help-38

[solar tinsel]

Wait

I’m dumb

81*4

[boreal zodiac]

Ummm, 36^2 is pretty big

[solar tinsel]

Didn’t do that

One sec

[solar tinsel]

Yes let me complete rq and then I’ll get back

[boreal zodiac]

ok

[solar tinsel]

I solved it

[boreal zodiac]

Great!

[solar tinsel]

Forgot to multiply the 81 by 4

Thanks for help

[boreal zodiac]

Sure thing

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[radiant vale]

i need help

[radiant vale]

how do u do this....

<@&286206848099549185>

anyone?

[brazen forge]

you just need to calculate this>

?

[radiant vale]

i need to use an identitiy

im not sure whihc

[brazen forge]

im not sure either then, which indentities you can choose from?

[river willow]

the cubic table identity

[radiant vale]

huh...

[river willow]

I mean, using those is going to be more trouble than just finding the values directly

[radiant vale]

yeha but its for an exam if i dont use the identity i wont get the marks

😭

[wraith hinge]

presumably they want u to use the last one

[river willow]

I didn't even see the last

[brazen forge]

i think its the 2nd to last one

[radiant vale]

okay ill see if i can solve it with that

[radiant vale]

oh

[wraith hinge]

ok

i mean

[brazen forge]

im not sure tbh

[wraith hinge]

its the direct equivalent

[brazen forge]

yeah

[wraith hinge]

,, a^3 + b^3 + c^3= (a+b+c)(a^2+b^2+c^2 -ab-bc-ca) +3abc

[wraith hinge]

just plug in the values u have for a b c

[radiant vale]

this isin't working

[wraith hinge]

well it should

what did u do

[radiant vale]

well

i substituted the values

[wraith hinge]

right so what's the miscalculation

,calc 9^3 +(-3)^3 + (-6)^3

[solid kiln]

Result:

486

[wraith hinge]

so lets see

[radiant vale]

oops

i did it wrong

i got a different number

wait nvm i got it from a friend

i got the answer

[wraith hinge]

lmao alright

[radiant vale]

how do i close

[wraith hinge]

.close

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[upper orchid]

someone help solve for 3rd√2 * 4th√3

[river willow]

what do you mean solve?

[upper orchid]

wait

[zinc moth]

bro is probably in the middle of an exam

💀

[river willow]

take the twelfth power of the original

[upper orchid]

no its a sample paper

[upper orchid]

whaa?

[river willow]

exactly what words mean

[upper orchid]

sorry didnt get it

[zinc moth]

yeah would work

[frail heron]

use calculator

[river willow]

take the original multiply it by itself 12 times

[zinc moth]

take (((2^1/3 * 3^1/4)^12)^1/12)

the 12s cancel out so the equality stands

while the inner 12 distribute to the power removing the roots

so this becomes (2^4*3^3)^1/12

[river willow]

now you have 12 solutions

[zinc moth]

💀

[upper orchid]

thank you sm

do i close it?

[river willow]

.close

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[quasi hinge]

hi

[quasi hinge]

cany anyone help with this

how do i form the integral here

[woven nova]

you can shift things

if we shift everything one unit to the left, we rotate around the x axis

so we can use the shell method

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[old lance]

How can I position a Tikzpicture next to a \subsection{}?

\documentclass[12pt]{extarticle}
\usepackage{pgfplots, tikz}
\usepackage{amsmath}


\begin{document}
\subsection*{Square}
\begin{tikzpicture} % I want this tikz picture to appear to the right of \subsection{Square}
\draw[fill=blue!20] (0,0) rectangle (1,1);
\draw[<->] (1.25,0) -- (1.25,1);
\node at (1.5,0.5) {$L$};
\draw[<->] (0,-0.25) -- (1,-0.25);
\node at (0.5,-0.5) {$L$};
\end{tikzpicture}
\end{document}

[solid kiln]

proofAd
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[rough goblet]

is this a latex issue?

[old lance]

I attached the image to explain a bit better, I wanna move the square next to the text

[rough goblet]

if so we have #bots for that and #latex-help i believe

[old lance]

oh so I should ask in #latex-help ?

mb I didnt see it

[rough goblet]

yea, i think so

[old lance]

alr ty

[rough goblet]

nah is fine

!done

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[old lance]

.close

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[marsh forum]

$\int \frac{dx}{\cos^3\left(x\right)\sqrt{\sin\left(2x\right)}}$

[solid kiln]

Why am. I here

[marsh forum]

any suggestions?

[light karma]

oh

ive done this before

[sick onyx]

u might wanna try rationalizing

[marsh forum]

hmm, OK

[light karma]

rotionalize?

no

open sin2x

[sick onyx]

oh yeha

that works better

[light karma]

then modify the whole equation in terms of sec and tan

[sick onyx]

yes yes that works

[light karma]

this should work

[marsh forum]

ah, or would half angle be a good idea here?

[sick onyx]

wdym half angle

use the double angle formula for sin 2x

[light karma]

2sinxcosx

yes

[sick onyx]

then tranform into tna

[marsh forum]

I meant sin(2x)=2tan(x)/sec^2(x)

[sick onyx]

tan

[sick onyx]

yes

[light karma]

then divide numerator and denominator by cos^4x

[sick onyx]

thats how it is supposed to be

[sick onyx]

let him do it now ig

[light karma]

yeah

mb

[marsh forum]

we don't need to express cos^3(x) as 1-tan^2(x)/(1+tan^2(x))?

[light karma]

nah

[sick onyx]

just place it as 1/sec^3 instead

put the thig from deno to numerator

yeah it gets solved further

[marsh forum]

ok, thanks!

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[old orbit]

I dont understand how to do this

[wraith hinge]

split them

like

2^ab can be written as 2^a * 2^b

then it is easy

[old orbit]

2^x can be written as 2^x

how else are u supposed to

"split them"

[tepid rock]

This is wrong though.

[cobalt cloak]

wouldnt that be 2^(a+b)

[old orbit]

!!

[tepid rock]

4^z = (3^y)^z = ((2^x)^y)^z

= 5

According to the question

[old orbit]

its wrong

I got the answers in front of me it should be D. 10

[cobalt cloak]

theres a +1 in the exponent of the question

so multiply it by 2

[tepid rock]

Obviously I am not giving you answers. I just told you how you should proceed.

This is part of the solution.

[devout drift]

How about taking log both sides

[cobalt cloak]

it would work but time consuming

[old orbit]

i would prefer that tbh

[devout drift]

We don't need the exact values just

The expression

For x,y,z

Right

[cobalt cloak]

no

[devout drift]

Yes

[tepid rock]

It's rather apparent even without logs, but if you want to use that explicitly then that's that.

[cobalt cloak]

you mean for what

oh yeah

i misunderstood u

[devout drift]

For 2^{xyz + 1}

[cobalt cloak]

srry

[devout drift]

That's fine

Well that

Or

$2^x = 3 \ (2^x)^y = 3^y$

[solid kiln]

ColdTee

[devout drift]

And similiarly raise to z

In the next step

[tepid rock]

👀

[devout drift]

Right

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[glass tide]

trying to figure out why the normal equations A^TAx = A^Tb always have a solution

https://math.stackexchange.com/a/2920412/1162831 in this answer they're saying that "the normal equations always have a solution" is equivalent to saying "the column space of A^T is contained in the column space of A^TA"

why is that? i don't see how it's equivalent

[left oriole]

the set of all A^Tb (where b varies over all possible vectors of that size) is precisely the column space of A^T

so if A^TAx = A^Tb has a solution for every b, that means precisely that every vector in the column space of A^T can be represented as A^TAx, and the latter is a vector in the column space of A^TA

[glass tide]

ohhhh i see it now!!! tyvm

.close

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[river willow]

you could write out all possibilities

[jolly isle]

Mayuka can get 0 or 1 tails, for Kai there are 4 possibilities, tail tail, head head, tail head, head tail which are 2, 0, 1, 1 tails, I thinks its 3/4 probobility for that reason but idk chat gpt gave different answer 1/4

[waxen galleon]

i have a general understanding of integrals/anti derivatives but as soon as i stumbled across solving the exercises from an exam paper (that has no solution) has been driving me nuts 😭, so the exercise i’m stuck on is the current one where

[austere cedar]

#❓how-to-get-help

[river willow]

how did you get 3/4?

[jolly isle]

I mean kai has 4 outcomes, 3 of which can satisfy the rule of getting the same amount of tails as mayuka thats why 3/4.

Am I so wrong?

[river willow]

yes

[jolly isle]

I hate probobility

[river willow]

consider this, both people flip one coin, what is the probability they both get the same number of tails

[jolly isle]

1/4?

[river willow]

how did you get that

just write down all possibilities

[jolly isle]

both people has a 1/2 chance to get tail, they are independent senarios so we multiply 1/2 * 1/2 which is 1/4

[river willow]

Mayuka kai1 kai2
000
001
010
011
100
101
110
111

these are the possibilities, let 1 denote a head

all of them have an equal chance of happening, 1/8

what's the probability that they have the same number of tails

[river willow]

they could alternatively both not get tails(total 0 tails). that also has a 1/4 chance of happening and they would have the same number of tails

[river willow]

so it's total of 1/2 in this scenario

[jolly isle]

but here they match in 2 senarios so 1/8 + 1/8 = 2/8 = 1/4 right?

[river willow]

for each scenario, write down the number of tails Mayuka and kai gets

[jolly isle]

a wait its 3/8?

[river willow]

yes

[jolly isle]

I should do these when I am not sleepy I feel like I have brain rot rn so stupid

anyway thank yoj meow

.close

[river willow]

np

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[high plank]

bro i don understnd this question

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[wraith arch]

what part do you not understand

[high plank]

from the so

i got k/3 dy

[wraith arch]

okay so if we integrate that

what do we get?

[high plank]

k/3 .x

[wraith arch]

k/3 y you mean

cuz it’s in dy

and from there we can just plug in y back

[high plank]

so dx =k/3 . y

so

do i sub in k

[wraith arch]

not exactly

from the substitution

we got what dy/dx is

[high plank]

dx = dy/(3x^2e^(x^3))

[wraith arch]

so if i multiply the dx over

that works too

if we plug that into dx

we can see that some terms cancel out

[high plank]

how do i get the e^x^3

[wraith arch]

so what would be our final integral after the substitution

[high plank]

k/3 .y

[wraith arch]

after the integration yes

so what is y

[high plank]

k?

and 0

[wraith arch]

not exactly

[high plank]

😭

[wraith arch]

y is a substitution we used yes

y = e^x^3

from the second line

[high plank]

ohhhh yhhhhhh

[wraith arch]

we just substitute that back in

then we get what they have there

[high plank]

that make alot of sense thank u

[high plank]

then they sub in k

but where odes the -1

come

[wraith arch]

because e^0 is 1

if i substitute in 0 for x

[high plank]

oh okkk thanks

[wraith arch]

ywyw

[high plank]

so do they rearange and u get 2

[wraith arch]

well you just have the set the two final equations equal

and k is 2

[high plank]

and also does this mean that the integreal of the equation is = 2/3(e^8 -1)

[high plank]

oh u just match it up

[wraith arch]

yes as long as k is 2

[wraith arch]

yep

[high plank]

ok thank u

[wraith arch]

we are just trying to find what k is

[high plank]

ty

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[amber hull]

How do i solve this question?

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[wraith hinge]

I'm working on reverse engineering some code for a Roblox game that i'm working on, it involves breaking doors determined by the players power, with each door in the sequence becoming stronger and stronger, although the already existing code has a multiplier for each door power, with only the first doors power actually being listed in code, here's the sequence of power it takes to break each door: 1, 12, 24, 40, 60, 84, 112, 144, 180, 220, 264, 312

[wraith hinge]

if i increase the starting doors power to "10" in order to break it, the sequence is similar: 10, 120, 240, 400, 600, 840, 1.12k, 1.44k, 1.8k, 2.2k, 2.64k, 3.12k

i'd like to find out what each number is consistently multiplied by in order to achieve a sequence like this so that i may edit the values on my own

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@wraith hinge Has your question been resolved?

[shrewd ridge]

that's pointless?

what you said suggests that the values are
n, 12n, 24n, 40n, 60n, 84n, 112n, 144n, 180n, 220n, 264n, 312n
you either mean you want to make it depend on the previous value instead so it becomes
12 2 1.67 1.50 1.40 1.33 1.29 1.25 1.22 1.20 1.18
or you want a formula for the sequence 12 2 1.67 1.50 1.40 1.33 1.29 1.25 1.22 1.20 1.18

either way makes no sense to want it

wait, you're saying the code doesn't have the sequence, only lists the first door

[wraith hinge]

the differences are 11, 12, 16, 20, 24… i see a pattern

[shrewd ridge]

yeah i found it too

2x(a+1)(a)

where x is the first door

but if a is 1, then it's x instead

[wraith hinge]

so? it’s just a piecewise function then

[shrewd ridge]

yes

[wraith hinge]

easy to code

something like function get_door_power(a){if(a = 1){return START_DOOR_POWER;}else{return 2*START_DOOR_POWER*(a+1)*a}}

javascript because i’m on a phone and it can be a single line

probably easy to translate into lua it is pretty simple

where START_DOOR_POWER is x

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[bright fjord]

Q: Given the following information, calculate the degrees of freedom that should be used in pooled standard deviation ‘t’ test.
s12 = 4,  s22 = 9, n1 = 18 and n2 = 24.

On the answer key, it says that the final answer is 18+24 = 42. Is it b/c of df = n1+n2? Also, would the solution be the same if I was asked the same Q but with different numbers? Just confused b/c I’ve seen other formulas where it says df = n-1 or df = n1+n2-2

<@&286206848099549185>

[bright fjord]

<@&286206848099549185>

.close

[warped mantle]

@bright fjord

need help still?

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[oak wasp]

whats the next step

[bold jay]

assuming all that's right, you might want to try to make a nice formula for the series

[oak wasp]

i can see the pattern i just dont know how to put it as a series

[bold jay]

or like...

well ok what's the pattern

[oak wasp]

((-1)^n n * n -1 )/(x+1)^n

something like that

[bold jay]

no i don't think so

[oak wasp]

oh

[bold jay]

well if 'something like that' is carrying a lot of weight then sure lol

just looking at the numerators you have in the derivatives

they go
10
-20
60
-240

what will the next one be?

[oak wasp]

that multiplied by 5

and positive

[bold jay]

yea so the first one is 10*1, second is 10*2, third is 10*3*2, then 10*4*3*2, then 10*5*4*3*2

pattern spotted?

[oak wasp]

oh yes

o so its some kind of factorial?

[bold jay]

yep

[oak wasp]

great

how would i write that

or

how do i put that stuff into a sequence

or i mean maclaurin series

[bold jay]

fix this up a little bit

once you know how to evaluate the derivatives at 0, constructing the series is np

[oak wasp]

((-1)^n 10*n!)/(x+1)^n

is this correct?

[bold jay]

it's easy to check

does it match this when n=4?

[oak wasp]

shucks no

hmmm

((-1)^n 10*n!)/(x+1)^1 + n

problem fixed?

[bold jay]

closer

[oak wasp]

i have no clue

thats the closest i can think of

[bold jay]

the negative signs are off

[oak wasp]

ohh

((-1)^n -10*n!)/(x+1)^1 + n

we start with a negative then

[bold jay]

yea that's fine

[oak wasp]

nice

now what

[bold jay]

(-1)^(n+1) also works

also write (x+1)^(1+n), not (x+1)^1 + n

[oak wasp]

ok got it

is that my answer?

[bold jay]

no

[oak wasp]

dang

[bold jay]

you want to fill these in

with a = 0

[oak wasp]

nice

now am i done?

[bold jay]

bruh

[oak wasp]

LOL

bruh im serious

[bold jay]

also one more little correction, you mean (-1)^n * -10*n!

[oak wasp]

wait a minute i think theres a problem

[bold jay]

the question is asking for a maclaurin series and an interval of convergence. how would the answer just be replacing a with 0 in the above expression

[oak wasp]

if i set it to zero its negative still

or i mean n as 0

[bold jay]

oh yea it works a little differently at 0

[bright fjord]

Hi, I figured it out. Thank you tho! Do lmk if you're free hehe. I have a quick Stats Q

[oak wasp]

shucks

now what

[bold jay]

we'll just treat when n = 0 separately

just say the thing we did holds for n >= 1

[bold jay]

so here

[oak wasp]

so instead n = 0 we do n = 1?

[bold jay]

you want to evaluate f^(n)(0) for each n

[bold jay]

ignore that for now, we just want to evaluate the derivatives at 0

[oak wasp]

okok

[bold jay]

that gives us the coefficients in the series

[oak wasp]

at 0 it just gives us some number/0

wait

im dumb

no

it gives us 1^n

[bold jay]

how

[oak wasp]

or i mean

[bold jay]

this is the "0th derivative" of f

(it's f itself)

[oak wasp]

o so thats 0/1

then the next is 10/1

then -20/1

and so on

[bold jay]

yea the next ones are just (-1)^(n+1)*10*n! for whatever n

[oak wasp]

yea

[bold jay]

i like (-1)^(n+1) more

[bold jay]

so you can fill this in now

[oak wasp]

okay

gim 1 to write

got it

[oak wasp]

so i chaged the sequence to ((-1)^(n+1) * 10n!)/(x+1)^(1+n)

o wait nice

now that gets rid of the negative at 0

[bold jay]

wait what is this

[oak wasp]

uh

the sequence that we came up with

but u said u liked whatever better so i changed it

[bold jay]

what sequence

[oak wasp]

or the pattern

the pattern is what i mean

[bold jay]

that looks like a formula for evaluating the derivatives

[oak wasp]

what LOL

[bold jay]

but we already did that

[oak wasp]

oh so i dont need that anymore?

[bold jay]

the sequence you want is f(0), f^(1)(0), f^(2)(0), f^(3)(0), ...

that's what these are

[oak wasp]

i see

okok

[bold jay]

f(0) = 0, and the rest are given by this

[oak wasp]

okay

got it

i think

whats the next step

[bold jay]

show what you've done so far

[oak wasp]

all i have written down is the pattern and the f(0)s

[bold jay]

not that important but this is... no

[oak wasp]

LOL ok

[bold jay]

this is ok but you don't need to write these out

just fill this in now

f^(n)(0) = (-1)^(n+1)*10*n! for all n >= 1

[oak wasp]

what does for all n >=1 mean

or where should i write that

i know what it means

[bold jay]

i would... under where you wrote 'pattern is' kill the summation

then write "so f^(n)(0) = (-1)^(n+1)*10*n! for all n >= 1"

[oak wasp]

Got it

now what

[bold jay]

what have i told you to do like 3 times already

[oak wasp]

sorry i might have misunderstood but i wrote what you told me to write on my paper

[bold jay]

ok what do you have now

[bold jay]

should say n >= 1

[oak wasp]

corrected

[bold jay]

anyway put something on the right hand side now

[oak wasp]

are you still willing to help me

.close

[trim jolt]

Channel closed

Closed by @oak wasp

Use .reopen if this was a mistake.

[bold jay]

sorry went afk

[oak wasp]

np

i still need help tho, same place where we left off if you dont mind

[trim jolt]

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[oak wasp]

im still here

[trim jolt]

@oak wasp Has your question been resolved?

[wise dagger]

where are you stuck?

[oak wasp]

hold on

i dont know how to continue

i havent found the series yet

[wise dagger]

i see

looks pretty good

first hint: since the given Maclurin series starts with n=0, and your pattern works with n≥1

you can pick the case when n=0 out and write the final sum starting with n=1

e.g. $\sum^{\infty}{n=0}\frac{f^{(n)}(0)}{n!}(x-0)^n=\frac{f^{(0)}(0)}{0!}(x-0)^0+\sum^{\infty}{n=1}\frac{f^{(n)}(0)}{n!}(x-0)^n$

Do you understand this part? @oak wasp

[oak wasp]

kind of

[solid kiln]

Biscuity

[wise dagger]

that is

take the case when n=0 out

and since we know that for this question:

$\frac{f^{(0)}(0)}{0!}(x-0)^0=\frac01(1)=0$

[solid kiln]

Biscuity

[wise dagger]

we just need this part of the series:

$\sum^{\infty}_{n=1}\frac{f^{(n)}(0)}{n!}(x-0)^n$

[solid kiln]

Biscuity

[wise dagger]

now we have starting number is n=1

[oak wasp]

o i kind of see now

[wise dagger]

we can directly use the pattern you've found

$\sum^{\infty}_{n=1}\frac{(-1)^{(n+1)}(10n!)}{n!}(x-0)^n$

[oak wasp]

i think the 10n! shouldnt b exponent no?

[solid kiln]

Biscuity

[wise dagger]

typo

[oak wasp]

okay j making sure lol

[wise dagger]

and then we can cancel out the n!

[oak wasp]

so just that and remove the n!s

[wise dagger]

$\sum^{\infty}_{n=1}(-1)^{(n+1)}(10)x^n$

[solid kiln]

Biscuity

[oak wasp]

okok

[wise dagger]

something like that

[oak wasp]

is that the full thing?

[wise dagger]

that should be the full thing

[oak wasp]

okay awesome

now i need to see the interval of convergence

which i havnet havea clue on

[wise dagger]

how did your book/notes/teacher/prof tell you about how to find the interval of convergence?

[oak wasp]

she literally did not teach us

so i really dont have a clue

we didnt even learn this maclaurin nor taylor series in class LOL

so mind helping me? LOL

[wise dagger]

https://m.youtube.com/watch?v=4L9dSZN5Nvg

[wise dagger]

try this

[oak wasp]

shucks

okay

thanks

wait sorry i have one more question

,rotate

[oak wasp]

how would i find the foci

[wise dagger]

sorry, for this, i don't know

you might wanna close this and start with a new channel since this is another question

[oak wasp]

okok

.close

[trim jolt]

Channel closed

Closed by @oak wasp

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[trim jolt]

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• After 15 minutes, feel free to ping <@&286206848099549185>.
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[hardy hull]

hi can someone help with question 7. I don’t understand how to do it

[zinc ginkgo]

did you draw a picture

[hardy hull]

yeah

[zinc ginkgo]

show it

[hardy hull]

ok

[zinc ginkgo]

do you see a triangle your line makes with the x-axis and y-axis?

your line being the hypotenuse

[hardy hull]

yeah

[zinc ginkgo]

do you know the side lengths of that triangle

[hardy hull]

yep

[zinc ginkgo]

what are they

[hardy hull]

both 3

[zinc ginkgo]

and do you know one angle in the triangle

[hardy hull]

yup

90

im also just confused what it is asking

[zinc ginkgo]

use side-angle-side or recognize this is a special triangle

[wraith hinge]

This is an isosceles triangle

Use its properties

[zinc ginkgo]

label the angle your line makes with the x-axis

[hardy hull]

inside or outside angle

[zinc ginkgo]

"positive direction"

[hardy hull]

so outside?

[zinc ginkgo]

i don't know what you consider outside

just draw and show

[wraith hinge]

where is positive direction?

towards the right, isnt it?

[hardy hull]

yup

ah I see

so the inside one

[wraith hinge]

yeah

[hardy hull]

ah it is 45

[wraith hinge]

Correct

[hardy hull]

alright i will quickly go over C. and if I dont understand I will come back and ask

[wraith hinge]

if you trigonometry, it will be a lot easier

[hardy hull]

yes thats true

(ive completely forgotten about trigonometry)

[wraith hinge]

you can find $tan\theta$

[solid kiln]

EinPest

[wraith hinge]

and hence $\theta$ it self

[solid kiln]

EinPest

[hardy hull]

O is opposite the hypoteneuse right

[wraith hinge]

can u make a diagram

what do you refer by O?

[hardy hull]

i actually dont remember

its been 3 months since ive done trigonometry and ive forgotten everything

[wraith hinge]

tan is perpendicular by base

[hardy hull]

thank u i’m just gonna watch a basic class on trigonometry on youtube

cause i don’t remember how

.close

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[old orbit]

do u study edexcel or caie?

[hardy hull]

what’s that

[old orbit]

oh nevermind then

the questions that you gave

it looks like its from the british curriculum

[hardy hull]

cambridge textbook

probably why

[old orbit]

yeah caie

stands from

cambridge international examinations

[old orbit]

for*

[hardy hull]

yeah probably caie then

[old orbit]

are u giving A levels?

[hardy hull]

whay is that 😭😭😭

lmao

[old orbit]

oh nevermind then

anyways thank you for ur time

[hardy hull]

lol no problem

.close

[old orbit]

.close

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[jagged light]

i cant reply 2 my preivous math help channel, somehow it got timed out

[hexed shard]

ok this is a cool question

the way id approach is is this

we can write p in factored form $p(x)=(x-k)(x-l)$

[solid kiln]

The Great D

[hexed shard]

if we manage to write k and l in terms of a and b

and then solve for (x-k) = (x-l)

you should be in a good spot

makes sense?

[jagged light]

yea

[hexed shard]

you can use the quad formula to write k and l in terms of a and b

[jagged light]

$(x-k)(x-l) = x^2-(k+l)x+kl$

[solid kiln]

Tenick

[jagged light]

so a=-(k+l) and b=kl ?

[hexed shard]

wait actually im confused, my bad

[jagged light]

no prob

i just noticed i mightve misunderstood the problem:

the $b$ in $x^2 + ax + b$ is a constant, not a coefficient, so maybe $b$ can also be negative?

[solid kiln]

Tenick

[hexed shard]

but it says that a and b are positive

[jagged light]

or wait nvm i think constants are also considered coefficient

[trim jolt]

@jagged light Has your question been resolved?

[wraith hinge]

you want to minimize $kl$ given that $(60 - k)(60 - l)$ is a square

[solid kiln]

ampl

[wraith hinge]

when do you think kl is small?

[trim thicket]

0,0

[wraith hinge]

when k and l are positive integers

[trim jolt]

@jagged light Has your question been resolved?

[jagged light]

k = l = 1?

[limpid sun]

has to have two distinct solutions, so k must not equal l

[trim jolt]

@jagged light Has your question been resolved?

[jagged light]

a = 46, b = 4
just tried different values

[proud maple]

Which question?

[jagged light]

this

[proud maple]

I don't think this is possible.

@jagged light

[jagged light]

really but i got the problem from alcumus, it's a website giving random problems (w/ solution afterwards if i got it correct, or after 2 mistakes, or if i give up)

[proud maple]

a^2 is not equal to 4b
try plugging in a^2 = 4b in p(60)

[atomic moat]

b is considered a coefficient because ax^1+bx^0 @jagged light

b is not a constant

[trim jolt]

@jagged light Has your question been resolved?

[patent birch]

@jagged light have you heard of vieta's?

[jagged light]

nah

[patent birch]

so im going to frame a separate problem

let me know if you get lost anywhere

imagine i have a quadratic

[jagged light]

oke

[patent birch]

with roots p and q

[jagged light]

ye

[patent birch]

suppose you know that in this quadratic, the coefficient of x^2 is 1

can you determine the quadratic?

for arbitrary p and q

[jagged light]

whats determine the quadratic mean

[patent birch]

like suppose p and q are 2 and 3

i want the quadratic x^2 - 5x + 6

because the roots are 2 and 3

[jagged light]

oh

[patent birch]

im basically asking you to reverse solve a quadratic equation

[jagged light]

$x^2-(p+q)x+pq$

[solid kiln]

Tenick

[patent birch]

brilliant

so vieta's formulas

basically says that for any quadratic ax^2 + bx + c

if the roots are p and q

p+q = -b/a

pq = c/a

which, if you let a=1, gives the result you just figured out

that is what i would do as a starting point for this problem

[jagged light]

makes sense

[patent birch]

because the condition that the roots need to be integers is very difficult to check if you select a and b

its much easier to select the roots and calculate a and b

its the difference between p+q and pq vs the quadratic formula 😬

[jagged light]

alright ill try that

[earnest granite]

I looked into it , and I think you have to work with restrictions

,help tex it says that p(60) is a perfect square so it can be expressed as k^{2}\ =\ 60^{2}+60x+b so we can say that k\ =\ +-\sqrt{60^{2}+60x+b} so our first restriction is 60^{2}+60a+b\ \ge0 then we got that the trinomyal x^{2}+ax+b has to two roots thus its discriminant has to be positive so we get a^{2}-4b>0 which is a system of inequalities then we find domain for b and we get the minimum value that satisfies the conditions.

,help

[solid kiln]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

Command tex it says that p(60) is a perfect square so it can be expressed as k^{2}\ =\ 60^{2}+60x+b so we can say that k\ =\ +-\sqrt{60^{2}+60x+b} so our first restriction is 60^{2}+60a+b\ \ge0 then we got that the trinomyal x^{2}+ax+b has to two roots thus its discriminant has to be positive so we get a^{2}-4b>0 which is a system of inequalities then we find domain for b and we get the minimum value that satisfies the conditions. not found!
Use the ,list command without arguments to see a list of commands.

Feb 16, 2024, 6:00 PM

[earnest granite]

,list

[solid kiln]

My commands!

Use ,ls to obtain a briefer listing, and use ,help <cmd>to view detailed help for a particular command, or ,help to view general help.

If you still have questions, talk to our friendly support team here.

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Selfridge
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[jagged light]

it's kind of hard 2 understand

so far i'm thinking $p, q < 0$

[solid kiln]

Tenick

Tenick

[jagged light]

because for some distinct roots $p$ and $q$,
$x^2 -(p+q)x + pq$ where $p + q = -b$ and $pq = c$

[solid kiln]

Tenick

[jagged light]

so to satisfy the condition that $x^2+ax+b$ have coefficients $a, b > 0$, $p$ and $q$ must be negative

[solid kiln]

Tenick

[trim jolt]

@jagged light Has your question been resolved?

[trim jolt]

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[graceful zenith]

need help with this

[graceful zenith]

been struggling with this for while now

[scarlet escarp]

what have you tried

[graceful zenith]

everything

the thing is idk much abt vector addition

[scarlet escarp]

i would suggest you look at a khan academy video

[graceful zenith]

alr

[graceful zenith]

but would u find PQ?

[scarlet escarp]

what are you asking

[graceful zenith]

would the distance pq be the answer for a) ?

[scarlet escarp]

the answer is yes, but i dont think you're thinking of the right distance

you want the length of the vector P + Q

so you first need to find what P + Q is

[graceful zenith]

ohk

thx i'll look into a vid

.close

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[patent gale]

How do I use the rational root theorem?

[viscid flower]

rational root theorem gives you a form of every potential rational root of the function

[split chasm]

you already have it in the first part of the question

[viscid flower]

oh

neither of us read the question

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[shadow cipher]

Is my part a correct and also how do i do part B without calculus little confused about that.

[bright quarry]

part a is correct you should say -x(sqrt(1-x^2))=-f(x)

[orchid wagon]

in other words, find the interval where f(x) is defined

[bright quarry]

this would be the domain

[shadow cipher]

They’re asking for the range tho

[bright quarry]

don’t believe you can find the max/min values without calc

what course is this

[shadow cipher]

1-x^2 >= 0, 1>=x^2, x<=+_1

[shadow cipher]

I guess we can use calculus also but my class has covered only derivatives in calc

[bright quarry]

abs(x)<1

or equal to

[shadow cipher]

The teacher said we can do it using functions also

[bright quarry]

that’s all u need

[shadow cipher]

Okay what’s that method?

[bright quarry]

notice the slope at the max/min values

what is that slope

at the peaks

[shadow cipher]

0

[bright quarry]

mhm

[orchid wagon]

misread, my bad

[bright quarry]

and what function models the slope

[bright quarry]

no worries

[shadow cipher]

The derivative of the function at a point gives the slope

[bright quarry]

do u know how to find the derivative function

youll need product rule and power rule

[shadow cipher]

Yeah I’ll just do that I’ll lyk the answer I’m getting

[bright quarry]

or only power rule

if ur good with algebra tricks

but i’d use product rule

[shadow cipher]

It’s x*(1-x^2)^1/2

[bright quarry]

mhm

that’s the function

what’s the derivative

[shadow cipher]

I applied power rule

[bright quarry]

u need product rule

[shadow cipher]

Do we have to apply chain rule also so we get -2x multiplied

[bright quarry]

and chain rule

do u know the product rule

[shadow cipher]

Yeah u’v+v’u

[bright quarry]

mhm

this is that scenario

we have two functions of x

have to use product rule

[shadow cipher]

Ok one sec just working thru that will send a photo

This is what I’ve learnt in product rule

[bright quarry]

in the future use parentheses for multiplication the cross looks like an x

and it gets confusing

[shadow cipher]

Oh okay will do

[bright quarry]

but yes

this is right

now we have the function that models the slope

we want the points where the slope =0

so what do we do

[shadow cipher]

We need to put in the x coordinate into the function that models the slope to find the slope

[bright quarry]

we set the derivative =0

because the slope is zero at those points

we can find the x coordinate where the slope=0

[shadow cipher]

Oh okay yeah

Let me just try the at

[bright quarry]

then use these x coordinates to find the y value

[shadow cipher]

That

[bright quarry]

by evaluating the function at that x coordinate

lmk what u got

i got

plus or minus

1/…

[shadow cipher]

How do we solve for x here?

This doesent look like a quadratic

Like the powers are wierd

[bright quarry]

rewrite the negative exponent

as a fraction

then get a common denominator

combine the two terms

then set the numerator =0

[shadow cipher]

Should i rewrite (1-x^2)^1/2 also

[bright quarry]

because for a fraction to equal zero

[shadow cipher]

In the form of sqr rt

[bright quarry]

the numerator must be zero

[bright quarry]

^-1/2=1/()^1/2

i didn’t want to write it out but yea

reciprocal

should be plus or minus 1/sqrt3

1-3x^2=0

which makes sense

because it’s less than 1

[bright quarry]

oh wait

i think it’s sqrt2

yea

sqrt2

i forgot to divide the 2 from the -2x^2 and the 1/2 on the bottom

so get a common denominator

by multiplying by the denominator on the right

it’ll work out nice

because sqrt(x)(sqrt(x))=x

[shadow cipher]

Is it the last step fine?

[bright quarry]

add the two terms

get a common denominator

the common denominator is sqrt(1-x^2)

so essentially multiply everything by that

to get rid of denominator

[shadow cipher]

Common denominator is 2*sqrt (1-x^2)

[bright quarry]

well the 2 cancels

-2x^2/2=-x^2

[shadow cipher]

,rotate

[bright quarry]

hmm

try multiplying both sides by sqrt(1-x^2)

since the right side is zero

u get rid of the denominator entirely

[shadow cipher]

,rotate

[shadow cipher]

Works?

[bright quarry]

no

the -2x^2 is attached to the right fraction

u cant add first

[bright quarry]

do this

[shadow cipher]

Multiplied both sides by root 1-x^2

[bright quarry]

no like from the beginning

before combining

[shadow cipher]

Oh okay so the first line

[bright quarry]

yes

[shadow cipher]

Even the (-2x^2) gets multiplied?

[bright quarry]

should get 1-2x^2=0

[bright quarry]

well yea

it’s the numerator

the right fraction gets multiplied

but

look

when u multiply the whole fraction by that

the sqrt cancels

because it’s in the denominator originally for that fraction

[shadow cipher]

Yeah

[bright quarry]

so the right fraction simplifies to -x^2

correct

does this make sense

[shadow cipher]

Yes

[bright quarry]

ok

then what’s the left term

[shadow cipher]

1-2x^2=-x^2

[bright quarry]

no

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

and the right fraction simplified to -x^2

so 1-x^2-x^2=1-2x^2=0

x=1/sqrt2

plus or minus

[shadow cipher]

Got it

We got the value of x now

That’s the point at which it’s the maximum

[bright quarry]

+1/sqrt2

yields the maximum

-1/sqrt2 yields the minimum

now just do f(1/sqrt2)

[shadow cipher]

a=-1/sqrt 2 <= y <= 1/sqrt2

[bright quarry]

and the first part asked if it was an odd function

so u know f(-1/sqrt2) is just the negation of what i get for

[bright quarry]

^

this

[shadow cipher]

1/sqrt 2 * under root 1-(1/sqrt 2) ^2

[bright quarry]

should be 1/2

so -1/2 and 1/2

[shadow cipher]

Yeah simplifies to that

1-1/2>=0

[bright quarry]

what

range is [-1/2,1/2]

[shadow cipher]

This simpliedies to 1/sqrt 2 * under root 1-1/2

[bright quarry]

yea

[shadow cipher]

Yes got that

[bright quarry]

then times 1/sqrt2 is 1/2

[shadow cipher]

So this is the and

Answer

[bright quarry]

no

what’s a

and what’s y

[shadow cipher]

Question was asking to find val a and b

A and B can also be called the minimum and maximum

And we found that

[bright quarry]

oh

[bright quarry]

well a and b are these ^

a=-1/2

b=1/2

remember it’s the y values

[shadow cipher]

Yeah got it

[bright quarry]

the x values were only used to find the y values

[shadow cipher]

Yep got it

Thanks for the help

[bright quarry]

ur welcome

[shadow cipher]

.close

[trim jolt]

Channel closed

Closed by @shadow cipher

Use .reopen if this was a mistake.

[trim jolt]

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[full dock]

$\frac{d}{dx}\left(e^{\cot\left(2x\right)}\right),\ \left(0<x<\frac{\pi}{2}\right)$, so i know how to differentiate this but what does the $\left(0<x<\frac{\pi}{2}\right)$ mean? Im guessing its a domain restriction or something but im not sure

[solid kiln]

water beam

[whole coral]

Yea domain restriction so you're not dividing by zero

[full dock]

so i dont have to worry about how i take the derivative?

[whole coral]

Nope you can just do it "normally"

[full dock]

okay

charrrr

i have another partial fractions thing to ask

what is that emoji

[whole coral]

[full dock]

$\int_{ }^{ }\frac{5y^{3}+y-3}{y^{4}+y^{2}}dy$

[solid kiln]

water beam

[full dock]

is the numerator going to have a lot of constants

[whole coral]

A few

It's not the worst of them all

[full dock]

not as bad as last time i hope?

[whole coral]

I don't think so

[full dock]

$\int_{ }^{ }\frac{5y^{3}+y-3}{y^{4}+y^{2}}dy=\int_{ }^{ }\frac{5y^{3}+y-3}{y^{2}\left(y^{2}+1\right)}dy$

[solid kiln]

water beam

[full dock]

so the y^2 will be a repeated factor i think

im not sure how to know when i should have an x attached to the constant

like for example

Ax + Bx, etc

in what situations do i add the x?

[whole coral]

Like consider the y^2 + 1, that one you can't factor

So then it'll become (Ax + B)/(y^2 + 1)

The y^2 is a repeated factor so you could either decompose it into A/y + B/y^2
[or notice that when you common denominator the above, that it's (Ay + B)/y^2]

[full dock]

hm okay

[full dock]

how do we know to have the B? why not just Ax on the numerator

[whole coral]

Similar reasoning to last time, if you try it out, in some cases it gets you "problems"

[full dock]

hmm ookay

[whole coral]

(at least this time, I'll save you from doing it all out again, you're safe... for now )

[full dock]

$\int_{ }^{ }\frac{A}{y}+\frac{B}{y^{2}}+\frac{Cx+D}{y^{2}+1}dy$

[solid kiln]

water beam

[full dock]

the last time i saw 4 constants was during that one problem

[whole coral]

Also this one looks more annoying, all things considered

[full dock]

nooooooooooooooooooooooooo

why would u say that