help-38

[mighty canyon]

Well... How can I know that you are actually want to get the answer or just doubt the software?

Think about the possibility that this is the assignment and you just say it is technicality question

I'm not your teacher.... so I have to be careful about this kind of question

I'm trying to help you

and you just ignore my suggestion on this question (and this can apply to other questions)

you just want the answer from my view

Stop and think for a second on what can you adjust your answer knowing that one of them is negative

or consider the different polynomial that I shown you

that should lead to the answer that this is possible

If I am being rude to you, I would not put the wall of text full of suggestions like that.

How can I demonstrate it without providing you the straight answer?

I know your point

But please understand that I also don't know you are student or not.

I have no way to check even with your words

I mean sure I can just say it is (4x+3)(-2x+9)

but you lost the point on how to think like that

because I keep saying that think about it for a second, you can reach there even with negative coefficient.

during that time, you could just check it...

but ok. It is my fault

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[strong horizon]

Can someone help me with this ode problem?

[strong horizon]

Does anyone understand it?

[mild thunder]

im trying to remember how to set these up but you do end up with the ay''+by'+cy=0 form iirc

[strong horizon]

Could I show you my work?

[mild thunder]

ye

[mild thunder]

it looks all good i only have small problem with lbs/ftslugs when you divide through the mass

but ive literally never used slugs so that might be fine

i just googled it, should be all good i think

[strong horizon]

slugs or the answer?

[mild thunder]

the answer

well i googled slugs but the answer should be good

[strong horizon]

thank you have you taken ode before?

[mild thunder]

ye it was a little bit ago

[strong horizon]

are you familiar with the undetermined coefficients method?

[mild thunder]

its vaugely remembered, are you having problems with it?

[strong horizon]

just wanted to see if my process to an answer made sense

[mild thunder]

yeah it looks good

[strong horizon]

Could we check this out?

[mild thunder]

ah this is one of the ones where you have to guess each part of the solution isnt it

the process is basically the same as earlier its just you have specific guess instead of solving with the characterstic iirc

so wed get some y=Ax+B+Ce^-2x
take a couple derivatives and plug it in the lhs and solve similar to the first

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@strong horizon Has your question been resolved?

[strong horizon]

I had a solution previously but I scratched it. Does this look better?

[strong horizon]

Anyone still on?

[mild thunder]

sorry i just spent like 20 mins trying to work through it, for some reason the e^x terms cancel when they usually shouldnt so im kinda confused. i might come up with something but itll be a minute

[mild thunder]

tho ill let you know, xe^x isnt the necessary guess here i dont think

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@strong horizon Has your question been resolved?

[inner ember]

Can you

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[eager lynx]

I just started learning pascals triangle
So far I know it smt about row 4

[vivid glade]

it's really convenient to think about counting

consider (a+b)^n for arbitrarily chosen a and b, with n a nonnegative integer: we have (a+b)^n = (a+b)(a+b)(a+b)...(a+b), n copies of this

when we begin distributing without combining terms, every term we get will be some combination of a's and b's, like aababbab..., where the i'th symbol was obtained by the i'th copy of (a+b)

that is, in each of these strings of a's and b's, each (a+b) contributes either an a or a b

there are n copies of (a+b), so if there are only k b's in a string, then the rest (n-k) of the entries in that string must be an a

when we rewrite each of these strings "aababbbaba..." by putting a's on the left and b's on the right, we get aaaaa...abbb...b = a^(n-k)b^k, provided there are k b's

so now we have a sum of these reordered strings, and it's clear that we can add all the ones that match and get some constant times a^(n-k)b^k

these constants are what we're proposing to be from pascal's triangle, but here's the counting aspect

the coefficient for a^(n-k)b^k is precisely the number of these strings aaa..abbb...b we get from expanding, and this number is equivalent to the number of ways to choose the k copies of (a+b) to contribute 'b' to the product

[eager lynx]

What exactly do you mean by "strings"?

[vivid glade]

by string I mean the product of some copies of a and b

e.g. (a+b)^3 = (a+b)(a+b)(a+b) = aaa + aab + aba + abb + baa + bab + bba + bbb

^ is a sum of 8 strings

but we simplify all the strings into like, a^3 + a^2b + a^2b + ab^2 + a^2b + ab^2 + ab^2 + b^3

[eager lynx]

But since it's to the power of 4 would that mean (A+b) (A+b) (A+b) (A+b)?

[vivid glade]

yeah, exactly

[eager lynx]

okay, sorry carry on

[vivid glade]

sure, I just chose 3 because it's not as trivial as 2 but not as much of a pain to type out as 4 lol

[vivid glade]

so when we simplify this junk further, we get the classical a^3 + 3a^2b + 3ab^2 + b^3

but notice that if we trace backwards in the steps, each coefficient of a^(n-k)b^k is equal to the number of strings we had that contained exactly k b's

and when we count the number of strings with n characters wherein exactly k of them are b's (the other n-k slots are forced to be a), if you're familiar with the choose function, this is 'nCk'

or $\binom nk$

[solid kiln]

Flip

[eager lynx]

So I'm currently trying to do what you are explaining actively with my current problem, and I am on (m+2)^4
But you use (a+b)^3
When you expanded to a^3 + a^2b + a^2b + ab^2 + a^2b + ab^2 + ab^2 + b^3
did you foil?
Does that mean I would foil (M+2)(M+2)(M+2)(M+2)

[vivid glade]

in grade school they call it foiling, but really it's called distributing. also, distributing (m+2)^4 is doing it the very long way, but the point of the exercise is to use this shortcut observation to make it much much faster

[eager lynx]

ohhh

I think i see

a^3 + a^2b + a^2b + ab^2 + a^2b + ab^2 + ab^2 + b^3 so with this I would only change the exponents right?

[vivid glade]

uh, no because I don't know how you would simply rewrite the exponents

[eager lynx]

I'm a little confused on the shortcut now.

[vivid glade]

here's the vision, determining the coefficient of m^2 in (m+2)^4 should be about as computationally taxing as determining the coefficient of m^4 in (m+2)^120

you would never do the latter by hand

[eager lynx]

Yep

[vivid glade]

uh but the result from this is that the coefficient of the term $a^{n-k}b^k$ in $(a+b)^n$ is $\binom nk=\frac{n!}{k!(n-k)!}$

[solid kiln]

Flip

[vivid glade]

this stuff is googlable, you'll find it under the binomial theorem that says $(a+b)^n=\sum_{k=0}^n\binom nka^{n-k}b^k$

[eager lynx]

so I would then use the formula n/k(n-k) What are those exclamation points for

[solid kiln]

Flip

[vivid glade]

they're denoting the factorial operator

$n! = n(n-1)(n-2)\cdots 2\cdot 1$ for $n\geq 1$, and $0!=1$

[solid kiln]

Flip

[vivid glade]

e.g. $\binom{120}4=\frac{120!}{4!116!}=\frac{120\cdot 119\cdot 118\cdot 117\cdot 116!}{(4\cdot 3\cdot 2\cdot 1)116!}=\frac{120\cdot 119\cdot 118\cdot 117}{4\cdot 3\cdot 2\cdot 1}$

[solid kiln]

Flip

[eager lynx]

Wow im ngl this is really complicated, rather than simple rules. The more you say thigns, the more I get lost

[vivid glade]

understandable

personally I disliked looking at pascal's triangle so I prefer the counting stuff

[vivid glade]

the upshot in this case is that, in expanding $(m+2)^4$, the term containing $m^2$ is $$\binom{4}{4-2}m^{4-2}2^2=2^2\binom{4}{2}m^2=(4\cdot 6)m^2=24m^2$$

[solid kiln]

Flip

[eager lynx]

what is the floating 4 above 4-2

is that divison

[vivid glade]

no, that's the 'choose' function as in here

[eager lynx]

oh

[vivid glade]

when saying it out loud, we say "n choose k"

"4 choose 4-2" i.e. "4 choose 2"

because it's the number of ways to choose k things from a collection of n

number of ways to pick a committee of k members from a population of n people, etc. etc.

additional stuff to look into

[eager lynx]

I'm really confused tbh. Are there any videos you can reccommend

[vivid glade]

anything about the binomial theorem is probably fine, companioned with combinations

[eager lynx]

alright thank you, sorry for the trouble, have a good day

.close

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[grave raptor]

Hello I've been working on this question and I've noticed that they've drawn their x and y components on for the normal force and got a different answer.

[grave raptor]

I drew the x and y components for the gravitational force

and I ended up using sin30 instead of tan30 in the answer key. I ended up getting a different answer and I was wondering why breaking down the x and y components for gravity didn't work

[wraith arch]

you usually dont want to break up gravity for this type of problem

since your car is in uniform circular motion in the x direction

splitting up gravity into the parallel and the perpendicular direction won't help

is the picture on the left the answer key?

cuz the force of friction does not point that way

[grave raptor]

yes

does it not?

[wraith arch]

it should point up or down the incline

depending on your speed

[grave raptor]

oh I see

then would the centripetal force just be made up of the x component of the

normal force?

[wraith arch]

mm not quite

its the x component of the normal force and the frictional force

like the FBD would be something like this

you would need to split both F_n and F_f up into x and y components

and then draw connections

the force of friction points down the incline here since we're trying to find the maximum speed

[solid kiln]

y0shi

[grave raptor]

oh I see

I should be using the normal force then

specifically for centripetal force type questions

[wraith arch]

wdym?

like splitting up the normal force?

[grave raptor]

yes

[wraith arch]

its only for this type of problem as its more convenient

the car is rotating around the x axis which is why we want everything in the x and y direction, not parallel and perpendicular

like what you did up there is a bit wrong since the car is not rotating parallel to the incline

it kinda depends on what connections we can draw

[grave raptor]

ohh okay thank you that makes sense

[wraith arch]

yw!

[grave raptor]

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[lusty burrow]

can someone state the type,domain,and range of this function? I thought it was an exponential growth function, domain is all real numbers, and range is all positive real numbers. But, at x=1 there’s a red dotted line does that mean that the function doesn’t go past there? If so, is it not an exponential growth function since exponential growth functions always have a range of positive real numbers?

[trail ingot]

yea, that's what the red line is supposed to mean

and yea i wouldn't call it an exponential function

it would be something like e^x + 1

not e^x

[lusty burrow]

ohh ok so then the range would (1, infinity)

[trail ingot]

yep

[lusty burrow]

ok tysmmm

[trail ingot]

np

[trail ingot]

[grim sparrow]

[lusty burrow]

.close

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[prisma glacier]

i need help proving that this is path connected

i figured it must be since for any 2 points you can draw a path following an arclength of the circle between those 2 points

but im having issues formalizing the argument

[last rose]

in ur setup are geodesics allowed or path-connected is only straight lines?

so if curved paths are allowed then yes, otherwise no

[prisma glacier]

geodesics should be fine

[unique minnow]

You could also find a continuous map whose image is $S^{n-1}$ from a path connected space.

[solid kiln]

Azyrashacorki

[prisma glacier]

that seems even more difficult

[unique minnow]

There's a very natural one that sends points in $R^n - {0}$ to $S^{n-1}$.

[solid kiln]

Azyrashacorki

[unique minnow]

Try with R^2 for instance.

In particular, how could you send every vector in R^2 to the unit sphere, unit being the keyword?'

[prisma glacier]

uh

if x is in R^2

then you can send it to the unit sphere by using x/|x|

wait

[unique minnow]

Well I guess R^2 - {0} actually but yeah

[prisma glacier]

then

wait how does this help

[unique minnow]

So if you can show this is continuous and surjective, and that R^2 - {0} is path connected, you should be able to conclude that the sphere is path connected.

And then this generalizes nicely to S^n

[prisma glacier]

em

this looks quite hard to do...

sorry, could i ask something

would this work instead

if i define f: [0,1] --> S^n where f = x + t(y-x)/|x+t(y-x)|, where y and x are points in S

the numerator would be a straight line from x to y, and then divided by the norm would create a unit vector

wait but i might divide by 0

[unique minnow]

I think that approach is fine, but I think you'd have to argue a bit differently if x and y are antipodal

[prisma glacier]

does division by 0 occur when they're antipodal

[unique minnow]

Yes

[prisma glacier]

uh

[unique minnow]

But if that is so, you can always choose some other point, make a path to there and and another from that point to the antipodal point.

[prisma glacier]

ah

you're right

ok then thats all

thanks for the help

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[hollow notch]

What is the answer to this question and what did I do wrong?

,rotate

[hollow notch]

<@&286206848099549185>

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@hollow notch Has your question been resolved?

[hollow notch]

<@&286206848099549185>

[mild oxide]

Differential equation?

@hollow notch can u clearly tell me what are u asking?

Try applying the substitution y=vx

I hope it will help

@hollow notch

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@hollow notch Has your question been resolved?

[hollow notch]

no its non-exact differential equation

[mild oxide]

Yea I saw it now wait imma do it then tell you

@hollow notch try xy=t I hope this will work

I am getting x^2y^2+3x^3=c it seems correct tho

@hollow notch did this help?

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[rapid horizon]

Hints

[left oriole]

which question

[trim jolt]

@rapid horizon Has your question been resolved?

[rapid horizon]

Both🙈

🥲

[mild oxide]

@rapid horizon yo for the help here

[rapid horizon]

Yo?

[mild oxide]

Yo

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@rapid horizon Has your question been resolved?

[rapid horizon]

<@&286206848099549185>

[mild oxide]

What bro

[rapid horizon]

.close

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[pseudo coyote]

27

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[lapis blade]

does fy-fy=0w or is 0w just indictative of 0

just wanting to know

[lethal gust]

Q12 guys help fast

[mild oxide]

Okok

1/cos n cos (n+1)

[lapis blade]

.close

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[mild oxide]

What

[lethal gust]

.reopen

[mild oxide]

.reopen

Harsh listen

[lethal gust]

Yeah speak

[mild oxide]

Do u know that cos x=(e^ix+e^-ix)/2

[lethal gust]

No

But I know telescopic sum rule

[mild oxide]

Okok then we need to approach in another way

Okok

Do u know that

[rapid horizon]

[mild oxide]

cos(n).cos(n+1)=(1/2)[cos(2n+1)+cos(1)]

Hello?

[lethal gust]

Wait

Let me think

How you got that

[mild oxide]

Listen leave it do u know that cos x is always less than one

[lethal gust]

Yeah

[mild oxide]

So listen now cos 0 degree is 1

[lethal gust]

Yeah

[mild oxide]

And cos 1 degree will also be close to 1 but less than 1

[lethal gust]

Yeah

[mild oxide]

So we can say that as there is cos in denominator

Denominator is always less than 1

[lethal gust]

Yeah

[mild oxide]

But numerator is 1 always

So 1/ less than 1 = greater than 1

[lethal gust]

Yeah yeah

[mild oxide]

So all this sum should be greater than 1

[lethal gust]

Yeah

[mild oxide]

But sin cannot become greater than 1

[lethal gust]

Obviously

[mild oxide]

So sin option gets eliminated

Now about cosec

[lethal gust]

Ok

[mild oxide]

U k cosec 0 degree is infinity

[lethal gust]

Yeas

[mild oxide]

So cosec 1 degree will be large

[lethal gust]

Yeas

[mild oxide]

Ok

But as angle goes till 45 degree

In cos

It should be not that much

So you can mark cosec 1

[lethal gust]

What

[mild oxide]

Listen cosec 1 is great

Ok

If you multiply with √2=1.414

It will lead to a bigger value

Wait let me reach home

I will give u trigonometric approach

[lethal gust]

<@&286206848099549185>

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[rapid horizon]

What will be the formula when ad-bc=1 what will be the formula then?

[nova spire]

?

is the set of such matrices with ad-bc = 0 instead even a group?

ah ad-bc = 1

[rapid horizon]

Sorry for edit

[nova spire]

how many did you find for ad-bc not 0?

[rapid horizon]

I don't know please give a hint

How to get the answer 48

[nova spire]

ok

what do you call the Field of integers mod 3?

or what's your notation for it

F3? Z/3Z?

[rapid horizon]

Z/3z

This one

[nova spire]

ok

Z/3Z is a field

[rapid horizon]

Yeah

[nova spire]

you might have seen that when K is a field

what type of object is K^n?

(like for example K= R, R^n... etc)

[rapid horizon]

3z is not a field

Sorry

It doesn't have multiplicative identity

[nova spire]

?

not 3Z

3Z = {3k, k in Z}

I asked about Z/3Z = {0,1,2}

the integers mod 3

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@rapid horizon Has your question been resolved?

[gilded pulsar]

When n is Prime Z/nZ is a field

Also if n is not a prime Z/nZ is not a field

[nova spire]

yes, though you're not the one I asked the question to...

anyway...

yes Z/3Z is a field

so what object does (Z/3Z)^n make?

[gilded pulsar]

I kinda wasn’t really into the convo and thought you had asked that in general oops

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[wind saffron]

Hello, I am studying affine geometry, particularly the theorem attached as an image. I asked ChatGPT to give me a numerical example of this (since the book doesn't include one). I would like to know if it is correct. It seems strange to me that it doesn't consider the vector (\overrightarrow{X_2X_3}) and (\overrightarrow{Y_2Y_3}). Also, in step 2, I don't understand how it was converted into matrix form. :c

Given the vertices:

\begin{itemize}
\item (X_1(1,1)), (X_2(2,3)), (X_3(4,2))
\item (Y_1(2,2)), (Y_2(3,5)), (Y_3(6,4))
\end{itemize}

\subsubsection*{Step 1: Calculate the vectors of the triangles}

For the triangle (X_1X_2X_3):

[
\overrightarrow{X_1X_2} = \begin{pmatrix} 2 - 1 \ 3 - 1 \end{pmatrix} = \begin{pmatrix} 1 \ 2 \end{pmatrix}, \quad \overrightarrow{X_1X_3} = \begin{pmatrix} 4 - 1 \ 2 - 1 \end{pmatrix} = \begin{pmatrix} 3 \ 1 \end{pmatrix}
]

For the triangle (Y_1Y_2Y_3):

[
\overrightarrow{Y_1Y_2} = \begin{pmatrix} 3 - 2 \ 5 - 2 \end{pmatrix} = \begin{pmatrix} 1 \ 3 \end{pmatrix}, \quad \overrightarrow{Y_1Y_3} = \begin{pmatrix} 6 - 2 \ 4 - 2 \end{pmatrix} = \begin{pmatrix} 4 \ 2 \end{pmatrix}
]

\subsubsection*{Step 2: Set up the system for the matrix (L)}

We seek the matrix (L) such that:

[
L \begin{pmatrix} 1 \ 2 \end{pmatrix} = \begin{pmatrix} 1 \ 3 \end{pmatrix}, \quad L \begin{pmatrix} 3 \ 1 \end{pmatrix} = \begin{pmatrix} 4 \ 2 \end{pmatrix}
]

This can be written in matrix form as:

[
L \begin{pmatrix} 1 & 3 \ 2 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 4 \ 3 & 2 \end{pmatrix}
]

[solid kiln]

Mαя

[wind saffron]

\subsubsection*{Step 3: Solve for (L)}

We multiply both sides by the inverse of ( \begin{pmatrix} 1 & 3 \ 2 & 1 \end{pmatrix} ):

[
L = \begin{pmatrix} 1 & 4 \ 3 & 2 \end{pmatrix} \begin{pmatrix} 1 & 3 \ 2 & 1 \end{pmatrix}^{-1}
]

Calculate the inverse of ( \begin{pmatrix} 1 & 3 \ 2 & 1 \end{pmatrix} ):

[
\begin{pmatrix} 1 & 3 \ 2 & 1 \end{pmatrix}^{-1} = \frac{1}{1 \cdot 1 - 3 \cdot 2} \begin{pmatrix} 1 & -3 \ -2 & 1 \end{pmatrix} = \frac{1}{-5} \begin{pmatrix} 1 & -3 \ -2 & 1 \end{pmatrix} = \begin{pmatrix} -\frac{1}{5} & \frac{3}{5} \ \frac{2}{5} & -\frac{1}{5} \end{pmatrix}
]

Multiplying the matrices to get (L):

[
L = \begin{pmatrix} 1 & 4 \ 3 & 2 \end{pmatrix} \begin{pmatrix} -\frac{1}{5} & \frac{3}{5} \ \frac{2}{5} & -\frac{1}{5} \end{pmatrix} = \begin{pmatrix} \frac{7}{5} & -\frac{1}{5} \ \frac{1}{5} & \frac{7}{5} \end{pmatrix}
]

\subsubsection*{Step 4: Find the translation (\mathbf{t})}

We calculate the translation (\mathbf{t}):

[
L \begin{pmatrix} 1 \ 1 \end{pmatrix} = \begin{pmatrix} \frac{7}{5} \cdot 1 + -\frac{1}{5} \cdot 1 \ \frac{1}{5} \cdot 1 + \frac{7}{5} \cdot 1 \end{pmatrix} = \begin{pmatrix} \frac{6}{5} \ \frac{8}{5} \end{pmatrix}
]

[
\mathbf{t} = \begin{pmatrix} 2 \ 2 \end{pmatrix} - \begin{pmatrix} \frac{6}{5} \ \frac{8}{5} \end{pmatrix} = \begin{pmatrix} \frac{10}{5} - \frac{6}{5} \ \frac{10}{5} - \frac{8}{5} \end{pmatrix} = \begin{pmatrix} \frac{4}{5} \ \frac{2}{5} \end{pmatrix}
]

\subsubsection*{Complete affinity}

The complete affinity is:

[
A(\mathbf{x}) = L \mathbf{x} + \mathbf{t} = \begin{pmatrix} \frac{7}{5} & -\frac{1}{5} \ \frac{1}{5} & \frac{7}{5} \end{pmatrix} \mathbf{x} + \begin{pmatrix} \frac{4}{5} \ \frac{2}{5} \end{pmatrix}
]

[solid kiln]

Mαя

[mild oxide]

Here for help

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<@&286206848099549185>

[mild oxide]

Yo

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[lime vigil]

prove with induction that for every natural n there exists:

[lime vigil]

I'm stuck at the part where I have to prove that if n is true then n + 1 is true

[burnt mulch]

Hint: $$\sum^{n+1}{k=1} a_k=a{n+1}+\sum^{n}_{k=1} a_k$

[solid kiln]

Civil Service Pigeon
Compile Error! Click the reaction for more information.
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[lime vigil]

the hint isnt helping me much ngl

wait 1 sec maybe it is

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[lime vigil]

I can't figure out how

[lime vigil]

it's supposed to be (6(n+1)^2 + n(n+1)(2n+1))/6

????

here's the full thing

this is making me rage fr

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[lime vigil]

.open

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✅

[lime vigil]

https://tenor.com/view/rage-korean-gif-11690731

<@&286206848099549185>

[ruby nexus]

this?

[lime vigil]

yes

how did the left side turn to the right side

[ruby nexus]

you just need to prove that left is equal to right... no like induction or smth?

[lime vigil]

no

not proving them

it's a part of an induction proof proving that for every natural n there exists this

[ruby nexus]

o, like 1^2 + 2^2 + 3^2 + ... + n^2

[woeful bramble]

(n+1)^2 = 6(n+1)^2/6 and mult the stuff out and add and pull out again?

[lime vigil]

?

[ruby nexus]

just a moment

[lime vigil]

here's the whole induction proof step by step (up to the point I can't figure out) if that helps

they check that it is true for n = 1, and then proceed to prove that if it is true for n then it is also true for n + 1

but the last part I can't figure out what was done

[ruby nexus]

can i write you how we studied?

[lime vigil]

oh

waot

wait

I'm dumb

[ruby nexus]

everything ok?

[lime vigil]

yeah it is said in the proof that if they prove that the left side is equal to the right side then it is proven

omg

thanks guys

[ruby nexus]

🙂

[woeful bramble]

the hard way

[lime vigil]

are you dividing by 0

[woeful bramble]

where?

ah no thats polynomdivision

[woeful bramble]

like you did division when you learned it in elementary school

you divide the first part multiply it back and subtract

and you need to get 0 each time or will have a rest at the end

[lime vigil]

first time seeing that

[woeful bramble]

e.g. at the start i divide 2n^3 by n thats 2n^2 and write that on the right side of the =
then i multiply back so 2n^2 * (n+1) thats 2n^3 + 2n^2 and i have to subtract that now from the whole polynom im dividing
there i get then 0 because 2n^3-2n^3 = 0 and 9n^2-2n^2 = 7n^2
and it begins again i divide 7n^2 by n that gives 7n and i write it on the right until i have sutracted everything and left side is 0 so to say

[dark crag]

yes. then you factor an n+1 from both expressions, it would become :
(n+1) ( 6(n+1) + n(2n+1) )/6

(n+1) ( 2n^2 + 7n + 6 ) / 6

dividing that part by n+2 wouls give you 2n+3 which is basically 2(n+1)+1

[lime vigil]

yeah it's all good I figured it out

I was confused because I thought that I should use algebra to get from the left side to the right side

but instead I was supposed to prove that both of them are equal

thanks tho

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[dark crag]

yw

[woeful bramble]

https://en.wikipedia.org/wiki/Polynomial_long_division
maybe that helps to understand what i did

[lime vigil]

what's the purpose of it tho

simplifying?

[woeful bramble]

to divide out a zero point

if you have an polynomial of degree 3 you cant use the midnight formula

you guess for example one zero point then divide that out by dividing by (x-a) if a is the zero point

then you have a polynomial of degree 2 and can use midnight formula for the two remaining ones

[solar wraith]

ig its called euclidean division ? probably he gonna know it by that name

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[finite tusk]

any hints on how to actually solve this?

[rigid vigil]

You can resolve the forces horizontally and vertically, ie add together their x and y components

[finite tusk]

like this?

[rigid vigil]

Why the sin and cos of theta+90?

It's simpler to simply use theta instead of theta+90 (as in your case, the cos(theta+90) would need a - sign)

And the same applies to the 60° instead of 30°, although that doesn't make a huge difference

[finite tusk]

okay but like what should i do now

[rigid vigil]

Also sin 60° = √(3)/2 preferably

Then you can solve the system of two equations with two variables

[rigid vigil]

To make things simpler, I would write $$\begin{cases} x: F \sin(30°) + 500\sin(\theta) = 0 \
y: F\cos(30°) + 500\cos(\theta)=750 \end{cases}$$

[solid kiln]

π=√g

[rigid vigil]

Because in your case, the x component with cos(theta+90) is wrong (missing a - sign)

[finite tusk]

is this right?

[rigid vigil]

That final line seems ok

[finite tusk]

and then what

[rigid vigil]

One way to do it is use cos x = √(1 - sin² x), and square everything so that there are no √ left. Then it's a second degree equation in sin x

Another way is to get it in the form sin a cos x - sin x cos a to get sin(a-x)

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[velvet juniper]

I am currently stuck trying to find line M-N's azimuth or bearing

[velvet juniper]

this is as far as i have gotten. am i supposed to use the reference bearing of 104* 21’ to find line MNs azimuth?

<@&286206848099549185>

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[wispy comet]

I have absolutely no idea, and it's so confusing.

Ik the trigonometric ratios:
sin = y/r
cos = x/r
sec = r/x
etc.

But I don't know how to match it up and it's confusing

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[wispy comet]

@atomic onyx Help

[atomic onyx]

number 4 should be 6,4

[wispy comet]

Oh-

You're right!

Thank you so much

You're very samrt bc that one is so confusing

[wispy comet]

And the other ones?

[atomic onyx]

you should figure out the rest of them with the first answer

quite easily

[wispy comet]

Kk I see now. Thanks so much!!

[wispy comet]

I found out the other values

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[terse haven]

"Gwen is packing for a two day trip and needs to pack socks. She has 5 pairs of white socks and 3 pairs of black socks in her drawer. Find (label) all missing probabilities then find the probability that the two socks she picks without looking are both white. Must show all steps."

[terse haven]

can someone point in the right direction step by step

i think i know what to do but not sure

[wraith arch]

so lets go step by step

what is the probability of picking a black sock as the first sock

[terse haven]

3?

[wraith arch]

well yeah there are 3 black socks

and there are a total of 8 socks

so the probability would just be 3/8

[terse haven]

ooh

5/8 probability of first white socks?

[wraith arch]

yep!

so now when we pick the second pair

our total would be down to 7 pairs in total

since we're doing this without replacement i assume

[terse haven]

mhm

[wraith arch]

so for the black pair branch

what would be the probability of getting another black pair?

[terse haven]

2/7?

[wraith arch]

mhm yeah thats it

and white?

[terse haven]

i wanna stay 5/7

[wraith arch]

yeah thats it

now for the white branch, we use the same process

[terse haven]

4/7?

for white

[wraith arch]

yep

[terse haven]

3/7 for black?

[wraith arch]

yep!

so what would be the probability that both pairs she picks are white?

[terse haven]

5/14?

[wraith arch]

yep thats it

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@terse haven Has your question been resolved?

[terse haven]

idk how im suppose to turn this into a venn diagram

best i got

[gloomy basalt]

I think you're misunderstanding what "no even numbers are odd" means

[terse haven]

i am aware

im struggling with this one

[gloomy basalt]

As an example, when we say "some odd numbers are negative", that means that there is some intersection between the set of odd numbers and the set of negative numbers

Does that make sense to you?

[terse haven]

50/50

i know to put 4 in Even, but idk what to put in Intersection, Odd, and the Universal

if the universal is even needed

[gloomy basalt]

The key fact is, when something belongs to two sets (in other words, has two properties) then it's placed in the intersection of those sets

When something doesn't belong to the two sets, it's placed anywhere other than the intersection

[terse haven]

uh huh

so do i just 0 in everything

[gloomy basalt]

No that wouldn't make sense

0 has just one place

So, when we say no even numbers are odd, that means the numbers we're talking about are even, but not odd

[terse haven]

can you give me an example drawing/image

[gloomy basalt]

Notice how the person who belongs to both sets is placed in the intersection, while the person who belongs to only one set is not in the intersection

[terse haven]

so 4 in even and 0 in intersection

nothing in odd

thats all im getting out of this

[gloomy basalt]

Is 4 an even number but not odd?

Is 0 an even number and an odd number?

This is what you're implying when you put 4 and 0 there

[terse haven]

yes

0 is not even or odd

[gloomy basalt]

Why not

What does even number mean to you? What does odd number mean to you?

[terse haven]

because the creator of math said so

listen idk

[terse haven]

it doesnt mean anything to mean

idk how that answers the question

[gloomy basalt]

The most sensible definition I found was "an even number is a number that can be divided by 2 WITHOUT REMAINDERS", and an odd number is one which can't

So for example, can 2 be divided by 2 without remainders? Yes, therefore 2 is even

Can 3 be divided by 2 without remainders? No, therefore 3 is odd

[terse haven]

final result

.close

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[earnest nymph]

hi can I have help with this, I have no idea what to do

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@earnest nymph Has your question been resolved?

[earnest nymph]

<@&286206848099549185>

[mild oxide]

Yo

[earnest nymph]

yo

[mild oxide]

Yo

What help u need

[earnest nymph]

idk how to do this q

[mild oxide]

Ok

U need AB vector

[earnest nymph]

how do I get it from the abs val

[mild oxide]

Okok lemme see

Can u send clearly

Like complete

Ques

[earnest nymph]

thats it

[mild oxide]

Listen bro I can't see what on right

[earnest nymph]

diagram belongs to the other q

[earnest nymph]

oh thats weird

[mild oxide]

AB vector . AC

Now I saw

[wraith arch]

hint: the definition of dot product

[mild oxide]

100

[wraith arch]

!nosols

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[mild oxide]

Okok sry

Bro listen @earnest nymph bro make a triangle

First

[earnest nymph]

aight

i have a triangle

[mild oxide]

A right angled triangle

As B=90

[earnest nymph]

yep

[mild oxide]

Mark ABC

Yus

Now u know what is dot product right

[earnest nymph]

yea

[mild oxide]

We need angle b/w AB And AC

Cos of it

[earnest nymph]

how do I find it

[mild oxide]

Mark angle a as theta

Did u mark?

[earnest nymph]

yep

[mild oxide]

Now find cos of it

That is cos theya

Theta*

[earnest nymph]

like an actual value?

[mild oxide]

Hint cos theta=base/hypotenuse

[earnest nymph]

yea BC/10

duck

[mild oxide]

BC?

[earnest nymph]

qait

im an idiot

10/AC

[mild oxide]

Np

Yus

Now u know what is dot product

Tell me that theta is angle b/w which two sides

[mild oxide]

Yus

[earnest nymph]

and sub costheta in?

[mild oxide]

Yes

[earnest nymph]

so heres my q

[mild oxide]

Theta angle b/w ab and ac

[earnest nymph]

you cross out the abs ac and the denominator ac right?

[mild oxide]

Yus

Cz angle is ratio of numbers

That is cos

[earnest nymph]

is the AC denominator a scalar cause its just trig?

[mild oxide]

Scalar cz cos of any angle is a ratio

[earnest nymph]

aight

thats why its 100

thanks mate

[mild oxide]

Welcome

[earnest nymph]

.close

[mild oxide]

.close

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[earnest nymph]

lmao

[mild oxide]

Cya

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[gentle dune]

Hi, I'm trying to determine the Length/Height/Width of various surfaces in M.C. Escher's Relativity in 3d
I am using the height of a stair (highlighted in yellow) as a unit of measurement

the surfaces I'm trying to measure are highlighted in Red, Green, and Blue.
the colors are just to differentiate overlapping boxes

I think this is three point perspective with the vanishing points in a equilateral triangle.
I tried to find a perspective grid in hopes that would solve my issue, but can't seem to find one that fits.

Any advice about how to go about finding these measurements is appreciated c:
(Sorry some of the highlights are so crooked ^^;)

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[gentle dune]

Any <@&286206848099549185> available? Thanks for your time ^^

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@gentle dune Has your question been resolved?

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@gentle dune Has your question been resolved?

[sage jacinth]

Are u trying to measure surface area of highlighted parts,if yes then consider a stair as unit length and compare

Still waiting

[gentle dune]

don't need the area, just the lengths of the sides.

i have been trying to use a stair to compare

[sage jacinth]

U took 2 minutes in typing this

[gentle dune]

its 7am

[sage jacinth]

Yes 👍

[sage jacinth]

7:30 pm here

[gentle dune]

i just woke up sorry if im slow to reply

[sage jacinth]

Np

I think you got the solution

[gentle dune]

i have been having issues trying to use the stairs to compare. Im not very familiar with perspective grids

the bottom two staircases were easy, but the upper ones im unsure about

[sage jacinth]

Use pc notepad😅

U don't need to use only stairs,after calculating length of door in form of stairs,u can use door also as a unit

[gentle dune]

not sure what you mean by this part.

using the doors as a unit might help a lot

i think i am still confused by the way things shift as they move towards the vanishing point

[sage jacinth]

Draw straight lines using that so u can compare lengths easily

[sage jacinth]

I have one more trick calculate height of men moving on stairs

[sage jacinth]

Many people are there so this may easy to find other things

[gentle dune]

thank you, i will tryt this out later after work

[sage jacinth]

Ok ,if u face any problem then ,dm is open

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[jovial surge]

Pre Calc : Converges and Diverges

If r = - 1/2 in a geometric series, does that mean that this particular series Diverge?

[clear cloud]

If |r| < 1 it converges

[jovial surge]

But I thought that

• r < 1 diverges

When does it diverge?

[clear cloud]

No, if |r| > 1 it diverges

Otherwise it converges

[jovial surge]

Ohh ok so then if r = -2/3
This will converge?

And if it's diverge, then it can't be a infinite series sum right?

Can Artmetic sequences be infinite sums?

Wait what

Is there a formula for that too?

[clear cloud]

If you have a_n+1 = a_n + r the if r > 0 the sum is +inf and if r<0 the sum is -inf

[jovial surge]

What does a_n mean

[clear cloud]

The n-term of the sequence

[jovial surge]

And this is for Arthmetic sequences only?

[dawn hull]

can anyone help me with some calc homework, im tryna do with some online tutor but she takes too long cuz she got other people pls help i have the test and hw due tmrw and i will be cooked if I dont submit the hw 🙏

[clear cloud]

Open a new channel, its an occupied one

[clear cloud]

Its different for geometric indeed

[jovial surge]

Ok

[dawn hull]

ok ty

[jovial surge]

I think I understand now, and it's only the absolute value of r?

For geometric

Series

[clear cloud]

Strictly less

Than 1

Converges

Otherwise it diverges even for r = 1

[jovial surge]

Ohh ok diverge when r is 1

[clear cloud]

Rare to study it, useless type of geometric sequence but if you have to know, you know now

[jovial surge]

Ok! Thank u so much, this helps 😁👍

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[rough maple]

can someone help me

[hallow spruce]

!show

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[spare yarrow]

oh, I might have an idea

it would be simpler to have a function in terms of y and x right?

[unique minnow]

Graphically, what does it mean if dy/dx doesn't exist?

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@worthy sky Has your question been resolved?

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[bold matrix]

Hello

[bold matrix]

I’m not sure how to substitute this into b^2-4ac

Would I put brackets around the xk-3x for example? For b

[marsh forum]

,rotate

[marsh forum]

uh, if the roots are identical D=0

[bold matrix]

Oh yeah right, it said 2 identical roots that’s what mislead me thx

[trim jolt]

@bold matrix Has your question been resolved?

[bold matrix]

<@&286206848099549185>

[undone sierra]

didnt the other person answer it for u

[bold matrix]

No that wasn’t my question, just a mistake they corrected for me. My question is up above the rotate command

[undone sierra]

its abt substitution?

[bold matrix]

Yeah

[undone sierra]

ok well u have to keep the expression factorised

dont foil it

k - 3 is the whole b value

[bold matrix]

Ah right and I just do it from there?

Ok and so (2-k) is the c value

[undone sierra]

yes

[bold matrix]

K thanks!

[undone sierra]

np

[bold matrix]

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[blazing geode]

I’m trying to find the period of it

It looks like tan2x

Ahhh

Ahh

Aghhhh

Ahhh

Furry animal

I’m turning into a furry animal

Ah

[tender vigil]

wtf

[blazing geode]

Furry animal

[tender vigil]

do you know what the period of sin2x is

[blazing geode]

Ahhh

For sure

[tender vigil]

what is it

[blazing geode]

You’re doubting my ability

I’m sad

[tender vigil]

🙂

[blazing geode]

It is pi

[tender vigil]

yes

and cos2x?

[blazing geode]

1

[tender vigil]

pi

[blazing geode]

Yes

[tender vigil]

ok

[blazing geode]

Furry animal!

[tender vigil]

so then sin(2(x+pi))/(2+cos(2(x+pi))) is what

[blazing geode]

is it relevant

[tender vigil]

yes

[blazing geode]

Oh it is the same thing

[tender vigil]

yes

that means the period of f is less than or equal to pi

bc it repeats after pi

so either the period is pi/some # or it is pi

u should convince urself that it is pi

u can test like pi/2, pi/3,…

[blazing geode]

I can’t

[blazing geode]

Awww

[tender vigil]

try using desmos to graph the function

u should see that it repeats over a period of pi

[blazing geode]

I can’t

It is illegal

[tender vigil]

😦

[blazing geode]

I was thinking

If there’s some brilliant way

To solve it

[blazing geode]

So

[tender vigil]

intuitively

[blazing geode]

Can I see the period

[tender vigil]

yea ig intuitively

[blazing geode]

How

[tender vigil]

f is a composition of periodici fcts with the same period

[blazing geode]

“Intuitively”, like a taking a shower

[tender vigil]

im talking about cos2x and sin2x

my bad i type short messages

so ur period should be a multiple of the period of cos2x or sin2x

[blazing geode]

But

But

[tender vigil]

but we already showed that it repeats after pi

therefore the period is pi

[blazing geode]

It is y= sin2theta and x=cos2theta +2

The graph is being shifted left by 2 unit!

Would it changes the period

How do you know the period is still pi

[tender vigil]

no

the graph is being shifted up by 2 units

think about the graph of cosine

the wave

if u moved it up 2 units its period doesnt change

it would only change if it got stretched horizontally

[blazing geode]

Why

How do you know the graph of it

What

[tender vigil]

i just know the basic trig graphs

they just look like a wave

[blazing geode]

Can I consider it as tan

Ahhh

Ahhh

Ahh

I want a brilliant method

Approach

Like algebra stuff

Convert it into a single tri function

Ah

I want it so bad

If I were newton or gauss, I would be able to do the conversion

[tender vigil]

ur right bro

i cant help w that

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@blazing geode Has your question been resolved?

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[gleaming locust]

how do you solve it

[shrewd ridge]

is that length of AC?

[gleaming locust]

no it should be the angle

[shrewd ridge]

ah ok

[gleaming locust]

I js forgot like the formula

ur finding x

[shrewd ridge]

maybe there's more information before the picture?

[gleaming locust]

no

u know how to do this one frownyfrog lol

it has to do with major minor arcs

im pretty sure

did bro dip

@shrewd ridge helo

dont do this to me

someone help 😭

[shrewd ridge]

i'm not pretending

i don't know geometry (or calculus)

123 is next to the arc

ok i get it, there's only one distance for C, that makes 123

i don't know how to solve but you're right that's enough information

[thorn spire]

I might be able to help?

123 degrees is a specific amount of the arc

wait 123 is what degree

is it the degree of the sector from the tangeant point of a and b?

[shrewd ridge]

this should work, idk if that's the intended way

[trim jolt]

@gleaming locust Has your question been resolved?

[gleaming locust]

what is going on!

[gleaming locust]

123 is the degree for the arc

between a and b

[gleaming locust]

what is this

[shrewd ridge]

it's my solution

[gleaming locust]

what did u get for x

[shrewd ridge]

28.5°

[gleaming locust]

it's wrong

[shrewd ridge]

no 24.5

[gleaming locust]

oh

wow u got it right

thank you frownyfrog

worth the 35 minutes this time 😄

.close

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[gleaming locust]

.reopen

[trim jolt]

✅

[gleaming locust]

@shrewd ridge hey can u tell me how u did it

cuz idk what u did in the image

[shrewd ridge]

it's just cos and sin

[gleaming locust]

ik but like

[shrewd ridge]

61.5 is the angle in the triangle near o
cos 61.5 is r / oC

which means oC is r / cos 61.5

it's just asking the calculator to do it for you

[gleaming locust]

ohhh I think I get it now

ima try

[shrewd ridge]

#help-35 is empty

[gleaming locust]

frog

can u help me

or busy

[shrewd ridge]

i don't like geometry sorry

[gleaming locust]

allg

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[marble wharf]

<@&268886789983436800>

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[graceful zenith]

Need help with this

,rccw

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[mild oxide]

Average form of Newton's cooling law

Try this

@graceful zenith

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[kind pecan]

I'm stuck on these 2 questions. Can someone help pls? I need to find the TSA

[trim jolt]

Please don't occupy multiple help channels.

[kind pecan]

<@&286206848099549185>

[cosmic lance]

Just a min

[kind pecan]

Ok

[cosmic lance]

Shape c without cut out = 2pi *4 * 10

80pi sq m

[kind pecan]

Thx but how is it 10?

[cosmic lance]

6+4

[kind pecan]

Ok

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[pastel pecan]

my course is telling me that 4+5(5+9)/2^2+3=74/7 but i dont get why. please help

[trail ingot]

.-.

me neither

[marble wharf]

well presumably its supposed to be (4+5(5+9))/(2^2+3)

[trail ingot]

ah

[marble wharf]

=(4+5*14)/(4+3)=(4+70)/7=74/7

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[crimson pine]

how would you do this?

[dire radish]

Pr means probability? Right

[trim jolt]

@crimson pine Has your question been resolved?

[crimson pine]

yeah

[cerulean spruce]

Might just have to do guess and check

[crimson pine]

well no

not guess and check

whatever

.close

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[flint basin]

is it fine if i ask a physics question?

[split chasm]

probably, whether people will be able to assist may depend on the amount of physics vs math knowledge required

[flint basin]

How to find halflife of this sample from the graph?

[hasty yew]

Exponential decay

[split chasm]

the worked solution seems to be there, which part do you have an issue with

[flint basin]

Do u have to add the background radiation back again after finiding the half life?

[split chasm]

not quite, half life is the end result

the left part. identifies how many ticks the radioactive material contributes (38/min)

when half of the substance decays, the radioactive material would only contribute 19/min

that is added to the background amount of 14

so that after half decays you'd expect to have (19+14) = 33counts/min

and then you read the value there on the graph

[flint basin]

background radiation?

[split chasm]

no

look on the graph where the y-coord is 33/min

[flint basin]

who do u add background radiation which is 14 later on?

[split chasm]

because its always there

that part doesn't decay

[flint basin]

so that is the halflife of the isotope?

[split chasm]

wdym by "that"

[flint basin]

1.5

[split chasm]

yes. 1.5 days is the half life