help-36

[tranquil pine]

It should be

[limpid torrent]

👍

.close

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[limpid torrent]

don't know what I'm doing on this one

[final saddle]

@limpid torrent Has your question been resolved?

[native spindle]

righto, do you know how to integrate using polar coordinates?

[limpid torrent]

1/2 integral from a to b r^2 dtheta

[native spindle]

nice. Now, what's the area of the white portion above the x axis and below the curve?

do you know how to calculate that?

[limpid torrent]

I'm guessing its seperate integrals with 0 to pi + pi to 2pi and 2pi to 3pi?

[native spindle]

you could do that

and you're on the right track

but there's a better way

[limpid torrent]

3pi/2

to 3pi

[native spindle]

maybe go from 2pi to 3pi

and then subtract the area from 0 to pi

[limpid torrent]

yeah that sounds better

[native spindle]

give that a try

lmk what you got from it

[limpid torrent]

wouldn't it just be from 2pi to 3pi?

[native spindle]

no, because that integral on its own would include the white portion under the curve and above the x axis

[limpid torrent]

so 3pi^3

B

[native spindle]

yep

that's what I got

here's an image explaining the regions you integrate over

when you integrate over theta, the area you are computing always comes from the origin

imagine you are trying to colour an image

but your pen is tied to a nail in the center of the page

that's integration using polar coordinates

it's a gross simplification, but I'm sure it gets the idea across

@limpid torrent in case you're not here rn

[limpid torrent]

ok thanks

[native spindle]

nw

was it correct?

[limpid torrent]

not sure yet

didn't finish the rest of the assignment

[native spindle]

I see

good luck with the rest of it

[limpid torrent]

but thank you

[native spindle]

no worries

[limpid torrent]

👍

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[obsidian veldt]

Given are two parallelograms ABCD and AECF with common diagonal
diagonal AC, where E and F lie inside the parallelogram ABCD.
Show and Proof:
The circumcircles of the triangles AEB, BFC, CED and DFA have a point
in common.

[obsidian veldt]

The centre of the circumcircle of an obtuse triangle lies outside the triangle.

I drew a sketch and recognised the parallel sides and equal angles, maybe this will work?

Is someone so friendly, and could help me to find a solution?

[long mesa]

hi

help ples hihi

[tired walrus]

@long mesa open your own channel. read #❓how-to-get-help for instructions.

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[obsidian veldt]

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[tender surge]

Hey have a question about the domain in this expression. 𝑔(𝑥) = sin(𝑥 +𝑎)+𝑏. Because there are no numerical values stated would that mean that the domain could be anything? If so, then would I write (-infinity, infinity)

[tulip coyote]

Generally if no domain is given, it's implied to be "wherever it makes sense"

[tender surge]

Oh ok, i'm a little confused because the math problem asks for two values. It allows me to input infinity in as a value, therefore would I just put in infinity and negative infinity?

[tulip coyote]

Personally I would - do you have a picture of how the entire problem looks?

[tender surge]

Yeah sure do

One sec

[tender surge]

Got it

[tulip coyote]

This one or another one? That doesn't seem to match up to your original question?

[tender surge]

My bad it sent the wrong one through

[tulip coyote]

Ahh yea I see - yea, $(-\infty, \infty)$ or all of $\bR$ would be fine imo

[soft zealot]

@tulip coyote

[tulip coyote]

Makes sense for any choice of real number

[tender surge]

Thanks for that

That solved it

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[drowsy hawk]

How do I solve b

[rich tide]

Are those comma separated numbers the coordinates?

[warped spear]

pretty sure it’s decimals

[drowsy hawk]

Decimals yea

[rich tide]

okay

[drowsy hawk]

I live in Europe xD

[rich tide]

are you familiar with law of cosines

[drowsy hawk]

Yea this chapter is teaching us that

[rich tide]

That's what will get you the answer

[drowsy hawk]

But don't I need to know an angle to use it

[final saddle]

@drowsy hawk Has your question been resolved?

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@drowsy hawk Has your question been resolved?

[final saddle]

@drowsy hawk Has your question been resolved?

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@drowsy hawk Has your question been resolved?

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@drowsy hawk Has your question been resolved?

[timid peak]

this guy has been waiting for an answer for like 9 hours

[cinder linden]

@drowsy hawk okay so

I havent used law of cosines in a while

so my method is probably not the best but it should work

to find the angle b you want to find the length A and C

then you can use A, C to find the angle in between them

you are given A, so find C

use the law of cosines

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[versed fractal]

need a hand with this one

[proven coral]

can you show work for your answers

[versed fractal]

not much i just entered the values from above

[proven coral]

do you know the product and quotient rules

[versed fractal]

yes

[proven coral]

apply them here

[versed fractal]

but theres no derivative

[proven coral]

the apostrophe represents the derivative

f'(x) = the derivative of f(x)

[versed fractal]

can you help me withthe first one

[proven coral]

apply the product rule

f and g are your functions

[versed fractal]

did 5 x 8 and got 40

[proven coral]

that's f'(5)*g'(5)

not (f(5)*g(5))'

[versed fractal]

so 8 x 2?

[proven coral]

no

apply the product rule

[rare cradle]

=16.

[versed fractal]

i guess i dont know the product rule

.close

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[reef shore]

i need help with some geometry questions

[fierce orchid]

[reef shore]

Hello?

[tranquil pine]

welp, remember the trig identity, sin(x) = cos (90-x)

(in degrees)

[reef shore]

Thanks

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[languid adder]

what is this

[dry light]

Do you know the work formula?

[languid adder]

ye

mag of force times mag of displacemnt cos angle

[dry light]

Well that should answer your question

You have force and displacement

[languid adder]

so d?

[dry light]

Yeah

[languid adder]

alr

thanks

[dry light]

Power deals with work over time

[languid adder]

yea

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[ripe gull]

Sketch the region bounded by the graphs of the algebraic functions and find the area of the region.

[ripe gull]

y=x, y=2-x, y=0

I’m confused on this

[magic sparrow]

have you sketched the region?

[ripe gull]

Yes

[magic sparrow]

what do you see?

[ripe gull]

Triangle

[magic sparrow]

yeah

and how do you find the area between curves?

normally it is the upper curve - lower curve yes?

but in this case, do you have a true upper or lower curve?

[ripe gull]

Yeah

No

[magic sparrow]

okay, so what might you do?

[ripe gull]

left curve - right curve?

[magic sparrow]

you could do that if you wanted to

it would probably be easier to split the bounds of integration

instead of going from 0 to 2

go from 0 to 1

and from 1 to 2

first use the left function

[ripe gull]

Wait why

[magic sparrow]

Look at the graph you sketched

if you find the area under the "left function" from 0 to 1

and add to it the area under the "right function" from 1 to 2

aren't you getting the total area you wanted?

[ripe gull]

I can’t see what you mean

[magic sparrow]

https://www.desmos.com/calculator/oevsiexmfg

look here

see how I split the area in half

find the left half

it would be the area between the upper red curve

and y=0

then add the right half

it would be the area between the upper blue curve

and y=0

[ripe gull]

Ohh

[magic sparrow]

the left half would be between 0 and 1

the right between 1 and 2

[ripe gull]

Would it be y=x and y=0 on the left

So it’s x-0?

Then u find definite integral of x

[magic sparrow]

mhm

[ripe gull]

Ok wait

Just for future reference

[magic sparrow]

because isn't the area between a curve and the x-axis just the definite integral of that curve anyways

[ripe gull]

Yeah

If y= 1

Like everything stayed the same but y=1

Instead of y=0

Would u still split into half

And do y=x and y=1 on left half

So x-1?

Def integral

[magic sparrow]

sorry, I can't tell what you are asking

[ripe gull]

So if the graph were to be

[magic sparrow]

if you instead found the area between

y=1 and y=x

between 0 and 1

the upper function would actually become y=1

[ripe gull]

Wait what

[magic sparrow]

if that is what you are asking

[ripe gull]

Oh yeah

Y=1 is upper function

[magic sparrow]

for this new hypothetical yeah

[ripe gull]

So 1-x

And for the right half

[magic sparrow]

for this new question you just brought up yes, but not for the original question you asked

[ripe gull]

It’s 1-(2-x)

[magic sparrow]

sure

[ripe gull]

Yup I’m just trying to understand more

[magic sparrow]

you can think of it like this

[ripe gull]

So then u add the definite integral of 1-x + definite integral of 1-(2-x)

[magic sparrow]

the reason upper - lower gives you the area in between the curves

is because taking just the upper

would give you the entire area down to the x-axis

and taking just the lower

would give you the area below the lower function down to the x-axis

the area in between the two functions

would be the difference

upper - lower

[ripe gull]

Yea

For left to right u do right-left, right?

To find in between

[magic sparrow]

well

if you wanted to find the area in between two functions

from left to right

the "upper function" would become the one furthest from the y-axis

so yes the rightmost function

[ripe gull]

For this is 4-y^2 the right function

Cause it’s on the right

[magic sparrow]

yes

[ripe gull]

Also for left to right

You plug the y values intercepts into the definite integral?

[magic sparrow]

yes

[ripe gull]

I’m confused why you can’t just go from left to right

From -5 to 4

[magic sparrow]

because you no longer have functions of x

the bounds mean

x=-5

and x=4

but

if you are doing an integral with respect to the y-axis

your bounds should be y=

not x=

[ripe gull]

Ohh

[magic sparrow]

finding area under something like y=x^2 from 0 to 1

[ripe gull]

Ur right

[magic sparrow]

is saying from x=0 to x=1

dx

[ripe gull]

That’s respect to x right

[magic sparrow]

yes

[ripe gull]

What if u made these equations into y=sqrt(4-x)

It would be respect to x right?

[magic sparrow]

you could yes

[ripe gull]

y=x+2

[magic sparrow]

then your issue would be no clear upper function

and you would have to split the bounds again

but if done correctly it would give the same area

[ripe gull]

How do you split it?

I’m confused about the splitting stuff

[magic sparrow]

well again the point is to get upper - lower

so looking at this

on the left side of x=0

the linear function is higher

[ripe gull]

Yes

[magic sparrow]

and on the right side of x=0

the quadratic is higher

so we would split it there

because otherwise we can't just say "upper-lower"

[ripe gull]

Yea

[magic sparrow]

because one isn't always above the other one

over the entire intervals

[ripe gull]

Ohh yeah I get what u mean

But why does the answer key say this

From -5 to 0

Why is it adding the top function and bottom

In the First A=

[magic sparrow]

because

graph y=sqrt(4-x)

sqrt is purely positive

so the lower function isn't y=sqrt(4-x)

it is -sqrt(4-x)

and when you subtract that

it becomes addition

[ripe gull]

Oh

Wait I can’t have a - sqrt?

[magic sparrow]

what is the square root of -4 ?

[ripe gull]

Wait isn’t it just -sqrt(4-x)

It’s on outside right?

[magic sparrow]

yes

the negative sign is on the outside

[ripe gull]

Like -sqrt(4-4

Then it’s 0

Wait

[magic sparrow]

in this picture

that sideways parabola

is the equation x=4-y^2

now to rewrite that

we get y^2=4-x

this does not equal y=sqrt(4-x)

in fact it equals

y= + or - sqrt(4-x)

and if you were to graph that

you would see that taking the positive sign

would make the top half

[ripe gull]

Ohhhhhhhhhhhhhhhhh

[magic sparrow]

and the negative sign would make the bottom half

[ripe gull]

Damn

[magic sparrow]

and then you would conclude that your " lower " function is

y=-sqrt(4-x)

and subtracting a negative makes addition

[ripe gull]

Yeah ur right it’s +-

[magic sparrow]

https://tenor.com/view/its-not-real-dwayne-johnson-seven-bucks-the-rock-its-not-the-truth-gif-20130196

[ripe gull]

What about the other one tho

[ripe gull]

Lol

[magic sparrow]

the other what?

[ripe gull]

Where did the 2 come from in the second part?

[magic sparrow]

well

you are finding the area shaded there

cant you see

so like we just decided

the top part is y=sqrt(4-x)

and the bottom part is y=-sqrt(4-x)

they are symmetric

they have the same area

[ripe gull]

O

[magic sparrow]

so you can just do 1

and multiply by 2

[ripe gull]

Ihhh

Math so confusing sometimes ngl

[magic sparrow]

integrals are fun once you get the hang of them

[ripe gull]

Ik how to solve integrals

But creating the equations are something else

[magic sparrow]

you'll get the hang of it

[ripe gull]

Thanks

[magic sparrow]

np

[ripe gull]

Appreciate the help

!

[magic sparrow]

yeah np

DM me if you ever need

[ripe gull]

Alright I’ll ask you if I need help again on other math problems

Do you also help with physics?

[magic sparrow]

sparingly

[ripe gull]

It’s just hs physics 1 no calculus yet

Alr thanks for the help

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[mild root]

I need help

On how to figure this out

Cause that’s too many words for my brain to process at this time of day

[supple copper]

wow that is ridiculously wordy

basically

a country is funding another country

and its represented by that function

where f(x) is the total funding in billions

and x is years after 2008

so they're asking in what year was the funding $10.5 billion

any idea on what you should do to find that?

[mild root]

I have no idea

[supple copper]

well what is f(x)?

hint i just told you

[mild root]

Total funding

In billions

[supple copper]

ye

and they're asking when the total funding in billions was 10.5

any ideas now?

[mild root]

Not really, I’m sitting in a dark room trying to finish this quickly

[supple copper]

alright you want to replace f(x) with 10.5

[mild root]

Ok

[supple copper]

and solving for x will get you the years after 2008

also maybe turn on the light then

lol

[mild root]

I’m supposed to be sleeping, but I forgot this is due tonight

And I’m trying to do the last 2 problems

😭

[mild root]

So, 10.5 = -1.335(10.5)^2 + 5.16(10.5) + 5.517 ?

[tranquil pine]

No, you'd only replace f(x) with 10.5

you are trying to solve for the x value when f(x) is equal to 10.5

[mild root]

Oh

[mild root]

Idk how to set that up

[tranquil pine]

Here, you've replaced both f(x) and x with 10.5

[mild root]

Oh

[tranquil pine]

You'd only replace f(x), because you know what the value of f(x) is

you are solving for x.

[mild root]

Oh

😭

This is too hard

[tranquil pine]

welp you know f(x) = -1.335x^2 + 5.16x + 5.517 right

as Kihei alr explained

you'd replace f(x) with 10.5

for example, if i had a random equation, lets say a=b+c, and i want to replace a with 20, the new equation would be 20=b+c

[mild root]

I’m not near a calculator

[mild root]

Ok

Can someone solve it for me?

.close

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[robust mulch]

wassup party people, answer key says erorr<.0043, i get .0021

[robust mulch]

my error term

i should do this

i have verified 200sin(0.5) is the maximum over the interval

[final saddle]

@robust mulch Has your question been resolved?

[final saddle]

@robust mulch Has your question been resolved?

[robust mulch]

lol ill figure it out then

.close

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[amber wadi]

Straight up

[frigid briar]

please use this channel for math related issue

[wide dragon]

answer is y-axis 🧠

[teal kayak]

ask a maths related question or face closure

[ashen shadow]

,w straight up

[soft zealot]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by 𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛#6985

[ashen shadow]

And check #❓how-to-get-help @amber wadi

[final saddle]

@amber wadi Has your question been resolved?

[teal kayak]

.close

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[lavish briar]

Hi how do u factor this

[fierce orchid]

I would just use polynomial division

[lavish briar]

I forgot how to do that

How do you do it again ;-;;

Is it the synthetic division

[tranquil pine]

use horner

[lavish briar]

What is horner?

[tranquil pine]

factorization method

[lavish briar]

Ohhhh

Alright I'll go search it up

Thank you

.close

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[short jewel]

Is ^3sqrt-0.064 = 0.4?

[short jewel]

and why does the texit bot says its false

[proven coral]

what is that supposed to be

$\sqrt[3]{-0.064}$?

[soft zealot]

a disappointing son

[short jewel]

Yes

[proven coral]

texit says its false because its false

[short jewel]

Why?

[charred solar]

if you cube a positive number, would you get a negative?

[short jewel]

we're supposed to find the square root

[proven coral]

you put a ^3 in front of your "sqrt" and said yes when i displayed a cube root

what is your question

[charred solar]

send a screen shot or picture of your question so we can help you better

[short jewel]

my classmates solution are 5 so im really confused

[tranquil pine]

if you want to make it easier put the decimal in fraction form

[proven coral]

what is your question

the original question

[tranquil pine]

there should not be a 5 in there at all

[proven coral]

for future reference we are not sitting next to you and we have no idea what the context is, so provide full context next time

your problem is $10\cdot \sqrt[3]{-0.064}$

[soft zealot]

a disappointing son

[tranquil pine]

theres a reason you got it wrong

[short jewel]

No

Thats number 10

[tranquil pine]

try looking at the question again

[short jewel]

the 10th question

[tranquil pine]

the negative sign

[short jewel]

FUCK i meant it 0.04

My classmates solution are 0.04

[proven coral]

your classmates are wrong

[proven coral]

going back to here

[short jewel]

No

[charred solar]

yep, so we know our answer would have to be negative right?

[short jewel]

Sooo

-0.4?

Texit bot says its still false

[charred solar]

did you put it in wrong?

[charred solar]

[short jewel]

Bot is bricked?

[charred solar]

wait i put it in wrong

[short jewel]

no negative in the board tho

[charred solar]

[short jewel]

NAH

Wolframalpha bricked?

[short jewel]

No negative

[short jewel]

its supposed to be ^3sqrt0.064 tho

man

Math is so

[charred solar]

wait wdym no negative? in your question it has a negative sign

[short jewel]

Cube root

only -0.064 has negative

[short jewel]

U put negative before the cube root

wth

In #bots it gave a different solution

[charred solar]

i put it in the cube root but mathway put it outside

[tranquil pine]

@short jewel if you multiply -0.4 three times you’ll get -0.064 but if you multiply 0.4 three times you’ll get 0.064, its very simple.

[short jewel]

im not using the #bots ever again lol

[short jewel]

yeah

Im gonna use the wolframalpha website one

okay mine is solved

Ty

.close

[charred solar]

np :)

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[craggy plume]

!status

[final saddle]

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

[broken jasper]

2

[craggy plume]

what have you done so far?

[broken jasper]

I thought the question said that 15000 people voted in favor of the proposal, so I just did 3x = 15000 where x was the number of people who voted against it

but then I realized that it said 15000 MORE people voted in favor than against and now I don't know how to go about this

[craggy plume]

the ratio x:3x is a correct deduction

to find the x that satisfies this you'll have to use the 15000 more piece of information

can you create an equation based off of it?

[broken jasper]

y + 15000 = x, where x is the number of people who voted for and y is against

[tired walrus]

there is a way to do this which hides all the algebra under the rug in a way

(votes for) : (votes against) = 3 : 1
therefore
(votes for) : (votes against) : (difference) = 3 : 1 : 2
and we know that difference = 15,000

@broken jasper

[broken jasper]

woah

ty both of u, I got it now

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[tranquil pine]

what trig identities do you use to get this?

[tulip coyote]

Addition formula, $\cos(x + \phi) = \cos(x)\cos(\phi) - \sin(x)\sin(\phi)$

[soft zealot]

@tulip coyote

[tranquil pine]

yeah i got that but i dont know where to go from here

[tulip coyote]

Substitute that in and then equate coefficients

[tranquil pine]

oh what

ohhhh

phi is a constant im dum lol

thank you

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[vital crag]

!show

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@sly hamlet Has your question been resolved?

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[tranquil pine]

how do i do 8b? Im not sure how to rationalise the denominator

[rustic wedge]

Expand the squares using identies (a+b)^2, and (a-b)^2

You'll get two terms

[tranquil pine]

38 and - 34

not sure what to do afterwards tho

[rustic wedge]

what the..

Are sure u r apply the identities right? @tranquil pine

[tranquil pine]

wait a min let me check once more

might have messed it up

[rustic wedge]

U arent supposed to get rationals

[tranquil pine]

yes i messed it up

i got it

thanks

[rustic wedge]

Nice 👍

Np

[tranquil pine]

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[rancid ravine]

the series An has the following properties:

a1 + a2 = -32
a3 + a4 = 0

the last number in the series is 60, how do we find the number of slots in the series?

[void valley]

is it arithmetic?

[rancid ravine]

yep

my bad didnt mention that

[void valley]

find a1 and d

firstly

then a1 + (n-1)d = 60, solve for n

[rancid ravine]

how do i find a1? I can only get it in a representation of a2

a1 = -32 - a2

then d = a2 - (-32 - a2)

[void valley]

use formula for n-th term

a2 = a1 + d

etc.

[rancid ravine]

so

d = 2a2 + 32

-32 - a2 + (n-1)(2a2 + 32) = 60

i am confusion

@void valley

[void valley]

look

a1 + a2 = -32
a1 + a1 + d = -32
2a1 + d = -32

a3 + a4 = 0
a1 + 2d + a1 + 3d = 0
2a1 + 5d = 0

Finally:

2a1 + d = -32
2a1 + 5d = 0

can you solve this?

[rancid ravine]

oh yeah

aight thanks

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[uncut aurora]

Is my calculator wrong or something

[uncut aurora]

I swear I've done something stupid just tried it on my other calculator

[modest birch]

what do you expect

[uncut aurora]

210

[modest birch]

cos^-1(cos(x)) does not necessarily equal x

,w cos 150 degrees

[soft zealot]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Saccharine#3780

[uncut aurora]

Wait so what should i do for this

[modest birch]

for what?

[uncut aurora]

Q6

Oh wait

360-150=210

Dw

,close

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[frank coyote]

the question is identify where the function w=exp(z^3+6z^2-15z+7) is not conformal
I know that in order to be conformal, the function has to be analytic and its derivative is nonzero
in order to be analytic, CR equations Ux=Vy and Uy=-Vx needs to be met
so I need to convert it from exponential form into cartesian form to get the U and V parts

[frank coyote]

so how to convert e^(z^3+6z^2-15z+7) or exp(z^3+6z^2-15z+7) into cartesian form?

[tranquil pine]

use euler's formula i suppose

[frank coyote]

but there are really multiple terms on the exponent...

[tranquil pine]

actually i will stop talking since i am quite unsure

[frank coyote]

I can make it w=e^(z)^(3) * e^(6z^2) * e^(-15z) * e^7 but not any further..

[frank coyote]

it's fine thanks tho!

[tranquil pine]

you can check where it is not conformal where the Cauchy-Riemann equations are not satisfied i suppose? I guess you can can compute the partial derivatives of U and V with respect to x and y and check where the equations are not satisfied with that, but I am not quite sure so i will leave someone more knowledgeable to explain

oh x and y are the real values after u convert of course

[frank coyote]

yes I would like to but I just can't separate the U and V part from e to a polynomial

because in my knowledge for example z=x^2+ix has a U of x^2 and a V of x

so like only in cartesian form I can find the U and the V

[frank coyote]

actually it's this simple lol

thank you tho!!

[tranquil pine]

oh noice

glad u managed to figure it out :3

[frank coyote]

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[broken stag]

source: https://www.youtube.com/watch?v=sKTyse15UCQ&list=PLAVG7GMBpcYB-Yp_8WjMRm-A-6FyG9MMX&index=9

can someone explain how the person got that? like pre-substituition?

been trying to read about laurent series(and trying to find worked examples with explanation), but i am still clueless

[broken stag]

what i have been getting, at least from my reading is
Laurent series is a series, where the powers are negative, and there are a few terms where the coefficient does not go to 0, and these are unique to each function

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[bold hinge]

Hi can anyone help me find the half life of his graph?

[broken stag]

do you know the definition of half life?

iirc, its the time it takes for something to go to half its value

[bold hinge]

Yeah

[broken stag]

you probably might need to check the graph scaling against a ruler and estimate the decimal

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[proud spoke]

need help with b

[proud spoke]

i got 38m/s for a

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

@proud spoke Has your question been resolved?

[proud spoke]

<@&286206848099549185>

hello?

[vital crag]

Did you see this

[proud spoke]

yes but i have no idea how to do it so i didn't start

for question a i solved for derivatve and plugged in t(8) to get velocity and acceleration

[vital crag]

You solve for t when s(t) = 60

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[next pagoda]

I want to rewrite this using other variables. This is a trick I learned in another book about the chain rule

But I'm worried I might be doing it wrong.
I want to let u = 7f(x) , v = 3f(u) , y = f(v)

I think?

Does that look correct so far?

[spring haven]

yea

[next pagoda]

now

follow up

The question says to let f(0) = 0 and f'(0) = 2.

At what point in using the chain rule would I use this?

Would I use it at all when doing the chain rule?

Or, would I use it when finding F'(0) or something? ( I feel like this makes less sense)

[next pagoda]

^

[spring haven]

try finding F'(x) in terms of the functions you just defined

[next pagoda]

For the first term, where u' = d/du (7f(x)), what would f(x) be considered ?

[spring haven]

$u = 7f(x) , v = 3f(u) , y = f(v), F'(x) = \frac{dy}{dx} = \frac{dy}{dv} \frac{dv}{du} \frac{du}{dx}$

[soft zealot]

kheerii

[spring haven]

does that make sense?

[next pagoda]

Yup

But how do I handle u' = [7f(x))]?

[spring haven]

what?

[next pagoda]

u' is another notation for the derivative. Like dy/dx

[spring haven]

yes

i know

$u' = (7f(x))' = 7f'(x)$

[soft zealot]

kheerii

[next pagoda]

so, F'(x) = 7f'(x) * 3f'(u) * f'(v) ?

I'm supposed to be able to sub back in stuff

[spring haven]

yes

[spring haven]

you can

just sub in the values of u and v

[next pagoda]

So like 3f'(u) = 3f'(7f(x))

[spring haven]

no

[spring haven]

into this

$F'(x) = 7f'(x) \cdot 3f'(7f(x)) \cdot f'(3f(7f(x)))$

[soft zealot]

kheerii

[spring haven]

I know it looks bad

but

notice you have the values of f(0) and f'(0)

and you want F'(0)

so just put in 0 for x

$F'(0) = 7f'(0) \cdot 3f'(7f(0)) \cdot f'(3f(7f(0)))$

[soft zealot]

kheerii

[next pagoda]

heck yeah

Honestly, this way for me is way easier to think about than what my professor wanted us to write it

it's the same idea but I like this better

[spring haven]

how did he tell you to do it

[next pagoda]

So, instead of writing out like u , f(u), he would write f'(g(x)) * g'(x). Where you leave g(x) alone inside the first term.

idk if I like it

[spring haven]

it's the same thing

[next pagoda]

yeah yeah

ik it is

but the way you write it is different

right?

[spring haven]

well

[spring haven]

the second half of this is the same

as what we wrote

dy/dx = dy/du * du/dx

[next pagoda]

right

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[final saddle]

Please don't occupy multiple help channels.

[digital pebble]

anyone idea on how to solve this?

.close

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[sharp laurel]

How do I find this?

[sharp laurel]

I plugged in:
f'(-1) and got positive
f'(1) and got negative
f'(3) and got positive

[strange shore]

are you trying values randomly or are you trying to find c with right steps ?

[sharp laurel]

So with using the number line, I identified the intervals in which it is increasing/decreasing. I use values that are to the left and right of my critical points

Increasing: (-Infinity, 0)U(2, Infinity)
Decreasing: (0, 2)

Though I may be missing something and would love feedback

[strange shore]

,w roots of 24x^3 - 96x^2

[soft zealot]

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Requested by Ｈｅｒｅｌｓ#1420

[strange shore]

why wolframalpha says 4 is a root of this tho

anyway

you studied the sign of f'(x) too ?

well I see your mistake

your derivative is wrong

I think

Ah nah im tired its correct

Yea the roots are 0 and 2

Now you should study the sign of f'(x)

[sharp laurel]

huh

[grim badger]

That's not all the roots

[strange shore]

-2 too but welp

[grim badger]

That's $x = \pm\sqrt{4}$

[soft zealot]

dldh06

[sharp laurel]

OH

dang it i keep forgetting

+- square roots

jesus

my roots are: -2, 0, 2 yes?

[strange shore]

yea

[grim badger]

How I remember is, if I manually write the sqrt, then it's +-. If the sqrt exists then it's just +

[sharp laurel]

ahh i see thanks a bunch

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[proud glacier]

Why is $\mathbb{Z}_{57}$ not a field?!?

[soft zealot]

normalAtmosphericPa=101,325

[proud glacier]

It forms an abelian group

[silk saddle]

57 aint prime

[proud glacier]

oops

[silk saddle]

you can check yourself

[proud glacier]

I already did

it "feels" so prime lol

[silk saddle]

I'm pretty sure U(57) does not have order 56

Lmao

Recall a field requires every nonzero element to be a unit

[proud glacier]

yeah

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[vapid urchin]

idk man, my formula is screwed

[vapid urchin]

isnt it 50+50e^-0.6t

plug in 0, 1, 2 for t?

[magic sparrow]

yeah, it gives you the formula, percentage of knowledge = P = ..... and the formula has variables Q, k, and t. It tells you what to let Q and k equal, so do that, and then it wants you to find out what the percentage of knowledge is at various times "t", so plug in those times "t" into the equation for "t" and this will give you what you want to know

[vapid urchin]

sup austin

thats what i did

plugged it into the calc

and got an error code

error: syntax

[magic sparrow]

then include better parenthesis

[vapid urchin]

where

on the 50?

50+(50)e^-0.6(t)

[magic sparrow]

idk, I am not your calculators syntax error lol. obviously you should be able to plug it into the calculator

so if something is going wrong, just be more careful

[vapid urchin]

lol

keep getting error code

gonna burn my calculator

with my mini lighter thats a flame thrower

[grim badger]

Are you legit typing in t?

Or replacing t with 0, 1, 2?

Or even better, can you send an image of the calc screen?

Of what you are typing in exactly

[grim badger]

I honestly can't tell from that, without seeing what your key presses are

[magic sparrow]

you might just be hitting the minus sign

instead of the negative sign

[grim badger]

That could be it

The negative sign, if I recall, is on the same row as the 0

[vapid urchin]

you gotta be fucking kidding me

a minus

and negative

are different?

LOL

cmon bro

[grim badger]

It's a few pixels shorter on the ti

And a bit higher, I think

So you can tell the difference between minus and negative

Like x - -3, people know it's minus negative 3, but on a calculator, they make it more visible

[vapid urchin]

thanks austin

and d

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[vapid urchin]

.reopen

[final saddle]

✅

[vapid urchin]

apparently my answers were wrong

oh i see it

.close

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[half mirage]

Why did they take critical point as 1 here ?

[dry light]

It tells you

[half mirage]

yeah, but should I equate (x^3-X^2) part to 0, out of 12(x^3-X^2) ?

[dry light]

The derivative is 12x^3 - 12x

Just factor out 12x

[half mirage]

derivative is 12x^3 - 12x^2

[dry light]

My bad just factor out 12x^2

Your critical points will still be 0 and 1

[half mirage]

12x^2(x-1) ?

[dry light]

Yeah

Then using zero product theorem:

12x^2 = 0
(x-1) = 0

[half mirage]

understood now

Many thanks @dry light 👍

.close

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Please don't occupy multiple help channels.

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@lofty flicker Has your question been resolved?

[wispy charm]

Do need help for both the questions?

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[lofty flicker]

.close

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[onyx sand]

how would i find b and d?

[sleek wave]

hint: lcm(a, b) = (a * b) / gcd(a, b)

[onyx sand]

ok i found the gcd of a,b

is it just that simple formula?

[sleek wave]

yes

[onyx sand]

i’m trying to think of a way to logically explain that formula

before i use it

since the lcm is the biggest factor that both divide, multiplying both and dividing by its greatest factor will give us the lowest multiple?

[sleek wave]

yeah

basically

cuz like

[onyx sand]

is this the main way to do it

because i don’t see myself doing this without a calculator on the midterm

[sleek wave]

its the easiest imo but theres also prime factorization

dyk what that is

[onyx sand]

yes

[sleek wave]

so then ye

[onyx sand]

but none of us know how to apply it 🥲

[sleek wave]

what course r u twkkng

taking*

[onyx sand]

intro to proofs ):

[sleek wave]

oh dam

[sleek wave]

like as in its just hard to use

if so u can learn division rules

like there are methods to figure out if a number is divisible by say 17 without a cslculator

[onyx sand]

i think we may have done it with euclidean algorithm?

[sleek wave]

yeah

just do that

[sleek wave]

or like what part do u not know how to apply is what im adking

[onyx sand]

is it the stuff in the blue?

we just talked about the topic no examples

[sleek wave]

rhats euclidean algorithm yeah

so what i would do is prime factorization

[onyx sand]

yes

[sleek wave]

i can explain it more in depth like with examples if u want

[onyx sand]

sure

[sleek wave]

okay so yk how every positive integer can be broken down into prime factors

[onyx sand]

that’s something else

i know it is writing out the term as a list of prime factors

but i’m not sure how to do that on such large number

[sleek wave]

like for me id just test first smaller numbers like 2 and 3

so like lets say 782

first id test if it is divisible by 2 and it is and u get 391

[onyx sand]

yeah

[sleek wave]

391 isnt divisible by 2 so ill check 3

the way u do that is adding up all the digits and seeing if its divisible by 3 so

3+9+1 = 13 and so it is not divisible by 3

it isnt divisible by 5 cuz its not snmitliple of 5

it isnt divisible by 7 (complicatdd)

it is however divisible by 17:
take the number formed by the last digits of 391 (91) and then multiply the rest of the number (3) by 2, if u get 0 or a number divsible by 17 its divisible by 17

and then u get 23 which is prime

so 782 = 2 * 23*17

[onyx sand]

does adding the digits only work for 3

[sleek wave]

yes i can send u a link for all divisibllity rumes

[onyx sand]

in my example i have 2(1336)

would that be as factored as possible, or can i pull another 2

[sleek wave]

i would do like

1336/2 = 668
668/2 = 344
344/2 = 172
172/2 = 86
86/2 = 43
43 is prine

and then u have to add another two

[onyx sand]

ah ok i see

[sleek wave]

so 2^6*43

also sorry im taking a while to respond im at school rn (free period)

and this mf keeps talking like crazy

[sleek wave]

yeah

[onyx sand]

it’s fine lol i’m in lecture too

and now same for other number?

[sleek wave]

yep

it just happens to be a rlly nice number

[onyx sand]

would this be the lcm

i have a feeling i’m combining the wrong numbers

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[strange rune]

$-3n^2 -nlog(n) = \Omega(n^2)$

[soft zealot]

Benjamin

[strange rune]

?

what is the definition of omega?

$f(n) = \Omega(g(n)) \iff .........$

[soft zealot]

Benjamin

[meager marten]

i bascially need to make functions smaller and smaller that will simplify to n^2

there was orignally an (n/8)^3 but since it was positive i removed it now idk where to go from your equation

[strange rune]

is f from positive numbers to positive numbers?

,w -3n² - nlog(n)

[soft zealot]

Results provided by WolframAlpha

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Requested by Benjamin#3570

[strange rune]

,w plot -3x²-xlog(x)

[soft zealot]

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Requested by Benjamin#3570

[strange rune]

this is f right?

so?

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[oak kraken]

So I am new to linearization but I understand it is derived from the point slope formula

My only question is: why is y equal to L(x)?

I think that the rest makes sense to me:
y1 = f(a)
x1 = a
m = f’(a)

Also is there a practical reason why y1, also know as f(a) is moved to the beginning of the right side rather than the end of the right side?

[hybrid heath]

L(x) is just the function for the line