help-17

[sturdy geyser]

Pls

[tame raptor]

ok

[sturdy geyser]

#help-17 message

@tame raptor

Wtf

S

@tame raptor Hello?

Still here?

[tame raptor]

solution 1=-1/5

solution 2=-2/3

Good night bro

[sturdy geyser]

I need steps

Bro

[tame raptor]

i gotta sleep

[sturdy geyser]

[tame raptor]

bro

cmon

you got internet in your house

why don't you search it on youtube

[sturdy geyser]

YouTube is confusing

[tame raptor]

its not

[sturdy geyser]

If I can’t get 1 part then I’m stuck

Because I cannot ask the person created it

[tame raptor]

destributive property

go learn that thing

i can't explain it to you

[sturdy geyser]

: (

[tame raptor]

https://www.youtube.com/watch?v=v-6MShC82ow

here

Good night

Bye

[vocal sleet]

@sturdy geyser Has your question been resolved?

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[fathom hedge]

how is b injectiev?

ah nvm im a fuckwit

I get how d is injective

sorry

holy ahsit

okay restart

question 28d

f is NOT injective, because multiple inputs can lead to the same output

for example, x = 1, and x = 0, both lead to f(x) = 0

however in terms of surjectivity

i don't see how it isnt surjective?

I even graphed it out

ahh

no nvm i dont see it

nvm i’m literally brain dead

i get it

.close

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[minor blade]

Is there a way to prove that this has max and min without differentiation?

[boreal badge]

Proving sin(cosx/x^2) has a max and min?

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@minor blade Has your question been resolved?

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[ocean crypt]

HI this is a puzzle and i can't figure it out... Does anyone have any idea how to solve the question? We need to figure out what the letters say

[near hearth]

they give you a^-1, K and C. Input into the given equation.
For example at Q:
.. = 21(16-2)Mod26 = 294Mod26 = 8 corresponding to letter I

[vocal sleet]

@ocean crypt Has your question been resolved?

[languid berry]

caesar cipher goes hard

waitt

no

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[crimson jetty]

🤨

!show

[vocal sleet]

Show your work, and if possible, explain where you are stuck.

[crimson jetty]

@chilly pawn.

[chilly pawn]

!status

[vocal sleet]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[chilly pawn]

1

[crimson jetty]

hint: you want to factor the x

and have it on one side

=============
You know how to solve equations of the form
Mx = c

where M is a constant matrix
c is a constant vector
x is a variable vector

so try putting it in this form

[chilly pawn]

So i replace rhs with c

Mx is done

[crimson jetty]

you cannot because c needs to be a constant vector

which 4x + thing isnt

[crimson jetty]

so try again with this in mind (i edited the above message to make it clearer)

[chilly pawn]

Send help

[crimson jetty]

see my comment above

[chilly pawn]

I’m ultra confused

[crimson jetty]

can't just do that.

or maybe u can, but i wouldnt advise it

i dont think u can.

============================
also, as an aside (M | c) = x

I dont like this notation with the augmented matrix, cant make sense of it.

[crimson jetty]

youve also confused 4x with x

because youre apparently using x = (x; y; z)

But in any case, I don't know if this approach works.

[chilly pawn]

I know it doesn’t but i didn’t understood your previous comment

Can you clarify further?

[crimson jetty]

=============
You know how to solve equations of the form
Mx = c

where M is a constant matrix
c is a constant vector
x is a variable vector

are we on the same page with regards to this point?

[crimson jetty]

before i go onto it actually, my 2nd point for why its wrong is

you've misused x

ones a scalar, ones a vector

1. dont reuse letters
2. remember whats a scalar, whats a vector

@chilly pawn that make sense?

[crimson jetty]

ok thats 10x better

thats how ur supposed to do it

except for some reason you put x on the left

it should be on the right

[chilly pawn]

You are right

[crimson jetty]

but yh this is the standard approach, lets say

[chilly pawn]

Ok

What does the infinitely many solutions mean

Det = 0?

Rank(null) > 1?

[crimson jetty]

so you want to consider this question for Ax = b in general, lets say

First of all, consider Ax = 0

Oh, before I even get there - let me just point out this question generalises beyond square matrices

so while det might be useful for square, you probably want to think about it generally

[crimson jetty]

Are you aware that we can think of this in terms of the linear transformation T(x) = Ax

Then ker T = null A.

(@ me when ur back)

[chilly pawn]

Yes

[crimson jetty]

Then as you've, spotted, this has something to do with

dim ker T

the dimension of the kernel

or rank(null) (im not so familiar with this notation)

but its > 0, you're interested in not > 1

===
if dim ker T = 0
This tells you the kernel is 0 dimensional. ie, ker T = {0}

If dim ker T > 0, then the kernel contains infinite points (it must be a line or a bigger subspace)

===
If that all makes sense, this characterizes the number of solutions to Ax = 0. Either 1 or infinite, depending on the dimension of the kernel.

Which makes progress towards your originial question of when Ax = b has infinite solutions

[chilly pawn]

🤔

[crimson jetty]

lmk if anything didnt make sense, or we can continue

[chilly pawn]

I think we may assume it’s a square matrix

Prolly it’s easier with determinant no?

[crimson jetty]

yes, we can get there once we come to a conclusion

but not quite yet

the determinant is just another way of checking the dimension of the kernel

[crimson jetty]

T(x) = 0 has infinite solutions iff dim ker T > 0

but what about T(x) = b

if you think about it geometrically, the image of T could be a plane in 3d space say

Then we could have no solutions if b isn't on that plane

so for an affine equation, you also need to check for this, basically

1. b is in the image of A
2. A has ker > 0. (if A is square, then det A = 0)

[chilly pawn]

Ok we can check 2

But i don’t know what image is

Ah image is being in the column space of the basis?

[crimson jetty]

column space of the matrix? yes

im not sure if there is a faster check for now

oh right, thats fairly fast tbh

it corresponds to your row reduction being consistent or inconsistent

inconsistent -> no solutions

[crimson jetty]

Actually, now I think of it, you should hopefully have been told how RREF form of a matrix characterizes how many solutions the linear system has

if there's an inconsistent row -> no solutions
if you have a row of 0's (and no inconsistent rows) -> infinite solutions
otherwise (in the case of a square matrix, identity matrix) -> one solution

[chilly pawn]

One solution

[crimson jetty]

is that a bot

you shouldnt rely on it.

[chilly pawn]

Alright one sec xd

[crimson jetty]

it should be clear to you why, if you reread it carefully and check the original question

[chilly pawn]

True

Do i need to reduce

[fiery wasp]

you have to augment the matrix with b

https://en.wikipedia.org/wiki/Augmented_matrix

oh wait

maybe you're doing something different

like, for there to be infinitely many solutions to Mx = b, you must first have det M = 0, because otherwise M is invertible and there is a unique solution for every b. so you need to find what values of k make the determinant 0. then, for these values, you have to check when b is in the column space of (A - 4I)

whenever the determinant of (A -4I) is zero and b is in the column space of (A - 4I)x = b, you will have infinitely many solutions (and vice versa)

[vocal sleet]

@chilly pawn Has your question been resolved?

[chilly pawn]

😮

[chilly pawn]

How to do det = 0 with this polynomial

[chilly pawn]

This is from here

[fiery wasp]

i got something else for the determinant , i think

yeah

it should be much simpler

i would recommend reducing the matrix a bit, first @chilly pawn

it makes it easier to compute

you can row reduce it to an upper triangular matrix pretty easily, and then the determinant is just the product of the diagonal entries

[chilly pawn]

😮

[chilly pawn]

[fiery wasp]

[chilly pawn]

Ok I’m stupid

You are right

One sec

@fiery wasp

[fiery wasp]

, so k = 0 or k = -3

so the equation can only have infinitely many solutions if k is one of these numbers

now you have to check if the vector b is in the span of the columns of (A - 4I) for each value

[chilly pawn]

What should i do with this?

[fiery wasp]

to check if b is in the span of the columns of a matrix M, you form the augmented matrix [M | b] and row reduce it

if you get a row like [0 0 0 | a] with a nonzero, then b is not in the span of the columns of M

[fiery wasp]

pog. so if k = 0, you see that the system is inconsistent

sine the last row is [0 0 0 | -9]

so there are no solutions when k = 0

now just do k = -3

[chilly pawn]

K = -3 what

We get a zero row

[fiery wasp]

you know that the only possible way for the system to maybe have infinitely many solutions is when det (A - 4I) = 0. this means that k = 0 and k = -3. when k = 0, the system is inconsistent, so there is no x such that (A - 4I)x = b. but now you just have to check k = -3

yeah

but its all good otherwise

if i remember correctly from yesterday

so there are infinitely many solutions in this case

[chilly pawn]

🤔

[fiery wasp]

yeah

[chilly pawn]

https://media.discordapp.net/attachments/903463353417072641/1154017257693913108/image0.jpg

So only -3?

[fiery wasp]

yee

[chilly pawn]

Ty, kind sir

.close

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[fathom glen]

I'm doing a projection vector question and idk how to put |a| in equation one
Can someone solve and send me a pic?

[fathom glen]

Better quality pic

The answer to |a| for me is 17/10

[mental quail]

i need help

solution like this

[dull maple]

Open your own help channel. Refer #❓how-to-get-help

Delete the messages here.

[dull maple]

Projection is not b + a(vector).

Projection of vector $\vec{A}$ along vector $\vec{B}$ is equal to $\vec{A} \cdot \vu{B}$.

[twin meteor]

Enemagneto

[dull maple]

You need to find projection of b along a. That would be-

[twin meteor]

Enemagneto

Enemagneto

[vocal sleet]

@fathom glen Has your question been resolved?

[fathom glen]

Yes thatc

Look at my question i already solved most

I just don't know how to put 17/10 in eq 1

[dull maple]

You didn't solve most of it. You have only solved |a|.

Do dot product of a and b and divide by |a|.

[fathom glen]

Oh I think I get it

.close

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[glacial onyx]

$\ang{x_1,y_1,z_1} \cdot \ang{x_2, y_2, z_2} = x_1x_2 + y_1y_2 + z_1z_2$

[twin meteor]

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

[grand frigate]

What are you trying to find?

and fix the formatting first

[vocal sleet]

@glacial onyx Has your question been resolved?

[dull maple]

Go to #chill. Not here.

[vocal sleet]

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[elfin ridge]

The rectangle shown at right has been broken into four smaller rectangles. The areas of three of the smaller rectangles are shown in the diagram. Find the area of the fourth one.

[elfin ridge]

PLEASE EXPLAIN

[grand frigate]

Maybe try labelling each sides 🙂

or just wait for someone else, i don't have a paint app

can't draw

[elfin ridge]

k

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@elfin ridge Has your question been resolved?

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[ocean zinc]

How do I calculate lim n tends to infinity (1 + (1/3n) )^n?

[ocean zinc]

I know lim n tends to infinity (1 + (1/n))^n = e

but I'm not sure how to use it

[lapis niche]

$\lim_{n \to +\infty} (1 + \frac{1}{3n})^n$

[twin meteor]

Learath2

[lapis niche]

Is this the question? and is n integer?

[ocean zinc]

Yeah that’s the question

n is a natural number

It’s for series

I mean sequences

[lapis niche]

u = 3n might help you see

[ocean zinc]

My prof hasn’t explained why lim n tends to infinity (1 + (1/n))^n = e

She just told us that it is

And asked us to use it

[lapis niche]

That’s usually a given in most calc 1 courses

It’s kinda the definition of e 😄

Well one definition of e

[ocean zinc]

Oh okay

[heavy yoke]

here's a proof from my calc textbook

[ocean zinc]

Ohh so will the answer be e^(1/3)?

[ocean zinc]

Thank you!

.close

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[void relic]

what does note that w* = w^2

[exotic mason]

omega^* means the complex conjugate of omega.

And it's pointing out that omega squared is the same thing as the complex conjugate. Because it is--the only difference between the two is the imaginary part is negative in omega^2 vs in omega.

Interestingly z_3 is labeled incorrectly on the diagram.

It's written correctly in the box next to the diagram, so I assume it's just a mistake.

[vocal sleet]

@void relic Has your question been resolved?

[void relic]

@exotic mason thank you for the help I understand it better now

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[harsh zenith]

Hello I need quick help about basic equation. IS that i'm not pretty sure of what i'm doing, and i'd like to know a bit more.

The following question is can I say that 1/4x = 0.25x ?

[lapis niche]

Sure

[harsh zenith]

Oh so I can right?

[lapis niche]

1/4 = 0.25

[harsh zenith]

Yea I wasnt sure

because maths rules etc etc

we never know, but thanks alot you're really fast

[lapis niche]

Just be careful, the x is still being multiplied by 0.25

[harsh zenith]

yea

[lapis niche]

It doesn’t become like a digit 😄

[harsh zenith]

yup

Thanks alot hehe

[lapis niche]

(0.25)x is how I usually write it, just to make it easier on the eyes

[harsh zenith]

.close

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[azure pasture]

hi, i'm working on continuity and I need some help explaining why this is DNE for this problem

[azure pasture]

the question is f(x) = |x|, a = 3

I do know that
f(3) = |3|
f(3) = 3

[flat whale]

Show the entire original question and answer

|x| is continuous so you're leaving something out

[azure pasture]

and i do know that then when we find the limit of x approaching 3 is going to be:
lim x -> -3 = |x| -> 2
lim x -> 3 = |x| -> 3

the whole question is to find out: f(x) = |x|, a = 3 is continuous or not

i think i'm confused because i don't know how it's DNE

[hard atlas]

who says that it is

[azure pasture]

my teacher's answer sheet

i just want to understand how he got that answer

[flat whale]

.

[azure pasture]

the entire original question is:

use the definition of continuity along with the properties of limits to show that the function is continuous at the number a
f(x) = | x |, a = 3

[flat whale]

Screenshot where the answer to that question says DNE

Or take a picture

[flat whale]

Yea no

[azure pasture]

is he wrong lol

[flat whale]

|x| is not the same as [[x]]

No you are

[azure pasture]

oh

can you please explain

[flat whale]

https://math.libretexts.org/Bookshelves/Combinatorics_and_Discrete_Mathematics/Elementary_Number_Theory_(Clark)/01%3A_Chapters/1.04%3A_The_Floor_and_Ceiling_of_a_Real_Number
[[x]] is the same as 1.4.1

[azure pasture]

got it, i just realized i never learned this

thanks for clarifying that

so after reading up on it a little more, i do understand the general idea of floor and ceilings, but why does the answer still not exist

is it because there are two answers technically

[flat whale]

Do you know what it's called when two one sided limits at one point aren't equal

[azure pasture]

no i haven't learned

[flat whale]

The limit does not exist

Or DNE for short

[azure pasture]

i understand

i want to know why tho

[flat whale]

Why what

It's the definition

[azure pasture]

is this ur explanation to why it's dne

[flat whale]

DNE covers more cases than that but yes

[azure pasture]

i dont mean to sound rude or annoying to u but i didnt understand ur explanation thats why

i just started my prof's lesson & he doesnt explain well either since he doesnt speak good eng so thats why i came on here to get a better explanation since i dont understand his reasoning

[flat whale]

https://tutorial.math.lamar.edu/classes/calci/onesidedlimits.aspx

[azure pasture]

but thanks for ur help

.close

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[vestal depot]

Hey could someone help me with this question I couldn't really find anything that would help me online

[cyan shadow]

what is going on in this photo?

what's the Australian pyramid mean

[lone linden]

Probably finding the consecutive differences

Quadratic sequence -> second differences are constant

[vocal sleet]

@vestal depot Has your question been resolved?

[vocal sleet]

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[dire coral]

cosec theta plus cot theta upon cosec theta minus cot theta is equal to cosec theta plus cot theta whole square is equal to 1 plus 2cotsquare theta plus 2cosec theta cot theta

[dire coral]

my handwriting is not v good sry

[lapis niche]

$csc\theta + cot\theta = (csc\theta + cot\theta)^2 = 1 + 2cot^2 \theta + 2(csc\theta)(cot\theta)$

[twin meteor]

Learath2

[lapis niche]

This is what I could make out, is this it?

[dire coral]

yess

[spiral inlet]

no, it isn't, right? Your first expression had a denominator

[dire coral]

oh op i didnt see tha

t

math is drivin me insane sry

[lapis niche]

$\frac{csc\theta + cot\theta}{csc\theta - cot\theta}= (csc\theta + cot\theta)^2 = 1 + 2cot^2 \theta + 2(csc\theta)(cot\theta)$

[twin meteor]

Learath2

[dire coral]

now its correct

[vocal sleet]

@dire coral Has your question been resolved?

[vocal sleet]

@dire coral Has your question been resolved?

[vocal sleet]

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[full vessel]

I know that the domain of f-1 is not just the doman and that something effects it but I am not sure what

could someone explain iy yo me

[halcyon wing]

,rccw

[formal rock]

@full vessel well

What do you know about domain and range?

[full vessel]

I know yhat the domain of f is the range of f inverse

[full vessel]

and that the domain and range usually dont effect each other…?

[formal rock]

Okay

What else?

[full vessel]

errrmmm

im not surr what ur looking for 😭

.close

[vocal sleet]

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[misty sky]

bruh

wait

im super confused on how to do these two problems

my teacher hasnt covered this 😭 can someone plz help

im so lost

[exotic mason]

You'll basically plug it into a graphing calculator or a spreadsheet (Excel, Google Sheets, etc) and tell it to do a linear regression.

[misty sky]

does desmos work

[exotic mason]

Probably

[misty sky]

okay let me get it rq

plug in the table right?

[exotic mason]

Yeah

[misty sky]

ok one sec

[karmic imp]

I don't think desmos can do linear approximations

[exotic mason]

https://help.desmos.com/hc/en-us/articles/4406972958733-Regressions

[peak axle]

Desmos can do linear regression but idk how to do it with desmos lol

That link ^^

[exotic mason]

Me neither, but there's a help article.

[misty sky]

its not

like

its not puttinj jn the dots

😭

[exotic mason]

Oh, it probably is. They're just ... very far up.

[misty sky]

wait nvm

YES lmfao found them

[karmic imp]

I see

[misty sky]

idk what tf to do w this information tho

[karmic imp]

Zoom in

Or change the scale

[misty sky]

i hate desmos

[exotic mason]

You can click the zoom fit icon to automatically adjust to the data.

[misty sky]

ill use one of the ones yall suggested

oh

ok

[karmic imp]

If you hit the wrench, you can set the scale

[karmic imp]

That last point looks way too wrong

[misty sky]

yeah

i put whats on the paper

[karmic imp]

But there are such things as outliers

[exotic mason]

The last point fits the data that was sent, yeah.

[misty sky]

idk what to do

i need to find the equartion of the line of best fit

[karmic imp]

Follow this

Type in y ~ mx_1 + b

[exotic mason]

Gotta be y_1 too

[misty sky]

where do i put the y_1

[exotic mason]

In place of the y

[misty sky]

OKAY we're gettign somewhere

[exotic mason]

Yup! And it did the thing. You've got the equation of a line it figured out.

Desmos is actually kind of good at this. Neat.

[misty sky]

how do i put it on paper

y=723.214x+31867.5

?

[exotic mason]

Yep

[misty sky]

im just plugging in y=mx+b 😭

really

OMG

and thats the answer?

[karmic imp]

Yes

[misty sky]

no way

[exotic mason]

Well, it does tell you to talk about what the slope means, and what the y-intercept means, and such.

But that's the equation for sure.

[karmic imp]

But not the answer to the full question

[misty sky]

okay let me put the equation

how do i interpret the slope and y-intercept in this situation 😭

[soft walrus]

Desmos has been a life-saver whenever I'm interpolating polynomials

[karmic imp]

Did you look at example 4?

[misty sky]

im so sorry usually id know more but my teacher didnt go over this

wait one sec

y=723.214(11)+31867.5
to find the average because its 11 years apart right

[exotic mason]

I've used it like twice ever. I'm a dinosaur I suppose. In grad school we used Mathematica and Maple, and then I latched onto Geogebra after I started teaching and never learned anything else.

[misty sky]

?

[exotic mason]

Seems right.

[soft walrus]

oh don't get me wrong Wolfram Mathematica is good too, and im ngl idek what Maple is

[misty sky]

39822.854

?

should i round

[exotic mason]

I did physics, and the theorists liked Mathematica and the Experimentalists liked Maple. I think it's more engineer-y.

[exotic mason]

If I were reporting the answer I'd round but it doesn't say you have to. shrugs

[misty sky]

umm ill just keep this answer

okay thank you so much

how do i interpret tho

idk how tf to do that

[exotic mason]

(Best practice would be to round, to be clear, because there's no way this process will get you meaningful precision to three decimal places.)

[misty sky]

since it's money i'd keep 2 decimal points right

.85

[exotic mason]

Sure, I suppose that works. Like I said, I'm not grading it so I don't know how weird your grader is gonna be.

So, what does the slope mean here, anyway?

[misty sky]

she grades weird asf so she'll find a mistake either way

OHH so i say like

723.214 per month???

no but that doesnt add up

[exotic mason]

It's rise over run, right? So it's a change in money, but how much time does going from 0 to 1 indicate?

In other words, what unit is x in?

[misty sky]

uhh

0 to 1 is 1 year

x is 11 because 24-13 is 11

IM SO confused

[exotic mason]

So it's not per month, it's change in tuition per year.

[vocal sleet]

@misty sky Has your question been resolved?

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[alpine mica]

Hey this was on a quiz and im looking as i do quiz corrections and wondering if this problem is even solvable

[crimson jetty]

likely typo

3, 7 isnt on it.

[spiral inlet]

they may have meant a tangent line that passes through (3,7)

but as written, no

[peak axle]

I guess they could have meant (3,9) but that’s just a wild guess

Or find a tangent line to x^2 that passes through (3,7) but I never remember doing a problem like that

[cyan shadow]

i bet its (3, 9) intended

[peak axle]

If this is calc 1 I’m guessing it’s supposed to be (3,9) but no idea lol

[vocal sleet]

@alpine mica Has your question been resolved?

[alpine mica]

so it was no solution otherwise yes?

This was on a quiz man they shouldnt have two errors since i have a problem just like this where it is not solvable

[vocal sleet]

@alpine mica Has your question been resolved?

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[terse forum]

$f\left(x\right)=\sqrt{-\left(x-4\right)}$

[twin meteor]

Akira 🍉

[terse forum]

Why does my teacher's answer sheet state that $f(4) - f(-5) = -8$ when Desmos calculates it as -3?

[twin meteor]

Akira 🍉

[karmic imp]

Show the answer key

[honest pollen]

-3 makes sense as an answer, it might be an error on your teacher's behalf

[terse forum]

3d

[karmic imp]

Could have been a horribly written 3 that looks like an 8

[terse forum]

no wait shit i think i got the wrong equation I mean im not exactly sure how would find an equation from graph

this one

[karmic imp]

Can you send a screenshot of the question you are doing? Like not just the graph but everything else relating to the question?

[terse forum]

sure

[honest pollen]

find f(4) and f(-5) on this graph, then subtract them
f(4) = -3
f(-5) = 3
-3 - 3 = -6 ...
Yeah... still not a -8

[terse forum]

for f(4) I found -3 and for f(-5) is 3 so my final will be (-3) -(3)= -6

i think there's something wrong I did

[honest pollen]

you sure this is the correct answer key? Even 3c looks off, teacher might've given you guys wrong answer key

[terse forum]

it's section 3

[outer warren]

its just a horridlywriten -6

[honest pollen]

that's a possibility too lol

[terse forum]

it looks like an eight for me

negative eight

[outer warren]

hence horridly

[honest pollen]

Email your teacher about it, it looks like they have made a mistake with the answer key

[karmic imp]

C and D don't match the key here, the blue text

[terse forum]

okay

[honest pollen]

you answers are correct, it's the answer key that's wrong for these questions

[terse forum]

my answers are written in blue do think it's right?

[karmic imp]

Yeah your answers look right so far

[terse forum]

so which one should I let him the answers are incorrect?

for c and d right?

am I doing it right or no

I think my teacher is just mistaken

thanks for the help though

.close

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[nocturne isle]

hii i need help with this

[nocturne isle]

this is my working

and no i cant find the ans

[dusty igloo]

@nocturne isle

[lapis niche]

Wouldn't just the two horizontal tangents work? 😄

Ah gradient 2

[dusty igloo]

k

[nocturne isle]

yes

like

itd so confusing

[lapis niche]

Implicit differentiation to the rescue?

[nocturne isle]

but this chapter isnt related to dydx

so i doubt thats the case

[lapis niche]

Hm, well it'd be easier with it 😄

Let me take a look at what you tried

[nocturne isle]

ok sure thanks

[vocal sleet]

@nocturne isle Has your question been resolved?

[lapis niche]

I don't get what you are trying to do with (b+3)/(a+5) = 2, a line through the center? but why would you want that to have slope 2?

You want a line through the center that's perpendicular to the tangent lines

[nocturne isle]

oh

yea i tried that

like gradient -1/2

and i still got it wrong

[lapis niche]

So $y + 3 = m(x + 5)$ with $m = -0.5$

[twin meteor]

Learath2

[lapis niche]

Then intersect the line with the circle

I get something like $(x+5)^2 + (\frac{x}{2} + \frac{5}{2})^2 = 80$

[twin meteor]

Learath2

[lapis niche]

Solved for x you get -13 and 3 as expected, so I think that's fine

I'll go get some sleep, good night

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[oak silo]

just wondering if someone could derive for a+(b/c) = (ac+b)/c, not sure exactly how it turns from one to the other

[restive rune]

hello I am new

[errant briar]

a+b/c = ac/c +b/c =(ac+b)/c

[royal kestrel]

^ multiply a by c/c to get ac/c

[royal kestrel]

if you need help w a question just go to #❓how-to-get-help

[oak silo]

ah understood ok

.close

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[boreal escarp]

i cant figure out whats wrong?

[honest pollen]

I can't see what's wrong either, after checking four different times I think nothing's wrong with your graph, could be something else

[vocal sleet]

@boreal escarp Has your question been resolved?

[boreal escarp]

.close

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[brave ruin]

So, from where I ended up, do you think it'd appropriate at this point to do a u sub?

I'm sure there was a point where I could have rewritten something earlier.

OR

did I start this problem wrong?

I wrote down in my notes from lecture that I won't always be doing this method

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[high saddle]

What does it mean to "show" it

If it's just f(x)

[pale perch]

say $f(x)=a_0+a_1 x+a_2 x^2+...+a_{m-1} x^{m-1}+a_m x^m$

[twin meteor]

AℤØ

[pale perch]

or something like that

similar with g(x)

when its the case that $m<n$ or such then it'd be better to write for $g(x)= b_0+b_1 x+b_2 x^2+...+b_{m-1} x^{m-1}+b_m x^m+b_{m+1} x^{m+1}+...+b_{n-1} x^{n-1}+b_n x^n$

[twin meteor]

AℤØ

[pale perch]

as long as you can show what you need to

[high saddle]

I know that the condition for f(x)/g(x) = 0 should be c/+-∞

How do i incorporate that

Sorry I'm really confused by this problem

I kind of don't understand why g(x) was written that way, why the m-1, m, m+1 then n-1, n?

[pale perch]

if you multiply the numerator and denominator by 1/x^n, then youll find that all the terms for f will tend to 0 since theyll all be some a/x^c, and in the denominator youll just be left with the coefficient of x^n, so itll be 0/b_n

[pale perch]

it was more to illustrate that n>m but i suppose its not strictly needed here

[high saddle]

Hmm

[pale perch]

b and c will follow a similar logic, just with different things being left over

at least thats one way to do it

[high saddle]

Okay, i have to go now so I'll close this. Thank you for your help though i still need digesting

.close

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[tidal cedar]

Hi i need help with (3x-4)(x-4)(x+5)

[tidal cedar]

i got 15x^3 -80x^2+80x but apparently it is wrong

[ornate ember]

can you show your steps?

[tidal cedar]

yes

also dramallama used to be my username!

[ornate ember]

hahaha that's really funny

[tidal cedar]

(3x^2-12x-4x+16)(x+5)

theb (x+5)(3x^2-16x+16)

[ornate ember]

looks good so far

[tidal cedar]

then x(15x^2-80x+80)

then 15x^3-80x^2+80x

i already did the 5

[ornate ember]

wait hold on

[rugged orchid]

(a + b)c ≠ a(bc)

[ornate ember]

yeah that

[tidal cedar]

when it became then x(15x^2-80x+80)

[rugged orchid]

(a + b)c = ac + bc

[tidal cedar]

?

i dont know what that is

[ornate ember]

In other words, it should be this:
x(3x^2 - 16x + 16) + 5(3x^2 - 16x + 16)

you need to still do FOIL

[tidal cedar]

didnt i already foiL?

when it became then x(15x^2-80x+80)

[rugged orchid]

You just multiplied 5 into the bracket

Doing a(bc)

a = x

b = 5

[tidal cedar]

yes then cant i multiply the x in now?

[rugged orchid]

c whatever is the right side bracket

You can’t

(a + b)c ≠ a(bc)

(x+5)(3x^2-16x+16) ≠ x(5)(3x^2-16x+16)

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[vast shale]

<@&286206848099549185>

[flat whale]

You really need to stop

[vast shale]

sorry i really dont understand what i did

[flat whale]

You've been told multiple times

!15m

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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.

[vast shale]

oh ok i undserstand now i will be back in 15 min

[flat whale]

@vast shale you also haven't even asked a question

Read the message

Don't open a channel for nothing for helpers to help with

[vast shale]

oh im very sorry

[vocal sleet]

@vast shale Has your question been resolved?

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[brave ruin]

If I wanted to factor the denominator here, how would you approach it?

[errant briar]

Is this an integration question?

[brave ruin]

Cause I guess I could do like (a^2+x^2)^1/2*(a^2+x^2)

[brave ruin]

it is yes

But I didn't think that part was important

[errant briar]

Substitute x=a tantheta

[brave ruin]

really

I thought this was a partial fraction decomposition problem

you're saying use this one right

[errant briar]

Yes

[brave ruin]

I can just rewrite the fractional exponent as a root

but I am still curious

would you be able to factor out the denominator ?

[errant briar]

Factoring won't work here, i think

[brave ruin]

gotcha

So, it would just be (2^root(a^2+x^2))^3

in the denominator

[errant briar]

Yes

[brave ruin]

okie

ty

nice

.close

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[sly sierra]

it's a decreasing function...

so on any interval of the form [x,infinity) it's bounded by the value at x

[thin vale]

you can tell a function is increasing without taking the derivative

if it has the property that when a<b f(a)>f(b)

the function is decreasing

maybe you can show that this function has that property

[sly sierra]

e^-a < e^-b if and only if e^(b-a) < 1 if and only if b-a < 0 if and only if b < a

[vocal sleet]

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[broken nimbus]

Hi there I was wondering if I could get some feedback for my explanation of the signifigance of the chain rule. The importance of the chain rule is that it allows us to find the input with respect to the input of the composite function within the function as oppose to the derivative being with respect to the output of the composite function within the function, which would not be useful as it would not be the original value we plug in for x, hence it doesn't provide any useful info as such.

[broken nimbus]

To summarize what I said if we have a composite function like f(g(x)) we want to know the instataneous slope with respect to whatever we input for x.

If the instataneous slope was with respect to g(x) its fine, however it doesn't provide any useful info as such as the instantaneous slope of the function would be with respect to the output of the composite function, and because we cannot control what the output of the composite function is it wouldn't provide any useful info as such.

[vast shale]

It's a very good explanation in my eyes, it's pretty clear.

[paper depot]

honestly overkill imo but im afraid to ask for more details.

[broken nimbus]

thank you!!

[vast shale]

However, you could rephrase it.
exempli gratia

The chain rule allows us to find the derivative of a composite function with respect to its input, rather than with respect to its output. This is important because it allows us to study the behavior of composite functions in a much more meaningful way.

[broken nimbus]

I tested it out with values and stuff.

[broken nimbus]

Ik, its overkill, but I just want to understand I fully understand it and can then make it more concise

im explaining it to myself

so overkill is fine.

[trim cedar]

whoever assigned you that is committing a crime

[broken nimbus]

nobody assigned me.

I love understanding things myself.

and so I rlly try to ensure I understand the concept yk. Like I can explain it to a five year old. Cus I find maths to be beautiful.

When there is a theorem or formula I try to elarn it. Although my knowledge is fairly little compared to many here, I am doing what I can.

Sure it takes extra time, but I'd rather ensure I understand the concept even if its overkill or more work. Idc, bcus I enjoy it and don't see it as work.

[meager hearth]

Chill man

No need to justify your love for math

[meager hearth]

Us bro

[vocal sleet]

@broken nimbus Has your question been resolved?

[broken nimbus]

💕

[vocal sleet]

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[dawn latch]

can i get some help

[dawn latch]

ik a is (-inf, inf)

but on b

dont i use y`

to find it

[honest pollen]

Yes

You need to find the derivative and see where it's equal to zero

[dawn latch]

can i get help finding the derivative

[honest pollen]

I can help with that

[dawn latch]

do u distribute the x^2/3

[honest pollen]

Yes

[dawn latch]

or do we use the derivative of it first

[honest pollen]

Distribute first

[dawn latch]

okay

so it wold be

idk wat it would be lol

[honest pollen]

lol

well

x^2/3 times x

what does that make

(x has a power of 1, or a power of 3/3)

[dawn latch]

x^5/3

[honest pollen]

Yep

so x^2/3 times 1/5 x

what does that make

[dawn latch]

x^5/3 / 5

[honest pollen]

correct

Now x^2/3 times the other number (1/2)

[dawn latch]

x^2/3 / 2\

oops ignore the \

[honest pollen]

yep, you got it right

so when you distribute you get x^5/2 / 5 + x/2

Do you know the power rule?

[dawn latch]

i forgot it

[honest pollen]

Ok, let me explain it

so say you have x^2

[dawn latch]

yea

[honest pollen]

and you wanted to know the derivative of that

[dawn latch]

ohhh

the nx^n-1

that thing

[honest pollen]

yepp

[dawn latch]

okayokay i remember

[honest pollen]

yeah, you could use this rule here to find the derivative

try it and lemme know what you get

[dawn latch]

hmm okay

so

its weird

but i got

5/3x^0/3 / 5 + 2/3x^-1/3 / 2

[honest pollen]

Yeah... idk what you did 😅
hold on...

Rewrite them this way

[dawn latch]

oh

[dawn latch]

i did not do that

LOL

[honest pollen]

Yeah, it's easier when you put the fraction out lol

try again and lemme know what you get

[dawn latch]

wait

so i just like multiply them

[honest pollen]

You multiply them, then subtract 1 from the power

It's easier to subtract 3/3 from the power (3/3 is the same as 1)

[dawn latch]

wait can i get a visual on that

idk how ur multiplying then subtracting 1

can u do it on just one so i can try the other

[honest pollen]

Yeah, first thing multiply the coefficient by the power, then subtract 1 from the power (I subtracted 3/3 rather than 1 to make it easier to subtract fraction)

I can help you with the (5/3)/5 and (2/3)/2 stuff as well

After you get this part though

[dawn latch]

okay so its (5/3)/5 x^2/3 + (2/3)/2 X^-1/3

[honest pollen]

correct

That's the derivative, you can simplify it more though

[dawn latch]

how do i simplify it more

[honest pollen]

Let me make a visual

might take a moment

[dawn latch]

okay

[honest pollen]

Okay, so first of all, multiplying (5/3)/5 by 1 should give you the same number

And we know that (1/5) / (1/5) equals 1

When you multiply both sides by 1/5, the bottom will be equal to 1

The top will be easy to find then

@dawn latch that's how you simply the fractions

[dawn latch]

o.0

wat about the x and the negative fraction exponent

[honest pollen]

Once you get the fractions set, I'll show you how to deal with negative exponents

(making another visual)

[dawn latch]

so do i simplify the fractions leaving out the x and their exponents for now

[honest pollen]

Yes, do the fractions first, don't change x

as long as the x is outside the fraction

[dawn latch]

si u git

so i got

5/15 and 2/6

(5/15)x^2/3 + (2/6)x^-1/3

[honest pollen]

Correct, try to simplify 2/6

and 5/15

[dawn latch]

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

[honest pollen]

correct

I'll show you how to get rid of the -1/3 next

[dawn latch]

okay

[honest pollen]

Now that the fractions are simplified, you can move the x^-1/3 in

[dawn latch]

and (x^2/3)/3

[honest pollen]

You can do that with the other x if you want, up to you

[dawn latch]

so i need to only do it to one

[honest pollen]

You don't need to do it to only one, but doing it here will help you get rid of the -1/3

I want you to think of this again

You can replace the 1 with x^1/3 over x^1/3

x^-1/3 times x^1/3 will give you x^0, and any number to the power of zero equals 1

[dawn latch]

wait wat LOL

wat about the 1/3x^2/3 one

[honest pollen]

That's another part of the equation, you can simplify every part individually, then put them together

[dawn latch]

oh i see

[honest pollen]

As long as they're separated by a + or - you can work on them individually

You should get something like this

[dawn latch]

wait

i thought

[dawn latch]

this was the answer

how did it go to the denominator

[honest pollen]

Look back at this image, multiplying it by 1 got you the same value

[honest pollen]

So I replaced the 1 with x^1/3 / x^1/3

[honest pollen]

And multiplied again

The top numbers in the fraction multiplied by top numbers, to get 1

Bottom numbers by bottom to get 3x^1/3

In other words... if you multiply an exponent by -1 you can move it to the other side of the fraction

[honest pollen]

They're both the same answer, just written differently

[dawn latch]

ohhh right

u moved it to the bototm to get rid of the negative

[honest pollen]

yepp

I just tried showing you why that method works

When you combine that with the other part of the equation you get this

[dawn latch]

okay i see

now how would we add that

lol

[honest pollen]

Both numbers have a common factor of 1/3

You can take out the 1/3 to simplify it more

1/3 ( x^2/3 + 1/x^1/3 )

And remember, you only need the critical points

in other words, a value for x that would make the equation equal to zero

[dawn latch]

wats next

[honest pollen]

See if you can plug zero in for x, what would you get

[dawn latch]

is this the final

[dawn latch]

or this

[honest pollen]

They're both the same equation, but this one is easier to work with

[dawn latch]

okay

imma try the 1/3 one

and just set it equal to 0

[dawn latch]

x=0 or the whole thing equal =

0

[honest pollen]

Yeah, look here, you want a value where this whole thing will be equal to zero

If the numbers inside the parenthesis equal to zero

then this equation will be equal to zero

[dawn latch]

i see

how would i get the other x

[honest pollen]

finding the x value or are you asking how I got the other x in the equation

[dawn latch]

finding the other x value

[honest pollen]

Look at it visually first, you know you want the signs to be opposite (one to be negative and another to be positive) to get zero in the parenthesis, so it has to be a negative number

Plugging zero in for x will give you a divide by zero error

So let's start with plugging -1 for x to get an idea of what we might get

Use a calculator if you'd like for this part

[dawn latch]

its because

i asked others wat they got

and they said x=0 and x=-1

im wondering how they got the -1

[honest pollen]

If x=-1, then plugging -1 in for x here would get you zero

[honest pollen]

but I'm not sure about the x=0, that doesn't seem like an answer

[dawn latch]

and another person agreed that they gor it

[honest pollen]

I graphed our derivative function, at 0 you have a divide by zero error so it can't be right

But -1 does seem to be the answer

[dawn latch]

oh

so theres only 1 crital point

[honest pollen]

hold on let me graph the original equation

-1 is the right answer... but...

[dawn latch]

im so lostttttttt

do u want to see an example

[honest pollen]

It seems that 0 is not defined in the derivative function but here it is in the original equation

[honest pollen]

ah right, I forgot about the limit stuff 🤦‍♂️

[dawn latch]

wait

did we mess up

[honest pollen]

Nah, you got it all right

x=0 is a critical point

x=-1 is one too

[dawn latch]

im confused on how we get x=-1

[honest pollen]

go back up here

[dawn latch]

okay i ahve this written

as the final

then the x=0 underneath

[honest pollen]

same equation as ours just written differently

[honest pollen]

yeah, it's the derivative, and if you tried plugging in points to find where it's equal to 0, you'd see that -1 would set it to zero so there's a critical point there

[dawn latch]

can u give a visual

for the -1 lol

[honest pollen]

yep jimmy a minute, I'll see what I can do

You want to see where the equation equals zero

[dawn latch]

yea

thats how we got x=0 right

[honest pollen]

The 1/3 is easy to get rid of at first

[dawn latch]

u divided it right

[honest pollen]

Kinda

[dawn latch]

divided 1/3 to both sides

[honest pollen]

yes

[dawn latch]

which will make it still equal to 0

[honest pollen]

Now, you want to see what values for x will set this to 0

Fun fact, x^1/2 equals sqrt(x)

Let me explain visually

[dawn latch]

lmao ty

yk me so well

[honest pollen]

So, this has nothing to do with the equation, but it's like a simple rule that's helpful

[dawn latch]

oo okay ty

[honest pollen]

If you had x^1/3, it'd be equal to the cube root of x

[dawn latch]

but we dont cuz its 2/3

[honest pollen]

x^2/3 can also be rewritten as (x^1/3) ^ 2

exponent to the power of an exponent

I mean–

lemme get a pic