help-33

[still temple]

no problem

when you done with a problem dont forget to .close

[merry ivy]

is the formula used geometric?

[still temple]

yeah the one were using is geometic, you may also hear it refured to as exponentinal

[merry ivy]

the formula i wrote down is
A n = a1 (r^ni1)

[still temple]

the alebraic forumla is

$C_n = 1.1^n*9$

[elfin berry]

aspwil

[still temple]

that looks fantasitc, one sec

[merry ivy]

?

should i close?

[still temple]

no we dont do homework for people, we explain how to do homework. if you need help see #❓how-to-get-help

[still temple]

do you have no more qeuestions?

[merry ivy]

im going to close so i can give this channel to you

.close

[marsh citrus]

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[wind chasm]

looks good to me

[crystal plover]

A bit complicated to me

And what I got is just 9e^(3x)+14

[wind chasm]

2nd derivative?

[crystal plover]

Yup

[wind chasm]

huh

[tacit fjord]

yh it should not be a lot

i dont know wth has happened there

[tacit fjord]

???

[wind chasm]

wait where did ln come from

[tacit fjord]

how the heck are there logs

[wind chasm]

in the first step

[wind chasm]

that was me just now

cramming for this test

[crystal plover]

Well

[wind chasm]

there shudn't b logs in this problem

[tacit fjord]

no

its just plain wrong

$x^2$

[elfin berry]

Shuri2060

[tacit fjord]

The derivative of this is 2x

[wind chasm]

first derivative is 3e^(3x) + 14x

[crystal plover]

$$\frac{d}{dx}(e^{3x}) = 3e^{3x}$$

[tacit fjord]

how the hell do you get logs

[elfin berry]

skypirate

[tacit fjord]

you're confusing things

The derivative of 2^x is (ln 2)2^x

just no.

[wind chasm]

u don't use logs when dealing with e as the base

[tacit fjord]

you're just mixing things up

and as a result somehow confused on this problem

[wind chasm]

I think he gets it

[tacit fjord]

$$y' = (e^{3x}+7x^2)'$$
$$= (e^{3x})'+(7x^2)'$$
$$= (e^{3x})'+7(x^2)'$$

[elfin berry]

Shuri2060

[wind chasm]

On a side note can any1 solve my question on related rates in help-10

[tacit fjord]

?

https://i.imgur.com/6tKfNGl.png

You realise this line is nonsense?

How do you even get it

y is certainly not equal to this

[wind chasm]

ok

[tacit fjord]

You just randomly decided to multiply the first thing by 3

and the next thing by ln7x and 0

yes?

[wind chasm]

@last magnet derivative of e^3x is 3e^3x right?

u agree?

then u have 7x^2 afterwards

just use power rule then

it becomes 14x

[tacit fjord]

It doesn't matter what method you use, as long as it's correct

Right now you have something which is anything but

Why did you randomly decide to differentiate

you write y =

Only on the next line you start writing y' =

[wind chasm]

ok so now u have 3e^3x +7x

[tacit fjord]

Do you see the inconsistency here to begin with?

Alright, next, assuming the typo

[wind chasm]

they're asking for 2nd derivative

[tacit fjord]

That isn't how to differentiate 7x^2

[wind chasm]

9e^3x +7

think that's it

[tacit fjord]

Look man

there's no point giving the answer

[wind chasm]

I'm not I'm tryna explain it :/

[tacit fjord]

When they have a fundamental misunderstanding of how derivatives work

$$(x^2)'$$

[elfin berry]

Shuri2060

[tacit fjord]

First you need to recognise this MUST be differentiated using the power rule

There is literally no other way

What you are trying to propose is plain wrong.

[wind chasm]

https://tenor.com/view/office-michael-scott-thank-you-gif-5278681

[tacit fjord]

Show how then

[wind chasm]

no

just take the derivative of y'

that's all the 2nd derivative is

[tacit fjord]

There is no regular way

and non-regular way

there is only one way

$$x^2 = e^{\ln x^2} = e^{2\ln x}$$

[elfin berry]

Shuri2060

[tacit fjord]

If you are proposing something like this

It's just... totally unnecessary

and you're doing it wrong even so

[wind chasm]

https://www.youtube.com/watch?v=RtGjBRIwONA @last magnet I'd reccomend this video

[tacit fjord]

If you want to differentiate this, then apply the chain rule properly

The answer is definitely not what you wrote.

Or maybe, just recognise the power rule

x^2 -> 2x

$$y' = (e^{3x}+7x^2)'$$
$$= (e^{3x})'+(7x^2)'$$
$$= (e^{3x})'+7(x^2)'$$
$$= (3e^{3x})+7(2x)$$
$$= 3e^{3x}+14x$$

[elfin berry]

Shuri2060

[tacit fjord]

So this is what you should have written down for the 1st derivative

[wind chasm]

. . .

[tacit fjord]

wrong and wrong

[marsh citrus]

@last magnet Has your question been resolved?

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[marsh citrus]

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[sonic roost]

13-18

[hexed solar]

use formula $slope= \frac{y_2 -y_1}{x_2-x_1}$

[elfin berry]

BrownMunde

[sonic roost]

Thanks

.close

[marsh citrus]

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[whole sleet]

Is that (z1/z2)/z3 or z1/(z2/z3)?
Note they are different

I'm guessing by your work, it should be (z1/z2)/z3

Anyway, you can write the division in rectangular form. See this source:
https://www.cuemath.com/numbers/division-of-complex-numbers/

[marsh citrus]

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[marsh citrus]

@still temple Has your question been resolved?

[civic moss]

Multiply both sides by 3

@still temple

[still temple]

so ur saying multiply 3 to cancel itself then 5

or u multiply 2

?

[civic moss]

Look

$$3 \cdot \frac{2x + 5}3 = 3 \cdot 7$$

[elfin berry]

abs_0

[civic moss]

$$2x + 5 = 21$$

[elfin berry]

abs_0

[still temple]

oh u mutplpy the 7?

alr

so when u mutply the 3 it canel itself right

[civic moss]

Right

Yeah 3 times something divided by 3 equals that thing

Think about it

[still temple]

im confused but either way i get it

ill try to do the problem now

[civic moss]

Ok

R u in algebra rn

[still temple]

yes im in gr 10

Is that correct

[marsh citrus]

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[still temple]

Can anyone help find the range pls

[still temple]

I got -8<y<10

[marsh citrus]

@still temple Has your question been resolved?

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[humble mortar]

I’m in 10th grade and I finished a unit on that a while ago

[marsh citrus]

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[dreamy fulcrum]

what am i doing wrong?

oops, wrong screenshot. Meant to attach this:

[opaque wolf]

Might they possibly want f(x) in the limit box

[dreamy fulcrum]

just tried it, I got an error saying that the only symbols allowed are numbers and x

and-+ and whatever

[opaque wolf]

by f(x) i mean each piecewise function

So as the limit approaches 0 from the left, which function would be used to find the limit

[dreamy fulcrum]

oh i see, instead of the derivative

[opaque wolf]

then evalulate what it approaches

Maybe

[dreamy fulcrum]

let me see

no, unfortunately

[remote elbow]

Are you sure you didn't mix up the left and right hand side?

[dreamy fulcrum]

I've tried both to make sure

[opaque wolf]

do you know what else could possibly be the issue

[remote elbow]

Show what you did

[dreamy fulcrum]

i tried the one above, and this:

along with:

[opaque wolf]

this is weird it looks like you have the concept down but i have no clue what they want to input

[remote elbow]

Again make sure you understand limits approaching from the left and right

[dreamy fulcrum]

Regardless of which way the limit approaches from, the derivative should be the same, right?

[dreamy fulcrum]

yeah, i feel the same way

[noble nebula]

In this task, you're literally proving it is not

aren't you?

[noble nebula]

I'd suggest checking the order

[remote elbow]

You lucked into the correct derivative. The first step is understanding which function in the piecewise to use for the left hand and right hand limits

[opaque wolf]

Remavas how did you speedrun active

[noble nebula]

yes

Honest answer: I don't know

I spent a couple of days with these help channels

[opaque wolf]

Both you and Gamer Dio got it out of nowhere

I guess I have to get on the help channel grind for that ethos

[noble nebula]

Probably 🤷

I have about 3k messages

maybe that's the threshold

[opaque wolf]

@dreamy fulcrum ^

[dreamy fulcrum]

oh

[opaque wolf]

try putting in -5 into the 2nd box

[dreamy fulcrum]

Oh. i see the problem then

it doesn't like that either

how

[opaque wolf]

can you see the answer? I have no clue what's happening

[dreamy fulcrum]

by order, you mean, which equation goes to which box?

[noble nebula]

yes

from 0-, use the 9x^2 + 2x part

[dreamy fulcrum]

I've tried that as well. It doesn't accept it

[noble nebula]

https://tenor.com/view/nick-young-question-marks-what-excuse-me-huh-gif-4486363

Yeah this is weird

[dreamy fulcrum]

alright im giving up

will ask prof tmrw

thank you guys though! i appreciate it!!

.close

[marsh citrus]

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[marsh citrus]

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[stray edge]

im not suire hwo to use the cross product

to find area

ive found qp x qr

to be [-7, 7, 3]

[cunning citrus]

Norm of cross product gives area of the parallelogram spanned by those vectors

[stray edge]

how?

[cunning citrus]

compare with the trig formula for area

$A=\frac{1}{2}ab\sin(C)$

[elfin berry]

Mosh

[cunning citrus]

$a=\norm{u} \
b=\norm{v} \
C=\theta \implies A=\frac{1}{2}\norm{u}\norm{v}\sin(\theta)=\frac{1}{2}\norm{u\cross v}$

[elfin berry]

Mosh

[stray edge]

why does |u x v| = |u||v|sin(theta)

[cunning citrus]

the norm of uxv is just that

Google a proof/look for the proof in your text

[marsh citrus]

@stray edge Has your question been resolved?

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[lost stream]

I need help with this question

[lost stream]

I don't understand what they are trying to ask me

[warped goblet]

It's the gray region

The intersection of A' and B'

[still temple]

No

[lost stream]

IV

[still temple]

Well I thought II is what he meant by grey

The background is grey yea

[warped goblet]

.

[lost stream]

so

[still temple]

How is it the interestcion of a and b

[warped goblet]

A prime

And B prime

[lost stream]

it's intersection of A' and B' because they dont want anything in a or b just outside of it

and IV happens to be the universal set

[still temple]

Ya

[lost stream]

Ok good

what would the difference be if it was a union?

[still temple]

That's his answer

He didn't ask to define that

He wanted to answer of the question

The IV part on the background

[warped goblet]

His answer was A' union B'

Which is wrong

[lost stream]

ya i'm asking what would be the difference

[warped goblet]

Oh

[lost stream]

if it was a union

like how do u notice it is a intersection

instead of a union

[warped goblet]

Union of A' and B' includes the parts where A' doesn't intersect with B'

Intersection is a region where both of them occur at the same time

[still temple]

Mb

[warped goblet]

Basically A union B is a region where either A or B occurs, but A intersect B is a region where both A and B must occur

[still temple]

The intersection here is the II region ig

[lost stream]

I get your first explainatioin

[warped goblet]

Yea A intersect B is II

[lost stream]

ur second one about intersection sounds similair to the first tho

[still temple]

And then what is IV

[lost stream]

sorry

I don't get your second definition

[warped goblet]

A' intersect B'

[still temple]

Ya

[warped goblet]

A union B is a region where either A occurs, A and B occurs, or B occurs

A intersect B is a region where only A and B occurs at the same time

[warped goblet]

For example A union B is the area of both of these circles, but A intersect B is just the area of II

[lost stream]

ohh

so union would've been the whole thing

nah I get it dw

thank you

!

[warped goblet]

Yep

[lost stream]

If you don't mind I will post a problem question later on

if you'd like to help

[warped goblet]

Sure if I'm around

[lost stream]

Ok so this question I got n(H')-n(F)

would that just mean my answer would be -9

[warped goblet]

Idk what's the full question but it's basically the number of elements which are not in the set of H minus the number of elements in the set of F

[lost stream]

since it doesnt want whats in the H set

should I get my intersected ones and -?

[marsh citrus]

@lost stream Has your question been resolved?

[karmic jay]

@lost stream for nH' get values in F which doesnt have mutuality with H and for F take all values of F

and yeah it makes the solution negstive

[lost stream]

ok I got -9

do u reckon that is right @karmic jay

[karmic jay]

No

[lost stream]

how come?

You said get the values of F minus the ones that have a mutuality with H

[karmic jay]

that doesnt have*

[lost stream]

ya my fault that doesnt have a mutuality

the only ones mutual are the J heart, Q heart and K heart

and since they dont want nothing in H

I got J spade, Q spade, K spade, J diamonds, Q diamond, K diamond, J clubs, Q clubs, K clubs

which is 9

@karmic jay

[karmic jay]

And subsract it by nF

[lost stream]

ya which is 0?

wait what

Nf is those that I mentioned

[marsh citrus]

@lost stream Has your question been resolved?

[marsh citrus]

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[marsh citrus]

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[wooden schooner]

This set up is giving me -16pi instead of 16 pi

<@&286206848099549185>

[marsh citrus]

@wooden schooner Has your question been resolved?

[marsh citrus]

@wooden schooner Has your question been resolved?

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[sharp jay]

If I've proved the first statement, how can I prove the second?

[marsh citrus]

@sharp jay Has your question been resolved?

[plush elk]

You prove a general statement, that is P(D)((e^rx)u(x))=(e^rx) P(D+rI)(u(x)) where D(f(x))=f’(x) and I(f(x))=f(x) P is a polynomial

This can be done by proving inductively that D^k((e^rx)u(x))=(e^rx)(D+rI)^k(u(x))

Then P(D) is just a linear combination of D^k for some k, for the same reason , P(D+rI) is also a linear combination of (D+rI)^k for those k with the same coefficients. Add those equations for each k together you get this general result

[sharp jay]

sorry i understand almost nothing of what was said

this is where im at

[plush elk]

What?

[sharp jay]

thats my proof for the first part

and then i bs the second part bc idk how to prove it with the summation

[plush elk]

For example,D(f(x))=f’(x),D^2(f(x))=f’’(x). Given a polynomial P(t)=2t^2+4t+5

P(D)(f(x))=2f’’(x)+4f’(x)+5f(x)

Got it now?

[sharp jay]

oh ok

are we talking about the second part or first part?

like is my first part right?

[plush elk]

P(t)=2t then P(D+rI)(f(x))=2(D+rI)(f(x))=2f’(x)+2rf(x)

No

I am talking about the statement I gave you

And your problem is just a special case of that statement

You see P(t)=Π(t-r_j) in this case

[sharp jay]

ok

[plush elk]

You are proving P(D)((e^rx)u(x))=(e^rx) P(D+rI)(u(x)) where P(t)=Π(t-r_j)

But you can prove it for any polynomial P just as what I said

[sharp jay]

t or d/dt?

[plush elk]

No No, t is just a indeterminate, you can choose any symbol for example s, P(s)=3s^2+7s-6 if you like

[sharp jay]

so in this case t=d/dt?

[plush elk]

No

You are not listening at all

D=d/dx

[sharp jay]

im listening i j dont understand im sorry

[sharp jay]

yes

[plush elk]

Wrote it here, the third and the fourth line, I told you what D and I mean

If P(s)=Σa_j s^j, and you have proven using induction that for any j, D^j((e^rx)u(x))=(e^rx)(D+rI)^j(u(x)), then multiply a_j on both sides, and add those equations together you have your result.

[sharp jay]

i have no clue where all these letters are coming from

[plush elk]

Then I am done, cause I gave every definition, I don’t know why you refuse to understand

[sharp jay]

youre trolling

[plush elk]

I don’t know where I could possibly make it more clear

[sharp jay]

then youre just a terrible teacher

[plush elk]

Where exactly, what symbol you don’t understand?

[sharp jay]

where the hell is a_j and s^j coming from

[plush elk]

Polynomial my god

You don’t know what polynomials are? Like p(s)=4s^3-7 is a polynomial

[sharp jay]

yes i know what a polynomial is

[plush elk]

a_3 =4, a_0=-7 this case

[sharp jay]

i never knew this notation. i understand now.

[plush elk]

Okay please continue, since your confusion is cleared

[sharp jay]

s^j?

[plush elk]

?

[sharp jay]

i just would rather use the symbols from the problem bc idk how any of them relate

[plush elk]

S is the indeterminate

[sharp jay]

right but i still dont know what that means

[plush elk]

You can use any letter you want as I said

You had problem with t so I used s instead

[sharp jay]

oh

s is the variable of the polynomial

[plush elk]

Yeah. Use whatever you want, G, p(G)=3G+17

[sharp jay]

i prefer t

[plush elk]

Okay

[sharp jay]

ok so take me step by step

[plush elk]

Replace every s with t

[sharp jay]

lets start by are you proving statement 1 or 2 rn?

[plush elk]

No your question one and two are both results of that statement I gave. I am telling you how to prove the statement

[sharp jay]

oh so youre proving both simultaneously?

[plush elk]

Your question 1 is the case where P(t)=t-r_j, your question 2 is the case where P(t)=Π(t-r_j)

[sharp jay]

oh ok

[plush elk]

I am proving the statement, your questions are going to be solved instantly simultaneously

[sharp jay]

ok

so what is the statement again?

[plush elk]

Begin with my first message here

[sharp jay]

so can you explain P(D)

[plush elk]

Here

[sharp jay]

and is e^rx e^{rx} or x e^r

[plush elk]

e^(rx)

[sharp jay]

D(P) then? why is it P(D)?

[plush elk]

No, no. I didn’t do anything to the polynomial, I didn’t take derivative of it, nothing

I use that example again:

P(t)=2t^2+4t+5 then

P(D)=2D^2+4D+5I

Just like D is a map from functions to functions , mapping f(x) to f’(x)

[sharp jay]

whats the i or l on the end

[plush elk]

P(D)=2D^2+4D+5I is also a map from functions to functions, mapping f(x) to 2f’’(x)+4f’(x)+5f(x)

[plush elk]

Again, here, my first message

I is the identity map, mapping f(x) to itself

If(x)=f(x)

[sharp jay]

gotcha

may i ask what the point is of even including it then if it does nothing

[plush elk]

What do you mean nothing? It solved your questions as corollary

It’s actually useful sometimes in ODE

[sharp jay]

are you sure there isn't a simpler way to prove this? we didn't go over any of this stuff :/

[plush elk]

But I don’t study analysis, I just happen to see that formula of a MIT ODE lecture video when I was a high school student

Why are you reluctant? It didn’t generalize much, just replaced P(t)=Π(t-r_j) with any polynomial P

[sharp jay]

its just hard to read in this type of font

im not grasping much of it

[plush elk]

It didn’t bring up any new idea new result

Just rephrased your question in a more clear and general form

[sharp jay]

i dont understand how to prove your second line tho "by induction" or whatever

[plush elk]

If that’s true for k , you prove it’s true of k+1

[sharp jay]

and i have no idea how to do that

is D^k the kth derivative

[plush elk]

That is if D^k((e^rx)u(x))=(e^rx)(D+rI)^k(u(x)), you show that D^(k+1)((e^rx)u(x))=(e^rx)(D+rI)^(k+1)(u(x))

Yes

Just D^(k+1)=D(D^k)

Apply D on both sides of D^k((e^rx)u(x))=(e^rx)(D+rI)^k(u(x))

[sharp jay]

apply?

oh like take the derivative of both sides?

[plush elk]

D^k((e^rx)u(x))=(e^rx)(D+rI)^k(u(x)) therefore

D[D^k((e^rx)u(x))]=D[(e^rx)(D+rI)^k(u(x))]

Yes

[sharp jay]

like product rule?

[plush elk]

Yeah, product rule on right hand side

[sharp jay]

oh i was gona do that with e^rx u

[plush elk]

D((e^rx)v(x)) where v(x)=(D+rI)^k(u(x))

The left hand side is D^(k+1)((e^rx)u(x))

[sharp jay]

dont know how to take derivative of (D+rI)^k u

[plush elk]

You don’t need to

[sharp jay]

im lost then

[plush elk]

I feel like I am feeding you the answer. D[(e^rx)(D+rI)^k(u(x))]= [(e^rx)(D+rI)^k(u(x))]’=(e^rx) [(D+rI)^k(u(x))]’+(re^rx)[(D+rI)^k(u(x)]=(e^rx) D[(D+rI)^k(u(x))]+(e^rx) rI[(D+rI)^k(u(x))]=(e^rx)(D+rI)(D+rI)^k(u(x))=(e^rx)(D+rI)^(k+1)(u(x))

[sharp jay]

its ok im just trying to understand this. i wont even answer the question with what youre giving me bc i dont get it. i think i j give up.

[plush elk]

I feel so helpless, nothing I say can make you read it. No matter how detailed I explain, no matter how many extra explanations I want to add, there is nothing I can do to make you accept it.

[sharp jay]

im clearly not as advanced as you are

i dont want you to think im denying your explanation. i truly cannot comprehend. but its not your fault.

[plush elk]

Many people just hate to see anything they are not familiar with , unless it comes from their superior, if their school force them to learn

Otherwise there’s no reason for them to learn anything at all

[sharp jay]

i hate seeing it bc i dont understand it

i want to learn it but i cant learn what youre telling me in an hour

[plush elk]

I am willing to explain it in any way you like. Anything you still don’t understand, I am happy to add extra explanations, give examples. But I am not the one that’s reluctant

I am willing to go through it as many times as you need but you have quitted.

[sharp jay]

i j feel bad wasting your time

i cant learn through this

[plush elk]

Yeah many people just keep saying they can’t. Anyway, if you change your mind I am here available.

[sharp jay]

i appreciate it. if my prof gives a solution, maybe ill compare it with yours and see if it helps.

[sharp jay]

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[vagrant mural]

How do I integrate things with stuff underneath the fracture?

[long arch]

you have to spot a reverse chainrule i think its called

if you have the integral of f'(x)/f(x) then it evaluates to ln(f(x))

[vagrant mural]

Can u show me on the 1st question

[long arch]

and the first integral is (f'(x)/2)/f(x) so if you factor out that constant of 1/2 you get it in the right form

[vagrant mural]

And I'll do the 2nd

Cause I don't understand it with words. I tried to integrate the top and bottom separately but thats wrong.

I think.

[long arch]

idk how to explain it what background knowledge do you have

can you do u substitutions and stuff

[vagrant mural]

I could but I'm bad

So the du for c would be the denominator?

[long arch]

let u = the denominator for part b and solve from there

[vagrant mural]

Can u physically show me.

[long arch]

gimme a sec to do it

[vagrant mural]

Ok thank you very much

[marsh citrus]

@vagrant mural Has your question been resolved?

[vagrant mural]

Are u sure u did this right?

How can u get logs in there?

[long arch]

integral of 1/u is ln(u)

[vagrant mural]

Nice

Thanks

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[ornate dragon]

How to calculate this with mean value theorem I know it's a Riemann series and there is well definite methods to this but I want to use the theorem.

[plush elk]

It doesn’t converge

$\frac{1}{k+n} \geq \int_{n}^{n+1} \frac{1}{k+x} dx=In(k+n+1)-In(k+n)$ so choose any integer m greater than k, the sum of first n terms is greater than sum of first m terms +In(k+n)-In(k+m) it goes to infinity when n approaches infinity.

[elfin berry]

Cogwheels of the mind

[ornate dragon]

tnx so much so the intervall is [k, k+1] an the function is f(x)=ln(x+n)

[plush elk]

Better not use the same symbol

When the interval is [n,n+1] the function is 1/(x+k)

[dry trail]

Seeing it as a Riemann sum, it appears to converge though, mmm what is happening?

[plush elk]

?

It doesn’t

[dry trail]

The sum is from k=0 to n right?

[plush elk]

You replace n with n^s, it converges iff n>1

[plush elk]

No,limit

[dry trail]

Mmm but I think the sum is on k

[plush elk]

Unless positive infinity count as answer

Shit…

I was so dumb

In(2) right?

[ornate dragon]

yes

[dry trail]

Yeah that is what I thought as well

[plush elk]

$\int_{n}^{2n+1} \frac{1}{x} dx \leq$ that sum $\leq \int_{n-1}^{2n} \frac{1}{x} dx$

[elfin berry]

Cogwheels of the mind

[ornate dragon]

you used the rectangle method?

[plush elk]

So $In(\frac{2n+1}{n})\leq$ that sum $\leq In(\frac{2n}{n-1})$

[elfin berry]

Cogwheels of the mind

[plush elk]

Then take limit of all

[plush elk]

$\frac{1}{n} \geq \int_{n}^{n+1} \frac{1}{x} dx$ and $\frac{1}{n} \leq \int_{n-1}^{n} \frac{1}{x} dx$

[elfin berry]

Cogwheels of the mind

[ornate dragon]

ok tnx

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[worthy owl]

really quickly. do u put restrictions on the numerator? (operations on rational expressions)

[burnt abyss]

nope

just the denominator needs to be ≠0

[worthy owl]

thank you!!

[burnt abyss]

np

[tacit trail]

I meantheres case like

$$\frac{\frac{a}{b}}{c}$$
c ≠ 0
b ≠ 0

[elfin berry]

Pluton

[tacit trail]

But thats still denominator

[marsh citrus]

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[stark knoll]

Obtain the range of the function....

I was trying to figure this by myself

And everything was okay... I am not on my house so i cannot show the procedure that I took...

But I thought something like... try to just leave the X on RHS....

And to get rid of that annoying square root of the bottom... I elevated everything to the square... but I couldnt solve the entire thing...

So... is there a better method or something that works for everything or it is just that my knowledge didnt let me solve the entire thing but my procedure was correct?

I also tried other things like factorizing some terms and expressing everything like a two variable quadratic ecuation.... but nothing worked

I am fine with just being shown the way

Of... quitting safely a square root from the denominator

quitting -> getting rid of, sorry idk english

[marsh citrus]

@stark knoll Has your question been resolved?

[civic moss]

Hm

[stark knoll]

In all my tries I always finished with x = some thing with a trinomium on the denominator that has an -x on it

[civic moss]

@stark knoll couldn’t you just reason through this one

You can’t plug in any x that is -3 or lower

At -2.999 it’s -5.9/0.00001 which approaches -infinity

And as x approaches infinity, the function becomes sqrt(x) which goes to infinity

So the range would be (-inf, inf)

But I’m sure a lot of that reasoning didn’t make sense lol @stark knoll

[stark knoll]

hmmmm

Yeah I understand

I was searching for a more algebrai... Idk english... A way just using algebra things, ik I could just graph the thing or try values... but if there isnt a clear way I suppose we can close the case

thx u so much anyways @civic moss !! <3

If u for any reason come out with something while sleeping or something like that because those things happen... lmk

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[oak fern]

I would like to compute this :

[oak fern]

$\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} ln(1+\frac{k}{n})$

[elfin berry]

Hmz

[cunning citrus]

what have you tried?

[oak fern]

I was thinking about using limit sum inversion

So I need to prove the uniform convergence

[noble nebula]

Where is Riemann when you need him

[oak fern]

Riemann for ln as f and (0,1) for (a,b)

[noble nebula]

If by ln you mean ln(x), no

[oak fern]

Arh

[noble nebula]

But very close.

[oak fern]

ln(x+1)

[noble nebula]

from 0 to 1. That's how I would do it.

[oak fern]

Okay I’ll try

Thanks

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[opaque seal]

Can someone explain to me how to get the answer of 54.3 in question b and 26 in question c

[opaque seal]

<@&286206848099549185>

[marsh citrus]

@opaque seal Has your question been resolved?

[opaque seal]

<@&286206848099549185>

[mystic minnow]

the farmers are supplying 31 million liters. They are selling all of those liters (either to people or the government) at a price of 1.75
Revenue = price x number sold

[marsh citrus]

@opaque seal Has your question been resolved?

[opaque seal]

@mystic minnow Ty sm! What about question c)?

[mystic minnow]

if the demand is for 5 mil, and the farmers are supplying 31 million, how many liters does the government have to buy?

[opaque seal]

@mystic minnow omg sorry for asking dumb questions. I’m just reviewing my old assignments and I forgot how to calculate it. It sucks that it doesn’t show solutions

[mystic minnow]

not a dumb question. Sometimes things don't click right away. It happens 🤷‍♂️ it helps to have someone readjust your perspective

[opaque seal]

@mystic minnow God bless you🙏🏻🙏🏻🙏🏻

[feral relic]

Pls tell the answer for this question plsss

<@&286206848099549185> pls

Hello

[mystic minnow]

see #❓how-to-get-help to open your own channel and get help

[feral relic]

Why?

[mystic minnow]

because that's how the server works.

[feral relic]

I can't open my own channel

After seeing how to get help

Zybikron

[mystic minnow]

• Pick a channel in the “Available” section and post your problem. The bot will move your channel to the “Occupied” section, and the channel will be reserved for you until your question is answered. Stick to one channel and don't post the same question in multiple channels. Don't ask for help in other channels if no one is responding in the one you have posted your question in.

[feral relic]

I can't understand that

What should I do

[sour relic]

go to available math help channels

occupy one

[feral relic]

How to occupy?

It is telling you don't have permission

[sour relic]

???

[feral relic]

To send messages

[sour relic]

send screenshot

[sour relic]

dude these are math help hidden

I said math help available

[feral relic]

Send screenshot pls

[marsh citrus]

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[marsh citrus]

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[gentle willow]

It's an equilateral triangle.
How to prove ∠Q1 = ∠Q2 = 60
And ∠B1 = ∠B2 = 30

[gentle willow]

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[deep fossil]

for this problem

[deep fossil]

should I just use the half angle formula for tan

oh no

[next ravine]

for cos

cos(x/2) = ?

[deep fossil]

or I make the reference triangle and use the half angle of cosine

[next ravine]

You can try that

[deep fossil]

so would it be cos(sqrt1+12/13over2)

[next ravine]

I haven't done the problem yet, but if you show work and every step then I can look

[deep fossil]

okay:)

look right?

[next ravine]

yup

[deep fossil]

awesome!:) I get it now thanks for your help I just hadnt done a problem like that yet

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[feral relic]

Hello

[fickle fractal]

?

[feral relic]

Hello

[fickle fractal]

erm

what help you need?

[feral relic]

Do you solve my help?

[fickle fractal]

well what is your problem in maths?

[feral relic]

This is pls solve

[fickle fractal]

are we trying to find D or x?

WTF

[feral relic]

We want to find the general solution

[marsh citrus]

@feral relic Has your question been resolved?

[plush elk]

You write cos(x) as (e^ix+e-ix)/2

Then use a formula

$P(D)(e^{λx}u(x))=e^{λx}P(D+λI)(u(x))$

[elfin berry]

Cogwheels of the mind

[plush elk]

Where P is a polynomial, D and I are two operators: D(f(x))=f’(x), I(f(x))=f(x)

[whole sleet]

What is D4 supposed to be?

Is that D⁴?

[marsh citrus]

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[feral relic]

Yes

[marsh citrus]

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[marsh citrus]

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✅

[marsh citrus]

@still temple Has your question been resolved?

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[vivid axle]

Hiya peeps!

Anyone able to help me with this binomial expansion Q. Struggling to well...expand it to put it simply. Looking at past examples, they make sense but this one isn't clicking as quickly as the others. Mostly down to the power.

I've attempted to make a >= 0 by just making it (a+1)^1/n and then using the binomial expansion but it's not exactly clicking as to...well...getting the final answer. Would appreciate the help!

[marsh citrus]

@vivid axle Has your question been resolved?

[marsh citrus]

@vivid axle Has your question been resolved?

[plucky shoal]

I think the hint can help out a lot. \begin{align*} a^{1/n} &\leq 1 + \dfrac{a-1}{n} \ \iff a &\leq \left(1+ \dfrac{a-1}{n}\right)^n & \textrm{by hint} \ &\leq 1 + a -1 = a & \textrm{by Bernoulli's inequality} \end{align*}

[elfin berry]

1345631

[vivid axle]

So that's using the hint. But do we not need to derive 1+(a-1)/n from a^1/n at all?

[plucky shoal]

No. We do not need to derive it. Here is some more detail. For 8 marks, we would need to say:
Let $a \in \mathbb{R}$ with $a \geq 1$ be arbitrary and let $n \in \mathbb{N}$ be arbitrary. Now, note that both $a^{1/n} \geq 0$ and $1 + \dfrac{a-1}{n} \geq 0$. Thus, we can apply the hint.

The point is that we don't know that the inequality printed holds. But it being true is equivalent to the second line I wrote above being true, by the hint. Then, by Bernoulli's inequality, we reach $a \leq a$, which is true. So then, we can start from the last line, and go back the reverse way. In our answer we just present the answer in the reverse way to present a logically sound argument.

[elfin berry]

1345631

[vivid axle]

I'm surprised that there was no need to derive it, as in lecture we've used the binomial expansion in order to get a term that (once completed with the algebra of limits) tends to a value

But I'll take it if it's that simple, but the question as a whole is 8 marks, theres one other part using the sandwich (or squeeze) theorem below

[plucky shoal]

You can make your solution more complicated so that you can be more confident that you obtain all marking points. For example, do not apply Bernoulli's inequality (note: this is derived from the Binomial theorem). Instead, use the Binomial theorem itself, and then derive the fact that Bernoulli's inequality holds from that.

You can start from (1+ (a-1)/n)^n = ... [apply Binomial theorem]
Truncate some terms, which are all non-negative, to get (1 + (a-1)/n)^n >= a.
Use the hint.

If you look carefully, this is just the logic I presented above, but in the reverse way. I hope this can help elucidate the question a bit more.

[vivid axle]

That makes sense now that you've explained it in a lot more detail than I was expecting lol

Now onto the next part of the question...to see if I can actually solve the sandwich theorem.

Thanks for the help!

[plucky shoal]

Sorry for explaining it in more detail than you were expecting. In hindsight, I should have left more things for you to think about, so that you derive more benefit from thinking about it on your own. I apologise for this.

Anyway, good luck for the rest of the Q.

[vivid axle]

No need to apologise for explaining it too much, means that I can see every step that you've taken rather than being left in the dark (I already experience that enough in further applied maths)

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[still temple]

How can I approach this???

I'm pretty lost

Distance usually to me means some point minus another one

[trail flare]

I’m going to assume this is the shortest distance because otherwise (unless the point lies on the line) the distance is variable

Fine the equation of the line

Then find a general form for the vector between the line and point

And consider the fact that must be perpendicular to the direction vector of the line

(I would check if the point lies on the line first honestly)

[still temple]

Is this right?

I'll get the cross product, then divide by the norm of u

[still temple]

For ur attention

[trail flare]

Personally I think the best way is to write a general eqn of the line (in one vector) and then the scalar product of that and the direction vector is 0

That lets you solve for t which essentially solves the problem

[still temple]

So I'm teaching myself here.....

Ah NM. I'm done for a bit

I'm more confused rn lol

Tyty

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[marble lodge]

I need help with this

[marble lodge]

I tried making a matrix and tried solving it with the conditions given but idk how to proceed

all I know is the dimension for a slew symmetric matrix is n(n-1)/2 but i don’t think that will work here

[wind wadi]

What does your matrix look like?

[marble lodge]

i’ll dm u

[wind wadi]

Especially, what kind of structure does it need to have to satisfy A^T = -A?

[marble lodge]

it’s a skew symmetric matrix

so all the diagonals are gonna be 0

and aij = -aji

[wind wadi]

yes

[marble lodge]

so what do i do

[wind wadi]

write the matrix like this including the zero-diagonal and spot a pattern

then to find a basis you can try to row reduce it

[marsh citrus]

@marble lodge Has your question been resolved?

[marble lodge]

how would you do that

[brittle anchor]

❌

[marsh citrus]

@marble lodge Has your question been resolved?

[marsh citrus]

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[still temple]

hi.. can someone help me please: for question number 3

[still temple]

this is what i have so far:

im stuck on the regions increasing/decreasing, regions concave up and down part

thanks in advance

[twin girder]

For increasing/decreasing just solve log(x)+1>=0?

[still temple]

i got x>1/10

is that right

and what would i put on the number line?

[twin girder]

Not 1/10 no

[still temple]

i searched up the answer on google and said that its Increasing on: (1/e, infinity) and decreasing on (0, 1/e)

[twin girder]

Indeed

[still temple]

but i have no idea how they found that

like idk what i should do to get the answer

[twin girder]

Solve this?

That is the derivative you found?

[still temple]

so do i just make the derivative = 0?

i did that and then got x=1/e

[twin girder]

That works since continious, yes

Now you need to check a value above and below 1/e to figure out sign

[still temple]

was the number line i drew above correct? or should i only have 1/e and -1/e on it?

also can i input -1 for the value that is below 1/e

[twin girder]

What number line?

[twin girder]

Do this and you have when its increasing and decreasing

[still temple]

ok

[still temple]

what nuymbers can i input tho

[twin girder]

What is that supposed to be an answer to?

Any value greater than 1/e

And any value less than 1/e

[still temple]

like to help me see where increasing and decreasing

[twin girder]

Yeah no idea what that is supposed to do

[still temple]

okay so i put 2 for the value greater than 1/e and got a positive number

[twin girder]

You only need a single

Like I said

And a single value less than 1/e

[still temple]

what do u mean

[still temple]

and for the value thats less can i use -2?

[twin girder]

Is your function defined at -2?

[twin girder]

You said you checked 2 values - you only need to check a single

[still temple]

no i meant that i used the number two as the value

[twin girder]

Oh sure

[still temple]

like i did this: f(2) = In(2) +1

[still temple]

im not sure😭

idk what to use for the value thats less than 1/e

hello?\

[twin girder]

A value less than 1/e

And ofc you need your function to be defined

[still temple]

ik but what number could i use idk what a value is less than 1/e

[twin girder]

What about 1/10000000000

Or whatever

There are inf amount of numbers less than 1/e

[still temple]

okay i got a negative number as the result

now what

[twin girder]

What does it mean for derivative to be negative?

[still temple]

that its decreasing

i think

[twin girder]

Indeed

Since our derivative is continious it is less than 0 for all values in (0,1/e)

[still temple]

ohh

so basically thats the region its decreasing on

[twin girder]

Yes

[still temple]

so do i say that the derivative is decreasing or that the function is decreasing

[twin girder]

The function is decreasing at a point if its derivative at said point is less than 0

[still temple]

ohhh okok

how can i solve the concvity

[twin girder]

Exact same idea as you just did

[still temple]

.close

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[kind echo]

hi

[kind echo]

trying to figure out how 6/2:7/3:8/4 became 9:7:6

[twin girder]

Multiply num in them all by 3

[kind echo]

why 3 tho?

[twin girder]

Because now they are all integers

[kind echo]

i didn’t understand

[twin girder]

7/3 not an integer

Multiply by 3 and it is

[kind echo]

ohhh

but we can multiply with any number then?

[twin girder]

Yes

2:1 same as 4:2

2nd one is just 1st multiplied by 2

[kind echo]

yes true

okay i get it

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[fluid talon]

hello im working on a math project and im being asked to get the height of a pile and the only information I have on the pile is the angle of repose and the density of it any help? 35° and 90lb/ft^3

[marsh citrus]

@fluid talon Has your question been resolved?

[fluid talon]

<@&286206848099549185>

okay so all ive gotten now i have to set up a trig function but i dont know how to solve it for height

Tan 35° = X/20

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[shut wadi]

howdyyy, does anyone know how to find the perimeter/area of parameterized surfaces? the problem specifically is:

"Suppose a surface has fundamental form I = du^2 + (u2+a2) dv^2 (constant a). Consider a triangle bounded by u = av, u = -av, v = 1.
a. Find its perimeter.
b. Find its area."

subject: differential geometry/tensors

[shut wadi]

my first thought was line integral over the square root of the first fundamental form, but i don't know what z' would be in this case

[elfin berry]

ashreaver

[marsh citrus]

@shut wadi Has your question been resolved?

[shut wadi]

<@&286206848099549185> hi hi, i have a differential geometry question

[shut wadi]

[fluid meteor]

not quite sure if its the correct way, but maybe try to calculate your metric tensor.

you than have to take the squareroot if the determinant of this tensor. Put your parameterization into f.

I guess f will be I here, and multiply I(parameterization)* the squareroot

Integrate this product and this is your surface integral. At least if your surface is 2 dimensional and lives in 3 dimensions. I assume that is the case here

[marsh citrus]

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[shut wadi]

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[marsh citrus]

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[old sable]

what'll be the reasoning that is necessary to set up the ODE?

[still temple]

well at any time t the net acceleration of the rocket ( the rate of change of velocity) is dependent on t (the time passed in seconds from launch) as 7t m/sec^2

therefore dv/dt = 7t

[old sable]

I get that and arrived at this $$\frac{7t^3}{6} + c_1 t + c_2 $$. But now what?

[elfin berry]

hiro

[still temple]

wait did you integrate twice?

this is displacement as a function of time right?

[old sable]

I don't this the question gives me enough information to calculate $c1$ and $c2$.
Yes, I just integrated it twice

[elfin berry]

hiro

[still temple]

I get what you're saying

but you're doing an indefinite integration

you need to do a definite integration or at t = 0 the velocity is also zero so you get c1 = 0 from there

[crystal wraith]

$\frac{dv}{dt} = a$\
\
$dv = adt$\

$\int_0^v{dv} = \int_0^{10}{7t}dt$

[still temple]

and also at t = 0 the displacement ( in this case height ) from the ground is also zero therefore c2=0 as well

[crystal wraith]

Something like this.

[still temple]

dv = adt

[crystal wraith]

Yes...

Mb

[still temple]

now take the limits as shown in the photo

[elfin berry]

What the hell am I doing here?

[still temple]

but also hiro there are 2 parts

[crystal wraith]

Yes, this is just a way you should start.

[still temple]

first you gotta integrate this twice to get how high it reaches until the engine stops

and then you can use the pre existing 0 = U^2 - 2gS and solve for S

where S is extra height travelled after reaching the point where the engine is turned off

[old sable]

💡 Got it!

[still temple]

and U is the velocity at that point

[old sable]

Thanks guys

[still temple]

NP bro

[old sable]

Phew, just setting up the ODE is the most of the weightlifting xD

[still temple]

do you need to setup one equation for both the situations?

[old sable]

I'm good. It's the general feeling whenever I sit down to practice ODEs

[still temple]

Yeah, well wait until you encounter PDE's

[old sable]

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[distant meadow]

what is this 3n−7−(−n−6)

[distant meadow]

3n−7−(−n−6)

[novel fjord]

an expression

[distant meadow]

does it have a end

[novel fjord]

what

[distant meadow]

is there an answer

whats 3n

[novel fjord]

a term

[distant meadow]

coul d 3n be 30

[novel fjord]

if n=10

[distant meadow]

then it would be 30?

[novel fjord]

if n=10

n is just a variable

[distant meadow]

310

[novel fjord]

if you're not given a value for n, keep it as a variable

[distant meadow]

yea

[novel fjord]

are you just looking to simplify the expression?

[distant meadow]

yea

[novel fjord]

cause unless you're told that n = something, your simplification is gonna be in terms of n

you just need to distribute the negative and combine like terms

[distant meadow]

i see

ok

[sour basin]

Do you know -(-x) =x

[marsh citrus]

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[young mirage]

does anyone know how to do this?

[young mirage]

I am trying surface integrals

but idk how to continue with this

[marsh citrus]

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[hazy lion]

so I am not confused by the problem so much as the mechanics

i feel this should be straightforward but im choking a little on the exact method

any hints?

[gloomy zealot]

no idea what any of that even is mate, but surely u can multiply it by 0 to have a predictable value

[marsh citrus]

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[hazy lion]

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[marsh citrus]

@foggy trail Has your question been resolved?

[inner acorn]

am i allowed to ask question in here?

[proper zodiac]

#❓how-to-get-help read

[inner acorn]

thanks

[marsh citrus]

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[queen pelican]

just need a brief explanation

of what it'd be

[marsh citrus]

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[merry ivy]

I’m not sure if I did my math right

<@&286206848099549185>

[marsh citrus]

@merry ivy Has your question been resolved?

[dreamy trench]

,w ((3,5,0),(8,-23,0),(-2,5,25)) (x,y,z)^T = (0,0,0)^T

I need a column vector lmao

[proper zodiac]

,w row echelon {{3,5,0},{8,-23,-109},{-2,5,25}}

Wolfy for some reason likes braces for matrices

[dreamy trench]

Ah that makes sense

[proper zodiac]

Oh you copied matrix wrong lol

[elfin berry]

Results provided by WolframAlpha

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Requested by iCaird#6258

[merry ivy]

that doesn't look like my matrix. do you where i messed up?

[smoky siren]

It would have been cleaner imo if you started with R_1 + R_3 -> R_1

[merry ivy]

like at the beginning?