/sqt/

But couldn't you argue the same with rows if you just rotate it 180 degrees?

You could but that's not how it's usually done. The circle itself is invariant under rotation around its centre, so it makes sense to consider circular permutations as those permutations with a similar property. Suppose you stand at the centre of the circle, then there's no natural way to pick a direction and face that way, so you'll treat all directions as the same. If you consider a row of chairs, there is typically a natural way to look the chairs: you stand so that you're facing the occupants of the chairs. When you're in this position, a 180 degree rotation will obviously look different.

abc, bca, cab are different permutations but in a circle they give the same “order”.

How do i answer question 49?

find a function for the distance away from point A at time t then find the absolute minimum of the function.

minimize the distance formula

Find the formula for the line through the point A and a point on the line X (in terms of t).
Calculate the dot product of that line with X, and set it equal to 0.
Solve for t.

Simple statistics & probability question:

Let's say that I'm playing a dice game. I have to choose a number between 0 and 99 and bet on the lucky number being higher or lower than the number I chose.

If I lose, the next bet will be increased by a percentage until I either win or run out of balance.

How do I calculate the probability of running out of balance (the amount that I would have to bet is higher than the balance itself)?

e.g

Betting that the lucky number is going to be less than 5. What's the probability of losing the total balance of $3.000.000, if my initial bet is $50 and each time I lose the bet gets increased by 5.84%

Betting that the lucky number is going to be more than 24. What's the probability of losing the total balance of $3.000.000, if my initial bet is $250 and each time I lose the bet gets increased by 300%

Which is more risky? This one I can tell just by calculating the probability of running out of balance.

Any help will be appreciated. I got stuck trying to make a probability-of-getting-rekt calculator and Google isn't helping. I can calculate the number of times that it takes to run out of balance but got stuck on getting the probability itself, I know it's a very basic question.

Insufficient information. How am I supposed to determine the range of possible numbers? Is the range 0 to 99? If so the game is broken because you can just bet that it'll be higher than 0 every time and always win.

You are right, I forgot to mention that part.

On the dice game the profits that you can get will be multiplied depending on the risk you're taking. So if you bet $5000 and you bet that the number will be higher than 1 (98% chance of winning) you'll get a multiplier of 1.01x so $50 if you win that bet.

If you do the same but you bet that the number will be higher than 2 (97% chance of winning) you will get a multiplier of 1.02x so $100 if you win. 1% of winning change = 99x profits if you win.

You cannot bet that the number will be higher/lower than 0 or 99. It has to be at least 1 or 98.

Researching on this I found that this is exactly how dice games online do it. The range is 0 to 99, but you cannot bet that the number will be higher than 0.

>On the dice game the profits that you can get will be multiplied depending on the risk you're taking.
So there is an equal chance of victory no matter the choice?

Nope, it's random. You'll of course have better change at winning if you choose something like a number higher than 1 (estimated 98% change of winning) but if your bet was correct you will get just 0.01 of return.

If you take a huge risk like betting that the number will bet less than 1 (estimated 1% change of winning), if you win you'll get 99x times what your bet was.

So for the initial question one of the things that I have to figure out is depending on the type of bet that I make, what's the probability of a negative result repeating over and over again until there's not enough balance to make the next bet.

We have Ax = b

If A has a left inverse, there is at most one solution that would be equal to Xb, where X is the left inverse. If A is only right invertible, and Z is the right inverse, there is at least one solution Zb. Why is it at most one for left inverse but at least one for right inverse?

Why isn’t the answer 3?

>Why is it at most one for left inverse but at least one for right inverse?
Because there's no reason for there to be any solution in the case of a left inverse. You can't prove that AXb=b since X is only a left inverse, but if you assume there is such a solution x to Ax=b in the first place, then applying the left inverse you get XAx=x=Xb. And similarly any other solution Ay=b would also satisfy XAy=y=Xb so the solution is unique IF it exists, so there's at most one.

With the right inverse, you can simply compute AZb=b and so you know there's at least one solution.

>Why isn’t the answer 3?
Why would the answer be 3?

Oh ok thanks, that makes sense

does it look square to you? is 3y*3y=12y^2?

3*4 are 12

4 I meant to say

Easy.

What are these guys on about?
reddit.com/r/badmathematics/comments/764cu9/one_of_my_tutoring_students_has_this_question_on/

You have to go back.

If I have a number of items, and a couple criteria to rank them by, what are some better methods of getting the overall "best" one than simply sorting them for each criterion, then summing up the results for each item and sorting the sums?

I am having a brainfart. I already have a partition coefficient and I am trying to find the molar concentration that would be in the octanol [Co] and aqueous [Cw] phases.

How do I do this? P = 38.905

[latex]text test[/latex]

[math]text test[/math]

>Polar decomposition theorem
[math]T=B{^T}\begin{bmatrix}{2/5}&{-1}&{0}\\{{11/5}}&{2}&{0}\\{0}&{0}&{1}\end{bmatrix}B[/math]
>find U, V and R
ok, I know this:
[math]
T=R\cdot U=V\cdot R
\\
U{^2}=T{^T}\cdot T
\\
V{^2}=T\cdot T{^T}
[/math]
well
[math]
U{^2}=T{^T}\cdot T=B{^T}\begin{bmatrix}{5}&{4}&{0}\\{{4}}&{5}&{0}\\{0}&{0}&{1}\end{bmatrix}B
[/math]
I need to diagonalize the matrix (I barely remember how), but... what do I do with the bases B and B^T?

Project the point to the line:
[math] \frac{ \langle (-1,2,3) , (1,1,3) \rangle }{ \lVert (1,1,3) \rVert ^2 } (1,1,3) [/math]

oops
replace (1,1,3) by (0,1,0)

Hello friends, could someone explain why we're able to just take a vector [math]\vec{a}[/math], take it apart into vectors [math]\vec{a_x}[/math] and [math]\vec{a_y}[/math], and just do fucking this [math]\vec{a}=\sqrt{{a_x}^2+{a_y}^2}[/math]? Where did vectors go!?

I have an understanding that this is possible because we can write [math]\vec{a}=a_x\vec{i}+a_y\vec{j}[/math] and since we're trying to get the magnitude of the vector [math]\vec{a}[/math], it works if we ignore [math]\vec{i}[/math] and [math]\vec{j}[/math] since all the do is indicate the direction, therefore removing that leaves us with the magnitude of [math]\vec{a}[/math]'s components, not being able to result in anything else but the magnitude of the vector [math]\vec{a}[/math] itself.

Is my understanding correct, friends?

>Hello friends, could someone explain why we're able to just take a vector a⃗ , take it apart into vectors ax→ and ay→, and just do fucking this a⃗ =ax2+ay2−−−−−−−−√?
You can't. \arrow{a} is a vector, the right hand side is a scalar.

But I keep seeing it written like [math]\vec{a}=\lvert{a}\rvert=\sqrt{{a_x}^2+{a_y}^2}[/math] in my class?

>But I keep seeing it written like a⃗ =∣a∣=ax2+ay2−−−−−−−−√ in my class?
Then your seeing it wrong or someone's writing it wrong. The second equality there is true but the first is false.

I'm trying to write a java program to find the longest word in a sentence.

I feel like this should do it, but it only gives me back the first word of a sentence, regardless of length.

The i value seems to be working how I want it, it will print the first pass that i = 3 and then i = -1, then break. which is how I wanted to set it up.

I've been doing okay so far in my cs course, but loops have been giving me a lot of trouble; especially the nested ones, which is why I try to use them as little as I can get away with.

Forgot to add, when I was talking about the i I was talking about when I enter a sentence like "aaa aaaa" it will give those values for i

Right, one more thing regarding vectors, I have this written in my notes [math]\vec{a^0}=\dfrac{\vec{a}}{\lvert{\vec{a}}\rvert}=\dfrac{a_x\vec{i}+a_y\vec{j}}{a}=\dfrac{a_x}{a}\vec{i}+\dfrac{a_y}{a}\vec{j}=\dfrac{a\cdot cos\alpha}{a}\vec{i}+\dfrac{a\cdot cos\beta}{a}\vec{j}=cos\alpha\cdot \vec{i}+cos\beta\cdot \vec{j}[/math]. Right, I get that and all, but how is [math]a_x\vec{i}=a\cdot cos\alpha[/math] possible, IF [math]cos\alpha=\dfrac{a_x\vec{i}}{\vec{a}}[/math]? How did it a vector magically become a scalar? Did I miscopy something again?

A vector [math] \vec{a} [/math] in the plane [math] \mathbb{R}^3 [/math] is just a pair of numbers [math] (a_1,a_2) [/math] . [math] a_1 [/math] and [math] a_2 [/math] are Numbers, not vectors. But you can consider the vectors [math] (a1,0) , (0,a2) [/math] which sum up to [math] \vec{a} [/math] .
This thing you wrote: [math] \vec{a} = \sqrt{a_1^2+a_2^2} [/math] is not true. [math] \sqrt{a_1^2+a_2^2} [/math] is the Length of the vector, not the vector itself, and it is usually denoted by [math] \lVert a \rVert [/math] .

Then tell your teacher that he should be fired.

[math] \vec{a^0} := \frac{\vec{a}}{|\vec{a}|} [/math] is the vector which points at the same direction as [math] \vec{a} [/math] but has length 1.
This means that it lies on the unit circle. A point on the unit circle is described by [math] (\cos{\theta},\sin{\theta}) [/math] for some [math] \theta \in [0,2\pi) [/math] , pretty much by definition.
It's not (cos,cos), it is (cos,sin).

And you are correct, [math] a_x \vec{i} = (a_x,0) [/math] is NOT equal to [math] |\vec{a}| \cos{\theta} [/math] . It is [math] a_x [/math] that is equal to [math] |\vec{a}| \cos{\theta} [/math] .
Did your professor write that shit like that or did you just copied incorrectly?

Maybe solve the eigenvalues cubic equation? Then the vector base will be the eigenvector. Not sure.

I thought this looked like continuum mechanics, and sure enough. continuummechanics.org/polardecomposition.html
If you still need help with this I'll try to look at it in a while.

>engineers trying to do linear algebra

>/sqt/
it's related

yes {1,1,9} and...?

Can you post the whole problem? It's a lot harder without knowing what B is supposed to be.

Need a graph isomorphic to its own dual. Is two vertices joined by a single line good?

No. Its dual only has a single vertex.

>Mean Square Error(MSE)
Isn't it just exactly what it says it is-- take each of your X values minus the average of your X values, square each of the results, and then sum all those squared results together?

why does it hurt more to kick a big rock than a pebble
pls explne in terms of physics, forces

...

>Can you post the whole problem?
t-that's the whole problem ;_;
>It's a lot harder without knowing what B is supposed to be.
basis (standard/cartesian i guess)

From a more general standpoint, this code is way more complicated than it needs to be. You don't need a for loop and two variables here; you could get by with just a while loop and one variable at most. Honestly, you may be better served by starting over and keeping in mind exactly what you want the code to do:

1) Search for the first space in the sentence
2) If there is a space, see if the word at the start of the sentence is longer than the current longest word, and process accordingly
3) If there is not a space, you have reached the last word in the sentence, process it accordingly and terminate.

But, if you want to stick this code, a few issues stand out:

First of all, you have j = sl.indexOf(" ",i) and i = sl.indexOf(" ", i). This works fine on the first iteration, but once you set i = sl.indexOf(" ",i), it is no longer going to do what you intend it to (i.e., find the first space in the remainder of the sentence). On the next loop, it will start from the ith letter in the sentence and search for the next space in the sentence, which is not necessarily the first space in the sentence.

Second, think about what temp = sl.substring(i,j) is doing, particularly in the context of what sl is during later iterations (i.e., once you have run sl = sl.substring((j+1)) ). Once you have removed words that you have already checked, the next word is at the start of sl.

Once this error is corrected, your code doesn't properly handle the last word in the sentence. When you go through the loop and there are two words left, you process the first word, then set i = sl.indexOf(" ", i). But when there is only one word left, i is then going to become -1, unless there is a space at the very end of the sentence. So you don't do anything for the last word.

thanks my nig

what's the consensus on atmospheric sciences/meteorology as a career?

Think recursively, nigga.

First call:
Penis dick ass

Second recursive call:
dick ass

Third recursive call:
ass

You'll want to calculate the length the first word, and then compare it to the max value returned by your recursive call (as that contains the largest word in the substring that it handles).

This is actually an interesting little question I haven't encountered. I might introduce when I interview SWE candidates. I will stipulate it must be recursive, of course.

>liberal
why was using this word necessary?

Thanks. I appreciate your input.

I'm kind of corned about the type of loop; we've only covered for loops so far, so that's the only one I believe we're allowed to use for this project.

I did manage to get it working properly earlier. I know there is probably a more eloquent and less code used way to get the same result, but I feel like it should be fine for someone just starting out in CS. It does give me the correct result each time I run it. I feel like I've learned a lot more about for loops today.

Now I just need to figure out what the next half of the program is supposed to do. It says the other half(other method) of the program is supposed to return the "largest" word in the string. I have no idea what largest is supposed to mean, though. My teacher still hasn't responded to my question about it.

Np mang, the instinct will come with time. One thing that might help is to comment more thoroughly. It can be really helpful when you are first starting out to basically write out in English what you want each part of the program to do before you write the actual code—that will help you write programs that are efficient and clean, instead of the frequently messy result that you get when you just sort of figure out the solution as you go along.

Also, make sure you test it really thoroughly—not just on stuff like "aaa aaaaa" but also on things where you have lots of words, a sentence that's all spaces, a sentence that starts and/or ends with a space, etc.

Can you give more detail on what the instructions for the program were?

Basically, we've just went over loops the past week or two. So all the things we're doing now is related to them.

The assignment was the make a class with two methods, one method was to find the longest word in the string, which was the one I figured out.

The second method "returns the largest word in s in alphabetical order." That is the exact wording from the assignment. Nothing else is given. The alphabetical part I think is just if two words are the "largest" then it will use whichever comes first. The whole description is kind of confusing.

It could be a typo, that happens a lot since they push the grunt work off to the teachers assistants and other people that aren't the teacher to make the assignments and grade things. But usually it's just stuff like the instructions say it's homework 4, even though it should be homework 5 since we already did 4. Stuff like that. Usually nothing this big.

Any chemfags here? Did I fuck this part up on my exam? Since the ratio of moles of gas is 1:1, each change should cancel out.... R-right?

Well, it could also be that you are supposed to:

a) return a list of the largest words, in alphabetical order
b) sort the letters in the largest word into alphabetical order and return it

I think your best bet is just to get clarification from the professor/TAs/someone else in the class, unless you're right up against the deadline

It's not due until Thursday. I just like to get them done ahead of time so I won't have to worry about it.

We're currently on fall break, and I don't have to go back until Thursday. Hoping he'll respond tomorrow; not sure how regularly he checks his email.

It's kind of frustrating, because the homework and stuff is all the same for everyone taking that specific class, regardless of their proff. So a lot of times he doesn't know what our stuff will cover and can't cover the stuff we'll need. Not saying he's a bad proff, just that the way it's set up makes it difficult for him to plan it out.

There's been more than one time he's had to go over something about homework before class starts because we didn't cover it sufficiently in class.

Hey degree/wisdom based question here. Thinking about pursuing chemistry materials science, am I going to be rich? I always thought it was cutting edge and was less gay than engineering. I'm looking at UCLA and acceptance rate from transfer in for that major is like 28%... Any thoughts?

In haskell, this is just

getLongest = foldl1 biggest . words
___where biggest a b = if (length a) > (length b) then a else b

The reaction is exothermic so heating it should drive the reaction backwards to the reactants. Your reasoning is correct for number 10 though

>java
pleb

lads is there a simple way to more easily compute the 20th derivative of sin(x^4) at 0 than manually taking the derivative 20 times and evaluating

>where biggest a b = if (length a) > (length b) then a else b
plen
getLongest = foldl1 (maximum length) . words

taylor series

I don't get why light being the same speed for all observers means that for the moving observer the light bouncing off the mirror occurred simultaneously with the origin.

Help, I'm writing an excercise where I'm using both the Lagrangian and the Laplace transform, afaik both are denoted by [math]\mathcal{L}[/math]

How would you solve this?

Use different styles
[math] L [/math]
[math] \mathscr L [/math]
[math] \mathfrak L [/math]
[math] \mathcal L [/math]

my default mathcal{L} gives me [math]\mathscr{L}[/math]
mathscr{L} gives me error
How do I get [math]\mathcal{L}[/math] then?
I'm using the package amsmath

also I'm very new to latex so sorry

See: tex.stackexchange.com/questions/58098/what-are-all-the-font-styles-i-can-use-in-math-mode
For different fonts and package dependencies.

if i wear 37 dB SNR ear protection at x distance from the audio source, at what distance y does a person without ear protection hear about the same volume as me?

probably not
wolframalpha.com/input/?i=20th derivative sin(x^4)

thanks

There is. See Using the power series for sin(x), with x^4 in place of x,
[math]f(x) = \sin(x^4) = x^4 - x^{12} / 3! + x^{20} / 5! + .. [/math]

The series expansion of f around a given point is unique as f is analytic, so [math] f^{ \left( 20 \right) }(0) / 20! = 1 / 5! [/math]

Oh he wanted to calculate it at 0. I thought he wanted a closed a expression.

>what do I do with the bases B and B^T?
Pretend they don't exist?

I have a physics exam tomorrow and I'm going through some past exams now. What the fuck is pic related even asking and what's the general strategy to solve it?

nigga this so ez

So what's the answer? Is it e?

You start with the wave equation and build up from there:
[eqn] \nabla^2 \psi = \frac{1}{c^2} \frac{\partial \psi}{\partial t} [/eqn]
The rest is trivial math so I'll let you do it.

I haven't studied physics since highschool, even though I'm studuying math.
x is there to describe a family waves, right? Like, at x=1 then the wave is like this, at 2 like this, etc. , right?
If yes, then the families with the same periods are obviously I&III and II&IV.

u sound like professor material desu

*notices your partial derivative* OwO what's this?

en.wikipedia.org/wiki/Wave_equation#Derivation_of_the_wave_equation

56-43=12

i dont understand man