help-33

[marsh citrus]

@austere bramble Has your question been resolved?

[twilit arrow]

maybe check out some of the programming/computer science servers in #old-network instead

[marsh citrus]

@austere bramble Has your question been resolved?

[main idol]

or find a programming language specific server

@austere bramble

[marsh citrus]

@austere bramble Has your question been resolved?

[marsh citrus]

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[torpid wing]

why did we integrate once for each power of 1/s ?

[marsh citrus]

@torpid wing Has your question been resolved?

[twin cape]

what is L and s?

[torpid wing]

Bruh

It's inverse laplace

[twin cape]

ok, like I'm not used to this notation, you don't need to be an asshole

[torpid wing]

Uh I wasn't

[marsh citrus]

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[tropic prawn]

what does k represent in y = mx + k

[novel juniper]

The y-intercept

[tropic prawn]

.close

[marsh citrus]

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[stiff maple]

how do you get from 4 to 5. my friend told me thats what your supposed to do but i dont get the steps behind it. -

[late geode]

,rotate

[late geode]

multiplied both sides to the equation by x^2

[stiff maple]

where did the 1 go though

and how is the 0 still 0

if its by both sides

[late geode]

what happens when you multiply 1 by x^2

[stoic slate]

If you multiply by 0 you get 0, that's why it's still 0

[marsh citrus]

@stiff maple Has your question been resolved?

[marsh citrus]

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[green oriole]

need help with this

[marsh citrus]

@green oriole Has your question been resolved?

[green oriole]

<@&286206848099549185>

[green oriole]

.close

[marsh citrus]

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[urban bobcat]

does anyone know how to solve this? "deside X from the matrix if A = ... "

[urban bobcat]

i have started solving the equation and have gotten to (A^T -7I) (X-B^T) = 0 but how do i solve out X from here?

[marsh citrus]

@urban bobcat Has your question been resolved?

[urban bobcat]

<@&286206848099549185>

[marsh citrus]

@urban bobcat Has your question been resolved?

[crystal wraith]

Oh I thought you closed it?

Wait you opened a new one with the exact same problem so it moves up lol.

[crystal wraith]

Well do you think A^T-7I is singular?

[urban bobcat]

what do you mean?

[crystal wraith]

Do you think A^T-7I is invertible?

[urban bobcat]

ohh hmm well maybe not

[crystal wraith]

Maybe

There's a way to know for sure.

[urban bobcat]

oh okay

[crystal wraith]

Like

You know A matrix already...

[urban bobcat]

yes

[crystal wraith]

So the transpose matrix too...

[urban bobcat]

so can i just move over A^T -7I?

[crystal wraith]

If you know it has an inverse, you can multiply by its inverse.

I'm not sure what you mean by "move over"

[urban bobcat]

okay well how do i know if it has an inverse?

[crystal wraith]

I thought you knew.

But alright

A square matrix M is invertible if it's NOT singular, and by that I mean det(M) is not 0.

Because, M^{-1} = adj(M)/|M|

So |M| shouldn't be zero.

[urban bobcat]

okay

[crystal wraith]

Once you multiply a matrix by its inverse you have Identity matrix.

And that (identity matrix) on multiplication with any other matrix is that other matrix itself.

[urban bobcat]

okay it is inversable so can i multiply both sides with it then?

[crystal wraith]

What do you get when you multiply by the inverse?

[urban bobcat]

no wait

doesnt it just become X =B^T?

[crystal wraith]

Precisely.

[urban bobcat]

okay i thin i understand then

so now i just pu in the value on B^T?

[crystal wraith]

Yep.

[urban bobcat]

okay so the answer is X = (5 -1, 8 3)?

[crystal wraith]

yes.

[urban bobcat]

okay yay! do you have time for another quick question?

[crystal wraith]

Maybe not.

[urban bobcat]

okay thats fine thank you anyways!

[frigid rampart]

You can ask me if you want, i might be able to answer

Just ping me though

[urban bobcat]

okay thank you!

this is the question it says "let X be an unknown matrix. deside X if we know that ..."

is it always easier to solve the equation first and then put in the values? @frigid rampart

[frigid rampart]

Well yeah, i always solve and simplify the given equation is variable form and then substitute numerical values at the end

[urban bobcat]

okay so my first thought is to move over XC so i can factore out X?

[frigid rampart]

Im not really sure you can do that with matrices

[urban bobcat]

oh okay

[frigid rampart]

But hold on lemme check

[urban bobcat]

okay!

[wind whale]

yes you can

[urban bobcat]

okay great!

[wind whale]

you just don't have to mix this up with your usual algebra

the order matters and you have to keep it that way

AB is not the same as BA for example

[frigid rampart]

I think you cannot apply commutative property here

Rest of the concepts are valid

[wind whale]

that's what I said

[urban bobcat]

yes okay!

so we have X(A-C) = B?

[wind whale]

yes now it's the previous problem all over again

you'll want to see if A-C has an inverse

[urban bobcat]

yes right!

[wind whale]

if it does you can multiply by that (post multiply)

you'll get A directly from there

[urban bobcat]

okay!

i tried checking and i am pretty sure it has an inverse becuse i did not get 0

so i multiply both sides with (A-C)^-1?

[frigid rampart]

Yes

[urban bobcat]

okay but does that mean that i get X = B(A-C)^-1?

[wind whale]

yes

[frigid rampart]

Ginny

Im confused over this now

[urban bobcat]

okay! so do i put in values now or can i simplify more?

[wind whale]

you can't simplify it any more, it's about time you use the values

[urban bobcat]

okay!

[frigid rampart]

How are you able to check if (A-C) is invertible or not?, cuz we need to check the det value and that is something which only exists for square matrices

So whats going on here

[wind whale]

ACTUALLY

you are right, i saw the A and B values of the previous problem...

while the equation for this one

[urban bobcat]

hmm okay so i can´t check if it is invertable if it is a 3x3?

[wind whale]

it has an inverse if it's 3 * 3

but that's exactly why you can't see it here because the matrices aren't 3*3

they aren't square matrix

[urban bobcat]

ohh yea right

[frigid rampart]

There isnt a way to know for sure if you can invert it or not

[urban bobcat]

so i can multiply both sides with it since i cant check if it is inverable?

[frigid rampart]

So i guess ur gonna have to assume it is

Invertible*

[urban bobcat]

oh okay, is there no other way to solve out X without using the inverse?

[frigid rampart]

Not really

Thats the best method i can see

[urban bobcat]

okay

[frigid rampart]

Assume it is invertible

And try and check the answer

I think that if it wasnt invertible, X could be any matrix

It would be like multiplying a number to 0 but in this case you are not sure if it is 0

[urban bobcat]

ohh yea okay

but how do i calulate the inverse of A-C?

[frigid rampart]

Umm

I dont really remember how to do it

[urban bobcat]

when i tried looking it up it said it does not have an inverse

[frigid rampart]

I guess that is because we do not have a well defined determinant value or the matrix

So inverse for these matrices arent defined as well

So X could be any matrix

[urban bobcat]

hmm okay weird

[frigid rampart]

You should ask your teacher at your institution

Maybe the question is wrong

Im not really sure about this

[urban bobcat]

it also says in the question that gauss elimination and transposing is my friend

[frigid rampart]

Hmm i dont know what gauss elimination is

I guess i cant help you any further

[urban bobcat]

yes, even if i use tranposing i still have to find the inverse so i will have to ask my teacher

but thank you for the help!

[frigid rampart]

Okay

Could you like dm me the solution when you find out

Im actually curious to find out

[urban bobcat]

yes sure haha!

[frigid rampart]

Okay thanks ill be waiting

[urban bobcat]

it will prob be on monday tho!

[frigid rampart]

No problem

Add me in your friends list

If thats okay

[urban bobcat]

yes of course!

[urban bobcat]

.close

[marsh citrus]

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[fluid sparrow]

If I have an equation system with two planes in 3D space, what type of objects would the solutions be?
A line if they intersect and a plane if they’re ”the same”? Anything else?

[main idol]

https://www.mathwarehouse.com/algebra/planes/systems/three-variable-equations.php

and also
https://i.stack.imgur.com/XDenC.jpg

https://math.stackexchange.com/questions/3886213/arrangements-of-solutions-to-n-linear-equations-with-n-unknowns

[marsh citrus]

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[marsh citrus]

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[main idol]

!msgdel

[marsh citrus]

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[marsh citrus]

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[tepid holly]

hello

[tepid holly]

can someone explain me this how do we go from up to down ?

[boreal rose]

$(x+\frac{p}{2})^2 + (q-\frac{p^2}{4}) = x^2 + px + \frac{p^2}{4} + q - \frac{p^2}{4}$

[tepid holly]

yeah but what we did to x²+px+q

to get (x+p/2)²+(q-p²/4)

[boreal rose]

it's a technique called complete the square or sm

[tepid holly]

show the technique

[elfin berry]

nosqldb

[boreal rose]

@tepid holly this is an example of the technique

cuz we get back x^2 + px + q

[tepid holly]

yeah but u go from down to up

not from up to down

how i go from up to down

[boreal rose]

okay

observe when you square

(a+b)^2 = a^2 + 2ab + b^2

so we have a simliar form with x^2 + px + q

a = x, and for 2ab to be equal to px

b = p/2

now we are unable to get the q term, but that's okay

we can just add it

we get an extra term also p^2/4 which we can subtract

[tepid holly]

i still dont understand

[boreal rose]

what do you not understand

[tepid holly]

how you go from up to down

[boreal rose]

well we make an observation first

that we have a simliar form to (x+a)^2 = x^2 + 2ax + a^2

[tepid holly]

Ok

a=p/2 and what

[boreal rose]

now

we notice that we have an extra a^2 term

p^2/4

[marsh citrus]

@tepid holly Has your question been resolved?

[marsh citrus]

@tepid holly Has your question been resolved?

[marsh citrus]

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[torpid owl]

Confused on #12.

[torpid owl]

Number #10 should be 57.81 ft.

Number #11 should be 201.97ft.

When solving for #12, I got the ft. of 412.67

[cunning fiber]

the ball goes higher than the fence so it clears it

assuming those are right ofc

[torpid owl]

But the fence is 8 feet high.

Does that matter in regards to the problem?

[cunning fiber]

oh wait I can't read

it's 400 ft away and 8 ft high

[torpid owl]

yeah

[cunning fiber]

wait you got 412.67 ft?

[torpid owl]

Yes.

[cunning fiber]

and that's at 400 ft away right?

I mean it's the same logic

ball goes higher than fence

[torpid owl]

I believe so.

So does the 8ft matter?

[cunning fiber]

I mean that's the height the ball needs to be higher than

[torpid owl]

and it passes that when it travelled 400ft?

[cunning fiber]

assuming your calculations are correct, yes.

[torpid owl]

That was the worse typo I've ever seen for "correct".

Thanks for the help as always.

.close

[marsh citrus]

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[cunning fiber]

keyboard might have had a spasm or two lol

[marsh citrus]

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[inland lichen]

I was working on this problem and got this but can't seem to understand the alternate solution that my prof showed in his answers.

[marsh citrus]

@inland lichen Has your question been resolved?

[marsh citrus]

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[eternal flint]

square root of A=?

[exotic salmon]

What is $d|2^{2024}$

[elfin berry]

Solomaniac

[exotic salmon]

@eternal flint

[marsh citrus]

@eternal flint Has your question been resolved?

[desert dirge]

i presume its all the positive integers that divide 2^2024

[verbal peak]

Hint: $A=\sum_{d|2^{2024}}(\frac{1}{d+2^{2012}}+\frac{d}{2^{2024}+2^{1012}d})$

[elfin berry]

Caroline

[marsh citrus]

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[eternal flint]

.reopen

[marsh citrus]

✅

[eternal flint]

I dont get these

my math level is lower

[main idol]

Get what exactly

[eternal flint]

first of all, What does mean that sigma(?) symbol

[still temple]

lmao

[eternal flint]

I know it like pluses

[nocturne fiber]

sum across each divisor of 2^(2024)

[still temple]

its sum sign

[eternal flint]

all divisor

[eternal flint]

okey

so is it 2+2+2+2+... for 2024 times?

[main idol]

What is "it"

[eternal flint]

This one

Idk where to start or what is progress in "sum"

[verbal peak]

No

It's sum over divisors of fractions that are given

[eternal flint]

Can u give an example?

I do not get it clearly

[verbal peak]

Calculate this for example: $\sum_{d|15}{d}=?$

[elfin berry]

Caroline

[eternal flint]

1/3 + 1/5= 8/15 ?

[main idol]

Forgot 1

[eternal flint]

where should I put 1

[main idol]

Do you know what divides means

[verbal peak]

You shouldn't divide anything by 15 now

[verbal peak]

This sum is just sum of divisors of 15

[eternal flint]

yes

[main idol]

Write all the divisor of 15

[eternal flint]

at least I guess, english is not my first lang

1 3 5 15

than is it 1/1 + 1/3 + 1/5 + 1/15 ?

ow

1+3+5+15?

[main idol]

Yes

[eternal flint]

what are the pros if I use this

[eternal flint]

and It is really more complex

[eternal flint]

gotcha

[verbal peak]

Now you understand notation in the task

[verbal peak]

Here's a hint i gave you, do you have any ideas how to get to it?

[eternal flint]

I will try to find d firstly

than I will write it down (the numberal value)

and multiply in parentheses?

[eternal flint]

should I all write these one by one?

[verbal peak]

Okay, here's anoher hint - if d is divisor of n, then n/d is divisor of n as well

Do you see why?

[eternal flint]

It looks logical

I got it

but this is too high

2^2024 + 2^2023 + 2^2022 + 2^2021..... 2^1 + 2^0

how can I get these sum

[eternal flint]

like in 15, 5 is a divisor and 15/5 is 3 , 15/3 is 5 and It must be divisor

heey

[verbal peak]

Yes

[verbal peak]

Not the sum you need to find

[eternal flint]

what I need to find

[verbal peak]

Like you said

[verbal peak]

And that's a hint I want you to get to

[eternal flint]

square root of (sum 2^2024 and something to add)

or square root for 2^2024?

[verbal peak]

do you understand where did this come from?

[eternal flint]

multiply by 2 is gone, but why? where did it go and how u add that to there?

[eternal flint]

no

[verbal peak]

Coming back to this

If you replace divisor "d" in every fraction with "2^2024/d", sum doesn't change

[verbal peak]

Because of this fact

[eternal flint]

hmm

[verbal peak]

Try splitting A into two sums

First sum stays the same, in another - replace d with 2^2024/d

[eternal flint]

where is another sum

is it fractions?

[verbal peak]

$A=\sum_{d|2^{2024}}\frac{1}{d+2^{2012}}+\sum_{d|2^{2024}}\frac{1}{d+2^{2012}}$

[elfin berry]

Caroline

[verbal peak]

Agree?

[eternal flint]

wait for a sec

[verbal peak]

[eternal flint]

2^2024/d divided by 2^2024+ 20^2012 multiply(?) 2^2024/d

am I writing in true one

[verbal peak]

Can you type in latex? It's hard to read this

(Or paper)

[eternal flint]

how to type in latex

[eternal flint]

my device doesnt have camera

it is computer

[verbal peak]

https://latexeditor.lagrida.com/ use this site for example

[eternal flint]

okey

that was really hard

[verbal peak]

It's not really true

[eternal flint]

AAAAAHHH

let me learn the truth

how u make it same as first fraction

[verbal peak]

So you agree with this

[eternal flint]

no I dont get how u find that

I put 2^2024/d in all "d"

[eternal flint]

and I found this

[verbal peak]

I just splitted 2*S into S + S

[eternal flint]

2S is S+S

[verbal peak]

Exactly

[verbal peak]

That's why it's true

[eternal flint]

ow

okey

but I have one more question

[eternal flint]

we find it because first sum was multiplyed by 2

[verbal peak]

Yes

[eternal flint]

but how we found that

and I did not replace anything

d to 2^2024/d

[verbal peak]

replace d with 2^2024/d in second sum

[eternal flint]

okey

[eternal flint]

I can replace all d with that

there is no finishes

[verbal peak]

Yes

I said why you can do this earlier

[eternal flint]

why was that

[eternal flint]

is it true?

and what should ı do next

[verbal peak]

Convert it into single sum

[verbal peak]

Like this

[eternal flint]

square root of A is the just first sum?

and fraction

[eternal flint]

square of first sumfraction

[verbal peak]

No, you have to evaluate it first

[eternal flint]

like what?

what do you mean by evaluate

haaaa

Is it like

no

it isnt

[verbal peak]

Find numerical value

[eternal flint]

how?

it is equals to what

how can I solva equation

[verbal peak]

Turn sum of two fraction into one fraction

[eternal flint]

fraction is 2/ d+2^2012 ?

just fraction

before there is the sum

[verbal peak]

No

[eternal flint]

why

how it will be

[verbal peak]

$\frac{1}{d+2^{1012}}+\frac{d}{2^{2024}+2^{1012}d} = ?$

[elfin berry]

Caroline

[eternal flint]

we recome the sameplace

sorry

I couldnt learn this right now

thanks a lot

and I need to go right now

If it is fine for you, I want to talk about this tomorrow

If not, it is fine for me, thanks a lot

[verbal peak]

Sure, no problem

[eternal flint]

see ya

[marsh citrus]

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[onyx marten]

Stuck in this integer

[onyx marten]

Dont know even if im going in the right path

[nocturne glen]

continue simplifying first

[onyx marten]

common factor??

[nocturne glen]

you have a fraction over a fraction

[onyx marten]

then cancel 1+cos

[nocturne glen]

first get it to a single fraction

[onyx marten]

alr

Whats the best way tu simplify that further

[iron meadow]

i think both the num and denominator have a factor of (1+cos u)

[onyx marten]

mmm numerator doesnt seem to have that factor

[iron meadow]

sin^2(cos u + 1)

factor out the sin^2

[onyx marten]

aaaa

right

didnt see that

[nocturne glen]

yeah it comes out very nicely

and then you just usub

[onyx marten]

got it

thanksss

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[main idol]

<@&268886789983436800> spam

[marsh citrus]

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[idle solar]

Anyone know where the -1 came from?

Shouldn't it be (sinx+cosx)(sin + cos) +1

[spark otter]

a^2 - b^2 =

[still temple]

At step 2 on the numerator we added 1 and thus had to substract 1 as well

[spark otter]

not the place OP's pointing at (look at blue arrow)

[idle solar]

Yes the 1 was turned to sin^2 + cos^2

[idle solar]

Can you elaborate?

I dont think i know that

[still temple]

My bad

[spark otter]

sin(x) + cos(x)= a

b = 1

[spark otter]

the numerator BEFORE the change is a^2 - b^2

[idle solar]

Oooh

Got it

X^2 - y^2 = (x - y)(x + y)

Right?

[spark otter]

yes

[idle solar]

Awesome

Ty guys

[idle solar]

All good

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[heady elbow]

Guys can someone help me with this? I already answered some of the questions but I’m not sure so can you check for me? And I don’t know how to find the measure of the diagonal (OP) and the leg (SP) so can someone help me with these?
The first photo is the whole question/problem
The second photo is my answer for questions 1-3.
The third photo is my solving for x and the bases (SO) and (AP)
The fourth photo shows my answers to angle questions 6-7.

[marsh citrus]

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[timid drum]

How do i mathematicly prove that he is right :

Kittens
Mr. Felix: My love, how many kittens are in our new litter?
Mrs. Felix: Can't you count? Four, my sweetheart! Mr. Felix: How many are boys?
Mrs. Felix: It's hard to say. I don't know yet.
Mr. Felix: It's not too difficult to understand; for each kitten, there's a fifty-fifty chance it's a girl. So, of course, it's most likely two boys and two girls!
Is Mr. Felix right? Why or why not?

[zinc sparrow]

each kittens gender is independent from the others

so even tho it’s 50 50 per kitten it doesn’t mean a 50 50 of the kittens

so for all boys it would be like (1/4)^2

and once you’ve done all 4 arrangements you add them up

[static quarry]

two boys and two girls is the most likely outcome, but far from guaranteed

[golden willow]

the chance of one kitten being a boy or girl is 1/2
the chance of another kitten being that same gender is also 1/2,
1/2 * 1/2 is 1/4
the chance of another kitten being the opposite gender of them two is 1/2
the chance of another kitten also being opposite gender as the first two is 1/2
again, 1/2 * 1/2 is 1/4

Now, the combining those
1/4 * 1/4 is 1/16

[marsh citrus]

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[glad lotus]

Any help with this one, I've been trying but I haven't got anywhere

[calm harbor]

[marsh citrus]

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@glad lotus Has your question been resolved?

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[low solar]

how did 6615 and 6000 become 21/20

[fathom ridge]

,w gcd(6615, 6000)

[elfin berry]

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[fathom ridge]

divide by 15

[low solar]

oh

simplified

my bad

im too dumb

[fathom ridge]

yeah

and square root I assume

[low solar]

wdym square root

oh

yea

mb

[low solar]

for this i will transfer the 100 that side

so 21/20 x 100 = 1+r

right?

[fathom ridge]

yes

[marsh citrus]

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[eternal flint]

closest Z to f(5)

[marsh citrus]

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[errant tulip]

can anyone help me with linear algebra?

[floral linden]

for part a, put the x and y values from the equation of line m into the equation of the plane, and solve for s

[errant tulip]

i did the part a im struggling with b and c

sorry forgot to mention

s=2sqrt3

and then the point is x=3sqrt3 y=sqrt3 z=-2sqrt3

[floral linden]

for part b, find the vector LP where P is the point of intersection and L is a point on the line. then, the dot product of vector LP and the direction vector of L = 0, so you can solve for the parameter and find the magnitude of LP

[errant tulip]

do i just take any L

[floral linden]

L is a point on the line L so you take the vector eqn of line L

[errant tulip]

so just (1,1,0) right

[floral linden]

no, (t, t, 0) since x and y can be anything, not just 1

t is some parameter, similar to how s is a parameter for line M

[errant tulip]

do i not need to find a unit vector of l?

[floral linden]

nah

essentially, you're finding a vector that "joins" the intersection point and the line

and trying to find for which value of t is this perpendicular to the line

because if you think about it then the point of shortest distance would be perpendicular to the line

[errant tulip]

so i just find dot product of (t,t,0) and (3sqrt3 ,sqrt3,-2sqrt3)

and just equate it to 0?

or am i tripping

wait no im so stupid

nvm

.close

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[fickle shell]

https://i.imgur.com/5ioRkr1.png

[fickle shell]

u = 2x-3

x = (u + 3)/2

(u + 3)(u)^5/2

(u^6)(3u^5)/2

(u^6)/2 + (3u^5)/2

Integrates to:

(u^5)/10 + (3u^4)/8

Then sub in 2x-3 for u

That was my answer

What did I do wrong

[marsh citrus]

@fickle shell Has your question been resolved?

[low ruin]

You need calculate du/dx.

So du/dx = something

Therefore du/something = dx

Once you get rid of dx should be all good

[marsh citrus]

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[still temple]

find the limit without using L'Hopital's rule:
lim x>0 (1+arctg4x)^1/x

[novel juniper]

mind typing that in TeX?

[sinful glen]

it's two ways, one is sorta shortcut other is a standard way

although the shortcut is derived from the standard

take the log

[elfin berry]

NEON

[fathom ridge]

you can just approximate arctan 4x as 4x and use the limit definition of e?

[sinful glen]

it'll go 1 raised to infinity form

[fathom ridge]

yeah thats why you use the limit definition of e

[still temple]

yeah i knew the answer has to be e i just don't know how to change arctg 4x into just 4x

[sinful glen]

you can subtract 1 from inside the power, evaluate the limit and put the result to power of e

[novel juniper]

alternatively use the fact that $\lim_{x\rightarrow0}\left(\frac{\arctan\left(ax\right)}{ax}\right)=1$

[elfin berry]

Why am. I here

[fathom ridge]

for x close to 0 arctan x = x

¯_(ツ)_/¯

same for sin and tan

[still temple]

i see

[elfin berry]

NEON

[fathom ridge]

so it should just be e^4 from here?

[sinful glen]

yeah

[still temple]

ohhh i see it now

thank you

[fathom ridge]

,w limit as x goes to 0 (1 + arctan(4x))^(1/x)

[elfin berry]

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[fathom ridge]

yup we gucci

[still temple]

thanks a lot

.close

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[celest tendon]

how do i do a pythagoras theorem on a trapezoid

[devout mauve]

wdym

[celest tendon]

pythagoras theorem

on a trapezoid

[exotic salmon]

wdym ?

[celest tendon]

uhh

is there no pythagoras theorem on trapezoids?

[devout mauve]

in what sense do you want a pythagoras theorem

can you make a sketch?

[celest tendon]

uhh

im also comfused hold up

something like this

[exotic salmon]

Why do you want to apply pythagoras to this ?

[celest tendon]

i dont know i have a math test tommorow and apperantly we are gonna have to do pythagoras theorem on trapezoids

so thats why im asking

the book and stuff we wrote down barely helps me

oh nevermind i found out how

[still temple]

how?

[celest tendon]

google

this is correct right?

[devout mauve]

well this is still just "normal" pythagoras inside a right triangle

but yes

[still temple]

bruh so
💀 nvm
yes

[celest tendon]

u can close the channel now idk how to do it

[devout mauve]

write .close

[celest tendon]

.close

[marsh citrus]

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[gloomy ridge]

Can someone check my work? im really not confident about this. about vector spaces maps and basis.

[marsh citrus]

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[spring stirrup]

how would i go about solving this question

[spring stirrup]

this is my teachers but i dont understand

Why is he solving for t

[sinful glen]

he's solving for t to find the time when the tank is 60% full

[spring stirrup]

Ohh

Ok

But here he uses the chain rule right?

But the question asks to use product rule

[sinful glen]

they're the same thing as far as names are concerned

some textbooks call it an onion

[spring stirrup]

ohh ok

Tysm

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[willow salmon]

I understand the 2T for part c

but idk why is it cos theta

[still temple]

the tangential component of the tension would not apply force on the pulley

its like airplane flew just above your head and you didnt feel anything 💀

[marsh citrus]

@willow salmon Has your question been resolved?

[willow salmon]

So it's not 2T cos theta?

<@&286206848099549185>

[still temple]

so it would be 2tcostheta

sintheta would be tangential

[marsh citrus]

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[willow salmon]

Ohhh

Thank you

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[still temple]

I've dermined f'(x) to be 2x

using variable "a" for x

so our corresponding point should be (a, a^2)

our slope I've determined to be 2(1)=2

so then the formula I'm using is

$2a=\frac{a^2+3}{a-1}$

[elfin berry]

wyldinwilliam

[still temple]

$2a^2-2a=a^2+3$

[elfin berry]

wyldinwilliam

[still temple]

$a^2-2a-3=0$

[elfin berry]

wyldinwilliam

[still temple]

$(a-3)(a+1)=0$

[elfin berry]

wyldinwilliam

[still temple]

$$a=3$$
$$a=-1$$

[elfin berry]

wyldinwilliam

[still temple]

substituting x back in we have

two x values

$$x=3$$
$$x=-1$$

[elfin berry]

wyldinwilliam

[still temple]

now, we need an eqn for both lines

$9=2(3)+b$

[elfin berry]

wyldinwilliam

[still temple]

$b=3$

[elfin berry]

wyldinwilliam

[still temple]

$y=2x+3$

[elfin berry]

wyldinwilliam

[still temple]

is the first equation

but this is wrong apparently

so I need some clarification

.close

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[pastel pollen]

i need help with this question

[whole sleet]

What are your "Axioms of Order"?

Teachers really can't just refer to these with normal names, huh?

[pastel pollen]

😭

[marsh citrus]

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@pastel pollen Has your question been resolved?

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[dense meadow]

It takes person 1 X hours to travel A kilometres. If person 1 increases their speed by b kilometres per hour, the journey will take her C hours less time. Find X.

How do i solve this?

[stable kite]

send a pic of the problem

[exotic salmon]

Dude, it's basic unitary method.

$X - \frac{A}{\frac{A}{X}+b} = C$

[elfin berry]

Solomaniac

[exotic salmon]

@dense meadow

[dense meadow]

oh

thanks

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[smoky marsh]

gimme a sec

ok so im getting this as my answer but the textbook says this

[marsh citrus]

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[plush glen]

can someone help me draw a system of unchanging rings with more than 4 circles.
A system of unchanging rings must satisfy three unchanging rules:

If two circles meet at a marked point, they share exactly two marked points.
If two marked points lie on a common circle, they must share exactly two common circles.
We can get from any circle to any other by some number of hops along marked points.

[marsh citrus]

@plush glen Has your question been resolved?

[plush glen]

<@&286206848099549185>

[plush glen]

<@&286206848099549185>

please

goddammit

[neat nebula]

is this the system that you want ?

or does this drawing describe what you want?

[plush glen]

i need something with 5 rings tho

i made this but idk if this is correct

[neat nebula]

I think they need to meet in the centre

[plush glen]

can you help?

[neat nebula]

I will try, I have no experience in regards to circles but I will do my best

[plush glen]

thanks

[neat nebula]

So I would start by just making a circle to base the rest of them on it

So we have the equation (x-h)^2+(y-k)^2=r^2

[plush glen]

Ok

[neat nebula]

for the middle circle which it's middle lies on the point (h,k)

btw should the system be generalised?

[plush glen]

Yeah

[neat nebula]

ok so we need then 4 circles to have the same radius as the middle circle but their midpoint on the outer edge of the middle circle right ?

[plush glen]

And this too

[plush glen]

Yeah

[neat nebula]

hmmmm so which shape are we trying to describe ?

[plush glen]

Thats where im confused

[neat nebula]

oh ok so just have to draw it

[plush glen]

Guess so

[neat nebula]

right so I came up with something that I don;t know if it counts but let me show you

[plush glen]

Thanks

I appreciate it

Lemme take a look

[neat nebula]

cool

[plush glen]

It says if the circle meets they must share two points

Ut some circle only share one

[neat nebula]

hmmm yes the outer ones

[plush glen]

I think this might be it

[neat nebula]

yes I think this is correct

[plush glen]

ok

thanks anyways

[neat nebula]

sorry for not being helpful I am not very proficient on circle geometry

[plush glen]

dont worry

actually

i have one more question if you dont mind

[neat nebula]

oh yes

please, is it also geometry ?

[plush glen]

kinda?

[neat nebula]

ok send it

[plush glen]

im stuck here

thanks man, i really do

[neat nebula]

So in picture 1c we need 15*(2)^(1/2)

is that correct ?

[plush glen]

as in thread length?

[neat nebula]

yes

[neat nebula]

if this is true I figured it out

[plush glen]

show me

im a really confused on this one

[neat nebula]

so we have our width 2n-1 which is the "base of our shape"

[plush glen]

ok

[neat nebula]

for every block of our width we need a minimum of 2^(1/2) string

so to get the minimum amount of string for the base we have the width times 2^(1/2) so for the base we have (2n-1)*2^(1/2)

to that we add n numbers of teeth which are two blocks which is 2*2^(1/2) pre n

thus we have (2n-1)*2^(1/2) + 2**2^(1/2)n

[plush glen]

oh wow

[neat nebula]

in written form

[plush glen]

wiat

sorry to bust your bubble but

in a i had to proof that to stitch every square i would need at least mn(sqrt2+1)-1

[plush glen]

and it contradicts this

i think

[neat nebula]

yes but we can fix it easily

[plush glen]

how so?

[neat nebula]

so the minimum ammount of string needed per little square is mn(2^(1/2)+1)-1 so for th little square we have m = 1 ,n = 1 so we get 1( 2^(1/2)+1)-1 = 2^(1/2) ????

[plush glen]

oh right

mb

sorry

[neat nebula]

for what ?

[plush glen]

silly mistake

its 3 am here lol

[neat nebula]

Oh ok, I am in europe so it is like 6.30 p.m here

[plush glen]

welp thanks man, appreciate everything

i really mean it

.close

[marsh citrus]

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[marsh citrus]

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[bronze mason]

Can someone help me with this

[bronze mason]

I want to say that dim(kerf in kerg) = dim(kerf) + dim(kerg) - dim(kerf + kerg)

I open the mathstack but I still struck with why dim(kerf) and dim(kerg) = n-1

[devout mauve]

they are maps V->K. linearly independent in particular means nonzero. so image has at least one dimension. then rank nullity

[marsh citrus]

@bronze mason Has your question been resolved?

[marsh citrus]

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[marsh citrus]

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[nova totem]

Dont even know where to start

[nova totem]

i just kinda guessed on why 13000 maxmizes expected profit

[runic temple]

multiply (expected profit)(probability of it happening)

[nova totem]

but idk how they got to that conclusion

[nova totem]

is that where the 13000 comes from?

hold on

[runic temple]

(16000-13000)(0.6)

[nova totem]

where did the 0.6 come from

[runic temple]

probability of winning the bid

[nova totem]

wait

$$f\left(x\right)=\left(16000-x\right)\left(x-10000\right)\left(\frac{1}{15000-10000}\right)$$

[elfin berry]

Umbraleviathan

[runic temple]

yes, its expected value

[nova totem]

i hate stats

thanks

[runic temple]

yw

[nova totem]

gonan keep this open a bit

[runic temple]

.coose

[nova totem]

.clasdsaois

.cksow

asc.slcow

cac's.clca

asc.clcssoe

.clsoe

.clsoe

.close

[marsh citrus]

Channel closed

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[runic temple]

when umbra said "cac's.clca", that was so powerful ✊ 😔

[marsh citrus]

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[native stream]

It says it’s incorrect. I followed the formula of the chain rule?

[hardy slate]

just a small mistake at the end

it should be (14xe^x - 4y^3)(-sin(t)) + (7x^2 e^y - 12xy^2)cos(t)

so you'll need to expand that

and then also actually you probably want it all in terms of t

so sub out x and y

[native stream]

so for y, do i replace it with sin(t) and x for cos(t)?

also is it 14xe^x or 14xe^y?

[marsh citrus]

@native stream Has your question been resolved?

[native stream]

.close

[marsh citrus]

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[marsh citrus]

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[hasty talon]

i have a question

[hasty talon]

<@&286206848099549185>

i dont see the mistake

[desert dirge]

!15m

[marsh citrus]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[desert dirge]

tut

[craggy stump]

maybe use < instead of <=

[hasty talon]

i'll try, buai dont think so

cuz it is also =0 and =1

[craggy stump]

i cant remember exactly what happens, but isnt there a way to determine if the signs are < or <=

[hasty talon]

ik its <= tho

[hasty talon]

.

[hasty talon]

.

it has to be something else

hey

Domain is X axis, right ?

[craggy stump]

domain is the x axis yes

range is the y axis

[hasty talon]

im out of ideas then

[craggy stump]

try using <

[hasty talon]

if i submit it, thats it

so i have to be 100% sure

[craggy stump]

oh mb

[hasty talon]

no, its alright

[craggy stump]

i mean it seems like it is <=

[hasty talon]

yea

[craggy stump]

i remember < to have some sort of symbol on the line

[hasty talon]

and even if i click on the graph, it shows it

[craggy stump]

gimme a sec

ill look it up

[hasty talon]

take your time

thank you

[craggy stump]

yeah no i think it is <

cuz i graphed it on desmos

wait nvm

its actually weird

[hasty talon]

i dont think the website will take any answer

i changed the answer for another question but it just keeps telling me to change my answer

oh wait, nvm

mb

it works

it just doesnt show the answer unlike the first question, my bad

[craggy stump]

i recall working on this in doing asymptotes

ah

ive just forgot how to determine the signs

[craggy stump]

all good

[hasty talon]

thank you

[craggy stump]

np

were done right??

alright

.close

aw

[craggy stump]

oh yeah i remembered, so if something like that was to happen, (lets say 0<x<1) the y value shouldn't be defined, since thats outside of the domain

[hasty talon]

Y wasnt in the equation but thanks

[marsh citrus]

@hasty talon Has your question been resolved?

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[smoky nest]

Need some help with this + like another page if that’s okay

[marsh citrus]

@smoky nest Has your question been resolved?

[pure bloom]

What are you having trouble with?

[marsh citrus]

@smoky nest Has your question been resolved?

[smoky nest]

I want to know what story I could make up to represent this graph, I’m having trouble coming up with a good story and an accurate story too the graph.

[pure bloom]

Gotcha.

Ok so if you want to make a story to fit the graph you need to observe certain aspects about the graph

Like this graph has a negative slope, so the y value decreases with larger x values

And if you're talking about making a story to fit the graph you want to relate the real world and the graph

[smoky nest]

I was thinking of like instead of gaining money they were loosing money by the day.

[pure bloom]

Yeah that could work

[smoky nest]

and since it’s negative I wanted it to be somewhat similar to that

but I just don’t have the creativity currently to think of what to say 😭

I don’t want to do like gambling but idk if it would make sense on how they were losing money everyday

Sorry if it’s over complicated

[pure bloom]

You could do like a bill, subscription, or some kind of daily purchase

Or what I was thinking was you could have money and then the amount of an item you want to purchase

So the more items you purchase the less money you have

You can do it however you want there's a lot of room for creativity here

[smoky nest]

I feel like that’s more accurate than bills

[pure bloom]

Yeah

[smoky nest]

But what would they purchase everyday?

Like everyday needs?

[pure bloom]

Since it's -2 you'd need to find something that you'd buy for 2$ each day...

[marsh citrus]

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[smoky nest]

Okay, thank you so much

[fickle compass]

I think u opened a channel without knowing-

U might wanna close it if u dont need it

[smoky nest]

Sorry I think i did something

Thank

You