help-0

[fallen tartan]

.close

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[willow stump]

hey guys

[willow stump]

how do I rotate e^x

by 45 degrees?

like from this

to something a bit like this

I tried using a rotation matrix but I don't know how that works

what do I plug into desmos for it to work?

[vale wigeon]

$\frac{x-y}{\sqrt{2}} = e^{\frac{x+y}{\sqrt{2}}}$

[ocean seal]

Kanga Gang Annihilator Ann

[willow stump]

ah ok I just messed up the matrix multiplication

thanks

wait this isn't occupied anymore

how do I make it not occupied

thanks

doesn't that free the room?

.close

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[zinc tiger]

how do i solve 2e)

[zinc tiger]

do i elevate everything with e^

e ln ( ) = e ln2 ?

[lone heart]

@zinc tiger Has your question been resolved?

[median dirge]

Hi... what do you have to do in this question?

[raven rover]

To solve for x: You can use log properties to combine the LHS into one logarithm, and then exponentiate both sides

[raven rover]

Exponentiate: this ^^

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[zinc tiger]

can you display it on paper how i may do so

.reopen

[lone heart]

✅

[raven rover]

I'd try it yourself first

[zinc tiger]

okay, im right on it

[raven rover]

As long as you're careful and you do the algebra correctly, the math will reward you with the correct answer :)

$\ln{x} - \ln(x - 1) = \ln{2}$

[ocean seal]

Shen

[zinc tiger]

something is incorrect i think

[raven rover]

Your combining of the logs is incorrect

Use Rule 2

Going backwards

[lone heart]

@zinc tiger Has your question been resolved?

[zinc tiger]

i made ln (x -1 ) into ln (x/1) in accordance with rule number 2

don't know where my formulation went wrong though

what about this

[raw shard]

@zinc tiger $ln(x-1) \neq ln(x)$

[ocean seal]

quantum

[raw shard]

where did you even come up with this

$ln(x) - ln(x-1) = ln\left(\frac{x}{x-1}\right)$

[zinc tiger]

idk im like whicked smart

[ocean seal]

quantum

[raw shard]

i’m not going to help you if you aren’t gonna take this seriously

[zinc tiger]

i am looking at your solution rn

im taking it seriously

[zinc tiger]

im confused what happened to ln2 at the right of the equation

with the line drawn over the = symbol

[raw shard]

i was only focusing on the left side

because that’s what you messed up

$ln(x) - ln(x-1) = ln\left(\frac{x}{x-1}\right) = ln(2)$

[ocean seal]

quantum

[zinc tiger]

thats legit what i did, almost

and you made it seem like im a psychopath for formulating that

no im joking, thanks for the help

dp o eøevate both sides with e^ now?

[raw shard]

i just didn’t know how you basically said x-1 = x lol

yes

[raw shard]

$\frac{x}{x-1} = 2$

[ocean seal]

quantum

[zinc tiger]

thats what i meant^ but it looks wrong to me still

why is there an x under and above

[raw shard]

you had $\ln\left(\frac{x}{x-1}\right) = \ln(2)$

[ocean seal]

quantum

[raw shard]

so do both sides to the power of e

to get $\frac{x}{x-1} = 2$

[ocean seal]

quantum

[zinc tiger]

yeah, but this solution seems ill

[raw shard]

if ln(x) = ln(y), wouldn’t you say x = y?

what do you mean

there’s nothing wrong with this

[zinc tiger]

like the value of x looks undefined but i guess its satisfactory

whateva

[raw shard]

so what would you do now

or can you do this yourself

[zinc tiger]

no no i appreciate your help, thank you

.close

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[alpine sable]

Show that if this function harmonic or not: $Re(\frac{z}{z^3-1})$

[ocean seal]

Cüneyt

[spring mountain]

hiii, help please?

[zealous bloom]

get your own channel

[spring mountain]

my bad, sorry

[lone heart]

@alpine sable Has your question been resolved?

[vale wigeon]

it's the real part of an obviously complex analytic function

[alpine sable]

Yes I understand, but it has been too long to solve (calculation progress), so, is there any shortcut that I missed or something?

A real number could be harmonic and could not be harmonic as well so we need to open it I guess

[vale wigeon]

??

do you have the result that Re(f(z)) is harmonic for any analytic f(z)

it follows directly ish from cauchy riemann

[alpine sable]

I thought the condition is

$u_{xx}+u_{yy}=0$

[ocean seal]

Cüneyt

[vale wigeon]

yes, and as i said if u = Re(f) with f analytic then it follows from cauchy riemann

[alpine sable]

I am not asking differentialibility

What does cauchy riemann have to do with this question I don't get it

[vale wigeon]

$u_x = v_y$ and $u_y = -v_x$ imply $u_{xx} + u_{yy} = 0$

[ocean seal]

Kanga Gang Annihilator Ann

[alpine sable]

I am gonna check the definiton again

thanks anyways

.close

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[ocean seal]

miku

[vocal slate]

Here is what I have tried:

this means $v \in S_1$ by definition

[ocean seal]

miku

miku

[vocal slate]

But the reason I'm unsure is because I read somewhere that the basis of some set should also be minimal

I tried proving $B$ is a basis of $S_1$
by proving

1. $B$ spans $S_1$
2. $B$ is linearly independent

[ocean seal]

miku

[vocal slate]

but nothing to do with "minimal" or "maximal" (to my understanding)

[lone heart]

@vocal slate Has your question been resolved?

[vocal slate]

i resign this question

.close

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[gleaming tree]

I’m just wondering how the monthly payment got. thanks!

[silver flare]

I think whst they did

Was they toom .0519

Multiply it by 27500

Then add what u get to 27500

Then divide by 15

Thats wring

Wrong

Dont listen to me

[gleaming tree]

ahahaaha thanks for trying tho

[median dirge]

Hi

@gleaming tree

[gleaming tree]

hiii!

[empty ridge]

@gleaming tree
The mechanics is that:
Loan(t+1) = Loan(t) * (1+interest_rate/12) - Montly_payment
The idea is that you Loan(15)=0

[median dirge]

https://emicalculator.net/

Scroll down

And you'll find out

How do they do it

[empty ridge]

It is a goal_seek in excel or if you have a HP-12c,
PV = 27500
n = 15
i = 5.19/12 (edited)
FV = 0
PMT

[median dirge]

EMI is just monthly payment

[empty ridge]

This is aproximate, since the timing of the payment matters (number of days in the month, if you pay in the begining of the month or end)...

[gleaming tree]

OHH NICEE THANK YOUU!!

[median dirge]

You're welcome!

[gleaming tree]

thank you tooo!! i appreciate you guys help!

thanks!!

ill close this now so others can use it thanks again!

.close

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[alpine sable]

I haven't had this solved since yesterday

no help with it

Do I use the second derivative test or first

because when I do the second derivative test its undefined for the critical points

<@&286206848099549185>

[empty ridge]

Hey @alpine sable, I don't think you need to algebrically solve this... Intuitively, the cube root is "scalling down" what is inside it and you can pretend to know whats going on... First draw what is inside the integral and then reason on what will happen to the integral value as x changes... That would be my aproach to this.

[alpine sable]

I dont understand

@empty ridge

[empty ridge]

@alpine sable, You know how to solve this for (1-x^2), right? I mean, how to draw the graph of this function and then reason about what would happen to the integral.

[alpine sable]

not really

im in calc 1

this is the end of derivative calc/start of integral calculus

@empty ridge

[empty ridge]

@alpine sable, ok. What have you learned about integrals?

I mean, so far.

[alpine sable]

Riemann sums

FTC

and how the part being integrated is F'(x)

so

F'(x) = f(x)

thats the approach I took to this problem

[empty ridge]

And the thing about integrals being the area under a curve, right?

[alpine sable]

yeah ofc

and a bunch of complex analysis

[empty ridge]

I would use this idea to draw the integral graph.

[alpine sable]

how

Im so lost

[empty ridge]

You know how to plot the graph of the function inside the integral right?

[alpine sable]

uhh

not really

[empty ridge]

it starts at cubic_root(-3) and ends at cubit_root(-3) and reaches a maximum at x=1,y = cubic_root(0) = 0; right? It will be a distorted parabola like thing

[alpine sable]

why is it cubic root (-3)

I don't know where it has a maximum

I was taking the second derivative

and I was getting undefined

[empty ridge]

the function inside the integral is a cubic_root (1 -x^2), right? At x = -2, this function will be valued cubic_root(1 - (-2)^2) = cubit_root(-3)

[alpine sable]

oh

yeah ok

so yes

maximum I think should be at +1 or -1

because those are the critical points

[empty ridge]

yes, you are right...

[alpine sable]

but my second derivative is undefined

when I plug those values in

[empty ridge]

sorry, i said something wrong...

[alpine sable]

also, Im not sure how to find where its increasing/ decreasing

[empty ridge]

Let me start again...

This will have a maximum cubic_root (1 -x^2) at x=0, y = 1

[alpine sable]

how

if the critical points are +-1

[empty ridge]

The critical points are based on the derivative... Let be step back, and explain my plan to solve this

My idea is to use the information about what is inside the integral to reason about the shape of the integral.

[alpine sable]

Ok

[empty ridge]

What is inside the integral is a parabola like thing: cubic__root( 1 - x^2). This function on the interval [-2, 2] will go from [-3 --> 1 --> -3].

So, it will cross the x-axix at x = -1 and x = 1.

What all this information say about the integral?

First, from -3 to -1, the thing inside the integral will have a negative value, right?

[alpine sable]

Yes

[empty ridge]

That is, we will be accumulating a negative thing.

So the integral will be decreasing in value (acruing to more negative sum)

at x=-1, the function inside turns positive, so the integral will start to become less negative, increasing in value... this will go on until x = 1....

and then, for x> 1, the function inside will become negative again, and the integral starts to decrease.

[alpine sable]

How so

Wouldn’t it be positive?

[empty ridge]

For x > 1, x^2 > 1 --> 1 - x^2 < 0

[alpine sable]

Ok

How would I know the Mac

Maximum

Though

[empty ridge]

How would you reason about it?

This is the thing inside

@alpine sable

[alpine sable]

Uh

[empty ridge]

From this, you can infer the shape of the integral. Given an derivative you can infer what would be happening (not exactly) with the function.

[alpine sable]

How would I do it without graphing software

Like on a test

[empty ridge]

From -2 to -1 the integral derivative (what is inside, would be negative, so the integral ....)

[alpine sable]

Would be positive?

Wait

[empty ridge]

You would have to solve the integral or reason about it intuitively.

Building the intuition now may be hard, but it will give you more flexibility to reason about complex situations and infer what to expect from the solutions of problems

[alpine sable]

I still don’t understand the maximum

[empty ridge]

Analytically what characterize a maximum?

[alpine sable]

Local maximum or absolute?

[empty ridge]

Local (and absolute in a sense)

[alpine sable]

All of the other values are less than that value

[empty ridge]

And in terms of derivatives?

[alpine sable]

First derivative test

Or second

[empty ridge]

Good, both.

[alpine sable]

But

Second is undefined

For the critical points

[alpine sable]

Like I said earlier

[empty ridge]

I'm speaking of a general case.

The graph I posted is for the thing inside the integral (that is, for the derivative of the F, by the FTC)

So, if I can pinpoint the condition for critical points from the thing inside the integral (the F') I wil be able to infer the maximum of F.

[alpine sable]

Wdym

[empty ridge]

The problem gave you a function F(s) that is defined as an integral of another function f(s) = cubicroot ...

By the FTC, the F'(s) = f(s) right?

@alpine sable, does it make sense?

@alpine sable Do you want to start again from the beggining? I am sorry if I'm confusing you.

[alpine sable]

Sure

If possible

[empty ridge]

Ok!

@alpine sable Your objective is to draw a graph and indicate some information about it. To draw this graph, we need to know where it starts. What is F(-2)?

[alpine sable]

integral of the cube root of 3

-3

@empty ridge

[empty ridge]

@alpine sable Nope, -3 is the derivative of F at -2. (Why? Since F(s) is an integral, by the FTC, what is inside the integral is the derivative of F)

The integral from -2 to -2 would result in a 0 sum (imagine the riemman sum from -2 to -2)

So, F(-2) = 0, because it would be the result of a integral from -2 to -2 (a zero area region).

[alpine sable]

ah ok

is there any way you can explain this over voice?

I feel like I'd understand it a lot faster

[empty ridge]

I'm sorry, my english sucks... I'm better at writting than at speaking...

[alpine sable]

gotcha thats fine

[empty ridge]

And my accent would confuse you more lol

[alpine sable]

its ok

[empty ridge]

You also know that F'(-2) = -3 right?

[alpine sable]

yes

[empty ridge]

This means that the graph should be decreasing

[alpine sable]

yes

[empty ridge]

It would keep decresing until F'(x) >= 0, right?

When would that happen?

Remember that what is inside the integral is the derivative of F (using the FTC).

[alpine sable]

yeah

when would what happen

[empty ridge]

When would F'(x) >= 0?

[alpine sable]

-1,1

[-1,1]

[empty ridge]

Great, so the graph of F would be decreasing from -2 until -1, right?

[alpine sable]

yes

[empty ridge]

at -1, you have reached one of the critial points. Does it make sense?

[alpine sable]

yes

[empty ridge]

From that point until 1, F would increase in value, since F'(x) >= 0 for x in [-1,1]

[alpine sable]

yeah

[empty ridge]

Then at 1, what will happen?

[alpine sable]

decreases

below x axis

F'(x) right

[empty ridge]

At 1 it would reach another critical value, right (it would stop increasing, and will start decresing from 1 to 2)

[empty ridge]

Right!

[alpine sable]

how is it increasing

on

-1,1

its decreasing from the maximum until 2

is it not

[empty ridge]

Since F'(x) >= 0 for x in [-1,1], F will be increasing in that interval, right?

From [1,2]. F'(x) <= 0, and therefore F will be decreasing.

In a table:

 x       | F'               |F                    

-2 -> -1 | Negative | Decreasing
-1 -> 1 | Positive | Increasing
1 -> 2 | Negative | Decreasing

@alpine sable does it make sense?

[alpine sable]

yes

thanks

[empty ridge]

@alpine sable Ok, now we have a problem. I'm thinking how you could infer the value of the integral at the critical points...

@alpine sable Where is the function concave up and down?

[alpine sable]

between [-1,1]

its concave up

[empty ridge]

@alpine sable This is true for the F', not for F.

[alpine sable]

Im not sure how to infer it

[empty ridge]

@alpine sable How you normaly determine the concavity of a function?

[alpine sable]

second derivative

[empty ridge]

So, using the information from the first derivative (what is inside the integral) you can determine the changes in concavity of F

[alpine sable]

yes

[empty ridge]

Then, that's it. You've done it!

[alpine sable]

nice thanks

.close

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[empty ridge]

I've just realized that I didn't know what I was doing, I was trying to explain how to solve this problem, but I'm not sure anymore that I know how to solve this. My approach was to use the information on the thing inside the integral to infer the shape of F.
But when I check on wolfram alpha, nothing made sense anymore.
My question: Is this a problem on complex analysis? Or is there a "real analysis" solution? Should I go back and study calc 1 again?

[lone heart]

@empty ridge Has your question been resolved?

[empty ridge]

.close

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[dry echo]

This is my work

Is whatever I've done so far wrong?

[lone heart]

@dry echo Has your question been resolved?

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[alpine sable]

i have the answers but idk how its formed, the answers are on the right side and the actual question is on the left, the dropdown box choices are vertical and horizontal, i just dont understand what the answers are saying, i just need someone to put it in form for me

[alpine sable]

and for the rest of the 9 questions its basically the same process

[hallow hinge]

i think i can explain this

so there are two points H is (-5, -3), K is (5, 3)

they make a line segment

[alpine sable]

ok

ty

[hallow hinge]

you need to find the length of the line

[alpine sable]

10 apart

[hallow hinge]

yeah do u know how you would find that though?

[alpine sable]

uh

[hallow hinge]

oh wait

ok nvm

its not because the distance between x1 and x2 is 10

[alpine sable]

hm

i dont know

x1 and x2 8 apart

-5 and 3 are 8 apart

[hallow hinge]

of fuck sorry im being really confusing

[alpine sable]

lol np

[hallow hinge]

ok so the answer k is wrong lol

[alpine sable]

alr

[hallow hinge]

the length is sqrt(136)

ill explain why tho

[alpine sable]

ok

[hallow hinge]

so basically what you can do when trying to find the length of a diagonal line is you can make an imaginary triangle like this:

you know about pythagorean theorem right?

[alpine sable]

yes

[hallow hinge]

so can you solve it knowing that?

[alpine sable]

yep

thank you

[hallow hinge]

np

yeah the answer key thought that the black line had a distance of 0 which doesn't make sense

[alpine sable]

yea lol

[hallow hinge]

it has a distance of 6

pro tip is don't rely on formulas like that

[alpine sable]

alright

ty

[hallow hinge]

do you need help with transformations of HK?

[alpine sable]

nope im good my teacher said i can skip that

so i just did half

[hallow hinge]

so that was the only problem?

[alpine sable]

yes

the other stuff i dont need to do

[hallow hinge]

cool

.close

type .close when ur done so the channel becomes available again @alpine sable

so how i got sqrt(136) is this:

a^2 + b^2 = c^2

10^2 + 6^2 = c^2

136 = c^2

sqrt(136) = c

so the length is ~11.662

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[rare verge]

help lol

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@rare verge Has your question been resolved?

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[grizzled badge]

Can someone check my work? I’ve been practicing and wanna know if I did this right

[glass lichen]

Total area, not signed area for the last question

Hint: They specifically said sum or difference of integrals

[grizzled badge]

So like f(a1)+f(a2)?

[glass lichen]

no

sign of $\int_b^c f(x)\dd{x}$ is what?

[ocean seal]

Mosh

[grizzled badge]

Negative area right?

[glass lichen]

yes

so $-\int_b^c f(x)\dd{x}$ has positive area

[ocean seal]

Mosh

[grizzled badge]

Ahhh

[glass lichen]

so you add up all the intervals ensuring they individually have pos area

[grizzled badge]

Gimmie 1 sec and tell me if it’s right?

[glass lichen]

alternatively and I Ieave this up to you to confirm, $\int_a^e\abs{f(x)}\dd{x}$, however in all practical uses that's very hard to use

[ocean seal]

Mosh

[glass lichen]

yes

[grizzled badge]

Sweet!! I have some more questions if you wanna check those for me

Also did the ones I circle were they right? I just wanna make sure so I don’t get confused

[glass lichen]

yes

[grizzled badge]

did i do this right?

[glass lichen]

why is B a left-sum?

[grizzled badge]

oh i giess i just looked at it as an over estimate and got confused

its a right hand

[glass lichen]

yep

[grizzled badge]

so i,) is just C?

i.)*

[glass lichen]

yes.

[grizzled badge]

do the others look correct?

[glass lichen]

double check iii

[grizzled badge]

maybe for iii.)

yeah

A

[glass lichen]

yes.

[grizzled badge]

im pretty sure i got this next one right but im not entirely confident

@glass lichen you still there? sorry if I'm bugging you lmao

.close

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[supple tundra]

Not sure about this last part. From a sketch of the function, it seems that it's possible for there to be 1, 2 or 3 roots depending on the values of $p,q$. But not sure how to utilize $4p^3+27q^2$

[ocean seal]

azeem321

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[dark pelican]

@jagged imp @remote heron

original channel lcosed for some reason

[jagged imp]

you deleted your original message

[dark pelican]

yeah i tried to add a better one

whatever

i just picked between the two that sounded right and i managed to get it right thanks

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[rough salmon]

what is the blending function for the 5th control point of a degree 4 bezier curve?

[bleak ridge]

That's a good question

Based on a little bit of looking around

I'd say it's

,w BernsteinBasis[4,4,x]

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[bleak ridge]

Funny wolfram moment

,w indefinite integral (d/dx BernsteinBasis[4,4,x])

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[bleak ridge]

Ok there you go x^4

For [0,1]

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@rough salmon Has your question been resolved?

[rough salmon]

okayy

thx a looot

@bleak ridge

[bleak ridge]

Np

[rough salmon]

ah

i was graphing the curve n when i plug in t^4 it aint working, turns out it's the 4th control point's blending func tat was wrong

breh

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[cosmic oak]

Hi can anyone help me with this question

[buoyant kayak]

what specifically?

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[hallow bluff]

Hi! I've had to solve for the x intercept with this and got (-7.22, 0).
Im wondering how can I now use this information to find the y intercept?

[wary stream]

So you understand that for the x intercept, y is equal to 0. If it is the y intercept, what does that mean for the x value?

[hallow bluff]

The x value is -7.22

The issue is that if I set y to 0 and solve from there like normal it is only checking it.

Ohh wait you need to make x zero this time dont I

[wary stream]

Yes this

[hallow bluff]

Ok thank you. And if I am showing my work I should change the f(x) to just y?

[wary stream]

Yeah, you can

[hallow bluff]

Thank you so much my friend. I will return in maybe 30ish minutes. Should I close the channel do you think or no?

[wary stream]

You can close it. There are multiple free channels

[hallow bluff]

Ok!

.close

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[hallow bluff]

.reopen

[lone heart]

✅

[hallow bluff]

Hold on, and you know what I should do with the 4 at this point?

[wary stream]

So you plug in x = 0, correct?

Just find the value of y

And that's the y intercept

[hallow bluff]

Yes however I am at a dilemma

I will show you

If I were to add 4 on both sides, I would just divide by three like normal?

[wary stream]

You just simplify the right side

[hallow bluff]

Adding like terms? I got -3 that way.

[wary stream]

You did math wrong

PEMDAS

[hallow bluff]

Oooh ok

I appreciate it

.close

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[alpine sable]

how can i represent this function using heavisides?

[alpine sable]

so far i have u(t-3)*2(t-3) for the first part up to t = 5

but idk what to do after that

[bleak ridge]

You could prob do some fancy like

Layering of functions

So of course the first layer is H(t-3)*2(t-3)

Then you could multiply that by H(5-t)

And add on H(t-5)

Idk itd be nasty

H(t-3) * 2(t-3) * H(5-t) + H(t-5)

Then a final multiplication by H(8-t)

H(8-t) * (H(t-3) * 2(t-3) * H(5-t) + H(t-5))

Lets see

,w Heaviside[8-t] * (Heaviside[t-3] * 2(t-3) * Heaviside[5-t] + Heaviside[t-5])

[ocean seal]

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[bleak ridge]

Oh no

[alpine sable]

lmao

[bleak ridge]

Thats not right

Hmm

Lemme go through the steps visually

,w Heaviside[t-3] * 2(t-3)

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[bleak ridge]

So yeah that looks right

[alpine sable]

yah

[bleak ridge]

,w plot Heaviside[t-3] * 2(t-3) * Heaviside[5-t]

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[bleak ridge]

Hmm

,w plot Heaviside[5-t]

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[bleak ridge]

Strange

Oh

Its just not plotting right

Wolfram moment

,w plot (Heaviside[5-t] * Heaviside[t-3] * 2(t-3)) from 0 to 10

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[bleak ridge]

Ok so yeah

,w plot (Heaviside[t-3] * 2(t-3) * Heaviside[5-t] + Heaviside[t-5]) from 0 to 10

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[bleak ridge]

Oh

Makes sense

,w plot (Heaviside[t-3] * 2(t-3) * Heaviside[5-t] + 4 * Heaviside[t-5]) from 0 to 10

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[bleak ridge]

,w plot (Heaviside[8-t] * (Heaviside[t-3] * 2(t-3) * Heaviside[5-t] + 4 * Heaviside[t-5])) from 0 to 10

[ocean seal]

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[bleak ridge]

Well there you go

There is one more consideration though

,w (Heaviside[8-8] * (Heaviside[8-3] * 2(8-3) * Heaviside[5-8] + 4 * Heaviside[8-5]))

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[alpine sable]

ahh i see thx alot!

[bleak ridge]

I dont think that its the exact value you want at t=8

[alpine sable]

seems to be ok on desmos

[bleak ridge]

Desmos supports heaviside?

[alpine sable]

u have to define it first

[bleak ridge]

Oh

I mean the technical definition of heaviside is more like

{t<0:0, t=0: 0.5, t>0: 1}
I think

But if you define it like that it should be fine

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[gritty gull]

https://gyazo.com/31c4dcab5d839540d4f77b72c7c09c72
parking garage to terminal A is expressed as 60r
and terminal A to terminal B is expressed as (r + 8.1)45

[gritty gull]

how would i write an equation that relates to both of these equations?

[vale wigeon]

if you know how far you traveled from garage to term A, and you know how far you traveled from term A to term B, how do you find the total distance traveled?

[gritty gull]

NVM LOL

https://gyazo.com/147243873f6770296a7e5beae15af91f

= 3,950

which would be my answer

however

i have no clue on how to rewrite this

in order to fit this equation

do you think (60r + 8.1)45 = 3,950 would work

(the reason why 8.1's in there is because the trip from terminal a to terminal b travels 8.1 ft per second faster)

or maybe

[lone heart]

@gritty gull Has your question been resolved?

[vale wigeon]

do you think (60r + 8.1)45 = 3,950 would work
no

@gritty gull i was expecting the very simple answer of "you add the distances"

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[warm shoal]

Hey im trying to solve a limit with L'hopitals and I understand now that the limit doesn't exist but I would like to see the error in my ways

[warm shoal]

I am sending a picture right now

the original question was the limit as x approaches -3 of sinx/(x-pi)(x+3)

My thought process was to try and force an indeterminate form by multiplying by (x+3)/(x+3)

[velvet pelican]

it's already indeterminate tho

[warm shoal]

and this provides a definite answer at the end which would suggest the graph has a removable discontinuity

[velvet pelican]

substituting in -3 gives sin(-3)/0

[warm shoal]

its not indeterminate initally

any number / 0 is undefined

not indeterminate

[velvet pelican]

ok so the limit doesnt exist

[warm shoal]

no its not that simple

the limit doesnt exist because the two-sided limits dont converge

its not simply because you plug in -3 and it doesnt work

what I dont understand is that mathematically my answer produces a definite number as the answer to the limit

but in reality it doesnt exist

[velvet pelican]

so when you used lhopital the second time

it wasn't in the form 0/0

since you still had sin(x)

[warm shoal]

well that explains it

thank you mr frsoty

how do i close this

.close

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[gritty gull]

OHHHHHHHH

[gritty gull]

so it wouldve been 60r + (r + 8.1)45 = 3,950

my gosh

im so dumb

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[rough owl]

Compute the regular equal payment needed to discharge both principal and interest of the loan. Suppose money borrowed is worth 7.5% compounded monthly for 25 years.

[rough owl]

Total Contract Price (TCP) = 6,890,000

Loan = ?

what is the formula of loan

how do i get the loan?

<@&286206848099549185>

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[barren glade]

3a

it requires the use of p+q concept

[agile crest]

anyone who can help me?

[solar merlin]

use an available channel. when you post in an occupied channel that someone's already asking for help for, it's like shitting in a bathroom stall someone's already pissing in

try looking up ''algebraic long division''

there's a lot of tutorials on it

[agile crest]

oh

[solar merlin]

@barren glade anyway, with the identity i posted above, you should be able to solve it. just plug in the values

[barren glade]

@solar merlin yup thanks

.close

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[light crane]

Hi! currently doing a wave equation with dirichlet boundary condition exercise, but i dont know how to get the initial displacement from the graph

[light crane]

initial displacement is u(x,0) but given the graph is u(x,t) against x im stuck there

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[odd rover]

<@&286206848099549185>

someone pls??

[raw shard]

don’t immediately ping helpers

you have to wait at least 15 minutes

[odd rover]

oh

i am sorry, i am new..i went through the rules and stuff now. will make sure next time 👍

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[raw shard]

[odd rover]

.close

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[odd rover]

Hi!
I wanted a bit of help with this question related to Circles. Can you some please guide me with what can be done to prove whats given?
Thanks!

[odd rover]

😭 😭 😭

@raw shard sorry to ping but can you help me with this please?

[raw shard]

no i don’t know how to do stuff like this

[odd rover]

://

thanks anyway

have a good day

[grizzled sundial]

Hey guys what's the difference between a stationary point of inflection and a point of inflection?

[raw shard]

#❓how-to-get-help

this isn’t your channel

[grizzled sundial]

ohh sorry

[odd rover]

@covert agate you seem to know circles mate. can you possibly help with these pls

[raw shard]

don’t ping random people asking them for help

[odd rover]

💀

[odd rover]

<@&286206848099549185> someone pls help

[covert agate]

i would just use coordinate geometry lmao

[covert agate]

so notice O is the circumcentre of PQR right

[odd rover]

mhmm

[covert agate]

and you extend the angle bisector of ∠ PQR and perpendicular bisector of QR to S

just prove they intersect the same point

[odd rover]

angle bisector of PQR??

[covert agate]

QPR sorry

[odd rover]

same point as what?

[covert agate]

angle bisector PQR and perpendicular bisector QR

[odd rover]

hmm

[covert agate]

prove the intersection lies on circle

idk there should be a better approach? not sure tho

[odd rover]

i was thinking to do this by proving that os is also the radius

[covert agate]

it is

[odd rover]

so OS=OR/OQ

[odd rover]

in the figure it is

but have to prove it

😭

[covert agate]

well uh

i don’t think you have to?

[odd rover]

it says that i have to prove its on the circumference

[covert agate]

there are infinitely many O’s satisfying ‘lying on the perpendicular bisector of QR’

one of those O’s is the circumcentre

no other information about O means you can, WLOG, assume O = circumcentre

[odd rover]

hmm

okay thanks

i am trying to figure this out

[covert agate]

oh

i think i have an approach?

if S wasn’t on circle, OQ not equal to OS and OS not equal to OR

prove this and you should ne good

[odd rover]

i suppose it needs to be proved that OQ=OS and/or OS=OR

[covert agate]

yea

i think consider a point on the perpendicular bisector of QR and inside/outside the circle?

proof by cases

then show it’s not the angle bisector of QPR

[odd rover]

sighs

well thanks

.close

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[ivory forge]

this is the qn and answer my teacher gave
will it not be wrong if we take n=2?

[glass lichen]

wdym?

[ivory forge]

if n=2 then nx^n-2 would be

0^0

[alpine sable]

There's a minus sign missing

[glass lichen]

????

where did you get 2=0 from?

[ivory forge]

?

[glass lichen]

Oh wait, sorry just woke up

you never plug in x=0.

its a limit

you dont care about what happens at x=0

[ivory forge]

oh okk got it thanks

.close

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[alpine sable]

@ivory forge ln(x) *x^n=ln(x) *1/(x^-n)

[ivory forge]

?

[alpine sable]

First part of the second line

[ivory forge]

oh i meant to write ln(x)/1/x^n

🤦‍♂️😕.

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[chrome dune]

I keep get 13

Y = z+1
And z = 2
So, y = 2+1, right?

X-2×2+1=10

X-2×2=9

X=13

But there's no 13 in the choices

[tawny pivot]

y=3

now try

or put brackets around (2+1)

[chrome dune]

Okay

=16

Right? @tawny pivot

[alpine sable]

Yes

[chrome dune]

Okay, thx

.close

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[woven otter]

How do I convert and approximate a long decimals that has been rounded into fraction?
I once converted 0.2571428571 to 27/105 through trial and error, but I wonder if there's another short way of doing it
What I am now looking for is 0.2608695652 into fraction

[raw shard]

try 26/100 at first

then turn it into like 2600/10000

so then if you add 1 it does very little

,calc 2601/10000

[ocean seal]

Result:

0.2601

[raw shard]

oh darn that’s right

oh i see what you want

you want it with like 2 or 3 digit numbers

[woven otter]

I could go for that, but I know that'll affect the coming equations badly

maybe exact if I'm lucky

calculators only go up to certain decimal place
so I was thinking that 0.2608695652 can be converted into a nice 2-3 digits fraction xx/xxx, and not 2608695652/10000000000

I was thinking there would be a straightforward method instead of trial and error

[gray isle]

i suppose you could consider
1/0.2608695652 and see if that gets you something nicer

,calc 1/0.2608695652

[ocean seal]

Result:

3.8333333335889

[azure merlin]

In the given circle, O is the centre of the circle, DB is diameter and angle AOB = 120°. Find the measure of angle ADB and angle ACB.​

[fair crater]

I think this is a hard problem in approximation

[gray isle]

which i suppose you could approximate as 3.8333...

and convert that to a fraction and take its reciprocal

the same approach could work for

[ocean seal]

Result:

3.8333333335889

[gray isle]

,calc 1/0.2571428571

[ocean seal]

Result:

3.888888889537

[fair crater]

You could try the approach of
https://en.wikipedia.org/wiki/Continued_fraction#Best_rational_approximations

[gray isle]

which could be approximated as 3.8...
which as a fraction would be 35/9
and its reciprocal 9/35

[fair crater]

This is fine for back-of-the-envelope but is handwavy. I think approximation theory formalises this nicely

[gray isle]

yeh, it depends on what you're given and what you're allowed to use

[fair crater]

Though in the end I suppose - it depends on what you want to do. There's some really interesting stuff like https://math.stackexchange.com/questions/3749872/is-22-7-the-closest-to-pi-among-fractions-of-denominator-at-most-50

[woven otter]

hmm alright, I'll have a go with it, thanks

.close

[lone heart]

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[cunning bloom]

Could someone help with finding a function f:A-->B where f (the function) is bijective

[crisp grove]

didn't someone already answer?

[cunning bloom]

no

i did get help before but i just gave up and tried another way

but now we are back here sooo

any idea 🙂

[crisp grove]

like you can find a bijection from {0,1,..} to {2,3..}

use it to handle the zero,

[lone heart]

@cunning bloom Has your question been resolved?

[lone heart]

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[tawny fable]

Hey there

[tawny fable]

I have a question in linear algebra

what does p(-2) = 0 mean?

It means when -2 is the input the equation outputs 0?

when trying to figure out if it is a subspace of V

I should run some examples of equations from R4[x]

and try to see if there is closure for addition and multiplication?

like, it there a easier way? what if it was R10[x]?

anyone ?

<@&286206848099549185>

[chrome salmon]

No need

[tawny fable]

Hi Sam

what do you mean no need?

[chrome salmon]

Just use p(x) and q(x) notation

You don't have to show example

[tawny fable]

I don't think I fully understand that.
As we learned, we need to show 3 axioms hold:

1. 0v is in the vector space
2. closure to addition
3. closure to multiplication

If we want to prove it is a subset of a vector space

and If we want to show it is not, we need to give a counter example

[tawny fable]

but I don't think I just understand what do you mean by that. and also, what is the connection between that and homogeneous and non-homogeneous linear equations

[ocean seal]

it's Sam

[tawny fable]

which is closure to addition

but what about the 0v vector

[chrome salmon]

If p(x)=0 then ofc p(-2)=0

[tawny fable]

but how do you know that p(x) = 0?

[chrome salmon]

Idk that's one of the element of R4[x]

[crisp grove]

where's the question?

[tawny fable]

here

[chrome salmon]

.

[chrome salmon]

That's cap

Let me type ot for you @crisp grove

[crisp grove]

no I can see it now

[crisp grove]

"tap to see attachment" seems to fix it

[chrome salmon]

Alright

Yk 0 in R4[x]

[tawny fable]

I just dont understand if that means

that the x's are 0

or the coefficients

I just don't have the image of that in my head

idk why

I guess it is confusing because we say it is a vector now and I am used to see it as an equation

[chrome salmon]

The coefficients

How do you write p(x) generally

Using coefficient a,b,c,d

[tawny fable]

a(x)^4 + b(x)^3 + ... d(x)^1 + e

isnt that with e?

when the degree is 4

?

[chrome salmon]

if all a,b,c,d,e are equal to 0 then does p belong to R_4[x]

[tawny fable]

well I guess it is

[chrome salmon]

Yeah so p(x)=0 does make sense

[tawny fable]

Does it also mean that 0v is inside every polinome?

[chrome salmon]

I don't get what you mean

"Inside"

[tawny fable]

I mean

0v C R_n[x] for every n C N

[chrome salmon]

Oh yeah

[tawny fable]

So in the question

what they mean is that I look at the group of all the polynomials s.t. p(-2) = 0

[chrome salmon]

Yes

[tawny fable]

does it mean that p(x) in this case is linear depentent?

[chrome salmon]

To what tho

[tawny fable]

idk

[chrome salmon]

{1,x,x²,x³,x⁴}?

[tawny fable]

i am thinking it can only be multiplications of a certain number

in order to make p(-2) = 0 true

[chrome salmon]

For an example p(x)=x²-2

p(x)=2x-4 is in set too

[tawny fable]

but it should be from degree 4 isnt it?

[chrome salmon]

Less than

[tawny fable]

oh

it includes all degrees in it then?

[tawny fable]

isnt that p(-2) = (-2)^2 - 2 = -4?

[chrome salmon]

Deg(p) =< 4

[chrome salmon]

Edit to -4

[tawny fable]

idk why it is so confusing for me...

so for showing 0v C U1 in this exercise

[chrome salmon]

[tawny fable]

I say 0v C R4[x] for every x in R
-2 is in R
==> 0v C p?

[chrome salmon]

U_1

[tawny fable]

yeah

3.(a)

[chrome salmon]

Foe every x?

[tawny fable]

oh

x is not mentioned...

nvm

the how do I explain that 0v C p?

then

0*(-2)^4 + 0*(-2)^3 + 0*(-2)^2 + 0*(-2)^1 +0 = 0?

= 0v?

[chrome salmon]

Yeah

See p(x) = ax⁴+bx³+cx²+dx+e for a,b,c,d,e in R

That means p(x)=0 is in R_4[x]

Since a,b,c,d,e can be =0

Understand what R_4[x] is

p(x)=e is also a thing

[tawny fable]

It is all the polynomials with the degree <= 4

as we said

[chrome salmon]

Yes

So it can be p(x)=e

Right

[tawny fable]

for instance we have all vectors like
v = [1,2,3,4,5]
u = [0,0,0,0,5]
etc..

[chrome salmon]

Nah don't think that way

[tawny fable]

oh

then what way should I look at it?

[chrome salmon]

The basis we are working here is {x⁴,x³,x²,x,1}

[tawny fable]

yup

this I get

[chrome salmon]

We aren't working on R⁴

[ocean seal]

it's Sam

[tawny fable]

how do you know you can use distributivity here?

[ocean seal]

it's Sam

[chrome salmon]

p and q are polynomials

[tawny fable]

1 is when e also equals 1?

[chrome salmon]

Yes

[ocean seal]

it's Sam

[tawny fable]

and x^2 is when c = 1 and all the rest are 0?

[chrome salmon]

Yes

When all are 0 , we get 0 in R_4[x]

[tawny fable]

So can you explain to me what is exactly p(2) = 0?

@chrome salmon

Or any of the other exercises

[chrome salmon]

p is a polynomial whose zero is 2

For example x-2 , x²-4 , 5x-10 ,...

Set of all such polynomials is U_1

U_1 = {x-2 , x²-4 , 5x-10 , x³-4x,....}

This won't contain polynomials like x-3 , x²+1

[tawny fable]

But isn’t -2(-2) = 4

Should it be 0?

[chrome salmon]

So what

x²-4 = 2²-4 =0

You asked me for p(2)=0

[tawny fable]

Oh, I meant -2

Yeah I get it now

So basically all the functions of the type of polynomials that sum up to 0 when inputs -2?

[chrome salmon]

Yes

If you mean p(-2) by sum up

[tawny fable]

yeah

@chrome salmon

first of all thank you

So to see if I understand it correctly

if I look at (b)

does it mean that e must be 0?