help-39

[hollow cobalt]

We have the assumption that ax^2sin(x) + b(cos(x) - 2x) is the zero function

Equivalently, that expression is equal to 0 for all real x

Your goal right now is to show whether or not a = 0 and b = 0 is forced by that assumption

[small drum]

I pluged pi/4

[hollow cobalt]

What about x = 0 though

[small drum]

I guess I didnt understand 😦

[hollow cobalt]

Also the first line is incorrect

Hm? It looks like you have understood everything so far though

[small drum]

we are searching for a non zero sol

[hollow cobalt]

You have ax^2sin(x) + b(cos(x) - 2x) = 0 for all x, so the first line should rather be ax^2sin(x) = b(2x - cos(x))

[hollow cobalt]

Or trying to show that a and b must be 0

[hollow cobalt]

Anyway, try plugging x = 0 into ax^2sin(x) + b(cos(x) - 2x) = 0

[hollow cobalt]

Yeah

So b = 0

And now we have ax^2sin(x) = 0 for all x

Come up with a value for x which will make x^2sin(x) nonzero and that will be pretty much it

Because once you plug that value for x in, a = 0 will be implied

[small drum]

and does that prove that the only case where af1 = bf2 is when a = b = 0?

[hollow cobalt]

Yes, except we started with af1 + bf2 = 0 rather than af1 = bf2

[small drum]

so to finish the proof I can pick pi/2

[hollow cobalt]

Yup

[small drum]

which makes the x^2sin(x) non zero and equal zero so a = 0

its still weird to me that we find b using a case

but I guess Ill get used to that

[hollow cobalt]

In actuality you don't even need to see whether a = b = 0 is an actual solution, so the other values of x can be ignored

In general you will always know a = b = 0 is a solution because that's how linear combinations work

So you only need to show that a = b = 0 is implied and that's enough

[small drum]

yeah but how can we know there arent any other sols?

[hollow cobalt]

That's what we showed though

If a solution exists, then a = 0 and b = 0 must be true for that solution

Meaning a different solution wouldn't exist

It would mean we made a wrong step somewhere if there existed a nonzero solution

But we didn't

[small drum]

its still a bit hard for me to grasp but I think I kind of get it
Thank you so much for the help

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[glad valley]

I need help solving this derivatives

[pearl pond]

@glad valley Has your question been resolved?

[glad valley]

<@&286206848099549185>

Geez where are all the helpers today?

<@&286206848099549185>

[pearl pond]

@glad valley Has your question been resolved?

[prime apex]

Hello

[hidden dust]

All of them?

[glad valley]

No, no, nevermind

Already got that fixed

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[rigid belfry]

Hey guys, can you check if i did it right?

[rough stream]

No idea how you got from line 3 to line 4

[rigid belfry]

bruhh

[rough stream]

Your goal is to factor numerator and denominator

[edgy stone]

you can't just remove x like that

[regal herald]

tut tut, illegal moves

[rigid belfry]

Then is this better?

no wait

BRUH

[rough stream]

Your problem is arbitarily removing x. Can't just choose random terms to remove like that

I'll give you a smaller problem, that will help with the larger one.
Can you factor 3x^2 - 7x +2?

[rigid belfry]

(x-2)(3x-1)?

[rough stream]

Cool! Same with the denominator, x^2 + x - 6. What's that, when factored?

[rigid belfry]

ohh

I'm sorry if i enraged 3 of yall oops

[rough stream]

Huh? You're good lol

[rigid belfry]

I am trying to learn it

(x-2)(x+3)

huh okkkk

[rough stream]

Now that you don't have seperate terms, cancelling a factor is legal

[rigid belfry]

what

oh read illegal

so this?

[edgy stone]

yes

[rigid belfry]

is this it???

[edgy stone]

well you just have to find this now

should be simple

[rigid belfry]

woah

thank you guys.

sorry for the trouble

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[devout shale]

<@&286206848099549185>

<@&286206848099549185>

[regal herald]

youre an impatient one, arent you

[devout shale]

im sorry i j have something to go to

myb

[sacred crest]

You do realize that annoying helpers won’t make them want to help you

[devout shale]

ik its myb but im in a rushh im sorryy

[sacred crest]

Well no matter the rush, don’t be so entitled. Don’t invade other channels, dont overping, don’t follow helpers into other channels. You’re not entitled to help any more than anyone else just cause you’re in a hurry. Stick to the rules

[devout shale]

sorry

[pearl pond]

@devout shale Has your question been resolved?

[pearl pond]

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[fierce burrow]

I get how to setup the integral, but I'm confused on how to actually computate it/solve it

[pearl pond]

@fierce burrow Has your question been resolved?

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<@&286206848099549185>

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[fierce burrow]

I've set up the integral, but I don't know how to solve it from there

[fierce burrow]

Here’s my work so far:

[leaden wadi]

Try expanding the exponent part.

[pearl pond]

@fierce burrow Has your question been resolved?

[fierce burrow]

Could I just do the square root as is because everything there can be squared, or is that not possible and I just have to expand it out?

Sorry for the late response

[pearl pond]

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[cold swift]

what do you need

use regular long division instead

its more better than dealing with fractions in synthetic

[smoky saffron]

oh ok

[cold swift]

but you just give the first result in the synthetic division a term wiuth a degree -1 than the leading term

so the first term would be 6x

mb 6x^2

and you would keep that 6x^2 outside of the fraction

since it is already 'divided'

[smoky saffron]

wait

sorry im bad at this butr

what do i do knowing this information @cold swift

im not logical

[cold swift]

the top thing is your solution

thats it

[smoky saffron]

oh ok

ohhhhh

ur right

🤔

it dont factor either

@cold swift do yk

what to do

[cold swift]

wat

hold on did you get remainder

[smoky saffron]

i got 0 as remainder

[cold swift]

alr then its just 2x^2 +11x -4

[smoky saffron]

OHHHHHHHH

bc i divided it

[cold swift]

if you want to check do (2x^2 +11x -4)(3x-2)

its the inverse

[smoky saffron]

i got the answer

[cold swift]

nice

[smoky saffron]

ok i just never did long division in these questions before so i was confused but thanks

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[lyric oriole]

hi I'm not sure if I'm missing something here

[lyric oriole]

or is this site bugged

I need to give the range of this function https://i.imgur.com/rX9zOee.png

This is the description:
"There are two curves on the x y coordinate plane. The first curve begins at the closed point (−3, −3), then goes up and to the right to the open point (−2, −2). The second curve starts at the closed point (−2, −1), goes up and to the right, reaches a maximum at the approximate point (1.5, 3), goes down and right, and ends at a closed point at approximately (2, 2.75)."

apparently [-3,3] is incorrect

so is [-3,3)

it says the maximum is at an "approximate point (1.5, 3)" what am I supposed to do with that

I guess approximate means it could be a little higher

[west notch]

there is a gap

[lyric oriole]

but how do I write that in notation without overshooting it

there is no real gap though is there?

-1 has a value

sorry -2*

oh right

[west notch]

the domain doesn't have a gap

but the range does

[lyric oriole]

it will never be -1 to -2

right

you know I thought this review was dumb but maybe I really need it lmao

thanks

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[full plover]

hi

[full plover]

lh

nvm

[hollow condor]

?

[pearl pond]

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[smoky saffron]

does this mean

[smoky saffron]

wait

so i multiply numerator and denominator by sqrt(x+6) + 1

and then when im done i multiply the numerator and denominator by 2+ sqrt (x+9) ?

The answer is -2 ?

but how

[west sapphire]

where did that (2 + sqrt(x+9)) come from in the denominator

[smoky saffron]

i was trying to rationalize it

[west sapphire]

but you didn't multiply by (2 + sqrt(x+9)), you multiplied by (sqrt(x+6) + 1)

also you're not drawing your square roots properly, some of them are only covering the x when they should be covering x + 9 or x + 6 or whatever

makes it hard to read and very error prone

[smoky saffron]

oh ok ur right

so

what do i do after this step?

@west sapphire

because the x+5 becomes 0 if u plug in -5

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[agile condor]

Not sure how to do this one

[pearl pond]

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[agile condor]

.close

[naive dirge]

Looks like b?

By elimination

[orchid sentinel]

What does this q mean?

[naive dirge]

When the input is real number, so as the output

[naive dirge]

Just my interpretation of course

[orchid sentinel]

Ohh

[vale turtle]

why B?

reason

[earnest stratus]

What's the domain for log(x) ?

[pearl pond]

@vale turtle Has your question been resolved?

[vale turtle]

(0,infinity)

all real numbers

[pearl pond]

@vale turtle Has your question been resolved?

[vale turtle]

Close

.close

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[torn willow]

Can someone help me with this question pls

[sinful apex]

!status

[pearl pond]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

@torn willow Has your question been resolved?

[torn willow]

1

<@&286206848099549185>

[dense goblet]

hint: basil and dorothy need to be late the same number (either 3 or 4) of times to make a whole number of pennies
test each case

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[real geyser]

why is (c)= -2?

[real geyser]

why not 1 though

its close

unlike -2

[light helm]

the - in superscript indicates the left hand/side limit

follow the graph from a bit left of x = 0, towards where x=0 and see what y value you're approaching

[real geyser]

thank you

forgor the line

mb

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[tame ore]

How to do part 1

[pearl pond]

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@tame ore Has your question been resolved?

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[queen rock]

Can anyone help me out with this matrix proof? I have to prove that A^2-B^2 is also symmetrical if A and B are both symmetrical. I was thinking of using transposed matrix for this but not sure how to start.

[queen rock]

@summer imp you mind helping me out or are you busy

[summer imp]

Hmm

You can probably use the transpose here

Since it distributes over addition

You want to show that $(A^2-B^2)^T = A^2 - B^2$

[jolly parrot]

Azyrashacorki

[queen rock]

Yep

With ith rows and jth columns

But I suck at that

[summer imp]

There's a few ways to show this though.

Lost battery in my tablet, but I think you could expand the difference of squares and exploit the fact that $AB=BA$ if $A$ and $B$ are symmetric

[jolly parrot]

Azyrashacorki

[queen rock]

hmm

what do you mean with expanding the difference of squares because i don't think it's called that in my language.

[midnight haven]

use the definition of a symmetric matrix

[queen rock]

yeah $A=A^T$ but im not sure how to actually do it as a proper proof

[jolly parrot]

Bbg

[midnight haven]

Let $C = A^2 - B^2$

[jolly parrot]

45

[midnight haven]

NTS that $C = C^T$

[jolly parrot]

45

[midnight haven]

use the properties of the transpose and how it commutes with powers and distributes over addition to show this

@queen rock

[queen rock]

ok lemme try, i'll hit you back with a picture if im stuck

[midnight haven]

ok

[queen rock]

@midnight haven not done yet, but is it correct till this point?

[midnight haven]

You’re over complicating

[queen rock]

I was taught to do it like this

Just not good at it

@midnight haven So, is it correct or am I doing something wrong?

Or how would you do it

[pearl pond]

@queen rock Has your question been resolved?

[queen rock]

<@&286206848099549185> can anyone else help me?

[pearl pond]

@queen rock Has your question been resolved?

[finite bridge]

$A^2 - B^2 = AA - BB$ which due to symmetry is $A^T A - B^T B$.\
If $A^T A$ is symmetric and $B^T B$, then $A^T A - B^T B$ will also be symmetric (obvious right?).
So just show that $A^T A$ is always symmetric

[jolly parrot]

TRAMPELTIER

[queen rock]

ok thanks, i'll try it like this

[pearl pond]

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[hollow basalt]

so if i have the curve

r(t) = <t^3 + 3t, t^2 + 1, ln(1 + 2t)> 0 ≤ t ≤ π,

and i want to find the tangent line perpendicular to the plane

15x + 4y + .4z = 10 

then i want to find the derivative

r'(t) = <3t^2 + 3, 2t, 2/2t+1>

and the normal of the plane is <15, 4, .4>
so now my question is how are you supposed to find a tangent vector perpendicular to the plane's normal?
so I tried dot product:

(3t^2 + 3)15 + 8t + (2/2t+1)4 = 0

but i have no clue how to solve for t? is there another way or am i doing something wrong

[pearl pond]

@hollow basalt Has your question been resolved?

[hollow basalt]

<@&286206848099549185>

[rough stream]

You want to find a tangent vector perpendicular to the plane, no?

So, the tangent vector will be in the same direction as the normal.

[pearl pond]

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[hollow basalt]

oh right im dumb 😆 got that mixed up

[pearl pond]

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[dusk otter]

Can anyone help with these two questions?

[dusk otter]

This is the information that I gathered so far about it, not much

[pearl pond]

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[dusk otter]

YuP!

[pearl pond]

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[quiet pulsar]

I tried this process to get each coeff. but I got wrong

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[indigo trench]

do any of yall know how to do this (Linear Algebra)

[indigo trench]

like honestly how can i represent this as a matrix

cuz then it would be easy to find the dimensions

[cursive wraith]

you don't need to write it as a matrix

try to think first of the image of such a linear map

and to do that, maybe recall what an orthogonal projection does

[indigo trench]

it would be like like a vector parralel to the subspace right

[cursive wraith]

so which subspace are we making the vector "parallel to" in this case?

[indigo trench]

guess we never defined that

ma fault

[cursive wraith]

well it's defined here isn't it?

[indigo trench]

O

is the subspace R

?

[cursive wraith]

R is not a subspace of R^n

it's not even a subset of R^n

[indigo trench]

so parralel to the set formed by R^n?

[cursive wraith]

wait

well not at all here

what is the projection done onto here?

[indigo trench]

the nonzero vector v

that is in

R^n

[cursive wraith]

this is not a precise enough subspace

An orthogonal projection is done onto a subspace V

[indigo trench]

but what is this subspace V

[cursive wraith]

well

the sentence "T is an orthogonal projection onto some vector v" should effectively replace that

so what HAS to be V?

Clearly V has to contain which vector according to this?

[indigo trench]

The vectors in T?

like if we imagined it a matrix, the columns in T

[cursive wraith]

don't think about T as a matrix

really there are not a lot of vectors we introduced here is there?

[indigo trench]

nah not rlly

[cursive wraith]

so intuitively, what does V have to contain?

[indigo trench]

all the vectors that are orthogonal...

[cursive wraith]

that's the opposite of the vectors we want

that's the subspace that is Parallel to what we want

it's not the subspace that is onto

[indigo trench]

WAIT

could it contain all the vectors that are kernals

[cursive wraith]

?

[indigo trench]

nvm i thought i went somewhere

[cursive wraith]

no, the kernel IS the parallel subspace

we want the ONTO subspace

what does "ONTO" mean?

[indigo trench]

projection

right

[cursive wraith]

yes

so it's the Image right?

it's what remains after the projection

[indigo trench]

hmmmmm

[cursive wraith]

so, what vector remains after the projection T?

by definition of T, which vector is unchanged by T?

[indigo trench]

the 0 vector?

[cursive wraith]

sure but another one maybe?

[indigo trench]

or the vector v

?

[cursive wraith]

Yes that's what I wanted

v is unchanged by T

by definition of projection ONTO v

so V contains v

[indigo trench]

oh😭

[cursive wraith]

take a stab at what the subspace V is

if it has to contain v

[indigo trench]

V is the subspace containing the image?

[cursive wraith]

V is the image

but

[cursive wraith]

based on this information

what is V?

what's the first subspace of R^n that comes to mind that contains v?

[indigo trench]

V right?

[cursive wraith]

We want another expression of V

when you think of any vector x

what subspace do you immediately think of that contains x?

maybe even take it as small as possible

[indigo trench]

maybe a line containing all vectors spanned by x?

[cursive wraith]

that's it

the SPAN

of x

[indigo trench]

OH

[cursive wraith]

and indeed when there's only 1 vector that spans a subspace, it's a line

[indigo trench]

so is V the span of the v vector?

in R^n?

[cursive wraith]

yes

[indigo trench]

ohhhhh

[cursive wraith]

so with what we said

Im(T) = ?

[indigo trench]

= to the span of vector v in R^n?

[cursive wraith]

yes

[indigo trench]

so then

would the dimension

be just n?

of that image

[cursive wraith]

you said that when we span with only one vector, it's a line

btw the dimension of R^n is n

so if dim(V) was n then V = R^n.... not that possible

[indigo trench]

but we have more than a line here right

or do we not

[cursive wraith]

You said V = span(v)

[indigo trench]

O

so

its one vector

which means

[cursive wraith]

Its basis is only one vector

[indigo trench]

which means dim (im(T) = 1

[cursive wraith]

But the subspace is an infinity of vectors

[indigo trench]

ahhhhh

then for the nullity

in part b

[cursive wraith]

Yes

[indigo trench]

is that

n-1

?

[cursive wraith]

Yes, and why?

[indigo trench]

bec of the rank nullity theorom and we know m = n

[cursive wraith]

Yes

[indigo trench]

so rank + nulllity = n and on and on and on

ok i see

u a lifesaver broski

[pearl pond]

@indigo trench Has your question been resolved?

[pearl pond]

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[mellow fox]

Topic: Length of Parametric Arcs

2 Questions:

1. How is the graph formed? (I tried desmos but it didn't turn out the same)
   which part of the graph is x=cos^3(t) and which is y=sin^3(t)?
   and
2. How are we able to get just 1/4 of the curve and multiply by 4? Won't finding the arc length from 0 to pi/2 give the parts indicated in green and not 1/4 of the graph?

thank you in advance!

[pearl pond]

@mellow fox Has your question been resolved?

[mellow fox]

<@&286206848099549185>

[pearl pond]

@mellow fox Has your question been resolved?

[spiral pivot]

@mellow fox https://www.desmos.com/calculator/t9mwge1bzp

It seems to graph it fine for me

[mellow fox]

oooh, I forgot that parametrics are graphed differently! thank you!

[spiral pivot]

No worries

This might also help clear up your question about which part is the x and y

And if you replace the bounds with 0 to π/2 you should also see why it works

[mellow fox]

yep, it makes sense now! tysm for your help!

.close

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[eager tusk]

Can I get help on this problem?

[stoic path]

Use integration by parts

[eager tusk]

10 is the one

That’s my work

I did

[stoic path]

So are you having problems with this integral?

[eager tusk]

The integration was incorrect?

[stoic path]

No it's correct

Where are you stuck tho

[eager tusk]

Well i thought I was done

but it says it's incorrect

[novel narwhal]

The circled integral is incorrect.

[eager tusk]

is it supposed to be (1/1+x^2) (2)

[stoic path]

There's a 3x² multiplied on the right side, i forgot to enclose it within the circle

[novel narwhal]

Differentiating ln(3x^2) doesn't give you the integrand.

it gives you 6x/(3x^2)

[stoic path]

Ig you're supposed to integrate (3x^2)/(1+x^2) but it won't be ln(3x^2) tho

[eager tusk]

Oh so I understand some of that

let me try it

so I integrate (3x^2/1+x^2)

[novel narwhal]

Yep. You can take the 3 outside the integral, then add 1 and minus 1 in the numerator. Then you can split it up and get:

[jolly parrot]

Αρχιμήδης

[novel narwhal]

Now you've got a standard integral.

[eager tusk]

how in the world

where does the x^2 go?

[novel narwhal]

\begin{align*}
\frac{x^2}{1+x^2} &= \frac{x^2+1-1}{1+x^2}\
&= \frac{x^2+1}{1+x^2}-\frac{1}{1+x^2}\
&=1-\frac{1}{1+x^2}
\end{align*}

[jolly parrot]

Αρχιμήδης

[novel narwhal]

This was wrong.

[eager tusk]

ahhhhh ok

I see it now

so do i integrate that now?

[novel narwhal]

Yep

[eager tusk]

3x^2arctan(x) - x - arctan(1 + x^2)

[novel narwhal]

Almost. The fraction goes to arctan(x)

[eager tusk]

ohhh

how strange I thought it went ln (x)

[jolly parrot]

Αρχιμήδης

[novel narwhal]

So you'd need an x on top to get a ln function.

[novel narwhal]

Did you need help with anything else, or are we able to close this now?

[eager tusk]

I got thank you very much

.close

[pearl pond]

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[pearl pond]

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[lone nebula]

Hi, i am not sure how to continue

[lone nebula]

the first image is the exercise

and this is the solution in book

[burnt minnow]

how did you get 6/25?

[lone nebula]

i havent finnished the exercise

it needs a and b

6/25*a^2-3 b^2-1

i think something like that

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[cunning stump]

I'm pretty sure I made a mistake somewhere

But I've been looking for ages and can't find it

Could someone help me

[fallen cipher]

You seem to have forgotten x

[cunning stump]

Where?

Ooh

In the integral

Alright

Thanks

.close

[pearl pond]

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[fallen cipher]

No, not in the integral. In the algebra!

[cunning stump]

Oh yah mb

Brain a lil fried

[fallen cipher]

Oh. It's in the second image you sent. It should be $2xe^{2x}$.

[jolly parrot]

Van21st

[cunning stump]

Yah gave me the correct answer now

Thanks again

.close

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[opaque crown]

hi i need help with this question using combinations and permutations

[tall flint]

,rccw

[opaque crown]

ye sorry bout the rotation

@tall flint can you help me?

[tall flint]

!status

[pearl pond]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[opaque crown]

i know i have to use comb and perm

but i dont know how

[limpid roost]

(ABC)DEFGHIJ

[opaque crown]

????

why?

how?

[tall flint]

!nosols

[pearl pond]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[limpid roost]

Whoops my bad

[opaque crown]

BRO IM NOT ASKING FOR ANSWERS JUST HELP

I WANT TO KNOW WHAT IM SUPPOSED TO DO TO GET ANSWER

NOT THE DIRECT ANSWER

[limpid roost]

So we have 10 flowers

[tall flint]

I didn't say you were. It was pretty clearly directed at EndTimes

[limpid roost]

Let's label them A through J

[opaque crown]

how?

i didnt even understand what he said

[opaque crown]

ok got it

[limpid roost]

And 3 of them are always together

[opaque crown]

yep

[limpid roost]

Let's say ABC are always together

[opaque crown]

yep

[limpid roost]

So we have (ABC)DEFGHIJ

[opaque crown]

yes

[limpid roost]

Now notice that if we want to keep the group ABC together, we can treat it as a single element

[opaque crown]

how do we solve this using combinations and permutations?

cus i need to solve using that

[limpid roost]

Listen.

[opaque crown]

ok...

continue

[limpid roost]

Now notice that if we want to keep the group ABC together, we can treat it as a single element

[opaque crown]

yes

[limpid roost]

So if we treat ABC as a single element, how many permutations of (ABC)DEFGHIJ do we have

[opaque crown]

what do you mean permutations?

[limpid roost]

Are you familiar with the concept of a permutation?

[opaque crown]

bro can't we just use the formula

n!/(n-r)! * r!

is it possible?

[limpid roost]

Not quite. Because we are just arranging elements in a row

[opaque crown]

ok then well i don't rly get what your saying so........

sorry

.close

[pearl pond]

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[limpid roost]

It's just 8!

[pearl pond]

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[tawny canyon]

3x + 1?

[pearl pond]

Please don't occupy multiple help channels.

[burnt minnow]

?

what's the objective?

[tawny canyon]

3x + 1

[eternal reef]

ig he's talking about the collatz conjecture

[tawny canyon]

=?

[cosmic charm]

what do you want from it, do you want to cook it or what

[tawny canyon]

I want to devour it

[tulip ore]

if you dont know about it, you can devour it here: https://www.youtube.com/watch?v=094y1Z2wpJg

[pearl pond]

@tawny canyon Has your question been resolved?

[pearl pond]

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[untold pollen]

I need help in resolving problems with matrices...
so, I have these two matrices, it says, given the matrices A and B, 3x3:

[untold pollen]

AND

determine, if possible, the matrix X, 3x3, given that:
X+A=2(X-B)
the resolution for this is
X+A=2(X-B) = X+A=2X-2B = X-2X=-A-2B = -X=-A-2B = X=A+2B

so, what I want to know is, how was this determined???
what rules do I gotta follow or how can I determine these types of situations?

[vocal dove]

wdym?

[untold pollen]

like, what was the way to resolve this?

how do you describe it?

please help me asap

I need to know and research it, so I can use these formulas for my exam

[midnight haven]

didnt they just solve it like a linear equation..

[daring arch]

just simplify the equation and group the similar terms, you'll get the required answer

[untold pollen]

but how do i learn to do that?

what do I have to follow?

[daring arch]

you dont know to solve linear equations?

2(x-1)=0
2x-2=0
2x=2
x=1

you know this right?

its the same thing

[untold pollen]

unfortunately not, what do i look up on google to learn it quickly?

[daring arch]

look up articles for solving linear equations either by Khan Academy or LibreTexts

should help you

[pearl pond]

@untold pollen Has your question been resolved?

[untold pollen]

I'm gonna need more help soon

[pearl pond]

@untold pollen Has your question been resolved?

[pearl pond]

@untold pollen Has your question been resolved?

[untold pollen]

do I solve this 5(X+B) = 4x+A, in the same manner as the previous one??

replace the X's position?

as in 5X-4X = -5B+A????

PLEASE HELP

<@&286206848099549185>

[deft swift]

In order to solve those kind of problems you need to make it in the format of X = something so you're on the right path.
5(X+B) = 4x+A
=> 5X + 5B = 4X + A
=> X = A - 5B
and now you just compute that

X =
-4 -5 -5
-5 -4 -5
-5 -5 -4

[untold pollen]

is F = {3} a diagonal matrix and a scalar matrix? since there is no more dimension and other elements, it probably can't right?

and a symmetric matrix? I think it can be a symmetric matrix, since there are no other elements

[untold pollen]

why is it X = A-5B??? and is it not 5X-4X = -5B+A before???

<@&286206848099549185>

@deft swift @daring arch please help guys

[daring arch]

thats right

[daring arch]

since you need to find the Matrix X, you have to get a single value of X

5X-4X, both are having same term i.e, X so you can simplify them further

[daring arch]

yes F={3} is indeed a diagonal and scalar matrix

and yes, it is also a symmetric matrix

since the transpose is same as the matrix

[pearl pond]

@untold pollen Has your question been resolved?

[pearl pond]

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[hexed oriole]

Looking for help with this differential equation

[hexed oriole]

This is what I’ve done, however I cannot seem to progress from here

(2nd picture should be a -v on the left not -x)

[pearl pond]

@hexed oriole Has your question been resolved?

[rough stream]

Is that the right problem?

[pearl pond]

@hexed oriole Has your question been resolved?

[hexed oriole]

wym? Like is the question worded right?

[pearl pond]

@hexed oriole Has your question been resolved?

[pearl pond]

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[cerulean smelt]

(3, 5) is a point on the graph of y = f(x). Find the corresponding point on the graph of each of the following relations.

a) y = 3f(-x + 1) + 2

[cerulean smelt]

I transformed (x, y) -> (-x + 1, 3y + 2)

but im not sure if that's correct

[lost flax]

Why did you choose 3y+2 instead of 3y?

[cerulean smelt]

sorry, forgot to include it in the question, there is + 2

fixed it

[lost flax]

I think this is not correct

You have to reverse the process for the x

[cerulean smelt]

I first factored the -, so
y = 3f[-(x - 1)] + 2

so that would be one to the right?

[lost flax]

It is correct

What I meant is that, if it were y = f(2x) instead of y = 3f(-x + 1) + 2, then the transformation would have been (x,y) -> (x/2,y)

[cerulean smelt]

ohh yeah

[lost flax]

But in this case the inverse is the same

-(-x+1) + 1 = x

Which is a coincidence

[cerulean smelt]

alright, so (3, 5) -> (1/-1 (3) + 1, 3(5) + 2) -> (-4, 17)

is that right?

[lost flax]

No, -3+1 is -2, not -4

If you change that then it is correct

You can also check if it is correct easily
Write y = 3f(-x + 1) + 2 and then substitute (x,y) with the point you obtained

So 17 = 3f(--4+1) + 2 = 3f(5) + 2 is unrelated to f(3) = 5, which you were given

But 17 = 3f(--2+1) + 2 = 3f(3) + 2 is the same as f(3) = 5

[cerulean smelt]

ahh that makes sense, thank you!

so (-2, 17) is the final point?

[lost flax]

Yes

[cerulean smelt]

aight

.close

[pearl pond]

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[nimble osprey]

i have a dejavu

[pearl pond]

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[pearl pond]

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[restive bluff]

Did this one yesterday as well and trying to redo it

Here is my work one sec

Is there any potential error here?

Because I believe that there is some step I got wrong here since my final calculation resulted in 141pi

when I was supposed to get 9pi

ignore me 🤦‍♂️

I forgot a y in the -y + 7

.close

[pearl pond]

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[restive bluff]

yeah Idk how I can be so stupid

[pearl pond]

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[dreamy cairn]

struggling to work my way through this

i know what the graph of cosine looks like

however im not sure how to apply this to asymptotes

[summer imp]

What is the question asking of you?

[dreamy cairn]

oh sorry its just asking for all horizontal asymptotes

[summer imp]

Ok, so you want behaviour for large x, have you tried computing the limits at infinity/-infinity?

[dreamy cairn]

large x? im a little unfamiliar with what you mean

i generally know how to compute limits with inf but we just started asymptotes

[dreamy cairn]

i see

so im thinking i can loosely graph infiinity but

my only problem si this

i know the graph of cosine

but not cos(1/x)

[subtle ginkgo]

you only have to compute it, you know how to compute limits ?

[dreamy cairn]

im not sure if my prof hsa really shown us the easy way to compute limits for trig functions

hes only shown us how to solve it if we know what the graph looks like

hmm

is it like this tho

plugging inf into 1/x gives 0

cos(0) = 1

4 * 1 ?

[subtle ginkgo]

i see , so it was primitive approach, ok i show you example

[dreamy cairn]

for the pos infinity

[subtle ginkgo]

wait

:

2 mins plz

[dreamy cairn]

kk

[subtle ginkgo]

I deliberately chose this example to be similar to yours

[dreamy cairn]

so it is kinda what i said ?

[subtle ginkgo]

y = 0 is both side horizontal asymptote

[subtle ginkgo]

you have to take a paper now and try to do the same with your function, and you paste photo here

to show me

ok?

just behave analogously

in same way as i did for you

[subtle ginkgo]

your given function behaves very alike,

[subtle ginkgo]

yes but you need to write all formally

[dreamy cairn]

kk one sec

[subtle ginkgo]

ok)

,rccw

[dreamy cairn]

oh i doxxed myself

[subtle ginkgo]

all is ok except , two moments, when you write symbol of an infinmity, remove it from your work

[dreamy cairn]

well ill delete it after

where? on the side?

i just did it to clarify the sections i was working on

[subtle ginkgo]

you used symbol of an infinity

ian vwery informal way

just remove it this small part ok ?

[dreamy cairn]

im not sure where youre referencing

[subtle ginkgo]

you wrote

infimity insisde cosinus

correct?

[dreamy cairn]

but isnt that how you do limits? you plug in whats being approached

[subtle ginkgo]

we do , but we do not write it

[dreamy cairn]

ah i see

[subtle ginkgo]

remove just this one moment

in yoru limits

professor wil be glad

i show you also , how i wud write:

[subtle ginkgo]

this way, is absolutely ok and enough

[dreamy cairn]

my professor just gets mad when we dont rewrite the limit each time

unless were plugging in

[subtle ginkgo]

but if you put zero there

that means

you computed limit

so for what to write it again ?

you do not need to write

lim 0 = 0

that is very simple limit, , you do not need to worry about it

[dreamy cairn]

ill rewrite it

[subtle ginkgo]

ok

[dreamy cairn]

is there anyone who can delete my first image

i dont really want my full name showing

[subtle ginkgo]

you can

[dreamy cairn]

it wont let me delete the bots image

[subtle ginkgo]

so maybe moderator, i cant remove it

[dreamy cairn]

<@&268886789983436800>

[analog imp]

I got you home dawg

[dreamy cairn]

i appreciate you

[analog imp]

stay frosty homeslice

[subtle ginkgo]

I accept it ) but in a book, when you write lim 4cos(1/x) = lim 4cos0 = 4x1 = 4, the second limit is not necessary

writing cos0 there, means, you used your limit in place = lim 4cos(1/x)

so if you write lim there, is not needed

write like me: lim4cos(1/x) = 4cos0 = 4x1 = 4

[dreamy cairn]

kk

[subtle ginkgo]

eventuall you can write it as : lim 4cos(1/x) = 4 lim cos(1/x) = 4 cos0 = 4 x 1 = 4

[dreamy cairn]

what would i do if it was 4sin(x) instead

4sin(inf) is obv not an exact value

[subtle ginkgo]

it wud be harfer thing because

[dreamy cairn]

but would it just be equal to sin(1) because of the range 1 and -1

[subtle ginkgo]

lim sin x does not exisit

sif x goes to infinity

so you wued need to prove

that such limit does not exist

did yoru teache show such examples ?

[dreamy cairn]

no

[subtle ginkgo]

so i suppose, you do not need to worry about such situaitons

[dreamy cairn]

we did do 3arctan(x)

[subtle ginkgo]

but geoemtrically, if you look at the grpah, you wil see that the graph is not clsoe to any line

3arctan x is nice

it reaches

3tiems Pi/2

in plus

and

3times (-pi/2) in minus infitniy

[dreamy cairn]

sorry i mean 3arctan(x^2)

[subtle ginkgo]

so 3times Pi/2 n both directions

[dreamy cairn]

but why pi/2

he showed us the graph for it but

im confused on if i should memorize the graph or not

[subtle ginkgo]

you should

[dreamy cairn]

are there any other ones i should

[subtle ginkgo]

well all grpahs of elementary functions, main ones

these are basic ones and you should know them

[dreamy cairn]

ah i see

alrighty i think im getting the general gist now

thank you for the help i appreciate it

[subtle ginkgo]

yvw 🙂 Good Luck!

[pearl pond]

@dreamy cairn Has your question been resolved?

[pearl pond]

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[cinder brook]

f(x)f(y) - af(xy) = x + y
find all a to make this equation work

[cinder brook]

i solved one part but i dont know what i did wrong to miss the other

nvm i found it

instead of f(0)^2 = af(0) i somehow got f(0)^2 = af(0)^2

from there i just get f(0) = 0 or a = f(0)

.close

[pearl pond]

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[pearl pond]

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[outer echo]

how would I solve this problem?

[regal herald]

get it into ref

[outer echo]

\[
  \begin{bmatrix}
  1 & 4 & -2 & -3 & 1\\
  0 & 0 & 1 & 4 & -4\\
  0 & 0 & 0 & 0 & 0
  \end{bmatrix}
\]

[jolly parrot]

Price

[regal herald]

well, tis consistent, you can do more though

a+4b+5d=-7
c+4d=-4

oh dear, a c should be a d

[outer echo]

to clarify this is a b c and d right

[regal herald]

sorry about that, yes the above is right

a=-7-4b-5d
c=4-4d
much simpler
(7-4y-5t, y, 4-4t, t)

[outer echo]

c = 4 - 4d

so is that like saying:

\[
c = 4 - 4d:\\
\begin{bmatrix}
-2\\
1\\
0
\end{bmatrix}
= 4 - 4 \times
\begin{bmatrix}
-3\\
4\\
0
\end{bmatrix}
\]

[jolly parrot]

Price

[regal herald]

not really
c and d arent vectors really

rather entries of a vector

if the matrix of the first 4 columns is A, the thing its acting on is x and the result is the last column Y

then Ax=Y

and x=(a,b,c,d)

[regal herald]

so here each number in the column is the coefficient of a,b,c or d in their corresponding equation, 1 row is one equation

and the last number is the value of said equation

[outer echo]

so 1 is the coefficient of a, 4 is the coefficient of b, and c and d have two coefficients?

or am i reading it wrong

oh wait yeah i think i am reading it wrong

[regal herald]

i can write this augmented matrix as 3 equations

x+4y-2z-3t=1
z+4t=-4
-x-4y-z-9t=11

we make a matrix by this:

[\begin{bmatrix}
1 & 4 & -2 & -3\
0 & 0 & 1 & 4\
-1 & -4 & -1 & -9
\end{bmatrix}

\begin{bmatrix}
x\
y\
z\
t
\end{bmatrix}

=
\begin{bmatrix}
1\
-4\
11
\end{bmatrix}]
a

oop

maybe you can salvage that haha

hmm

[outer echo]

yee that makes sense

[regal herald]

one step closer

interesting

[jolly parrot]

AℤØ
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[regal herald]

simultaneously worse and better

but i think you get the idea lol

[outer echo]

\[
  \begin{bmatrix}
    1 & 4 & -2 & -3\\
    0 & 0 & 1 & 4\\
    -1 & -4 & -1 & -9
  \end{bmatrix}
  \begin{bmatrix}
  x\\
  y\\
  z\\
  t
  \end{bmatrix}
  =
  \begin{bmatrix}
  1\\
  -4\\
  11
  \end{bmatrix}
\]

[jolly parrot]

Price

[outer echo]

like this?

[regal herald]

exactly so

that x,y,z,t would be the a,b,c,d i was doing

[pearl pond]

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[pearl pond]

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[muted night]

Find all positive integers $n$ such that there exist a positive integer $m$ and distinct prime numbers $p,q$ such that $1<p<q, q-p\mid m,$ and $p,q\mid n^m+1.$

[jolly parrot]

BeeReallyYum

[muted night]

i figured that all numbers $n\equiv 5\pmod 6$ are solutions but I couldn't prove that there are other solutions or that they are indeed the only ones

[jolly parrot]

BeeReallyYum

[pearl pond]

@muted night Has your question been resolved?

[muted night]

<@&286206848099549185>

[pearl pond]

@muted night Has your question been resolved?

[muted night]

<@&286206848099549185>

[pearl pond]

@muted night Has your question been resolved?

[muted night]

<@&286206848099549185>

[pearl pond]

@muted night Has your question been resolved?

[pearl pond]

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[sharp mulch]

how is this not right??

[sharp mulch]

-1/(14(7x-2)^2) is the integral

[light helm]

show what you did after

it looks like you messed up a sign upon evaluation / simplification

[sharp mulch]

wdym after

the integral?

[light helm]

you should be subtracting
-1/(14(7x-2)^2) evaluated at x=2

however it looks like you just evaluated at x=2

[sharp mulch]

because the top bound goes to 0

[light helm]

which leaves you with
$$\red{-} F(2)$$

[jolly parrot]

ℝαμΩℕωⅤ

[light helm]

not F(2)

what you've calculated was just F(2)

[sharp mulch]

F(2) is -1/2016

[light helm]

yes

[sharp mulch]

but that isnt correct

[light helm]

yes

you're not reading what i'm saying

$F(x)\eval_a^b = \what$

[jolly parrot]

ℝαμΩℕωⅤ

[light helm]

in general

[sharp mulch]

oh my 😭

i got it its just 1/2016

sorry for wasting your time

[pearl pond]

@sharp mulch Has your question been resolved?

[pearl pond]

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[upper hornet]

help

[upper hornet]

derivative of 1/sin^2(5x)

[marsh steeple]

do u know what 1/sin(x) is?

[upper hornet]

cosecant?

[marsh steeple]

yep

try changing it to that and see what u can do with that

[upper hornet]

chain rule?

.close

[pearl pond]

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[flat kayak]

not sure where to even start

[flat kayak]

or what to do

[pearl pond]

@flat kayak Has your question been resolved?

[edgy stone]

integrate twice and call it a day

[flat kayak]

huh

[plush bramble]

Integral of second derivative = first derivative

[flat kayak]

yea but idk where to even start

lawl

[flat kayak]

.close

[pearl pond]

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[brittle harbor]

Can anyone explain to me how we solve this type of problem with absolute value functions?

[near echo]

if the absolute values are equal, they are either equal
y - 1 = y + 2
or one is equal to the negative of the other
y - 1 = -(y + 2)

from there just find the y values that satisfy at least 1 of those two equations

[pearl pond]

@brittle harbor Has your question been resolved?

[pearl pond]

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[jolly parrot]

drinking water

[pearl pond]

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[versed venture]

.close

[pearl pond]

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[dusk otter]

Would anyone be avaible to check over 4 math questions just need to know if they are correct as there is no solution

[plush bramble]

Just upload one at a time

[dusk otter]

Yup gotcha ! Ty I appreciate it!

Oh I actually need to ask 5

This quesiton is just referrring to that equation at the top there

[plush bramble]

Work looks fine if all your numbers are correct