help-0

[wild pendant]

so we can try and make the lower bound 0

that doesn't really make any progress, but when we find an antiderivative, that would make the computation a tad easier

I am seeing lots of oppotunities for trig substitution, because of the constant-x^2 terms leading to pythagorean identities and possibly nice simplifications

furthermore, the substitution rsin(theta)=x might be a good start, because the upper bound becomes alpha and the lower bound is still 0 (if you take my advice on the even function aspect)

so I will leave that as my suggested first more

*move

also, that nasty radicle on the right looks very sexy after you mess around with the trig post sub

try it out

I think my sub will make the integral look way less horrible

[copper tangle]

ok tyty

[wild pendant]

np

I thought I might be in the same shit as you, but I found a decent way

[copper tangle]

Why sub in r sin theta?

[little drum]

Cause terms like $\sqrt{R^2 - x^2}$ and $\frac{x}{\sqrt{R^2 - x^2}}$ are too appealing :o

[wild pendant]

first, makes the radicles become rcos and 2, makes the bounds alpha only

[ocean seal]

Ansh

[wild pendant]

elaborating on what ansh said

[copper tangle]

Oh okay.

So just sub all x for R sin theta?

[wild pendant]

yep

[copper tangle]

I mean like why introduce a new variable

[wild pendant]

and notice all of the 1+trig^2 shit you get

its a goldmine for trig identity practice

can latex work on wolframalpha?

[little drum]

Yes it does, ig

[wild pendant]

ig?

[copper tangle]

wait. did you mean x= R sin alpha?

or theta

now im confused

[wild pendant]

theta

sry

the upper bound should be alpha

but I have been using my noggin and not writting out the steps on paper

so you (are kinda) forced to investigate regardless

[copper tangle]

ok

[copper tangle]

^

[wild pendant]

damn, I want the advanced math channel someday

would be really helpful with my real analysis self study

but idk how to get it

[copper tangle]

So basically integrate by parts and u subsitute?

[little drum]

Cause... it makes your integrand looks less messy

[wild pendant]

no

[little drum]

Simplifies your integrand for your convenience

[copper tangle]

oh

[wild pendant]

no need for parts

everything in multiples of trig

[copper tangle]

so just sub

ok

[wild pendant]

sub only and see what you get

In my head, I think I got tangent theta

but check for yourself

shit

scrap that

[copper tangle]

hu

h

[wild pendant]

I think it might be like 1-cos(a)/cos(theta) or something like that

i'll check

[ocean seal]

KaiML

[copper tangle]

like then I wont be able to relate it back to alpha

what

what

[wild pendant]

theta=a

[copper tangle]

Oh

[wild pendant]

your bounds are in the variable theta now

no need for worry

[copper tangle]

ook

[wild pendant]

we have determined that in the u sub step 🙂

[copper tangle]

waitwaitwait

i am integrating with dx

not d(alpha)

[wild pendant]

no, dtheta

[copper tangle]

where the dx go

[wild pendant]

ansh, you got that?

it looks something nice, like this

[copper tangle]

woah

[wild pendant]

probably

[copper tangle]

I cannot read that :P

[wild pendant]

I suck at math (sorta) so I fuck up big time sometimes

4pi times the integral of cos(theta)-cos(alpha) from 0 to alpha all with respect to theta

[copper tangle]

ok

the first 4 lines?

[little drum]

$2\pi \int_{-\alpha}^{\alpha} R^2(\cos \theta - \cos \alpha ) d \theta$

[wild pendant]

you got that?

sweet,... wait

you got an R constant?

[copper tangle]

yea

[wild pendant]

where did that sneek in?

well,idk if I am wrong

I care if you are making progress

so I can leave it to you from here, kai?

[copper tangle]

uh

[wild pendant]

*idc not idk

[ocean seal]

Ansh

[copper tangle]

You think you could type out this for me?

then I can read it and I think ill be fine

[wild pendant]

same as ansh accept I got no R constant

*except

[copper tangle]

as in like the whole thing

so i get the steps

[wild pendant]

ok

let me write it out in full (might take a bit)

[copper tangle]

ok

ty

ping me when ur done k?

[wild pendant]

@copper tangle wake tf up

[copper tangle]

ok

tysm

[wild pendant]

does this make it clear?

[copper tangle]

yep

[wild pendant]

also, getting advanced access was trivial (just press a green tick) ahaha

[copper tangle]

lol

[wild pendant]

ok, I hope you learned that you just to not shit your pants when you get something like this. Consider the context of your course and maybe you need just one move to crumble everything apart ( or not , hehe)

[copper tangle]

lol

Ok

[wild pendant]

sweet

[copper tangle]

I have learned

[wild pendant]

case closed

[copper tangle]

Bye

.close

[lone heart]

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[dire juniper]

i know the measure of B is 95, from that how will i solve for arc ABC?

[little drum]

The angle subtended by an arc of a circle at its center is twice the angle it subtends anywhere on the circle's circumference.

or for easier understanding, the angle subtended by a chord of a circle at its centre is double the angle subtended by it at the corresponding arc

So, you can figure m(ABC) right?

[dire juniper]

cant get the gist of that wait 😅

[little drum]

basically here, m(AOB) = 2 x m(ACB)

[dire juniper]

ohhhh

so arc ABC is equal to twice the measure of angle D

am i right

[little drum]

hmm :c I don't really know what m ABC there means

[dire juniper]

i think it pertains to the bigger angle of arc ABC

[little drum]

is why I'm not too sure but I can say, yes, since angle B is 95°, AC is surely above the center and so angle AOC subtended at the center has to be double of ADC

[dire juniper]

like this maybe?

say Y is the center

[little drum]

ohh

yes, that'd be twice angle D yes

[dire juniper]

thanks for the help, didnt see the part where we could use angle D for the property of inscribed circles

thanks, rlly appreciate it

[little drum]

Ah.. you can use cyclic quadrilateral properties

to figure D

[dire juniper]

yep did that prior

[lone heart]

@dire juniper Has your question been resolved?

[lone heart]

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[forest knot]

Can someone help me in finding a solution for Xn . I dont know what the next step is

[alpine sable]

What are Cn and dn?

[forest knot]

So Xn is a first order linear recurrence relation

Cn, dn are variables for each term, they change

but they are constant

idk how to explain it

They differ between each n, but they are not like a variable that can be anything, they are constant (respective to their n value)

[alpine sable]

So x1=d1

[forest knot]

The answer is supposed to be
But, i dont know how to get my line to be like α-1, since im missing a c_n at the end, and i dont know how to get from line α-1 -> α either

[forest knot]

no, its c2d1

i think

[alpine sable]

x0=0

[forest knot]

oh wait

yes

its d1

mb

[alpine sable]

What is the product doing

j is not used

@forest knot

[forest knot]

sooo

from what i unerstand

since C1 does not exist

J serves to like , displace i by 1

Wait

my bad

i cut off the Cj didnt notice

[alpine sable]

I don't understand the problem

What is confusing you

[forest knot]

How to get from this line

to this line

And then how to get to this line

[alpine sable]

Rearrange the last term

Write d1 first

[forest knot]

What

[alpine sable]

Here

The term before 0

Rearrange it

[forest knot]

like so?

[alpine sable]

Write it as d1c2c3...cn

[alpine sable]

You flipped the order

[forest knot]

i am confused 😕

i thought thats what u wanted

[little drum]

Your pfp looks so cute and so fitting at this moment

Lmao

[forest knot]

xD

[alpine sable]

your sum is correct

You just have to rearrange it

Instead of writing cn*cn-1*... *d1

[forest knot]

but im not multipying

im adding

how can i mix a term with c2, with c3. They are seperate so i should be adding no?

[little drum]

that's... wrong!?

[forest knot]

is it?

[alpine sable]

.

[little drum]

Yeah obv... you can look at d_1 sitting in the deepest nest and still coming out with only one coefficient factor?

[forest knot]

😮

This part is proper right?

[little drum]

basically it's pretty intuitive I think... just look at each of those d_i coming out of their nests one by one

[little drum]

Yeah except some bracket issues

[little drum]

for example look at d_1 ... it jumps out of the deepest nest, the immediate factor being c_2 and the last factor that it comes across is c_n

so d_1 in your summation would have coefficient factor : (c_2)(c_3)(c_4)(c_5)...(c_n)

similarly, look at d_2 ... it jumps out of the second deepest nest, immediate factor : c_3, last one c_n
hence, d_2 comes along with (c_3)(c_4)(c_5)...(c_n)

[forest knot]

Is this whats happening?

[little drum]

Again, since I'm doing math... I can't help but notice here that any $d_i$ would come along with $c_{i+1} \cdot c_{i+2} \cdot \cdot \cdot c_n$

[ocean seal]

Ansh

[forest knot]

yes i notice that

[little drum]

which can be written as

ahem

*elegantly written as

[forest knot]

The product of Ci+1 to n, times di?

[little drum]

$\prod_{j=i+1}^n c_j$

[ocean seal]

Ansh

[forest knot]

yeah, thats the factor part

for some di

[little drum]

And so, your summation just ends up being the sum of d_i times this factor

[forest knot]

why

[little drum]

where i goes from 1 to n

[forest knot]

shouldnt it be

[little drum]

wasn't it you who wrote this?

[forest knot]

yes

[little drum]

Isn't that clearly $\prod_{i=2}^{n} c_i \cdot d_1 + \prod_{i=3}^{n} c_i \cdot d_2 + ...$

[ocean seal]

Ansh

[forest knot]

Thats how i understand it

[forest knot]

yes

[forest knot]

But from here .

[forest knot]

to here

i dont get that jump

[little drum]

Oh sorry lmao

lemme rephrase it

[little drum]

this ... whole thing

ends up

being the sum of

[little drum]

this expression

from i = 1 to n

[forest knot]

yeah

[little drum]

So,

$x_n = \sum_{i=1}^{n} (d_i \prod_{j=i+1}^{n} c_j )$

[forest knot]

To represent this code mathamatically, shouldnt u take the sum of i=1->n; (Di * product(j=i+1->n) of cj)

but isnt this taking the sum of Di, and multiplying the total sum by the product thing

[ocean seal]

Ansh

[little drum]

oof

there :o

[forest knot]

😮

its the same thing??

the brackets are not required?

[little drum]

Honestly I don't know 😂 is why I put the brackets there

[forest knot]

yeah it makes sense to me w/ the brackets, but i thought the answer was taking the sum of just Di and multiplying the total sum of Di, with each factor, and i couldnt see how it came to fruition

damn

brackets was the problem

[little drum]

Wait

lmao

Actually here,

since j starts from i + 1, it is understood that the product is associated with the summation

[forest knot]

oh

i understand now xD

[little drum]

Good luck +1 !! Are you reading recurrence for Olympiad purposes?

[forest knot]

No 😄

olympiad is for big bois, im just trying to pass my discrete math course

.close

[lone heart]

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[forest knot]

Exam in 2 days, worth 100% of my mark 😦 its gonna be 4 questions in 3 hours 💀

[lone heart]

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[alpine sable]

I'm trying to prove signed curvature (for curves in R^2) is unchanged by orientation-preserving reparametrizations, but I'm stuck at the very beginning.

We define "signed curvature" for a curve x : I->R^2 as the quantity <a(t), R_90(v(t))> / |v(t)|^3 , where a(t) = x''(t), v(t) = x'(t), and R_90 is a rotation by 90 degrees.

So I first started with a reparametrization y : I -> R^2 so that y = x ○ s for some continuous s : I->I. Of course, we assume s' > 0 so orientation is preserved. Then with the chain rule I get some really long and ugly inner product, and I'm not sure how I'm supposed to the fact s' > 0 to get anywhere. In particular, I get < s''(t)*v(s(t)) + s'(t)^2 a(s(t)), R_90 (s'(t) * v(s(t)) > / | v(t) |^3.

[lone heart]

@alpine sable Has your question been resolved?

[alpine sable]

.close

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[formal spoke]

How

[lone heart]

@formal spoke Has your question been resolved?

[formal spoke]

.close

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[buoyant arrow]

how would you solve this system?

[alpine sable]

<@&286206848099549185> is there a helper that can help with me grasping the concepts of Precalculus ? If possible can someone dm pls

[buoyant arrow]

dont @ helpers in the first 15 min if you ask an question and do it in the right channel

[echo socket]

Is this a diophantine equation or x,y and z are just any real numbers?

[buoyant arrow]

I'm still learning the beginning of linear algebra so yes

[echo socket]

Well we're given three unknowns and 2 linear equations, as far as I'm concerned there are infinitely many solutions to this

Not sure on how to use linear algebra here, but what you could do it parameterise the unknowns

[buoyant arrow]

i need to give a answer in a vector

so i need an (x,y,z)

my z= 1/2+y/2
x = 5 + 3y-(1/2+y/2)

[echo socket]

I guess you need to write this as the equation A*v = c, where A is 3 by 2 matrix, v is (x, y, z) vector and c is (5, -1)

But I'm not really familiar with non-square matrices so idk

[buoyant arrow]

i need to solve the line where plane 1 = plane 2

[lone heart]

@buoyant arrow Has your question been resolved?

[buoyant arrow]

.solved

.close

[lone heart]

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[lone heart]

@warped iron Has your question been resolved?

[true pulsar]

Go through each answer choice to see if it's correct

First, can you deduce any more information about the triangle?

[lone heart]

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[quaint mica]

help pls

[quaint mica]

integral of dx/srt36-4x^2

[vague holly]

what do you have

[quaint mica]

arcsin(x/3)+C

[vague holly]

almost right

[quaint mica]

Using this rule

I know there’s another way to do it

But I’m trying to use this rule here

[vague holly]

well, what's the problem

[quaint mica]

I m not sure where 1/3 comes into play

[placid zinc]

You'd need to carry out a u-sub in order to get the correct form

Are you comfortable with such?

[vague holly]

how would that help

[quaint mica]

Ah got ya

[vague holly]

You need a trig sub

[placid zinc]

Or wait, my bad you don't need to do that. Just gotta factor out the 4

∫ dx / 2√[9 - x²]

That's where the 3 comes into play

[vague holly]

yep, then it's just taking an antiderivative

not forgetting the chain rule

[quaint mica]

But if you factor put the 4

U get 1/2

Not 1/3

And that’s not right

Helllo

[vague holly]

why is that not right

[quaint mica]

lol woops sorry

.close

[lone heart]

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[shy briar]

What should be my thought process for the next step?

[shy briar]

I did the basic step, and induction step. But now I have to prove it. But I don’t know how to start

[frigid geyser]

thats it for mathematical induction

[shy briar]

i am done?

[frigid geyser]

oh your induction step isnt actually stepping

umm

so for the induction step you assume that it is true when $m = n$

[ocean seal]

xdk1235

[frigid geyser]

and then show that it is also true when $m = n+1$

[ocean seal]

xdk1235

[frigid geyser]

oh is that a cursive n

anyway just pick some random letter it doesnt particularly matter

[shy briar]

right so I wrote down the equation now I have to rearrange it and solve with algebra

[frigid geyser]

yeah so this is good

now you cant really do much from here so use the inductive hypothesis

(that the statement is true for n)

[shy briar]

assume its true

and then?

[frigid geyser]

use that

you assume it so that you can use it

[shy briar]

so I subtitue that summation to n to (n/3+1)

[frigid geyser]

ummmm

well not exactly substitute but yeah

apply $\sum_{k=1}^n \frac{1}{k} \le \frac{n}{3} + 1$

[ocean seal]

xdk1235

[shy briar]

I get a bit confused on how to "apply it"

This is fine?

[frigid geyser]

You have to prove that the sum is less than or equal to m+1/3 + 1

So dont write that part yet

And for where you have <=> write less than or equal to instead

[shy briar]

ah right

I can not simplify further, even using a simplify calculator

Did I go wrong somwhere?

[lone heart]

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[ornate rover]

A side of a square has a length of 2 units what is the area of the square

[bleak ridge]

[ornate rover]

?

[wary stream]

What's the formula for area of a square?

[ornate rover]

Check your dm

[wary stream]

Don't dm me

I'm not here to do your work for you or help privately

[ornate rover]

I took a screenshot cause I can’t type it

[wary stream]

You can post it here

[wary stream]

As I asked, what's the formula for area of a square?

[lone heart]

@ornate rover Has your question been resolved?

[lone heart]

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[sly bolt]

https://i.imgur.com/w2WVb11.png
what are the brackets around the left expression? I know <u,v> is the inner product, but what is |<u, v>|?

[sly bolt]

*Cauchy–Schwarz inequality

[sonic kiln]

It's just the absolute value of <u,v>.

or the module if it's in C

[sly bolt]

module?

[sonic kiln]

Is you vector-space defined over complex numbers?

The module of a complex number is kinda like the absolute value of a real number

uh that might just be the french name though

Ah it's modulus in english

[sly bolt]

i see

thanks

[sonic kiln]

no problem

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[fair saffron]

What is a derivative and how can i get derivative of a number?

[alpine sable]

Derivative is the rate of change of one variable with another. Derivative of a number (constant) is always 0.

[fair saffron]

Thx

.close

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[keen shell]

My solution:-

I just need a hint on how to approach the problem

[fair saffron]

Bro this is closed, more people will see if you use #help-10 or #help-11

[keen shell]

Thanks

.close

[fair saffron]

Just srnd your question on #help-10

[keen shell]

K

[lone heart]

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[lone heart]

@atomic plover Has your question been resolved?

[sly mantle]

.close

[lone heart]

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[supple tundra]

guess u didnt learn much huh

[wide parcel]

self reflection not evaluated
Yeah

[alpine sable]

SATT

can be used jus subtracting the 180-angles cant if be used for all?

[alpine sable]

?

[wide parcel]

the top right of this page literally says "self reflection not evaluated"

[alpine sable]

yeah

this is for understanding still gotta do it

help pls

[alpine sable]

taken help channel

[lone heart]

@alpine sable Has your question been resolved?

[lone heart]

@alpine sable Has your question been resolved?

[lone heart]

@alpine sable Has your question been resolved?

[tacit arch]

#❓how-to-get-help

[lone heart]

@alpine sable Has your question been resolved?

[limpid dune]

@alpine sabledude its been 3 hours

lock this shit

have some compassion

@heavy otter lock this please, this guy has been occupying this channel for 7 hours

[alpine sable]

ok ill just send the question again other time later

needed help understanding still do but ye ill send another time

.close

[lone heart]

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[alpine sable]

https://media.discordapp.net/attachments/639282516997701652/933455516976418836/unknown.png?width=582&height=298

[alpine sable]

I posted the question and my work (find the indefinite integral). My work is wrong I guess

This is the answer

https://media.discordapp.net/attachments/639282516997701652/933454866347601931/unknown.png?width=119&height=40

where did I mess up?

I don't see how they got (-1/14) without the variable 't'

[proven rune]

ur du is wrong

if u = 1-t^2
then du = -2t dt

@alpine sable

[alpine sable]

sorry thats what i meant to write but that doesnt change my answer

cuz i never plugged

wait

ok yea that doesnt change what i wrote

[proven rune]

see extreme right side u have written wrong

ok good

[alpine sable]

yea ik

i accidentally put the 2 there

but idk how they just excluded the t

[proven rune]

from here t dt = -du/2

[alpine sable]

OH SHIT

i didn't see the 't' in the problem

lmao

[proven rune]

sub that
u get integ of (u)^6 . -du/2

[alpine sable]

im blind as a bad holy hell

[proven rune]

bruh

[alpine sable]

bat*

jesus christ thanks man

[proven rune]

ok now solve it

[alpine sable]

have a good one

[proven rune]

np bro

[alpine sable]

.close

[lone heart]

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[thorny minnow]

Can someone help with this

[thorny minnow]

I’ve tried like 15 times and can’t figure it out

What

Ok

Cool bro

Then do it

[woeful pulsar]

<@&268886789983436800> this seems pretty irrelevant

[thorny minnow]

Can someone like actually help me with the problem

[tall wing]

pretty irrelevant is putting it lightly

[sly mantle]

thanks

[woeful pulsar]

@thorny minnow do you have other similar questions? I haven't seen a question of this sort before

[tall wing]

banned, thanks

[woeful pulsar]

yeah i forgot the word "explicit"

[thorny minnow]

Yeah but I know how to do them

It’s just this one I have no clue how to do

[woeful pulsar]

a "write f(x) in exponential form"

It seems the question wants you to write x=....^f(x)+....

[thorny minnow]

Like for this the answer is y=-((1/2)^((x-3)/2) +1)

[woeful pulsar]

oh that explains

so you want the inverse of this function

right?

[thorny minnow]

Yeah pretty much

[woeful pulsar]

so what we want to do is find the inverse

so we start with $x=-2\log_3\left(\frac{1}{2}y+1\right)-1$

[ocean seal]

Element118

[woeful pulsar]

and try to make y the subject

[thorny minnow]

Yep

And then add 1 and divide by -2 right?

[lone heart]

@thorny minnow Has your question been resolved?

[lone heart]

@thorny minnow Has your question been resolved?

[lone heart]

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[narrow flower]

I have a doubt in Equation of 3D Cylinder

[narrow flower]

I want to know the equation of cylinder with base cross section as x^2+y^2=9

and Hight 10

I initially thought that the equation should be x^2+y^2+z=9, where 0<z<10

But it didn't worked, I graphed in graphing software

[pale kestrel]

(x - a)^2 + (y - b)^2 = r^2

that's it.

[narrow flower]

I know that equation of infinite cylinder is x^2+y^2=r^2

But I want to limit it

like bound the hight

not make it infinite

[pale kestrel]

Ok, then you just set that as a separate condition

[narrow flower]

which condition?

[pale kestrel]

0 < z < 10

[narrow flower]

should I set 0<r<10

[pale kestrel]

???

[narrow flower]

ok wait let me graph

[pale kestrel]

what.

You realise it's impossible to write it as 1 equation

[narrow flower]

I will graph the condition and check

[narrow flower]

ya

[pale kestrel]

Or even if it is, you kinda don't want to

[narrow flower]

it is not working

[pale kestrel]

what?

what is a

?????

[narrow flower]

Idk

in geogebra it came by default

I typed that 0<r<10

[pale kestrel]

well i dont use geogebra so i cant help u there

[narrow flower]

by default it is coming

what software u use?

[pale kestrel]

i dont plot 3d cylinders

[narrow flower]

ohh

[pale kestrel]

i use desmos

[narrow flower]

ok

is this same method applicable for 2D Cartesian polar equations?

so that if I add z^2 it becomes a 3D Body?

[little drum]

why you using 3D calculator btw :o

Ooo.. I see.. then x^2 + y^2 = r^2 with an additional |z - 5| < 5 should suffice

[narrow flower]

let me plot

.close

[lone heart]

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[alpine sable]

When using u-sub (e^(2x)), can you sub it in to both?

or just pick one?

[proven rune]

ah m8 its u

[alpine sable]

yo

[proven rune]

take e^2x = u

[alpine sable]

ye

but can i sub in U for both times

or just one

[proven rune]

so du = e^2x . 2 dx

e^2x . dx = du/2

[proven rune]

obv both

[alpine sable]

ok i figured

[candid turret]

i would take u= e^{2x}+1

[proven rune]

oh yeah that works better

[alpine sable]

then im left with u(u+1)

yea

i guess so

[proven rune]

@alpine sable follow this

more easier

[alpine sable]

way easier

gee

[proven rune]

ye m8

sorry

[alpine sable]

no worries lol

[alpine sable]

I think u = e^2x would be better

no

i disagree

much cleaner than without -1

[proven rune]

nah m8

[alpine sable]

well

i got this

then this

then i did (1/2)ln(e^4) - (1/2)ln(e^0)

but the book got a dif answer.

[pale kestrel]

hmm

$$\int^2_0\frac{e^{2x}}{e^{2x}+1}\dd{x}$$

[ocean seal]

Shuri2060

[pale kestrel]

u = e^{2x} + 1

u' = 2e^{2x} = 2(u-1)

[alpine sable]

u' = (1/2)e^{2x}

[pale kestrel]

$$=\int^{x=2}_{x=0}\frac{u - 1}{u}\dv{x}{u}\dd{u}$$

[alpine sable]

not 2?

[ocean seal]

Shuri2060

[pale kestrel]

ay?

heck no.

[alpine sable]

aw shit im thinking of anti derivative

[pale kestrel]

$$=\int^{x=2}_{x=0}\frac{1}{2u}\dd{u}$$

[ocean seal]

Shuri2060

[pale kestrel]

$$=\int^{u=e^4+1}_{u=2}\frac{1}{2u}\dd{u}$$

[ocean seal]

Shuri2060

[pale kestrel]

$$=\frac{1}{2}\left[\ln u\right]^{u=e^4+1}_{u=2}$$

[ocean seal]

Shuri2060

[pale kestrel]

$$=\frac{1}{2}\ln\frac{e^4+1}{2}$$

[ocean seal]

Shuri2060

[alpine sable]

why did you put u - 1 here

isnt that what you use the du for

just multiply the integral by 1/2

and that turns to a 1

[pale kestrel]

uh what

[alpine sable]

why did you put u - 1 there

[alpine sable]

on the numerator

[pale kestrel]

cus thats what e^2x is

its u-1

[alpine sable]

ok well we get the same thing anyway

[pale kestrel]

???

bruh i just did substitution

and dx = (dx/du) du

[alpine sable]

bro dw about it ty tho

got it

.close

[lone heart]

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[supple tundra]

which part bud

[glad sluice]

heh? do you mean Q6?

can you do the first part? that's just completing the square

hint: you can rewrite the function as x^2 - 4x + 4 - 4 + k

you forgot the k

2nd part seems suspicious

if f(x) is defined for x >= p then there are many possibilities

however if we want to find the range while ignoring the domain

then there is one definite answer

also this seems wrong

try again

why u forget k

ok

for 2nd part

do you understand that the graph of f(x) is an upward parabola?

so it should have a minimum y value somewhere

and then it tends to infinity

positive infinity because upward parabola

so what's this minimum y value

y u forget k

k-4 is a better representation but who cares

anyhow

yes so minimum value of y is k-4

and maximum value?

yes great

so what would be the range?

you should also understand why it covers every number between the minimum and maximum value though

it's obvious because its continuous in this case

range is the set of values for y

no

[glad sluice]

.

what was the minimum value of y we determined?

also y is the same as f

because we're plotting f(x)

right

so what is the minimum value of f?

so what would the range of f be?

great job!

but you can write it in set notation if you like

$[k-4, +\infty)$

[ocean seal]

Peppermint Demon

[glad sluice]

idk what notation you're supposed to use though

why?

yes but for iii they have asked for the domain

smallest value of p
x >= p is the domain

so you should look for the minimum value of x rather than y

correct

can you do further?

so minimum value of p?

ye

can you do the iv one?

$f^{-1} (x)$ is the function for which $f(f^{-1} (x)) = x$

[ocean seal]

Peppermint Demon

[glad sluice]

so try substituting $x \to f^{-1} (x)$ in $f(x)$

[ocean seal]

Peppermint Demon

[glad sluice]

then find $f^{-1} (x)$ using quadratic formula, but you should only take 1 value

[ocean seal]

Peppermint Demon

[glad sluice]

this is the definition of inverse function

[glad sluice]

.

hm?

yes

would be better if you used this form

= what

how

you have to replace the x in f(x) as well

f(x) = (x-2)^2 - 4 + k

replace every x by f^-1 (x)

that is what i mean by substituting

[glad sluice]

.

f(x) also has a x

yep

now use the definition

[glad sluice]

this

read the definition carefully

f(f^-1 (x)) = x

here also you have f(f^-1 (x)) on the left hand side

so can you simplify the left hand side?

yes

yes now solve for f^(-1) (x)

in terms of x

$2 \pm \sqrt{x+4-k}$

[ocean seal]

Peppermint Demon

[glad sluice]

right now this gives 2 solutions

however we only want one

so choose, whether positive or negative square root

hint: look to the domain

which we found out in iii

sure?

great

now can you calculate the domain for this?

$f^{-1} (x)= 2 + \sqrt{x+4-k}$

[ocean seal]

Peppermint Demon

[glad sluice]

square root is undefined for negative numbers

use this

you have to find the domain for x in terms of k

domain is always for x

[pale kestrel]

domain is always for the parameter of the function

f(x) means f takes x as input / parameter

k must be a constant

wait

i dont think so

[glad sluice]

the thing in the square root always has to be positive or zero for the function to be defined

[pale kestrel]

if k = 5, what you said

isn't good for x = 0

[glad sluice]

not necessarily

you're adding 4-k to x

doesn't mean x has to be greater than 4-k

yupp

k-4

yes

also it can be equal

because square root of 0 is perfectly defined

now notice this nice thing about inverse functions that
domain of f^-1 (x) is the same as the range of f(x)

this is a consequence of the definition itself if you think about it

nah man I have to go do some work sorryy

someone else will though

better if you repost the questions

[lone heart]

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[pine monolith]

Help

[pine monolith]

Welp

Lol

Ok so I got no clue how to do dis

Can any person help

[glad sluice]

try applying vieta's relations

[pine monolith]

Idk that

U would have to teach me

Lol

I had been trying to do this problem since last weekend

Pls helpp

Plssssss

I cant stay in suspense for long☹

.end

.stop

.close

[lone heart]

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[keen sonnet]

how is 4 wrong

i got

x+1-x+3
= x-x+3+1
= 4

[wintry lotus]

You may have went backwards

[proven rune]

x+3-x-1/x2+4x+3

= 2/x2+4x+3

@keen sonnet

[keen sonnet]

is it 2

i thought it mightve been

-2

oh wait

i did do it backwards

ty

[lone heart]

@keen sonnet Has your question been resolved?

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[shut parcel]

.reopen

how do i solve for x if my equation is 1008.01 (1+x*90/365)

[shut parcel]

@crisp iron

[buoyant kayak]

that's an expression, not an equation

[shut parcel]

im solving for x

[buoyant kayak]

you're not solving for anything

you have an expression

[shut parcel]

so i cant solve this

[buoyant kayak]

not until you equate it to something

[shut parcel]

idk but thats what my teacher wrote

[buoyant kayak]

unfortunately i have no idea what your problem is and i'm not your teacher

[shut parcel]

i guess it was (1+x*90/365) = 1008.01

[buoyant kayak]

what do you think your first step should be in isolating x?

$\frac{1+90x}{365}=1008.01$

[ocean seal]

a disappointing son

[shut parcel]

minus 1 on both sides

combine x with 90/365

[buoyant kayak]

you wouldn't subtract 1 just yet

try getting rid of the denominator on the left side first

[shut parcel]

90/365 would be 0.246…

so 0.246…x

[buoyant kayak]

not quite how that works

[shut parcel]

sorry

[buoyant kayak]

if i gave you $\frac{x}{4}=2$, how would you solve for x?

[ocean seal]

a disappointing son

[shut parcel]

since its 90/365, would i do the opposite which is multiply

[buoyant kayak]

it's not just 90/365

[buoyant kayak]

answer this

[shut parcel]

x4

x=8

[buoyant kayak]

same concept with your fraction

$\frac{1+90x}{365}=1008.01$

[ocean seal]

a disappointing son

[buoyant kayak]

what's the first step?

[shut parcel]

*365

[buoyant kayak]

right

[shut parcel]

1008.01*365

[buoyant kayak]

right

and that equals 1+90x

now what?

[shut parcel]

-1 on both sides

[buoyant kayak]

right

[shut parcel]

/90

[buoyant kayak]

right

now you have x by itself

👍

[shut parcel]

thank u

i understand it but when it has lots of numbers, i forget or it confuses me loool

thank u again

[lone heart]

@shut parcel Has your question been resolved?

[shut parcel]

@buoyant kayak what if the equation was 1008.01 (1+x*90/365) = 1017.26

minus 1008.01 first from both side

[lone heart]

Channel closed

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Use .reopen if this was a mistake.

[shut parcel]

.reopen

[lone heart]

✅

[shut parcel]

idk if it was closed or not but @buoyant kayak what if the equation was 1008.01 (1+x*90/365) = 1017.26

minus 1008.01 first from both side right

[buoyant kayak]

if that's multiplication, no

[shut parcel]

divide first right

[buoyant kayak]

yep

then it's just the same steps as the last problem

[shut parcel]

just practically do the opposite

im so lost

im getting a diff answer from the answer key

the answer for x is 3.72%

i divide 1008.01 on both sides so 1017.26/1008.01

is that right

[buoyant kayak]

yes

[shut parcel]

then multiply by 365

[buoyant kayak]

3.72 is not correct

[buoyant kayak]

yep

[shut parcel]

we’re using the s = p (1+rt) method

[buoyant kayak]

oh

[shut parcel]

yeah

idk why she’s showing us the mathematically method but it doesnt work

[buoyant kayak]

is it $1008.01(1+\frac{90x}{365})=1017.26$

[ocean seal]

a disappointing son

[shut parcel]

let me show u

lol idk if tht helps

[buoyant kayak]

yeah i misunderstood the equation you sent

[buoyant kayak]

it's this, so before you multiply by 365 you subtract 1

[shut parcel]

wait so we dont solve it how we usually solve mathematically?

i thought u said before subtracting by 1, u multiply by 365

what i did first is calculate 1000(1+0.0325*90/365) so its easier for me to understand

then 1008.01 (1+x*90/365) = 1017.26

what do i do next after

P (1+rt) = S is the same as 1008.01 (1+x90/365) = 1017.26

[lone heart]

@shut parcel Has your question been resolved?

[buoyant kayak]

yeah i misunderstood your equation

i thought it was $1008.01(\frac{1+90x}{365})=1017.26$

[ocean seal]

a disappointing son

[buoyant kayak]

but it was $1008.01(1+\frac{90x}{365})=1017.26$

[ocean seal]

a disappointing son

[buoyant kayak]

in this one, after dividing by 1008.01, you'd multiply by 365 first to get rid of the common denominator under both those terms

[buoyant kayak]

but in this one, you technically could do the same thing, but you'd have 365+90x=(1017.26)(365) because you would have to multiply the 1 by 365 as well

or you could just subtract the 1 first, then multiply by 365

[lone heart]

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[dusk hamlet]

$L={ww^R | w \in {a,b}^*}$

[ocean seal]

WizzardKid

[dusk hamlet]

Is this language regular?

[lone heart]

@dusk hamlet Has your question been resolved?

[lone heart]

@dusk hamlet Has your question been resolved?

[lone heart]

@dusk hamlet Has your question been resolved?

[lone heart]

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[unkempt umbra]

yeah it is

nice one, took me a sec

[lone heart]

@alpine sable Has your question been resolved?

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[small trellis]

Is the first one a mistake?

Oh, but it says that it can be done like this too

But wouldn't that mean that the 2x part in first image should still be just x?

[vale wigeon]

no

antiderivatives are defined only up to an additive constant

log(2x) and log(x) differ bby exactly that much

specifically, log(2x) = log(x) + log(2)

[small trellis]

so 1/2 ln(2x) == ln(x)/2?

[alpine sable]

No

That's 1/4*ln(x²)

1/2ln(2x)=(ln(2)+ln(x))/2

[small trellis]

But this seems to work if the 1/2 on the right side is taken outside of the [] for the calculation with the limits

like in the images

[vale wigeon]

+C

where C = log(2)/2

[small trellis]

And we can ignore that for this question?

[vale wigeon]

yes

[small trellis]

So it's + C for ln(x)/2 but the constant is included when it's 1/2 ln(2x), correct?

[vale wigeon]

...you're overthinking it

[small trellis]

So, as you said, both those things are equal if I add C to the second one?

[small trellis]

Please, just tell me if I'm right

I can understand

[vale wigeon]

you're saying things that might kind of be technically correct but you're overthinking it

[small trellis]

It's ok

this is nice to understand

[daring marlin]

hii how do u write the answer to a numerical expression?

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@small trellis Has your question been resolved?

[lone heart]

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[proven rune]

in what

<@&268886789983436800> we got a madlad here

bro this is a maths server

not a roblox one

pls close the channel by typing .close

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✅

[tiny minnow]

do .close plz

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@alpine sable Has your question been resolved?

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[alpine sable]

I need help taking the primitive function out of 5e^(0.5x)

[alpine sable]

In my book it says e^(kx) will become (e^(kx))/k when k isn't 0

So that's what i did. getting 10e^(0.5x)

[marsh rapids]

yes

[alpine sable]

But the calculator says different

[marsh rapids]

what does it say

[alpine sable]

It gives me (5e^(0.5x))/2

wolfram alpha gives the same answer

[marsh rapids]

are you sure you took the primitive

yep derivative

[alpine sable]

Oh wow

[solar merlin]

lol

[alpine sable]

I forgot haha

Well thanks haha

[marsh rapids]

have a good day

[alpine sable]

.close

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[alpine sable]

small=x
medium=2x
large=2x+120

x+2x+2x+120=1620
5x+120=1620
5x=1500
x=300ml

small=300ml
medium=600ml
Large= 720ml

[alpine sable]

I made the question into an equation then substituted x=300 into the original equations.

[drifting hull]

and what is your question?

[alpine sable]

How much is in each cup

[drifting hull]

As you calculated: small=300ml, medium=600ml,Large= 720ml