinct states appear in the data:
1 = a
2 = b
[>] state coding:
[alphabet] [label] [long label]
1 a a a
2 b b b
[>] 1 sequences in the data set
[>] min/max sequence length: 27/27
[>] 1 sequence(s) - min/max length: 27/27
[>] max. depth L=3, nmin=2, ymin=0.001
[L] [nodes]
0 1
1 2
2 4
3 7
[>] computing sequence(s) likelihood ... (0.024 secs)
[>] total time: 0.312 secs
[>] pruning results:
[L] [nodes] [pruned]
3 7 2
2 4 0
1 2 0
[>] computing sequence(s) likelihood ... (0.159 secs)
[>] 1 sequence(s) - min/max length: 27/27
[>] max. context length: L=3
[>] 2pruned nodes removed
[>] found 10 distinct context(s)
[>] total time: 0.007 secs
[1] [2] [3] [4] [5] [6] [7] [8] [9] [10]
[1] 0.4814815 0.6153846 0.625 0.5714286 0.75 0.75 0.5714286 0.75 0.75 0.5714286
[11] [12] [13] [14] [15] [16] [17] [18] [19]
[1] 0.75 0.25 0.6666667 0.5 0.5 0.4285714 0.3333333 0.4285714 0.6666667
[20] [21] [22] [23] [24] [25] [26] [27]
[1] 0.3333333 0.5714286 0.25 0.8 0.75 0.4285714 0.6666667 0.6666667
[>] 1 sequence(s) - min/max length: 27/27
[>] max. context length: L=3
[>] 2pruned nodes removed
[>] found 10 distinct context(s)
[>] total time: 0.007 secs
prob
[1] 7.58916e-08
PST documentation built on May 29, 2017, 5:16 p.m. | 1,694 | 5,421 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2019-04 | latest | en | 0.850555 |
https://indoxxiid.com/qa/question-how-many-unique-values-can-2-bits-hold.html | 1,618,284,856,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038072082.26/warc/CC-MAIN-20210413031741-20210413061741-00267.warc.gz | 427,294,056 | 9,144 | # Question: How Many Unique Values Can 2 Bits Hold?
## How many items can be represented with 2 bits?
A 2-bit system uses combinations of numbers up to two place values (11).
There are four options: 00, 01, 10 and 11.
A 1-bit image can have 2 colours, a 4-bit image can have 16, an 8-bit image can have 256, and a 16-bit image can have 65,536..
## What is the maximum number of values that can be represented in 7 bits?
Binary number representationLength of bit string (b)Number of possible values (N)6647128825695128 more rows
## What happens to values above 255 in binary?
An example of an 8-bit overflow occurs in the binary sum 11111111 + 1 (denary: 255 + 1). … Overflow errors happen when the largest number that a register can hold is exceeded. The number of bits that it can handle is called the word size . Most CPUs use a much bigger word size than 8 bits.
## How many bytes is 8 numbers?
Eight bytes contain 64 bits of information, so you can store 2^64 ~ 10^20 unique items using those bits. Those items can easily be interpreted as the integers from 0 to 2^64 – 1 . So you cannot store 302 decimal digits in 8 bytes; most numbers between 0 and 10^303 – 1 cannot be so represented.
## What is a group of 16 bits called?
hextetIn computing, a hextet is a sixteen-bit aggregation, or four nibbles. As a nibble typically is notated in hexadecimal format, a hextet consists of 4 hexadecimal digits. A hextet is the unofficial name for each of the 8 blocks in an IPv6 address.
## What is the largest number you can represent in 5 bits?
31Therefore, range of 5 bit unsigned binary number is from 0 to (25-1) which is equal from minimum value 0 (i.e., 00000) to maximum value 31 (i.e., 11111).
## How do you write 2 in binary?
When you get to “two”, you find that there is no single solitary digit that stands for “two” in base-two math. Instead, you put a “1” in the twos column and a “0” in the units column, indicating “1 two and 0 ones”. The base-ten “two” (210) is written in binary as 102.
## How many numbers can 64 bits represent?
As a recap, remember that the maximum number stored in a 64 bit register / variable is 2^64 – 1 = 18446744073709551615 (a 20 digit number).
## What’s the largest decimal value you can represent in binary with just 8 bits?
The largest number you can represent with 8 bits is 11111111, or 255 in decimal notation. Since 00000000 is the smallest, you can represent 256 things with a byte. (Remember, a bite is just a pattern.
## How many unique values can 8 bits represent?
256 different numbersFor example, an 8-bit number scheme can represent 256 different numbers.
## What is the largest binary number that can be expressed with 32 bits?
2,147,483,647The number 2,147,483,647 (or hexadecimal 7FFFFFFF16) is the maximum positive value for a 32-bit signed binary integer in computing.
## What happens every time you increase the number of bits by one?
Every time you increase the number of bits you’re working with by just one bit, you double the number of values that can be encoded. This is extremely relevant to digital imaging, since binary numbers are used internally by the computer to represent images.
## How many values can a bit hold?
two possible valuesThe name is a portmanteau of binary digit. The bit represents a logical state with one of two possible values. These values are most commonly represented as either “1”or”0″, but other representations such as true/false, yes/no, +/−, or on/off are common.
## How many unique numbers can you store in three bits?
One bit can store two values. 0 and 1 Two bits can strore four values 00, 01, 10, and 11 three bits can store eight values 000,001,010,011, 100,101,110 and 111.
## What’s the largest value you can represent in binary with just 3 bits?
7Explanation: The largest decimal number that we can represent with 3 bits is 7, if binary number system is unsigned that means you can’t represent any negative number in this system. Because all three bits are used in this system. The binary number is 111, which is equal to 7 in decimal.
## How many things can 3 bits represent?
One bit can store two values. 0 and 1 Two bits can strore four values 00, 01, 10, and 11 three bits can store eight values 000,001,010,011, 100,101,110 and 111. it means the combination of all the bits represents one value like 010 represents 2 and 101 represents 5.
## What is a group of 16 bits?
Each 1 or 0 in a binary number iscalled a bit. From there, a group of 4bits is called a nibble, and 8-bitsmakes a byte. … It could be 16-bits, 32, 64, or even more.
## What is 16 bits called?
2. There’s no universal name for 16-bit or 32-bit units of measurement. The term ‘word’ is used to describe the number of bits processed at a time by a program or operating system. So, in a 16-bit CPU, the word length is 16 bits. In a 32-bit CPU, the word length is 32 bits.
## How many values can 16 bits represent?
A 16-bit integer can store 216 (or 65,536) distinct values. In an unsigned representation, these values are the integers between 0 and 65,535; using two’s complement, possible values range from −32,768 to 32,767.
## What is the largest hexadecimal number that can be held in one word?
4,294,967,295The number 4,294,967,295, equivalent to the hexadecimal value FFFF,FFFF16, is the maximum value for a 32-bit unsigned integer in computing.
## What’s the smallest decimal number that you can represent with 3 bits?
Answer and Explanation: The smallest decimal number that you can represent with three bits is either 0 or -4. | 1,448 | 5,554 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2021-17 | longest | en | 0.899055 |
https://mathoverflow.net/questions/292417/what-is-the-value-of-this-simple-game-with-primes | 1,555,945,694,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578555187.49/warc/CC-MAIN-20190422135420-20190422161420-00376.warc.gz | 482,792,360 | 30,201 | # What is the value of this simple game with primes?
Consider the following game. Alice selects an integer $n$ from $[1,b]$, while Bob selects an integer $m$ from $(a,b]$ (for concreteness, you may choose $a=10^{10}$ and $b=10^{1000}$). Alice wins if $m-n$ is a positive prime, Bob wins otherwise. Both players are allowed to use mixed strategies (that is, select their integers at random according to any distribution they want). What is the value of this game? That is, what is the maximal winning probability $p^*$ Alice can achieve, no matter how Bob plays?
My preliminary calculation shows that, for $a$ fixed and $b$ large, $O((\log b)^{-2}) \leq p^* \leq O((\log b)^{-1})$. The problem is to significantly narrow the gap between these estimates.
Update (some details): If Bob selects $m$ uniformly at random from $(a,b]$, then, whatever $n$ Alice selects, her winning probability is at most $\frac{\pi(b-n)}{b-a} \leq \frac{\pi(b)}{b-a}$, where $\pi$ is the prime counting function, which is $O((\log b)^{-1})$ if $a$ fixed and $b$ large.
Next, let Alice select $n$ from $(2,b]$ with probability $q(n)=\frac{\log^2 n-\log^2(n-1)}{\log^2 b-\log^2 2}.$ Then, whatever $m$ Bob selects, Alice wins with probability $\sum\limits_{p\leq m}q(m-p)$, where the summation is over all primes $p$ not exceeding $m$. Because $q$ is non-increasing, $\sum\limits_{p\leq m}q(m-p) \geq \sum\limits_{i=0}^{\pi(m)-1}q(m-i)=\frac{\log^2 m-\log^2(m-\pi(m))}{\log^2 b-\log^2 2} > \frac{\log^2 m-\log^2(m-\pi(m))}{\log^2 b}=\frac{2}{\log^2 b}\int\limits_{m-\pi(m)}^m \frac{\log x}{x}dx > \frac{2}{\log^2 b}\int\limits_{m-\pi(m)}^m \frac{\log m}{m}dx = \frac{2}{\log^2 b}\pi(m) \frac{\log m}{m} > \frac{2}{\log^2 b}.$
• As a first step, one may consider a simpler game: Alice wins with probability $(\log(m-n))^{-1}$ if $m-n>2$, and $0$ otherwise (that is, replace primes with random set of integers with approximately the same density). What is $p^*$ in this case? – Bogdan Feb 7 '18 at 17:50
• What about negative primes? Does Alice win if, for example, $m-n=-13$? It would also help the reader if you showed your "preliminary calculation", to avoid duplication of effort. – Greg Martin Feb 7 '18 at 18:04
• In the randomized version, Alice can pick $1$ and win with probability at least $1/\log(b)$. – Julian Rosen Feb 7 '18 at 23:27
• In response to Greg Martin's comment, I have clarified that only positive prime counts, and displayed the calculation. – Bogdan Feb 8 '18 at 10:49
• [−1.5,1] in Bounty description should of course be [−1.5,-1]. – Bogdan Feb 12 '18 at 10:04 | 841 | 2,566 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2019-18 | latest | en | 0.765749 |
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# Earth Exploration Toolbook: Exploring Air Quality in Aura NO2 Data
In this chapter, students will explore relationships between air quality and population density using the image visualization tool, Google Earth. You will learn how to download NO2 data and analyze them to develop a conceptual understanding of how... (View More)
Audience: Middle school, High school
Materials Cost: Free
# Modeling Hot and Cold Planets: Activity A Modeling Hot and Cold Planets
In this activity, student teams design small-scale physical models of hot and cold planets, (Venus and Mars), and learn that small scale models allow researchers to determine how much larger systems function. There is both a team challenge and... (View More)
Audience: High school
Materials Cost: \$10 - \$20 per group of students
# Using Mathematical Models to Investigate Planetary Habitability: Activity B Making a Simple Mathematical Mode
In this activity, students build a simple computer model to determine the black body surface temperature of planets in our solar system: Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune and Pluto. Experiments altering the luminosity and... (View More)
Audience: High school
Materials Cost: Free per group of students
# Using Mathematical Models to Investigate Planetary Habitability: Activity C The Role of Actual Data in Mathematical Models
Students explore how mathematical descriptions of the physical environment can be fine-tuned through testing using data. In this activity, student teams obtain satellite data measuring the Earth's albedo, and then input this data into a... (View More)
Audience: High school
Materials Cost: Free per student
# Using Mathematical Models to Investigate Planetary Habitability: Activity A Finding a Mathematical Description of a Physical Relationship
In this activity, student teams learn about research design and design a controlled experiment exploring the relationship between a hypothetical planet, an energy source, and distance. They analyze the data and derive an equation to describe the... (View More)
Audience: High school
Materials Cost: Free per student
# Modeling your Water Balance
Students create a physical model illustrating soil water balance using drinking glasses to represent the soil column, and explain how the model can be used to interpret data and form predictions. Using data from the GLOBE Data Server, they calculate... (View More)
# Time Warp
In this inquiry investigation, students conclude that the motion of the Earth is linked to the changes we observe such as the length of the day. Students learn about the reason behind the Earth's time zones. An optional water clock and sand clock... (View More)
Keywords: Time; Orbit; Time zones
1 | 669 | 3,294 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2019-04 | latest | en | 0.801898 |
https://forums.civfanatics.com/threads/zones-of-control.682034/ | 1,685,240,897,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224643462.13/warc/CC-MAIN-20230528015553-20230528045553-00472.warc.gz | 295,334,441 | 21,225 | # [C3C]Zones of Control
#### AnthonyBoscia
##### Emperor
Is there any documentation that shows exactly how zone of control attacks are calculated? It seems, anecdotally to me anyway, that the unit with the ZoC uses its attack value rolled against the moving unit's defense value. Neither the manual nor the civilopedia spell this out precisely. How does bombard strength factor into this? In unmodded Civ 3, the radar artillery has a ZoC, but no attack value. Has this ever been properly tested or confirmed? Thank you in advance for your input.
#### Vuldacon
##### Dedicated to Excellence
Supporter
Anthony... Not exactly sure but because only 1 health is knocked off the Unit by all Units that have ZOC, it appears that the ZOC programing is set up to only injure another Unit by 1 health regardless of the attacking Power of the ZOC Unit.
#### Flintlock
##### King
It just so happens I was recently reading the game's logic for zone of control. The game does roll to apply damage based on the intercepting unit's strength and the moving unit's defense. The specific formula for the probability to take damage is: 1 - (D*16 / (D*16 + A)), where D is the moving unit's defense strength and A is the intercepting unit's attack or bombard strength, whichever is higher. The game will always select the strongest available unit to try to do damage.
#### AnthonyBoscia
##### Emperor
Very interesting. Thanks for checking this. That explains why high offense units are usually the ones to attack over artillery units.
#### Civinator
##### Blue Lion
Supporter
It just so happens I was recently reading the game's logic for zone of control. The game does roll to apply damage based on the intercepting unit's strength and the moving unit's defense. The specific formula for the probability to take damage is: 1 - (D*16 / (D*16 + A)), where D is the moving unit's defense strength and A is the intercepting unit's attack or bombard strength, whichever is higher. The game will always select the strongest available unit to try to do damage.
Your great formula is now a part of the CCM Ingame Help.
#### Alekseyev_
##### Warlord
You continue to be our hero, Flintlock
#### Vuldacon
##### Dedicated to Excellence
Supporter
Flintlock... I find it interesting that only 1 health point is reduced from the Moving Unit's Health Bar regardless of the Strength of the Defending SOC Unit.
There must be additional programing that limits the Damage to the moving Unit.
#### Theov
##### Deity
Flintlock... I find it interesting that only 1 health point is reduced from the Moving Unit's Health Bar regardless of the Strength of the Defending SOC Unit.
There must be additional programing that limits the Damage to the moving Unit.
The unit attacks one time.
The outcome is binary and costs one HP.
#### Vuldacon
##### Dedicated to Excellence
Supporter
The unit attacks one time.
The outcome is binary and costs one HP.
This was what I meant in Post #2. There is really no point to try to calculate ZOC Defending Units Strengths vs the Moving Units. One HP is removed.
That said, I have Not tested an extremely low strength ZOC Unit against an extremely high strength Moving Unit.
#### Alekseyev_
##### Warlord
What? Of course there is a point, Flintlock gave the formula. A unit with higher defense is less likely to lose the hit point, a unit with more attack is more likely to take it. And especially the defensive strength matters a lot as it is multiplied by 16.
#### Vuldacon
##### Dedicated to Excellence
Supporter
Alekseyev_ ... If you just want to know if 1 Hit Point will be removed by a Defending ZOC Unit, then the formula will help of course.
My Point is that when damage is done, only 1 Hit Point damage is done regardless of the Strengths of the Units.
Last edited:
#### Alekseyev_
##### Warlord
Yes, because no combat takes place. A unit moves past an enemy unit with ZOC, and then the game checks this formula to see if 1 HP is removed or not. It's not a normal combat or bombardement attack taking place, it's an entirely different function that only runs a single time, taking 1 HP or 0, with higher attack causing higher chances of 1.
#### Vuldacon
##### Dedicated to Excellence
Supporter
Combat has nothing to do with what I am saying... IF a Unit has Extreme Power, then 1 Hit should remove more than 1 Hit Point from another Unit
That is the Programing for ZOC Not considering the Power of a Hit from an Extremely Powerful Unit that would Remove much more than 1 Hit Point.
#### Flintlock
##### King
Combat has nothing to do with what I am saying... IF a Unit has Extreme Power, then 1 Hit should remove more than 1 Hit Point from another Unit
That is the Programing for ZOC Not considering the Power of a Hit from an Extremely Powerful Unit that would Remove much more than 1 Hit Point.
I've been thinking about this. I'm working on ZoC anyway right now, so might as well cover as many requests as I can. It would be possible, easy in fact, to insert some code that runs right after the game's base ZoC logic and applies additional damage depending on the units' stats. The question is how much additional damage should be done. When the intercepting unit uses its bombard strength, the natural thing to do would be to use its rate of fire to give it multiple chances to do damage. However when a unit uses its attack strength, I'm not sure what would make the most sense. I'm thinking of giving the unit an additional chance to do damage for each movement point it has over the defending unit. That would make sense since attack units with ZoC generally have high movement (ex. cavalry, modern armor), and the zone of control is presumably supposed to simulate hit-and-run attacks.
#### Vuldacon
##### Dedicated to Excellence
Supporter
Flintlock... ZOC for the Attacking Units:
Rate of Fire using Bombard Strength and Movement Point using Attack Strength would be most appropriate.
#### Ozymandias
##### In Terra Fantasia
Supporter
From "civ-wiki.de" -
"A unit that has a control zone has the chance to attack enemy units that cross it during the enemy round without any danger of its own.
• A single attack is carried out with a fraction of the attack strength of the unit with control zone against the defensive value of the moving unit. This has about sixteenths of the normal value (here more detailed investigations are necessary). The defense value of the moving units does not enjoy bonuses from the terrain.
• This chance only comes into play if the movement of the enemy unit starts in the control zone and leads to another field of the control zone.
• A unit with a control zone can use them as often as desired in one round.
• All units in a settlement (size 1 to 6) with a city wall automatically have a control zone. If the city grows to size 7 or higher, the city wall loses this effect.
• The control zone has no effect on units that have only one health point."
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4K | 1,597 | 7,018 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2023-23 | latest | en | 0.930979 |
https://www.gradesaver.com/textbooks/science/physics/college-physics-4th-edition/chapter-5-problems-page-188/55 | 1,679,808,070,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945433.92/warc/CC-MAIN-20230326044821-20230326074821-00585.warc.gz | 908,936,382 | 12,817 | ## College Physics (4th Edition)
Published by McGraw-Hill Education
# Chapter 5 - Problems - Page 188: 55
#### Answer
The radial acceleration component is $2.45~m/s^2$ The tangential acceleration component is $2.54~m/s^2$ The tension in the string is $11.9~N$
#### Work Step by Step
We can find the radial acceleration component: $a_c = \frac{v^2}{r}$ $a_c = \frac{(1.40~m/s)^2}{0.800~m}$ $a_c = 2.45~m/s^2$ The radial acceleration component is $2.45~m/s^2$ We can find the tangential acceleration component: $mg~sin~\theta = ma_t$ $a_t = g~sin~\theta$ $a_t = (9.80~m/s^2)~sin~15.0^{\circ}$ $a_t = 2.54~m/s^2$ The tangential acceleration component is $2.54~m/s^2$ We can consider the net force on the bob directed along the line toward the center of the circle. We can find the tension in the string: $\sum~F = \frac{mv^2}{r}$ $T - mg~cos~\theta = \frac{mv^2}{r}$ $T = mg~cos~\theta + \frac{mv^2}{r}$ $T = (1.00~kg)(9.80~m/s^2)~cos~15.0^{\circ} + \frac{(1.00~kg)(1.40~m/s)^2}{0.800~m}$ $T = 11.9~N$ The tension in the string is $11.9~N$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 432 | 1,203 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2023-14 | latest | en | 0.783332 |
https://www.physicsforums.com/threads/basic-conceptual-questions.209815/ | 1,508,193,240,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187820466.2/warc/CC-MAIN-20171016214209-20171016234209-00711.warc.gz | 1,222,613,064 | 14,535 | # Basic Conceptual Questions
1. Jan 19, 2008
### ian_durrant
I'm working on some homework and I've become stummped on two questions (one of them is wrong, the other is right). The bolded answer is the answer I've chosen.
1. The problem statement, all variables and given/known data
Question 3 10 points Save
Which one of the following quantities is a vector quantity?
A. the age of the earth
B. the mass of a freight train
C. the earth's pull on your body
D. the temperature of hot cup of coffee
E. the number of people attending a soccer game
Question 4
Two vectors A and B are added together to form a vector C. The relationship between the magnitudes of the vectors is given by A + B = C. Which one of the following statements concerning these vectors is true?
A. A and B must be displacements.
B. A and B must have equal lengths.
C. A and B must point in opposite directions.
D. A and B must point in the same direction.
E. A and B must be at right angles to each other.
Anyone have any thoughts?
thanks for any help
2. Jan 19, 2008
### plutoisacomet
Hello and welcome. Your questions will not be answered directly due to the rules of the forum. I see both questions are vector questions.
For question 3, look at each selection and see if you can assign it a Magnitude and a Direction. If so, you have a Vector. If not, you have a scalar.
For Questiion 4 Just apply your difinition of a vector and the qualities of a vector and you will easily answer the question yourself.
Good luck | 374 | 1,510 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2017-43 | longest | en | 0.942479 |
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# The line represented by equation y=x is the perpendicular
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The line represented by equation y=x is the perpendicular [#permalink]
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27 Oct 2012, 20:28
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The line represented by equation y=x is the perpendicular bisector of line segment AB. If A has the coordinates (-3,3), what are the coordinates of B?
[Reveal] Spoiler:
Hello Friends,
I would like you to help me with coordinate geometry. This is a problem from MGMAT.
I understand that the perpendicular bisector slope is 1 so the slope os segment AB is −1. so by substituting the values we get b.
y=mx+b
3=-1(-3)+b
0=b
line containing segment AB is y=-x
For these two lines
y=x———(1)
y=-x———(2)
so x=-x
then how to proceed?
how can we determine x=0 and y=0 and find midpoints.
The answer is B(3,-3)
Can you put some light?
Please explain step by step as my maths is pretty weak.
Thank you
Last edited by Bunuel on 29 Oct 2012, 06:51, edited 1 time in total.
Renamed the topic, edited the question and moved to PS forum.
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Re: coordinate geometry [#permalink]
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27 Oct 2012, 23:24
1
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pranav123 wrote:
Hello Friends,
I would like you to help me with coordinate geometry. This is a problem from MGMAT. The line is represented by equation y = x is the perpendicular bisector of line segment AB. If AB has the coordinates (-3,3), what are the coordinates of B?
I understand that the perpendicular bisector slope is 1 so the slope os segment AB is −1. so by substituting the values we get b.
y=mx+b
3=-1(-3)+b
0=b
line containing segment AB is y=-x
For these two lines
y=x———(1)
y=-x———(2)
so x=-x
then how to proceed?
how can we determine x=0 and y=0 and find midpoints.
The answer is B(3,-3)
Can you put some light?
Please explain step by step as my maths is pretty weak.
Thank you
From the equations, you can see that the line y=-x is bisected at the origin by y = x. So the midpoint of segment AB is (0,0).
Midpoint of a line segment having points $$A(x_1,y_1) and B(x_2,y_2)$$ is $$(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$$
We have the values of the point A and the midpoint. So point be can be easily found.
Kudos Please... If my post helped.
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Re: The line represented by equation y=x is the perpendicular [#permalink]
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29 Oct 2012, 06:54
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Expert's post
pranav123 wrote:
The line represented by equation y=x is the perpendicular bisector of line segment AB. If A has the coordinates (-3,3), what are the coordinates of B?
[Reveal] Spoiler:
Hello Friends,
I would like you to help me with coordinate geometry. This is a problem from MGMAT.
I understand that the perpendicular bisector slope is 1 so the slope os segment AB is −1. so by substituting the values we get b.
y=mx+b
3=-1(-3)+b
0=b
line containing segment AB is y=-x
For these two lines
y=x———(1)
y=-x———(2)
so x=-x
then how to proceed?
how can we determine x=0 and y=0 and find midpoints.
The answer is B(3,-3)
Can you put some light?
Please explain step by step as my maths is pretty weak.
Thank you
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Re: The line represented by equation y=x is the perpendicular [#permalink] 29 Oct 2012, 06:54
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# The line represented by equation y=x is the perpendicular
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We come from the future
# The Coriolis Effect Part II: As the Hurricane Turns
The Coriolis Effect pushes objects clockwise in the northern hemisphere and counterclockwise in the southern hemisphere. And yet hurricanes spin in exactly the opposite direction. Why?
Yesterday we saw how the movement of the earth affects certain objects. When seen from above, the objects are seen to move in a straight line. From the general human point of view – on that's stuck on a rotating planet – the objects appear to drift to the right or left as the world literally turns out from under them. In the northern hemisphere objects appear, from the thrower's point of view, to curve to the right as they are thrown. In the southern hemisphere they appear to curve to the left.
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The mechanics of this are best understood when walked through physically. You'll need some object to mark a spot on the floor, and if possible, a friend. Put a book or a plate or some object down on the floor a few feet away from you. This is a low pressure system. A low pressure system is created when air is heated enough to lift upwards, creating a relative vacuum below. Higher pressure air from around the system rushes in to fill the vacuum. That's you.
(To get into character, picture yourself wooshing through these palm trees.)
We'll pretend that you're in the northern hemisphere. This means the earth turns to the left under you, and your course appears to be diverted to the right. The low pressure system is pulling you in. Take a slow step towards the low pressure system, and have your friend pull on your right sleeve. If you're alone, imagine that someone is turning you sharply to the right. If you've done it correctly, you'll be a little closer to the low pressure system, but you'll have turned to the right, so the system will be on your left. | 493 | 2,331 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2021-17 | latest | en | 0.951461 |
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Ball_mill_Finish Mill Ball Charge Calculation - ScribdBall Mill . t 50 40 30 25 20 17 15. % 0 0 15 20 22 23 20 100. 0 0 26.2 34.9 38.4 40.2 34.9 174.6. 10/21/2012 t 0 0 15.72 20.94 23.04 24.12 20.94 104.76 60 104.76. 60% % 0 0 15 20 22 23 20 100. 80% t 0 0 21 28 31 32 28 140 80. Chamber 1 - 11 m L x Φ 4.4 m 50 40 30 25 20 17 15. Total. 0 0 26.2 34.9 38.4 40.2 34.9cement ball mill grinding media calculation,cement ball mill grinding media calculation,Calculations of media charge - Jyoti Ceramic Industries Pvt. Ltd.Calculations of media charge. To calculate media charge for batch mill. M = 0.000929 x D2 x L where : M = Grinding media charge in kgs. D = Internal Dia of the Mill in cms. after lining. L = Internal length of the mill in cms. after lining. To calculate grinding media charge for continuous type ball mill, M = 0.000676 x D2 x L
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### Untitled - International Cement Review
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Aug 25, 2016 . This is a simple video slideshow, if you want to know more details, please click on our website spellightbaptistschool , we will provide a p.
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### how to calculate 90 mm ball charge cement mill
Large balls are used at the inlet, to crush clinker nodules (which can be overmm in diameter). Ball diameter .. how to calculate 90 mm ball charge cement mill They are the only supplier offering a wide range of grinding media . (cement, mineral, …) in a mill. . He will also calculate the mill's optimum filling . Read More >>.
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Sep 15, 2005 . Keywords: ceramic ball milling media, mass wear rate model, theory developing, alumina ceramic ball, experiment testing. 1. Introduction. Since ball mill was discovered, it has become a most common and non-fungible grinding methods in many industries such as steel, ceramic, cement, glass and.
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### Calculation of the power draw of dry multi-compartment ball mills
May 6, 2004 . In this paper, a new approach for the calculation of the power draw of cement grinding ball mills is proposed. . Vp. – Fraction of mill volume loaded with balls. fCs. – Fraction of critical speed. Ss. – Ball size factor. To determine the power that a dry grinding needs, full grate discharge mill. Equation 1 is.
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### Correlations for the Grindability of the Ball Mill As . - Semantic Scholar
Finally the calculated values of the fines in terms of the grindability of the mill obtained through the . rubber, textiles, sintering, cement and line, powders for the detergent industry, pulverized coal for power . balls, time of grinding, particle density and speed of the ball mill (rpm) have been considered for the present.
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The Ball Mill Abrasion (BMA) test was developed by the University of Queensland Materials Performance group to simulate the high stress abrasion conditions experienced in production ball mills. It gives accurate predictions of the service life of wear resistant alloys in mill liner and grinding media applications. In this paper.
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### Grinding media sorting and balls | Magotteaux
Grinding media are the means used to crush or grind material (cement, mineral, …) in a mill. Grinding media sorting is performed when the ball load wears out. . He will also calculate the mill's optimum filling degree according to several parameters such as mill type and size, material to be ground, specific operating.
### Improvement in Performance of Ball Mill in Cement Grinding - IJESC
Clinker is grinder with impact and friction forces inside the cylinder. Grinding media which is usually called as grinding balls are made up of high chromium steel. A cement mill is the comminution machine which is used to grind the hard and big sized clinker that comes from kiln section of cement industry into the fine optimal.
### Breakage Characteristics of Heat-Treated Line . - MDPI
Jan 12, 2018 . Keywords: ball mill; line; population balance model; back calculation; grinding. 1. Introduction . Cement and steel industries account for a high percentage of this usage, and in Korea, the steel and cement .. The grinding media were stainless steel balls with a diameter of 2.54 cm and the mill.
### Process Diagnostic Studies for Cement Mill Optimization
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If it is also assumed that the mean overall values of Si for a mixture of balls is the weighted mean of Si values for each ball size, equations are derived for calculating this mean value. As an example, the results are used in a mill simulation to show the quantitative effect of different ball mixes in a two-compartment cement mill.
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desired particle size. Grinding mills are classified by the media used for grinding. The more common types of mills are: Ball mills - These mills use forged steel balls up to 5 inches (127 mm) in diameter to crush and . The required clutch torque is calculated from the power rating of the mill motor, clutch shaft rpm and an.
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### The cement mill - Understanding Cement
Cement clinker is usually ground using a ball mill. This is essentially a large rotating drum containing grinding media - normally steel balls. As the drum rotates, the motion of the balls crushes the clinker. The drum rotates approximately once every couple of seconds. The drum is generally divided into two or three chambers,. | 2,086 | 8,759 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2019-09 | latest | en | 0.663064 |
http://dahl.mines.edu/courses/dahl/alg/Ans5_1c.html | 1,726,350,038,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651580.74/warc/CC-MAIN-20240914193334-20240914223334-00758.warc.gz | 9,040,714 | 2,551 | ### (2) is the correct answer.
Because:
x3y-y3x = x2 -y2 = (x-y)(x+y) = x+y
x2y-xy2 x-y (x-y) | 55 | 121 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2024-38 | latest | en | 0.539672 |
http://www.cactuspear.org/cgi-bin/main.cgi?dir=mathematics&page=pythagorean | 1,563,244,796,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195524475.48/warc/CC-MAIN-20190716015213-20190716041213-00031.warc.gz | 192,248,588 | 1,947 | ## the pythagorean theorem
Euclid of Alexandria proved that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the two sides. Here are two pictorial demonstrations. The first is a version of a Chinese proof by dissection that dates from 200 BC to 200 AD. In the diagram below at left, we wish to show that the area of the red square is equal to the sum of the areas of the yellow and blue squares.
We construct the figure at the upper right by removing the red square from the figure at left and filling in the large square that circumscribes the yellow and blue squares. This is accomplished by adding three gray triangles, each of them congruent to the gray triangle in the figure at left. The figure at the bottom right is obtained by removing the yellow and blue triangles from the figure at left, and adding three more gray triangles.
The second demonstration is by tessellation:
Left: We've tiled the plane in two ways. First with blue and yellow squares whose sides represent the sides of a right triangle. Second with diagonal squares, one of them pictured in red, whose sides are equal to the hypotenuse of the original triangle.
We see that any given portion of a plane tiled in this way contains the same number of red, yellow, and blue squares, and the result is established.
Right: We can rearrange the pieces of the complete yellow and blue squares to form a complete red square.
First, move the triangular part of the yellow square that is outside the red square to the lower left part of the red square, as shown.
Next, move the large trianglar portion of the blue square that is outside the red square to the upper left part of the red square, as shown.
Finally, move the remaining small blue triangle that is outside the red square to the upper right part of the red square, as shown. The rearranged portions of the yellow and blue squares completely fill the red square, and the result is established.
Source: The Pythagorean Theorem: A 4,000 Year History, by Eli Maor
cactuspear home | 446 | 2,051 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2019-30 | latest | en | 0.925233 |
https://www.jiskha.com/questions/933897/If-fixed-costs-are-350-000-the-unit-selling-price-is-29-and-the-unit-variable | 1,539,696,180,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583510749.55/warc/CC-MAIN-20181016113811-20181016135311-00348.warc.gz | 979,034,958 | 4,860 | # Accounting
If fixed costs are \$350,000, the unit selling price is \$29, and the unit variable costs are \$20, what is the break-even sales (units) if the variable costs are decreased by \$4?
26,924 units
12,069 units
21,875 units
38,889 units
1. 0
1. 12,069
posted by Benjamin
2. 26,924
posted by Kaylee
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More Similar Questions | 714 | 2,838 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2018-43 | latest | en | 0.893177 |
https://www.piramalfinance.com/vidya/detailed-information-on-how-to-manage-debt-to-income-ratio/ | 1,713,497,283,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817253.5/warc/CC-MAIN-20240419013002-20240419043002-00448.warc.gz | 849,726,436 | 11,814 | How To?
# Detailed Information On How To Manage Debt To Income Ratio
Personal Finance
08-11-2023
Debt-to-income ratio (DTI) compares the monthly loan payments to that the monthly salary. It indicates the part of the net monthly income (pre-tax). It helps to pay off the rent, mortgage, credit card, and other things.
A person’s debt-to-income ratio (DTI) impacts their overall financial health. They can decide whether or not they should apply for credit by using the DTI calculator to assess their comfort level with their current debt. The ratio is expressed as a percentage. Lenders use it to assess your ability to handle your monthly payments. Whether one is fully capable of repaying a debt.
Higher DTI ratios are typically associated with riskier borrowers.
## What elements comprise a DTI ratio?
Mortgage lenders calculate a Debt-to-income ratio using two factors: A front-end ratio and a back-end ratio. Let’s see how it’s done:
• The ratio of front-end. It is also known as the housing ratio. It shows what proportion of your gross monthly income would go into housing costs. Such as your payment of debt and property taxes. Also, home-owners insurance and HOA dues.
• The ratio of back-end: Indicates the percentage of your income required to fulfil all of your monthly debt commitments, in addition to your mortgage and housing costs. This includes any revolving debt that appears on your credit reports. Such as credit card debt and vehicle loans. Child support, student loans, and child custody obligations as well.
## Calculation Of Debt-To-Income Ratio
Some simple steps to follow to calculate the Debt-to-income ratio are:
• Count up all of your monthly expenses. These payments could be minimum credit card payments. Auto, student, or personal loan payments. Alimony payments or child support payments. Any other debt repayments that appear on your credit report every month.
• You can calculate it by dividing your monthly debt payments by your gross monthly income.
• Your DTI ratio can be calculated by converting the number into a percentage.
Remember, this figure does not include other regular monthly expenses like daily utilities, insurance payments, medical costs, daycare, etc. These budget systems won’t be taken into account by your lender while assessing how much money to lend you.
## Why Is This Ratio Important?
A high Debt-to-income ratio can indicate that a person has too much debt. It can be relative to their monthly income. Creditors that have low debt-to-income ratios are much better. They can manage their monthly loan payments. As per result, before giving any loan or credit to a potential borrower, banks want to see low DTI percentages. The proper front-end ratio, according to lenders, should be no higher than 28%. The back-end ratio, which takes into account all costs, should be no higher than 36%. Because you’ll be paying off more debt, lowering your credit use ratio will also assist raise your credit score. The maximum DTI ratio a creditor can have yet approved for credit is 43%, as per general rules.
## Consequences Of High Debt-To-Income Ratio
Your financial life may be impacted in several negative ways. If your debt-to-income ratio exceeds the acknowledged benchmark of 43%. Some of them are:
• Less financial mobility – You have less money to save, invest, or spend. If a sizable amount of your salary is going toward paying off debt-restricted
• Confined eligibility for mortgages- If your debt-to-income ratio is over 43%. You might not be eligible for an Eligible Credit. It might only be approved for more stringent or costlier mortgages.
• Worse conditions when you borrow money or apply for credit. When your debt-to-income ratio is high, you will be viewed as a riskier debtor. Lenders may impose stricter restrictions. Higher interest rates and harsher punishments. It is for missed or late payments. Also, when approving loans or financing for risky borrowers.
## Managing The Ratio And Keeping It Low
How should one attend this? Here are a series of steps which can help.
• Create a budget to keep accurate accounting records. Cut back on pointless purchases. Put more money toward paying off debt.
• Make a strategy to pay off your debts.
• Reduce the cost of your debt. Look into strategies to lower your rates if you have high-interest credit cards. Call your credit card provider first. See whether the interest rate can be lowered. Check If your account is sufficient with no credits. Are you consistently make your bill payments on time? You might have more success trying this approach.
• Don’t accumulate more debt. Don’t use your credit cards for important expenditures. Take out new loans for them. This is crucial before and during a property purchase.
Maybe your debt-to-income ratio doesn’t directly affect your credit score. But lenders or credit issuers will probably ask for your income. This is when you apply. Your debt-to-income ratio will be considered just as your credit score will be one consideration in their examination of your application. Due to this, keeping a healthy debt-to-income ratio might be just as crucial. It is for obtaining a loan or credit as having a high credit score.
## Conclusion
Keeping high credit is as important as keeping low debt. For instance, make sure the debt-to-income ratio is low. It is to maintain stability. Maintain a manageable DTI ratio. It indicates where you can handle the debt. This may increase your ability to obtain financial goods. Be very smart, especially when it comes to taking any financial decision. Check your progress each month. Do this by tallying up your debt-to-income ratio. You may maintain your reasons to continue your debt modestly. Do this by observing your DTI decline. Don’t lose hope. Make the best out of the scenario. For more information on relatable topics, visit Primal Finance.
Know More
Know More | 1,210 | 5,895 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-18 | latest | en | 0.951945 |
https://social.msdn.microsoft.com/Forums/en-US/fd6a09c6-8225-4460-8160-25595ff39c1b/how-to-find-a-specific-coordinate-for-small-basic?forum=smallbasic | 1,600,628,151,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400198287.23/warc/CC-MAIN-20200920161009-20200920191009-00372.warc.gz | 640,965,914 | 15,153 | # How to Find a Specific Coordinate for Small Basic?
• ### Question
• Hi and thanks for reading my request.
Currently, I am running into problems trying to figure out what the X and Y coordinates are for moving object named Ball. I have published the small basic game so the code can be seen by importing it into your small basic program by importing FGZ797.
To shoot the bullet, you must press the key 'A.'
Is there any coordinate or command to find out the coordinates of the moving object and apply it? I am using Shape.Animation to make the Ball move, however, I would like the ball to stop when the bullet enters anywhere in the Ball's terrain. Is there any possible way to do this with the current code I have? I am very new to small basic and would greatly appreciate any intended help.
My main concern really is what should I put into the If command here highlighted in bold. I am unsure on what to do to get the desired effect of the ball stopping when the bullet "hits."
If (Bulletx = x) Then
Hit = "true"
If Hit = "true" Then
Shapes.Animate(Ball,x,y,0)
Sound.PlayClickAndWait()
Program.Delay(200)
GraphicsWindow.Clear()
Ship()
Endif
Endif
Once again, thanks for all the help. Any support would be appreciated.
Thursday, March 18, 2010 5:55 PM
• You need to know the position of the target and bullet, X and Y coordinates of both.
You can then check for a hit by checking if they overlap, or perhaps easier is to check the distance between the objects.
If the bullet is at (bX,bY) and the target is at (tX,tY), then the distance between them is
dist = Math.SquareRoot((bX-tX)*(bX-tX) + (bY-tY)*(bY-tY))
Then, in your if you need:
If (dist < distMin) Then
Hit = "True"
...
where distMin is some variable holding a minimum distance before they collide.
Also, note that shapes are drawn from the top left corner, so if you want to position an object at (X,Y) and its width is W and height is H, then you need:
Shapes.Move(shape,X-W/2,Y-H/2)
This is using the Shapes.Move , which will be much easier than Shapes.Animate, because you always know where the shape is since had to give it the coordinates - this is how all of the action type games are made in SmallBasic, the Animate isn't really very interactive as you discovered, since we don't know where it is while it is moving.
Also, looking at your code, you should consider having a main game loop, where all the movement update is done and just calls subroutines to do specific tasks and act on the variable changes triggered by the event calls.
You are well on the way though.
This is some of the ideas in your code - by no means finished in any way FGZ797-0.
Thursday, March 18, 2010 7:54 PM
### All replies
• You need to know the position of the target and bullet, X and Y coordinates of both.
You can then check for a hit by checking if they overlap, or perhaps easier is to check the distance between the objects.
If the bullet is at (bX,bY) and the target is at (tX,tY), then the distance between them is
dist = Math.SquareRoot((bX-tX)*(bX-tX) + (bY-tY)*(bY-tY))
Then, in your if you need:
If (dist < distMin) Then
Hit = "True"
...
where distMin is some variable holding a minimum distance before they collide.
Also, note that shapes are drawn from the top left corner, so if you want to position an object at (X,Y) and its width is W and height is H, then you need:
Shapes.Move(shape,X-W/2,Y-H/2)
This is using the Shapes.Move , which will be much easier than Shapes.Animate, because you always know where the shape is since had to give it the coordinates - this is how all of the action type games are made in SmallBasic, the Animate isn't really very interactive as you discovered, since we don't know where it is while it is moving.
Also, looking at your code, you should consider having a main game loop, where all the movement update is done and just calls subroutines to do specific tasks and act on the variable changes triggered by the event calls.
You are well on the way though.
This is some of the ideas in your code - by no means finished in any way FGZ797-0.
Thursday, March 18, 2010 7:54 PM
• I love litdev's solutions because so often they go beyond programing basic and include some branch of science.
I think another approuch or slight variation of his(?) solution is to split one move or animation into a series of moves. If you where to split it into 2 moves/animations you could check locations (Moves -1) or 1 times. If you were to split the move into 100 moves using a simple for loop, for example you would have 99 checks and a much better chance of trapping a collision. At some point the number of splits would become unporductive and an interesting question would be to determine that point.
If you want to crank it up a level or two, what if you where to take the faster object, in this case the bullet, and plot it's location by a fraction of a second. Again the smaller the time interval the more accurate the collision trap. Applying some physics, you should also be able to plot the direction and speed of the ball after the collision. How you would do this we would need Litdev's science brain!
Thursday, March 18, 2010 8:36 PM | 1,251 | 5,171 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2020-40 | latest | en | 0.919541 |
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Keep reading to learn more about Word search solver cheat and how to use it. Math can be difficult for some students, but with the right tools, it can be conquered.
## The Best Word search solver cheat
Word search solver cheat can be found online or in mathematical textbooks. Algebra is one of the most important and valuable subjects any student can learn. But it can also be one of the hardest. That’s why it’s so important to have a good understanding of basic algebraic concepts before you even step foot into your first math class. When you’re first learning algebra, the best way to go about it is to break down each problem into smaller, more manageable pieces. This will make it easier to understand how each piece fits together and how they relate to each other. Another great way to make sure you understand what you’re doing when you’re solving algebra problems is to use a math calculator. They allow you to break down your problems into easy-to-understand steps and will save you time and frustration in the long run. If you find that you’re having a hard time with algebra, don’t hesitate to ask for help. It might just be that you need a refresher on some of the basics before diving into more complex problems.
If that leaves you with an imaginary number, then that is your factor. You can also check to see if one of the roots is a perfect square (the square root of a perfect square is a perfect cube). There are many ways to factor quadratics: - 1st Degree - 2nd Degree - 3rd Degree - 4th Degree - 5th Degree - 6th Degree Factoring quadratics is also called graphing quadratics. To graph a quadratic, set up a coordinate system (x axis, y axis) and plot points on the graph from left to right at intervals of . The coordinates must be in increasing order (horizontal) and must start at the origin. The slope of a line is defined by the ratio of its rise to its run. If a point has an absolute value greater than 1, it will move rightward (positive x direction). If it has an absolute value less than 1, it will move downward (negative x direction). If it has an absolute value of 0, it will stay put (no
The HCF can also be used to simplify a problem by eliminating one or more smaller factors from the numerator or denominator. . . . The HCF can also be used to simplify a problem by eliminating one or more smaller factors from the numerator or denominator.
The matrix 3x3 is sometimes referred to as the “cross product” of three vectors. The following diagram illustrates a 3x3 matrix. The numbers in the matrix indicate which planes are being crossed. For instance, if row 1 is on the top left and row 2 is on the top right, then these two rows are being crossed. Similarly, if row 1 is on the bottom left and row 2 is on the bottom right, then these two rows are being crossed. In general, if any two rows are on opposite sides of a given plane, then both rows will be crossed by that plane. For example, a 3x3 with row 1 and column 2 on opposite sides of the x-axis will be crossed by all three planes: xy (row 1), yz (row 2) and zxy (row 3). A 3x3 with row 1 and column 2 both above or below the y-axis will only be crossed by one plane: xy (row 1). The numbers in each column indicate which submatrices they belong to. For example, if row 1 belongs to column 2 and row 3 belongs to column 4, then those two rows belong to submatrices C2 and C4, respectively. Likewise, if any three columns have their numbers in common, then they belong to submatrices C2xC3 and
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posted by on .
Suppose that a firm is the only domestic producer of a commodity and that
there are no imports of the good. The firm’s total cost and demand curves are
given by the following two equations:
TC = 6Q + ·05Q2 : Q = 360 - 20P
q) The government wishes to impose a maximum price of \$14 on the
commodity. Assuming that the firm is a profit maximiser, what
quantity will it produce and what will be the level of its profits.
p.s. originally, p=15 and q=60
• economics-micro - ,
Hummm.
This firm should act like a monopolist. Your P=15 and Q=60 are what the monopolist would do sans any government intervention.
With a price cap below what the monopolist would charge, simply plug 14 into the demand equation and solve for Q. Total revenue will be 14*Q. Plug the Q into the total cost equation then calculate profit. | 213 | 849 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2017-22 | latest | en | 0.928886 |
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How do I sum across dimensions other than time?
Hello
Can't see how this was done in the allocations video. I have a list for products, and a line item for sales. I want to introduce a line item that returns the total sales for all products against every product. 0x80070490 Beyond allocation use, is there a way to introduce a line item that returns the value at the next level up a multi-level hierarchy? - Assume we have a three level product hierarchy: Product>Product Group>Total Products. Such a function would enable me to return the % contribution that each Product contributes to its Product Group, and the contribution that each Product Group contributes to Total Products.
thanks
iosman
2 REPLIES 2
Community Boss
Re: How do I sum across dimensions other than time?
Hi iosman,
This is certainly possible. Simply create a new line item called Total Sales, without any dimensions and with the formula as Sales.
Then create a third line item called % of Total Sales with the formula as Sales/Total Sales
This works because your products hierarchy has a top level called Total Products
Please let me know if this helps!
Regards,
Anirudh
Highlighted
Certified Master Anaplanner
Re: How do I sum across dimensions other than time?
Hi Iosman,
Unfortunately I have a very complicated answer for your very simple question, but it does work. It will require a few more line items though! If I understand correctly, you want to see the percentage of each product relative to its parent. As in the following screenshot:
Sidenote: I have a rather standard way to determine which level of the hierarchy the cell is refering to. I use this for filtering tables where I want to show a certain level, while I want to use synchronize selection on the dashboard (selecting levels in a view, disables the synchronize selection function), but it can also be useful for these kinds of formulas. In order for this to work, you need to know how many levels the list will have and it will require maintenance if you add another level.
Here are the steps to create this result:
1. Create a line item "Ratio", set the value to 1 and set the summary to 'MIN'
2. Create a line item "Lvl 2", set the value to set the dimension to Product Group and set the summary to 'Ratiowith parameters: "Ratio / Ratio"
3. Create a line item "lvl 3", set the value to 3 and set the summary to 'Ratio' with parameters: "Lvl 2 / Ratio"You now have a line item that shows a 1 when it is on total products, a 2 when it is on a product group and a 3 when it is on a product:
4. Add a Product Group line item with format Product Group which refers to the parent of product PARENT(ITEM(Product))
5. Add a "Sales product group" line item which does a look up on Sales using the Product Group line item: Sales[LOOKUP: Product group]
6. Add the last line item "Sales %" with as summary "FORMULA":
IF 'Lvl 3' = 1 THEN 1 ELSE IF 'Lvl 3' = 2 THEN Sales / Sales[SELECT: Product Group.All Products] ELSE Sales / Sales product group
7. The formula checks the level and determines its calculation based on that level so,
1. If 'lvl 3' = 1 that means the current product dimension is on Total Products, thus 100%.
2. If 'lvl 3' = 2 that means the current product dimension is on Product Group, thus Sales devided by the total sales.
3. If 'lvl 3' = 3 that means the current product dimension is on Product, thus Sales devided by the sales of the relative product group.
Hope this is not too complicated!
KR,
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http://www.enotes.com/homework-help/find-empirical-formula-compound-306758 | 1,462,492,426,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461861700245.92/warc/CC-MAIN-20160428164140-00130-ip-10-239-7-51.ec2.internal.warc.gz | 491,084,355 | 11,558 | # Find the empirical formula of the following compound.A 2.203 sample of an organic compound was extracted from a plant. When it was burned in oxygen, the hydrogen in the compound was converted to...
Find the empirical formula of the following compound.
A 2.203 sample of an organic compound was extracted from a plant. When it was burned in oxygen, the hydrogen in the compound was converted to 1.32g of water and the carbon was oxidised to 3.23g of carbon dioxide.
Asked on
### 1 Answer |Add Yours
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The mass of the organic sample that was taken from the plant is 2.203 g. It was burned in oxygen and the products were 1.32 of water and 3.23 g of carbon dioxide. The empirical formula of the organic compound has to be determined based on this information.
It is assumed that the compound consists of only hydrogen and carbon.
2 g of hydrogen reacts with 16 g of oxygen to form 18 g of water following the chemical reaction 2H + O --> H2O
12 g of carbon reacts with 32 g of oxygen to form 44 g of CO2 following the chemical reaction C + O2 --> CO2
1.32 g of water is produced when (1.32/18)*2 = 11/75 g of hydrogen is burned. 3.23 g of carbon dioxide is produced when (3.23/44)*12 = 969/1100 g of carbon is burned. The sum of the mass of hydrogen and that of carbon dioxide is 1.027 g. This is not equal to the mass of the sample given. The data provided is not consistent.
We’ve answered 320,440 questions. We can answer yours, too. | 368 | 1,454 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2016-18 | longest | en | 0.973582 |
https://www.doubtnut.com/qna/645610851 | 1,720,806,855,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514450.42/warc/CC-MAIN-20240712161324-20240712191324-00655.warc.gz | 592,472,828 | 36,737 | # The total kinetic energy of a body of mass 10 kg and radius 2 m moving with a velocity of 15 m/s without slipping is 2250 joule. The radius of gyration is
A
1m
B
2m
C
1.89m
D
1.71m
Video Solution
Text Solution
Verified by Experts
## Total kinetic energy of the doby is 12mv2(1+K2R2) 2250=12×10×(15)7(1+K2(2)2) 2250×210×225=1+K24 ⇒1×4=K2 K=2m
|
Updated on:21/07/2023
### Knowledge Check
• Question 1 - Select One
## The total kinetic energy of a body of mass 10 kg and radius 0.5 m moving with a velocity of 2 m/s without slipping is 32.8 joule. The radius of gyration of the body is
A0.25 m
B0.2 m
C0.5 m
D0.4 m
• Question 2 - Select One
## A circular disc of mass 0.41 kg and radius 10 m rolls without slipping with a velocity of 2 m/s. The total kinetic energy of disc is
A0.41 J
B1.23 J
C0.82 J
D2.4 J
• Question 3 - Select One
## The M.I. of a wheel of mass 8 kg and radius of gyration 25 cm is
A5kgm2
B1.5kgm2
C2.5kgm2
D0.5kgm2
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Doubtnut is the perfect NEET and IIT JEE preparation App. Get solutions for NEET and IIT JEE previous years papers, along with chapter wise NEET MCQ solutions. Get all the study material in Hindi medium and English medium for IIT JEE and NEET preparation | 620 | 1,942 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2024-30 | latest | en | 0.81649 |
https://study.com/academy/answer/a-piano-tuner-hears-one-beat-every-2-0-s-when-trying-to-adjust-two-strings-one-of-which-is-sounding-440-hz-a-how-far-off-in-frequency-is-the-other-string.html | 1,576,406,443,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541307813.73/warc/CC-MAIN-20191215094447-20191215122447-00423.warc.gz | 542,016,303 | 23,452 | # A piano tuner hears one beat every 2.0 s when trying to adjust two strings, one of which is...
## Question:
A piano tuner hears one beat every {eq}2.0 \ s {/eq} when trying to adjust two strings, one of which is sounding {eq}440 \ Hz {/eq}.
(a) How far off in frequency is the other string?
## Beat frequency
Beats occur when two sound waves of slightly different frequencies {eq}f_h {/eq} and {eq}f_l {/eq} interferes where {eq}f_h(f_l) {/eq} is the higher (lower) frequency. The beat frequency, or the number of beats per unit of time, is
{eq}f_\text{beat} = f_h - f_l. {/eq}
The beat frequency is {eq}f_\text{beat} = 1 \ \text{beat}/2 \ \text{s} = 0.5 \ \text{Hz} {/eq}. The question is how far off is the frequency of the other string. This is simply the beat frequency 0.5 Hz, i.e. the other string is 0.5 Hz greater or smaller than the 440 Hz string. | 267 | 865 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2019-51 | longest | en | 0.875856 |
https://www.numbersaplenty.com/8333712391488 | 1,606,329,727,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141184123.9/warc/CC-MAIN-20201125183823-20201125213823-00210.warc.gz | 802,254,864 | 3,431 | Search a number
8333712391488 = 2631127915130071
BaseRepresentation
bin111100101000101011111…
…1101110111110101000000
31002111200202202010100201120
41321101113331313311000
52043014420333011423
625420242110432240
71520043106354341
oct171212775676500
932450682110646
108333712391488
112723342427200
12b27167502680
13485b33c599c3
1420b4d4cb40c8
15e6ba4110ee3
hex79457f77d40
8333712391488 has 336 divisors, whose sum is σ = 24706500833280. Its totient is φ = 2476805760000.
The previous prime is 8333712391427. The next prime is 8333712391513. The reversal of 8333712391488 is 8841932173338.
It is a Cunningham number, because it is equal to 28868172-1.
It is a congruent number.
It is an unprimeable number.
It is a polite number, since it can be written in 47 ways as a sum of consecutive naturals, for example, 277119493 + ... + 277149563.
It is an arithmetic number, because the mean of its divisors is an integer number (73531252480).
Almost surely, 28333712391488 is an apocalyptic number.
8333712391488 is a gapful number since it is divisible by the number (88) formed by its first and last digit.
It is an amenable number.
It is a practical number, because each smaller number is the sum of distinct divisors of 8333712391488, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (12353250416640).
8333712391488 is an abundant number, since it is smaller than the sum of its proper divisors (16372788441792).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
8333712391488 is a wasteful number, since it uses less digits than its factorization.
8333712391488 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 30338 (or 30317 counting only the distinct ones).
The product of its digits is 20901888, while the sum is 60.
The spelling of 8333712391488 in words is "eight trillion, three hundred thirty-three billion, seven hundred twelve million, three hundred ninety-one thousand, four hundred eighty-eight". | 596 | 2,061 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2020-50 | latest | en | 0.841756 |
https://otfried.org/courses/cs206/pp4.html | 1,686,126,035,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224653631.71/warc/CC-MAIN-20230607074914-20230607104914-00533.warc.gz | 505,980,386 | 4,438 | # Programming project 4
Download the archive pp4.zip. It contains all the files for this project.
In this project you will implement an abstract data type to represent a continuous piecewise linear function.
Our ADT is called PieceWiseLinear, and can be used as follows:
>>> from pwlf import PieceWiseLinear
>>> f = PieceWiseLinear(1, 3, 7, 5)
The constructor takes four arguments $$x_{0}, y_{0}, x_{1}, y_{1}$$, and creates a piecewise linear function $$f$$ over the domain $$[x_{0}, x_{1}]$$, with $$f(x_{0}) = y_{0}$$ and $$f(x_{1}) = y_{1}$$.
In the example above, the function is defined on the interval $$[1, 7]$$, with $$f(1) = 3$$ and $$f(7) = 5$$.
We can obtain the domain of a function using the domain method:
>>> f.domain()
(1, 7)
And, of course, we can evaluate the function $$f(x)$$ for various values of $$x$$. When $$x$$ is not in the domain, we get an error, of course:
>>> f(1)
3.0
>>> f(3)
3.6666666666666665
>>> f(7)
5.0
>>> f(0)
ValueError: argument is not in domain
>>> f(8)
ValueError: argument is not in domain
Note: evaluating a function with the function call syntax works if we define the magic __call__ method for our objects.
To make more interesting piecewise linear functions (with more than one piece), we can use the join method. It returns a new function (the object itself remains unchanged), which combines two functions:
>>> g = PieceWiseLinear(7, 5, 13, 2)
>>> f1 = f.join(g)
>>> f1.domain()
(1, 13)
>>> for x in [1, 3, 7, 8, 13]:
... print(f1(x))
...
3.0
3.6666666666666665
5.0
4.5
2.0
Note that join is only allowed when the domains are consecutive, and the combined function will be continuous:
>>> g1 = PieceWiseLinear(12, 2, 19, 3)
>>> f1.join(g1)
ValueError: domains are not contiguous
>>> g2 = PieceWiseLinear(13, 3, 19, 3)
>>> f1.join(g2)
ValueError: discontinuity at connection point
Using join, we can build piecewise linear functions with many pieces:
>>> g3 = PieceWiseLinear(13, 2, 19, 3)
>>> g4 = PieceWiseLinear(19, 3, 21, -5)
>>> f2 = f1.join(g3.join(g4))
>>> f2.domain()
(1, 21)
Of course we provide a nice string conversion for PieceWiseLinear objects:
>>> print(f)
(1,3)..(7,5)
>>> print(f1)
(1,3)..(7,5)..(13,2)
>>> print(f2)
(1,3)..(7,5)..(13,2)..(19,3)..(21,-5)
Note: To format numbers in the way that our test scripts except it, you must use the %g formatting specifier. You should format the coordinates of each breakpoint $$(x,y)$$ using the expression
"(%g,%g)" % (x,y)
We can also do arithmetic operations on our functions. First, we can multiply them with a number:
>>> print(f2)
(1,3)..(7,5)..(13,2)..(19,3)..(21,-5)
>>> f3 = 3 * f2
>>> print(f3)
(1,9)..(7,15)..(13,6)..(19,9)..(21,-15)
>>> f2(10)
3.5
>>> f3(10)
10.5
Note: we can make this work my defining the magic method __rmul__ on the PieceWiseLinear object. The expression n * f calls the method f.__rmul__(n).
Note that the function f2 is not modified by the multiplication in any way.
Second, we can add or subtract a constant to a function:
>>> f4 = f3 - 10
>>> f5 = f3 + 99
>>> print(f4)
(1,-1)..(7,5)..(13,-4)..(19,-1)..(21,-25)
>>> print(f5)
(1,108)..(7,114)..(13,105)..(19,108)..(21,84)
To make this work, you need to implement the method add_number in the given template. (It will be called from the magic methods __add__ and __sub__ after checking the type of the right hand side.) The second argument is a multiplication factor for the constant, for addition it will be +1, for subtraction it will be -1.
Note that again the function f3 is not modified by the addition/subtraction in any way.
Finally, it is possible to add or subtract two PieceWiseLinear objects. For this to work, the domains of the two functions need to overlap in an interval. It is not necessary for the two functions to have the same domain, they also do not need to have the same breakpoints:
>>> print(f4)
(1,-1)..(7,5)..(13,-4)..(19,-1)..(21,-25)
>>> g5 = PieceWiseLinear(0, 0, 15, 8)
>>> f6 = f4 + g5
>>> print(f6)
(1,-0.466667)..(7,8.73333)..(13,2.93333)..(15,5)
>>> f6.domain()
(1, 15)
Note that the domain of the sum is the intersection of the two domains. The intersection cannot just be a point:
>>> g6 = PieceWiseLinear(15, 0, 16, 1)
>>> f6 + g6
ValueError: domains do not overlap
To make this work, you need to implement the method add_pwlf in the template. Again, the second argument is a multiplication factor, and will be either +1 or -1.
Start from the template code pwlf.py, which already defines all the right methods and provides some error checking of arguments.
When you are done, you can run pwlf.py to see the following test output:
\$ python3 pwlf.py
f1 = (1,-1)..(3,1)
f2 = (3,1)..(7,-5)
f1(1) = -1
f1(2) = 0
f1(3) = 1
f2(3) = 1
f2(5) = -2
f2(7) = -5
f = (1,-1)..(3,1)..(7,-5)
f(1) = -1
f(2) = 0
f(3) = 1
f(5) = -2
f(7) = -5
Domain of f1 = (1, 3), domain of f2 = (3, 7), domain of f = (1, 7)
g1 = f + 2 = (1,1)..(3,3)..(7,-3)
g2 = f - 6 = (1,-7)..(3,-5)..(7,-11)
g3 = 3 * f = (1,-3)..(3,3)..(7,-15)
h1 = 5 * f + 3 = (1,-2)..(3,8)..(7,-22)
h2 = 0.5 * f - 2 = (1,-2.5)..(3,-1.5)..(7,-4.5)
g = h1 + h2 = (1,-4.5)..(3,6.5)..(7,-26.5)
d = (0,0)..(2,19)..(6,12)
e1 = g + d = (1,5)..(2,20)..(3,23.75)..(6,-6.25)
e2 = g - d = (1,-14)..(2,-18)..(3,-10.75)..(6,-30.25)
g(1) = -4.5, d(1) = 9.5, e1(1) = 5, e2(1) = -14
g(2) = 1, d(2) = 19, e1(2) = 20, e2(2) = -18
g(3) = 6.5, d(3) = 17.25, e1(3) = 23.75, e2(3) = -10.75
g(4) = -1.75, d(4) = 15.5, e1(4) = 13.75, e2(4) = -17.25
g(5) = -10, d(5) = 13.75, e1(5) = 3.75, e2(5) = -23.75
g(6) = -18.25, d(6) = 12, e1(6) = -6.25, e2(6) = -30.25
For more thorough testing, including testing of the error messages, use the unit test script test_pwlf.py. | 2,118 | 5,659 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2023-23 | latest | en | 0.71594 |
https://www.mersenneforum.org/showthread.php?s=7396581f095b7ddcad6989ba1c694b1d&t=20102 | 1,653,103,925,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662534773.36/warc/CC-MAIN-20220521014358-20220521044358-00405.warc.gz | 1,033,643,687 | 12,054 | mersenneforum.org GMP-ECM 7.0-dev assertion failed!
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2015-03-08, 11:07 #1 Antonio "Antonio Key" Sep 2011 UK 32·59 Posts GMP-ECM 7.0-dev assertion failed! Was working my way through factoring 85*2^67692-1 when this occurred: - Code: F:\ECM> F:\ECM>ecm -v -pm1 -maxmem 12288 -inp num2ecm.txt 500000 GMP-ECM 7.0-dev [configured with GMP 6.0.0, --enable-asm-redc] [P-1] Input number is (85*2^67692-1)/(3*7*179*486271481*563592288920887*6600048163432517) (20337 digits) Using REDC Using lmax = 16384 with NTT which takes about 409MB of memory Using B1=500000, B2=658722972, polynomial x^1, x0=2500857976 P = 39585, l = 16384, s_1 = 8064, k = s_2 = 2, m_1 = 3 Probability of finding a factor of n digits: (Use -go parameter to specify known factors in P-1) 20 25 30 35 40 45 50 55 60 65 0.14 0.026 0.0034 0.00036 3e-005 2.1e-006 1.3e-007 6.9e-009 3.1e-010 1.3e-011 Step 1 took 213565ms Computing F from factored S_1Assertion failed! Program: F:\ECM\ecm.exe File: pm1fs2.c, Line 1857 Expression: (__builtin_constant_p (1UL) && (1UL) == 0 ? ((F[deg])->_mp_size < 0 ? -1 : (F[deg])->_mp_size > 0) : __gmpz_cmp_ui (F[deg],1UL)) == 0 This application has requested the Runtime to terminate it in an unusual way. Please contact the application's support team for more information. F:\ECM> Since this had worked up to finding the last factor I reran the previous P-1 test: - Code: F:\ECM>ecm -v -pm1 -maxmem 12288 -inp num2ecm.txt 500000 GMP-ECM 7.0-dev [configured with GMP 6.0.0, --enable-asm-redc] [P-1] Input number is (85*2^67692-1)/(3*7*179*486271481*563592288920887) (20353 digits) Using REDC Using lmax = 16384 with NTT which takes about 409MB of memory Using B1=500000, B2=658722972, polynomial x^1, x0=1998205761 P = 39585, l = 16384, s_1 = 8064, k = s_2 = 2, m_1 = 3 Probability of finding a factor of n digits: (Use -go parameter to specify known factors in P-1) 20 25 30 35 40 45 50 55 60 65 0.14 0.026 0.0034 0.00036 3e-005 2.1e-006 1.3e-007 6.9e-009 3.1e-010 1.3e-011 Step 1 took 140697ms Computing F from factored S_1 took 61449ms Computing h took 7426ms Computing DCT-I of h took 780ms Multi-point evaluation 1 of 2: Computing g_i took 26427ms Computing g*h took 1607ms Computing gcd of coefficients and N took 15958ms Multi-point evaluation 2 of 2: Computing g_i took 26427ms Computing g*h took 1607ms Computing gcd of coefficients and N took 15943ms Step 2 took 176094ms So everything appears to be OK up to removing the last factor found, very strange! I am running ECM on the number without a problem at the moment. Code: GMP-ECM 7.0-dev [configured with GMP 6.0.0, --enable-asm-redc] [ECM] Using B1=50000, B2=15446350, polynomial x^2, 3 threads ____________________________________________________________________________ Curves Complete | Average seconds/curve | Runtime | ETA -----------------|---------------------------|---------------|-------------- 36 of 220 | Stg1 187.6s | Stg2 42.10s | 0d 00:49:45 | 0d 03:56:40 (Running on Windows 7 sp1, 16GB ram on a i5 3570k)
2015-03-08, 11:10 #2
xilman
Bamboozled!
"๐บ๐๐ท๐ท๐ญ"
May 2003
Down not across
32·5·251 Posts
Quote:
Originally Posted by Antonio Was working my way through factoring 85*2^67692-1 when this occurred: - ... (Running on Windows 7 sp1, 16GB ram on a i5 3570k)
Please report this to the ECM dev team.
Paul
2015-03-08, 19:50 #3 jwaltos Apr 2012 Oh oh. 1110000112 Posts Along with reporting to the ECM development team, what build version were you using please? There seem to have been a few different versions over the last little while.
2015-03-09, 07:24 #4
Antonio
"Antonio Key"
Sep 2011
UK
21316 Posts
Quote:
Originally Posted by jwaltos Along with reporting to the ECM development team, what build version were you using please? There seem to have been a few different versions over the last little while.
The original failure was noticed with GMP-ECM 7.0.0 svn2540 configured with GMP 6.0.0
I went back and tried an earlier version with the following results: -
Code:
f:\yafu\ecm>ecmx -pm1 -maxmem 12288 -inp num2ecm.txt 110000
GMP-ECM 6.4.4 [configured with MPIR 2.6.0] [P-1]
Input number is (85*2^67692-1)/(3*7*179*486271481*563592288920887*6600048163432517) (20337 digits)
Using B1=110000, B2=44251182, polynomial x^1, x0=452485962
Step 1 took 56628ms
Step 2 took 30062ms
f:\yafu\ecm>ecmy -pm1 -maxmem 12288 -inp num2ecm.txt 110000
GMP-ECM 6.4.4 [configured with GMP 6.0.0, --enable-asm-redc] [P-1]
Input number is (85*2^67692-1)/(3*7*179*486271481*563592288920887*6600048163432517) (20337 digits)
Using B1=110000, B2=44251182, polynomial x^1, x0=3499163316
Step 1 took 47876ms
Assertion failed!
Program: f:\yafu\ecm\ecmy.exe
File: pm1fs2.c, Line 1851
Expression: (__builtin_constant_p (1UL) && (1UL) == 0 ? ((F[deg])->_mp_size < 0 ? -1 : (F[deg])->_mp_size > 0) : __gmpz_cmp_ui (F[deg],1UL)) == 0
This application has requested the Runtime to terminate it in an unusual way.
Please contact the application's support team for more information.
So it looks like it may be a problem with GMP 6.0.0
2015-03-09, 08:56 #5 jwaltos Apr 2012 Oh oh. 11×41 Posts You are probably correct. I recall in an earlier post that GMP works best within a Linux environment whereas MPIR works best within Windows.
2015-03-09, 11:46 #6 wreck "Bo Chen" Oct 2005 Wuhan,China 22·43 Posts I met some strange behavior when using B1=3e9 on R323 with -maxmem 1000 Here is the log Code: GMP-ECM 7.0-dev [configured with MPIR 2.6.0,--enable--openmp] [ECM] Input number is (10^323-1)/(9*207...) (271 digits) [Fri Nov 28 17:52:02 2014] Computing batch product (of zu bits) of primes below B1=0 took -1073741824ms Using MODMULN [mulredc:0,sqrredc:1] Using B1=3000000000,B2=84996584711230,polynomial Dickson(30),sigma=2:2092172894349633986 dF=131072,k=438,d=1345890,d2=11,i0=2219 ... Step 1 took 209618ms Using 32 small primes for NTT Estimated memory usage: 735M ... Step 2 took 25928380ms ... It took 1000 seconds when using B1=11e7, so there seems something wrong of the B1=3e9.
2015-03-09, 12:05 #7
xilman
Bamboozled!
"๐บ๐๐ท๐ท๐ญ"
May 2003
Down not across
32·5·251 Posts
Quote:
Originally Posted by wreck I met some strange behavior when using B1=3e9 on R323 with -maxmem 1000 Here is the log Code: GMP-ECM 7.0-dev [configured with MPIR 2.6.0,--enable--openmp] [ECM] Input number is (10^323-1)/(9*207...) (271 digits) [Fri Nov 28 17:52:02 2014] Computing batch product (of zu bits) of primes below B1=0 took -1073741824ms Using MODMULN [mulredc:0,sqrredc:1] Using B1=3000000000,B2=84996584711230,polynomial Dickson(30),sigma=2:2092172894349633986 dF=131072,k=438,d=1345890,d2=11,i0=2219 ... Step 1 took 209618ms Using 32 small primes for NTT Estimated memory usage: 735M ... Step 2 took 25928380ms ... It took 1000 seconds when using B1=11e7, so there seems something wrong of the B1=3e9.
What took 1000 seconds? Stage 1, stage 2 or the entire computation?
If stage 1 alone took 1000 seconds something indeed looks strange as the subsequent run took only 1/5 as long whereas 3e9/11e7 = 27.3
OTOH, 25928380 / 1000 / 27.3 = 950 which is in line with expectations.
Likewise (25928380 + 209618) / 1000 / 27.3 = 957, also consistent.
Paul
2015-03-09, 12:49 #8 wreck "Bo Chen" Oct 2005 Wuhan,China 22·43 Posts I mean 1000 seconds when using B1=11e7 for stage 1. I suppose the stage 1 takes longer time when B1 get larger,this is why I think there are something wrong. The binary is download from gilchrist's site.
2015-03-10, 04:57 #9
ATH
Einyen
Dec 2003
Denmark
24×32×23 Posts
Quote:
Originally Posted by Antonio So it looks like it may be a problem with GMP 6.0.0
Your number worked for me with SVN2655 both with GMP 6.0.0a and MPIR 2.7.0. You can try these if you want, they are compiled for a Sandy Bridge cpu:
ecm2655-mpir.zip
ecm2655-gmp.zip
2015-03-10, 08:21 #10
Antonio
"Antonio Key"
Sep 2011
UK
32·59 Posts
Quote:
Originally Posted by ATH Your number worked for me with SVN2655 both with GMP 6.0.0a and MPIR 2.7.0. You can try these if you want, they are compiled for a Sandy Bridge cpu: ecm2655-mpir.zip ecm2655-gmp.zip
Thanks - those worked for me too.
Taking the hint, I recompiled and tuned for my Ivy Bridge using the latest version (SVN2657) + GMP 6.0.0a and that's working as well.
Similar Threads Thread Thread Starter Forum Replies Last Post bsquared GMP-ECM 4 2013-03-01 15:52 RickC Hardware 8 2010-10-28 03:31 Andi47 GMP-ECM 2 2009-08-04 14:11 yoyo GMP-ECM 8 2009-01-16 20:57 musos Information & Answers 1 2008-12-12 14:29
All times are UTC. The time now is 03:32.
Sat May 21 03:32:05 UTC 2022 up 37 days, 1:33, 0 users, load averages: 1.06, 1.25, 1.32 | 3,165 | 8,663 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2022-21 | latest | en | 0.663485 |
https://mail.haskell.org/pipermail/haskell-cafe/2010-December/087609.html | 1,701,602,525,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100499.43/warc/CC-MAIN-20231203094028-20231203124028-00666.warc.gz | 434,855,572 | 2,010 | Daniel Peebles pumpkingod at gmail.com
Thu Dec 23 18:52:07 CET 2010
```Fair enough :) that'll teach me to hypothesize something without thinking
about it! I guess I could amend my coinductive proof:
http://hpaste.org/paste/42516/mirrormirror_with_bottom#p42517
does that cover bottom-ness adequately? I can't say I've thought through it
terribly carefully.
On Thu, Dec 23, 2010 at 12:30 PM, Ryan Ingram <ryani.spam at gmail.com> wrote:
> On Thu, Dec 23, 2010 at 8:19 AM, Daniel Peebles <pumpkingod at gmail.com>
> wrote:
> > Simulating bottoms wouldn't be too hard, but I don't think the statement
> is
> > even true in the presence of bottoms, is it?
>
> Isn't it?
>
> data Tree a = Tip | Node (Tree a) a (Tree a)
>
> mirror :: Tree a -> Tree a
> mirror Tip = Tip
> mirror (Node x y z) = Node (mirror z) y (mirror x)
>
> --
>
> -- base cases
> mirror (mirror _|_)
> = mirror _|_
> = _|_
>
> mirror (mirror Tip)
> = mirror Tip
> = Tip
>
> -- inductive case
> mirror (mirror (Node x y z))
> = mirror (Node (mirror z) y (mirror x))
> = Node (mirror (mirror x)) y (mirror (mirror z))
> -- induction
> = Node x y z
>
> -- ryan
>
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https://eng.libretexts.org/Bookshelves/Computer_Science/Programming_and_Computation_Fundamentals/Book%3A_Programming_Fundamentals_-_A_Modular_Structured_Approach_using_C___(Busbee)/16%3A_Counting_Loops/16.05%3A_Practice_15_Counting_Loops | 1,685,951,721,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224651325.38/warc/CC-MAIN-20230605053432-20230605083432-00021.warc.gz | 266,330,596 | 29,214 | # 16.5: Practice 15 Counting Loops
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## Learning Objectives
With 100% accuracy during a: memory building activity, exercises, lab assignment, problems, or timed quiz/exam; the student is expected to:
1. Define the terms on the definitions as listed in the modules associated with this chapter.
2. Identify which selection control structures are commonly used a counting loops.
3. Be able to write pseudo code or flowcharting for the for control structure.
4. Be able to write C++ source code for a for control structure.
5. When feasible, be able to convert C++ source code from while loop acting like a counting loop to a for loop and and vice versa.
## Exercises
Questions
1. Only for loops can be counting loops.
2. The integer data type has modular arithmetic attributes.
3. The escape code of \n is part of formatting output.
4. Nested for loops is not allowed in the C++ programming language.
5. Counting loops use all four of the loop attributes.
1. False
2. True
3. True
4. False
5. True
## Lab Assignment
#### Creating a Folder or Sub-Folder for Chapter 15 Files
Depending on your compiler/IDE, you should decide where to download and store source code files for processing. Prudence dictates that you create these folders as needed prior to downloading source code files. A suggested sub-folder for the Bloodshed Dev-C++ 5 compiler/IDE might be named:
• Chapter_15 within the folder named: Cpp_Source_Code_Files
If you have not done so, please create the folder(s) and/or sub-folder(s) as appropriate.
Download and store the following file(s) to your storage device in the appropriate folder(s). You may need to right click on the link and select "Save Target As" in order to download the file.
#### Detailed Lab Instructions
Read and follow the directions below carefully, and perform the steps in the order listed.
• Compile and run the Lab_15a.cpp source code file. Understand how it works.
• Copy the source code file Lab_15a.cpp naming it: Lab_15b.cpp
• Convert the code that is counting (all four attributes) to a for loop.
• Build (compile and run) your program.
• After you have successfully written this program, if you are taking this course for college credit, follow the instructions from your professor/instructor for submitting it for grading.
## Problems
#### Problem 15a - Instructions
Using proper C++ syntax, convert the following for loop to a while loop.
C++ source code
for (x = 0; x < 10; x++)
{
cout << "Having fun!";
}
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https://stackoverflow.com/questions/8921999/how-to-find-and-return-a-duplicate-value-in-array/8922278 | 1,726,041,377,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651344.44/warc/CC-MAIN-20240911052223-20240911082223-00748.warc.gz | 505,190,161 | 69,115 | How to find and return a duplicate value in array
`arr` is array of strings:
``````["hello", "world", "stack", "overflow", "hello", "again"]
``````
What would be an easy and elegant way to check if `arr` has duplicates, and if so, return one of them (no matter which)?
Examples:
``````["A", "B", "C", "B", "A"] # => "A" or "B"
["A", "B", "C"] # => nil
``````
• `arr == arr.uniq` would be an easy and elegant way to check if `arr` has duplicates, however, it doesn't provide which were duplicated. Commented May 3, 2020 at 12:56
``````a = ["A", "B", "C", "B", "A"]
a.detect{ |e| a.count(e) > 1 }
``````
I know this isn't very elegant answer, but I love it. It's beautiful one liner code. And works perfectly fine unless you need to process huge data set.
Looking for faster solution? Here you go!
``````def find_one_using_hash_map(array)
map = {}
dup = nil
array.each do |v|
map[v] = (map[v] || 0 ) + 1
if map[v] > 1
dup = v
break
end
end
return dup
end
``````
It's linear, O(n), but now needs to manage multiple lines-of-code, needs test cases, etc.
If you need an even faster solution, maybe try C instead.
And here is the gist comparing different solutions: https://gist.github.com/naveed-ahmad/8f0b926ffccf5fbd206a1cc58ce9743e
• Except quadratic for something that can be solved in linear time. Commented Mar 28, 2013 at 7:47
• Providing O(n^2) solutions for linear problems is not the way to go.
– tdgs
Commented May 3, 2013 at 9:18
• @jasonmp85 - True; however, that's only considering big-O runtime. in practice, unless you're writing this code for some huge scaling data (and if so, you can actually just use C or Python), the provided answer is far more elegant/readable, and isnt' going to run that much slower compared to a linear time solution. furthermore, in theory, the linear time solution requires linear space, which may not be available Commented May 25, 2013 at 6:10
• @Kalanamith you can get duplicated values using this `a.select {|e| a.count(e) > 1}.uniq` Commented Jul 12, 2013 at 16:34
• The problem with the "detect" method is that it stops when it finds the first duplicate, and doesn't give you all the dups. Commented Jan 26, 2014 at 22:57
You can do this in a few ways, with the first option being the fastest:
``````ary = ["A", "B", "C", "B", "A"]
ary.group_by{ |e| e }.select { |k, v| v.size > 1 }.map(&:first)
ary.sort.chunk{ |e| e }.select { |e, chunk| chunk.size > 1 }.map(&:first)
``````
And a O(N^2) option (i.e. less efficient):
``````ary.select{ |e| ary.count(e) > 1 }.uniq
``````
• The first two are much more efficient for large arrays. The last one is O(n*n) so it can get slow. I needed to use this for an array with ~20k elements and the first two returned almost instantly. I had to cancel the third one because it was taking so long. Thanks!! Commented Feb 4, 2013 at 5:15
• Just an observation but the first two that end with .map(&:first) could just end with .keys as that part is just pulling the keys on a hash. Commented Mar 2, 2014 at 8:20
• @engineerDave that depends on the ruby version being used. 1.8.7 would require &:first or even {|k,_| k } without ActiveSupport. Commented Nov 30, 2014 at 17:01
• here are some benchmarks gist.github.com/equivalent/3c9a4c9d07fff79062a3 in performance the winner is clearly `group_by.select` Commented Jul 13, 2015 at 13:54
• If you're using Ruby > 2.1, you can use: `ary.group_by(&:itself)`. :-) Commented Jan 3, 2017 at 13:43
Simply find the first instance where the index of the object (counting from the left) does not equal the index of the object (counting from the right).
``````arr.detect {|e| arr.rindex(e) != arr.index(e) }
``````
If there are no duplicates, the return value will be nil.
I believe this is the fastest solution posted in the thread so far, as well, since it doesn't rely on the creation of additional objects, and `#index` and `#rindex` are implemented in C. The big-O runtime is N^2 and thus slower than Sergio's, but the wall time could be much faster due to the the fact that the "slow" parts run in C.
• I like this solution, but it will only return the first duplicate. To find all duplicates: `arr.find_all {|e| arr.rindex(e) != arr.index(e) }.uniq`
– Josh
Commented Jan 7, 2015 at 4:01
• Nor does your answer show how to find if there are any triplicates, or whether one can draw elements from the array to spell "CAT". Commented Jul 11, 2015 at 5:09
• @bruno077 How is this linear time? Commented Mar 13, 2016 at 23:21
• @chris Great answer, but I think you can do a bit better with this: `arr.detect.with_index { |e, idx| idx != arr.rindex(e) }`. Using `with_index` should remove the necessity for the first `index` search. Commented Sep 15, 2016 at 12:57
• How would you adapt this to a 2D array, comparing duplicates in a column? Commented Sep 16, 2016 at 8:27
`detect` only finds one duplicate. `find_all` will find them all:
``````a = ["A", "B", "C", "B", "A"]
a.find_all { |e| a.count(e) > 1 }
``````
• The question is very specific that only one duplicate is to be returned. Imo, showing how to find all duplicates is fine, but only as an aside to an answer that answers the question asked, which you have not done. btw, it is agonizingly inefficient to invoke `count` for every element in the array. (A counting hash, for example, is much more efficient; e.g, construct `h = {"A"=>2, "B"=>2, "C"=> 1 }` then `h.select { |k,v| v > 1 }.keys #=> ["A", "B"]`. Commented Jul 11, 2015 at 5:04
Here are two more ways of finding a duplicate.
Use a set
``````require 'set'
def find_a_dup_using_set(arr)
s = Set.new
end
find_a_dup_using_set arr
#=> "hello"
``````
Use `select` in place of `find` to return an array of all duplicates.
Use `Array#difference`
``````class Array
def difference(other)
h = other.each_with_object(Hash.new(0)) { |e,h| h[e] += 1 }
reject { |e| h[e] > 0 && h[e] -= 1 }
end
end
def find_a_dup_using_difference(arr)
arr.difference(arr.uniq).first
end
find_a_dup_using_difference arr
#=> "hello"
``````
Drop `.first` to return an array of all duplicates.
Both methods return `nil` if there are no duplicates.
I proposed that `Array#difference` be added to the Ruby core. More information is in my answer here.
Benchmark
Let's compare suggested methods. First, we need an array for testing:
``````CAPS = ('AAA'..'ZZZ').to_a.first(10_000)
def test_array(nelements, ndups)
arr = CAPS[0, nelements-ndups]
arr = arr.concat(arr[0,ndups]).shuffle
end
``````
and a method to run the benchmarks for different test arrays:
``````require 'fruity'
def benchmark(nelements, ndups)
arr = test_array nelements, ndups
puts "\n#{ndups} duplicates\n"
compare(
Naveed: -> {arr.detect{|e| arr.count(e) > 1}},
Sergio: -> {(arr.inject(Hash.new(0)) {|h,e| h[e] += 1; h}.find {|k,v| v > 1} ||
[nil]).first },
Ryan: -> {(arr.group_by{|e| e}.find {|k,v| v.size > 1} ||
[nil]).first},
Chris: -> {arr.detect {|e| arr.rindex(e) != arr.index(e)} },
Cary_set: -> {find_a_dup_using_set(arr)},
Cary_diff: -> {find_a_dup_using_difference(arr)}
)
end
``````
I did not include @JjP's answer because only one duplicate is to be returned, and when his/her answer is modified to do that it is the same as @Naveed's earlier answer. Nor did I include @Marin's answer, which, while posted before @Naveed's answer, returned all duplicates rather than just one (a minor point but there's no point evaluating both, as they are identical when return just one duplicate).
I also modified other answers that returned all duplicates to return just the first one found, but that should have essentially no effect on performance, as they computed all duplicates before selecting one.
The results for each benchmark are listed from fastest to slowest:
First suppose the array contains 100 elements:
``````benchmark(100, 0)
0 duplicates
Running each test 64 times. Test will take about 2 seconds.
Cary_set is similar to Cary_diff
Cary_diff is similar to Ryan
Ryan is similar to Sergio
Sergio is faster than Chris by 4x ± 1.0
Chris is faster than Naveed by 2x ± 1.0
benchmark(100, 1)
1 duplicates
Running each test 128 times. Test will take about 2 seconds.
Cary_set is similar to Cary_diff
Cary_diff is faster than Ryan by 2x ± 1.0
Ryan is similar to Sergio
Sergio is faster than Chris by 2x ± 1.0
Chris is faster than Naveed by 2x ± 1.0
benchmark(100, 10)
10 duplicates
Running each test 1024 times. Test will take about 3 seconds.
Chris is faster than Naveed by 2x ± 1.0
Naveed is faster than Cary_diff by 2x ± 1.0 (results differ: AAC vs AAF)
Cary_diff is similar to Cary_set
Cary_set is faster than Sergio by 3x ± 1.0 (results differ: AAF vs AAC)
Sergio is similar to Ryan
``````
Now consider an array with 10,000 elements:
``````benchmark(10000, 0)
0 duplicates
Running each test once. Test will take about 4 minutes.
Ryan is similar to Sergio
Sergio is similar to Cary_set
Cary_set is similar to Cary_diff
Cary_diff is faster than Chris by 400x ± 100.0
Chris is faster than Naveed by 3x ± 0.1
benchmark(10000, 1)
1 duplicates
Running each test once. Test will take about 1 second.
Cary_set is similar to Cary_diff
Cary_diff is similar to Sergio
Sergio is similar to Ryan
Ryan is faster than Chris by 2x ± 1.0
Chris is faster than Naveed by 2x ± 1.0
benchmark(10000, 10)
10 duplicates
Running each test once. Test will take about 11 seconds.
Cary_set is similar to Cary_diff
Cary_diff is faster than Sergio by 3x ± 1.0 (results differ: AAE vs AAA)
Sergio is similar to Ryan
Ryan is faster than Chris by 20x ± 10.0
Chris is faster than Naveed by 3x ± 1.0
benchmark(10000, 100)
100 duplicates
Cary_set is similar to Cary_diff
Cary_diff is faster than Sergio by 11x ± 10.0 (results differ: ADG vs ACL)
Sergio is similar to Ryan
Ryan is similar to Chris
Chris is faster than Naveed by 3x ± 1.0
``````
Note that `find_a_dup_using_difference(arr)` would be much more efficient if `Array#difference` were implemented in C, which would be the case if it were added to the Ruby core.
Conclusion
Many of the answers are reasonable but using a Set is the clear best choice. It is fastest in the medium-hard cases, joint fastest in the hardest and only in computationally trivial cases - when your choice won't matter anyway - can it be beaten.
The one very special case in which you might pick Chris' solution would be if you want to use the method to separately de-duplicate thousands of small arrays and expect to find a duplicate typically less than 10 items in. This will be a bit faster as it avoids the small additional overhead of creating the Set.
• Excellent solution. It's not quite as obvious what's going on at first as some of the methods, but it should run in truly linear time, at the expense of a bit of memory. Commented Jul 11, 2015 at 6:51
• With find_a_dup_using_set, I get the Set back, instead of one of the duplicates. Also I can't find "find.with_object" in Ruby docs anywhere. Commented Oct 6, 2016 at 22:44
• @Scottj, thanks for the catch! It's interesting that no one caught that before now. I fixed it. That's Enumerable#find chained to Enumerator#with_object. I'll update the benchmarks, adding your solution and others. Commented Oct 6, 2016 at 23:08
• Excellent comparison @CarySwoveland Commented Dec 27, 2016 at 1:33
Ruby 2.7 introduced `Enumerable#tally`
And you can use it this way:
``````ary = ["A", "B", "C", "B", "A", "A"]
ary.tally.select { |_, count| count > 1 }.keys
# => ["A", "B"]
``````
``````ary = ["A", "B", "C"]
ary.tally.select { |_, count| count > 1 }.keys
# => []
``````
Ruby 2.7 also introduced `Enumerable#filter_map`, it's possible to combine these methods
``````ary = ["A", "B", "C", "B", "A", "A"]
ary.tally.filter_map { |el, count| el if count > 1 }
# => ["A", "B"]
``````
• This is what I ended up going to as its the only one thats actually a good answer in 2022. Commented Apr 19, 2022 at 15:39
• This is really great. Thanks. I ended up making an initializer that adds a `.duplicates` method onto `Array` so we can just call `["A", "B", "C", "B", "A"].duplicates #=> ["A", "B"]`. Commented Nov 29, 2022 at 20:26
Alas most of the answers are `O(n^2)`.
Here is an `O(n)` solution,
``````a = %w{the quick brown fox jumps over the lazy dog}
h = Hash.new(0)
a.find { |each| (h[each] += 1) == 2 } # => 'the"
``````
What is the complexity of this?
• Runs in `O(n)` and breaks on first match
• Uses `O(n)` memory, but only the minimal amount
Now, depending on how frequent duplicates are in your array these runtimes might actually become even better. For example if the array of size `O(n)` has been sampled from a population of `k << n` different elements only the complexity for both runtime and space becomes `O(k)`, however it is more likely that the original poster is validating input and wants to make sure there are no duplicates. In that case both runtime and memory complexity `O(n)` since we expect the elements to have no repetitions for the majority of inputs.
Ruby Array objects have a great method, `select`.
``````select {|item| block } → new_ary
select → an_enumerator
``````
The first form is what interests you here. It allows you to select objects which pass a test.
Ruby Array objects have another method, `count`.
``````count → int
count(obj) → int
count { |item| block } → int
``````
In this case, you are interested in duplicates (objects which appear more than once in the array). The appropriate test is `a.count(obj) > 1`.
If `a = ["A", "B", "C", "B", "A"]`, then
``````a.select{|item| a.count(item) > 1}.uniq
=> ["A", "B"]
``````
You state that you only want one object. So pick one.
• I like this one a lot, but you have to throw a uniq on the end or you'll get `["A", "B", "B", "A"]` Commented Dec 4, 2012 at 20:28
• Great answer. This is exactly what I was looking for. As @Joeyjoejoejr pointed out. I have submitted an edit to put `.uniq` on array. Commented Feb 12, 2013 at 8:07
• This is hugely inefficient. Not only do you find all duplicates and then throw away all but one, you invoke `count` for each element of the array, which is wasteful and unnecessary. See my comment on JjP's answer. Commented Jul 11, 2015 at 5:16
• Thanks for running the benchmarks. It is useful to see how the different solutions compare in running time. Elegant answers are readable but often not the most efficient. Commented Jul 12, 2015 at 1:29
find_all() returns an `array` containing all elements of `enum` for which `block` is not `false`.
To get `duplicate` elements
``````>> arr = ["A", "B", "C", "B", "A"]
>> arr.find_all { |x| arr.count(x) > 1 }
=> ["A", "B", "B", "A"]
``````
Or duplicate `uniq` elements
``````>> arr.find_all { |x| arr.count(x) > 1 }.uniq
=> ["A", "B"]
``````
Something like this will work
``````arr = ["A", "B", "C", "B", "A"]
arr.inject(Hash.new(0)) { |h,e| h[e] += 1; h }.
select { |k,v| v > 1 }.
collect { |x| x.first }
``````
That is, put all values to a hash where key is the element of array and value is number of occurences. Then select all elements which occur more than once. Easy.
I know this thread is about Ruby specifically, but I landed here looking for how to do this within the context of Ruby on Rails with ActiveRecord and thought I would share my solution too.
``````class ActiveRecordClass < ActiveRecord::Base
#has two columns, a primary key (id) and an email_address (string)
end
``````
The above returns an array of all email addresses that are duplicated in this example's database table (which in Rails would be "active_record_classes").
``````a = ["A", "B", "C", "B", "A"]
a.each_with_object(Hash.new(0)) {|i,hash| hash[i] += 1}.select{|_, count| count > 1}.keys
``````
This is a `O(n)` procedure.
Alternatively you can do either of the following lines. Also O(n) but only one iteration
``````a.each_with_object(Hash.new(0).merge dup: []){|x,h| h[:dup] << x if (h[x] += 1) == 2}[:dup]
a.inject(Hash.new(0).merge dup: []){|h,x| h[:dup] << x if (h[x] += 1) == 2;h}[:dup]
``````
This code will return list of duplicated values. Hash keys are used as an efficient way of checking which values have already been seen. Based on whether value has been seen, the original array `ary` is partitioned into 2 arrays: first containing unique values and second containing duplicates.
``````ary = ["hello", "world", "stack", "overflow", "hello", "again"]
hash={}
arr.partition { |v| hash.has_key?(v) ? false : hash[v]=0 }.last.uniq
=> ["hello"]
``````
You can further shorten it - albeit at a cost of slightly more complex syntax - to this form:
``````hash={}
arr.partition { |v| !hash.has_key?(v) && hash[v]=0 }.last.uniq
``````
Here is my take on it on a big set of data - such as a legacy dBase table to find duplicate parts
``````# Assuming ps is an array of 20000 part numbers & we want to find duplicates
# actually had to it recently.
# having a result hash with part number and number of times part is
# duplicated is much more convenient in the real world application
# Takes about 6 seconds to run on my data set
# - not too bad for an export script handling 20000 parts
h = {};
h = {} # result hash
ps.select{ |e|
ct = ps.count(e)
h[e] = ct if ct > 1
}; nil # so that the huge result of select doesn't print in the console
``````
``````r = [1, 2, 3, 5, 1, 2, 3, 1, 2, 1]
r.group_by(&:itself).map { |k, v| v.size > 1 ? [k] + [v.size] : nil }.compact.sort_by(&:last).map(&:first)
``````
`each_with_object` is your friend!
``````input = [:bla,:blubb,:bleh,:bla,:bleh,:bla,:blubb,:brrr]
# to get the counts of the elements in the array:
> input.each_with_object({}){|x,h| h[x] ||= 0; h[x] += 1}
=> {:bla=>3, :blubb=>2, :bleh=>2, :brrr=>1}
# to get only the counts of the non-unique elements in the array:
> input.each_with_object({}){|x,h| h[x] ||= 0; h[x] += 1}.reject{|k,v| v < 2}
=> {:bla=>3, :blubb=>2, :bleh=>2}
``````
``````a = ["A", "B", "C", "B", "A"]
b = a.select {|e| a.count(e) > 1}.uniq
c = a - b
d = b + c
``````
Results
`````` d
=> ["A", "B", "C"]
``````
If you are comparing two different arrays (instead of one against itself) a very fast way is to use the intersect operator `&` provided by Ruby's Array class.
``````# Given
a = ['a', 'b', 'c', 'd']
b = ['e', 'f', 'c', 'd']
# Then this...
a & b # => ['c', 'd']
``````
• That finds items that exist in both arrays, not duplicates in one array. Commented Apr 25, 2018 at 11:43
• Thanks for pointing that out. I've changed the wording in my answer. I'll leave it here because it's already proven helpful for some people coming from search. Commented May 1, 2018 at 19:28
This runs very quickly (iterated through 2.3mil ids, took less than a second to push dups into their own array)
Had to do this at work with 2.3 mil IDs I imported into a file, I imported list as sorted, also can be sorted by ruby.
``````list = CSV.read(path).flatten.sort
dup_list = []
list.each_with_index do |id, index|
dup_list.push(id) if id == list[index +1]
end
dup_list.to_set.to_a
``````
``````def duplicates_in_array(array)
hash = {}
duplicates_hash = {}
array.each do |v|
hash[v] = (hash[v] || 0 ) + 1
end
hash.keys.each do |hk|
duplicates_hash[hk] = hash[hk] if hash[hk] > 1
end
return duplicates_hash
end
``````
This will return a hash containing each duplicate in the array, and the amount of time it is duplicated
for example:
``````array = [1,2,2,4,5,6,7,7,7,7]
duplicates_in_array(array)
=> {2=>2, 7=>4}
``````
I needed to find out how many duplicates there were and what they were so I wrote a function building off of what Naveed had posted earlier:
``````def print_duplicates(array)
puts "Array count: #{array.count}"
map = {}
total_dups = 0
array.each do |v|
map[v] = (map[v] || 0 ) + 1
end
map.each do |k, v|
if v != 1
puts "#{k} appears #{v} times"
total_dups += 1
end
end
puts "Total items that are duplicated: #{total_dups}"
end
``````
• This isn't an answer to the original question. Commented Apr 19, 2022 at 15:39
1. Let's create duplication method that take array of elements as input
2. In the method body, let's create 2 new array objects one is seen and another one is duplicate
3. finally lets iterate through each object in given array and for every iteration lets find that object existed in seen array.
4. if object existed in the seen_array, then it is considered as duplicate object and push that object into duplication_array
5. if object not-existed in the seen, then it is considered as unique object and push that object into seen_array
let's demonstrate in Code Implementation
``````def duplication given_array
seen_objects = []
duplication_objects = []
given_array.each do |element|
duplication_objects << element if seen_objects.include?(element)
seen_objects << element
end
duplication_objects
end
``````
Now call duplication method and output return result -
``````dup_elements = duplication [1,2,3,4,4,5,6,6]
puts dup_elements.inspect
``````
• Code-only answers are generally frowned upon on this site. Could you please edit your answer to include some comments or explanation of your code? Explanations should answer questions like: What does it do? How does it do it? Where does it go? How does it solve OP's problem? See: How to anwser. Thanks! Commented Oct 21, 2019 at 17:17
``` [1,2,3].uniq!.nil? => true ``` ``` [1,2,3,3].uniq!.nil? => false ```
Notice the above is destructive
• this does not return duplicated values Commented Jan 16, 2019 at 13:14 | 6,307 | 21,327 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2024-38 | latest | en | 0.892196 |
https://www.cgaa.org/articles/how-many-cards-can-fit-in-an-elite-trainer-box | 1,670,325,424,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711077.50/warc/CC-MAIN-20221206092907-20221206122907-00161.warc.gz | 742,763,443 | 16,514 | # How many cards can fit in an elite trainer box?
Category: How
Author: Dennis Reeves
Published: 2022-03-04
Views: 943
## How many cards can fit in an elite trainer box?
If you are talking about the size of an elite trainer box, then the answer is quite a few. An elite trainer box is about the size of a regular deck of cards, so it can easily hold around 100 cards. However, if you are talking about how many cards can fit inside an elite trainer box without being too snug, then the answer is around 80 cards.
## How many cards can fit in an elite trainer box if they are double sleeved?
Assuming you are referring to the Pokemon Elite Trainer Box, then the answer is 120 cards. This is because the box comes with 8 dividers, and each divider can hold 15 double-sleeved cards.
## How many cards can fit in an elite trainer box if they are triple sleeved?
Assuming you are referring to the dimensions of the Elite Trainer Box released in conjunction with the Sun & Moon expansion, each box is 13.4 x 9.1 x 2.9 inches, or 340 x 231 x 74 mm. If we conservatively estimate that each card sleeve adds 0.1 mm of thickness to a card, then that means each box can theoretically hold up to339 triple-sleeved cards.
However, it's important to keep in mind that the Elite Trainer Box was not designed with the idea of holding triple-sleeved cards in mind. As such, trying to stuff that many cards into the box will likely result in a less than ideal experience. The box lid will likely not close properly, and the cards may be difficult to remove from the box.
If you're looking to store a large number of triple-sleeved cards, we recommend investing in a storage solution specifically designed for that purpose, such as a card binder or box.
## How many cards can fit in an elite trainer box if they are quadruple sleeved?
If you are talking about an elite trainer box that is meant for the storage of Pokemon cards, then the answer is that it can hold up to 210 cards if they are quadruple sleeved. This is because the box itself is divided into two compartments, each of which can hold up to 105 cards. Of course, if you want to be able to actually close the box, you will need to leave some extra room for the lid, so approximately 200 cards is the maximum that can be stored in an elite trainer box if they are quadruple sleeved.
## How many cards can fit in an elite trainer box if they are quintuple sleeved?
Assuming you are referring to the original Elite Trainer Box released in 2016, each card slot is approximately 2.5” x 3.5”. This gives the card slot a surface area of 8.75 in². Each card in a quintuple sleeve will have a surface area of 11.625 in². This means that each card will occupy 1.325 in² more space than the card slot. When this is multiplied by the number of card slots in the box (36), we get a total of 47.4 in² of space that will be occupied by cards. This means that the maximum number of quintuple sleeved cards that could fit in the box would be 352.
## How many cards can fit in an elite trainer box if they are septuple sleeved?
Assuming that an elite trainer box is the largest size box produced by Ultra Pro, then the dimensions of the box are 15 x 11 x 3 inches. If the cards are septuple sleeved, meaning that each card is in seven sleeves, then the thickness of each card would be approximately 0.11 inches. This means that the box can theoretically hold approximately 2,727 cards.
Of course, this number is not exact because it does not account for the space taken up by the dividers between the card rows, nor does it account for the fact that the cards might not fit perfectly into the box. However, it does give a rough estimate of how many cards can be stored in an elite trainer box when septuple sleeved.
## How many cards can fit in an elite trainer box if they are nonuple slee
Assuming you are referring to nonuple sleeved cards, the total number of cards that can fit in an Elite Trainer Box would be 180. This is because each card sleeve can hold 9 cards, and there are 20 card sleeves in an Elite Trainer Box. 9 times 20 equals 180.
## Related Questions
### Where can I find the Elite Trainer box in the center?
The Elite Trainer box can be found near the Video Games and Consoles section.
### What is included in the Pokémon TCG shining fates Elite Trainer box?
The Pokémon TCG: Shining Fates Elite Trainer box includes 10 booster packs of the new expansion, plus 1 foil promo card featuring Eevee VMAX 65 card sleeves featuring Gigantamax Eevee.
### How much is the Pokemon TCG Elite Trainer box?
The Pokemon TCG Elite Trainer box costs \$39.99
### What comes in the sword and shield—battle styles Elite Trainer box?
Sword and shield decks come with 8 booster packs, 45 Energy cards, plus a framed collector’s card with artwork of one of the eight Pokémon-EX.
### What is the champion's path Elite Trainer box?
The Pokémon TCG: Champion's Path Elite Trainer Box is a prize worth fighting for. It features the following items: -Two 60-card decks of cards -1 custom dice bag -1 rules sheet -1 strategy guide
### What comes in the Elite 4 trainer box plus?
The Sword & Shield Elite Trainer box plus comes with Zacian and Zamazenta, as well as a gold promo card.
### Where can I find the Pokémon TCG shining fates Elite Trainer box?
The Pokémon TCG Shining Fates Elite Trainer Box is available in the Pokémon Center and where Pokémon TCG products are sold.
### What's in the Pokemon TCG Hidden Fates Elite Trainer box?
The Pokémon TCG: Hidden Fates Elite Trainer box includes everything you need to get started playing the game, including: - 31 card deck consisting of powerful new Pokémon and their Trainer cards - 5 dice - 6 indicator condition markers - 48 counter sheet tiles - 2 player's guide - 1 rulebook
### What is Pokémon TCG shining fates product review?
The Shining Fates product review is a detailed evaluation of the Pokémon TCG Shining Fates set. The review includes an analysis of the cards, gameplay, and costs involved with this set.
### What is the value of the Shining fates trainer card?
The Shining Fates trainer card is worth around \$18 in general, but can be found for slightly less in slightly played condition.
### What is the Best Pokemon TCG box to buy?
The Pokémon TCG: Shining Fates Elite Trainer box is the best option for buyers. It offers great value for money, with a large number of cards and accessories included.
### Where can I find the Pokemon TCG Elite Trainer box?
The Pokémon TCG Elite Trainer box is available in the Pokémon Center and where Pokémon TCG products are sold.
### What's in the Elite Trainer box?
8 Sword & Shield booster packs A selection of Energy cards 65 card sleeves featuring the Legendary Pokémon Zacian or Zamazenta
### What comes in the Pokemon TCG sword and shield Elite Trainer box?
This Box includes 65 card sleeves featuring Gigantamax Single Strike Urshifu or Gigantamax Rapid Strike Urshifu.
### What is a battle style trainer card?
Battle style Trainer cards are cards that provide support for either Single Strike or Rapid Strike attacks, helping you to build a deck around your preferred style.
### How many Energy cards are in sword and shield battle styles?
There are 45 energy cards in sword and shield battle styles.
### What is included in the Pokémon TCG shining fates Elite Trainer box?
This box contains 10 Pokémon TCG: Shining Fates booster packs, 1 foil promo card featuring Eevee VMAX 65 card sleeves featuring Gigantamax Eevee.
### What comes in the Elite 4 trainer box plus?
Inside the Elite 4 trainer box plus, you'll find a gold promo card alongside 12 booster packs (3 each from Sword & Shield, Rebel Clash, Darkness Ablaze, and Vivid Voltage). | 1,774 | 7,735 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2022-49 | latest | en | 0.95355 |
https://kr.mathworks.com/matlabcentral/cody/problems/2605-find-out-value-of-polynomial-at-different-value/solutions/506396 | 1,591,269,843,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347439928.61/warc/CC-MAIN-20200604094848-20200604124848-00115.warc.gz | 415,339,819 | 15,648 | Cody
# Problem 2605. Find out value of polynomial at different value.
Solution 506396
Submitted on 29 Sep 2014 by James
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 = [1 8]; n=0; y_correct = 8; assert(isequal(your_fcn_name(x, n),y_correct))
ans = 8
2 Pass
%% x = [1 0]; n=0; y_correct = 0; assert(isequal(your_fcn_name(x, n),y_correct))
ans = 0
3 Pass
%% x = [1 0 5]; n=1; y_correct = 6; assert(isequal(your_fcn_name(x, n),y_correct))
ans = 6 | 199 | 582 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2020-24 | latest | en | 0.639072 |
https://pinoybix.org/2015/01/mcqs-in-antennas-part9.html | 1,718,417,377,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861578.89/warc/CC-MAIN-20240614235857-20240615025857-00421.warc.gz | 408,292,080 | 81,142 | You dont have javascript enabled! Please enable it! MCQ in Antennas Part 9 | ECE Board Exam
# MCQ in Antennas Part 9 | ECE Board Exam
This is the Multiples Choice Questions Part 9 of the Series in Antennas as one of the Communications Engineering topic. In Preparation for the ECE Board Exam make sure to expose yourself and familiarize in each and every questions compiled here taken from various sources including but not limited to past Board Examination Questions in Electronic System and Technologies, Communications Books, Journals and other Communications References.
#### Online Questions and Answers Topic Outline
• MCQ in Basic considerations
• MCQ in Wire Radiators in Space
• MCQ in Current and Voltage Distributions
• MCQ in Resonant and Non-resonant Antennas
• MCQ in Antenna Terms and Definition
• MCQ in Antenna Gain and Resistance
• MCQ in Bandwidth, Beamwidth and Polarization
• MCQ in Grounded and Ungrounded Antennas
• MCQ in Grounding Systems and Matching Systems
• MCQ in Antenna Types
• MCQ in Directional and Non-directional Antennas
• MCQ in Microwave Antennas
• MCQ in Wideband and Special Purpose Antennas
#### Continue Practice Exam Test Questions Part 9 of the Series
401. An ungrounded antenna near the ground
a. Acts as a single antenna of twice the height
b. Is unlikely to need an earth mat
c. Acts as an antenna array
d. Must be horizontally polarized
Solution:
402. One of the following consists of non-resonant antennas:
a. The rhombic antenna
b. The folded dipole
c. The end-fire array
Solution:
403. One of the following is very useful as a multiband HF receiving antenna. This is the:
a. Conical horn
b. Folded dipole
c. Log-periodic
d. Square loop
Solution:
404. Which of the following antennas is best excited from a waveguide?
a. Biconical
b. Horn
c. Helical
d. Discone
Solution:
405. Indicate which of the following reasons for using a counterpoise with antennas is false:
a. Impossibility of a good ground connection
b. Protection of personnel working underneath
c. Provision of an earth for the antenna
d. Rockiness of the ground itself
Solution:
406. One of the following is not a reason for the use of an antenna coupler:
a. To make the antenna look resistive
b. To provide the output amplifier with the correct load impedance
c. To discriminate against harmonics
d. To prevent reradiation of the local oscillator
Solution:
407. Indicate the antenna that is not wideband
a. Discone
b. Folded dipole
c. Helical
d. Marconi
Solution:
408. Indicate which one of the following reasons for the use of an earth mat with antennas is false:
a. Impossibility of a good ground connection
b. Provision of an earth for the antenna
c. Protection of personnel working underneath
d. Improvement of the radiation pattern of the antenna
Solution:
409. Show which of the following terms does not apply to the Yagi-Uda array
a. Good bandwidth
b. Parasitic elements
c. Folded dipole
d. High gain
Solution:
410. An antenna that is circularly polarized is the
a. Helical
b. Small circular loop
c. Parabolic reflector
d. Yagi-Uda
Solution:
411. The standard reference antenna for the directive gain is the
a. Infinitesimal dipole
b. Isotropic antenna
c. Elementary doublet
d. Half-wave dipole
Solution:
412. Top loading is sometimes used with an antenna in order to increase its
a. Effective height
b. Bandwidth
c. Beamwidth
d. Input capacitance
Solution:
413. Cassegrain feed is used with a parabolic reflector to
a. Increase the gain of the system
b. Increase the beamwidth of the system
c. Reduce the size of the main reflector
d. Allow the feed to be places at a convenient point
Solution:
414. Zoning is used with a dielectric antenna in order to
a. Reduce the bulk of the lens
b. Increase the bandwidth of the lens
c. Permit pin-point focusing
d. Correct the curvature of the wavefront from a horn that is too short
Solution:
415. A helical antenna is used for satellite tracking because of its
a. Circular polarization
b. Maneuverability
d. Good front-to-back ratio
Solution:
416. The discone antenna is
a. A useful direction-finding antenna
b. Used as a radar receiving antenna
c. Circularly polarized like other circular antennas
d. Useful as a UHF receiving antenna
Solution:
417. One of the following is not omnidirectional antenna
a. Half-wave dipole
b. Log-periodic
c. Discone
d. Marconi
Solution:
418. The polarization of a discone antenna is ______________.
a. Horizontal
b. Vertical
c. Omni
d. Directional
Solution:
419. _____________ is the horizontal pointing angle of an antenna.
a. Right angle
b. Angle of elevation
c. Bandwidth
d. Azimuth
Solution:
420. Which is a properly terminated antenna?
a. Rhombic
b. Hertz
c. Marconi
d. Dipole
Solution:
421. ____________ is a device that detects both vertically and horizontally polarized signals simultaneously.
a. Crystal
b. Orthomode transducer
c. Light transducer
d. Optoisolator
Solution:
422. How much does the radiated power of an antenna increases if its current increased by 3.3 times?
a. 6.6 times
b. 3.3 times
c. 10.89 times
d. 9.9 times
Solution:
423. What do you call the energy that was not radiated into space or completely transmitted?
a. Incident waves
b. Captured waves
c. Standing waves
d. Modulated waves
Solution:
424. What is the estimated medium wind loading in the Philippines for antenna tower design?
a. 200 kph
b. 250 kph
c. 300 kph
d. 100 kph
Solution:
425. The minimum number of turns a helix antenna must have
a. 4
b. 5
c. 3
d. 6
Solution:
426. When testing transmitter to prevent interfering with other stations, which type of antenna must be used?
a. Dummy antenna
b. Herztian antenna
c. None
d. Void antenna
Solution:
a. Antenna
b. Transmitter
c. Transmission line
d. Transceiver
Solution:
428. Determine the gain of a 6 ft parabolic dish operating at 1800 MHz
a. 15.5 dB
b. 30 dB
c. 11.2 dB
d. 28.17 dB
Solution:
429. Radiation characteristic of a dipole
a. figure of eight
b. omnidirectional
c. bi-directional
d. unidirectional
Solution:
430. An antenna which is not resonant at particular frequencies and so can be used over a wide band of frequencies is called
a. Aperiodic
b. Cassegrain
d. Boresight
Solution:
431. Two wires that are bent 90 degrees apart.
a. Rhombic
b. Hertz
c. Dipole
d. Log-periodic
Solution:
432. Harmonic suppressor connected to an antenna
a. Tank circuit
b. M-derived filter
c. Low-pass filter
d. High-pass filter
Solution:
433. Theoretical gain of a Herztian dipole
a. 0 dB
b. 1.76 dB
c. 2.15 dB
d. 3 dB
Solution:
434. A helical antenna is used for satellite tracking because of
a. Maneuverability
b. Good front-to-back
c. Circular polarization
Solution:
435. A convenient method of determining antenna impedance
a. Reactance circle
b. Stub matching
c. Smith chart
d. Trial and error
Solution:
436. Unity gain antenna.
a. Half-wave dipole
b. Rhombic
c. Dummy
d. Isotropic
Solution:
437. EIRP stands for ______________________.
a. Effective isotropic reflected power
b. Effective isotropic refracted power
c. Efficient and ideal radiated power
Solution:
438. Which of the following refers to the smallest beam of satellite antennaโs radiation pattern?
a. Global beam
b. Zoom beam
c. Spot beam
d. Hemispheric beam
Solution:
439. A region in front of a parabolic antenna
a. Transmission zone
b. Fraunhofer
c. Fresnel
d. All of these
Solution:
440. An antenna that can only receive a television signal.
a. Isotropic antenna
b. Reference antenna
c. TVRO
d. Yagi antenna
Solution:
441. Radiation pattern of a discone
a. Figure of eight
b. Bi-directional
c. Omnidirectional
d. Unidirectional
Solution:
442. Radio wave concentration in the direction of the signal emitted by a directional antenna.
b. Transmitted signal
Solution:
443. The reflector and director of an antenna array are considered as:
a. Transcendental elements
b. Feed-points
c. Driven elements
d. Parasitic elements
Solution:
444. An electronic equipment used to measure standing wave ratio:
a. Altimeter
b. Multimeter
c. Reflectometer
d. Wavemeter
Solution:
445. The product of the power supplied to the antenna and its gain relative to a half-wave dipole in a given direction.
a. Rated power
b. ERP
c. Peak envelope power
d. Carrier power
Solution:
446. What makes an antenna physically long electronically short?
d. All of these
Solution:
447. The capture area of an antenna is directly proportional to the
a. Distance between transmitter and receiver
b. Power density of the signal
c. Gain of the antenna
d. Frequency of the received signal
Solution:
448. A type of an undergrounded antenna is a/an ________________.
a. Hertz
b. Isotropic
c. Parabolic
d. Marconi
Solution:
449. What is meant by antenna gain?
a. The ratio of the signal in the forward direction to the signal in the backward direction
b. The ratio of the amount of power produced by the antenna compared to the output power of the transmitter
c. The final amplifier gain minus the transmission line losses (including any phasing lines present)
d. The numeric ratio relating the radiated signal strength of an antenna to that of another antenna
Solution:
450. It consists of a number of dipoles of equal size, equally spaces along a straight line with all dipoles fed in the same phase from the same source.
a. End-fire array
b. Yagi antenna
c. Log-periodic antenna
Solution:
#### Questions and Answers in Antennas Series
Following is the list of multiple choice questions in this brand new series:
MCQ in Antennas
PART 1: MCQ from Number 1 โ 50 Answer key: PART 1
PART 2: MCQ from Number 51 โ 100 Answer key: PART 2
PART 3: MCQ from Number 101 โ 150 Answer key: PART 3
PART 4: MCQ from Number 151 โ 200 Answer key: PART 4
PART 5: MCQ from Number 201 โ 250 Answer key: PART 5
PART 6: MCQ from Number 251 โ 300 Answer key: PART 6
PART 7: MCQ from Number 301 โ 350 Answer key: PART 7
PART 8: MCQ from Number 351 โ 400 Answer key: PART 8
PART 9: MCQ from Number 401 โ 450 Answer key: PART 9
PART 10: MCQ from Number 451 โ 500 Answer key: PART 10
### Complete List of MCQ in Communications Engineering per topic
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• Reading time:3 mins read
• Post category:C Programs
In this programming tutorial, you will be able to learn how to swap two numbers. There are 4 methods explained.
In order to understand the programming examples, you should have some knowledge of the following C-Programming concepts:
Swap Two Numbers Using Temporary Variable
``````#include<stdio.h>
int main(int argc, char const *argv[])
{
int number1,number2,temp;
printf("\nSwapping of two numbers in c\nEnter two numbers \n");
scanf("%d%d",&number1,&number2);
printf("Number 1: %d\nNumber 2: %d",number1,number2);
//swapping
temp=number1;
number1=number2;
number2=temp;
printf("\nAfter swapping");
printf("\nNumber 1: %d\nNumber 2: %d\n",number1,number2);
return 0;
}
C Program to swap two numbers``````
In the above programming example, the user is asked to enter 2 numbers. The numbers are stored in `number1` and `number2`.
By using a temporary variable numbers entered by the user are swapped. i.e.
`temp=number1;number1=number2;number2=temp;`
After swapping two numbers `number1` and `number2` are printed onto the console.
Swap Two Numbers Without Third Variable by Addition and Subtraction
``````#include<stdio.h>
int main(int argc, char const *argv[])
{
int number1,number2;
printf("\nSwapping of two numbers in c\nEnter two numbers \n");
scanf("%d%d",&number1,&number2);
printf("Number 1: %d\nNumber 2: %d",number1,number2);
//swapping
number1=number1+number2;
number2=number1-number2;
number1=number1-number2;
printf("\nAfter swapping");
printf("\nNumber 1: %d\nNumber 2: %d\n",number1,number2);
return 0;
}``````
In the above programming example, the user is asked to enter 2 numbers. The numbers are stored in `number1` and `number2`.
By using the addition and subtraction operator the numbers entered by the user are swapped. i.e.
`number1=number1+number2;number2=number1-number2;number1=number1-number2;`
After swapping two numbers `number1` and `number2` are printed onto the console.
Swap Two Numbers without Third Variable by Multiplication and Division
``````#include<stdio.h>
int main(int argc, char const *argv[])
{
int number1,number2;
printf("\nSwapping of two numbers in c\nEnter two numbers \n");
scanf("%d%d",&number1,&number2);
printf("Number 1: %d\nNumber 2: %d",number1,number2);
//swapping
number1=number1*number2;
number2=number1/number2;
number1=number1/number2;
printf("\nAfter swapping");
printf("\nNumber 1: %d\nNumber 2: %d\n",number1,number2);
return 0;
}``````
In the above programming example, the user is asked to enter 2 numbers. The numbers are stored in `number1` and `number2`.
By using a multiplication and division operator the numbers entered by the user are swapped. i.e.
`number1=number1*number2;number2=number1/number2;number1=number1/number2;`
After swapping two numbers `number1` and `number2` are printed onto the console.
Swap Two Numbers without Third Variable by Xor Operation
``````#include<stdio.h>
int main(int argc, char const *argv[])
{
int number1,number2;
printf("\nSwapping of two numbers in c\nEnter two numbers \n");
scanf("%d%d",&number1,&number2);
printf("Number 1: %d\nNumber 2: %d",number1,number2);
//swapping
number1=number1^number2;
number2=number1^number2;
number1=number1^number2;
printf("\nAfter swapping");
printf("\nNumber 1: %d\nNumber 2: %d\n",number1,number2);
return 0;
}``````
In the above programming example, the user is asked to enter 2 numbers. The numbers are stored in `number1` and `number2`.
By using a xor operator the numbers entered by the user are swapped. i.e.
`number1=number1^number2;number2=number1^number2;number1=number1^number2;`
After swapping two numbers `number1` and `number2` are printed onto the console. | 1,041 | 3,709 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2022-49 | longest | en | 0.79745 |
https://necteo.com/burrumbuttock/what-is-a-polynomial-example.php | 1,716,693,671,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058861.60/warc/CC-MAIN-20240526013241-20240526043241-00437.warc.gz | 367,910,703 | 9,704 | What is this polynomial series? Mathematics Stack Exchange Polynomial codes for error detection Example Another example of multiplied by any polynomial can't produce a polynomial with an odd number of terms:
## Polynomials Calcworkshop
C# Polynomial Example CenterSpace. Definition of a polynomial: Learn to identify if a polynomial is a monomial, binomial, or trinomial., Choosing the best trendline for your data. Polynomial. A polynomial The following example shows an Order 2 polynomial trendline.
A quadratic polynomial is a polynomial of degree 2. A univariate quadratic polynomial has the form f(x)=a_2x^2+a_1x+a_0. An equation involving a quadratic polynomial What is an example of non-polynomial and non-linear system of equation with 3 variables? What is a polynomial and what is not a polynomial? What is a polynomial function?
Polynomial codes for error detection Example Another example of multiplied by any polynomial can't produce a polynomial with an odd number of terms: Learn how to add, subtract, and multiply polynomial expressions. For example, write (2x+3)(x-1) as 2xВІ+x-3.
Let us look at an example of polynomial interpolation to gain some intuitive between 1 and 2, so that interpolation with a quadratic polynomial will yield This unit is a brief introduction to the world of Polynomials. We will add, subtract, multiply, and even start factoring a polynomial.
What is a polynomial? Polynomials: Definitions & Evaluation. That last example above emphasizes that it is the variable portion of a term which must have a What are polynomials? How are they built? What can you do with them? And why are they important? Keep on reading to find out! For example, 3x, x 2/3,
translate between polynomial representations •Example = 6 3 + 7 •Fast Fourier Transform (FFT) takes advantage of the Polynomial regression Example: For input data: x = {0, 1, 2 The code listed below is good for up to 10000 data points and fits an order-5 polynomial,
Here are the steps required for Dividing by a Polynomial Containing More Than One Term (Long Division): Example 3 – Divide: Polynomial regression is an overdetermined system of equations that uses least squares as a method of For example, a 2nd degree polynomial
What is a Quadratic Polynomial? 10 Surefire Video Examples! Facebook Tweet Pin Shares 55. This lesson is all about Quadratic Polynomials in standard form. This unit is a brief introduction to the world of Polynomials. We will add, subtract, multiply, and even start factoring a polynomial.
Example showing how to create and manipulate Polynomial objects. Choosing the best trendline for your data. Polynomial. A polynomial The following example shows an Order 2 polynomial trendline
Polynomial regression Example: For input data: x = {0, 1, 2 The code listed below is good for up to 10000 data points and fits an order-5 polynomial, A polynomial function is a function that can be defined by evaluating a polynomial. More precisely, a function f of one argument from a given domain is a polynomial
What Are Polynomials? Introduction to Polynomials . Share Flipboard Email Print Noel Henderson / Getty Images Math. Example of Polynomial in a Equation. Cyclic Redundancy Code (CRC) Polynomial Selection For Embedded Networks For example, [Press92] lists 16-bit polynomials and polynomials are so that they can
Learn how to add, subtract, and multiply polynomial expressions. For example, write (2x+3)(x-1) as 2xВІ+x-3. Adding and Subtracting Polynomials. 1 hr 34 min 35 Examples. Introduction to Video: Adding and Subtracting Polynomials; What is a Polynomial? and How to Classify
Polynomial Models MATLAB & Simulink. Cyclic Redundancy Code (CRC) Polynomial Selection For Embedded Networks For example, [Press92] lists 16-bit polynomials and polynomials are so that they can, 3. Graphs of polynomial functions We have met some of the basic polynomials already. For example, f(x) = 2is a constant function and f(x) = 2x+1 is a linear function..
### Polynomial Trending Investopedia
What is polynomial Definition and Meaning - Math Dictionary. Remarks. For polynomials in one variable, finding the factors is equivalent to finding the roots: \$x=a\$ is a root of a polynomial \$p(x)\$ if and only if \$(x-a)\$ is a, Polynomial codes for error detection Example Another example of multiplied by any polynomial can't produce a polynomial with an odd number of terms:.
What Is an Example of a Prime Polynomial? Reference.com. Constant polynomial Polynomial containing only constant terms is a Constant polynomial. Example : -45, 56, -98 are constant polynomials. Types of Polynomials., In this section we look at factoring polynomials a topic that will appear in pretty much every chapter in this course and so is vital that For example, 2, 3, 5,.
### COMPLEX POLYNOMIALS The Library of Congress
define polynomial with an example.. Brainly.in. In this guide, polynomials are described in terms of their degree. For example, a third-degree (cubic) polynomial is given by. https://en.m.wikipedia.org/wiki/Sonine_polynomial Polynomials - more definitions and examples of polynomials and polynomial equations.
Polynomial regression is an overdetermined system of equations that uses least squares as a method of For example, a 2nd degree polynomial For example, the polynomial x 4 + 1, which is irreducible in the field of rational numbers, can be factored into. in the field of real numbers and into.
What is this polynomial series? [closed] Ask Question. For example, in row \$2\$, all coefficients are \$1\$ since the only coefficient in row \$1\$ was \$1\$. Constant & Linear Polynomials Constant polynomials A constant polynomial is the same thing as a constant function. is an example of a linear polynomial.
That means, for example, that 2x means two times x, or twice x. If x is 7, then 2x is 14. The parts of a polynomial separated by plus or minus signs are called "terms". What is a polynomial? Polynomials: Definitions & Evaluation. That last example above emphasizes that it is the variable portion of a term which must have a
In this section we look at factoring polynomials a topic that will appear in pretty much every chapter in this course and so is vital that For example, 2, 3, 5, What is this polynomial series? [closed] Ask Question. For example, in row \$2\$, all coefficients are \$1\$ since the only coefficient in row \$1\$ was \$1\$.
Remarks. For polynomials in one variable, finding the factors is equivalent to finding the roots: \$x=a\$ is a root of a polynomial \$p(x)\$ if and only if \$(x-a)\$ is a example of a polynomial this one has 3 terms: Polynomial comes from poly-(meaning "many") and -nomial (in this case meaning "term")
Here are the steps required for Dividing by a Polynomial Containing More Than One Term (Long Division): Example 3 – Divide: We present a few examples of polynomial subtraction below, and they will make the above idea clear to you: Simple Subtraction Examples Example 1:
Definition of a polynomial: Learn to identify if a polynomial is a monomial, binomial, or trinomial. I'm trying to understand algorithm complexity, and a lot of algorithms are classified as polynomial. I couldn't find an exact definition anywhere. I assume it is the
Constant polynomial Polynomial containing only constant terms is a Constant polynomial. Example : -45, 56, -98 are constant polynomials. Types of Polynomials. Polynomial regression is an overdetermined system of equations that uses least squares as a method of For example, a 2nd degree polynomial
What is an example of non-polynomial and non-linear system of equation with 3 variables? What is a polynomial and what is not a polynomial? What is a polynomial function? Polynomial interpolation is a method of estimating values between known data points. When graphical data contains a gap, but data is available on either side of the
What is this polynomial series? [closed] Ask Question. For example, in row \$2\$, all coefficients are \$1\$ since the only coefficient in row \$1\$ was \$1\$. Adding and Subtracting Polynomials. 1 hr 34 min 35 Examples. Introduction to Video: Adding and Subtracting Polynomials; What is a Polynomial? and How to Classify
Among the five non-collapsed answers as of writing this: One correctly answers a totally different question. Namely, “What are examples of a zero degree polynomial?” You've probably used a polynomial in your head more than once when shopping. For example, you might want to know how much three pounds of flour, two dozen eggs and
## What Is an Example of a Prime Polynomial? Reference.com
What is a non-polynomial? Quora. Among the five non-collapsed answers as of writing this: One correctly answers a totally different question. Namely, “What are examples of a zero degree polynomial?”, We present a few examples of polynomial subtraction below, and they will make the above idea clear to you: Simple Subtraction Examples Example 1:.
### What is a non-polynomial? Quora
Polynomials Algebra I Math Khan Academy. What is a Quadratic Polynomial? 10 Surefire Video Examples! Facebook Tweet Pin Shares 55. This lesson is all about Quadratic Polynomials in standard form., Choosing the best trendline for your data. Polynomial. A polynomial The following example shows an Order 2 polynomial trendline.
What is polynomial time complexity? Update Cancel. ad by What is an example of how to convert an algorithm of X time complexity and Y space complexity into W time What is this polynomial series? [closed] Ask Question. For example, in row \$2\$, all coefficients are \$1\$ since the only coefficient in row \$1\$ was \$1\$.
I'm trying to understand algorithm complexity, and a lot of algorithms are classified as polynomial. I couldn't find an exact definition anywhere. I assume it is the Note: If the only factors a polynomial are 1 and itself, then that polynomial is prime. To learn all about prime polynomials, check out this tutorial!
For example, they are used to form polynomial equations, For example, polynomial trending would be apparent on the graph that shows the relationship between the Remarks. For polynomials in one variable, finding the factors is equivalent to finding the roots: \$x=a\$ is a root of a polynomial \$p(x)\$ if and only if \$(x-a)\$ is a
You've probably used a polynomial in your head more than once when shopping. For example, you might want to know how much three pounds of flour, two dozen eggs and What is this polynomial series? [closed] Ask Question. For example, in row \$2\$, all coefficients are \$1\$ since the only coefficient in row \$1\$ was \$1\$.
Here are some examples of polynomial functions and the language we use to describe them: A polynomial as oppose to the monomial is a sum of monomials where each monomial is called a term. The degree of the polynomial is the greatest degree of its terms.
A degree in a polynomial function is the Degree of a Polynomial Function. In the following three examples, one can see how these polynomial degrees are Constant & Linear Polynomials Constant polynomials A constant polynomial is the same thing as a constant function. is an example of a linear polynomial.
For example, the polynomial x 4 + 1, which is irreducible in the field of rational numbers, can be factored into. in the field of real numbers and into. We present a few examples of polynomial subtraction below, and they will make the above idea clear to you: Simple Subtraction Examples Example 1:
In mathematics, a polynomial is an expression consisting of variables (also called indeterminates) and coefficients, that involves only the operations of additi… Polynomial inequalities show up in real world applications when intervals of numbers are involved. In this lesson, you'll learn what polynomial...
Definition of a polynomial: Learn to identify if a polynomial is a monomial, binomial, or trinomial. 21/06/2018В В· How to Solve Polynomials. factor out a term common to the second two terms in the polynomial. For example, the second two terms in the polynomial +
Shown above is a simple example of the polynomial, and this is how polynomials are usually expressed. The term having the largest value of exponent (2 in this case This unit is a brief introduction to the world of Polynomials. We will add, subtract, multiply, and even start factoring a polynomial.
### Polynomial regression Rosetta Code
Definition of a Polynomial Basic-mathematics.com. Choosing the best trendline for your data. Polynomial. A polynomial The following example shows an Order 2 polynomial trendline, 3. Graphs of polynomial functions We have met some of the basic polynomials already. For example, f(x) = 2is a constant function and f(x) = 2x+1 is a linear function..
### What Is an Example of a Prime Polynomial? Reference.com
Definition of a Polynomial Basic-mathematics.com. 21/06/2018В В· How to Solve Polynomials. factor out a term common to the second two terms in the polynomial. For example, the second two terms in the polynomial + https://en.wikipedia.org/wiki/Chebyshev_polynomials What Are Polynomial Models? Polynomial Model Structure. For example, n a =3 and n b =2 correspond to the following model: A (s) = s 4 + a 1 s 3 + a 2 s 2 + a 3 B.
What Are Polynomials? Introduction to Polynomials . Share Flipboard Email Print Noel Henderson / Getty Images Math. Example of Polynomial in a Equation. In mathematics, a polynomial is an expression consisting of variables (also called indeterminates) and coefficients, that involves only the operations of additi…
Example showing how to create and manipulate Polynomial objects. What Are Polynomials? Introduction to Polynomials . Share Flipboard Email Print Noel Henderson / Getty Images Math. Example of Polynomial in a Equation.
Polynomial Equation & Problems with Solution. A polynomial equation is an expression containing two or more Algebraic terms. Taken an example here – 5x 2 y 2 + 7y 2 + 9 Polynomial regression Example: For input data: x = {0, 1, 2 The code listed below is good for up to 10000 data points and fits an order-5 polynomial,
Learn what is polynomial? Definition and meaning on easycalculation math dictionary. This unit is a brief introduction to the world of Polynomials. We will add, subtract, multiply, and even start factoring a polynomial.
In this section we look at factoring polynomials a topic that will appear in pretty much every chapter in For example, 2, 3 So factor the polynomial in \ A quadratic polynomial is a polynomial of degree 2. A univariate quadratic polynomial has the form f(x)=a_2x^2+a_1x+a_0. An equation involving a quadratic polynomial
Here are the steps required for Dividing by a Polynomial Containing More Than One Term (Long Division): Example 3 – Divide: What is polynomial time complexity? Update Cancel. ad by What is an example of how to convert an algorithm of X time complexity and Y space complexity into W time
Constant polynomial Polynomial containing only constant terms is a Constant polynomial. Example : -45, 56, -98 are constant polynomials. Types of Polynomials. Here are some examples of polynomial functions and the language we use to describe them:
Let us look at an example of polynomial interpolation to gain some intuitive between 1 and 2, so that interpolation with a quadratic polynomial will yield Here are some examples of polynomial functions and the language we use to describe them:
Cyclic Redundancy Code (CRC) Polynomial Selection For Embedded Networks For example, [Press92] lists 16-bit polynomials and polynomials are so that they can Polynomial.h #ifndef POLYNOMIAL_H #define POLYNOMIAL_H #include using namespace std; class Polynomial { friend bool operator==(const Polynomial & lhs
The figures below give a scatterplot of the raw data and then another scatterplot with lines pertaining to a linear fit and a quadratic fit overlayed. Constant & Linear Polynomials Constant polynomials A constant polynomial is the same thing as a constant function. is an example of a linear polynomial.
Let us look at an example of polynomial interpolation to gain some intuitive between 1 and 2, so that interpolation with a quadratic polynomial will yield Choosing the best trendline for your data. Polynomial. A polynomial The following example shows an Order 2 polynomial trendline | 3,602 | 16,314 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2024-22 | latest | en | 0.879311 |
https://wikizero.com/index.php/en/Transcendental_function | 1,722,802,727,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640412404.14/warc/CC-MAIN-20240804195325-20240804225325-00650.warc.gz | 506,334,853 | 22,906 | Transcendental function
From Wikipedia the free encyclopedia
In mathematics, a transcendental function is an analytic function that does not satisfy a polynomial equation, in contrast to an algebraic function.[1][2] In other words, a transcendental function "transcends" algebra in that it cannot be expressed algebraically using a finite amount of terms.
Examples of transcendental functions include the exponential function, the logarithm, and the trigonometric functions.
Definition
Formally, an analytic function f (z) of one real or complex variable z is transcendental if it is algebraically independent of that variable.[3] This can be extended to functions of several variables.
History
The transcendental functions sine and cosine were tabulated from physical measurements in antiquity, as evidenced in Greece (Hipparchus) and India (jya and koti-jya). In describing Ptolemy's table of chords, an equivalent to a table of sines, Olaf Pedersen wrote:
The mathematical notion of continuity as an explicit concept is unknown to Ptolemy. That he, in fact, treats these functions as continuous appears from his unspoken presumption that it is possible to determine a value of the dependent variable corresponding to any value of the independent variable by the simple process of linear interpolation.[4]
A revolutionary understanding of these circular functions occurred in the 17th century and was explicated by Leonhard Euler in 1748 in his Introduction to the Analysis of the Infinite. These ancient transcendental functions became known as continuous functions through quadrature of the rectangular hyperbola xy = 1 by Grégoire de Saint-Vincent in 1647, two millennia after Archimedes had produced The Quadrature of the Parabola.
The area under the hyperbola was shown to have the scaling property of constant area for a constant ratio of bounds. The hyperbolic logarithm function so described was of limited service until 1748 when Leonhard Euler related it to functions where a constant is raised to a variable exponent, such as the exponential function where the constant base is e. By introducing these transcendental functions and noting the bijection property that implies an inverse function, some facility was provided for algebraic manipulations of the natural logarithm even if it is not an algebraic function.
The exponential function is written ${\displaystyle \exp(x)=e^{x}}$. Euler identified it with the infinite series ${\textstyle \sum _{k=0}^{\infty }x^{k}/k!}$, where k! denotes the factorial of k.
The even and odd terms of this series provide sums denoting cosh(x) and sinh(x), so that ${\displaystyle e^{x}=\cosh x+\sinh x.}$ These transcendental hyperbolic functions can be converted into circular functions sine and cosine by introducing (−1)k into the series, resulting in alternating series. After Euler, mathematicians view the sine and cosine this way to relate the transcendence to logarithm and exponent functions, often through Euler's formula in complex number arithmetic.
Examples
Let c be a positive constant. The following functions are transcendental:
{\displaystyle {\begin{aligned}f_{1}(x)&=x^{\pi }\\[2pt]f_{2}(x)&=c^{x}\\[2pt]f_{3}(x)&=x^{x}\\f_{4}(x)&=x^{\frac {1}{x}}={\sqrt[{x}]{x}}\\[2pt]f_{5}(x)&=\log _{c}x\\[2pt]f_{6}(x)&=\sin {x}\end{aligned}}}
For the second function ${\displaystyle f_{2}(x)}$, if we set ${\displaystyle c}$ equal to ${\displaystyle e}$, the base of the natural logarithm, then we get that ${\displaystyle e^{x}}$ is a transcendental function. Similarly, if we set ${\displaystyle c}$ equal to ${\displaystyle e}$ in ${\displaystyle f_{5}(x)}$, then we get that ${\displaystyle f_{5}(x)=\log _{e}x=\ln x}$ (that is, the natural logarithm) is a transcendental function.
Algebraic and transcendental functions
The most familiar transcendental functions are the logarithm, the exponential (with any non-trivial base), the trigonometric, and the hyperbolic functions, and the inverses of all of these. Less familiar are the special functions of analysis, such as the gamma, elliptic, and zeta functions, all of which are transcendental. The generalized hypergeometric and Bessel functions are transcendental in general, but algebraic for some special parameter values.
A function that is not transcendental is algebraic. Simple examples of algebraic functions are the rational functions and the square root function, but in general, algebraic functions cannot be defined as finite formulas of the elementary functions.[5]
The indefinite integral of many algebraic functions is transcendental. For example, the logarithm function arose from the reciprocal function in an effort to find the area of a hyperbolic sector.
Differential algebra examines how integration frequently creates functions that are algebraically independent of some class, such as when one takes polynomials with trigonometric functions as variables.
Transcendentally transcendental functions
Most familiar transcendental functions, including the special functions of mathematical physics, are solutions of algebraic differential equations. Those that are not, such as the gamma and the zeta functions, are called transcendentally transcendental or hypertranscendental functions.[6]
Exceptional set
If f is an algebraic function and ${\displaystyle \alpha }$ is an algebraic number then f (α) is also an algebraic number. The converse is not true: there are entire transcendental functions f such that f (α) is an algebraic number for any algebraic α.[7] For a given transcendental function the set of algebraic numbers giving algebraic results is called the exceptional set of that function.[8][9] Formally it is defined by:
${\displaystyle {\mathcal {E}}(f)=\left\{\alpha \in {\overline {\mathbb {Q} }}\,:\,f(\alpha )\in {\overline {\mathbb {Q} }}\right\}.}$
In many instances the exceptional set is fairly small. For example, ${\displaystyle {\mathcal {E}}(\exp )=\{0\},}$ this was proved by Lindemann in 1882. In particular exp(1) = e is transcendental. Also, since exp() = −1 is algebraic we know that cannot be algebraic. Since i is algebraic this implies that π is a transcendental number.
In general, finding the exceptional set of a function is a difficult problem, but if it can be calculated then it can often lead to results in transcendental number theory. Here are some other known exceptional sets:
• Klein's j-invariant ${\displaystyle {\mathcal {E}}(j)=\left\{\alpha \in {\mathcal {H}}\,:\,[\mathbb {Q} (\alpha ):\mathbb {Q} ]=2\right\},}$ where ${\displaystyle {\mathcal {H}}}$ is the upper half-plane, and ${\displaystyle [\mathbb {Q} (\alpha ):\mathbb {Q} ]}$ is the degree of the number field ${\displaystyle \mathbb {Q} (\alpha ).}$ This result is due to Theodor Schneider.[10]
• Exponential function in base 2: ${\displaystyle {\mathcal {E}}(2^{x})=\mathbb {Q} ,}$This result is a corollary of the Gelfond–Schneider theorem, which states that if ${\displaystyle \alpha \neq 0,1}$ is algebraic, and ${\displaystyle \beta }$ is algebraic and irrational then ${\displaystyle \alpha ^{\beta }}$ is transcendental. Thus the function 2x could be replaced by cx for any algebraic c not equal to 0 or 1. Indeed, we have: ${\displaystyle {\mathcal {E}}(x^{x})={\mathcal {E}}\left(x^{\frac {1}{x}}\right)=\mathbb {Q} \setminus \{0\}.}$
• A consequence of Schanuel's conjecture in transcendental number theory would be that ${\displaystyle {\mathcal {E}}\left(e^{e^{x}}\right)=\emptyset .}$
• A function with empty exceptional set that does not require assuming Schanuel's conjecture is ${\displaystyle f(x)=\exp(1+\pi x).}$
While calculating the exceptional set for a given function is not easy, it is known that given any subset of the algebraic numbers, say A, there is a transcendental function whose exceptional set is A.[11] The subset does not need to be proper, meaning that A can be the set of algebraic numbers. This directly implies that there exist transcendental functions that produce transcendental numbers only when given transcendental numbers. Alex Wilkie also proved that there exist transcendental functions for which first-order-logic proofs about their transcendence do not exist by providing an exemplary analytic function.[12]
Dimensional analysis
In dimensional analysis, transcendental functions are notable because they make sense only when their argument is dimensionless (possibly after algebraic reduction). Because of this, transcendental functions can be an easy-to-spot source of dimensional errors. For example, log(5 metres) is a nonsensical expression, unlike log(5 metres / 3 metres) or log(3) metres. One could attempt to apply a logarithmic identity to get log(5) + log(metres), which highlights the problem: applying a non-algebraic operation to a dimension creates meaningless results.
References
1. ^ Townsend, E.J. (1915). Functions of a Complex Variable. H. Holt. p. 300. OCLC 608083625.
2. ^ Hazewinkel, Michiel (1993). Encyclopedia of Mathematics. Vol. 9. pp. 236.
3. ^ Waldschmidt, M. (2000). Diophantine approximation on linear algebraic groups. Springer. ISBN 978-3-662-11569-5.
4. ^ Pedersen, Olaf (1974). Survey of the Almagest. Odense University Press. p. 84. ISBN 87-7492-087-1.
5. ^
6. ^ Rubel, Lee A. (November 1989). "A Survey of Transcendentally Transcendental Functions". The American Mathematical Monthly. 96 (9): 777–788. doi:10.1080/00029890.1989.11972282. JSTOR 2324840.
7. ^ van der Poorten, A.J. (1968). "Transcendental entire functions mapping every algebraic number field into itself". J. Austral. Math. Soc. 8 (2): 192–8. doi:10.1017/S144678870000522X. S2CID 121788380.
8. ^ Marques, D.; Lima, F.M.S. (2010). "Some transcendental functions that yield transcendental values for every algebraic entry". arXiv:1004.1668v1 [math.NT].
9. ^ Archinard, N. (2003). "Exceptional sets of hypergeometric series". Journal of Number Theory. 101 (2): 244–269. doi:10.1016/S0022-314X(03)00042-8.
10. ^ Schneider, T. (1937). "Arithmetische Untersuchungen elliptischer Integrale". Math. Annalen. 113: 1–13. doi:10.1007/BF01571618. S2CID 121073687.
11. ^ Waldschmidt, M. (2009). "Auxiliary functions in transcendental number theory". The Ramanujan Journal. 20 (3): 341–373. arXiv:0908.4024. doi:10.1007/s11139-009-9204-y. S2CID 122797406.
12. ^ Wilkie, A.J. (1998). "An algebraically conservative, transcendental function". Paris VII Preprints. 66. | 2,664 | 10,384 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 26, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2024-33 | latest | en | 0.919524 |
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# Begginer question
P: n/a I googled for it but found 0 results "|=" my question is: what does |= do? | is a bitwise or = is equal but what |= do? I am using C not C++ if it makes any difference. Nov 4 '06 #1
4 Replies
P: n/a fermineutron wrote: I googled for it but found 0 results "|=" my question is: what does |= do? | is a bitwise or = is equal but what |= do? I am using C not C++ if it makes any difference. You should have a good C book. Then you should read the book. -- Joe Wright "Everything should be made as simple as possible, but not simpler." --- Albert Einstein --- Nov 4 '06 #2
P: n/a fermineutron: >| is a bitwise or= is equal but what |= do? x = x + 5; x += 5; x = x | 5 x |= 5; -- Frederick Gotham Nov 4 '06 #3
P: n/a Frederick Gotham wrote: fermineutron: | is a bitwise or = is equal but what |= do? x = x + 5; x += 5; x = x | 5 x |= 5; Thanks Nov 4 '06 #4
P: n/a "fermineutron" San Diego Supercomputer Center <* We must do something. This is something. Therefore, we must do this. Nov 4 '06 #5
### This discussion thread is closed
Replies have been disabled for this discussion. | 376 | 1,267 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2020-10 | latest | en | 0.944134 |
https://centsai.com/life/health-and-money/cost-of-being-unvaccinated/?co_id=8142 | 1,656,951,823,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104432674.76/warc/CC-MAIN-20220704141714-20220704171714-00017.warc.gz | 199,962,995 | 37,415 | I’ve been thinking a lot lately about COVID-19. It drives many of my everyday decisions like sitting outside at restaurants in 90-degree Florida weather (“I’d rather sweat than die”), events that I choose not to attend, and interactions with other people.
I recently posted the message, below, on all of my social media channels to implore people to get vaccinated. It tells the true story of an unvaccinated person who recently died. Many people followed up with words like sad and unnecessary.
According to the Kaiser Family Foundation, COVID-19 hospitalizations in June and July cost the U.S. health system over \$2 billion dollars! Most of those hospitalized were unvaccinated individuals. Being a certified financial planner, I started to wonder what costs COVID patients, themselves, were incurring and decided to research the financial costs of being unvaccinated.
My thinking is that this information could motivate some people to get vaccinated. At the risk of touching what can, unfortunately, be a politically charged topic, even one life saved is a victory! An analogy is that many smokers will not be motivated to quit by seeing the long-term cost of cigarettes, but a few might be.
Below are 10 potential expenses for people who remain unvaccinated without a solid medical or other legal reason to do so.
Full Disclosure: It should be noted that some — but not all — of these financial impacts (the loss of a breadwinner’s income following death, medical bills, the cost of a funeral, wealth depletion, and financially stressed survivors) can occur whether someone who gets COVID-19 was vaccinated or not.
However, the risk of incurring debt from the virus and experiencing resulting financial stress is lower when you are vaccinated because you are less likely to die or need very expensive health care.
## Loss of a Breadwinner’s Income
When people die, their income stops. Period. Add in some time value of money calculations, and the loss is further magnified. Using simple math, losing the income of someone earning the 2021 U.S. median income of \$61,937 for 20 years is a loss of \$1,238,40 of potential income, excluding raises and promotions.
## Loss of a Job
Employers are starting to wield “sticks” instead of dangle “carrots” to get their workforce vaccinated. Some unvaccinated workers may lose their jobs as employers announce “zero tolerance” mandatory vaccination policies in conjunction with a required return to workplaces.
Most legal experts say this is permissible as long as employers make reasonable accommodations for workers who are unvaccinated for reasons such as disability and pregnancy.
## No “Sick Day” Paychecks
Delta Airlines announced in August that it will limit the number of paid sick days that unvaccinated employees will be allowed to take if they contract COVID-19.
It will also charge unvaccinated employees \$200 monthly health insurance premium surcharges starting November 1. Other employers are expected to enforce similar policies.
## Funeral and Burial Costs
Average funeral costs vary by state but on average cost between \$7,000 and \$9,000. According to Federal Reserve research, 36 percent of U.S. households in 2020 would have difficulty paying a \$400 emergency expense. Simply put, financially fragile families cannot afford a funeral without selling assets or incurring new debt.
## Crushing Medical and Other Debt
Fair Health reported that 15 percent to 20 percent of people who get COVID-19 and seek treatment may need a hospital stay and that the average hospital stay cost is \$73,300 for a patient without health insurance and \$38,221 for an insured patient.
Some bills for long hospital stays are well into six figures. Many waivers that initially exempted patients from out-of-pocket COVID-related expenses have expired and patients may be “on the hook” for costs that third-party insurers do not cover, including ongoing “long-COVID” medical bills that could last for years.
## Financially Stressed Survivors
Unvaccinated people who are hospitalized and/or die and incur debt as a result of COVID-19 put undue financial (as well as emotional) stress on their survivors.
Cases have been reported of families swamped by medical expenses, getting hit with balance billing statements despite Congressional bans, and declaring bankruptcy to escape mounting medical expenses.
When unvaccinated people lack health or disability insurance or die without life insurance, the situation is even more dire for their survivors, who may need to seek public assistance to get by.
## Wealth Depletion
As noted above, COVID-19 is expensive. Though commercial payers are expected to pay some expenses, so, too, will patients and their families. Some have turned to crowd funding their medical bills on the GoFundMe website.
Every dollar that is withdrawn from savings or unable to be earned due to death or COVID-19 treatment or quarantine represents a “hit” to wealth that an individual or family might otherwise accumulate.
## Possible Insurance Impacts
To date, the largest life insurance companies say they are not currently considering vaccine status during policy underwriting (except for those with high-risk conditions) or when deciding whether to pay a claim.
In addition, vaccination status does not affect premiums for existing policies. This could change as more becomes known about the virus. Policy applicants will likely be asked if they have had COVID. As noted above, unvaccinated workers may see their health insurance premiums increase.
The Wall Street Journal recently noted that Live Nation Entertainment, the world’s largest concert promoter, will require proof of vaccination or a negative COVID-19 test for entry at its events beginning in October.
## Loss of Money-Saving “Social Capital”
Social capital is the network of relationships that people have that they can “lean on” when times are tough. Especially when people have limited financial resources, social capital is invaluable.
People can drive you places, care for you or your children, and lend you money. It has been widely reported that people who are unvaccinated are losing friends.
From actress Jennifer Aniston to the neighbor next door, many people do not want to be around unvaccinated friends who increase their risk of breakthrough COVID. Bottom line: People may not be there for you when you need them most.
Finally, since vaccines are free and easy to obtain, some people are questioning the costs that unvaccinated people impose on everyone in society. They are tired of experiencing “financial fallout” as a consequence of other people’s choices.
This includes taxpayers paying for Medicare and Medicaid, policyholders paying higher health insurance premiums, and those facing long waits for “routine” medical services at overwhelmed hospitals. This perspective is not unique, however. Similar comments are also sometimes made about societal costs of other “voluntary” health-related practices such as smoking.
## The Bottom Line
It is now becoming more acceptable to debate, support, and even adopt, policies that deny care, limit benefits, or pass along COVID-related costs to unvaccinated people as a way to motivate them or as a form of “punishment.”
Unlike “lifestyle” factors (e.g., diet, smoking, physical activity) that primarily affect the health of one individual, vaccination status has the potential to affect large groups of people (e.g., a workplace) during a global pandemic.
Stay tuned — we are 18 months into the pandemic with many more health and financial impacts likely to come.
With many diverse and conflicting viewpoints about vaccination and ever-evolving research findings and government guidelines, it is likely to be a “bumpy ride.” Stay safe and be well. | 1,545 | 7,782 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2022-27 | latest | en | 0.972231 |
https://www.hackmath.net/en/word-math-problems/mathematical-olympiad?tag_id=109 | 1,611,146,994,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703520883.15/warc/CC-MAIN-20210120120242-20210120150242-00557.warc.gz | 814,899,525 | 11,435 | # Mathematical Olympiad + natural numbers - math problems
#### Number of problems found: 20
• Year 2018
The product of the three positive numbers is 2018. What are the numbers?
• Z9–I–1
In all nine fields of given shape to be filled natural numbers so that: • each of the numbers 2, 4, 6, and 8 is used at least once, • four of the inner square boxes containing the products of the numbers of adjacent cells of the outer square, • in the cir
• MO C–I–1 2018
An unknown number is divisible by just four numbers from the set {6, 15, 20, 21, 70}. Determine which ones.
• Z7-I-4 stars 4949
Write instead of stars digits so the next write of product of the two numbers to be valid: ∗ ∗ ∗ · ∗ ∗ ∗ ∗ ∗ ∗ ∗ 4 9 4 9 ∗ ∗ ∗ ∗ ∗ ∗ 4 ∗ ∗
• Six-digit primes
Find all six-digit prime numbers that contain each one of digits 1,2,4,5,7 and 8 just once. How many are they?
• Twos
Vojta started writing the number of this year 2019202020192020 into the workbook. .. And so he kept going. When he wrote 2020 digits, no longer enjoyed it. How many twos did he write?
• Star equation
Write digits instead of stars so that the sum of the written digits is odd and is true equality: 42 · ∗8 = 2 ∗∗∗
• Last digit
What is the last number of 2016 power of 2017
• MO 2016 Numerical axis
Cat's school use a special numerical axis. The distance between the numbers 1 and 2 is 1 cm, the distance between the numbers 2 and 3 is 3 cm, between the numbers 3 and 4 is 5 cm and so on, the distance between the next pair of natural numbers is always i
• Self-counting machine
The self-counting machine works exactly like a calculator. The innkeeper wanted to add several three-digit natural numbers on his own. On the first attempt, he got the result in 2224. To check, he added these numbers again and he got 2198. Therefore, he a
• Number train
The numbers 1,2,3,4,5,6,7,8 and 9 traveled by train. The train had three cars and each was carrying just three numbers. No. 1 rode in the first carriage, and in the last carriage was all odd numbers. The conductor calculated sum of the numbers in the firs
• Octahedron - sum
On each wall of a regular octahedron is written one of the numbers 1, 2, 3, 4, 5, 6, 7 and 8, wherein on different sides are different numbers. For each wall John make the sum of the numbers written of three adjacent walls. Thus got eight sums, which also
• Alarm clock
The old watchmaker has a unique digital alarm in its collection that rings whenever the sum of digits of the alarm is equal to 21. Find out when the alarm clock will ring. What is their number? List all options . ..
• Z9–I–4 MO 2017
Numbers 1, 2, 3, 4, 5, 6, 7, 8 and 9 were prepared for a train journey with three wagons. They wanted to sit out so that three numbers were seated in each carriage and the largest of each of the three was equal to the sum of the remaining two. The conduct
• Coloured numbers
Mussel wrote four different natural numbers with coloured markers: red, blue, green and yellow. When the red number divides by blue, it gets the green number as an incomplete proportion, and yellow represents the remainder after this division. When it div
• Clubhouse
There were only chairs and table in the clubhouse. Each chair had four legs, and the table was triple. Scouts came to the clubhouse. Everyone sat on their chair, two chairs were left unoccupied, and the number of legs in the room was 101. How many chairs
• Z9-I-4
Kate thought a five-digit integer. She wrote the sum of this number and its half at the first line to the workbook. On the second line wrote a total of this number and its one fifth. On the third row, she wrote a sum of this number and its one nines. Fina
• MO Z8-I-1 2018
Fero and David meet daily in the elevator. One morning they found that if they multiply their current age, they get 238. If they did the same after four years, this product would be 378. Determine the sum of the current ages of Fero and David.
• Shepherd
Kuba makes a deal with a shepherd to take care of his sheep. Shepherd said Kuba that after a year of service, he would receive twenty gold coins and one sheep. But Kuba resigned just after the seventh month of service. But shepherd rewarded him and paid h
• Squirrels
The squirrels discovered a bush with hazelnuts. The first squirrel plucked one nut, the second squirrel two nuts, the third squirrel three nuts. Each new squirrel always tore one nut more than the previous squirrel. When they plucked all the nuts from the
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https://www.shaalaa.com/question-bank-solutions/using-the-rolle-s-theorem-determine-the-values-of-x-at-which-the-tangent-is-parallel-to-the-x-axis-for-the-following-functions-f-x-x2-2xx-2-x-1-6-mean-value-theorem_232925 | 1,679,993,306,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948817.15/warc/CC-MAIN-20230328073515-20230328103515-00303.warc.gz | 1,105,330,328 | 10,014 | Tamil Nadu Board of Secondary EducationHSC Arts Class 12th
# Using the Rolle’s theorem, determine the values of x at which the tangent is parallel to the x-axis for the following functions: f(x)=x2-2xx+2,x∈[-1,6] - Mathematics
Sum
Using the Rolle’s theorem, determine the values of x at which the tangent is parallel to the x-axis for the following functions:
f(x) = (x^2 - 2x)/(x + 2), x ∈ [-1, 6]
#### Solution
f(– 1) = (1 + 2)/(-1 + 2) = 3
f(6) = (36 - 12)/8 = 24/8 = 3
⇒ f(– 1) = 3 = f(6)
f(x) is continuous on [– 1, 6]
f(x) is differentiable on (– 1, 6)
Now, f'(x ) = ((x + 2)(2x - 2) - (x^2 - 2x)(1))/(x + 2)^2
= (x^2 + 4x - 4)/(x + 2)^2
Since the tangent is parallel to the x-axis.
f'(x) = 0
⇒ x2 + 4x – 4 = 0
⇒ x = - (4 +- sqrt(16 + 16))/2
x = - (4 +- 4sqrt(2))/2
= - 2 +- 2sqrt(2)
x = - 2 +- 2sqrt(2) ∈ (-1, 6)
Concept: Mean Value Theorem
Is there an error in this question or solution?
#### APPEARS IN
Tamil Nadu Board Samacheer Kalvi Class 12th Mathematics Volume 1 and 2 Answers Guide
Chapter 7 Applications of Differential Calculus
Exercise 7.3 | Q 2. (ii) | Page 21
Share | 441 | 1,110 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2023-14 | latest | en | 0.759046 |
https://questions.llc/questions/777402 | 1,702,065,679,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100769.54/warc/CC-MAIN-20231208180539-20231208210539-00810.warc.gz | 543,968,025 | 6,636 | # Write the equation of the tangent line to the graph of the function at the indicated point. Check the reasonableness of your answer by graphing both the function and the tangent line.
Y=(x-2)/(15-x^2)
x=-4
HOW DO WE DO THIS
## To find the equation of the tangent line to the graph of a function at a given point, we can follow these steps:
1. Find the derivative of the function.
2. Evaluate the derivative at the given x-coordinate to find the slope of the tangent line.
3. Use the point-slope form of a linear equation to form the equation of the tangent line using the slope and the given point.
Now let's apply these steps to the specific problem:
1. Find the derivative of the function Y = (x-2)/(15-x^2):
- To simplify the calculation, we can rewrite the function using a common denominator:
Y = (x-2)/[(sqrt(15)-x)(sqrt(15)+x)].
- Apply the quotient rule to find the derivative:
Y' = [((sqrt(15)-x)(sqrt(15)+x)) - (x-2)(2x)]/[(sqrt(15)-x)(sqrt(15)+x)]^2.
- Simplify the expression inside the brackets:
Y' = [(15 - x^2 + 2x - 2x^2) - (2x^2 - 4x)]/[(sqrt(15)-x)(sqrt(15)+x)]^2.
Y' = (15 - 3x^2)/[(sqrt(15)-x)(sqrt(15)+x)]^2.
- Simplify further if desired.
2. Evaluate the derivative at the given x-coordinate, x = -4:
Y'(-4) = (15 - 3(-4)^2)/[(sqrt(15)-(-4))(sqrt(15)+(-4))]^2.
Y'(-4) = (15 - 3(16))/[(sqrt(15)+4)(sqrt(15)-4)]^2.
Y'(-4) = (15 - 48)/[(sqrt(15)+4)(sqrt(15)-4)]^2.
3. Use the point-slope form of a linear equation and the given point (-4, Y(-4)) = (-4, Y(-4)) to write the equation of the tangent line:
Y - Y(-4) = Y'(-4)(x - (-4)).
Y - Y(-4) = Y'(-4)(x + 4).
Now you have an equation of the tangent line in point-slope form. To check the reasonableness of the answer, you can graph both the original function and the tangent line. | 574 | 1,761 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.75 | 5 | CC-MAIN-2023-50 | latest | en | 0.778451 |
https://www.mql5.com/en/forum/133687 | 1,493,311,530,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917122619.60/warc/CC-MAIN-20170423031202-00243-ip-10-145-167-34.ec2.internal.warc.gz | 949,887,479 | 17,531 | # Double Weighted Moving Average ... Missing?
16
2011.05.24 13:52
Hi All:
We have DEMA, TEMA, AMA and etc on MQL4 Code Base but not the Double Weighted Moving Average which I think is very useful as well.
Can someone shed some light on coding this Double Weighted Moving Average?
In addition to that, I not so sure about the calculation phase but is the Linear Weighted Moving Average on MetaTrader is the same as the common Weight Moving Average that we can found on other platforms such as Ninja and TradeStation? Can someone confirm on this?
Regards,
James
16
2011.05.25 06:27
Can anyone help?
TIA!
2584
2011.05.25 09:51
3827:
Hi All:
We have DEMA, TEMA, AMA and etc on MQL4 Code Base but not the Double Weighted Moving Average which I think is very useful as well.
Can someone shed some light on coding this Double Weighted Moving Average?
In addition to that, I not so sure about the calculation phase but is the Linear Weighted Moving Average on MetaTrader is the same as the common Weight Moving Average that we can found on other platforms such as Ninja and TradeStation? Can someone confirm on this?
Regards,
James
I never heard about a double weighted moving average what is the formula for calculating a double weighted moving average ? Are you sure you are not talking about a double exponential moving average ? Also known as DEMA.
16
2011.05.25 11:16
No, it is different than the DEMA.
It is the standard weighted moving average 'weighted' one more time. Pretty common in my trading circle.
TIA!
2584
2011.05.25 16:38
so is it an average of an average ?
16
2011.05.25 17:53
Yes, just like DEMA. You average the EMA one more time. Double it. For TEMA, you average the DEMA one more time and make it 3 times in average.
For Double Weighted Moving Average, take the standard Weighted Moving Average and average it one more time with the same period/length.
Can you pls help?
Many TIA!
14765
2011.05.25 19:36
```lwmaBuf[]
dwmaBuf[]
...
double LWMA(double array[], int per, int bar)
{
double Sum = 0;
double Weight = 0;
for(int i = 0;i < per;i++)
{
Weight+= (per - i);
Sum += array[bar+i]*(per - i);
}
if(Weight>0) double lwma = Sum/Weight;
else lwma = 0;
return(lwma);
}
start(){...
for(...){
lwmaBuf[shift] = LWMA(Close, MAperiod, shift);
dwmaBuf[shift] = LWMA(lwmaBuf, MAperiod, shift);
}
```
16
2011.05.26 05:24
WHRoeder, pls pardon that my programming skill is poor. Can you post a full version of the code? Pls. Many thanks to you!
14765
2011.05.26 16:25
Take ANY indicator and modify. No slaves here, learn to code or pay someone. | 704 | 2,576 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2017-17 | longest | en | 0.931253 |
http://www.formuladirectory.com/user/formula/25 | 1,550,932,632,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550249504746.91/warc/CC-MAIN-20190223142639-20190223164639-00387.warc.gz | 334,068,362 | 4,721 | HOSTING A TOTAL OF 318 FORMULAS WITH CALCULATORS
## Perimeter of a parallelogram
The perimeter of parallelogram is the total distance around the outside, which can be found by adding together the length of each side. In the case of a parallelogram, each pair of opposite sides is the same length, so the perimeter is twice the base plus twice the side length.
## $2\left[l+b\right]$
Perimeter of a parallelogram (P) = 2 × (length(l) + Breadth(b))
ENTER THE VARIABLES TO BE USED IN THE FORMULA
Similar formulas which you may find interesting. | 137 | 547 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2019-09 | latest | en | 0.802171 |
https://yamol.tw/item_essay-7.%5B15%5D+Suppose+we+have+12+integers+60+40..-471848.htm | 1,695,656,731,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233509023.57/warc/CC-MAIN-20230925151539-20230925181539-00149.warc.gz | 1,197,920,835 | 25,543 | # 中山◆電機◆資料結構題庫
【非選題】
7.[15] Suppose we have 12 integers: 60, 40, 25, 10, 50, 80, 15, 65, 30, 90, 20, 55. Construct a binary search tree by inserting the integers, starting from 60, one after another. Please answer the following questions for the binary search tree:
【題組】7.5 [3] What is the height of the binary tree? Let one-node tree have a depth of 1. | 126 | 355 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2023-40 | latest | en | 0.835653 |
http://www.planetside.co.uk/forums/index.php/topic,23135.0.html?PHPSESSID=k92acvj98gpu85nfkot1nhn524 | 1,527,266,707,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867140.87/warc/CC-MAIN-20180525160652-20180525180652-00607.warc.gz | 433,786,254 | 9,521 | ### Author Topic: micro export to 3dMax problem (Read 1076 times)
#### dorianvan
• Member
• Posts: 377
##### micro export to 3dMax problem
« on: May 08, 2017, 11:54:58 PM »
I have a problem with the geometry that I exported with the micro exporter. I have a beach scene with a water line in terragen, so the slope is downward into the water. I export orthographic straight down, then merge the obj into 3dMax, then I place a flat plane where the water line goes. The problem is the geometry doesn't give me a beach like in terragen. Any ideas?
-Dorian
#### paq
• Terragen Alpha Tester (Win)
• Posts: 192
##### Re: micro export to 3dMax problem
« Reply #1 on: May 09, 2017, 02:47:11 AM »
Maybe because of the planet curvature ? I think Lake object in Terragen is not planar either.
Just an idea
Gameloft
#### dorianvan
• Member
• Posts: 377
##### Re: micro export to 3dMax problem
« Reply #2 on: May 09, 2017, 03:37:31 AM »
That thought crossed my mind too, but that doesn't seem likely because this is such a tiny area compared to the earth. This is a quite a bit tipped as you can see. These are the exact same camera angles, but the merged obj seems to be tipped to the right. I need this to be precise because I will be doing water with another program and need an accurately level obj mesh. I must be missing something.
-Dorian
#### Dune
• Terragen Alpha Tester (Win)
• Posts: 13542
• Corkscrew Bird
##### Re: micro export to 3dMax problem
« Reply #3 on: May 09, 2017, 06:07:45 AM »
Problem might be that the actual beach is maybe 1-2K from 0/0/0. So a plane perpendicular to Y would have an angle, and the exporter would probably export on this angle too. But I'm not familiar with exporting. Can you relocate in Max to the actual location used in TG? Or use a sphere?
In case you still haven't seen enough of my work: www.ulco-art.nl
#### dorianvan
• Member
• Posts: 377
##### Re: micro export to 3dMax problem
« Reply #4 on: May 09, 2017, 12:27:35 PM »
I suppose the subject of this thread really should be "Does moving DEM mess with micro exporter data?"
« Last Edit: May 09, 2017, 08:57:45 PM by dorianvan »
-Dorian
#### dorianvan
• Member
• Posts: 377
##### Re: micro export to 3dMax problem
« Reply #5 on: May 10, 2017, 02:21:07 PM »
It appears that the DEM shouldn't be moved when you need to export obj meshes. It seems to tilt geometry and flatten it out some, and make it very touchy to do anything with. The non-moved DEM seemed to work great.
I wonder if there is a way to counter this problem so the DEM can be moved when also needing to export.
Maybe the exporter can't deal yet with moved DEM's, or it's a bug...
« Last Edit: May 10, 2017, 06:26:08 PM by dorianvan »
-Dorian
#### Oshyan
• Planetside Staff
• Posts: 12141
• Holy snagging ducks!
##### Re: micro export to 3dMax problem
« Reply #6 on: May 11, 2017, 12:38:11 AM »
Reading this thread now, I realize the "tilt" you're talking about is more subtle, something you might not really notice in a render but would show up in export. In this case I would check the options on the Displacement tab of the Heightfield Shader node. Particularly the Displacement Direction and Flatten Surface First options. See if changing any of those has an effect.
- Oshyan
#### dorianvan
• Member
• Posts: 377
##### Re: micro export to 3dMax problem
« Reply #7 on: May 11, 2017, 03:22:45 AM »
It actually has tilted quite a bit for such a tiny piece of real estate (height at 200m), and it's the wrong way tilt too! My camera is orthographic aiming straight down 90 degrees. Plus, it flatted the terrain out some and is very "touchy" in 3dMax. At the start, Ulco had relocated the DEM to a different location because of some reason I can't recall, and then he did his work and I tried to export a piece. Didn't work. However, as a test, when it was moved back to the original geo-referenced location, the exported obj worked great, none of those problems.
The question is, has the micro exporter been tested to see if it exports correctly on a moved DEM tiff? We'd rather not move it if we don't have to, because everything is just right now.
-Dorian
#### Dune
• Terragen Alpha Tester (Win)
• Posts: 13542
• Corkscrew Bird
##### Re: micro export to 3dMax problem
« Reply #8 on: May 11, 2017, 05:39:35 AM »
The area is near equator, and the reason for moving the DEM was that trees were all on a nasty angle, and camera's have to be adjusted every time you rotate. Pretty awkward.
In case you still haven't seen enough of my work: www.ulco-art.nl
#### Matt
• Planetside Staff
• Posts: 2978
• I'm the crazy one
##### Re: micro export to 3dMax problem
« Reply #9 on: May 11, 2017, 09:59:14 PM »
The micro exporter simply exports the rendered geometry. There isn't anything shader-specific that could change this. Whatever is exported is the surface produced by the shader and it will be the same in the Terragen render.
When working on a planetary scale the surface will deviate from a plane, increasingly as you move away from the origin. Also, by default the Heightfield Shader displaces along the local "up" vector, but this changes in space as the planet has a curved surface. If Ulco moved the height field it's going to be different from before. But there are options in the Displacement tab to change the displacement direction to the Y axis and to flatten surface to Y = 0. If you of change both of those things you can simulate a planar world. Alternatively you can render the terrain using a Plane instead of a Planet.
Matt
Just because milk is white doesn't mean that clouds are made of milk.
#### Dune
• Terragen Alpha Tester (Win)
• Posts: 13542
• Corkscrew Bird
##### Re: micro export to 3dMax problem
« Reply #10 on: May 12, 2017, 06:11:55 AM »
Thanks Matt! I put all on two planes and it works perfectly. Off to the next challenge...
In case you still haven't seen enough of my work: www.ulco-art.nl
#### dorianvan
• Member
• Posts: 377
##### Re: micro export to 3dMax problem
« Reply #11 on: May 12, 2017, 03:46:58 PM »
Yes thanks Matt! And thanks Ulco for understanding what Matt/Oshyan were saying (sometimes above my head). Also just solved my multi-layered polygon issue with your helping Klaus. It would be great if TG didn't create doubled up polygons at the bucket borders though. Maybe you could program it so the exporter will automatically adjust to the size of the image rendered so that won't happen, who would ever want doubled polys anyway.
-Dorian
#### Oshyan
• Planetside Staff
• Posts: 12141
• Holy snagging ducks!
##### Re: micro export to 3dMax problem
« Reply #12 on: May 14, 2017, 03:24:28 AM »
We have intentions to improve the export process in the future. Limiting it to a single thread is a pretty non-ideal option though (it makes it a lot slower), so hopefully we can find another option to improve the quality of the exported geometry.
- Oshyan
anything | 1,859 | 6,889 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2018-22 | latest | en | 0.89896 |
https://www.ammarksman.com/what-is-a-grid-square-on-a-map/ | 1,701,269,614,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100112.41/warc/CC-MAIN-20231129141108-20231129171108-00213.warc.gz | 734,051,028 | 10,220 | # What is a grid square on a map?
## What is a grid square on a map?
A grid of squares helps the map-reader to locate a place. The vertical lines are called eastings. They are numbered – the numbers increase to the east. The horizontal lines are called northings as the numbers increase in an northerly direction.
How many grid squares are in the United States?
There are 488 grid squares in the contiguous 48 states. Each 2° by 1° area is about 100 by 70 miles. Each major field “DN” contains 10×10 or 100 numbered grid squares, e.g. DN00 – DN99. Each square is further divided by adding two letters, as in CN97xn, about 4 by 3 miles.
What is the size of a grid square?
When the term ‘grid square’ is used, it can refer to a square with a side length of 10 km (6 mi), 1 km, 100 m (328 ft), 10 m or 1 m, depending on the precision of the coordinates provided.
### Why are the grid lines numbered?
Lines numbers respect the writing mode of the document and so in a right-to-left language for example, column line 1 will be on the right of the grid. The image below shows the line numbers of the grid, assuming the language is left-to-right.
What is the size of a grid square on a map?
approximately 70 × 100 miles
An instrument of the Maidenhead Locator System (named after the town outside London where it was first conceived by a meeting of European VHF managers in 1980), a grid square measures 1° latitude by 2° longitude and measures approximately 70 × 100 miles in the continental US.
What is the United States national grid based on?
USNG relies on the familiar Universal Transverse Mercator (UTM) coordinate system and is applied not only in the United States but also worldwide. USNG is a nonproprietary alphanumeric referencing system derived from the Military Grid Reference System (MGRS).
#### What is the US national grid system?
The United States National Grid (USNG) is a multi-purpose location system of grid references used in the United States. It provides a nationally consistent “language of location”, optimized for local applications, in a compact, user friendly format.
What is my QRA locator?
The QRA locator, also called QTH locator in some publications, is an obsolete geographic coordinate system used by amateur radio operators in Europe before the introduction of the Maidenhead Locator System.
Is the map of Hawaii based on a grid square?
The map of Hawaii is based on the grid square map information from ARRL. U.S. Grid Square Map Maidenhead Created Date 5/29/2013 1:43:17 PM
## How big is a grid square in miles?
*An instrument of the Maidenhead Locator System (named after the town outside London where it was first conceived by a meeting of European VHF managers in 1980), a grid square measures 1° latitude by 2° longitude and measures approximately 70 x 100 miles in the continental US.
What is the location of the invisible grid?
Information current as of November 2011. An invisible grid, based on 1° latitude by 2° longitude, encompasses the globe and is used for geographic location and identification during communications using your VHF and UHF radios. As HF operators collect QSL cards from around the world, VHF/UHF operators collect grid squares.
Is the ICOM logo a trademark or registered trademark?
©2013 Icom America Inc. The Icom logo is a registered trademark of Icom Inc. 41390 U.S. Grid Square Map Maidenhead | 773 | 3,384 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2023-50 | latest | en | 0.917928 |
https://rudeboybert.github.io/SDS192/ | 1,713,079,990,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816875.61/warc/CC-MAIN-20240414064633-20240414094633-00168.warc.gz | 485,997,484 | 26,123 | # Schedule
Topics:
1. Data visualization (pink): Grammar of Graphics, Five Named Graphs (5NG), color theory.
2. Working with data (blue): data wrangling, importing, and formatting
3. Maps and spatial data (green): Maps and geospatial data.
4. Learning how to learn new data science tools (yellow): SQL, TBD.
Note that while topics and topics dates may change, all problem sets (PS), project, and midterm dates will not.
# Lec 38: Fri 12/10
## Announcements
• Today: the in-class data assistants will hold office hours
• Final project due date/time is Friday 12/17 at 2pm (not 9pm)
## Today’s Topics/Activities
### 1. In-class exercise
• Work on Final Project
# Lec 37: Wed 12/8
## Announcements
• ModernDive will always be free, always available at ModernDive.com
• For Friday’s lecture the in-class data assistants will hold office hours
• Finals week:
• I’ve posted office hours for next week
• Final project due date/time is Friday 12/17 at 2pm (not 9pm)
• Spinelli center notes:
• Friday 12/10 is final Spinelli drop-in tutoring hours
• Thank you to the in-class data assistants Marium, Emma, Sunni, and Swaha
• Thank you to the Spinelli Center tutors
• Reflection exercise
• Time to fill out course evaluations
## Today’s Topics/Activities
### 1. In-class exercise
• Work on Final Project
# Lec 36: Mon 12/6
## Announcements
• Go over final project submission instructions
• Wednesday is final lecture I’ll be present; on Friday the Spinelli data assistants will hold in-class office hours.
• Note on coding style
## Today’s Topics/Activities
### 1. In-class exercise
• Test two useful packages below:
• patchwork: Combine two ggplots together
• janitor: Clean-up messy variable names
• Work on Final Project
library(tidyverse)
# 1. Combine two ggplots together using patchwork package:
library(patchwork)
plot_1 <- ggplot(mtcars) + geom_point(aes(mpg, disp))
plot_2 <- ggplot(mtcars) + geom_boxplot(aes(gear, disp, group = gear))
# Side-by-side:
plot_1 + plot_2
# On top of each other:
plot_1 / plot_2
# 2. Say we have a data frame with really messy names:
data_frame_ugly <- tibble(
asdf ?!? qwerty% = c(1, 2),
variable.name...NAMES = c(2,1)
)
data_frame_ugly
# You can clean them very easily using the clean_names() function from the
# janitor package
library(janitor)
data_frame_clean <- data_frame_ugly %>%
clean_names()
data_frame_clean
# Lec 35: Fri 12/3
## Announcements
• MTH/SDS tenure track search email
• Final project:
• Open Slack to #final_project channel
• Group leader: Create a Slack DM with all members AND myself and say “we’re a group”
• Submission details for Final Project to be posted on Monday.
## Today’s Topics/Activities
### 1. Chalk Talk
• dplyr and SQL are very similar. Both based on the same idea of database normalization (1970).
• Moral of the story: If you know the dplyr package for data wrangling, you can learn SQL very quickly.
### 2. In-class exercise
Perform 2-3 SQL queries to convince yourself that if you know dplyr, you can learn SQL very quickly. This exercise is based on Prof. Baumer’s lecture notes.
1. Setup
1. Install and open MySQLWorkbench; this whole process takes about 10 minutes and necessitates creating an account with Oracle.
2. Close the “Welcome to MySQL Workbench” message
3. Click the plus sign next to “MySQL Connections” to add a connection to a SQL database.
4. Setup a new connection" as shown in #general in Slack
5. Click the resulting “Playing with SQL” connection and input the password in #general in Slack
2. Running SQL code
1. Copy and paste the code below into the Query window
2. For each of the 10 code segments: highlight it and then run it by clicking the “lightning” icon.
# Lec 34: Wed 12/1
## Announcements
• Before chalk talk:
1. Download the following zip file: example_webpage.zip
2. Move example_webpage.zip to your SDS192 folder on your computer
3. Unzip example_webpage.zip. Windows users: be sure to “Extract all”
4. In the resulting example_webpage folder, double-click the RStudio Project example_webpage.Rproj icon to open RStudio Project mode
## Today’s topics/activities
### 1. Chalk talk
example_webpage is a portion of the R Markdown Websites code for this course webpage:
• Inputs: .Rmd and _site.yml files
• Output: webpage in docs/ folder, in particular the index.html mainpage
• “Deploying” your webpage: Many ways
### 2. In-class exercise
Today you’ll modify the source code for example_webpage and then deploying this webpage using Netlify drop:
1. Create an account on netlify.com using your GitHub account
• In index.Rmd change the author from "Albert Y. Kim" to you and change the title
• Build your R Markdown Website by going to the “Build” panel of RStudio -> Clicking “Build Website”.
You can also use the keyboard shortcuts:
• macOS: Command+Shift+B
• windows: Control+Shift+B
3. Deploy your R Markdown Website
• Go to Netlify Drop
• Drag-and-drop the docs/ folder output in your example_webpage RStudio Project folder.
• If you want to rename your webpage’s URL rather than use the default one you’ve been assigned: Click on “Domain settings” -> Click on the “…” next to your default site name -> Click on “Edit site name” -> Rename your site
# Lec 33: Mon 11/29
## Announcements
• Candidates for two new SDS faculty will be on campus this week and next
• For data science position: See “Meet the SDS Data Science New Faculty Candidates!” email sent to SDS student mailing list.
• For joint MTH/SDS position: To be confirmed soon.
• Because of this office hours are highly inconsistent and variable this week. However, they will always be confirmed and posted at least 24h in advance.
• Final project in groups of 2-3 will be assigned on Wednesday and due Fri 12/17 9pm (last day of exams). You can choose your group.
• PS07 posted
• To do before chalk talk
• Open the GitHub repo for the fivethirtyeight R package
## Today’s topics/activities
### 1. Chalk talk
GitHub: Theory and terminology
• What is git?
• What is GitHub?
• Terminology: Repo, local vs remote, clone, pull, commit/push
• Most important files in any repo: README.md
### 2. In-class exercise
• Work on MP3 (due tomorrow at 9pm), don’t forget to submit your Peer Evaluation Google Form.
• Work on PS07
# Lec 32: Mon 11/22
## Announcements
• MP3 now due Tuesday 11/30 at 9pm (after break). There will be no extensions past this due date/time.
• None
• Work on MP3
# Lec 31: Fri 11/19
## Announcements
• MP3 now due Tuesday 11/30 at 9pm (after break). There will be no extensions past this due date/time.
• None
• Work on MP3
# Lec 30: Wed 11/17
## Today’s topics/activities
### 1. Chalk talk
• Recap of all sf data frames seen so far in MP3 Project -> examples.Rmd
• Federal Information Processing Standard (FIPS) codes for counties
• Example: Looking up the database:
• 25XXX = Massachusetts counties
• 25105 = Hampshire County, Massachusetts
• From MP3 Project -> examples.Rmd -> Section 3 -> Look at contents of mass_pop_orig -> GEOID variable:
> mass_pop_orig
Simple feature collection with 14 features and 7 fields
Geometry type: MULTIPOLYGON
Dimension: XY
Bounding box: xmin: -73.50814 ymin: 41.23796 xmax: -69.92839 ymax: 42.88659
First 10 features:
GEOID NAME variable estimate moe
1 25017 Middlesex County, Massachusetts B01003_001 1600842 NA
2 25025 Suffolk County, Massachusetts B01003_001 796605 NA
3 25001 Barnstable County, Massachusetts B01003_001 213496 NA
4 25027 Worcester County, Massachusetts B01003_001 824772 NA
5 25011 Franklin County, Massachusetts B01003_001 70577 NA
6 25013 Hampden County, Massachusetts B01003_001 467871 NA
7 25015 Hampshire County, Massachusetts B01003_001 161032 NA
8 25021 Norfolk County, Massachusetts B01003_001 700437 NA
9 25005 Bristol County, Massachusetts B01003_001 561037 NA
10 25009 Essex County, Massachusetts B01003_001 783676 NA
• Work on MP3
# Lec 29: Mon 11/15
## Announcements
• Work on MP3 this Wed, Fri, and Mon before Thanksgiving break.
## Today’s topics/activities
### 1. Chalk talk
• Application Programmer Interfaces
• Choropleth maps. In particular, how you set the bins corresponding to the color gradient can affect how your map looks. As indicated here, there are two approaches:
• Equally sized interval bins
• Quantile based bins
### 2. In-class exercise
• Go over code in MP3 folder -> examples.Rmd -> Section 3 on “Choropleth maps using census data”
• You will need to register an API key from the census bureau. Carefully read warning message to do so.
# Lec 28: Fri 11/12
## Announcements
• Grad school panel on Mon 11/22 featuring SDS alumna. More info in #general
## Today’s topics/activities
### 2. In-class exercise
1. Work on PS06. This is direct practice for MP3.
2. Then start on MP3.
# Lec 27: Wed 11/10
## Announcements
• Check important Slack announcement in #general
• Mini-Project 3 posted.
## Today’s topics/activities
### 1. Chalk talk
1. Example solutions to examples.Rmd Section 1 exercise. Code posted in #mp3; screencast below
2. Shapefiles
### 2. In-class exercise
While you are free to work in any order you like, I suggest you:
1. Go over solutions to examples.Rmd -> Section 1 on “Converting data frames to sf objects”
2. Go code for Section 2 “Loading shapefiles into R”. This is direct practice for PS06.
3. Work on PS06. This is direct practice for MP3.
4. Then start on MP3.
# Lec 26: Mon 11/8
## Announcements
• Mini-project 3 (MP3) assigned on Wednesday, due Tuesday 11/23 at 9pm.
• Add yourselves to the #mp3 channel. Please ask all questions about MP3 in #mp3, not in #questions
• By Tuesday 5pm I will post the new groups (of two) in the #mp3 channel. Until #mp3 is due, you will sit next to your partner in class.
• Please reach out to your partner with a Slack DM before Wednesday’s lecture to coordinate meeting before lecture so you can sit next to each other.
• If you have seating restrictions due to hearing, sight, or mobility issues, please DM me.
• PS06 to be posted by this evening
• Compare the following London underground maps. As stated in this article, the transit map on the right sacrifices accuracy for clarity.
• The map of stations as they truly exist:
• The transit map inside the stations and trains. All lines are either straight or at 45 degrees and futhermore the geographic space is distorted.
## Today’s topics/activities
• Download MP3.zip
• Move MP3.zip to your SDS192 folder on your computer
• Unzip MP3.zip. Windows users: be sure to “Extract all”
• In the resulting MP3 folder, double-click the RStudio Project MP3.Rproj icon
• Verify that RStudio opens with MP3 written in the top-right
### 1. Chalk talk
• RStudio Projects
• sf package for static maps in ggplot2 and loading shapefiles. See Spatial Data Science for more.
### 2. In-class exercise
• In MP3 folder -> examples.Rmd -> Section 1 on “Converting data frames to sf objects”, do exercises
# Lec 25: Fri 11/5
No lecture today, instead optional in-class office hours:
• Sec02 Sabin-Reed 220: 9:25-10:25
• Sec01 Stoddard G2: 10:55-11:55
# Lec 24: Wed 11/3
## Announcements
• Slack note in #random
• No office hours tomorrow (Thursday). Instead, optional in-class office hours on Friday.
## Today’s topics/activities
### 1. Chalk talk
• Practice midterm posted in #midterms
• “When would you use left or right join?”
### 2. In-class exercise
• Open office hours
# Lec 23: Mon 11/1
## Announcements
• Practice midterm posted on Slack in #midterms; we’ll go over solutions on Wednesday
• Midterm II discussion: see midterms page
• Work on MP2
# Lec 22: Fri 10/29
## Announcements
• Useful RStudio cheatsheets: Go to RStudio menu bar on top -> Help -> Cheatsheets:
• Data Transformation with dplyr
• Data Visualization with ggplot2
## Today’s topics/activities
### 1. Chalk talk
• Install and then load the tidyverse package: An umbrella package that installs/loads many useful packages for data science all at once.
# Don't do all this:
library(ggplot2)
library(dplyr)
library(tidyr)
library(stringr)
library(tibble)
library(forcats)
library(purrr)
library(tidyverse)
• Work on MP2
# Lec 21: Wed 10/27
## Announcements
• Mid-Semester Assessment:
• Talk about Spring 2022 SDS courses
• Added “Tips & Tricks” tab to menu bar of course webpage
• Update to syllabus
• For a truly unique perspective on Data Visualization: Mona Chalabi @monachalabi. See video below:
• None
• Work on MP2
# Lec 20: Mon 10/25
## Announcements
• MP1 grades posted: See Slack #mp1 for details
• SDS is currently working hard to hire 3 new faculty who will start in July 2022; I’m chairing one committee and sitting on another. As a result for the next month
• My office hours will be highly variable; consult the calendars in the syllabus often
• There will unfortunately be lags in returning grading
• Sec01 Stoddard only: I won’t be able to stay past 12:05pm so that I can attend lunch meetings.
## Today’s topics/activities
### 1. Chalk talk
• Recap of Lec19: Why did we use inner_join() for solution to LC 3.20 on computing Available Seat Miles
• Importing spreadsheet data into R. Either Excel files or .csv Comma-Separated Values files. See image of example .csv file below.
• Data formats: “tidy” AKA long/tall/narrow format versue “non-tidy” AKA wide format
### 2. In-class exercise
• Go over ModernDive reading in schedule above
# Lec 19: Fri 10/22
## Today’s topics/activities
### 1. Chalk talk
• Pseudocode to compute Available Seat Miles
• Work on MP2
# Lec 18: Wed 10/20
## Announcements
• Slack:
• See #general Slack channel and give feedback on Spinelli tutors
• Practice making text look like code: create a DM with your project partner and let’s practice.
• Discuss Mini-Project 2 in full detail
• Feel free to message me on weekends, I just likely won’t respond.
## Today’s topics/activities
### 1. Chalk talk
• What is pseudocode?
• Types of joins:
• Copy code below to your classnotes.Rmd
• Refer to image (ignore semi_join())
### 2. In-class exercise
With your MP2 partner, practice data wranling! Complete ModernDive Learning Check 3.20: Using data in nycflights13 package, compute available seat miles for each airline separately:
1. Write out the pseudocode first
2. Then code it
# Lec 17: Mon 10/18
## Announcements
• Mini-Project 2 posted
• PS05 to be posted by this evening
• If you’re curious about my experiences in grad school, working at Google, switching to academia, and advice for aspiring data scientists, check out my appearance on episode #43 of the DataBytes podcast “To Google and Back.” Also available on Apple Podcasts and Google Play.
## Today’s topics/activities
### 1. Chalk talk
• Lec15 on group_by() and summarize(): difference between sum() and n() summary functions.
• Lec16 on mutate(): ifelse() function
• _join(), select(), and rename() functions
### 2. In-class exercise
• Go over ModernDive reading in schedule above
# Lec 16: Fri 10/15
## Announcements
• Mini-project 2 (MP2) assigned on Monday
• Add yourselves to the #mp2 channel. Please ask all questions about MP2 in #mp2, not in #questions
• By Sunday 5pm I will post the new groups (of two) in the #mp2 channel. Until #mp2 is due, you will sit next to your partner in class.
• Please reach out to your partner with a Slack DM before Monday’s lecture to coordinate meeting before lecture so you can sit next to each other.
• If you have seating restrictions due to hearing, sight, or mobility issues, please DM me.
## Today’s topics/activities
### 1. Chalk talk
• mutate() new columns/variables and arrange() i.e. sort rows
### 2. In-class exercise
• Go over ModernDive reading in schedule above
# Lec 15: Wed 10/13
## Announcements
• See Slack #general for info about presentation of SDS major
• PS04 (shorter) will be posted this afternoon
• Keyboard shortcuts for:
1. %>% in RStudio: command + shift + m on macOS, control + shift + m on Windows
2. Running code in RStudio: command + enter on macOS, control + enter on Windwos
3. Quickly jumping between apps: command + tab on macOS, alt + tab on Windows
4. Selecting many files at once: click first file, hold shift, click last file
5. Deleting files: command + delete on macOS, delete on Windows
## Today’s topics/activities
### 1. Chalk talk
• summarize() rows and group_by() %>% summarize()
### 2. In-class exercise
• Put finishing touches on MP1
• Go over ModernDive reading in schedule above
# Lec 14: Fri 10/8
## Announcements
• You are responsible for completing the ModernDive readings for Lec13 on the %>% operator and filter() before Wednesday’s lecture
• A shorter PS04 will be assigned on Wednesday, due on Monday 10/18 9pm
• None
• Work on MP1
# Lec 13: Wed 10/6
## Today’s topics/activities
### 1. Chalk talk
• Computer file theory
• What are folders/directories?
• How does R Markdown find the .ics file?
• What are .zip files? Special note for Windows users
• Computer file hygiene: Delete files you don’t need anymore
• Intro to data wrangling
• Pipe operator %>% pronounced “then”
• filter() rows that meet a certain criteria
### 2. In-class exercise
• Go over ModernDive reading in schedule above
# Lec 12: Mon 10/4
## Announcements
• Midterm:
• No talking about it until after 5pm today please; there is one or more students who need to take it.
• Why student ID and not name? For anonymized grading.
• Update to office hours on syllabus
• MP1:
• Lecture schedule for Wed, Fri, and Wed after break
• Post questions about MP1 in #mp1 on Slack
• Discussion on managing group dynamics:
• Life happens. If it does and it will affect you work, at the very least communicate and give your partner a heads up (text, Slack, etc.)
• What to do when issues arise?
• Don’t forget you’ll be filling out peer evaluation Google Form
## Today’s topics/activities
### 1. Chalk talk
• Trend lines via a geom_smooth() layer. Two types:
• Linear regression
• LOWESS: Locally Weighted Scatterplot Smoothing
• Example code:
library(ggplot2)
library(dplyr)
library(gapminder)
# 1. Recreate plot from PS02 but with no color:
gapminder_2007 <- gapminder %>%
filter(year == 2007)
# 1.a) Add LOESS smoother layer with geom_smooth()
ggplot(data = gapminder_2007, mapping = aes(x = gdpPercap, y = lifeExp, size = pop)) +
geom_point() +
geom_smooth()
# 1.b) Remove standard error bars by setting se = FALSE
ggplot(data = gapminder_2007, mapping = aes(x = gdpPercap, y = lifeExp, size = pop)) +
geom_point() +
geom_smooth(se = FALSE)
# 1.c) Change span of "smoothing" window by change the value of span
ggplot(data = gapminder_2007, mapping = aes(x = gdpPercap, y = lifeExp, size = pop)) +
geom_point() +
geom_smooth(se = FALSE, span = 0.25)
# 1.d) Force line to be straight. i.e. linear regression
ggplot(data = gapminder_2007, mapping = aes(x = gdpPercap, y = lifeExp, size = pop)) +
geom_point() +
geom_smooth(method = "lm", se = FALSE)
### 2. In-class exercise
• Copy the example code above to your classnotes.Rmd and go over the code
• Optional: Go over ModernDive reading in schedule above (this topic is covered in SDS 201/220 intro stats)
• Work on MP1
# Lec 11: Fri 10/1
## Announcements
• Open Slack at the start of every lecture
• Check for DM’s
• Check #midterms channel
• In order to not disadvantage students who take the midterm earlier
• I won’t be answering any Slack #midterms after 3pm today
• I’ve instructed the Friday Spinelli tutor not to answer questions about the midterm
## Today’s topics/activities
### 1. Chalk talk
• Go over practice Midterm I. Boxplot for question 3.c):
• Work on MP1
# Lec 10: Wed 9/29
## Announcements
• Sit next to your MP1 partner; your partner was posted in the #mp1 channel on Sunday 5pm.
• If have an Office of Disability Services accommodations letter and you haven’t already, please Slack DM it to me.
• Midterm I info posted
• Mini-Project 1 info posted
## Today’s topics/activities
### 2. In-class exercise
• With your partner, build a minimally viable product of your MP1
# Lec 09: Fri 9/24
## Announcements
• On Slack #general: new SDS student lounge in McConnell 209
• Additional resource: Prof. Ben Baumer’s book Modern Data Science with R used in his version of SDS 192.
• Mini-project 1 (MP1) assigned on Monday
• Slack demo of how to subscribe to a #channel: Adding yourselves to the #mp1 channel. Please ask all questions about MP1 there
• You will be assigned groups for MP1, MP2, and MP3. You can choose your groups for the final project.
• By Sunday 5pm I will post the groups (of two) in the #mp1 channel. Until #mp1 is due, you will sit next to your partner in class.
• Please reach out to your partner with a Slack DM before Monday’s lecture to coordinate meeting before lecture so you can sit next to each other.
## Today’s topics/activities
### 1. Chalk talk
• Recap of barplots: Exercise on pie charts vs barplots below
• Color theory
1. color vs fill aesthetics in ggplot2
2. Selecting an appropriate color palette from colorbrewer2.org
3. How does ggplot2 pick default colors? Using a color wheel
4. Also define colors in terms of hex codes
### 2. Exercise on pie charts vs barplots
Say the following piecharts represent results of an election poll at time points: A = September, B = October, and C = November. At each time point we present the proportion of the poll respondents who say they will support one of 5 candidates: 1 through 5.
Based on these 3 piecharts, answer the following questions:
1. At time point A, is candidate 5 doing better than candidate 4?
2. Did candidate 3 do better at time point B or time point C?
3. Who gained more support between time point A and time point B, candidate 2 or candidate 4?
Compare that to using barplots. Which do you prefer?
### 3. In-class exercise
• Go over ModernDive reading in schedule above.
• Changing default color and fill color aesthetics:
1. Copy and paste the code below into your classnotes.Rmd file
2. Change both the color of the scatterplot points and the fill of the bars. You can do this by selecting a palette from colorbrewer2.org or by setting them manually
3. Run colors() in your console to get English names of all colors in R
library(ggplot2)
library(dplyr)
library(nycflights13)
library(gapminder)
# 1. Recreate plot from PS02, but change default "color" palette of points:
gapminder_2007 <- gapminder %>%
filter(year == 2007)
ggplot(gapminder_2007, aes(x = gdpPercap, y = lifeExp, size = pop, color = continent)) +
geom_point() +
scale_color_brewer(palette = "Set1")
# 2.a) Recreate Figure 2.26 but change default "fill" color of bars by adding a
# palette layer:
ggplot(flights, aes(x = carrier, fill = origin)) +
geom_bar(position = position_dodge(preserve = "single")) +
scale_fill_brewer(palette = "Set1")
# 2.b) Recreate Figure 2.26 but change default "fill" color of bars by manually
# changing colors in a layer:
ggplot(flights, aes(x = carrier, fill = origin)) +
geom_bar(position = position_dodge(preserve = "single")) +
scale_fill_manual(values = c("darkorange", "forestgreen", "navyblue"))
# 2.c) Recreate Figure 2.26 but change default "fill" color of bars by manually
# changing colors in a layer using hex codes from:
# https://www.color-hex.com/color-palette/114219
ggplot(flights, aes(x = carrier, fill = origin)) +
geom_bar(position = position_dodge(preserve = "single")) +
scale_fill_manual(values = c("#dc323a", "#003f77", "#c4c1c1"))
# Lec 08: Wed 9/22
## Announcements
• Problem sets:
• PS03 posted
## Today’s topics/activities
### 1. Chalk talk
• Searching the internet effectively: a critical data science tool
• Wrapping-up boxplots:
• For a side-by-side boxplot, the x variable has to be categorical
• Summary statistics that are robust to outliers: median and IQR
• Why 1.5 x IQR?
• Barplots
• geom_bar() when counts are not pre-computed i.e. listed individually
• geom_col() when counts are pre-computed and saved in a variable
### 2. In-class exercise
• Go over ModernDive reading in schedule above.
# Lec 07: Mon 9/20
## Announcements
• Office of Disability Services is looking to hire a note taker for the class. If interested, see note in Slack #general.
• Problem sets:
• PS02 due today at 5pm
• PS03 to be posted by 6pm today
## Today’s topics/activities
### 1. Chalk talk
• Recap of histograms
• Facets to split a visualization by the values of another variable
• Default ordering of functions such as ggplot() where data = is assumed first and mapping = is assumed second
• Boxplots! Powerful, but tricky!
Say we want to study the distribution of the following 12 values which are pre-sorted:
1, 3, 5, 6, 7, 8, 9, 12, 13, 14, 15, 30
They have the following summary statistics. A summary statistic is a single numerical value summarizing many values. Examples include the immediately obvious mean AKA average and median. Other less immediately obvious examples include:
• Quartiles (1st, 2nd, and 3rd) that cut up the data into 4 parts, each containing roughly one quarter = 25% of the data
• Minimum & maximum
• Interquartile-range (IQR): the distance between the 3rd and 1st quartiles
Min. 1st Quartile Median = 2nd Quartile 3rd Quartile Max. IQR
1 5.5 8.5 13.5 30 8 = 13.5 - 5.5
Let’s compare the points and the corresponding boxplot side-by-side with the values on the $$y$$-axis matching:
### 2. In-class exercise
• Go over ModernDive reading in schedule above.
# Lec 06: Fri 9/17
## Today’s topics/activities
### 1. Chalk talk
• In-class demo of using RMarkdown features in a classnotes.Rmd file to save lecture code
• Take screenshots of your screen!
• Histograms for visualizing the distribution of a numerical variable
### 2. In-class exercise
• If you still haven’t been able to “Knit to PDF”, please ask for help
• Go over ModernDive reading in schedule above.
# Lec 05: Wed 9/15
## Announcements
• PS02 was posted after Monday’s lecture.
## Today’s topics/activities
### 1. Chalk talk
• Overplotting and two approaches for addressing it
• Linegraphs
### 2. In-class exercise
• Explore the different formatting tools in R Markdown: go to RStudio top menu bar -> Help -> Markdown quick reference.
• Sec01 in Stoddard: There was an typo in Step 8 in last lecture’s in-class exercise. If you weren’t able to Knit directly to PDF, please re-attempt Steps 8-9. Knitting directly to PDF, instead of Knitting to Word and then saving to PDF, is the preferred submission format for all problem sets. It will be less hassle for you and provide consistency for the graders.
• Go over ModernDive reading in schedule above.
# Lec 04: Mon 9/13
## Announcements
• Problem Set 02 due next Monday 5pm, now posted under Problem Sets
## Today’s topics/activities
### 1. Chalk talk
• Recap of previous lecture
• “Where can I save all the code I run in class?” In an R Markdown .Rmd file; R Markdown is a tool for reproducible research
Input: An .Rmd file Output: An .html, .docx, or .pdf file.
### 2. In-class exercise
In-class battle-testing and practicing for PS02:
1. At a couple of steps in this process, you will be asked to install packages. Say yes to all of them.
2. If at any point your code won’t knit, go through these 6 R Markdown Fixes first, then seek assistance. These 6 fixes will resolve 85% of issues.
3. Create new R Markdown .Rmd file:
• Go to RStudio menu bar -> File -> New File -> R Markdown
• Set “Title” to “My first R Markdown report” and “Author” as your name.
• Save this file as testing somewhere on your computer. This will create a file called testing.Rmd
4. Method 1: “Knit” a report to HTML:
• Click the arrow next to “Knit” -> “Knit to HTML”.
• An HTML webpage should pop up. However, it may be blocked by your browser. If so, in your browser’s URL bar, click on “Always allow pop-ups”.
5. Method 1: Publish HTML report on web:
• Click on blue “Publish” button on top right of the resulting pop-up html.
• Select RPubs.
• If you haven’t previously, create an account on Rpubs.com. If you have previously, login.
• Set “Title” to “My first R Markdown report” and “Slug” to “testing”
• You should end up with a webpage that looks like this one. This is live on the web!
6. Method 1: Update HTML report on web:
• Make some trivial change to your testing.Rmd file.
• “Re-knit” your report and make sure your trivial change is reflected.
• The blue “Publish” button should now read “Republish”
• Click “Update existing”
7. Method 2: “Knit” a report to Word
• Click the arrow next to “Knit” -> “Knit to Word”.
• Save the resulting Word document as a pdf file.
8. Only if you are a macOS user:
• Next to “Console” go to “Terminal”
• Run this line of code:
sudo chown -R whoami:admin /usr/local/bin
9. Method 3: “Knit” a report to PDF
• Run the following code in your console just once:
install.packages('tinytex')
tinytex::install_tinytex()
• Click the arrow next to “Knit” -> “Knit to PDF”.
# Lec 03: Fri 9/10
## Announcements
• Spinelli Center SDS drop-in tutoring hours now open! Get individual attention from SDS majors! In Sabin-Reed 301
• Sunday through Thursday 7-9pm
• Friday 2:35-3:30pm
• By popular request:
• Sec 01 in Stoddard G2 will now start 5 minutes later: 10:55 AM instead of 10:50AM
• Sec 02 in Sabin-Reed 220 will now end 5 minutes earlier: 10:35 AM instead of 10:40 AM
• I added extra instructions for Problem Set 01 after lecture, posted under Problem Sets
• Show don’t tell how to tag questions on gradescope
• ASA StatFest 2021 Sat 9/18 thru Sun 9/19 flyer and event webpage
• Sunday 9/19 at 11:50AM: Opportunities in Statistics & Data Science in Academia, Government, & Non-Profit featuring SDS’s Prof. Randi Garcia!
• Keynote address by Robert Santos, 116th President of the ASA, and President Biden’s nominee to serve as Director of the United States Census Bureau! If approved by the Senate, he would be the first Latinx Director of the Bureau!
## Today’s topics/activities
### 1. Chalk talk
• Recap of previous lecture
• Grammar of Graphics
• 5NG1: Scatterplots
• Next time:
• Question: Do I need to re-type my code in the Console every single time?
### 2. In-class exercise
• Go over ModernDive reading in schedule above.
# Lec 02: Wed 9/8
## Announcements
• Problem Set 01 due this Monday 5pm, posted under Problem Sets.
## Today’s topics/activities
### 1. Chalk talk
• Intro to Slack
• What is difference between R and RStudio?
• What are R packages?
### 2. In-class exercise
• Go over ModernDive reading in schedule above.
• You are responsible for completing a lecture’s readings before the next lecture. Ex: you are responsible to read all of ModernDive Chapter 1 before Wednesday.
• I teach lectures assuming you have not done the readings beforehand. However, if it suits your learning style better, please do read beforehand.
• While you don’t need to turn in your learning check answers, I highly recommend you still do them. The solutions are in Appendix D of the book.
# Lec 01: Fri 9/3
Welcome!
## Today’s topics/activities
• Course webpage: bit.ly/sds192kim
• My story
• “Knock on wood if you’re with me”
1. Data viz
2. Data wrangling
3. Maps
4. Websites
• Break!
• Executive summary of syllabus
• This weekend: Complete intro survey
## Code examples from class
# Data visualization
library(fivethirtyeight)
library(ggplot2)
library(dplyr)
year_bins <- c("'70-'74", "'75-'79", "'80-'84", "'85-'89", "'90-'94",
"'95-'99", "'00-'04", "'05-'09", "'10-'13")
bechdel <- bechdel %>%
mutate(five_year = cut(year, breaks = seq(1969, 2014, 5), labels = year_bins))
ggplot(bechdel, aes(x = five_year, fill = clean_test)) +
geom_bar(position = "fill", color = "black") +
labs(x = "Year", y = "Proportion", fill = "Bechdel Test") +
scale_fill_brewer(palette = "YlGnBu")
# Data Wranling
library(fec16)
addMarkers(lng=-72.64022, lat=42.31706, popup="Smith College") | 8,597 | 31,807 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-18 | latest | en | 0.694336 |
https://www.telematika.org/py/cb2nd_08_clustering/ | 1,611,811,676,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704835901.90/warc/CC-MAIN-20210128040619-20210128070619-00163.warc.gz | 1,031,330,232 | 3,775 | ## Jupyter Snippet CB2nd 08_clustering
Jupyter Snippet CB2nd 08_clustering
# 8.8. Detecting hidden structures in a dataset with clustering
from itertools import permutations
import numpy as np
import sklearn
import sklearn.decomposition as dec
import sklearn.cluster as clu
import sklearn.datasets as ds
import sklearn.model_selection as ms
import matplotlib.pyplot as plt
%matplotlib inline
X, y = ds.make_blobs(n_samples=200,
n_features=2,
centers=3,
cluster_std=1.5,
)
def relabel(cl):
"""Relabel a clustering with three clusters
to match the original classes."""
if np.max(cl) != 2:
return cl
perms = np.array(list(permutations((0, 1, 2))))
i = np.argmin([np.sum(np.abs(perm[cl] - y))
for perm in perms])
p = perms[i]
return p[cl]
def display_clustering(labels, title):
"""Plot the data points with the cluster
colors."""
# We relabel the classes when there are 3 clusters
labels = relabel(labels)
fig, axes = plt.subplots(1, 2, figsize=(8, 3),
sharey=True)
# Display the points with the true labels on the
# left, and with the clustering labels on the
# right.
for ax, c, title in zip(
axes,
[y, labels],
["True labels", title]):
ax.scatter(X[:, 0], X[:, 1], c=c, s=30,
linewidths=0, cmap=plt.cm.rainbow)
ax.set_title(title)
km = clu.KMeans()
km.fit(X)
display_clustering(km.labels_, "KMeans")
km = clu.KMeans(n_clusters=3)
km.fit(X)
display_clustering(km.labels_, "KMeans(3)")
fig, axes = plt.subplots(2, 3,
figsize=(10, 7),
sharex=True,
sharey=True)
axes[0, 0].scatter(X[:, 0], X[:, 1],
c=y, s=30,
linewidths=0,
cmap=plt.cm.rainbow)
axes[0, 0].set_title("True labels")
for ax, est in zip(axes.flat[1:], [
clu.SpectralClustering(3),
clu.AgglomerativeClustering(3),
clu.MeanShift(),
clu.AffinityPropagation(),
clu.DBSCAN(),
]):
est.fit(X)
c = relabel(est.labels_)
ax.scatter(X[:, 0], X[:, 1], c=c, s=30,
linewidths=0, cmap=plt.cm.rainbow)
ax.set_title(est.__class__.__name__)
# Fix the spacing between subplots.
fig.tight_layout() | 589 | 1,944 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2021-04 | latest | en | 0.42824 |
https://jiyuushikan.org/what-is-the-mass-of-1-mole-of-raindrops/ | 1,632,843,619,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060877.21/warc/CC-MAIN-20210928153533-20210928183533-00621.warc.gz | 388,429,036 | 6,036 | ## How many moles of raindrops are in the Pacific Ocean Round your answer to 2 substantial digits?
Answer Expert Verified The correct number of substantial digits is 2, because 50. mg has 2 signficant digits, so the answer need to be presented as 3.0 * 10^ 19 kg. Which rounded to two substantial digits is 24 moles of rain drops.
You are watching: What is the mass of 1 mole of raindrops
15.998 grams
5.4×109 g
2.4 x 1027 g
## How many type of moles of copper atoms have a mass equal to the mass of a penny?
Because a penny is created of 97.50% zinc and 2.50% copper, it therefore includes 2.4375 grams of zinc and also 0.0625 grams of copper. At a molar mass of 65.380 grams per mole for zinc and also 63.546 grams per mole for copper, a penny therefore consists of 0.037282 moles of zinc and 0.moles of copper.
185000 moles
## What is the mole variety of oxygen?
One mole of atoms of oxygen has actually a mass of 16 g, as 16 is the atomic weight of oxygen, and contains 6.02 X 1023 atoms of oxygen.
## How many kind of atoms are in a mole of oxygen?
One mole of oxygen atoms contains 6./b>×1023 oxygen atoms.
## How many atoms are in a mole?
While the mass of 1 mole of a certain …” Correct. While the mass of 1 mole of a details substance will vary, 1 mole of ANY substance will ALWAYS have actually approximately 6.atoms (rounded to 3 significant figures).
## What is a 1 mole?
The mole (symbol: mol) is the unit of measurement for amount of substance in the Internationwide System of Units (SI). It is identified as precisely 6.sup>23 particles, which might be atoms, molecules, ions, or electrons.
## What is Mole on humale body?
Moles are a prevalent form of skin expansion. They often show up as tiny, dark brown spots and also are caused by clusters of pigmented cells. Moles generally appear in the time of childhood and adolescence. Most civilization have 10 to 40 moles, some of which might readjust in appearance or fade away over time.
## What is the mole formula?
Avogadro’s number is a very vital relationship to remember: 1 mole = 6.022×1023 6.022 × 10 23 atoms, molecules, proloads, etc. To transform from moles to atoms, multiply the molar amount by Avogadro’s number. To convert from atoms to moles, divide the atom amount by Avogadro’s number (or multiply by its reciprocal).
## How did Avogadro discover the mole?
The term “Avogadro’s number” was first offered by French physicist Jean Baptiste Perrin. If you divide the charge on a mole of electrons by the charge on a single electron you obtain a value of Avogadro’s number of 6.x 1023 pposts per mole.
## Why is mole principle important?
Significance of Mole: atoms and molecules are incredibly little and the mole concept permits us to count atoms and also molecules by weighing macroscopic amounts of product.
114.818 grams
5.00 moles
## How do you transform moles to mass in grams?
In order to convert the moles of a substance to grams, you will certainly must multiply the mole worth of the substance by its molar mass.
## How do you find just how many type of grams are in a mole?
Divide the mass of the substance in grams by its molecular weight. This will offer you the number of moles of that substance that are in the mentioned mass. For 12 g of water, (25 g)/(18.015 g/mol) = 0.666 moles.
## How perform I calculate grams?
How to Calculate Grams & Milligrams
Write, or enter right into your calculator, the number of g which you want to convert to mg.Multiply the variety of g by 1,000. Example: 2.25 g X 1,000 = 2,250 mg.Reverse the process to convert mg to g. Divide the variety of milligrams by 1,000. Example: 1,250 mg split by 1,000 = 1.25 g.
## How many moles are in 12.0 grams of O2?
The answer is 31.9988. We assume you are converting in between grams O2 and mole. You can watch more details on each measurement unit: molecular weight of O2 or mol The SI base unit for amount of substance is the mole. 1 grams O2 is equal to 0. mole.
## How many kind of moles are in 100 grams of gold?
If you look at the table, gold has actually an atomic weight of 196.967 g/mol. 100g split by 196.967g/mol gives you . 507699 mol. We all remember Avogadro’s variety of 6.023x atoms/mol.
## How many type of atoms are in 2 moles of gold?
We know we have 2 moles of Au, so multiply 6.02 x 2. Answer is 12.04 x 10^23 atoms of Au.
## How many atoms of gold are in 1g of gold?
30.573 Billion-Billion Atoms
## How many kind of atoms are in 5g of gold?
of atoms = offered mass/molar mass x no. Of atoms = 5/197 x 6.022 x 10^23 = 0.25 x 6.022 x 10^23 = 1.5 x10^23 atoms.
## How many kind of atoms are in pure gold?
1 Answer. 0.650 g Au contain 1.99×1021atoms .
## How many type of gold atoms are in a 1 kg gold bar?
Therefore, 1 kg of a gold bar has actually 3.057 × 10^24 atoms of gold. | 1,314 | 4,806 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2021-39 | longest | en | 0.930667 |
https://answers.search.yahoo.com/search?p=V&fr2=piv-web&b=21&pz=10&bct=0&xargs=0 | 1,591,296,315,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347445880.79/warc/CC-MAIN-20200604161214-20200604191214-00004.warc.gz | 245,384,397 | 24,897 | # Yahoo Web Search
1. Sort by
1. ### Let {v_1...v_n} be a spanning set for the vector space V, and let v be any other vector in V.?
{v_1,...,v_n} is a spanning set of V. And v is a vector of V. There exist n numbers...
2 Answers · Science & Mathematics · 12/03/2009
2. ### Vectors : ||v||=1, what is the value of v?
v the vector? No one can say, could be anything, also it depends on how...3)) i-hat + 1/sqrt(3) j-hat + 1/sqrt(3) k-hat, and anything else. people often write v (no bold or arrow on top) for shorthand to be |v| which is the same thing as ||v...
4 Answers · Science & Mathematics · 30/07/2010
3. ### What if vectors are linearly independent and do not span V?
Since v₁, v₂, ..., vₘ are linearly independent, then the equation: a₁ v₁ + a₂...of any set of vectors. Now assume that {v₁, v₂, ..., vₘ₊₁, vₘ₊₁} is not linearly independent. Therefore...
3 Answers · Science & Mathematics · 06/04/2017
4. ### V= LxWxH solve for w?
V= LxWxH ... solve for w Okay, in order to solve for 'w', we...it by itself somehow. Let's rewrite this equation (formula) like this: v = (l)(w)(h) -----> *divide both sides by (l)(h) in order to isolate w...
3 Answers · Education & Reference · 21/09/2010
5. ### How to crochet V-sts?
For V-stitches, the number of chains for the foundation row must be a multiple of 2 + 1, then you add on the 3 stitches for turnng. So, for 6 V-stitches plus a border on each side, multiply 6 x 2 (= 12) + 1 (= 13) + 3 (= 16...
3 Answers · Games & Recreation · 19/02/2012
6. ### can someone define x=v_x*t?
v_x is velocity in the X direction. A better way to write this is Vx = |V| cos(theta), where |V| is the velocity magnitude (speed) and theta is...
1 Answers · Science & Mathematics · 07/02/2010
7. ### Is V=pie d cubed/4 dimensionally correct?
V= 4 / 3 * pi * r ^ 3 r = d / 2. so V = 4 / 3 * pi * (d /2 ) ^ 3 = 4... for your dimensions, if d is in metres, V is in cubic metres. If d is in inches, V...
1 Answers · Education & Reference · 05/11/2007
8. ### Let u = (1 1) and v = (1 2) and define U = span{u} and V = span{v}.?
So, I presume, U + V simply means the union of the two sets, whereas U ++ V...look like graphically. They are simply all multiples of u and v respectively, which form a line in the directions of u and v respectively. ...
2 Answers · Science & Mathematics · 19/07/2011
9. ### find the component of v, where u = 2i -j and w = i + 2j?
v= u + 2w v= 2i - j + 2(i + 2j) v = 2i -j + 2i + 4j v = 4i + 3j
2 Answers · Science & Mathematics · 25/06/2011
10. ### Find each quantity if v= 3i-5j and w=-2i+3j?
||v|| = √((3)² + (-5)²) = √(9 + 25) = √34 ||w|| = √((-2)² + (3)²) = √(4 + 9) = √13 ||v|| - ||w|| = √34 - √13 ≈ 2.2254
2 Answers · Science & Mathematics · 14/03/2013 | 960 | 2,744 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2020-24 | latest | en | 0.779273 |
https://www.convertunits.com/from/billion+cubic+foot/hour/to/cubic+yard/minute | 1,701,970,944,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100677.45/warc/CC-MAIN-20231207153748-20231207183748-00001.warc.gz | 782,520,643 | 12,682 | ## Convert billion cubic foot/hour to cubic yard/minute
billion cubic foot/hour cubic yard/minute
How many billion cubic foot/hour in 1 cubic yard/minute? The answer is 1.62E-6. We assume you are converting between billion cubic foot/hour and cubic yard/minute. You can view more details on each measurement unit: billion cubic foot/hour or cubic yard/minute The SI derived unit for volume flow rate is the cubic meter/second. 1 cubic meter/second is equal to 0.00012713279836558 billion cubic foot/hour, or 78.477036028136 cubic yard/minute. Note that rounding errors may occur, so always check the results. Use this page to learn how to convert between billion cubic feet/hour and cubic yards/minute. Type in your own numbers in the form to convert the units!
## Quick conversion chart of billion cubic foot/hour to cubic yard/minute
1 billion cubic foot/hour to cubic yard/minute = 617283.95062 cubic yard/minute
2 billion cubic foot/hour to cubic yard/minute = 1234567.90123 cubic yard/minute
3 billion cubic foot/hour to cubic yard/minute = 1851851.85185 cubic yard/minute
4 billion cubic foot/hour to cubic yard/minute = 2469135.80247 cubic yard/minute
5 billion cubic foot/hour to cubic yard/minute = 3086419.75309 cubic yard/minute
6 billion cubic foot/hour to cubic yard/minute = 3703703.7037 cubic yard/minute
7 billion cubic foot/hour to cubic yard/minute = 4320987.65432 cubic yard/minute
8 billion cubic foot/hour to cubic yard/minute = 4938271.60494 cubic yard/minute
9 billion cubic foot/hour to cubic yard/minute = 5555555.55556 cubic yard/minute
10 billion cubic foot/hour to cubic yard/minute = 6172839.50617 cubic yard/minute
## Want other units?
You can do the reverse unit conversion from cubic yard/minute to billion cubic foot/hour, or enter any two units below:
## Enter two units to convert
From: To:
## Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 70 kg, 150 lbs, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 599 | 2,335 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2023-50 | latest | en | 0.765508 |
https://community.filemaker.com/thread/119887 | 1,503,311,643,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886108264.79/warc/CC-MAIN-20170821095257-20170821115257-00234.warc.gz | 653,967,012 | 24,091 | 2 Replies Latest reply on Jul 20, 2011 2:07 PM by raybaudi
# Re: How to create a custom function using tail recursion technique
### Title
Re: How to create a custom function using tail recursion technique
### Post
I have a registration form with about 20 fixed filling areas (with empty filling squares in each filling area) and want to fill in each empty square with one character from my database information, e.g. a name like "JACKIE" and the form has only 4 spaces to fill in and I have to truncate the word "JACKIE" to "JACK" and the filling squares are large. I need to pad the word "JACK" with spaces like "J A C K" to fill in the form.
I can easily create a generic script to truncate the word "JACKIE" to "JACK" and separate the characters with spaces with a loop script step.
I read about the tail recursion but could not understand how to create a custom function with a recursive loop.
I wonder if any FM experts can help me to create a custom function e.g. fnPaddedWord (word; characters; spaces) so that all I need is to provide (1) the word to fill in (i.e. JACKIE), (2) the number of characters to fill in (i.e. 4) & (3) fixed amount of spaces to separate the characters to be able to fill in the required area as "J A C K"
• ###### 1. Re: Re: How to create a custom function using tail recursion technique
Recursive Calculations is one of the main reasons for devising a custom function.
// fnPaddedWord ( Word , Characters , Spaces )
// Word: the text to be padded
// Characters the number of characters from word to be padded. Remaining characters will be truncated
// Spaces: the number of spaces to insert between each nontruncated character
Case ( characters = 0 ; "" ;
IsEmpty ( Word ) ; "" ;
Left ( Word ; 1 ) & Substitute ( 10^Spaces - 1 ; 9 ; " " ) & Trim ( fnPaddedWord ( Right ( Word ; Length ( Word ) - 1 ; Characters - 1 ; Spaces ) )
)
How it works: 10^Spaces is 10 raised to the Spaces power. If Spaces = 2, this produces 100. Subtracting 1 produces a number with only the 9 digit and one such digit for every space (99 when spaces = 2, 999 when spaces = 3 and so forth). The substitute function replaces each 9 with a space to pad the characters.
The function has two conditions that will terminate the expression: If there are no more characters to Pad, or if the specified number of characters has been padded.
Also, since the function returns spaces after the last padded character, Trim is used to remove the trailing spaces.
• ###### 2. Re: Re: How to create a custom function using tail recursion technique
... or you could use calculated repeating fields, so you will be able to center the chars into their respective boxes.
For example:
Middle ( Extend ( word ) ; Get ( CalculationRepeatitionNumber ) ; 1 ) | 677 | 2,766 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2017-34 | longest | en | 0.806027 |
benedictheal.com | 1,620,840,722,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989766.27/warc/CC-MAIN-20210512162538-20210512192538-00418.warc.gz | 6,616,216 | 1,922 | ## Introduction
Pappus's Theorem starts with any two lines. On each, three arbitrary points are chosen. Taking two points from the top line, and two from the bottom, the point of intersection of the diagonals is found. This is done three times, giving three points2 Here the diagonals are coloured blue, green and brown.
The three points are collinear!
Another formulation is in terms of a hexagon, where alternate vertices lie are collinear. The hexagon therefore needs to be folded over itself.
Play with this diagram by moving the lines and points. The red line shows the collinear intersections of diagonals. You can reorder the points on the lines. | 140 | 657 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2021-21 | latest | en | 0.939133 |
https://www.iberlibro.com/9780137010592/Learn-Statistics-Levine-David-0137010591/plp | 1,511,347,048,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806543.24/warc/CC-MAIN-20171122084446-20171122104446-00002.warc.gz | 823,873,179 | 14,618 | # Even You Can Learn Statistics: A Guide for Everyone Who Has Ever Been Afraid of Statistics (2nd Edition)
## David M. Levine; David F. Stephan
3,35 valoración promedio
( 96 valoraciones por Goodreads )
Even You Can Learn Statistics: A Guide for Everyone Who Has Ever Been Afraid of Statisticsi s a practical, up-to-date introduction to statistics—for everyone! Thought you couldn’t learn statistics? You can—and you will! One easy step at a time, this fully updated book teaches you all the statistical techniques you’ll need for finance, quality, marketing, the social sciences, or anything else! Simple jargon-free explanations help you understand every technique. Practical examples and worked-out problems give you hands-on practice. Special sections present detailed instructions for developing statistical answers, using spreadsheet programs or any TI-83/TI-84 compatible calculator. This edition delivers new examples, more detailed problems and sample solutions, plus an all-new chapter on powerful multiple regression techniques. Hate math? No sweat. You’ll be amazed at how little you need. Like math? Optional “Equation Blackboard” sections reveal the mathematical foundations of statistics right before your eyes! You’ll learn how to:
· Construct and interpret statistical charts and tables with Excel or OpenOffice.org Calc 3
· Work with mean, median, mode, standard deviation, Z scores, skewness, and other descriptive statistics
· Use probability and probability distributions
· Work with sampling distributions and confidence intervals
· Test hypotheses with Z, t, chi-square, ANOVA, and other techniques
· Perform powerful regression analysis and modeling
· Use multiple regression to develop models that contain several independent variables
· Master specific statistical techniques for quality and Six Sigma programs
About the Web Site
Download practice files, templates, data sets, and sample spreadsheet models—including ready-to-use solutions for your own work! www.ftpress.com/youcanlearnstatistics2e
"Sinopsis" puede pertenecer a otra edición de este libro.
About the Author:
David M. Levine, a much-honored innovator in statistics education, is Professor Emeritus of Statistics and Computer Information Systems at Baruch College (CUNY). His bestselling books include Statistics for Managers Using Microsoft Excel, Basic Business Statistics, Business Statistics: A First Course, Statistics for Six Sigma Green Belts with Minitab and JMP, Six Sigma for Green Belts and Champions, and Design for Six Sigma for Green Belts and Champions.
David F. Stephan is an independent instructional technologist. As a computer information systems instructor at Baruch College (CUNY), he pioneered the use of computers in classrooms, devised interdisciplinary multimedia tools, and created techniques for teaching computer applications in a business context. The developer of PHStat2, the Pearson Education statistics add-in system for Microsoft Excel, he has coauthored several books with David M. Levine.
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## 1.Even You Can Learn Statistics: A Guide for Everyone Who Has Ever Been Afraid of Statistics (2nd Edit
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## 2.Even You Can Learn Statistics: A Guide for Everyone Who Has Ever Been Afraid of Statistics (2nd Edition)
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ISBN 10: 0137010591 ISBN 13: 9780137010592
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## 3.Even You Can Learn Statistics: A Guide for Everyone Who Has Ever Been Afraid of Statistics (2nd Edition)
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Destinos, gastos y plazos de envío | 1,174 | 4,862 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2017-47 | latest | en | 0.755548 |
https://crypto.stackexchange.com/questions/2569/how-does-one-implement-the-inverse-of-aes-mixcolumns | 1,702,043,387,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100745.32/warc/CC-MAIN-20231208112926-20231208142926-00173.warc.gz | 222,390,925 | 42,628 | # How does one implement the Inverse of AES' MixColumns
what will I do?
Because it said that I will used this:
During decryption the Mix Column the multiplication matrix is changed to:
0E 0B 0D 09
09 0E 0B 0D
0D 09 0E 0B
0B 0D 09 0E
How does one implement this maxtrix multiplication?
It's pretty much the same as the forward Mix Column direction; it's a series of multiplications in $GF(2^8)$, however, instead of multiplying by 1, 2 and 3, you're multiplying by 9, 11, 13 and 14.
The multiplication rule isn't that complex; however, it is a bit fancier than the quick rule we got for $\times 2$ and $\times 3$. If you're happy with doing a table lookup, this wikipedia page gives tables of values for $x\times 9$, $x\times 11$, $x \times 13$ and $x \times 14$.
Another way to approach it is to take the rule you already know for $x \times 2$, and use it several times, as in:
$x \times 9 = (((x \times 2) \times 2) \times 2) + x$
$x \times 11 = ((((x \times 2) \times 2) + x) \times 2) + x$
$x \times 13 = ((((x \times 2) + x) \times 2) \times 2) + x$
$x \times 14 = ((((x \times 2) + x) \times 2) + x) \times 2$
(where $+$ is addition in $GF(2^8)$; you know it better as "exclusive-or")
Further explination would require me to get into the technicalities of what multiplication in a finite field actually is; I'm not sure you're quite ready for that; if you think you might be, you might want to start in on this article.
• Thanks. It was very helpful. I have used the 2nd approach as I had implemented the multiplication by 2 and 3. Apr 3, 2019 at 8:27
For the MixColumn matrix $$M$$, it is true that $$M^4 = I$$. So, by performing this transformation trice you get its inverse ($$M^3 = M^{-1}$$). A little crazy, but it might be practical in certain situations.
• It would be very constructive if you convert the As I remember into a real knowledge with a reference or calculation. Also, you can give and compare the number of operation required to calculate the inverse and 3rd power. Mar 17, 2019 at 14:34 | 610 | 2,030 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 3, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2023-50 | longest | en | 0.921128 |
http://www.ask.com/math/mensuration-mathematics-790558118376426c | 1,464,142,596,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464049274059.91/warc/CC-MAIN-20160524002114-00188-ip-10-185-217-139.ec2.internal.warc.gz | 362,688,146 | 21,175 | Credit:Cultura RM/MatellyCollection Mix: SubjectsGetty Images
Q:
What is mensuration in mathematics?
A:
Mensuration is a branch of mathematics that deals with the measurement of areas and volumes of various geometrical figures. Figures such as cubes, cuboids, cylinders, cones and spheres have volume and area. Mensuration deals with the development of formulas to measure their areas and volumes.
Know More
The area of a cube is obtained using the formula A = 4 x length squared. The volume of a cube is obtained using the formula V = length cubed.
The area of a cuboid is obtained using the formula A = 2 [lb + lh + bh] where l is the length, b is the breadth and h is the height of the cuboid. The volume of the cuboid is arrived at using the formula V = length x breadth x height.
The curved surface area of a right circular cylinder can be obtained using A = 2πrh where h is the height of the cylinder and r is the radius of the base of the cylinder. The total surface area of the cylinder is obtained using A = 2πr [h+r], and the volume is arrived at using V = πr2h.
A cone has a curved surface area called the lateral surface area. The volume of a cone is one-third the volume of a cylinder with the same height and radius of the base.
Area is expressed in square units, while volume is expressed in cubic units.
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Geometry is defined as the area of mathematics dealing with points, lines, shapes and space. Geometry is important because the world is made up of different shapes and spaces. It is broken into plane geometry, flat shapes like lines, circles and triangles, and solid geometry, solid shapes like spheres and cubes.
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In mathematics, translation means moving an object from one location to another. It is a term often used in geometry. In translation, the object is moved without rotating, reflecting or resizing it.
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An intersection in mathematics usually refers to the point where two lines cross each other on a plot, but can also refer to shared numbers in data sets. It is possible for more than two lines to meet at an intersection, and it is also possible for a line to have more than one distinct intersection. | 484 | 2,218 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2016-22 | latest | en | 0.949727 |
https://socratic.org/questions/how-do-you-solve-for-x-in-3x-y-7 | 1,725,884,512,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651098.19/warc/CC-MAIN-20240909103148-20240909133148-00308.warc.gz | 528,042,864 | 6,210 | # How do you solve for x in 3x + y =7?
Mar 23, 2018
$x = \left(\frac{- y + 7}{3}\right)$
#### Explanation:
To begin isolating x, use algebraic manipulation to begin moving everything to the opposite side as the x term.
Begin by moving the y term.
$\left(3 x + y = 7\right) = \left(3 x = - y + 7\right)$
Next, divide both sides by 3
$\left(3 x = - y + 7\right) = \left(\frac{- y + 7}{3}\right)$
If you need the answer in slope intercept form just break the fractions apart and rewrite the $- \frac{y}{3}$ term as $- \frac{1}{3} y$
$x = - \frac{1}{3} y + \frac{7}{3}$ | 201 | 575 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 6, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2024-38 | latest | en | 0.693233 |
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In the historical development of set theory, it was mentioned that Russel demonstrated, the following classical definition is not sufficient since it leads to paradoxical constructs. If you are a beginning student of the set theory, the classical definition of Cantor is a good starting point, because it is highly intuitive.
## Definition: Set, Set Element, Empty Set (Cantor)
(Original, naive set definition of Cantor (1895))1
A set is a combination of well-distinguishable, mathematical objects. Let $$X$$ be a set.
• If an object $$x$$ belongs to the set $$X$$, it is called ist element and written as $$x\in X$$.
• We write $$x\notin X$$, if $$x$$ is not an element of the set $$X$$.
• If $X$ has no elements, we call $X$ empty, and write $X=\emptyset,$ otherwise non-empty and write $X\neq\emptyset.$
1 Nowadays, we use the Zermelo-Fraenkel axioms (ZFA) to define sets.
| | | | | created: 2014-03-22 15:09:49 | modified: 2019-09-07 08:12:09 | by: bookofproofs | references: [656], [7838] | 302 | 1,074 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2022-49 | latest | en | 0.835325 |
https://www.coursehero.com/tutors-problems/Other-Homework/554751-I-dont-understand-the-math-behind-the-below-The-mean-biomass-is-343/ | 1,531,967,873,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590443.0/warc/CC-MAIN-20180719012155-20180719032155-00563.warc.gz | 841,394,632 | 84,781 | View the step-by-step solution to:
# Trees and Carbon Table of Contents:) Introduction (2.) Carbon Cycle and Global Warming (3.) Carbon Sequestering in Trees (4.) Allometry (5.) Forest...
I don't understand the math behind the below.
The mean biomass is 3431
(b.) For this question, convert biomass per m2 to biomass per acre. Since 1 acre = 4047 m2, multiply the mean biomass per 100 m2 by 40.47. Thus average biomass per acre = ____________ kg/acre.
Trees and Carbon Table of Contents: (1.) Introduction (2.) Carbon Cycle and Global Warming (3.) Carbon Sequestering in Trees (4.) Allometry (5.) Forest Data: Carbon Analysis (6.) Activity Introduction To understand the environment, it is important to understand how organisms and their surroundings interact. Since all organisms use energy, we need to understand how energy can be used and transferred. Because all organisms are made of substances, it is equally important that we understand how chemicals are used and transported through an ecosystem. This exercise will help contribute to our understanding of the movements of compounds in ecosystems. The transport and transformation of substances in the environment are known collectively as biogeochemical cycles. These global cycles involve the circulation of elements and nutrients that sustain both the biological and physical aspects of the environment. For example, all known organisms on this planet depend on water to sustain them. They are constantly cycling water, consuming it on a regular basis either by itself or with nutrients, while expelling water (with waste products) at the same time. Besides being critical for the biosphere, water is also an extremely important part of the physical environment. When water vapor condenses to form clouds, more of the Sun's rays are reflected back into the atmosphere, usually cooling the climate. Conversely, water vapor is also an important greenhouse gas in the atmosphere, trapping heat in the infrared part of the spectrum in the lower atmosphere. Water is also involved in other biogeochemical cycles. The hydrologic cycle intersects with almost every other element cycles, as well as some of the geological cycles such as the sedimentary cycle. Carbon Cycle: Example In this and other activities, we are going to study how carbon cycles through our ecosystem and how mankind affects this cycle. It is important that we understand how carbon cycles through the ecosystem for two reasons. The first of these reasons is that all organic material contains carbon. From the smallest vitamin molecule all the way up to the long polymer chains of proteins and DNA, carbon provides the basis of all organic compounds. The second reason why we need to understand the carbon cycle is because of its effect on the physical environment. Carbon, in the form of carbon dioxide, is released as a waste product of oxidation. This means that it is released during the combustion of fossil fuels, as well as the respiration of organisms. As we will see later, this can have a tremendous effect on our climate, since carbon dioxide is a greenhouse gas.
Carbon has two phases in the carbon cycle: gaseous and solid. Its gaseous phase is mostly in the form of carbon dioxide, but it can also be found in compounds like methane and carbon monoxide. Carbon dioxide can be taken out of the atmosphere by photosynthesis in plants, which convert the carbon into a solid form (sugars) that can be stored or put back into the air during respiration. It can also be removed from the atmosphere by being absorbed by water, where it becomes available to water plants for photosynthesis as well as being available to form compounds such as calcium carbonate (chalk) or to be put back into the atmosphere when the water gets warmer. As we can see, the carbon cycle has reservoirs where it is stored as a solid. The diagram below shows some of these. In a cycle that has reached equilibrium, the rate at which carbon is removed from storage is equal to the amount that is being taken out of the atmosphere. The reason why many people are concerned about the carbon cycle is because mankind's intervention has caused this system to go grossly out of equilibrium. By burning fossil fuels, mankind has upset the balance of the cycle and greatly increased the rate at which carbon is returning to the gaseous phase. Is this a problem? In order to understand why it might be a problem, we need to understand more about the properties of carbon dioxide.
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If biomass given is = 3431kg/100m2 Then it will... View the full answer
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Browse Documents | 1,002 | 4,960 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2018-30 | longest | en | 0.905326 |
https://www.lmfdb.org/EllipticCurve/Q/1728/d/ | 1,632,734,227,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058415.93/warc/CC-MAIN-20210927090448-20210927120448-00177.warc.gz | 861,691,484 | 10,421 | # Properties
Label 1728.d Number of curves $3$ Conductor $1728$ CM no Rank $1$ Graph
# Learn more
Show commands: SageMath
sage: E = EllipticCurve("d1")
sage: E.isogeny_class()
## Elliptic curves in class 1728.d
sage: E.isogeny_class().curves
LMFDB label Cremona label Weierstrass coefficients j-invariant Discriminant Torsion structure Modular degree Faltings height Optimality
1728.d1 1728bb3 $$[0, 0, 0, -7884, 357264]$$ $$-1167051/512$$ $$-23776267862016$$ $$[]$$ $$3456$$ $$1.2738$$
1728.d2 1728bb1 $$[0, 0, 0, -204, -1136]$$ $$-132651/2$$ $$-14155776$$ $$[]$$ $$384$$ $$0.17521$$ $$\Gamma_0(N)$$-optimal
1728.d3 1728bb2 $$[0, 0, 0, 756, -5616]$$ $$9261/8$$ $$-41278242816$$ $$[]$$ $$1152$$ $$0.72452$$
## Rank
sage: E.rank()
The elliptic curves in class 1728.d have rank $$1$$.
## Complex multiplication
The elliptic curves in class 1728.d do not have complex multiplication.
## Modular form1728.2.a.d
sage: E.q_eigenform(10)
$$q - 3q^{5} + q^{7} - 3q^{11} + 4q^{13} + 2q^{19} + O(q^{20})$$
## Isogeny matrix
sage: E.isogeny_class().matrix()
The $$i,j$$ entry is the smallest degree of a cyclic isogeny between the $$i$$-th and $$j$$-th curve in the isogeny class, in the LMFDB numbering.
$$\left(\begin{array}{rrr} 1 & 9 & 3 \\ 9 & 1 & 3 \\ 3 & 3 & 1 \end{array}\right)$$
## Isogeny graph
sage: E.isogeny_graph().plot(edge_labels=True)
The vertices are labelled with LMFDB labels. | 530 | 1,409 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2021-39 | latest | en | 0.53852 |
https://www.coderforevers.com/c/c-program/calculate-compound-interest/ | 1,563,860,143,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195528869.90/warc/CC-MAIN-20190723043719-20190723065719-00509.warc.gz | 638,894,543 | 17,854 | # C Program to Calculate Compound Interest
113
views
## In this Program, you’ll learn how to calculate compound Interest.
To properly understand this Program Calculate Compound Interest you should know the following:
• Input/Output
• Arithmetic Operator
• Data Types
Below is the formula to Calculate the Compound Interest
Compound Interest: Principle * (1 + Rate / 100) time
or
In the above formula
```P is principle amount R is the rate and T is the time span```
#### Logic to Calculate the Compound Interest
Step by step how to find compound interest.
1. Input principle amount. Store it in some variable say principle.
2. Input time in some variable say time.
3. Input rate in some variable say rate.
4. Calculate compound interest using the above formula.
5. Finally, print the resultant value of CI.
#### Program to Calculate Compound Interest
``````#include <stdio.h>
#include <math.h>
int main()
{
float principle, rate, time, CI;
/* Input principle, time and rate */
printf("Enter principle (amount): ");
scanf("%f", &principle);
printf("Enter time: ");
scanf("%f", &time);
printf("Enter rate: ");
scanf("%f", &rate);
/* Calculate compound interest */
CI = principle* (pow((1 + rate / 100), time));
/* Print the resultant CI */
printf("Compound Interest = %f", CI);
return 0;
}``````
Output
``````Enter principle (amount): 1200
Enter time: 2
Enter rate: 5.4
Compound Interest = 1333.099243``````
Note: The `pow()` function computes the power of a number.
This Program can be used in making application for banks and calculators.
Ask your questions about how to Calculate the Compound Interest in C and clarify your/others doubts by commenting. Documentation.
Please write to us at [email protected] to report any issue with the above content or for feedback.
Related examples: | 436 | 1,813 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2019-30 | latest | en | 0.735641 |
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### 15013342 soalan-matematik-penilaian-2-tahun-3[1]
1. 1. SK (2) JALAN BATU TIGA TEST 2 / APRIL 2009 MATHEMATICS YEAR 3 100 1 HOURNAME: ………………………………………….. CLASS: …………………..…Answer all the questions.A)OBJECTIVE QUESTIONS. CIRCLE THE CORRECT ANSWER.1. 4568 = A Four thousand five hundred and eighty-six B Four hundred five thousand and sixty-eight C Four thousand five hundred and sixty-eight D Four thousand fifty six and eight2. 9427 = 9000 + 400 + ________ + 7 The missing number is A 2 C 200 B 20 D 20003. 86 953. What is the place value of 6? A ones C hundreds B tens D thousands4. Sofea has 496 flags. She gives to Adam 54 flags. After that she buys 48 flags. How many flags does Sofea have? A 490 C 394 B 598 D 502 ( 4 marks )B) CHOOSE AND UNDERLINE THE CORRECT ANSWERS.1. 9256 is ( more than / less than ) 9156.2. 6778 is ( more than / less than ) 6877.3. 5380 is more than( 5400 / 5300 )4. 8880 is less than ( 9990 / 8800 ). ( 4 marks )
2. 2. C) WRITE THE NUMBER WORDS OR NUMERALS Numeral Words s 1. 7568 2. Nine thousand five hundred and one 3. 1869 4. Ten thousand ( 4 marks )D) MATCH EACH OF THE FOLLOWING1. 2000 + 342 = 9999 – 502 =2. 8171 + 1326 = 9889 – 2100 =3. 5638 + 2151 = 8567 – 6225 =4. 6003 + 870 = 8987 – 2114 = ( 8 marks )E) ANSWER ALL THE QUESTIONS. WRITE YOUR ANSWER CLEARLY AND SHOW YOUR WORKING IN THE SPACES PROVIDED.1) 7236 in word is 2) Three thousand and forty-five in numeral is 1m 1m3) 2373 + 3608 = ___________ 4) 4208 + 1596 = ___________ 1m 1m
3. 3. 5) 6 894 + 3256 + 46 = 6) 45 687 + 3564 + 2564 = 1m 1m7) 8630 - 1761 = ___________ 8) 5883 - 2934 = ___________ 1m 1m9) 5482 - __________ = 652 10) What is the missing number? 856 + 2349 + ___________ = 5437 Fill in the blank. 1m 1m11) 6875 plus 5672 is equal to 12) Find the difference between 6754 and 2369 2m 2m13) 9 7 4 3 14) 2 2 3 1 - 5 2 5 2 3 7 - 4 4 + 2 8 2m
4. 4. 2m15)6 + 6 + 6 = 36 16) 3 X 7 = 21 X = 36 + 7 + 21= 2m 2m17)4 X 6 = __________ 18) 5 X = 35 5 X 8 = __________ X 7 = 49 2m 2m19) On school holiday, 2587 boys, 20) There were 6975 peoples in3956 girls and 1284 teachers Stadium Malawati during avisited National Science Centre. football match. On the next day,Find the total of them. there were 485 peoples less than first night. What is difference between first and second night? 2m 2m 30 marksDisediakan Oleh:Pn. Norizan Awang/April 2009 | 1,056 | 2,708 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2017-04 | latest | en | 0.614063 |
https://www.traditionaloven.com/building/masonry/cement/convert-japan-fun-cement-to-tonne-metric-t-cement.html | 1,542,432,166,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743282.69/warc/CC-MAIN-20181117040838-20181117062838-00240.warc.gz | 1,042,028,799 | 11,826 | Portland Cement 1 Japanese fun mass to Metric tonnes converter
# Portland cement conversion
## Amount: 1 Japanese fun (fun) of mass Equals: 0.000000027 Metric tonnes (t) in mass
Converting Japanese fun to Metric tonnes value in the Portland cement units scale.
TOGGLE : from Metric tonnes into Japanese fun in the other way around.
## Portland cement from Japanese fun to tonne (Metric) Conversion Results:
### Enter a New Japanese fun Amount of Portland cement to Convert From
* Whole numbers, decimals or fractions (ie: 6, 5.33, 17 3/8)
* Precision is how many numbers after decimal point (1 - 9)
Enter Amount :
Decimal Precision :
CONVERT : between other Portland cement measuring units - complete list.
Conversion calculator for webmasters.
## General Portland cement
General or common purpose Portland cement type (not any other weaker/cheaper cement replacement-version). It's the primary masonry binder hence bonding agent for mortars and concretes consisting of building sand, stones or other gravel aggregate, mixed with water.
By standard practice, when freshly poured, Portland cement has unit volume mass of 94 lbs/cu-ft - 1506 kg/m3 (but it becomes denser as the storage time is prolonged, when it gets compressed or vibrated; in such situations its weight per volume can increase to as high as 104 lbs/cu-ft). This calculator is based on the fresh form Portland cement w/ the standard mass properties of 94 pounds to 1 cubic foot.
Convert Portland cement measuring units between Japanese fun (fun) and Metric tonnes (t) but in the other reverse direction from Metric tonnes into Japanese fun.
conversion result for Portland cement: From Symbol Result To Symbol 1 Japanese fun fun = 0.000000027 Metric tonnes t
# Converter type: Portland cement measurements
This online Portland cement from fun into t converter is a handy tool not just for certified or experienced professionals.
First unit: Japanese fun (fun) is used for measuring mass.
Second: tonne (Metric) (t) is unit of mass.
## Portland cement per 0.000000027 t is equivalent to 1 what?
The Metric tonnes amount 0.000000027 t converts into 1 fun, one Japanese fun. It is the EQUAL Portland cement mass value of 1 Japanese fun but in the Metric tonnes mass unit alternative.
How to convert 2 Japanese fun (fun) of Portland cement into Metric tonnes (t)? Is there a calculation formula?
First divide the two units variables. Then multiply the result by 2 - for example:
2.66666666667E-8 * 2 (or divide it by / 0.5)
QUESTION:
1 fun of Portland cement = ? t
1 fun = 0.000000027 t of Portland cement
## Other applications for Portland cement units calculator ...
With the above mentioned two-units calculating service it provides, this Portland cement converter proved to be useful also as an online tool for:
1. practicing Japanese fun and Metric tonnes of Portland cement ( fun vs. t ) measuring values exchange.
2. Portland cement amounts conversion factors - between numerous unit pairs.
3. working with - how heavy is Portland cement - values and properties.
International unit symbols for these two Portland cement measurements are:
Abbreviation or prefix ( abbr. short brevis ), unit symbol, for Japanese fun is:
fun
Abbreviation or prefix ( abbr. ) brevis - short unit symbol for tonne (Metric) is:
t
### One Japanese fun of Portland cement converted to tonne (Metric) equals to 0.000000027 t
How many Metric tonnes of Portland cement are in 1 Japanese fun? The answer is: The change of 1 fun ( Japanese fun ) unit of Portland cement measure equals = to 0.000000027 t ( tonne (Metric) ) as the equivalent measure for the same Portland cement type.
In principle with any measuring task, switched on professional people always ensure, and their success depends on, they get the most precise conversion results everywhere and every-time. Not only whenever possible, it's always so. Often having only a good idea ( or more ideas ) might not be perfect nor good enough solution. If there is an exact known measure in fun - Japanese fun for portland cement amount, the rule is that the Japanese fun number gets converted into t - Metric tonnes or any other Portland cement unit absolutely exactly.
Conversion for how many Metric tonnes ( t ) of Portland cement are contained in a Japanese fun ( 1 fun ). Or, how much in Metric tonnes of Portland cement is in 1 Japanese fun? To link to this Portland cement Japanese fun to Metric tonnes online converter simply cut and paste the following.
The link to this tool will appear as: Portland cement from Japanese fun (fun) to Metric tonnes (t) conversion.
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# Hypothesis Testing of Soft Drink Filling Machine
A soft drink filling machine, when in perfect adjustment, fills the bottles with 12 ounces of soft drink. Any over filling or under filling results in the shutdown and readjustment of the machine. To determine whether or not the machine is properly adjusted, the correct set of hypotheses is H0: m = 12 Ha: m=/ 12
TRUE/ FALSE
#### Solution Summary
This brief solution identifies the validity of the decision rule.
\$2.19 | 113 | 501 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2018-22 | latest | en | 0.773324 |
http://www.edaboard.com/thread365469.html | 1,516,638,377,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891485.97/warc/CC-MAIN-20180122153557-20180122173557-00026.warc.gz | 441,397,716 | 20,036 | # Nonlinear model of amplifier simulation using SDD
1. ## [Moved]: How to simulate DC Nonlinear model?
Hi,
I have an equation that describes the linearity of my amplifier. and I want to make a plot based on the equation. How could I do that in ADS?
this the equation I have vo=a1+a2vi+a3vi^2, and I want to plot vo vs vi...
Please help,
thanks
•
2. ## Re: How to simulate DC Nonlinear model?
Can you create a lissajous figure? It's the natural choice.
Excellent for observing whether your amplifier introduces distortion.
3. ## Re: How to simulate DC Nonlinear model?
Originally Posted by Freestylesoap
I have an equation that describes the linearity of my amplifier.
and I want to make a plot based on the equation.
How could I do that in ADS?
this the equation I have vo=a1+a2vi+a3vi^2, and I want to plot vo vs vi...
It is very easy.
Use SDD(Symbolically Define Device) or Verilog-A.
http://edadocs.software.keysight.com...efined+Devices
http://edadocs.software.keysight.com...he+SDD+and+UCM
4. ## Re: How to simulate DC Nonlinear model?
Originally Posted by pancho_hideboo
Hopefully it's not too late to get back to you. Thank you for your help. The SDD example is really helpful.
I tried out the example, but it doesn't work as what I expect. Compression point, OIP2, and OIP3 do not match to my original amplifier.
I have the equations set up as follows.. K1, K2, and K3 are calculated base on gain, OIP2, and OIP3. Do I input anything wrong here?
•
5. ## Nonlinear model of amplifier simulation using SDD
Hi all,
I would like to come up with a DC nonlinear model of an amplifier. I calculated k1, k2, and k3 based on gain, OIP2, and OIP3 of the amp. And now, I need to simulate the output voltage equation vo=k1vi+k2vi^2+k3vi^3. I use SDD2P block to simulate(as shown in the pictures below), but I could not get similar OIP2 and OIP3 values by doing HB simulations. Does anyone know what I've done wrong in the simulation setup?
Thank you so much!
•
6. ## Re: Nonlinear model of amplifier simulation using SDD
Originally Posted by Freestylesoap
I calculated k1, k2, and k3 based on gain, OIP2, and OIP3 of the amp.
Show me equations for k1, k2 and k3 you use.
http://edadocs.software.keysight.com...pageId=5923594
And you use "vi" in SDD without defining it.
7. ## Re: How to simulate DC Nonlinear model?
Originally Posted by Freestylesoap
Hopefully it's not too late to get back to you.
You were online on 19th March 2017 at least.
Respond quickly.
8. ## Re: Nonlinear model of amplifier simulation using SDD
Originally Posted by pancho_hideboo
Show me equations for k1, k2 and k3 you use.
http://edadocs.software.keysight.com...pageId=5923594
And you use "vi" in SDD without defining it.
The main page keeps saying there's no new reply for me. That's why I didn't open the page. Sorry!
I use vi=_v1. I set up the variables as shown in picture. Really appreciate your help!!
9. ## Re: Nonlinear model of amplifier simulation using SDD
No.
I request equations for k1, k2 and k3.
•
10. ## Re: Nonlinear model of amplifier simulation using SDD
Originally Posted by pancho_hideboo
No.
I request equations for k1, k2 and k3.
vi=Acos(w1t)+Bcos(w1t); vo=k1vi+k2vi^2+k3vi^3
Gain=20log(K1) -> get K1
OIP2=2Po-IM2, IM2=10log((A^2*K2)/(2*z0))[dBm] -> know OIP2 and Po, so get K2
2OIP3=3Po-IM3, IM3=10log((A^3*K3*3/4)/(2*z0))[dBm] ->know OIP3 and Po, so get K3
Hope it makes sense.
11. ## Re: Nonlinear model of amplifier simulation using SDD
First, set k2=0.
Can you get reasonable results when k2=0 ?
I think your equations for k2 and k3 are very suspicious.
See http://www.designers-guide.org/Model...rf-systems.pdf
There are many definition of IM2.
And IM2 is out of band.
Show me definition of your IM2.
--[[ ]]-- | 1,080 | 3,753 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2018-05 | latest | en | 0.911799 |
https://www.lidolearning.com/questions/ph-bb-lakhmirs9-ch1-ex-q94/derive-the-following-equation-/ | 1,680,201,401,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949355.52/warc/CC-MAIN-20230330163823-20230330193823-00028.warc.gz | 952,944,129 | 12,749 | # Lakhmir Singh Solutions Class 9 Physics Solutions for Exercise in Chapter 1 - Motion
Question 94 Exercise
Derive the following equation of motion by graphical method \mathrm{v^2=u^2+2as} where the symbols have their usual meanings.
Consider the velocity-time graph of a body as shown below:
The initial velocity is u at the point A which changes from A to B at a uniform rate in time t. Let v be the final velocity which is equal to BC in the graph. OC represents the time t. A perpendicular is drawn CB from point C and AD is drawn parallel to OC. BE is perpendicular from point B to OE. The area under OABC is the distance travelled s.
Distance travelled, s = area of trapezium OABC
OA + CB = u + v
OC = t
Therefore, \mathrm{v^2=u^2+2as}
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https://aiuta.org/en/a-430-g-football-is-lying-on-the-ground-then-is-kicked-and-is-moving-at-25-ms-if-the-duration-of.42484.html | 1,563,424,731,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525500.21/warc/CC-MAIN-20190718042531-20190718064531-00309.warc.gz | 306,044,157 | 6,718 | Physics
# A 430 g football is lying on the ground, then is kicked and is moving at 25 m/s. If the duration of the impact was 0.0010 s, what was the average force on the ball?
#### BettieZumbrunnen
5 years ago
Tools used for this solution:
Impulse = change in momentum
Impulse = (force ) x (time applied)
Momentum = (mass) x (speed)
=================================
Momentum of the ball after the kick = (mass)x(speed) = 0.43kg x 25 m/s = 10.75 kg-m/s .
Impulse exerted by the kicker's toe = 10.75 kg-m/s .
(force) x (time) = 10.75 kg-m/s .
Force x 0.001 sec = 10.75 kg-m/s .
Divide each side by 0.001 sec :
Force = (10.75 kg-m/sec) / (0.001 sec) = 10,750 kg-m/s² = 10,750 newtons.
That's about 2,416 pounds from the kicker's toe, during that ¹/₁₀₀₀ of a second.
(1.21 tons !) | 261 | 787 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2019-30 | latest | en | 0.906896 |
https://en.wikipedia.org/wiki/Nimbers | 1,495,753,538,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608617.6/warc/CC-MAIN-20170525214603-20170525234603-00552.warc.gz | 740,488,383 | 12,789 | # Nimber
(Redirected from Nimbers)
In mathematics, the nimbers, also called Grundy numbers, are introduced in combinatorial game theory, where they are defined as the values of nim heaps. They arise in a much larger class of games because of the Sprague–Grundy theorem which states that every impartial game is equivalent to a nim heap of a certain size. The nimbers are the ordinal numbers endowed with a new nimber addition and nimber multiplication, which are distinct from ordinal addition and ordinal multiplication. The minimum excludant operation is applied to sets of nimbers.
## Contents
Nimber addition (also known as nim-addition) can be used to calculate the size of a single nim heap equivalent to a collection of nim heaps. It is defined recursively by
αβ = mex({α′ ⊕ β : α' < α} ∪ {αβ′ : β′ < β}),
where the minimum excludant mex(S) of a set S of ordinals is defined to be the smallest ordinal that is not an element of S.
For finite ordinals, the nim-sum is easily evaluated on a computer by taking the bitwise exclusive or (XOR, denoted by ) of the corresponding numbers. For example, the nim-sum of 7 and 14 can be found by writing 7 as 111 and 14 as 1110; the ones place adds to 1; the twos place adds to 2, which we replace with 0; the fours place adds to 2, which we replace with 0; the eights place adds to 1. So the nim-sum is written in binary as 1001, or in decimal as 9.
This property of addition follows from the fact that both mex and XOR yield a winning strategy for Nim and there can be only one such strategy; or it can be shown directly by induction: Let α and β be two finite ordinals, and assume that the nim-sum of all pairs with one of them reduced is already defined. The only number whose XOR with α is αβ is β, and vice versa; thus αβ is excluded. On the other hand, for any ordinal γ < αβ, XORing ξαβγ with all of α, β and γ must lead to a reduction for one of them (since the leading 1 in ξ must be present in at least one of the three); since ξγ = αβ > γ, we must have α > ξα = βγ or β > ξβ = αγ; thus γ is included as (βγ) ⊕ β or as α ⊕ (α ⊕ γ), and hence αβ is the minimum excluded ordinal.
## Multiplication
Nimber multiplication (nim-multiplication) is defined recursively by
α β = mex({αβ + α β′ + α' β′ : α′ < α, β′ < β}).
Except for the fact that nimbers form a proper class and not a set, the class of nimbers determines an algebraically closed field of characteristic 2. The nimber additive identity is the ordinal 0, and the nimber multiplicative identity is the ordinal 1. In keeping with the characteristic being 2, the nimber additive inverse of the ordinal α is α itself. The nimber multiplicative inverse of the nonzero ordinal α is given by 1/α = mex(S), where S is the smallest set of ordinals (nimbers) such that
1. 0 is an element of S;
2. if 0 < α′ < α and β is an element of S, then [1 + (α′ − α) β′] / α′ is also an element of S.
For all natural numbers n, the set of nimbers less than 22n form the Galois field GF(22n) of order 22n.
In particular, this implies that the set of finite nimbers is isomorphic to the direct limit as n → ∞ of the fields GF(22n). This subfield is not algebraically closed, since no other field GF(2k) (so with k not a power of 2) is contained in any of those fields, and therefore not in their direct limit; for instance the polynomial x3 + x + 1, which has a root in GF(23), does not have a root in the set of finite nimbers.
Just as in the case of nimber addition, there is a means of computing the nimber product of finite ordinals. This is determined by the rules that
1. The nimber product of distinct Fermat 2-powers (numbers of the form 22n) is equal to their ordinary product;
2. The nimber square of a Fermat 2-power x is equal to 3x/2 as evaluated under the ordinary multiplication of natural numbers.
The smallest algebraically closed field of nimbers is the set of nimbers less than the ordinal ωωω, where ω is the smallest infinite ordinal. It follows that as a nimber, ωωω is transcendental over the field.[1]
The following tables exhibit addition and multiplication among the first 16 nimbers.
This subset is closed under both operations, since 16 is of the form 22n. (If you prefer simple text tables, they are here.)
Nimber addition (sequence A003987 in the OEIS)
This is also the Cayley table of Z24 - or the table of bitwise XOR operations.
The small matrices show the single digits of the binary numbers.
Nimber multiplication (sequence A051775 in the OEIS)
The nonzero elements form the Cayley table of Z15.
The small matrices are permuted binary Walsh matrices.
Nimber multiplication of powers of two (sequence A223541 in the OEIS)
Calculating the nim-products of powers of two is a decisive point in the recursive algorithm of nimber-multiplication.
## Notes
1. ^ Conway 1976, p. 61. | 1,249 | 4,821 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2017-22 | latest | en | 0.950795 |
https://cstheory.stackexchange.com/questions/15012/how-hard-is-to-compute-delta-v?noredirect=1 | 1,563,282,587,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195524548.22/warc/CC-MAIN-20190716115717-20190716141717-00203.warc.gz | 347,149,428 | 36,679 | # How hard is to compute $\Delta_{|V|}$?
Let $G=(V,E)$ be a graph. Let $\Delta_k$ be the quantity defined in this question. Let $\mathcal{C}$ be the set of vertex covers of $G$. The following holds:
$$|\mathcal{C}| = 2^{|V|} - \sum_{k = 2}^{|V|} \Delta_k \cdot 2^{|V|-k}$$
Let us focus on $\Delta_{|V|}$.
## Question
How hard is to compute $\Delta_{|V|}$?
## Motivation
The quantity $$\Delta_{|V|}\ \ mod\ \ 2$$ is the parity of the number of vertex covers of $G$. If such quantity is $0$ then $|\mathcal{C}|$ is even, otherwise $|\mathcal{C}|$ is odd.
Therefore (forgive me if I'm wrong):
1. Being able to compute $\Delta_{|V|}$ in polynomial time implies $\mathbin{\oplus}$P = P.
2. $\Delta_{|V|}$ being $\#$P-hard to compute implies $\mathbin{\oplus}$P = PP.
It seems that in both cases we would get an important result. Clearly there is a third option: computing $\Delta_{|V|}$ may be of intermediate counting complexity, strictly harder than FP but strictly easier than $\#$P-hard.
## Update 08/01/2013 21:50
After reading again and more carefully T. Williams' answer to this question, my understanding is that such answer precisely proves that computing $\Delta_{|V|}$ is $\#$P-hard (because the constraint $[0,1,1,1]$ forces every node of $G$ to be present).
However, my second conclusion above does not hold: to conclude $\mathbin{\oplus}$P = PP, computing $\Delta_{|V|}\ \ mod\ \ 2$ should be $\#$P-hard. But the $\#$P-hardness of computing $\Delta_{|V|}$ is not known to imply that computing $\Delta_{|V|}\ \ mod\ \ 2$ is $\#$P-hard as well.
Only a more vague and harmless conclusion can be drawn, more or less along these lines: something which appear to be false from a practical point of view (i.e. Graph Isomorphism $\not\in$ P) would imply something already formally known to be true (i.e. computing $\Delta_{|V|}$ is $\#$P-hard).
• So do you still have a question ? I'm confused now. – Suresh Venkat Jan 9 '13 at 5:36
• You are right, there is no question any more... It may be deleted, or it may be rephrased. I'm less willing to delete it, because $\Delta_{|V|}\ \ mod\ \ 2$ being the parity of $|C|$ is for me an interesting fact, I don't know if it was already known previously. – Giorgio Camerani Jan 9 '13 at 8:44
As the OP pointed out in an update above, the second question about computing $\Delta_{|V|}$ is answered in this question.
The first question is about computing the parity of the number of vertex covers. This problem is $\bigoplus$P-hard even for planar 3-regular graphs. See Section 4.1 of this paper (particularly Corollary 4.2).
First, the notation for the corresponding Holant problem is Pl-$\bigoplus$Holant$([0,1,1]|[1,0,0,1])$. Since there is a 1-to-1 correspondence between vertex covers and independent sets, sometimes people also refer to the Holant expression for independent set (in this case, $\bigoplus$Holant$([1,1,0]|[1,0,0,1])$) as though it were the vertex cover problem.
Second, I don't know why they called this problem $\bigoplus$3/2-Bip-VC (that is, why "3/2-Bip"). As I said above, this Holant expression is for the counting of vertex covers (equivalently, independent sets) in 3-regular graphs. | 927 | 3,170 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2019-30 | longest | en | 0.870915 |
https://answerbun.com/engineering/abaqus-vcct-fatigue-consistently-underpredicts-disbond-growth-rate-2/ | 1,660,150,004,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571198.57/warc/CC-MAIN-20220810161541-20220810191541-00295.warc.gz | 122,792,374 | 14,676 | # Abaqus vcct fatigue consistently underpredicts disbond growth rate
Engineering Asked on December 30, 2021
I’m modelling a fatigue disbond using abaqus. I’ve understood the disbond growth rate is equal to:
$$dfrac{partial a}{partial N} = c_3 Delta G^{c_4}$$
I’ve set up an analysis with MMB specimen. I’ve set $$c_3 = 5.59times 10^{-20}$$ and $$c_4 = 4.33$$. The strain energy release rate is shown below:
Doing the math myself, I find:
da_dn_expected = c3 * energy_release ^ c4 = 5.59e-20 * 635 ^ 4.33 = 7.645e-8 [m/cycle]
Comparing that to the simulated results:
da_dn_found = element_width / cycles_to_next_node_release = 0.001 / 13713 = 7.29e-8 [m/cycle]
Which is a 5% difference with what I expected to find.
I’ve run a number of simulations with different mesh sizes as well as different settings and I’ve found the same difference amoung all of these results for all different crack lengths. I’ve added a figure below to illustrate:
Does anybody know what is causing this offset?
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1 Asked on September 1, 2021 | 855 | 3,016 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 3, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2022-33 | latest | en | 0.888032 |
https://www.evi.com/q/700000_kb_are_how_many_mb | 1,429,609,916,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1429246641266.56/warc/CC-MAIN-20150417045721-00125-ip-10-235-10-82.ec2.internal.warc.gz | 997,766,392 | 5,452 | # 700000 kb are how many mb
• 700,000 kilobytes is 667.57 mebibytes.
• tk10npubl tk10ncanl
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• 700000 kb is equal to how many mb | 904 | 2,642 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2015-18 | longest | en | 0.77287 |
https://www.vedantu.com/question-answer/a-stationary-cylinder-of-oxygen-used-in-a-class-11-physics-cbse-5f8918c90592a47a579fcde9 | 1,726,418,779,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651632.84/warc/CC-MAIN-20240915152239-20240915182239-00211.warc.gz | 959,359,134 | 28,311 | Courses
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# A stationary cylinder of oxygen used in a hospital has the following characteristics at room temperature 300k, gauge pressure$1 \cdot 38 \times {10^7}{\text{Pa}}$, volume$16L$. If the flow area, measured at atmospheric pressure, is constant $2 \cdot 4\dfrac{L}{{\min .}}$ the cylinder last nearly for,A) 5hB) 10hC) 15hD) 20h
Last updated date: 15th Sep 2024
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Hint:The pressure of the cylinder after the cylinder gets empty is equal to the atmospheric pressure as the cylinder is getting empty at room temperature. The volume of the cylinder is the flow of the cylinder in the time taken by the cylinder to get empty.
Formula used: The formula for the isothermal process is given by ${P_1}{V_1} = {P_2}{V_2}$ where ${P_1}$ is the initial pressure of the cylinder ${V_1}$ is the initial volume of the cylinder ${P_2}$ is the final pressure of the cylinder and ${V_2}$is the final volume of the cylinder.
Step by step solution:
It is given that the initial pressure is equal to $1 \cdot 38 \times {10^7}{\text{Pa}}$ also the initial volume is $16L$ also the
Final pressure will be atmospheric pressure of${P_2} = {10^5}Pa$ and final volume will be equal to ${V_2} = 2 \cdot 4t$ where t is the time taken to empty the cylinder.
In the isothermal process we get,
$\Rightarrow {P_1}{V_1} = {P_2}{V_2}$
Replace the value of${P_1}$,${P_2}$ and ${V_1}$ we get
$\Rightarrow \left( {1 \cdot 38 \times {{10}^7}} \right)\left( {16} \right) = \left( {{{10}^5}} \right){V_2}$
$\Rightarrow 2 \cdot 4t = \dfrac{{\left( {1 \cdot 38 \times {{10}^7}} \right)\left( {16} \right)}}{{\left( {{{10}^5}} \right)}}$
$\Rightarrow 2 \cdot 4t = \left( {1 \cdot 38 \times {{10}^2}} \right)\left( {16} \right)$
$\Rightarrow 2 \cdot 4t = 2208$
$\Rightarrow t = \dfrac{{2208}}{{2 \cdot 4}}$
$\Rightarrow t = 920\min .$
Converting minutes into hours.
$\Rightarrow t = \dfrac{{920}}{{60}}h$
$\Rightarrow t = 15.3h$
$\Rightarrow t \simeq 15h$
So the time taken by the cylinder to empty itself is$t = 15h$.
The correct option for this problem is option C.
Additional information:The isothermal temperature means that the temperature is always constant. The work done for isothermal process is given by $W = {P_1}{V_1}\ln \left( {\dfrac{{{V_2}}}{{{V_1}}}} \right)$ where W is the work done ${P_1}$ is the initial pressure ${V_1}$ is the initial volume of the cylinder and ${V_2}$ is the final volume of the cylinder.
Note:The relation for the isothermal process is based on the property of constant temperature. The ideal gas equation is the main equation from which the isothermal process relation of the body is taken out by keeping temperature as constant. | 840 | 2,764 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2024-38 | latest | en | 0.735121 |
https://bookdown.org/zhongyufei/Data-Handling-in-R/R-loop-description.html | 1,652,924,448,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662522741.25/warc/CC-MAIN-20220519010618-20220519040618-00424.warc.gz | 200,215,878 | 13,886 | ## 9.1 R循环介绍
### 9.1.1 简单示例
total <- 0
for(i in 1:100){
total <- total+i
}
print(paste0('1到100连续相加求和等于:',total))
#> [1] "1到100连续相加求和等于:5050"
# loop structure
# for (var in seq) {expr}
### 9.1.2 循环结构
R中有三种循环结构：
• for
for循环是大家使用最多的循环结构，for循环的迭代次数是固定的，并且事先知道。如最开始的示例中，1连续相加到100,共计加法次数是100次。
for循环示例如下：
library(tidyverse)
df <- tibble(
a = rnorm(10),
b = rnorm(10),
c = rnorm(10),
d = rnorm(10)
)
output <- vector("double", ncol(df)) # 1. output
for (i in seq_along(df)) { # 2. sequence
output[[i]] <- median(df[[i]]) # 3. body
}
output
#> [1] -0.2458 -0.2873 -0.0567 0.1443
seq_along可以?seq查看用法.
You might not have seen seq_along() before. It’s a safe version of the familiar 1:length(l), with an important difference: if you have a zero-length vector, seq_along() does the right thing:
#wrong
seq_along(c())
#> integer(0)
1:length(c())
#> [1] 1 0
# generates the integer sequence 1, 2, ..., length(along.with). (along.with is usually abbreviated to along, and seq_along is much faster.)
• while
readinteger <- function(){
}
while (response!=42) {
print("Sorry, the answer to whatever the question MUST be 42");
}
• Repeat
repeat循环与while循环类似。如下所示，直到满足if条件后才会跳出循环结构。
i <- 1
total <- 0
repeat{
total <- total+i
i <- i+1
if(i > 100){
print(paste0('连续相加求和等于:',total))
break
}
}
#> [1] "连续相加求和等于:5050"
### 9.1.3 循环控制
R中如何中断或退出循环呢？除了自然结束的for循环，while,repeat是如何结束的呢，在R中，我们可以通过break以及next控制循环，上一示例中我们已经看到break是如何跳出循环的。
• next 用法
for(i in letters[1:6] ){
if(i == "d"){
next
}
print(i)
}
#> [1] "a"
#> [1] "b"
#> [1] "c"
#> [1] "e"
#> [1] "f"
• break 用法
m=10
n=10
ctr=0
mymat = matrix(0,m,n)
for(i in 1:m) {
for(j in 1:n) {
if(i==j) {
break;
} else {
# you assign the values only when i<>j
mymat[i,j] = i*j
ctr=ctr+1
}
}
print(i*j)
}
#> [1] 1
#> [1] 4
#> [1] 9
#> [1] 16
#> [1] 25
#> [1] 36
#> [1] 49
#> [1] 64
#> [1] 81
#> [1] 100
# 共赋值多少次
print(ctr)
#> [1] 45
### 9.1.4 嵌套循环
# not run
v <- vector(length = 100)
for(i in 1:10){
for(j in 1:10){
v[i*j] = i * j
}
} | 930 | 1,993 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2022-21 | latest | en | 0.276157 |
http://math.montana.edu/courses/s216/index.html | 1,548,283,085,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547584415432.83/warc/CC-MAIN-20190123213748-20190123235748-00395.warc.gz | 146,127,086 | 9,823 | ## Contact Information
### Course Supervisor
Dr. Nicole Carnegie
Office: Wilson 2-242
Phone: 994-3123
e-mail
### Course Student Success Coordinator
Office: Wilson 2-263
Phone: 994-6557
email
## Course Description
Stat 216 is designed to engage you in the statistical investigation process from developing a research question and data collection methods to analyzing and communicating results. This course introduces basic descriptive and inferential statistics using both traditional (normal and t-distribution) and simulation approaches including confidence intervals and hypothesis testing on means (one-sample, two-sample, paired), proportions (one-sample, two-sample), regression and correlation. You will be exposed to numerous examples of real-world applications of statistics that are designed to help you develop a conceptual understanding of statistics. After taking this course, you should be able to:
• Understand and appreciate how statistics affects your daily life and the fundamental role of statistics in all disciplines;
• Evaluate statistics and statistical studies you encounter in your other courses;
• Critically read news stories based on statistical studies as an informed consumer of data;
• Assess the role of randomness and variability in different contexts;
• Use basic methods to conduct and analyze statistical studies;
• Evaluate and communicate answers to the four pillars of statistical inference: How strong is the evidence of an effect? What is the size of the effect? How broadly do the conclusions apply? Can we say what caused the observed difference?
### MUS Stat 216 Learning Outcomes
1. Understand how to describe the characteristics of a distribution.
2. Understand how data can be collected, and how data collection dictates the choice of statistical method and appropriate statistical inference.
3. Interpret and communicate the outcomes of estimation and hypothesis tests in the context of a problem.
4. To understand the scope of inference for a given dataset.
## Required Textbook
Introduction to Statistical Investigations (ISI) by Tintle, Chance, Cobb, Rossman, Roy, Swanson, and VanderStoep (Wiley, 2016). MSU negotiated a reduced price for the textbook available only through the MSU Bookstore (ISBN -9781119385943), and offers a custom e-textbook that includes videos for each section (videos are not required for the course). If you prefer to purchase the custom e-textbook alone, you may purchase it from the MSU Bookstore or here.
You may purchase either the print textbook (used or new) or an e-textbook, but you must have access to the textbook during each class period. Other materials, such as readings and assignments, will be downloaded from D2L. | 546 | 2,720 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2019-04 | latest | en | 0.874094 |
http://poj.org/problem?id=1306 | 1,498,559,390,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128321309.25/warc/CC-MAIN-20170627101436-20170627121436-00603.warc.gz | 314,701,266 | 3,535 | Online JudgeProblem SetAuthorsOnline ContestsUser
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Combinations
Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 9564 Accepted: 4415
Description
Computing the exact number of ways that N things can be taken M at a time can be a great challenge when N and/or M become very large. Challenges are the stuff of contests. Therefore, you are to make just such a computation given the following:
GIVEN: 5 <= N <= 100; 5 <= M <= 100; M <= N
Compute the EXACT value of: C = N! / (N-M)!M!
You may assume that the final value of C will fit in a 32-bit Pascal LongInt or a C long. For the record, the exact value of 100! is:
93,326,215,443,944,152,681,699,238,856,266,700,490,715,968,264,381,621, 468,592,963,895,217,599,993,229,915,608,941,463,976,156,518,286,253, 697,920,827,223,758,251,185,210,916,864,000,000,000,000,000,000,000,000
Input
The input to this program will be one or more lines each containing zero or more leading spaces, a value for N, one or more spaces, and a value for M. The last line of the input file will contain a dummy N, M pair with both values equal to zero. Your program should terminate when this line is read.
Output
The output from this program should be in the form:
N things taken M at a time is C exactly.
Sample Input
```100 6
20 5
18 6
0 0
```
Sample Output
```100 things taken 6 at a time is 1192052400 exactly.
20 things taken 5 at a time is 15504 exactly.
18 things taken 6 at a time is 18564 exactly.```
Source
[Submit] [Go Back] [Status] [Discuss] | 516 | 1,704 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2017-26 | longest | en | 0.819168 |
http://mathhelpforum.com/calculus/149278-limit-computation.html | 1,524,717,709,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125948064.80/warc/CC-MAIN-20180426031603-20180426051603-00548.warc.gz | 189,135,826 | 10,010 | 1. Limit computation
Hi,
How can i solve this next limit?
lim (cos(x/√n))^n as n->infinity
Thanks for any help...
2. Originally Posted by rebecca
Hi,
How can i solve this next limit?
lim (cos(x/√n))^n as n->infinity
Thanks for any help...
$\displaystyle \lim\limits_{n\to\infty}\cos^n\left(\frac{x}{\sqrt {n}}\right)\to 1^{\infty}$, an indeterminate form.
What you'll need to do at this stage is take natural logs of both sides, and then apply L'Hopitals rule.
Can you proceed?
3. I will continue from the point where mr. chris stopped...
$\displaystyle \lim_{\n\to\infty}cos^n(\frac{x}{\sqrt{n}})=e^\lim _{\n\to\infty}lncos^n(\frac{x}{\sqrt{n}})$
(the last pass is from continuous of function e)
Now, lets concentrate only in power of e.
$\displaystyle lim_{\n\to\infty}lncos^n(\frac{x}{\sqrt{n}})=lim_{ \n\to\infty}nlncos(\frac{x}{\sqrt{n}})$
And now we do a substitution: $\displaystyle n=\frac{1}{t}$.
$\displaystyle lim_{t\to 0}\frac{lncos({x}\sqrt{t})}{t}$
So we got a $\displaystyle \frac{0}{0}$ form.
Hence we can use l'Hôpital's rule.
$\displaystyle lim_{t\to 0}\frac{-xsin(x\sqrt(t))}{2cos(x\sqrt(t))\sqrt(t)}$
So we got again a $\displaystyle \frac{0}{0}$ form.
Hence we can use again l'Hôpital's rule.
$\displaystyle lim_{t\to 0}\frac{\frac{-x^2cos(x)\sqrt{t}}{2\sqrt{t}}}{\frac{-(xsin(x\sqrt(t))\sqrt(t)-cos(x\sqrt(t)))}{\sqrt{t}}}=\frac{-x^2}{2}$
So:
$\displaystyle \lim_{\n\to\infty}cos^n(\frac{x}{\sqrt{n}})=e^{\li m_{\n\to\infty}lncos^n(\frac{x}{\sqrt{n}})}=e^{\fr ac{-x^2}{2}}$ | 558 | 1,516 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2018-17 | latest | en | 0.555479 |
https://forums.khronos.org/showthread.php/5804-evaluating-side-effetcs-and-Logical-amp-amp-and-for-vectors?p=18571&mode=threaded | 1,506,372,947,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818693363.77/warc/CC-MAIN-20170925201601-20170925221601-00086.warc.gz | 655,712,930 | 10,238 | ## evaluating side-effetcs and Logical && and || for vectors
Since logical && and || need to be evaluated component-wise for vectors, are c[0] and c[2] the only components that need to be evaluated because only b[0] and b[2] were nonzero? What if the RHS of && is a complicated expression(e.g. b&&foo())? Is the statement about evaluating the RHS only when LHS is equal/unequal to 0 meant only apply when both operands are scalar?
int4 a;
int4 b=(int4)(1,0,1,0);
int4 c=(int4)(10,10,10,10);
a=(b&&--c);
//a is (-1,0,-1,0) but should c be now (9,10,9,10) or (9,9,9,9)?
The 1.0 spec states:
"And (&&) will only evaluate the right hand operand if the left hand operand compares unequal to 0. Or (||) will only evaluate the right hand operand if the left hand operand compares equal to 0. For built-in vector types, the operators are applied component-wise" | 249 | 856 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2017-39 | latest | en | 0.813094 |
http://www.mrexcel.com/forum/excel-questions/50528-pulling-information-one-worksheet-based-range.html | 1,398,281,535,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1398223203422.8/warc/CC-MAIN-20140423032003-00026-ip-10-147-4-33.ec2.internal.warc.gz | 727,363,338 | 16,344 | # Pulling Information From One Worksheet Based On A Range
This is a discussion on Pulling Information From One Worksheet Based On A Range within the Excel Questions forums, part of the Question Forums category; I have this workbook where I have various worksheets that have to work off a rate grid. This rate grid ...
1. ## Pulling Information From One Worksheet Based On A Range
I have this workbook where I have various worksheets that have to work off a rate grid. This rate grid is a separate worksheet and has four companies which has the maximum number of e-mails allowed and the rate associated for that amount of e-mails (or any under). So it
Volume company1 company2 company3 company4
250,000 \$0.0200 \$0.0240 \$0.0065 \$0.0020
500,000 \$0.0150 \$0.0170 \$0.0065 \$0.0020
1,000,000 \$0.0100 \$0.0121 \$0.0065 \$0.0020
2,500,000 \$0.0080 \$0.0121 \$0.0055 \$0.0020
5,000,000 \$0.0066 \$0.0067 \$0.0035 \$0.0020
7,500,000 \$0.0058 \$0.0067 \$0.0035 \$0.0020
10,000,000 \$0.0052 \$0.0067 \$0.0035 \$0.0020
15,000,000 \$0.0044 \$0.0067 \$0.0035 \$0.0020
20,000,000 \$0.0038 \$0.0067 \$0.0035 \$0.0020
25,000,000 \$0.0034 \$0.0067 \$0.0035 \$0.0020
30,000,000 \$0.0030 \$0.0067 \$0.0035 \$0.0020
35,000,000 \$0.0024 \$0.0067 \$0.0035 \$0.0020
40,000,000 \$0.0022 \$0.0067 \$0.0035 \$0.0020
So I have 4 other worksheets for each company. I have a certain amount of e-mails in each column that we expect to send out. So I need to pull in the correct rate based on this. So let's say the following:
In my company1 worksheet I have that I will be sending out 236,000 e-mails; it should pull \$0.0200. If I showed I was sending 500,000 then it would then switch to \$0.0150 and so on. Is there an easy way to do this?
Thanks!
Bobby
2. ## Re: Pulling Information From One Worksheet Based On A Range
It sounds like you need to use the vlookup function. It goes like this:
=vlookup(value, table range, col, type)
First I would give a name to the table with the rates (insert, name, define). The range name, whilst not absolutely necessary, makes reading the function a little clearer. Lets say you call that range "rates", and lets say that the number of emails you're looking up for company 1 is in cell b2 (on the company 1 sheet), then the formula looks like this:
=vlookup(b2,rates,2,false) - the word false can be left out - so =vlookup(b2,rates,2) will do the same. False tells excel to find the lowest closest match, true tells it to match exactly. Anyhow, the function will lookup the value in B2 in the first column of the "rates" table and return the value in the 2nd column. Change that 2 to a 3 and it will get the value in the third column i.e company 2 and so on. Hope it helps.
3. ## Re: Pulling Information From One Worksheet Based On A Range
******** ******************** ************************************************************************>
Microsoft Excel - Book22 ___Running: xl2000 : OS = Windows Windows 2000
(F)ile (E)dit (V)iew (I)nsert (O)ptions (T)ools (D)ata (W)indow (H)elp (A)bout
G2G6H6 =
A
B
C
D
E
F
G
H
1
Volumecompany1company2company3company4*Min*
2
250,000\$0.0200*\$0.0240*\$0.0065*\$0.0020**250,000*
3
500,000\$0.0150*\$0.0170*\$0.0065*\$0.0020****
4
1,000,000\$0.0100*\$0.0121*\$0.0065*\$0.0020**236,000500,000
5
2,500,000\$0.0080*\$0.0121*\$0.0055*\$0.0020**company1company1
6
5,000,000\$0.0066*\$0.0067*\$0.0035*\$0.0020**\$0.0200*\$0.0150*
7
7,500,000\$0.0058*\$0.0067*\$0.0035*\$0.0020****
8
10,000,000\$0.0052*\$0.0067*\$0.0035*\$0.0020****
9
15,000,000\$0.0044*\$0.0067*\$0.0035*\$0.0020****
10
20,000,000\$0.0038*\$0.0067*\$0.0035*\$0.0020****
11
25,000,000\$0.0034*\$0.0067*\$0.0035*\$0.0020****
12
30,000,000\$0.0030*\$0.0067*\$0.0035*\$0.0020****
13
35,000,000\$0.0024*\$0.0067*\$0.0035*\$0.0020****
14
40,000,000\$0.0022*\$0.0067*\$0.0035*\$0.0020****
Sheet1 *
[HtmlMaker 2.32] To see the formula in the cells just click on the cells hyperlink or click the Name box
PLEASE DO NOT QUOTE THIS TABLE IMAGE ON SAME PAGE! OTHEWISE, ERROR OF JavaScript OCCUR.
Formulas...
G2:
=MIN(A2:A14)
G6, copied to G7...
=VLOOKUP(MAX(G4,\$G\$2),\$A\$2:\$E\$14,MATCH(G5,\$A\$1:\$E\$1,0),1)
4. ## Re: Pulling Information From One Worksheet Based On A Range
That almost worked. But try doing company2 for 500,001. I am getting \$0.0170 when it should be \$0.0121. Any ideas?
Thanks,
Bobby
5. ## Re: Pulling Information From One Worksheet Based On A Range
Originally Posted by robertuva
That almost worked. But try doing company2 for 500,001. I am getting \$0.0170 when it should be \$0.0121. Any ideas?
Thanks,
Bobby
Specs are all important...
******** ******************** ************************************************************************>
Microsoft Excel - Book22 ___Running: xl2000 : OS = Windows Windows 2000
(F)ile (E)dit (V)iew (I)nsert (O)ptions (T)ools (D)ata (W)indow (H)elp (A)bout
G2G6H6 =
A
B
C
D
E
F
G
H
1
Volumecompany1company2company3company4*Min*
2
250,000\$0.0200*\$0.0240*\$0.0065*\$0.0020**250,000*
3
500,000\$0.0150*\$0.0170*\$0.0065*\$0.0020****
4
1,000,000\$0.0100*\$0.0121*\$0.0065*\$0.0020**236,000500,001
5
2,500,000\$0.0080*\$0.0121*\$0.0055*\$0.0020**company1company2
6
5,000,000\$0.0066*\$0.0067*\$0.0035*\$0.0020**\$0.0200*\$0.0121*
7
7,500,000\$0.0058*\$0.0067*\$0.0035*\$0.0020****
8
10,000,000\$0.0052*\$0.0067*\$0.0035*\$0.0020****
9
15,000,000\$0.0044*\$0.0067*\$0.0035*\$0.0020****
10
20,000,000\$0.0038*\$0.0067*\$0.0035*\$0.0020****
11
25,000,000\$0.0034*\$0.0067*\$0.0035*\$0.0020****
12
30,000,000\$0.0030*\$0.0067*\$0.0035*\$0.0020****
13
35,000,000\$0.0024*\$0.0067*\$0.0035*\$0.0020****
14
40,000,000\$0.0022*\$0.0067*\$0.0035*\$0.0020****
Sheet1 *
[HtmlMaker 2.32] To see the formula in the cells just click on the cells hyperlink or click the Name box
PLEASE DO NOT QUOTE THIS TABLE IMAGE ON SAME PAGE! OTHEWISE, ERROR OF JavaScript OCCUR.
G2:
=MIN(A2:A14)
G6, which is copied to G7...
=INDEX(\$B\$2:\$E\$14,MATCH(MAX(G4,G2),\$A\$2:\$A\$14)+(VLOOKUP(MAX(G4,G2),\$A\$2:\$A\$14,1)<>MAX(G4,G2)),MATCH(G5,\$B\$1:\$E\$1,0))
#### Posting Permissions
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• | 2,338 | 6,188 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2014-15 | latest | en | 0.873558 |
https://sellessay.com/2021/11/10/get-answer-evaluate-ymax-and-ymin-by-examining-one-time-period-between-pills/ | 1,643,391,021,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320306301.52/warc/CC-MAIN-20220128152530-20220128182530-00232.warc.gz | 545,937,568 | 20,826 | # [Get Answer] Evaluate ymax and ymin by examining one time period between pills.
If antiMegatherium the medicine against giant sloth is administered intravenously at a steady rate Show more If antiMegatherium the medicine against giant sloth is administered intravenously at a steady rate of q mg per day then the total amount of the drug in the patients body at time t in days x(t) mg satisfies the differential equation x + kx = q; x(0) = 0 (k > 0). (a) (Sep 11 10 pts): Find x(t) and find k supposing that when a patient receives a constant dose of q = 40 mg per day x(t ) 20 mg. Be sure to specify the units of k. (b) (Sep 11 6 pts): Suppose instead that the patient takes 4 pills of 10 mg each per day equally spaced. Denote by y(t) the total amount of the drug in the body. Make a qualitative sketch of the graph of the function y(t) for 24 hours assuming that we are in the steady state situation in which each successive period of 6 hours looks the same as the others. Label on the vertical axis the maximum and minimum levels of the drug in the body ymax and ymin. Draw a vertical segment to designate the jump in y each time the patient takes a pill. c) (Sep 11 10 pts): Its more convenient to take a pill than to receive a drug intravenously but the amount of the drug in the body varies more. Find out by how much in this problem. Evaluate ymax and ymin by examining one time period between pills. (You can compare these values to the baseline value of 20 mg.) Show less | 362 | 1,484 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2022-05 | latest | en | 0.909763 |
https://electronics.stackexchange.com/questions/157653/am-i-reading-my-multimeter-correctly | 1,674,946,352,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499695.59/warc/CC-MAIN-20230128220716-20230129010716-00617.warc.gz | 243,934,893 | 38,326 | # Am I reading my multimeter correctly?
I'm trying to wrap my head around the 20m and the 2000µ setting. On 20m I see 0.01 but on the 2000µ setting I see 005. Does this mean my circuit is using 5µA of current? Why the discrepancy?
## 4 Answers
This is normal for the accuracy of your meter, at 20m you read 10µA at the more accurate for your range 2000µ you read 5µA.
This multimeter looks suspiciously similar, look at its spec sheet. It is important to realize that every meter has an accuracy and it varies per range. At 2000µA the meter's specification is +/-1.0% +/-5 counts. (Check the manual that came with your meter for its own specifications).
This means your reading 005 can vary:
• min: 5µA - 1% - 5 counts = 5µA - 0.05µA - 5µA = -0.05µA
• max: 5µA + 1% + 5 counts = 5µA + 0.05µA + 5µA = 10.05µA
The same accuracy applies for the 20mA range. This means your reading 0.01 can vary:
• min: 0.01mA - 1% - 5 counts = 0.01mA - 0.0001mA - 0.05mA = -0.0401mA = -40.1µA
• max: 0.01mA + 1% + 5 counts = 0.01mA + 0.0001mA + 0.05mA = 0.0601mA = 60.1µA
This clearly shows that you always want to use the range that is as close as possible to the measured value. It also shows that the last digit can often only be used to view the trend of a measurement rather than measuring its absolute value.
This specification does not account for the burden voltage as mentioned in other answers. The internal resistance of your meter influences the circuit and with that may influence the measurement too. Unfortunately the higher the resolution of the range, the higer the burden voltage becomes. So if you switch your range to an accurate range, the meter influences the circuit with a relatively large internal resistance.
• Also note the Accuracies are guaranteed for 1 year, 23°C±5°C, less than 75%RH. Mar 1, 2015 at 21:45
With a 4 digit DVM the reading of 00.01 is the averaged up reading, equivalent to 5uA. (0.005ma). The leading 0's being shown are misleading though.
Depending on the circuit you measure, the internal burden resistance may influence the current you are trying to measure.
Especially on low current settings the burden resistance gets larger and has a bigger influence.
But as others have noted in 20m you are hitting the accuracy limit of your DMM. | 654 | 2,280 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2023-06 | latest | en | 0.904648 |
https://virtualnerd.com/geometry/quadrilaterals/ | 1,603,392,093,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107880014.26/warc/CC-MAIN-20201022170349-20201022200349-00064.warc.gz | 601,963,398 | 8,316 | ### More Specific Topics in Quadrilaterals
#### How Do You Find Values for Variables in the Angles of a Quadrilateral To Make it a Parallelogram?
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#### How Do You Use the Diagonals of a Rectangle to Find the Value of a Variable?
When you make diagonals inside a rectangle, those diagonals are congruent. With this information, you can find the value of a variable that's part of a measurement of a diagonal. This tutorial shows you the steps.
#### How Do You Use Variables to Name Coordinates for a Figure Placed on the Coordinate Plane?
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#### How Do You Write a Coordinate Proof?
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#### How Do You Find the Value for a Variable in the Angles of a Quadrilateral To Make it a Rhombus?
A parallelogram needs to have certain qualities in order to be a rhombus. In this tutorial, you'll use your knowledge of these shapes in order to find the value for a variable that will make a given parallelogram a rhombus.
#### How Do You Construct a Square?
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#### How Do You Classify Quadrilaterals?
Got a figure with four sides? Then you have a quadrilateral! But there are many special types of quadrilateral. Follow along as this tutorial shows you how to figure out the possible names for a given quadrilateral!
#### How Do You Find Measures of Missing Exterior Angles?
Looking for the missing measurements of exterior angles in a polygon? If you already have the other exterior angle measurements, you can use those to help you find your missing measurements! How? Remember, the sum of the exterior angles of ANY polygon is always 360 degrees. Check out this tutorial and see how to use this knowledge to find those missing measurements!
#### How Do You Find Missing Angles in a Quadrilateral With Variables?
Trying to figure out the measurements of the interior angles of a polygon? Then check out this tutorial! This tutorial shows you how to create an equation and solve it to find those missing measurements. Take a look!
#### How Do You Find a Missing Angle in a Quadrilateral?
Trying to figure out the measurements of the interior angles of a polygon? Then check out this tutorial! This tutorial shows you how to create an equation and solve it to find those missing measurements. Take a look!
#### What is a Rectangle?
A rectangle is one of the many fundamental shapes you'll see in math. Rectangles have special properties that can be very useful in helping you solve a problem. This tutorial introduces you to rectangles and explains their interesting qualities!
#### What is a Parallelogram?
A parallelogram is a special type of quadrilateral with some special properties. In this tutorial, take a look at parallelograms and learn what kinds of quadrilaterals can also be called parallelograms!
#### What is a Trapezoid?
A trapezoid is a special type of quadrilateral with some special properties. This tutorial introduces you to trapezoids and gives you a look at the special properties needed for a quadrilateral to be called a trapezoid. Check it out!
#### What is a Polygon?
Got a closed figure with three or more sides? Then you have a polygon! In this tutorial, you'll learn about the properties of a polygon, see the names of the most popular polygons, and learn how to identify polygons. Check it out!
The term quadrilateral is a really fancy sounding name for a certain kind of polygon. Did you know that there are special types of quadrilaterals? Watch this tutorial to learn about quadrilaterals and their special types.
#### What is a Square?
A square is one of the many fundamental shapes you’ll see in math. Squares have special properties that can be very useful in helping you solve a problem. This tutorial introduces you to squares and explains their interesting qualities!
#### What is a Rhombus?
A rhombus is a special kind of quadrilateral. Knowing about the special properties of a rhombus is important to identifying and using these special polygons. This tutorial introduces you to the rhombus and explains its unique qualities. Take a look!
#### What is a Regular Polygon?
Polygons come in all different shapes and sizes. Some polygons have special properties that allow them to be called regular polygons. In this tutorial, you'll see what it takes for a polygon to be given this special name. Check it out!
#### How Do You Find the Sum of the Interior Angles of a Polygon?
Polygons have all kinds of neat properties! For example, if you know the number of sides of a polygon, you can figure out the sum of the interior angles. That knowledge can be very useful when you're solving for a missing interior angle measurement. Check out this tutorial to learn how to find the sum of the interior angles of a polygon! | 1,114 | 5,309 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2020-45 | longest | en | 0.877632 |
https://lists.ringingworld.co.uk/pipermail/ringing-theory/2012-September/023836.html | 1,708,826,049,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474573.20/warc/CC-MAIN-20240225003942-20240225033942-00722.warc.gz | 369,445,097 | 2,086 | # [r-t] Minor
Hayden Charles hcharles at grandsire.co.uk
Sun Sep 2 12:47:32 UTC 2012
```Rob Weatherby wrote on 02/09/2012 12:35:
>
> Hi Guys,
> Could someone please explain this in beginners terms for me. I just don't understand the format.
> Can you help me out by explaining this in relation to the lead ends as pbs etc....
Have a look at (for example):
<http://www.heatons.fsnet.co.uk/Ringing/Conducting/Session5/Theory.htm>
This explains a lot of the notation.
- is a bob, s a single.
The headings refer to the positions of the 6, so W is the first lead end
where the 6 dodges 5-6 up, H is the fifth lead end where it dodges 6-5 down.
4ths means that the 6 makes the bob, In that it runs in, Out that it
runs out. S3rds - make 3rds at a single.
The numbers under the W and H column mean the number of calls in the
same position, and the 3* points to the note underneath (s - -) meaning
the call is a single in the first course and a bob in the next two.
Similarly the 6 gives the pattern of calls at the same position, s - - s
- - (each of these will be one course apart).
In the first touch the lead sequence starts:
s p p p p
b p p p p
b p p p b
p p p p b
p p p p b at which point you should have the lead end 132456
That should give enough to get started. Try writing out the individual
lead ends to check that you get the figures given.
Hayden Charles
``` | 398 | 1,379 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2024-10 | latest | en | 0.925097 |
https://physicscalc.com/physics/acceleration-calculator/ | 1,725,991,603,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651303.70/warc/CC-MAIN-20240910161250-20240910191250-00696.warc.gz | 439,825,527 | 13,892 | Created By : awnish
Reviewed By : Phani Ponnapalli
Last Updated : May 10, 2023
Acceleration Calculator is a free online tool that calculates the acceleration of an object effortlessly and quickly by taking the initial velocity, final velocity and time. Just you need to enter initial velocity (v), final velocity (u), and time (t) in the specified input sections of the calculator and press on the calculate button to find the acceleration in the blink of an eye.
Given:Speed difference
Initial speed
Final speed
Time
Given:Distance travelled
Initial speed
Distance
Time
Choose a Calculation:
Net force
Mass
Step by Step Process to Evaluate Acceleration
Below mentioned are the simple steps to find the acceleration. Follow these guiding principles and get the result for your numbers in a less amount of time. Steps are along the lines.
• Let us take the initial velocity, final velocity, and time.
• Evaluate the difference between final velocity and initial velocity.
• Divide the obtained difference number by given time to get the acceleration value.
Acceleration Formula
Acceleration is the rate of change of an object speed. Its an vector quantity and has magnitude and direction. The acceleration formula is given as
Acceleration = Change in velocity / Time taken = (v - u) / t.
Average Acceleration = Γêåv / Γêåt =( vf - vi) / (tf - ti)
Example
Question: A plane has a take off speed of 300 km/h. What is the acceleration in m/s2 of the plane if the plane started from rest and took 45 seconds to take off?
Solution:
Given that
Final velocity (v) = 300 km/h
Time taken (t) = 45 seconds
As plane started from rest, initial velocity u = 0
Acceleration = (v - u) / t
We have to convert 300 km/h to m/s as time is in seconds.
300 km/h = 300 * 1000 m / 3600 sec = 83.3 m/ sec
By substituting these values in the acceleration formula, we get
Acceleration = (83.3 - 0) / 45 = 83.3 / 45
= 1.85 m/ sec2
Acceleration of the plane = 1.85 m/ sec2
Physicscalc.Com has got concepts like friction, acceleration due to gravity, water pressure, gravity, and many more along with their relevant calculators all one under one roof.
Frequently Asked Questions on Acceleration Calculator
1. What is meant by Acceleration?
Acceleration is defined as the rate at which object changes its speed or velocity. In simple terms, object is accelerating means its velocity is changing. As per the newton's second law, the acceleration is directly proportional to the force acting on a body and inversely proportional to the mass.
2. What is initial and final velocity?
Initial velocity means the velocity of an object before it has the effect of acceleration. After accelerating the object for some amount of time, the velocity will be the final velocity.
3. What is the difference between acceleration and average acceleration?
Acceleration of an object is change in object velocity over an increment of time. This can change when there a change in object speed or direction. Average acceleration is the change of velocity over a period of time.
4. Is zero acceleration constant acceleration?
Zero acceleration means there is no change in velocity i.e body is moving with same velocity. while constant acceleration means the velocity of an object increases or decreases with sane amount in same time. Therefore, both zero acceleration and constant acceleration are not same.
5. What acceleration is needed to accelerate a car from 36 km/h to 72 km/h in 25 seconds?
Initial velocity = 36 km/h = 36 * 1000 m / 3600 s = 10 m/s
Final velocity = 72 km/h = 72 * 1000 m / 3600 s = 20 m/s
time = 25 seconds
Acceleration = (20 - 10) / 25
= 10 / 25 = 0.4 m/s2 | 856 | 3,672 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2024-38 | latest | en | 0.879667 |
https://www.sciencebuddies.org/stem-activities/its-about-timeto-make-a-sundial?class=AQX6XTFx1PKLM2vmZCPXfCP3HdUdY-aeWRrgLXmdBo2wtQmjrRdAf_FxdFa7rrx7yhfvMz4XK2VPRieHDtpzrEAQBOeyM5sasp_ChPXsT845-w | 1,618,319,354,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038072366.31/warc/CC-MAIN-20210413122252-20210413152252-00159.warc.gz | 1,086,185,094 | 18,002 | # It’s About Time….To Make A Sundial!
Create Announcement
This feature requires that you be logged in as a Google Classroom teacher and that you have an active class in Google Classroom.
Key Concepts
Earth’s rotation, time, sundial
## Introduction
Have you ever watched an old movie, and when a character asks what time it is the other characters all look at the sky, instead of checking their watches, or their cell phones? There’s no giant digital clock in the sky, those characters are using the position of the Sun in the sky to tell time, as people have done for generations.
The oldest known instruments for telling time, the sundial, allow us to track the position of the sun more accurately. Up until the early 19th century, sundials were the main instrument that people used to tell time. If they are correctly placed, sundials can be used to accurately tell time, down to the minute!
In this activity you will be making your own sundial, and using your body to track the movement of the Sun across our sky!
This activity is not appropriate for use as a science fair project. Good science fair projects have a stronger focus on controlling variables, taking accurate measurements, and analyzing data. To find a science fair project that is just right for you, browse our library of over 1,200 Science Fair Project Ideas or use the Topic Selection Wizard to get a personalized project recommendation.
## Background
For millennia, people have used sundials to tell the time of day, based on the apparent position of the Sun in the sky. There are many types of sundials, but in general each consists of a gnomon, a thin rod that casts a shadow onto a dial, a flat plate or platform. The apparent movement of the Sun across our sky is the result of the Earth’s rotation on its axis. As the Earth spins, the Sun appears to move across our sky – but really we’re the ones who are moving!
As the Sun’s position changes in our sky, the shadow it casts will align with lines marking each hour, indicating the time of day. The accuracy of a sundial is affected by a number of factors, including the fact that the angle of the Earth’s rotation isn’t perfectly perpendicular, and the Earth isn’t perfectly round. As a result, corrections have to made to sundials to account for these changes.
For today’s activity we’ll be making a simple sundial (using a clock to help us!), as well as tracking the position of the sun by observing our shadows.
## Materials
• Sidewalk chalk
• Tape measure or yardstick
• Pen or pencil
• Large concrete space with no shadows
• A clock
• A paper plate
• A plastic straw
• A ruler
• Markers or crayons
• Paperweight or a few small stones
• Sunny weather!
## Preparation
1. This activity works best if you start early in the day so you have a few hours of daylight to do your testing. We recommend starting at 9am and going until at least noon, or starting at noon and testing until at least 3pm.
## Procedure
Part 1
*This is a two-person activity, so be sure to ask a friend or parent for help!
1. Start by choosing a place where you will always stand during this activity. Make sure it is in the middle of the open concrete space, with no shadows nearby. Mark this space by using your chalk to outline your shoes.
2. Stand in your chosen spot, and have your helper use the sidewalk chalk to trace the outline of your shadow on the concrete.
4. Repeat Steps 2-3 every 30 minutes, each time marking the time of day at the top of your shadow.
Part 2
*This part of the activity works best if you start close to noon.
1. While you are waiting to trace your shadow, use a pencil or pen to carefully poke a hole through the center of your paper plate.
2. Check the time. Round up to the nearest hour, and write this number at the very edge of your plate with a crayon or marker. For example, if the clock says 9:45am, write ‘10’ on the plate. Use your ruler to draw a straight line from the number you wrote to the hole in the center of the plate.
3. Wait until the clock reads the hour that you wrote before proceeding to the next step (in the example from Step 6, you would wait 15 minutes until the clock reads 10:00am).
4. Take your plate and plastic straw outside. Put the plate on the ground and poke the straw through the hole you made. Slant the straw slightly toward the line you drew.
5. Carefully rotate the plate so that the shadow of the straw lines up with the line you drew. Do you think the shadow will stay in the same place all day? Why or why not?
6. Place the paperweight or stones on the very edges of the plate to hold it in place.
7. Every hour, check your sundial and the position of the shadow on your plate. If you started at 10am, note the position of the shadow at 11am and write ‘11’ on the edge of the plate where this shadow falls. Each time you check the sundial, write the hour on the edge of the plate where the shadow falls. Why do think the shadow is moving? What does your sundial remind you of?
## Observations and Results
For your sundial, you should have noticed something similar. At each hour, the shadow of the straw was in a different position, each time moving clockwise from the start position. After a few hours, you should have noticed that the sundial looks like the face of a clock, with the numbers evenly spaced out around the plate.
The reason for your shadow’s change in shape and position have to do with the earth’s rotation on its axis. As the Earth spins, the Sun appears to move across our sky. The Sun is highest in the sky at noon or midday, and at this point it casts its shortest shadow, because it is directly above us in the sky. In the morning and later in the afternoon, the Sun is off-center, and therefore casts a longer shadow.
The position of the shadow also changes as the Sun appears to move across our sky. You can see something similar if you shine a flashlight on your hand, and then move the flashlight. As you move the light, the position of your hand’s shadow will change with the movement of the flashlight. The position of the Sun in our sky is dictated by the speed of the Earth’s rotation – the Earth rotates on its axis at a speed of 460 meters/second, or approximately 1,000 miles/hour! When the Earth rotates 15 degrees on its axis, its just as though the Sun has moved 15 degrees across our sky. As a result of this movement, (and depending on where you live) the shadow cast by the sun moves approximately 30 degrees each hour, so that over the course of 12 hours, it travels a full 360 degrees around your sundial.
## Credits
Megan Arnett, PhD, Science Buddies
## Reviews
|
Science Buddies | | 1,495 | 6,635 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2021-17 | latest | en | 0.935945 |
https://www.johndcook.com/blog/2020/07/15/kolmogorov-superposition/ | 1,702,108,076,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100873.6/warc/CC-MAIN-20231209071722-20231209101722-00386.warc.gz | 929,789,077 | 14,632 | # Decomposing functions of many variables to functions of one variable
Suppose you have a computer that can evaluate and compose continuous functions of one real variable and can do addition. What kinds of functions could you compute with it? You could compute functions of one variable by definition, but could you bootstrap it to compute functions of two variables?
Here’s an example that shows this computer might be more useful than it seems at first glance. We said it can do addition, but can it multiply? Indeed it can [1].
We can decompose the function
into
where
So multiplication of two variables can be decomposed into the addition and composition of functions of one variable. What other functions of two variables can be decomposed this way? What about functions of three or more variables?
The Kolmogorov-Arnol’d theorem says that all continuous functions, of however many variables you like, can be decomposed this way.
Here is the original version of the Kolmogorov-Arnol’d theorem from 1957:
Let f : [0,1]nR be an arbitrary multivariate continuous function. Then it has the representation
where the univariate functions Φq and ψq,p are continuous and defined on the real line. Furthermore the ψq,p can be chosen independent of f.
Later refinements showed that it is possible to get by with a single outer function Φ depending on f , but independent of q [1].
The original theorem was not constructive, but later Jürgen Braun and Michael Griebel gave a constructive proof [2].
## References
1. Sidney A. Morris. “Hilbert 13: Are there any genuine continuous multivariate real-valued functions?” Journal: Bull. Amer. Math. Soc. DOI: https://doi.org/10.1090/bull/1698. Posted online July 6, 2020
2. Jürgen Braun, Michael Griebel. On a Constructive Proof of Kolmogorov’s Superposition Theorem. Constructive Approximation 30, 653 (2009). https://doi.org/10.1007/s00365-009-9054-2
## 5 thoughts on “Decomposing functions of many variables to functions of one variable”
1. Is it possible to explicity modify the example to extend the domain to all of R^2?
2. You can certainly use the extended reals, since the extended reals are homeomorphic to the unit interval.
You could compose with something like inverse tangent and rescale to map R to [0, 1], but I think you need continuity at ± infinity.
3. In the example, the domain of g_2(x) restricts the allowed values of x. Could there be a way to lift this restriction ?
4. For the Neural Networks community, Hilbert’s 13th problem (at the International Congress of Mathematicians in 1900 “if a certain specific function of three variables can be written as a multiple (yet finite) composition of continuous functions of just two variables” – see for example https://www.math.toronto.edu/~drorbn/Talks/Fields-0911/Hilbert13th.pdf ) was the Big Bang moment.
Perhaps in the 1960’s that Big Bang failed to ignite despite the presence of “flammable materials” (at least not without today’s hindsight) but definetely it did “Bang” Bigly [sic :)] in the 1980’s with the application of Backpropagation to multi-layer NN.
The last equation you cite represents more-or-less a 3-layer (1 input of dim N, 1 “hidden” of size 2N, 1 output of dim 1) and the psi’s to be independent of f. In today’s NN jargon the psi’s the non-linearities / sigmoidal functions. It was later adjusted for the case of multi-layer NN and was called “Kolmogorov’s Mapping Neural Network Existence Theorem” by Hecht-Nielsen ( https://cs.uwaterloo.ca/~y328yu/classics/Hecht-Nielsen.pdf )
I am not sure who was the first to link Kolmogorov’s theorem to NN, perhaps Hecht-Nielsen in 1989? But almost immediately enough momentum was gathered. See for example Kurkova’s https://www.cs.cas.cz/~vera/publications/journals/kolmogorov_relevant_small.pdf
The story is perhaps relevant to your next post on Math History ( https://www.johndcook.com/blog/2020/07/16/math-history/ ) as an Appendix perhaps titled “The Parallel and Sealed Worlds of Mathematics”? Lorentz, Sprecher and I guess lots of others were re-visiting Kolmogorov’s 1957 theorem since a few years after it was published while at the same time 1960’s NN research was “begging” for such an existence theorem which would theoretically, at least, ensure research funding.
As the NN field was also “begging” for a training algorithm for deeper NN while Backpropagation was around since the 1960’s as just another application of Chain Rule.
Given that today’s AI applications are mostly based on NN, I wonder what would have happened if it all started 20 years earlier. I guess nothing because even when NN were revived in 1990’s it took another 20/25 years to reach today’s level. Which is fair since other factors were more important, for example: computing power and price, the vast scale of data sharing via the Internet as a means of sharing training data and demand for “apps” and of course the current state of Capitalism. The accumulation of Capital, responsible for the financial crises, this time was translated to bigger risks, easy funding for start-ups meaning accelerated invention of new consumer needs so that it can satisfy them. So in this case NN is neither the chicken nor the egg but rather chicken food and it does not matter if it was revived in the 1960’s or 1990’s or 2010’s! – in my opinion.
I was also wondering about why the theorem was named after “Kolmogorov-Arnol’d” although it was published by Kolmogorov alone. Hecht-Nielsen’s paper cited above states that its discovery was a “friendly duel” between the two Soviet Mathematicians who set to resolve all remaining questions posed by Hilbert 50 years earlier. A duel won by Kolmogorov.
I have also posted some, perhaps overlapping, information about NN and the 1960’s in your other excellent post many years ago:
https://www.johndcook.com/blog/2016/10/14/the-big-deal-about-neural-networks/
bw,
Andreas Hadjiprocopis
5. In my previous comment I missed an additional connection for the K-A theorem, regarding the neuronal-like analogies of NN, i.e. the neurons, as the simple building blocks.
It’s difficult or impossible to pinpoint exactly the role of neurons in a monolithic ad-hoc transfer function in R^N.
Whereas in its equivalent *superposition* of *simple* functions in R^1, thanks to Kolmogorov-Arnol’d, this analogy/mapping is quite clear and immediately opens to the *distributivity* and massive scale of the brain, linking the ideas of D.Rumelhart, J.McCleland, Parallel Distributed Processing et al.
bw,
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