help-17

[cyan talon]

that already specifies it as a function also

@hushed flint

[cyan talon]

sure that works

gg

@hushed flint

[hushed flint]

So for T(ax) = aT(x) do I just choose any value for a

[cyan talon]

wym you choose a value

you have to show that the equality is true no matter what a is

[hushed flint]

Oh

[hushed flint]

So can I just use a x vector

T(ax1, ax2, ax3) = aT(x1,x2,x3)?

Oh I think I did it, thanks !

🙏

.close

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[abstract moon]

how do i obtain x and y?

[abstract moon]

i don't understand where squareroot of 2/2 was taken

<@&286206848099549185>

[wide fiber]

In general memorize the unit circle

But...

r = 1

So what's the Pythagorean Identity?

[nova delta]

Is it just me, or is there something wrong?

[wide fiber]

What's wrong

[nova delta]

The length of the hypotenuse. Aren’t the legs usually 1 and 1 and the hypotenuse is sqt(2)/2

[wide fiber]

No

[tall patio]

why would they have length 1?

[nova delta]

Ok I had to make sure

That’s the other triangle

But yea memorize the unit circle

[wide fiber]

@abstract moon

There's a way to derive it

Do you still need help?

[abstract moon]

yes please

[wide fiber]

Ok

So what's the Pythagorean Theorem

[abstract moon]

c² = a² + b²

[wide fiber]

Ok

What is c in this case?

[abstract moon]

hypotenuse which is 1 unit

[wide fiber]

Ok

Would you agree that x and y are the same at 45 degrees?

[abstract moon]

x and y are the points right?

[wide fiber]

"x" is the x-axis line

"y" is the y-axis line

[abstract moon]

like the sides of the triangle?

[wide fiber]

Yes

[abstract moon]

yes

[wide fiber]

Ok

So plug in what you know

[abstract moon]

i still don't get it

[wide fiber]

c² = a² + b²

so c = 1

a and b are the same

so you can make it 2a^2 or 2b^2

@abstract moon what's the new equation?

[vocal sleet]

@abstract moon Has your question been resolved?

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[scarlet bough]

what am i doing wrong😭😭😭

[scarlet bough]

part c

<@&286206848099549185> ?

[tall lintel]

2.24?

for (a)

[scarlet bough]

bro none of the helpers know maths

can u do it?

[tall lintel]

oh i found the area under the first peak

nvm sorry

[scarlet bough]

did u try part c?

[vocal sleet]

@scarlet bough Has your question been resolved?

[scarlet bough]

<@&286206848099549185> pls

[terse cargo]

guys

im not a helper

idk how i got it

misclick or smth

pls remove it

[flat whale]

you have to DM mods

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[tropic anchor]

Need help with 2c

[tropic anchor]

<@&286206848099549185>

<@&286206848099549185>

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[vast shale]

How do I do part iii?

I attempted it

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@vast shale Has your question been resolved?

[vast shale]

<@&286206848099549185> please I need help

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[shrewd cape]

I'm just lost

[thin vale]

Do you know what the order of a sequence is

[hybrid flicker]

order of a periodic sequence yes, do you know what it means?

[shrewd cape]

How many times it takes for the sequence to repeat itself

@hollow rampart

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[scenic wharf]

i know this is probably an easy problem but im slightly struggling on this

[pallid zenith]

how far have you gotten? what have you tried? where'd you get stuck?

[scenic wharf]

i got as far as writing potential equations and attempting to solve for x

For permiter i wrote 6x = 120, and for area i wrote 2x^2 = 1512

i however could not solve for x

it kept being way to high or low

[plucky eagle]

I'm a dumbass but since the area is like 1512 ft^2 then wouldn't the sides (or X) be square root of 1512?

[scenic wharf]

let me try that ill get back to ya

no

[plucky eagle]

so it's like 38.8x38.8 feet

[jovial bloom]

no because he's building a rectangular pen

not a square pen

[plucky eagle]

real

[scenic wharf]

but im solving for x

and one of the sides is 2x

so x will be constant

or

sorry

[plucky eagle]

but 2XxXxXx2X is 1512sqft

[scenic wharf]

it will consistently be a rectangle

2x(x) = 1512

thats the equation

the label of square feet doesnt change the 1512

because it would be labeled as ft^2

but ft is a label and not part of the equation

therefore

[plucky eagle]

This seems like the easiest shit but I'm high rn so I'm not gonna bother

[scenic wharf]

doesnt change number

[jovial bloom]

so when you solve this for x what do you get?

[scenic wharf]

27.5

[jovial bloom]

shouldn't that be fine then?

[scenic wharf]

when i put that back into area equation it works, but it doesnt work for perimeter

the perimeter equation is x + x + 2x + 2x or 6x = 120

[jovial bloom]

i don't believe that's the correct equation of the perimeter

[scenic wharf]

why not?

its each side added up

and he has 120 ft of fence

[jovial bloom]

from the picture

[scenic wharf]

to use

[jovial bloom]

i mean i think the question is poorly presented but it seems to me that they assume you know that the other side of the pen is enclosed by the building

and not another length of fence

[scenic wharf]

oh you are right

wait one moment

[jovial bloom]

actually it says it in the question

[scenic wharf]

it does

i forgot

give me one second ill see if the equation works now

it does it is less than 120 ft

which works

[jovial bloom]

yes

[scenic wharf]

.close

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[scenic wharf]

thanks for the help

[jovial bloom]

no problem

[vocal sleet]

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[lament linden]

lmfao

[vast shale]

😭

[lament linden]

sadge

calculus BC optimization problems

idk how to model this

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@lament linden Has your question been resolved?

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[candid nest]

Hello! Im working on factoring cubics, I have this question here a^3-9(a^2)b+27a(b^2)-27b^3. Im lost on how to even start factoring this.

[vast shale]

I dont think its factorable?

[candid nest]

Oh, could this not be simplified?

[vast shale]

How did you get to that?

I dont think you can factor the first equation at all

[candid nest]

its the differences in cubes?

[cyan shadow]

It is very factorable

thats right, you pull out the a-3b from the first one

then you can simplify that second part as well

[candid nest]

even further?

[vast shale]

nvm then, my brain isn't in factoring mode today

[cyan shadow]

yes

[candid nest]

But its not a perfect square?

[cyan shadow]

is it not?

whats (a-3b)^2

oh

its not because you messed up the first step

[drifting jackal]

Not a^2 + 3ab + 9b^2

[cyan shadow]

yeah its supposed ot factor into that, lemme see what they did wrong to get to it

[candid nest]

oops

it's just a plug in formula tho?

[drifting jackal]

You can't even apply this

Because it's not a difference of cubes

[candid nest]

oh

[drifting jackal]

a^3 - b^3 is difference of cubes

[cyan shadow]

i mean the answer is that ||its just (a-3b)^3|| but I don't really understand their workings

[candid nest]

How am I supposed to solve this then?

*factor this

[drifting jackal]

For me, if you know that the a terms, the power starts at 3 and decreases to 0, and the b terms goes from 0 to 3, so to me, that indicates binomial expanision

[candid nest]

teacher says its a difference of cubes,

just needs to be applied and formatted

[drifting jackal]

It's not a difference of cubes

To apply, sum/difference of cubes, it should be in these forms

[candid nest]

Then how should I solve this??

[drifting jackal]

You should ask your teacher for clarity

[candid nest]

.close

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[shut zinc]

i am trying to find the null space of this 3x4 matrix. i got one row to go to zeros but i cant figure out how to continue

[shut zinc]

<@&286206848099549185>

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[shut zinc]

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[cinder oyster]

hello

[cinder oyster]

So

I believe the center for this one will be -3,0

And for radius

I will do y^2=52 and get sq root

But here is problem i get

It says im wrong sadly

This should be the answer they said

.close nvm the teach said it was mistake

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[worn goblet]

I just want to know what R is

[worn goblet]

so what the bounds would be

what would y be?

0 <= x <= 2

[vast shale]

have ur first bounds be dependant on y

cuz otherwise u gotta split into 2 integrals

[worn goblet]

yeah

thats the point

[vast shale]

?

what does y vary from

oh

u want 2 integrals?

[worn goblet]

well

one integral (the first one) in terms of x

0 to 2

[vast shale]

why

???

do u need to have 2 seperate integrals or something?

[worn goblet]

yes

[vast shale]

okay thats weird but alright

[worn goblet]

look at the top

[vast shale]

no

i mean

[worn goblet]

of the image

[vast shale]

LIke

2 double integrals

[worn goblet]

oh

no

[vast shale]

okay

[worn goblet]

just one

[vast shale]

so why are u doing it like this

[worn goblet]

wym

[vast shale]

make it dxdy

not

dydx

[worn goblet]

why

why not dydx

[vast shale]

.

[worn goblet]

oh

ok...

[vast shale]

yeah

okay

[worn goblet]

wouldnt i still need to find the bounds of y tho

[vast shale]

so what does y vary from

what does y vary from

[worn goblet]

in terms of x?

[vast shale]

no

numbers

[worn goblet]

0 - 1

[vast shale]

okay

now what does the x value start at?

[worn goblet]

0

[vast shale]

whats the equation of x on the left side?

in terms of y

[worn goblet]

x = y+1

y+1

[vast shale]

urm

x=y+1 means the equation is x-1?

the y intercept isnt -1?

[worn goblet]

whaaa

the line i mean

the line on the right side of this rhombus

is bounded by the graph y=x-1

[vast shale]

what is the equation

of the line

circled in red

[worn goblet]

oh

y=x

[vast shale]

yes

[worn goblet]

i thought you meant the one on the right

mb

[vast shale]

what about the other line

thats the ending

[worn goblet]

y=x-1

[vast shale]

in terms of y

[worn goblet]

x=y+1

[vast shale]

yeah

now ur done

just solve the double integral

[worn goblet]

so 0 <= y <= 1

wait what

wait

im still confused

how would i arrange the bounds

for x

[vast shale]

D = [y, y+1] x [0, 1]

[worn goblet]

oh

y y+!

y <= x <= y+1

right

[vast shale]

yes

if u did dydx u would have 3 double integrals btw

[worn goblet]

i see

weir

weird*

ok

thank you

let me check if its right

[vast shale]

np

[worn goblet]

um

wait

like this?

[worn goblet]

^^^

[vast shale]

oh

i dont think that line

is y=x

gimme a sec lemme plot it

yea those arent the right equations

u gotta actually do the y=mx+b stuff

and figure them out

[worn goblet]

wym

here one sec

so the points

are

(0, 0)

(1, 0)

(2, 1)

(1, 1)

[vast shale]

oh wait it is

[worn goblet]

yeah...

but the answer was wrong

i have more attempts

so its ok

[vast shale]

1 sec lemme do it myself

[worn goblet]

im pretty sure it should be dydx

only cuz 90% of my textbook does dydx

and problems ive done earlier have done dydx

[vast shale]

yeah but that doesnt make sense cuz then u need two double integrals bec ur y bounds change half way

can i see the full q?

[worn goblet]

mk

i thought i could skip the transformation

but im gonna do it rn

fucking jacobians bruh

i dont want to do that shit

ah whatever

[vast shale]

wat result did u get here?

[worn goblet]

-1/4

ok so

u = y-x

v=y

how do i even do this transformation tf

[vast shale]

wait why is -1/4 wrong

[worn goblet]

let me retry it

im pretty sure i have inf attempts

[vast shale]

,integral from 0 to 1 integral from 0 to x of y^2-xy dy dx + integral from 1 to 2 integral from x-1 to 1 of y^2-xy dy dx

how do i do wolfram bot

[worn goblet]

oh shit its right

[vast shale]

💀

[worn goblet]

also idk

lmfao

what on earth did i input the first time

[vast shale]

bro i was redoing my work for the last 10mins

[worn goblet]

im sorry

[vast shale]

figuring out why tf that was wrong

[vast shale]

lmao all good

[worn goblet]

lol

[vast shale]

this is the alternative way btw

cuz u wanted to do dydx

[worn goblet]

yeah i see

multiple double integrals

[vast shale]

yea

[worn goblet]

i understand

well thank you

[vast shale]

np

[worn goblet]

.close

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[obsidian creek]

"determine whether the function represents a linear or non-linear function"
this is just simple y = mx + b, i'm just not sure how to put it in slope-intercept form

[vast shale]

what is slope intercept?

y=mx+b ?

or y-y1 = m(x-x1)?

[obsidian creek]

y = mx + b

OHH okay wait i think i understand it now, if you can just write -x as -1 can you rearrange it to be y = -1(x) + 3 cubed 8 ??

[vast shale]

3 cubed 8????

but yeah the rest is fine

cubed root of 8 is 3

so its just

y=-x+3

[obsidian creek]

my bad, cube root of 8

okay thank you thank you

.close

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[valid socket]

We can't say c is true because 0.1 as an epsilon is not close enough, right?

[vocal sleet]

@valid socket Has your question been resolved?

[flat whale]

Correct

[valid socket]

What's the type of area called where we don't account for the fact that area of a function below the graph is negative. I know geometric area is the one where we do consider it as negative.

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[worn goblet]

help please

[worn goblet]

obv you can rewrite the functions as u=1, u=3, v=2, v=3

i know that

.close

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[vast shale]

how me do

no know how to start

i tried doing f(10) then was like wait a min

now stucc

chatgpt is lost

[regal bane]

Note that if I take g of both sides, I get:
x = g(-x³ - x)

Can you differentiate that?

[vast shale]

ok thx

.close

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[vast shale]

wait no i cant

.reopen

[vocal sleet]

✅

[vast shale]

er like

i can but no see how that helps me

[regal bane]

Well, what did you get?

[vast shale]

g'(-x^3-x)= 1/(-3x^2-1)

[regal bane]

Yep that looks right

Is there any value of x that would set
-x³ - x = 10?

[vast shale]

hmmmmmmmmm

oh shit

im so braindead

i tried that earlier but kinda ignored the negative symbols

ty

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[molten ocean]

plshelp

[molten ocean]

PLEASE HELP

[raven owl]

Bro this is a 3 second google search

The 2nd option

[hasty pulsar]

First one frfr

I love plotting the x^3+arctanh(x) graph

Best record of news ever

[molten ocean]

the 2nd one?

i clicked the third one

help please

[vocal sleet]

@molten ocean Has your question been resolved?

[vocal sleet]

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[vocal sleet]

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[nocturne anvil]

<@&286206848099549185>

[pale perch]

!15m

[vocal sleet]

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.

[pale perch]

whats the issue

[nocturne anvil]

No one wanna come😭and I only used it once

[pale perch]

i meant with this

[nocturne anvil]

How to solve it😭

[pale perch]

just look at it, what is the pattern for x

[nocturne anvil]

Idk

[pale perch]

what happens when you move right one space

[nocturne anvil]

We deduct one

[pale perch]

then fill in the blanks

[nocturne anvil]

What abt this?

@warm bough

@vocal sleet

@reef grove

[wide fiber]

How far apart are AB on the x-axis?

How far apart are AB on the y-axis?

[nocturne anvil]

Ehh I don’t understand

[wide fiber]

Count

[nocturne anvil]

On the y axis b= 8 and a=5

On the x axis m=3 and a= 5

[wide fiber]

Try this instead

What point is A at?

[nocturne anvil]

9

[wide fiber]

In (x, y) notation

[nocturne anvil]

X= 5 y= 4

[wide fiber]

X is correct

How’d you find the y?

[nocturne anvil]

Are you talking about the first one or both

[wide fiber]

Both

Find the coordinates for both points

[nocturne anvil]

Y = -2

[wide fiber]

Remember… (x, y)

• x = Running to the ladder
• y = Climbing up or down

Perfect put it in (x, y) now

[nocturne anvil]

Y = -2

Is the first graph for the x axis and the second for the y axis?

[wide fiber]

No just ignore the different graphs for now

Plot point A and point B

(x, y)

[nocturne anvil]

A(x=5, y=-2)

[wide fiber]

A(5, -2) but correct

Same with B

[nocturne anvil]

B(x= -3, y= 8)

[wide fiber]

Perfect

Now which one is farther to the right?

[nocturne anvil]

A

[wide fiber]

Ok

#1 - The top left graph is asking the distance between A and B

You can look at it two ways

• Simply count the distance on the x-axis and y-axis
• Formula: (x2 - x1), (y2 - y1)

[nocturne anvil]

So which one is 1 A or B?

[wide fiber]

A is x2, B is x1

You can also just count though

[nocturne anvil]

I got x=8

[nocturne anvil]

And y= 16

[wide fiber]

Doing what?

[nocturne anvil]

Using the formula

[wide fiber]

Show me

[wide fiber]

X is correct

Why did you multiply y?

[nocturne anvil]

Ohh okay I got 6

[wide fiber]

-2

Subtract 8

[nocturne anvil]

Ahh I got -10

[wide fiber]

Awesome

But it’s asking distance so both are positive

[nocturne anvil]

Ahh how?

[wide fiber]

So #1 is 8 on the x-axis and 10 on the y-axis

Do you see how you could get that simply by counting

[nocturne anvil]

Ahh ok

[wide fiber]

?

[nocturne anvil]

Na I did the calculation

[wide fiber]

Yes try counting now

Anyways I gotta go so let me help you wrap this up

#2 is the midpoint, so just half both numbers

X = 8/2
Y = 10/2

[nocturne anvil]

X=6 and y =8?

[wide fiber]

Not subtracting

Dividing

If you had 8 slices of pizza and 2 friends, how many slices does each get?

[nocturne anvil]

X=4 and y = 5

[wide fiber]

Perfect

Now the easiest way to find M is counting

[nocturne anvil]

But why am I dividing 😭

[wide fiber]

Because you’re trying to find the midpoint

So the middle is half

[nocturne anvil]

Ahh

[wide fiber]

Ok so the easiest way to find M is counting

But you can look at it with a formula

A is (5, -2)

You’re moving left 4, and up 5

So (5-4, -2+5)

[nocturne anvil]

1, 3

[wide fiber]

Awesome

Do you see how you could’ve counted the small lines on the graph to find all those?

[nocturne anvil]

Yh

[wide fiber]

Alright

All good?

Can I close ticket

[nocturne anvil]

Na so what’s the answer?

😭

[wide fiber]

(1, 3)

[nocturne anvil]

Ayy

Bet thanks

[wide fiber]

Can I close ticket?

[nocturne anvil]

Yh

[wide fiber]

I’m gonna send you a resource I used from Algebra - Calc 3

Best YouTuber out there

[nocturne anvil]

Aight

[wide fiber]

.close

[vocal sleet]

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[alpine storm]

Consider a linear mapping T from the normed space U to the normed space V. We say that T is bounded if there exists some c >= 0 such that norm(Tu) <= c norm(u) for all u in U

[alpine storm]

Now my question is, considering this definition, can we ALWAYS make any linear mapping T bounded, by just choosing the right norm?

ive seen examples where the 1-norm makes T unbounded, while the infinity norm makes it bounded

[vocal sleet]

@alpine storm Has your question been resolved?

[cyan talon]

With finite dimensional U and V, you have that all norms are equivalent, so the boundedness of T shouldn't depend on the norms you take @alpine storm

[alpine storm]

how about in infinite dim

[paper depot]

something something derivative operator

[vocal sleet]

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[cursive fern]

can anyone help me with this probleM? Im stuck and i only did the kinematic diagram

[vocal sleet]

@cursive fern Has your question been resolved?

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[ivory violet]

can any1 explain why contradicitng means parrarel
-why not just randomely not parrarel and in any direction

[ivory violet]

.close

[vocal sleet]

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[ivory violet]

nvm

[vocal sleet]

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[ivory violet]

how would you do this Q

[ivory violet]

i got up to 9z = 0 on line 3 and 4

and -y + 3z + w = 0
and 2x + 2y + 2z = 0

[vocal sleet]

@ivory violet Has your question been resolved?

[vocal sleet]

@ivory violet Has your question been resolved?

[radiant pasture]

Hi @ivory violet, which method are you trying to use to solve this?

[vocal sleet]

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[vast shale]

.close

[vocal sleet]

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[paper grotto]

How do i do this?

[warped gazelle]

find the two graph y = f(x), y = g(x)

and make inequalities like y <= f(x) or y >= f(x)

and connect them with AND or OR

[vocal sleet]

@paper grotto Has your question been resolved?

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[tough lily]

I used the cas calculator to figure this out and it says it's undefined when the answer page says it's -4 X_X

[tough lily]

is this right?

[vast shale]

it wants you to find the limit

not the function evaluated at that point

[tough lily]

ooh.

how do I find the limit?

do I have to use the quotient law or is there another way

[vast shale]

how do u factor x^2-6x-7

[tough lily]

idk..

[round plover]

it says "by tabulating" in the question, so I think it just wants you to test points close to -1

[tough lily]

I think the closest i could get it to -4 is this

idk if it's right

[round plover]

so the limit is -4

[tough lily]

but I was supposed to know it's -4 without looking at answers

whenever it says tabulating am i supposed to assume that i need to plug in numbers

[round plover]

tabulating as in, make a table of values

otherwise you can try simplify the polynomials

[vocal sleet]

@tough lily Has your question been resolved?

[tough lily]

wait so here I simplify the polynomials?

[round plover]

it says tabulating again, but you can try that yes

[tough lily]

it doesnt have the cas sign like the other did

so I'm assuming i'm meant to do this by hand

[vocal sleet]

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[tough lily]

.reopen

[vocal sleet]

✅

[tough lily]

it worked by just substituting but i'm still kinda unsure about the first one

like how would I know it's -4

and not like -3

[round plover]

how did you do it by substituting

if you put in 2 or -2, the denominator is (+/-2)^2 - 4 = 0

[tough lily]

first one doesn't exist

since it comes out to 8/0

second one comes out to 0/0 so the answer is just 0

[round plover]

that's not how limits work

that doesn't work in general

[tough lily]

does it not

why did the book say that its the answer..

😭

[round plover]

just because it matches the answer doesn't mean the method works

substituting the values in and getting something like 8/0 could mean that the limit is infinite

or that it doesn't exist

0/0 gives even less information

the limit could be any real number, +/-inf, or DNE

[tough lily]

what's the method to be used then?

[round plover]

factorise the polynomials

[tough lily]

how do I factorise 😭

i forgot

[round plover]

the denominator is difference of two squares

https://www.youtube.com/watch?v=3kE_Q-JdlO4

[tough lily]

ty ill watch this

[tough lily]

(x-2)(x+2)??

idk

[round plover]

you can expand the brackets to check

[tough lily]

expand the brackets?

[round plover]

multiply them out

[tough lily]

-4

what do I do with -4

[round plover]

where did the x's go

[tough lily]

dont they cancel out?

[round plover]

no?

[tough lily]

lmao

what

x- and x+

isnt that just 0

or is it just x-4

[round plover]

use the FOIL method if you aren't familiar with expanding brackets

multiply them out

(x-2)(x+2)

Firsts, so x times x is x^2

then Outer, x times +2 is +2x

Inner, -2 times x is -2x

Last, -2 times +2 is -4

so the whole thing is x^2 + 2x - 2x -4 = x^2 - 4

[tough lily]

aaaa

i get it

i didnt know that meant foil

[round plover]

so yes, x^2-4 factorises to (x+2)(x-2)

now factorise the numerator, x^2 -4x + 4

[tough lily]

isnt it the same?

(x+2) (x+2)

[round plover]

yes

so we have $\frac{(x+2)(x+2)}{(x-2)(x+2)}$

[twin meteor]

Desync

[round plover]

simplify it

[tough lily]

i forgot how i even got it

wait

wait why is it not (x-2)(x+2)

[round plover]

oh I didn't check what you said

(x+2)(x+2) doesn't work, yeah

(x-2)(x+2) also isn't quite right

[tough lily]

is it ever (x+2)(x+2)?

in what cases would it be?

[round plover]

you can expand the brackets again

you get x^2+4x+4

but the numerator in this question is x^2**-**4x+4

[tough lily]

i dont know

[round plover]

as per the video, find two numbers that multiply to 4 and add to -4

[tough lily]

2?

2 and -2

[round plover]

2 times -2 is -4

2 plus -2 is 0

you want them to multiply to +4 and add to -4

[tough lily]

-2 + -2

[round plover]

yes

[tough lily]

aa

[round plover]

so the factorisation is (x-2)(x-2)

[tough lily]

yeah

wait

why -4

why does it have to add to -4

[round plover]

because the coefficient of x in the polynomial is -4

just expand the brackets out

[tough lily]

aaa okei

x^2-2x-2x+4?

[round plover]

yes

[tough lily]

ooh

it comes back to the original

answer

i mean

thingy

lmao

polynomial

[round plover]

so we have $\frac{x^2-4x+4}{x^2-4}=\frac{(x-2)(x-2)}{(x-2)(x+2)}=\frac{x-2}{x+2}$

[twin meteor]

Desync

[round plover]

now, plugging in 2 is valid, because we don't get a zero denominator

and we get 0/4, so the limit as x->2 is definitely 0

whereas in the original fraction, we had 0/0 which is an indeterminate form

lim x->-2 is still problematic, but you can still consider what the fraction does as you approach -2 from the two different sides

if x is getting closer to -2 from the positive side (x=-1.9, x=-1.99,...), then the numerator is negative

because it's x (which is negative), take away 2, to be even more negative

and the denominator is very small but positive

and because you're dividing something negative by something very small and positive, the fraction as a whole becomes very large and negative, so the right hand limit goes to minus infinity

from the left side, the numerator is still negative, but the denominator is now very small and negative, so we have something negative being divided by something very small and negative, so the fraction as a whole is very large and positive

so the left hand limit is positive infinity

because the left and right limits don't match, the limit as a whole does not exist

[tough lily]

rightt

I think I get it now? i think i just need more practice

i have a mid term next week friday so ill prob be back

on here if i need help

thank u tho <3

[round plover]

nw

remember to .close your channel if you don't have anything else for now

[tough lily]

yess

tyty

.close

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[vivid sapphire]

hi

[vivid sapphire]

guys I am not understanding this

Degree 4. Roots of multiplicity 2 at x = -1/2
and roots
of multiplicity 1 at x = 6 and x = −2. y-intercept
at (0,18).

I did this question like 18=c(x-1/2)^2 (x-6) (x+2)

and solved for C

but apparently I was wrong

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[haughty karma]

3.14159265359

[vocal sleet]

@haughty karma Has your question been resolved?

[vast shale]

did you need help? @haughty karma

[vocal sleet]

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[dreamy summit]

hello

i need some help to the domain of this function

[vocal sleet]

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[fierce lake]

Hello

I cant seem to get the answer of e^a of this limit. I keep getting infinity. Im using L, Hopitals rules and since it is in a discrimate form im taking the ln of both sides.

[fierce lake]

Ignore the last 2 lines

[lapis marten]

First to second line is wrong

It should be A^-1 * x

[fierce lake]

AH

[lapis marten]

Or just take A out of the limit

[fierce lake]

did i mess up on the derivative?

aaaaaah

i seee

[lapis marten]

No, but it was already incorrect

[fierce lake]

what was?

[lapis marten]

The functions you took the derivative of

[fierce lake]

wdym

this is wrong?

[lapis marten]

You took the derivative of Ax^-1 instead of A^-1 * x

Green circle is already wrong

[fierce lake]

i see

i see

let me try something here

Is this right?

[lapis marten]

The function is f(x) = x * (1/A), so the derivative should be 1/A

[fierce lake]

i dont understand

[fierce lake]

the original function is in green no?

[lapis marten]

You took the derivative with respect to A instead of with respect to x

[lapis marten]

The first line is correct btw

[fierce lake]

OOOOOh

okok

i understand now

wow im just rushing

thank you for helping

.close

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[whole geode]

I am trying to multiply and divide rational expressions and I am stuck on one problem. The first picture I sent is the equation.

The second picture I sent is what I simplified the equation up to. The equation is able to be simplified even more because I looked up to find the answer to check my answer and it’s simplified up to x-7/3x but i have no clue how they got that. How do I simplify it even further?

[whole geode]

the pictures won’t send so i’m just going to type it out

[tidal umbra]

i cant see none the of the equations

[whole geode]

nevermind they’re sending

[tidal umbra]

can you factor 2 from (2x+6)?

[whole geode]

i don’t think so that’s what i thought

[tidal umbra]

im telling you to factor it out

[whole geode]

i already tried that and it didn’t work. i use desmos to see if the graphs align and it doesn’t work

i can try again

if i factor the 2 out would it become (x+3)(x+3)?

[tidal umbra]

,, \frac{2\cancel{(x+3)}(x-7)}{6x\cancel{(x+3)}}

[twin meteor]

!Yajat!

[whole geode]

one moment

OMG now i get it

so does the 2 just go away because you factored it out?

wait no because

[tidal umbra]

it cancels the 6 in the denominator

and you get yout answe

r

[whole geode]

so you get 3x from dividing 6x by 2?

from the 2 in the numerator?

[tidal umbra]

,, \frac{6\times x}{2}

[twin meteor]

!Yajat!

[tidal umbra]

what do you think?

[whole geode]

one moment i’m looking through my notes

[tidal umbra]

what notes?

its simple division

[whole geode]

my math notes

and i know what 6 divided by 2 is in just confused why the 2 is still there after you factor it out of 2x+6

i mean this is why i’m here, i’m not good at math lmao

[terse cipher]

not everyone is great but everyone can try to be good

[whole geode]

i understand but yeah i’m trying, that’s why i’m asking help here

[fleet blade]

thw 2 times x+3 tome make 2x+6

[whole geode]

okay i think i see it now thanks

.close

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[languid wraith]

👍

[full hatch]

👍

[haughty narwhal]

👍

[vocal sleet]

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[tepid nymph]

how to solve this?

[nova delta]

1. Convert your given angles to standard degrees.
2. All triangles equal to 180 degrees. Subtract the degrees you have from that
3. Convert number 2 back to minutes and seconds

Or you can just subtract 180 degrees from ABC and ACB and remember that you’re working with a 60 limit instead of a 100 limit for each decimal place

[vocal sleet]

@tepid nymph Has your question been resolved?

[tepid nymph]

Okay but how im gonna do that i dont get the process

[nova delta]

Do you know how to convert minutes to degrees?

[tepid nymph]

yes

but 17 mins cant turn to a degree

oh wait

[vocal sleet]

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[tepid nymph]

im supposed to add ABC and ACB and im going to subtract the answer from 180?

.reopen

[vocal sleet]

✅

[tepid nymph]

i think i got it

but whats the releationship between ABC, ACB and BAC?

[vocal sleet]

@tepid nymph Has your question been resolved?

[tepid nymph]

okay i got it now thanks

.close

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Remember:
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[snow path]

the ones with green boarders are correct

[snow path]

am i fucking up brackets or something?

[obsidian stream]

you are probably missing absolute values if that's the error you are getting

I don't want to actually to solve any differential equation so I didn't check anything though

[snow path]

oh god

this is my final try and im so over this. thank you. hopefully you're right

[obsidian stream]

well, like I said I didn't check anything

[snow path]

i do not blame you

whatsoever

[obsidian stream]

but you have a logarithm and without absolute values you are changing the domain

[snow path]

what sick fuck would want to

yeye absent minded mistake

[somber portal]

assuming you're integrating the secant, you're indeed missing the absolute value on the log.
Should be ln(|sec7x+tan7x|) i believe

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[vale kindle]

The instructions are to “resolve the non-linear inequality”. I’m also asked to express the solution using interval notation and graph the solution.

I know the correct answer is the opposite of mine

[regal bane]

Use an arrow → for "this step logically proceeds to this step"

[vale kindle]

Okay

[regal bane]

= means "the number to the left and right are the same"

[vale kindle]

Should I do it again?

[vale kindle]

Right

[regal bane]

No that's just a notational thing lol. I can still read your work

Okay so
-6 / (2x + 3) > 0

That's true when 2x + 3 < 0

Because you need to cancel out the negative on top

We could also divide the negative out and instead solve:
6 / (2x + 3) < 0

This might be why you're getting exactly the wrong sign

[vale kindle]

I don't understand

I'm looking at the procedure offered by Photomath. They have -6/(2x+3)>0 and then "divide both sides of the inequality by -6 and flip the inequality sign" to get 1/(2x+3)<0

But I've never heard something similar

<@&286206848099549185> I'm sure this is easy, but I don't get it

[vocal sleet]

@vale kindle Has your question been resolved?

[regal bane]

Whenever you divide/multiply both sides by a negative, you have to flip the inequality

[vale kindle]

OMG

OMG

OMG

I figured it out

I figured it out

No

No

No

I got it

.close

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[sterile pecan]

i need help i need to find the x intercepts of 3(x-5)^2 -4 = -1 by factoring

[sterile pecan]

i am here and i don't know what to do with the parenthesis and my notes are confusing me

[raven owl]

You want to solve for x, so start by adding 4 to the right

Usually it helps to do PEMDAS backwards when solving for a variable like x, so start by doing addition/subtraction, then multiplication/divison, then exponents, then parenthesis

[sterile pecan]

that would make sense but i have to make it =0 at the end

that's what my notes say

[raven owl]

Can you send your notes? If that was the case you would be done at step 2 of your work

[sterile pecan]

this one already has a 0 at the end

so i can start with subtracting 8

[raven owl]

You can do the same thing in your problem. The right side being 0 doesnt matter, the goal of the problem is to find what x values make the equation true

[sterile pecan]

no this is something else

i have to find the x intercept(s) and the vertex, and graph it

with factoring

and i have a like $150 math book but idk what type of math this is so i can't look it up in that

[raven owl]

Is there like an original text you got this problem from, might help if I can see the actual question

[sterile pecan]

i already did 5

[raven owl]

Ok lets start by finding the vertex because its close to being in vertex form

[sterile pecan]

ok