help-0

[wind cloak]

Find the range for each individual part

(remember the restriction on x)

I won't be here to check the answer so ask WA

[pine surge]

Hmm.. i didnt get to can u send me detailed solution

[pine surge]

WA??

[wind cloak]

we're not allowed to do that

[wind cloak]

Wolfram Alpha

[pine surge]

It alright can u send me a link to vid on how to find range of mad function??

Mod

How you did this??

I didnt get it?

[lone heart]

@pine surge Has your question been resolved?

[lone heart]

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[livid axle]

I'm having trouble proving this theorem: The algebraic multiplicity of an eigenvalue is always greater than or equal to the dimension of the corresponding eigenspace.

[lone heart]

Please don't occupy multiple help channels.

[livid axle]

Sorry I didn't see that I've already opened a channel.

.close

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[safe cape]

can someone remind me why 2e = 0

[safe cape]

oh nvm I remember now

.close

[upbeat hornet]

what is e

[lone heart]

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[gray isle]

but 2e isn't 0...

[lone heart]

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[fallen estuary]

can someone help me with 10c?

[fallen estuary]

I think you have to use this

[rigid prawn]

so for f(x)=sqrt(c^2-x^2)+a f'(x)= -x/sqrt(c^2-x^2)

what is the answer to a?

[fallen estuary]

oh ok

[fallen estuary]

-x/sqrt(9-x^2)

but idk why the +4 gets removed

[barren portal]

f(x)= $\sqrt{36-(x-7)^2}$

[ocean seal]

dotdoc.

[fallen estuary]

right

[barren portal]

what shape is it?

[fallen estuary]

semicircle

i hope

[barren portal]

describe more

centre, radius?

[fallen estuary]

a semicircle with a radius of what i presume to be 6

and a center of 7

[barren portal]

point?

describe as points (x,y)

[fallen estuary]

Center is (7,0)

Edges are (1,0) (13,0)

[barren portal]

one way to see that is just to square both sides of f(x) right?

[barren portal]

^

[fallen estuary]

I guess yeah

[barren portal]

yes it is

[fallen estuary]

Wouldn’t it become 36 then

Instead of 6

[barren portal]

so you see how your circle is, and you want to describe it’s slope of tangent at any point

[fallen estuary]

yes

[barren portal]

Then general equation of a circle is $(x-h)^2+ (y-k)^2= r^2$ where (h,k) is the centre

[ocean seal]

dotdoc.

[fallen estuary]

oh yeah

[barren portal]

since slope of tangent is nothing but derivatives

we can use some circle geometry to figure the derivative out

notice this will only apply for function which a circular or semicircular

not a general way to find derivative for any f(x)

[fallen estuary]

what circle geometry do we use

[barren portal]

here, you can see the that the slope of tangent line and radius line have a relation m1* m2= -1

[fallen estuary]

yeah

[barren portal]

so if you work out radius line slope, then you pretty much found the derivative when you are dealing with circular functions

[barren portal]

so can you work out radius line slope in the case of this?

[fallen estuary]

hmm

lets see

[barren portal]

feel free to draw the ciricle

[fallen estuary]

i can just sub in any values to get a triangle no?

[barren portal]

i’m not sure what you mean

Can you elaborate?

[fallen estuary]

wait nvm

[barren portal]

any point in semi circle is $(x,\sqrt{36-(x-7)^2})$

[ocean seal]

dotdoc.

[barren portal]

right?

[fallen estuary]

is it like the example where I have any point (x, f(x) ) so I just divide f(x) by x

yes

[barren portal]

what’s the equation of a slope of line?

[fallen estuary]

y = mx + b

[barren portal]

That’s equation of line

let’s say i gave a line, how do you determine it’s slope?

you don’t know the equation, you need find the slope m

[fallen estuary]

you divide

rise / run

[barren portal]

how many points you need to find that?

[fallen estuary]

one

wait no

two

[barren portal]

yes

so you need to find the slope of radius line passing through centre of circle right?

[fallen estuary]

yes

[barren portal]

notice, you are not plugging making a general equation for slope, not just a specific value

what are those two points you have here?

[fallen estuary]

(x, f(x))

[barren portal]

to find the slope of radius line

[barren portal]

That’s just one point, that’s lying in circle.

[fallen estuary]

hmm

idk the other point

[barren portal]

you need one more to find the slope of the line

[barren portal]

what line are were trying to find the slope of?

[fallen estuary]

radius

oh

[barren portal]

which passes through?

[fallen estuary]

(7,0)

[barren portal]

exactly

so you have two points

now

find the slope

[fallen estuary]

ok so the rise is

[brittle timber]

Can anyone help me with 22

[barren portal]

occupied *

[fallen estuary]

the rise is 0 to f(x)

and the run is 7 to x

[brittle timber]

Ohkkk

[barren portal]

what’s slope now?

m=?

[fallen estuary]

yoooo hold up

[barren portal]

mb mb

[fallen estuary]

so the rise is just the formula

[barren portal]

The ratio

[fallen estuary]

and the run is 7 - x

so its f(x) / 7-x

then u just flip it

and make it negative

for the tangent

right?

[barren portal]

yes

[fallen estuary]

nice

yep that's correct

wait

shouldn't it be negative

[barren portal]

let me see

Your run is x-7

[fallen estuary]

ohh

shit

yeah that makes sense

[barren portal]

No matter which one your choose first to find slope, but divide correctly

[fallen estuary]

yeah

thanks for that

[barren portal]

notice that this method for finding derivative only applies if f(x) is semicircular

the whole notion of derivatives is to extend tangent line to more curves than ciricles

[fallen estuary]

yeah

im just revising this coz this might be in our test

[barren portal]

which book is that?

[fallen estuary]

cambridge maths extension 1 year 11

australia

[barren portal]

Good one

usually math presented is throw some derivative formulas and you are supposed to use it blindly

this sure is a good one.

[fallen estuary]

it is pretty good yeah

[barren portal]

gl

[fallen estuary]

🙏🏿

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[warped topaz]

https://i.imgur.com/FdmSWP1.png

[warped topaz]

I dont know if im doing this correctly

[barren portal]

What’s does two vectors being perpendicular shows about dot product?

[warped topaz]

Oh

That they are 0 when multiplied together

[barren portal]

What’s the definition of dot product?

[marsh rapids]

hence, find an equation for y

[warped topaz]

Okok, thats a much better way of doing it than I was hahaha

[warped topaz]

I think I can manage that 1 moment

[marsh rapids]

also in this case, there's a little trick that can save some time

[warped topaz]

https://i.imgur.com/w8xC65t.png

Yep I got it 😄

That was waaay easier than how I was doing it haha

[marsh rapids]

what was your method ?

[marsh rapids]

the vector (a, b), when rotated 90° counterclockwise, becomes the vector (exercise, spoiler is solution) ||(-b, a) ||. Hence, find a suitable y value

[warped topaz]

I found the length of v using √(-28^2) + (-4^2) and the same for w, then I multiplied them both together to get 112 - 4y which gave me (112 - 4y)/(20√2 + √(16 + y^2))

It was such a mess hahaha

[warped topaz]

Oh

Thats a good way too

[marsh rapids]

their lengths are totally irrelevant

[warped topaz]

Dont you need them for the dot product?

[marsh rapids]

this extra knowledge means I skip the equation part

[warped topaz]

Ahhh true

[marsh rapids]

if two vectors are orthogonal, then rescaling them doesn't change the fact their dot product is 0

so the norms are irrelevant

you can freely turn (-28, -4) into (7, 1) to make the numbers nicer

[warped topaz]

Ah ok, so it can be whatever length it doesnt matter

[marsh rapids]

avoid the making the length 0 still, then you lose information

[warped topaz]

Oh yeah true haha

Thank you guys I think I get it now!!

So helpful as always haha

❤️

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[rotund shoal]

Where did I mess up?

The answer sheet says that I am wrong

[rigid prawn]

which angle is A and which angle is y

[lone heart]

@rotund shoal Has your question been resolved?

[nimble fern]

i got 29.92, what does the answer sheet says?

[rotund shoal]

I got this too, I realised I didnt use the exact value of the angle

Answer sheet said 26.55

[nimble fern]

oh

you need to find AD, right?

[rotund shoal]

Sorry, angle A is ABC and angle y is CAD

[rotund shoal]

length of ad

corrext

[nimble fern]

because 26.55 should be length of CD

oh wait lol

my bad

i see what we did wrong

[nimble fern]

CAD is 90°-y, not y

thats why lol

me careless

[rotund shoal]

Why would it be 90-y?

[nimble fern]

your y that is evaluated is angle BAC

[lone heart]

@rotund shoal Has your question been resolved?

[rotund shoal]

Ohhh silly mistake on my part

Thanks for identifying

[lone heart]

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[vale pulsar]

.close

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[safe cape]

question says to integrade

[safe cape]

is it right?

[vale wigeon]

no

$\int -\frac{1}{x^2} \dd{x} \neq -\frac{\ln|x|}{x}$

[ocean seal]

Ann

[vale wigeon]

i believe you came here just a few minutes ago asking whether the derivative of ln(x)/x was 1/x^2
and you were told by multiple ppl that it wasn't

so now you step on the exact same rake

[safe cape]

how is -1/x^2 integrated

[vale wigeon]

$\int x^p \dd{x} = \frac{x^{p+1}}{p+1}$ still applies for $p=-2$.

[ocean seal]

Ann

[safe cape]

is the answer for this question correct?

the answer is the one in the circle

the left side is the question

since it has that ln x thing

[vale wigeon]

did we decide to abandon the previous one

[safe cape]

they're related so I figure I would ask

[vale wigeon]

i see no relation between the two (beyond them both being integral problems), and you did not acknowledge my response to your question of "how do I integrate -1/x^2?"

[safe cape]

I was curious about them both since they both have that 1/x pattern

[vale wigeon]

ok first off

1. that "pattern" is kind of a fluke

[safe cape]

would it be x?

[vale wigeon]

what's "it"?

the integral of -1/x^2? no, the integral of -1/x^2 is not x.

[safe cape]

-1/x^2 after being integrated

[vale wigeon]

no, the integral of -1/x^2 is not x.

[safe cape]

ok

[vale wigeon]

would you like me to take you through applying the integral power rule properly?

or can you manage on your own?

[safe cape]

wait I'll send my working to see what went wrong

wait no nvm

[safe cape]

yes

[vale wigeon]

ok

let me restate the rule:

$\int x^p \dd{x} = \frac{x^{p+1}}{p+1}+C$

[ocean seal]

Ann

[safe cape]

ok

[vale wigeon]

for the purposes of calculating $\int \frac{1}{x^2} \dd{x}$, what should we take $p$ as?

[ocean seal]

Ann

[safe cape]

0?

[vale wigeon]

what is x^0?

[safe cape]

good point

I have no idea

[vale wigeon]

1/x^2 equals x raised to what power?

[safe cape]

2?

[vale wigeon]

is 1/x^2 equal to x^2?

[safe cape]

no

-2?

[vale wigeon]

is 1/x^2 equal to x^-2?

[safe cape]

yes?

[vale wigeon]

are you asking me or secondguessing yourself?

[safe cape]

I'm not sure

[vale wigeon]

might want to review exponent laws after we are done here.

anyway, yes, 1/x^2 = x^(-2).

here is the rule a third time:

$\int x^p \dd{x} = \frac{x^{p+1}}{p+1}+C$

[ocean seal]

Ann

[vale wigeon]

when we replace p with -2 here, what do we get?

[safe cape]

-1

[vale wigeon]

the entire right-hand side collapses to only a single number -1?

[safe cape]

x^-1/-1

[vale wigeon]

right...

can you simplify $\frac{x^{-1}}{-1}$?

[ocean seal]

Ann

[safe cape]

1/-x

?

this is right?

[vale wigeon]

yes now it is

[safe cape]

omg

thanks!

anything else you would want to add

I take it as a no

tq

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[alpine sable]

Prove srqt(22) is irrational
My work so far :

**Assume sqrt(22) is rational
-> sqrt(22) = a/b, Where a/b is a in it's simplest form a and b cannot both be even
-> 22 = a^2 / b^2 -> 22b^2 = a^2

22b^2 = a^2 can be expressed as 2 * 11 * b^2 = a^2 due to 22b^2 being a composite number
-> a^2 contains a factor of 2 -> a^2 is even
Therefore a is also even as the square root of an even number is also even

-> a=2k due to it being a mutiple of 2

**
However upon trying to further on the proof I can't seem to prove b^2 is even.

My problem :
**-> 22b^2 = (2k)^2 -> 22b^2 = 4k^2 -> b^2 = (4k^2)/(22)

I'm not sure on how to further proof.**

[raven haven]

notice that

since a^2 is even

and you said a is even

a^2/2 is still even

so this means

$11b^2$ is even right?

[raven haven]

can b^2 be odd?

[alpine sable]

22b^2 is factorisable by 2 to get 11b^2

But how would I prove that in my proof

I see that it's even though

[raven haven]

.

i kind of gave you the result by accident

[alpine sable]

I see since 22b^2 = a^2
and 11b^2 = a^2/2
11b^2 must be even as a^2 /2 is still even

How could I extend it further to prove b is even

[raven haven]

uh

can an odd number multiplied by

an odd number

be even?

[alpine sable]

Odd number multiplied by an odd is odd

[raven haven]

so

can b^2 be odd?

[alpine sable]

No as for 11b^2 to be even, b^2 must be even for 11b^2 to not be an odd number

[raven haven]

so

does this answer your question?

[alpine sable]

Yea it does thanks bro ❤️

[raven haven]

can I close this channel?

[alpine sable]

yea

[raven haven]

.close

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[raven haven]

in the future, if you dont get a response here try #proofs-and-logic

[alpine sable]

alr thanks bro very much

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[vast sonnet]

This is equal to 1^infinity right?

[high urchin]

maybe

[vast sonnet]

Is that not just 1?

<@&268886789983436800>

Like ik it's wrong

But idk why

[arctic steeple]

dont ping mods for math help

[vast sonnet]

the guy was spamming

🙂

[arctic steeple]

Ah okay i see, checked the logs

my bad

[vast sonnet]

nw

[lone heart]

@vast sonnet Has your question been resolved?

[vast sonnet]

.close

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[void tinsel]

Can someone explain if there is a f'(x) of f(x)= 1/x

[pale kestrel]

yes...

[mellow grail]

No

[pale kestrel]

power rule?

(dont confuse me )

[mellow grail]

How power rule?

Shoe me

OH

This ain't integration

Sorry for doubting you

[raven haven]

[void tinsel]

i need to find the f''(x) of f(x) = ln(1+x)
i know that the f'(x) = 1/x

[pale kestrel]

^

[wide anchor]

$\frac{d}{dx}(x^n) = nx^{n-1}$

[ocean seal]

despairful_deltoid

[void tinsel]

?

owww, is it -1/x^2?

[pale kestrel]

yh

[void tinsel]

and that becomes 1/x^4

and so on

[pale kestrel]

wat.

[void tinsel]

the f'(x) of f(x) = -1/x^2

[coral thorn]

yes

[void tinsel]

allright thnx

.close

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[quasi cove]

Need help on this.

[quasi cove]

This is what I have. Am I missing something out?

[alpine sable]

lemme check

its correct i guess sir

wait

[quasi cove]

None of the Options work out 😫😫

[alpine sable]

wait

a = 2, b = 1

oke sir u can close now

thank yoi

you

.close

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[stable grotto]

and we should rotate R2 region about AB

my solution :

and correct answer is 13pi/45

why my solution is not true?

sry i am blind

i know where is mistake

.close

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[quasi cove]

.reopen

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✅

[quasi cove]

.close

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[quasi cove]

.reopen

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✅

[quasi cove]

@alpine sable I need some explanations man

Is there any other ways except guess and Check? You can’t just give the answer and close the Channel lol

!nosols

[lone heart]

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

[quasi cove]

Is guess and check the only way out?

Some work I did. I’m trying to steer clear of guess and check. Any idea how to get the answer without this method?

Shall create new channel

.close

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[midnight robin]

i honestly dont know how to solve this, plz help me out fellas

[minor needle]

the degree of the monomial is the sum of the exponents of all included variables

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[midnight robin]

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[slow wraith]

Since angle CBD = angle BDC, the triangle BDC is isosceles in C

Thus BC must equal CD

[turbid root]

alright thank you.

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[dark crag]

On this app how did it go from step one to step two?

[mortal trellis]

x^2-1=(x-1)(x+1)

[vale wigeon]

quietly dividing both sides by (x+1) after doing the stated cross-multiplication

recommend not using this app as it promulgates the concept of "moving" things between sides

which is Not Great from a learning standpoint

[tardy stag]

that's how I always thought about it

[lone heart]

@dark crag Has your question been resolved?

[lone heart]

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[alpine sable]

Can someone please explain to me how this simplification works

[azure remnant]

exapanding top term gives -2(x^3) + 2x(y^2) + 2(x^3) + 2x(y^2)

the -2x^3 cancels out

so it gives 4xy^2

[alpine sable]

ahh thank you 🙂

[azure remnant]

you can exapand the bottom term to x^4 + 2(x^2)(y^2) + y^4

[alpine sable]

yep. ty

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[azure remnant]

np

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[safe tartan]

how do you find the angle between a vector and the horizontal?

[slender gull]

Directional cosines.

[safe tartan]

could i just send the question actually

[slender gull]

That's what you should always do, actually.

[safe tartan]

vector AB i found to be -22i + 15j

and thus the speed is sqrt709 m/s

but not quite sure howt o find the positive vector of the jet ski after ten seconds

[slender gull]

Um. Isn't your vector AB wrong?

[safe tartan]

is it ?

[slender gull]

I'd say so.

[safe tartan]

oh yeah it is

18i+15j

[slender gull]

So the ship is at origin.
It might help if you draw the scenario out.

Atleast the points where the jet ski is to be found.

[safe tartan]

yep i have done that

[slender gull]

Share.

[safe tartan]

bit messy but

The diagram should be right ignore the labels

[slender gull]

If the final position is say C

You need BOC (O being origin)

Yes?

[safe tartan]

yes

[slender gull]

You know the coordinates of B and C both.

Yes?

[safe tartan]

wait the coordinates of C?

[slender gull]

Well

Isn't that part i. Of the same problem?

I thought you already knew how to do that one...

[safe tartan]

im asking about i)

i could do ii) just dot products and solve for theta

[slender gull]

I see.

Okay.

[safe tartan]

yeah sorry for the confusion

[slender gull]

It's alright.

Your original message stated something with angle, so ...

Anyways

Do you realise that AB vector is actually the velocity vector?

Of the jet ski.

[safe tartan]

yes

[slender gull]

Do you know how displacement and velocity relate?

[safe tartan]

derivative

[slender gull]

With respect to time*

[safe tartan]

yep

[slender gull]

So if you know the velocity.
Time span.
Would you be able to find the displacement?

[safe tartan]

i would need to integrate

[slender gull]

You would.

Though this integral is pretty easy to evaluate since the velocity vector is a constant.

[safe tartan]

so

ah yeah i got it

ty

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[tender anchor]

is it 1?

so i did

limx->inf

e^5x

e^(5x)+e^(-5x) (cancelled the /2 out)

= limx->inf e^(5x)
------- = 1?
e^(5x)

[devout summit]

You cancelled /2 by what?

[tender anchor]

idk i just thought u could get rid of it bc its a irrelivant constant

[devout summit]

Not at all irrelevant.

$\lim_{x\to a}cf(x)=c\lim_{x\to a}f(x)$

[ocean seal]

euclid31415

[devout summit]

Given that the limit exists

So, if your limit after removing 2 exists and is 1 then your original limit has to be (2)×1

[tender anchor]

o

[devout summit]

In this case though that 1/2 is in the denominator so c here equals 1/(1/2)=2

[tender anchor]

ah true

so then 2 is the final answer in that case?

[devout summit]

Yes

But you do understand why that e^{-5x} term didn't matter right?

[tender anchor]

yeah bc it its 1/e^5x

so its negatively decreasing as x->infinitiy right

[devout summit]

$\frac{e^{5x}}{e^{5x}+e^{-5x}}=\frac{1}{1+e^{-10x}}$

[ocean seal]

euclid31415

[devout summit]

This should make it sufficiently clear that the denominator on right side goes to 1 and hence 1/(denominator) must also go to 1/1=1

[tender anchor]

yeah that makes sense

thank you

[devout summit]

Also, you could say it worked because f(x) = o(g(x))

Where f(x) = e^{-5x} and g(x) = e^{5x}

[tender anchor]

ah okay

i fully understand y the answer is 2 now

and how it goes to 1 / (1 / 2 )

ty

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[gray isle]

!status

[lone heart]

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

[meager plinth]

1

I'm not sure how to start

[gray isle]

you're pretty much being asked for the slope between $(x_1, f(x_1))$ and $(x_2, f(x_2))$

[ocean seal]

ℝamonov

[meager plinth]

o

I see, ty

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[winter seal]

yoohoo

[winter seal]

So I'm trying to understand this example

The topic is integrating partial fractions btw

I understand everything

up until this part

[echo socket]

The moment they start setting values of x in?

[winter seal]

Yeah

I can sorta see how they get -2 for B

But I'm getting 4 for C and not 2

like they're saying

[echo socket]

Okay, let's see what happens when we plug in x = 2
2 * 2 + 4 = (A + C) * 2^2 + (-2A - 2) * 2 + 4
8 = 4A + 4C - 4A - 4 + 4
8 = 4C
C = 2

[winter seal]

Wait we plug it into the left side too?

[echo socket]

Of course

[winter seal]

I thought we just left it alone

that changes everything

hold on let me try to redo that

[echo socket]

There is another way to look at solving for A, B and C btw

[winter seal]

What is that?

[echo socket]

For two polynomials to be equal to each other, the corresponding coefficients (as in the coefficients of the same term) should be equal

[echo socket]

So here you could basically solve the system of equations
4 = -2B
2 = -2A + B
0 = A + C

And you would get the same answers

[winter seal]

I can't do the system of equations for my life

let me try to do it the original way

[lone heart]

@winter seal Has your question been resolved?

[winter seal]

I don't get it

Well

I get taht

And it makes sense now

But what doesn't make sense is how did they rearrange A??

[winter seal]

I'm so confusused

How did they rearrange for A?????????????

[tardy stag]

wdym

i don't see them rearranging for A here unless you mean the part where they went from A + C = 0 to A = -C?

[lone heart]

@winter seal Has your question been resolved?

[lone heart]

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[subtle sand]

How Poisson mixture models are constructed and derive any two
Poisson mixture models.

[lone heart]

@subtle sand Has your question been resolved?

[lone heart]

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[burnt sable]

I just have a question on how to find the equation to put within the integral, I was able to do it with a rectangle, but the circle is throwing me off. Just #5! I have F = 9800Ad currently, but don't know how to get the area (A) or distance (d) from my point.

[near apex]

Where's your point?

[burnt sable]

I'll send a picture of my current work

[near apex]

So, you are considering d from water surface ?

[burnt sable]

Yes

d is from water surface to delta y

do i use the circle equation to find A?

[near apex]

Sorry, i forgot this chat.

[near apex]

I think it'd be better to make this in terms of theta using trigonometry.

[burnt sable]

hmmm

how would I do that?

[near apex]

Wait. Let me send a picture.

It'd be hard to put it in words.

[burnt sable]

Ok for sure!

[near apex]

It's a hand drawn picture so i'm sorry if it's a bit hard to read/ make sense of.

[burnt sable]

that's ok!

[near apex]

You can integrate over theta from -90° to 90°.

[burnt sable]

ahhh!!

that makes it a lot easier

[near apex]

Yes.

[burnt sable]

awesome thank you a ton

.close

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[silver pilot]

can someone tell me what i did wrong?

[silver pilot]

i checked my answer with photomath and mathway and they gave me different answers

[lavish sedge]

because this is a forth degree equation

x^2 would be equal to 4

not x

do you understand what i mean

[silver pilot]

so can i not use the quadratic formula here?

or do i just change the numbers

[lavish sedge]

you can do it, but only when you set your variables to y

let me show ypu what i mean

[silver pilot]

ok

[lavish sedge]

i set y= x^2 because the quadratic formula only works for equations which have the highest power = 2

hope it made sense

[lone heart]

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

[silver pilot]

ok i did it and i think i get it

ty

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[short blaze]

help

[short blaze]

whats the answer

i dont really understand what do the | | mean in this situation

is it absolute value?

[alpine sable]

Yes

[short blaze]

so how do i go about solving this

[tardy stag]

,tex .exp rules

[ocean seal]

Hayley

[alpine sable]

Use Deez

You can start by cancelling some stuff inside the root.

[short blaze]

so what would a5 and 6th root cancel into?

a^5*

would (a+1)^6 just cancel into (a+1)

[alpine sable]

Well, let's not hurry to that

Do you see that a, a⁵ cancels out?

[ocean seal]

.fire007

[alpine sable]

Leaving a^4

[ebon sparrow]

No?

[alpine sable]

No?? Why?

My wording might be wrong but

You get it... what I mean

"a" cancels with a "a" of a⁵

[surreal meadow]

$\frac{a^5}{a} = \frac{a^4 \cdot a}{a} = a^4 \cdot \frac{a}{a}$

[ocean seal]

maximofs

[surreal meadow]

a/a = 1

so you get a^4

[lone heart]

@short blaze Has your question been resolved?

[short blaze]

what happened to the a + 1

[surreal meadow]

it’s all still there

i’m just focusing on the a^5/a part

did that make sense to you

[short blaze]

yeah

doesnt a^5/a just turn into a^4

[surreal meadow]

yes

[short blaze]

ok so what about the other parts

i dont really know the process to solve this problem

[surreal meadow]

so now we have $$\sqrt[6]{\frac{b^2}{a^4(a+1)^6}}$$

[short blaze]

one sec

original problem

what hapepened to the 6th root

[surreal meadow]

oh sixth root

[ocean seal]

maximofs

[short blaze]

ok so can we turn 6th root into 3rd root times 2nd root

[surreal meadow]

what do you mean

[short blaze]

cant we get rid of the b^2 by seperating the 6th root into 3rd root times 2nd root

then cancelling the 2nd roots

with the b^2

[surreal meadow]

what does “3rd root times 2nd root” mean

[short blaze]

turn 6 into 3 times 2

?

[surreal meadow]

ok

we should probably write this as an exponent is my point

$$\left(\frac{b^2}{a^4(a+1)^6}\right)^{\frac{1}{3}\frac{1}{2}}$$

[ocean seal]

maximofs

[short blaze]

huh

im not familiar

[surreal meadow]

now familiar with what

[short blaze]

ok so u turned the 6th root into exponents?

is that the only way to solve this ?

[surreal meadow]

the sixth root of x is the same as x^(1/6)

[short blaze]

yeah

[surreal meadow]

that’s all that i did

[short blaze]

ok so now what

[surreal meadow]

i’ll ask you the same question

now what

[short blaze]

idk

[surreal meadow]

you seemed to have an idea before

[short blaze]

so can you cancel 1/2 and b^2

and cancel the exponent in (a+1)^6

and cancel 1/2 with 4 in a^4

[surreal meadow]

let’s be cautious here

you have the right idea

but what is the square root of x^2

[short blaze]

x

?

[lone heart]

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[short blaze]

.reopen

[lone heart]

✅

[short blaze]

ok 1/2 and b^2 cancels to b

but then what do u do with the 1/3

ok i see so we would get the cube root of b/a^2 right

[surreal meadow]

it’s not x

sqrt(x^2) = |x|

[short blaze]

but then how do they get 1/a+1

[short blaze]

whats the difference?

[surreal meadow]

between x and |x|?

[short blaze]

yeah

[surreal meadow]

what is |-1|

[short blaze]

oh no negatives

wait so whenever u take an nth root of an exponent then u have to use absolute value signs?

[surreal meadow]

no

it’s specific to even roots

[short blaze]

okay

[surreal meadow]

even roots give you a positive value

(or 0)

[short blaze]

what if its an odd exponent

[surreal meadow]

they can give you a negative value

the cube root of -1 is -1

[short blaze]

so if we take the square root of b^2, we get the absolute value of b

then if we take the 6th root of a+1^6 we get absolute value of a+1

and the cube root of b and a^2 is cube root of absolute value of b over a^2?

do we need absolute value signs for a

wait i have a question

can you simplify cube root of a&2

a^2*

whered u go 😭

@surreal meadow

[surreal meadow]

it’s been 5 minutes

chill lol

the cube root of a^2 is just that

[short blaze]

it just stays the same?

[surreal meadow]

there’s no nice simplification

[short blaze]

okay so is everything i said correct

[surreal meadow]

no, let's just walk through this together

i mean

what you said is fine

but i don't know if you're going to end up with the right answer since you're doing cube roots, square roots, and 6th roots all independently

the important take aways are these

(\sqrt[6]{\frac{b^2}{a^4(a+1)^6}} = \left(\frac{b^2}{a^4(a+1)^6}\right)^{\frac{1}{2}\cdot\frac{1}{3}} = \left(\frac{|b|}{a^2|a+1|^3}\right)^{\frac{1}{3}} = \frac{1}{|a+1|}\left(\frac{|b|}{a^2}\right)^{\frac{1}{3}} )

[ocean seal]

maximofs

[surreal meadow]

if that was what you were describing then perfect

as long as you understand the process

[short blaze]

where did 1/|a+1| come from

[surreal meadow]

(xy)^(1/3) = (x)^(1/3) * (y)^(1/3)

we pulled out the 1/|a+1|^3 from the inside

[short blaze]

do we just assume theres a one there

[surreal meadow]

what

oh

a one where

[short blaze]

ok wait so when u converted from the 3rd step to the 4th step

[surreal meadow]

(\frac{|b|}{a^2|a+1|^3} = \frac{|b|}{a^2}\cdot\frac{1}{|a+1|^3})

[ocean seal]

maximofs

[short blaze]

where did the one come from tho

[surreal meadow]

(\frac{a}{b\cdot c} = \frac{a}{b}\cdot\frac{1}{c})

[ocean seal]

maximofs

[surreal meadow]

this is just fraction multiplication

[short blaze]

ok so u would have to break the fraction apart first

then simplify?

[surreal meadow]

yes

[short blaze]

so thats the final answer

ok i think i got it

thanks

.close

[lone heart]

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[raven field]

hi im hiaving trouble finding du/dx

[raven field]

heres my work, would du/dx = 0/1

since thats hte only answer i can think of

[tardy stag]

$\frac2x = 2x^\inv$

[ocean seal]

Hayley

[tardy stag]

,tex .diff rules

[raven field]

o

[ocean seal]

Hayley

[raven field]

uh

which rule is dat

[tardy stag]

power rule

[raven field]

o nice

tysm

.close

[lone heart]

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[deep topaz]

I need help with these 3 questions if anyone has time

[lone heart]

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

[deep topaz]

For the first two I've tried different formulas and values but still can't get the right answer

The third one I just don't know how to do it

[lone heart]

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

[deep topaz]

In order of the questions

Sorry it's kind of messy

Ah wait hold on

[wary stream]

My suggestion, you should focus on one question at a time. Sending all 3 creates a good amount of confusion

[deep topaz]

Ok

This one first then?

.close

[lone heart]

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[tribal wraith]

am i doing something wrong here, it said find the reference angle for 2.4 and i subtracted it with pi/2 and got 2.4

[wary stream]

More context is needed please

[tribal wraith]

I got a and c correct, but im confused about b

[wary stream]

How did you get 2.4 if you did 2.4 minus pi/2?

[tribal wraith]

i did 2.4-2pi which gave me 2.4, and then i subtracted that by 6.28 which gave me -3.88 and I added into 2pi where i got 2.4

sorry if i worded wrong

did i miss a step

sorry this is confusing

[tacit arch]

Enter 2.4 - pi/2 into a calculator

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[lament steppe]

hey i need help with letter D, i need to find the restrictions on the equation but idk how to do that

[lament steppe]

pls @ me so i can come back quick and respond to any questions 🫶🏾

[naive crystal]

What have you tried??

[lament steppe]

dw i got it ty!

,close

.close

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[regal lagoon]

I need to find the derivative of this, but I don’t know what I did wrong

[tawny tapir]

have u used chain rule?

[torn elk]

is this f(x)? I can't read well

[tawny tapir]

.'

[gleaming ridge]

Your work is correct!

You did a small mistake

[regal lagoon]

4x^9

[gleaming ridge]

The 5 is in the numerator

Not in the denominator

[regal lagoon]

Ohhhhh

Thank you

[gleaming ridge]

😋

How did you think of dropping it

[regal lagoon]

I think I just though to drop the whole thing to the bottom rather than just the parenthesis

[gleaming ridge]

it's not 5^(-1) right 😛

[gleaming ridge]

i see i see

[regal lagoon]

.close

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[tribal wraith]

I need some help with this, I don't know where to start or what to do

[vale wigeon]

if the number -21 was anything else, would you be able to do the problem?

[tribal wraith]

nvm i just got it, 92.73 sorry

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[dapper lynx]

Hey all

[dapper lynx]

Double integrals question

I followed it through to its logical conclusion and solved it - but I think I've interpreted something wrong...

That third, recursive part seems... well, maybe my teacher is just trying to prove a point

Ohhhh... I think I see a problem...

Among others, potentially XD

That inner integral is supposed to be from x^2 to 1, not from 0 to x^2

Is what I'm thinking. Beyond that, is my interpretation correct?

No... I was right the first time (for B)

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[civic willow]

I understand that this is probably not the typical question, but I was wondering if anyone knew of any cartesian or parametric functions that could describe something like this. My guess is they would require some paramentric integrals with variable bounds, like the euler spirrals, but I'm not sure.

[buoyant storm]

,rotate

[buoyant storm]

i will try

[civic willow]

Thank you, appreciate any efforts.

[buoyant storm]

my com died

sorry

more precisely, desmos died

[civic willow]

Wow thats

what did you try 😂

I could try graphing it in mathematica

[buoyant storm]

um

$(\int_{0}^{t}sin^2(s^2) ds, \int_{0}^{t}cos^2(s^2)ds)$

[ocean seal]

orthogonal_1

[buoyant storm]

idk i have no idea

i just played around with euler spirals

[civic willow]

Yeah I get the feeling that it'll be the approach somewheree

[buoyant storm]

i feel like i've seen this pattern somewhere before

[civic willow]

Its a bit similar to a lemniscate

but I'm not sure if that actually helps me at all

[tardy stag]

there's this which isn't really what you want but is something maybe sort of idk

[buoyant storm]

WOW

nice

[civic willow]

That is actually really good

[tardy stag]

i can't figure out what to tweak to make it more pointy

or otherwise change it

[civic willow]

I'lll start pllaying around with it too

[tardy stag]

oh just make it b sint for the y part, that helps a lot

[civic willow]

I did this

Yerah lol

this is always my

function manipulating strat

Sttart adding random letters

[tardy stag]

lmaoo

[civic willow]

absolutely crushed it

alright well thats rreally excellent stuff, can I ask how you came up with the original?

[buoyant storm]

indeed lol

[tardy stag]

this is why kids shouldn't have phones, then they'll spend hours just drawing random polar crap in their calculators

[tardy stag]

i remembered that i tried graphing random polar functions in my calculator and one of them was a figure eight

like.... fifteen years ago

then just thought about what i needed for that, after a few tries i had that

then added the +t/a parameter to make it move

[civic willow]

Yeah nice!

[tardy stag]

how
i don't think you understand how bored i was on the school bus

[buoyant storm]

loll

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@civic willow Has your question been resolved?

[tardy stag]

@civic willow if you want it to follow a path i figured that out too https://www.desmos.com/calculator/w42os0cjru

you do need the normal vector to the path though

no, this method breaks down with complicated graphs ugh

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[vivid heath]

for this question

i understand how to do the dot product

however im use to working with positive angles for this

whats the difference for getting the angle in this case when this is what im usually doing?

[violet flare]

put the vectos in cartesian space and compute normally. negative angle of -110 would be in the fourth quadrant

[vivid heath]

third quadrant?

[violet flare]

yes third sorry

[vivid heath]

np np

[vale wigeon]

@vivid heath draw your two vectors starting at the same point

the angle between them should become obvious if you do that correctly

[vivid heath]

Do I look at them from the other direction though ?

[vale wigeon]

you've managed to overthink it...

[vivid heath]

lol

[vale wigeon]

here literally ive overlaid them just now

do you see what the angle between vectors A and B is

[vivid heath]

110+35

145?

[vale wigeon]

bravo

[vivid heath]

I hate overthinking lol

Ty

[vivid heath]

can i have a negative dot product?

or since its absolute value i take the positive?

[vale wigeon]

what's "it"?

[vivid heath]

sorry

[vale wigeon]

yes, the dot product of two vectors can be negative.