help-13

[crimson sedge]

im not sure how to solve this

[exotic furnace]

Does “factor theorem” ring a bell?

[crimson sedge]

no

how can i find 2 values in one equation.

do i subtitude those roots and make them equal to 0?

[exotic furnace]

I think you’re on to something

Basically yes that’s the factor theorem: (x-k) is a factor of a polynomial P(x) iff P(k)=0

[crimson sedge]

oh ok

let x = 2
16 + 4p + q = 0
and let x = -1/2
-1/4 + 1/4p + q = 0

??

[crimson sedge]

i dont think i can use polynomial division in this case

[exotic furnace]

just one tiny typo

[crimson sedge]

where

[exotic furnace]

Not sure why you have the extra x in the second equation

[crimson sedge]

oh ok

[exotic furnace]

But yes this is correct

[crimson sedge]

so do i use simultaneous equatoin to solve p and q?

[exotic furnace]

Yep

Hmm, not what I got

[crimson sedge]

is 20/7 wrong?

oh wait

i got p = -13/3

[exotic furnace]

Ok I think you got it now

[crimson sedge]

ok when i subtitude i got 4/3 as my q

[exotic furnace]

Yep

[crimson sedge]

-22/3 is my final answer

C

👍

[exotic furnace]

Yep

gj

[crimson sedge]

thanks i got another question

(P-1)x^2 +4x + (p-5) >0

for this one

do i use discriminant to find the set of value in p?

b^2 - 4 a c

[exotic furnace]

Yes

[crimson sedge]

so what do i do after that

is the discriminant = 0?? or > 0?

how do i know which sign i use?

[exotic furnace]

I’m assuming this needs to be true for all x, right?

[crimson sedge]

yh

[exotic furnace]

Then how many roots should the quadratic have?

[crimson sedge]

im not sure

i think its 2 but i dont know why

[exotic furnace]

Well you know you can only have a root if the quadratic is equal to 0

But the quadratic needs to be always greater than zero

[crimson sedge]

why is it quadratic? isnt it the discriminant

if b^2 - 4ac > 0 -> 2 real roots
if lower -> no real roots
if same -> one real roots

[exotic furnace]

We need to understand how our quadratic behaves before we make conclusions about the discriminant

[crimson sedge]

oh how can i do that

[exotic furnace]

Again, roots happen where the quadratic is equal to zero, right?

[crimson sedge]

yh

[exotic furnace]

but according to the problem, the quadratic must be strictly greater than zero

[crimson sedge]

oh so theres no real root?

[exotic furnace]

Yes

[crimson sedge]

OHHHHHHHH

[exotic furnace]

So the discriminant is in fact negative

[crimson sedge]

i understand what u meant now

so discriminnat should be lower than 0

which means i can find set of value of p

[exotic furnace]

Ye

[crimson sedge]

in quadratic formula

oh wait

it is

p^2 - 6p + 9 < 0

(p -3)(p-3) < 0

p < 3

[crimson sedge]

did u get p<3?

[exotic furnace]

Nope

[crimson sedge]

is my equation correct tho?

(p -3)(p-3) < 0

[exotic furnace]

No

Might have to show your work

[crimson sedge]

ok let me organise my work again

dam i got 3+/- 2 root 2

i forgot to times by -

[exotic furnace]

Ok

But here’s the thing

[crimson sedge]

Oh

[exotic furnace]

If a is negative, this would open downward

So you need a to be positive

[crimson sedge]

is that mean p > 3 + 2 root 2?

[exotic furnace]

Yes

[crimson sedge]

wait i dont understand

whats wrong with going downward

oh wait if its lower than 0, the graph will go below the 0

right?

[exotic furnace]

Yes

[crimson sedge]

therefore we just count the postive parts..

thanks for helping

u are very patient teacher 👍

[exotic furnace]

No prob

[crimson sedge]

how do i close the chat now?

do i just leave it?

[cedar kiln]

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[inner belfry]

c=10
r=c/2
so summation will be using Σ(a,x) and for other functions like gamma etc, will be: X(v,e) so X is the function like gamma and v is the top of the function and e is the bottom of the function (note the equation at the top is still true)
c/r=x^c
x=x^c /2
v=c/e^x
vcr=(v+n)(c+n)(r+n)
vcr/θ=vcr^(Σ(vcr,0))
Σ(vcr,0)=k
Z=vcr^kn
Z_kn=Z/k^-2
Z_kn^2=Z*z
n=Σ(1,3)
z=n^1, so is z=2 or z≠2

help me solve this

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[inner belfry]

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<@&286206848099549185>

[frank quiver]

this is horrifying

[inner belfry]

ill try to write it

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[cedar kiln]

@inner belfry Has your question been resolved?

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[versed fulcrum]

Hello

[versed fulcrum]

Can someone help me with this problem

[long arrow]

can you sketch the graph of y = |x^2 - 2x| or y = x^2 - 2x?

[versed fulcrum]

Like this?

[long arrow]

yes

now just imagine the line y = c (where c is some real constant)

it's just straight line parallel to x-axis

[versed fulcrum]

Okay

[long arrow]

now try to specify the value of "c" such that it crosses your graph four times

[versed fulcrum]

Ohhh

I have to find the vertex

and reflect it

Which is going to be greater than 0 and less than 1

That makes sense

Thanks so much

[long arrow]

yep

(0,1)

[versed fulcrum]

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[solid juniper]

my friend gave me this problem... does there exist an uncountable set $W\subset \mathcal{P}(\bN)$ such that every set in $W$ is infinite and for all $A,B\in W$ with $A\neq B$, $A\cap B$ is finite? i say yes and i think i have a proof but it's nonconstructive and uses zorn's lemma. idk if it's possible but can anyone find an explicit example you can write down?

[crimson delta]

just make them all disjoint

[solid juniper]

sorry

missed a condition

[wraith dagger]

layla

[solid juniper]

ok there that looks right

[crimson delta]

oof

well the first obvious idea is to use rationals instead of naturals

at least I have seen that before for similar problems

and then you do something like for each real x choosing all rationals that satisfy something nice

ah

choose a sequence that converges to x

@solid juniper

[solid juniper]

ohhh interesting

[crimson delta]

you of course cant really write down the sets either but you at least have an idea of whats going on

[solid juniper]

it's constructive enough for me lol

thx

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[dire geode]

Fun problem

[solid juniper]

ur fun

[cedar kiln]

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[dreamy zenith]

stuck on part c of 265

[dreamy zenith]

$(\frac{y+3}{2})^{\frac{1}{3}}$

[wraith dagger]

putridplanet

[dreamy zenith]

was the answer to b

[humble karma]

You differentiate this expression with respect to y now.

So this would be a chain rule application here

[dreamy zenith]

$\frac{1}{3}(\frac{y+3}{2})^{\frac{-2}{3}}\cdot\frac{1}{2}\frac{dy}{dx}$

[humble karma]

You're missing the inside derivative

[dreamy zenith]

?

[humble karma]

Remember the chain rule multiplies everything by the derivative of the inside function

You're like off by a factor

[wraith dagger]

putridplanet

[humble karma]

You can just leave it as 1/2 here since you're considering y to be independent in this equation

So no dy/dx

[dreamy zenith]

$\frac{1}{6}(\frac{y+3}{2})^{\frac{-2}{3}}$

[wraith dagger]

putridplanet

[humble karma]

Yesss

That should be it

[dreamy zenith]

the answer saysjust 1/6

[humble karma]

Well it's evaluated at f(1)

So y = f(1) = 2(1)^3-3 = -1

[dreamy zenith]

so plug in -1 for y

[humble karma]

Yes

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[marsh pond]

not saying this is the most efficient way to do it, but is there any error with this?

[livid hound]

error in first step already

[marsh pond]

where?

you can't move x^3 to be power of ln argument?

[livid hound]

you'd need to raise the whole initial argument to that power

[marsh pond]

wdym?

[livid hound]

$$a\log(bc) = \log((bc)^a) \redneq \log(bc^a)$$

[wraith dagger]

ℝam()n()v

[marsh pond]

ooopsss

I keep forgetting

5x is not a single term

even tho usually it is

but not for ln

[marsh pond]

you mean like this?

[livid hound]

that would work

[marsh pond]

yuck lol

but then I could also make it into 2 ln's

like so

but what is the derivative of something raised to the power of a variable raised to another power?

chain rule for this? or something else

it's so much easier with something simple like d/dx( x^2 )using power rule

[livid hound]

exponential chain chain power

take things one step at a time

[marsh pond]

start with this?

it's correct tho, is it not?

[livid hound]

1 sec

[marsh pond]

d/dx(ln(2x)) = 1/(2x) * d/dx(2x)

OK

[livid hound]

you turned 5^x^3 into 5^x^2

[marsh pond]

its a 3, sorry too small

i noticed that too

i think my pen size is too thick

i will correct it to be more clear

[livid hound]

then it's currently ok

[marsh pond]

OK

OK

so then the question is for the d/dx of those values

chain rule?

[marsh pond]

typical scribblings at a university lecture ...

i'm picking up bad habits from watching profs and their horrendous writing skills

[livid hound]

a lot of chain rule will be needed, yes

[marsh pond]

OK

[digital cliff]

are you doing it this way for masochistic reasons?

[marsh pond]

so just curious, because this is not the best way to do it, I know lol

[livid hound]

apparently yes

[marsh pond]

just to see how it could be done

https://tenor.com/view/the-pain-only-makes-me-stronger-the-mitchells-vs-the-machines-the-pain-makes-me-powerful-the-pain-feels-good-mitchells-vs-machines-gif-21431785

this is the painless way to do it:

but no pain, no gain

d/dx[5^(x^3)]

where do I begin with this

[digital cliff]

e^(x^3ln5)

[marsh pond]

lmao

yeah I think I have just dipped my toes into MATH 9999

I'm still on MATH 205

[digital cliff]

its not that bad, its the same way youd differentiate 5^x, express it as e^xln5

[marsh pond]

I'm trying to benchpress too much here, gotta build up

I'll just stick to product rule and play it safe for now

[marsh pond]

I don't see this at all tbh

maybe I should know this tho

even for MATH 205

[digital cliff]

$5^x=e^{xln5}$ then the derivative would be $ln(5)\cdot e^{xln5}=ln(5)\cdot 5^x$

[wraith dagger]

AℤØ

[digital cliff]

this is just the same but the x has an exponent so the chain rule comes out differently

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[ebon iris]

help

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[crimson delta]

you are one step away

multiply out the last thing

well one more

thats one minus too much for x

but the rest is correct

[cedar kiln]

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[gleaming osprey]

Linear algebra question:

Why must the dimensions of the matrices match when adding or subtracting, but not when multiplying?

[gleaming osprey]

@gleaming zinc

[cedar kiln]

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[flint plinth]

because matrix addition/subtraction is done "elementwise" and matrix multiplication is not

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[royal loom]

Hi, I was hoping someone could check my work for this.

So I want to prove $\vec{a}$ is an accumulation point of $S$ $\iff$ $\exists {\vec{x_k}}\subset S\setminus {\vec{a}}$ s.t $\vec{x_k}\to \vec{a}$

[wraith dagger]

Austin

[royal loom]

so first going in the (=>) direction

if a is an accumulation point of S then for all epsilon >0 there is a ball of radius epsilon around a such that that ball is a subset of S

then we can let epsilon=1

epsilon=1/2

and so on

epsilon=1/n

and continue infinitely

from each of these balls we can choose an element for our sequence xk, so we just pick a single thing from the ball that is not a itself

and since these are balls with positive radius, there is always something other than a in every ball

then the sequence xk we just constructed is in S and doesn't contain a

and xk -> a

so QED

?

okay and secondly going in the (<=) direction

if x_k of elements in S disjoint a converges to a, then for all epsilon > 0 there exists a natural number M s.t for all n>=M |xk-a|<epsilon additionally since no xk=a then 0<|xk-a|<epsilon so for all epsilon neighboorhoods of a then we have these xk in S with infinitely many more terms n>=M left in the sequence contained inside this neighboorhood.

so QED again

?

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[long swan]

No

[cedar kiln]

@faint yoke Has your question been resolved?

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[wraith dagger]

Austin

[royal loom]

so first going in the (=>) direction
if a is an accumulation point of S then for all epsilon >0 there is a ball of radius epsilon around a such that that ball is a subset of S
then we can let epsilon=1
epsilon=1/2
and so on
epsilon=1/n
and continue infinitely
from each of these balls we can choose an element for our sequence xk, so we just pick a single thing from the ball that is not a itself
and since these are balls with positive radius, there is always something other than a in every ball
then the sequence xk we just constructed is in S and doesn't contain a
and xk -> a
so QED
?
okay and secondly going in the (<=) direction
Austin — Today at 6:02 PM
if x_k of elements in S disjoint a converges to a, then for all epsilon > 0 there exists a natural number M s.t for all n>=M |xk-a|<epsilon additionally since no xk=a then 0<|xk-a|<epsilon so for all epsilon neighboorhoods of a then we have these xk in S with infinitely many more terms n>=M left in the sequence contained inside this neighboorhood.
so QED again
?

please tag me if you respond

[cerulean sail]

Seems all good to me

[royal loom]

hey chartbit!

thanks

it seemed pretty solid to me too but sometimes I miss things (a lot of times)

[cerulean sail]

It can happen, but today you’re all good you have the idea for it

[royal loom]

Ty again! hope you've been doing well

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[dusk pilot]

Attempted answer in the image.

I'm almost certain what I wrote is correct.

[sly abyss]

send ur work

[royal loom]

you wrote your intervals incorrectly

[sly abyss]

why did u do infinity

it snot

[fossil sigil]

I'm new to this server and I need help with my home work

[sly abyss]

open a new channel

[dusk pilot]

Jodha you need to find an empty help channel they're above the taken ones

[cedar kiln]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[dusk pilot]

Welcome to the server though

[sly abyss]

You found the right answers.

However,

you wrote the interval incorrectly

this is correct

but you just need to simplfy fractions

and thats ur anwer

[dusk pilot]

Okay, I'm not sure how. If X is greater than -4/3 isn't that (-4/3,oo)

?

[sly abyss]

you literally had the first anser

idk why u added the infinitys

this doesnt even make sense, no offense

[royal loom]

^

[dusk pilot]

None taken

[royal loom]

it is just -4/3 to 1/3

[sly abyss]

yes

[royal loom]

not -4/3 to infinity

and -infinity to 1/3

that is literally just excluding the answer you want

[dusk pilot]

Interesting

[royal loom]

what you wrote down, those intervals, is every real number except for your answer

[sly abyss]

ye lol

[royal loom]

so you somehow managed that

XD

[dusk pilot]

So no U

[sly abyss]

you said this as ur asnwer

if this said greater than, then u would be almost corect

still said the wrong thing, but it would be closer.

it would be -infinity, -4/3 U 1/3, infinty

but in this case because it says less than its just -4/3, 1/3

[dusk pilot]

I think I get it now, since they're not going separate directions X is within the parameters of each and not to infinity

That's where I was confused because if they go opposite directions, than it becomes two separated intervals

Right?

[sly abyss]

ye

[dusk pilot]

Sorry this whole interval stuff is messing with me. Much appreciate friends.

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[blazing hedge]

please help with how to find a

[runic garnet]

Well ur given the equation

[sly abyss]

ye

[runic garnet]

What info is given to u?

@blazing hedge

[blazing hedge]

idk how to solve

the temp

wait t equal to 0 and then just solve?

[astral steppe]

T(t) is the temperature

just plug it in and solve for a

[blazing hedge]

is that possible, isn't t the time?

[runic garnet]

t = time

T is a function of the time, representing temperature of the bread in degrees celsius

so yes, set t = 0

and T(t) = 205

and solve for a

[blazing hedge]

OHH thank u so much

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[runic garnet]

[cedar kiln]

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[oblique prawn]

are units of derivative just (units of independent variable)/(units of dependent variable)

[royal loom]

I don't think so

for example if we take the derivative of velocity then our units will be something like m/s^2 but if we take derivative of position our units will be just m/s

so it seems it depends on the units of what you are differentaiting

unless I am misinterpreting your question

[oblique prawn]

ye cuz for position the independent variable is m and dependent is s so its just m/s

[dire thorn]

Wouldnt s be the independent variable

[violet flume]

youll pick up a 1/m where x is the variable youre differentiating with respect to, yea

so like $\qty[ \dv{y}{t} ] = \frac{\qty[y]}{T}$

[wraith dagger]

jan Niku

[violet flume]

and $\qty[ \dv{y}{x} ] = \frac{\qty[y]}{L}$

[wraith dagger]

jan Niku

[violet flume]

and $\qty[ \dv[2]{y}{t} ] = \frac{\qty[y]}{T^2}$

[wraith dagger]

jan Niku

[oblique prawn]

uhh what

[oblique prawn]

i meant the other way around

[oblique prawn]

i meant (dependent variable)/(independent variable) sorry

[violet flume]

say that y(t) is a position function

[oblique prawn]

ye

[violet flume]

then if you differentiate with respect to t, you pick up a 1/time

[oblique prawn]

ye

[violet flume]

and retain the original units

yea

👀

[oblique prawn]

ye so meters * 1/time

[violet flume]

yup

[oblique prawn]

so its true?

[violet flume]

i dont think what you originally said is the same as what im saying

but if what youre saying is what im saying then yes

[oblique prawn]

oh ok

ty

.close

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[shell relic]

not sure where to start

.close

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[frigid sinew]

i dont even know how to appraoch this

[loud patrol]

whats a formula that relates cos(x) and sin(x)

[cedar kiln]

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[cedar kiln]

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[civic coral]

am i right?

[waxen igloo]

hey

yes, that is correct.

[civic coral]

what if the d/dx is actually d/dsin

what would we get as an result?

[civic coral]

what is d/dsin f(x)

[cedar kiln]

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@civic coral Has your question been resolved?

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[raw wren]

x(x-1)(x+1) on both sides i did

[raw wren]

@tropic oxide

[raw wren]

an cancelled out bs i left with 1/6 vs 1/8

[tropic oxide]

you "cancelled out"?

what exactly did you cancel out??

[raw wren]

x(x-1)(x+1) this on both side

[tropic oxide]

????

both sides of WHAT?

[raw wren]

x(x-1)(x+1) this was on the both sides on numerator when i factorised the given term

then all 3 cancel out no?

[tropic oxide]

both sides OF WHAT?

it sounds like you completely ignored the fact that they weren't just talking about these fractions but about the REMAINDER of a DIVISION!

[raw wren]

no then remainder i get is 1 both side

[tropic oxide]

you are still treating this as some kind of equation...

[raw wren]

assuming a equal sign beteen A AND B

[tropic oxide]

(x^3 - x)/6 = (x^3 - x)/8?

this is just wrong

[raw wren]

no the equal sign is not rigid it can be greater or less after solving

[tropic oxide]

i do not know how to explain to you how wrong this is

[raw wren]

(x^3 - x)/6 ?? (x^3 - x)/8? something like this

[raw wren]

idk but some books advise to assume like this

[tropic oxide]

again

you're not looking at these fractions

they meant to talk about the REMAINDER of the DIVISION of x^3 - x by 6 and by 8

[cedar kiln]

@raw wren Has your question been resolved?

[raw wren]

but wont 1 be remainder it its 1/6

so both are equal this way

[tropic oxide]

no

also press ❌ on the bot so the channel doesnt close on you

[cedar kiln]

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[raw wren]

.reopen

[cedar kiln]

✅

[raw wren]

damn'

[raw wren]

can you tell your approach

[tropic oxide]

i'd notice the following:

x^3 - x = (x-1) * x * (x+1)

this is a product of 3 consecutive integers, therefore exactly one of them will be a multiple of 3.

and also, since x is odd, x-1 and x+1 will both be even, hence divisible by 2.

therefore the product is divisible by both 3 and 2, and so is divisible by 6 and the remainder mod 6 is zero.

on the other hand, out of x-1 and x+1, two consecutive even numbers, exactly one will be a multiple of 4.

and a multiple of 4 times a multiple of 2 makes a multiple of 8.

so x^3 - x is actually also divisible by 8.

so that remainder is also 0.

[cedar kiln]

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[raw wren]

.reopen

[cedar kiln]

✅

[raw wren]

thankyou ann

[raw wren]

great i understood it

just one thing so three consecutive number will always have mltiple of 2 and 3 right?

4 may or may not

[finite lion]

i think x(x+1)(x-1) always divides by 6 because of it spanning 3 numbers

and then it also must divide by 8 because of 4 and 2

so both of them are 0

so both of them are same

qed

[raw wren]

ya right

[finite lion]

calm down

juggernaut

do you know modular arithmetic

[raw wren]

nope

[finite lion]

that’s fine

well

basically

x-1, x, x+1 mod 3 = {0, 1, 2}

and that mod 4 = {0, 1, 2} or {0, 2, 3}

[raw wren]

how is it used here though

[finite lion]

which is sufficient

[finite lion]

mod 4 = 0 mean’s one number contributes 4, and the 2 means another number contributes the 2

and then the 1 and 3 is bad and poo poo we should kick them in the balls

[finite lion]

and this means there’s a number mod 3 = 0

which means someone’s contributing a 3

infact the number is always divisble by 24, (except for small cases ofc)

i think for x=3 onwards it’s always divisble by 24

[raw wren]

i dont understand this lol

[finite lion]

ok let me draw it out

visual explanation should be the most intutiigtiigive

[raw wren]

okay

[finite lion]

no matter what number you pick

you end up with something that’s remainder 0 when divisible by 3

something of that but 1

and something but that but 2

WAIT

I COLOURED IT WRONG

OOPS

but the idea is there

now the question says x is odd

so x must be a black number

wait

here

notice how every black number is neighboured by a red and green number

[raw wren]

yes

i understood

thanks for such a through explaination'

[finite lion]

np

[raw wren]

.close

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[lapis laurel]

I have a initial sum of money that has a monthly interest rate. I also make monthly contributions to the account. How do I represent this as a function?

[lapis laurel]

I know that if it only where contributions it would be a:
$$ \Sigma $$
and if it were only a interest rate:
$$ money \cdot interestRate^x $$

[wraith dagger]

N1x1T4

[zenith sail]

you're referring to an annuity @lapis laurel

[lapis laurel]

So by figuring out what annuity is I can answer my question?

[zenith sail]

I mean

you didn't really give any specifics but

that formula describes the situation you mentioned^

[lapis laurel]

ok, thank you for giving me directions about this

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[blazing zephyr]

how do you even start? idk what do do with the area of the triangle BDF

[finite lion]

draw a lot of lines and do a lot of halving

draw BE and and try working from there

[tropic oxide]

theres a very big brain way of doing it but i am not sure i should divulge it here

[finite lion]

ABD is half, DFC is eigth, BFC is quarter, so DFC should be around an eigth, ABCD is 7.5*8=60

just approximate it

7.5 looks kinda like 60

so its D

/j

[tropic oxide]

DFC should be around an eigth,
why around

oh hang on wait

[finite lion]

for humor

x is around x

[tropic oxide]

[finite lion]

i guess

wait that rhymes

[tropic oxide]

ok but yeah if you try to look at areas of triangles as fractions of the rectangle

[finite lion]

talk about mathematical literature

[tropic oxide]

you can actually get the problem done quite nicely

[finite lion]

i call this type of priblem

"half the shape until 天荒地老"

[blazing zephyr]

DF and CE is perpendicular right

[finite lion]

doesnt matter

[blazing zephyr]

candies are you chinese

[finite lion]

same base, same height --> same area

[finite lion]

yes

[blazing zephyr]

ok

[finite lion]

ok

[blazing zephyr]

i got it thx

.close

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[neat dune]

try factoring then squeeze theorem

[broken valley]

how to factor?

[cedar kiln]

@broken valley Has your question been resolved?

[neat dune]

Can't factor completely but u can factor the x terms

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[shut spoke]

i dont understand how to set the limits for the double integration

[tropic oxide]

go polar?

this region is a dead ringer for polar

[shut spoke]

ah yes it states to use polar

but im not quite sure how to start

[tropic oxide]

well

[shut spoke]

from my understanding the double itnergation limits are the regions

[tropic oxide]

your thing is bounded by the circles r=1 and r=2

[shut spoke]

so 0 => 9pi/16 and 1 => 2?

[tropic oxide]

why 9pi/16

it's 2pi/3 ...

[shut spoke]

oh

how can you tell?

i mean it doesnt really

excatly say it right

[tropic oxide]

y = -sqrt(3) x

-sqrt(3) is tan(2pi/3)

[shut spoke]

oh what

i see

so given 0 => 2pi/3 and 1 => 2

and dA is equal to r dr dtheta

Is this right?

wait no i need to do something using the 2 curve equations right?

[tawdry crest]

Ann are you good at calc?

I need help with calc, anyone?

[cedar kiln]

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@shut spoke Has your question been resolved?

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[tropic oxide]

,rccw

[tropic oxide]

!status

[cedar kiln]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[spring plover]

I have no idea how to begin

[tropic oxide]

i'd just calculate m^2 + n^2 lol

it falls into place very nicely

[spring plover]

So just square everything?

Oh

[tropic oxide]

m = a cos(θ) + b sin(θ)
n = a sin(θ) - b cos(θ)

yes, square both of these and add them together, and then simplify.

[spring plover]

Okok

I feel like a toddler

What's next teacher?

Or was i wrong

[opaque pivot]

no

[spring plover]

So I was wrong?

[opaque pivot]

given that $m = a\cos \theta + b\sin \theta$ , find $m^2$

[wraith dagger]

nalin

[opaque pivot]

and do similarly for n^2

and add them up

[spring plover]

Seperately ok lemme try that too

Hm?

Anyone care to show it to me please ;-;

[opaque pivot]

that's not

how it works

[spring plover]

Oh god

Teach me please

I def failed my test

[opaque pivot]

$(\alpha + \beta )^2 = \alpha ^2 + \beta ^2 + 2\alpha \beta$

[wraith dagger]

nalin

[spring plover]

Oh

[opaque pivot]

when you do $m^2$, the LHS is $(a\cos\theta + b\sin\theta)^2$

[wraith dagger]

nalin

[spring plover]

Oh so no squaring individually

[opaque pivot]

no

[spring plover]

Okok

[opaque pivot]

try now

[spring plover]

Yup

When you take of the bracket it would be a2 cos2 and so on right?

[opaque pivot]

yes

[spring plover]

Any more simplification or just that

?

[opaque pivot]

correct

well for simplification

you can write 2abcosθsinθ as absin2θ

[spring plover]

Alright alright

Thank you so much

[opaque pivot]

np

glad to help

[spring plover]

May I ask how to close the channel?

[opaque pivot]

.close

[spring plover]

.close

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[gritty marten]

Check the series (2)^2n*3^1-n for convergence

[gritty marten]

In simplified form it becomes 3*(4/3)^n

I'm not sure though

[cedar kiln]

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[cosmic quartz]

excersise 4

how many figures have surface 6?

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[chrome slate]

i was thinking about the sequence 1/(k^(1/k)), but i dont know if the ^1/p outside of the sum will cause any problems

[cedar kiln]

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@chrome slate Has your question been resolved?

[cedar kiln]

@chrome slate Has your question been resolved?

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@chrome slate Has your question been resolved?

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[valid adder]

Is this correct?

[valid adder]

This was original graph

<@&286206848099549185>

[dawn monolith]

The slope/gradient of line B is not correct, second point of line B is at (4, 0), not (0,4)

[valid adder]

oh wait

Is this how u do it?

[valid adder]

Or is it this way but change 4.0 to 0.4

[dawn monolith]

yes

[valid adder]

Okay thanks

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[plain badge]

so for T2

the matrix is symmetric right

which means that when we diagonalize it we get D = S^-1 A S

and the matrix S is formed by the eigen vectors found from matrix A

also since matrix A is symmetric , matrix S must also be orthogonal

but for some reason the matrix i found isn't orthogonal

i found
-2 -1 1
1 0 2
0 1 1

[cedar kiln]

@plain badge Has your question been resolved?

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@plain badge Has your question been resolved?

[cedar kiln]

@plain badge Has your question been resolved?

[violet flume]

@plain badge does it matter that theres a repeated eigenvalue?

i mean these are the two vectors that arent orthogonal

nope im dumb

[cedar kiln]

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[indigo steppe]

HI

[indigo steppe]

Can u help me ?

I need to determine whether f is even, odd or neither

[vapid sand]

to determine if it is odd or even

you have to let f(x) = f(-x)

and simplify it from there

[indigo steppe]

can u write it pls on paper >

?

[vapid sand]

:V

alright

[spice kraken]

bro stop asking for answers it's the third time

[vapid sand]

THIRD TIME?!?

[indigo steppe]

but idk the solution bro

what I can do

[spice kraken]

we'll lead you step by step

[opaque pivot]

use the hint that you're given and work on it

[indigo steppe]

i have 8 Q like that just show me the one others i will do formyself

[mystic cave]

So, is that the exact wording of the question? Is there any other description?

[indigo steppe]

The Q is this

[indigo steppe]

What I need to do here i dont get

[hollow trail]

1. if f(-x) = f(x), it's even
2. if f(-x) = -f(x), it's odd
3. if f(-x) is neither of those things, it's neither
   plug in -x for x in your function and simplify

[indigo steppe]

I know this one

the solution way idk

I need simplify this ya ?

[indigo steppe]

ok tell me what to do

first i need simplify ?

.close

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[spice kraken]

?

[cedar kiln]

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[upbeat hemlock]

I need help with question b). This question is asking to find the next two iterations using the Bisecton method. The answer I found was [-1.5,-1.25] however solutions say otherwise and I am a bit confused on how to do it.

[spice kraken]

show your work please

[spice kraken]

?

you're correct though

[upbeat hemlock]

okay thanks

solutions are saying this

theres an error when calculating f(a2) and f(b2)

just needed some reassurance if i got the question right or not

[spice kraken]

.

what

the solution is wrong

[spice kraken]

yeah

but the end result is correct though probably a typo

[upbeat hemlock]

does it matter which number comes first in the square brackets

i thought the number that would be more negative would be on the left

[spice kraken]

yea

the question is kinda strange writing [-2,-1]

[upbeat hemlock]

okay well thanks for the help

.close

[cedar kiln]

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[crimson sedge]

hi

[cedar kiln]

Please don't occupy multiple help channels.

[tropic oxide]

!onechannel

[cedar kiln]

Please stick to your channel.

[tropic oxide]

.close

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[magic moss]

hi

[tawny cedar]

Please can someone go through this with me

So I think that 3 ordinates means 2 strips so I split from 0 to 1 into 2 strips that makes y(0.25) and y(0.75) I think.
So I evaluated sqrt(sin(0.25)) and sqrt(sin(0.75)) and and I did 0.5 ( The sum of those 2) and got 0.0902 which doesn't seem right

For Simpsons rule I'm getting the complete wrong answer, I split it up into 2 again, I did 0.5/3 ( sqrt(sin(0)) + [4 × (sqrt(sin(0.5))] + [2( sqrt(sin(1)) ] + sqrt (sin 1) = 0.12833

Which is even worse off approximation, what am I doing wrong, can someone help me 🙏

[tawny cedar]

Hey

[magic moss]

its my first time in discord

[tawny cedar]

No problem

[magic moss]

what do you want

[tawny cedar]

Help with that question

My working is there too, but it is wrong I'm pretty sure

[magic moss]

it is calculus ?

[tawny cedar]

It's further maths a level

[tawny cedar]

Don't think so

@spice kraken

[spice kraken]

bro why ping me

[tawny cedar]

Because you may be able to help me

[spice kraken]

I don't have 8 hands ahhhhhhh

[tawny cedar]

I'll wait

It's all good

Sorry

[spice kraken]

I'm not familiar with SImpson's rule

but I can try

nvm Iｓｕｃｋ

[cedar kiln]

@tawny cedar Has your question been resolved?

[cedar kiln]

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[winter jungle]

What would be a way to go about this?

[cedar kiln]

@winter jungle Has your question been resolved?

[cedar kiln]

@winter jungle Has your question been resolved?

[winter jungle]

<@&286206848099549185>

This is all I have

[cedar kiln]

@winter jungle Has your question been resolved?

[cedar kiln]

@winter jungle Has your question been resolved?

[cedar kiln]

@winter jungle Has your question been resolved?

[cedar kiln]

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[cedar kiln]

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[nocturne nexus]

how would i even solve this 😭 i tried using hte power rule and using the instantenous rate of change formula but im stumped 😭

[cedar kiln]

@nocturne nexus Has your question been resolved?

[nocturne nexus]

<@&286206848099549185>

[astral steppe]

if you're asking about a) it should be right

[nocturne nexus]

its wrong 😭 ive also tried 2/3x^(1/3)

[astral steppe]

maybe remove the parenthesis around your exponent

it must be a format error

[nocturne nexus]

😭

[astral steppe]

2x/3 maybe

at this point it's just trial and error

[nocturne nexus]

ya

[astral steppe]

good luck 🤞

[nocturne nexus]

ty

[nocturne nexus]

lmao it was because i used x instead of a 💀😭

.close

[cedar kiln]

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[proper plaza]

Hello I appear to be stuck in my calculations as I know that the answer is 14 over 3 but in my calculations I have calculated that the answer is 8 over 3 and therefore I've made a miscalculation and I'm wondering where I've went wrong?

[ancient lodge]

Did you make a typo with 14 instead of 16

[proper plaza]

Yes probably, but my answer appears to be off by 8/3 so either way my answer is incorrect.

Triple checked what I put in the calculator it is for sure this exactly as written:

[ancient lodge]

You split the integral incorrectly

[quick zephyr]

I'm rusty af in calculus (or applied calculus) but shouldn't it be 0 to 1

and 1 to 3

for the limits

[ancient lodge]

$v(t)<0$ for $t \in (0,1) \cup (1,3)$ and $v(t)>0$ for $t \in (1,2)$

[wraith dagger]

Civil Service Pigeon

[ancient lodge]

This comes from the fact that $$|x|=\begin{cases} x, x \geq 0 \ -x, x<0 \end{cases}$$

[wraith dagger]

Civil Service Pigeon

[ancient lodge]

Btw, as a minor point, be careful with your variables - you're integrating with respect to t, not x

[quick zephyr]

Bro's a texit expert

[ancient lodge]

Yeah, but in the future, try not to jump in when someone else is helping - it's interrupting and makes the situation more complicated if two people are saying the same thing at once

Particularly for the original helper since they're (usually) continuously cut off

[quick zephyr]

I legit didn't see you there lol

[cedar kiln]

@proper plaza Has your question been resolved?

[proper plaza]

I'm not to familiar with Discrete Mathematics. I thought it split there because graph appears to reflect at the point 2 as the derivative flips values there. But it is a piecewise function?

[ancient lodge]

Where did you get it's discrete from?

It's just observing the graph: v is positive when the graph lies above the axis and v is negative when the graph is below the axis

[proper plaza]

So you just need to split the integral when it's under and above the graph. The point at which that happens is where it needs to be split?

[ancient lodge]

Yes, so you can rewrite without the absolute value using this

[proper plaza]

Yes I remember that you need to split an absolute value into two separate functions.

[proper plaza]

Remove the absolute values here? Or do I have to solve for multiple integrals?

[ancient lodge]

There should be 3 integrals

[proper plaza]

Bounds 1 to 1?

Am I missing it?

[ancient lodge]

should be from 0 to 1, 1 to 2, and 2 to 3

since the function changes sign at 1 and 2

[proper plaza]

Does this indicate that a change of slope in the line will mean that a bound must exist there?

1 to 1, positive area, evaluate the integral there.
1 to 2 under the x axis...

I am confused as to why it is split at 2 in the integral.

1 to 3 could be evaluated as a single integral because that portion is all under the x-axis?

[ancient lodge]

this isn't true, also it's the t-axis (since it's a function of t, not x)

1 to 1, positive area, evaluate the integral there.
If both bounds of an integral are the same, the integral is 0

so I'm not quite sure what you're trying to accomplish with this

[ancient lodge]

The slope of the curve is always changing

why are you bring bounds into this

[proper plaza]

I intended to say 0 to 1, my mistake. I guess I'm confused because I'm not sure really what the criteria is for setting those bounds.

[ancient lodge]

Ig from this, you can say that $$|v(t)|=\begin{cases} -2t^2+8t-6, v(t) \leq 0 \ 2t^2-8t+6, v(t)>0 \end{cases}$$

[wraith dagger]

Civil Service Pigeon

[ancient lodge]

does this clarify anything

Ig from this, you can say that $$|v(t)|=\begin{cases} -2t^2+8t-6, t \in [0,1] \cup [1,3] \ 2t^2-8t+6, t \in (1,2) \end{cases}$$

[wraith dagger]

Civil Service Pigeon

[proper plaza]

The advance notation that you're using is something I'm not very familiar with so I'm a little bit confused but I'm trying to follow. I know for certain were talking about whether or not t is in the set of either set you specified. But how that relates to the bounds. I'm not sure. I do see one expressions has the signs switched. I assume that is because we are making it a positive area for elements in that set specifically. But this piecewise function only contains 2 functions so I guess I'm confused. You took v(t) and took it's absolute value which allowed it to become a piecewise function and then this allows you to split it into an integral. How does the third integral fit into this?

[ancient lodge]

But this piecewise function only contains 2 functions so I guess I'm confused.

Would it help if I rewrote it like this

Ig from this, you can say that $$|v(t)|=\begin{cases} -2t^2+8t-6, t \in [0,1] \ 2t^2-8t+6, t \in (1,2) \ -2t^2+8t-6, t \in [1,3] \end{cases}$$

[wraith dagger]

Civil Service Pigeon

[proper plaza]

Yup I'm starting to get this now. It makes sense, because the function dips below the t-axis for a certain amount of time we have to make a case in the piecewise function to account for this gap when we take the integral because we want it to be positive. We have to update the two other functions to account for that as well.

[ancient lodge]

@proper plaza any other questions?

cause I'm gonna go sleep now

[proper plaza]

How exactly did we know that this was going to be a piecewise function? Was it just by taking a look at the graph? We'd have to assume to get the integral's area we need a positive area no matter what, since we know what the graph looks like of the polynomial we can easily say that that the function in the integral itself must be |v(t)| because we want the total area for both areas under the t-axis and above the t-axis. Am I on the right track here?

[ancient lodge]

How exactly did we know that this was going to be a piecewise function?

By definition, absolute value is a piecewise function.

so if you want to rewrite the function without explicitly using absolute value, you need to use the other representation, which is the piecewise representation

[proper plaza]

"We'd have to assume to get the integral's area we need a positive area no matter what" right?

Which is how we knew it was abs and then a piecewise function

[ancient lodge]

Well distance is always non-negative, so we need to turn any negative signed error into positive area

which is why we use absolute value

[ancient lodge]

and then this

[proper plaza]

Okay is it was purely because distance and this relationship isn't true? "We'd have to assume to get the integral's area we need a positive area no matter what"

The 3rd expression in the piecewise function appears to overlap the (1, 2) interval, does it matter if we switch it to (2, 3)?

[cedar kiln]

@proper plaza Has your question been resolved?

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[bright karma]

"Find dy/dx using logarithmic differentiation"

so far ive just ln'd both sides

[rain drift]

ok good so far!

so waht does that look like?

[rain drift]

yep so good. Now you can use some properties of logarithms

namely $ln(e^a) = a$ and $ln(a^b) = bln(a)$

[wraith dagger]

MellowDramaLlama

[bright karma]

oh

so the first part = 2y

hmmm

and im assuming i use that second property on the second part

[rain drift]

yep!

Also, it's not $ln(sin(x))^{ln(x)}$ but rather $ln(sin(x)^{ln(x)})$

[wraith dagger]

MellowDramaLlama

[rain drift]

that should help with the 2nd part

[bright karma]

ln(x)ln(sin)?

[rain drift]

nice work! 🙂

[bright karma]

2y = ln(x)ln(sin)

now divide by 2?

[rain drift]

yep!

So then you'll get $y = \frac{1}{2}[ln(x)ln(sin(x))]$

[wraith dagger]

MellowDramaLlama

[rain drift]

now you'll have to use product rule and chain rule for one part

[bright karma]

ok

wait i see

1 sec

[rain drift]

not quite. you don't have to do another ln

you can just take the derivativie now

$y' = \left(\frac{1}{2}\right) [ (ln(x))'ln(sin(x)) + ln(x)(ln(sin(x))'(sin(x))' ]$

[wraith dagger]

MellowDramaLlama

[bright karma]

oh

am i going down the wrong path

[rain drift]

yeah like I said don't take ln of both sides atfter this. You have y by itself and that was the goal

[bright karma]

ok