help-42

[hybrid flint]

No

[bright lagoon]

your mom asked you to do this and you're not taking calculus?

[hybrid flint]

Yea

Ok give me a sec

[bright lagoon]

why

[hybrid flint]

I’ll take a pic of the msg

[bright lagoon]

differentiate x(t) twice and differentiate y(t) twice. no, I am mistaken

[hybrid flint]

Why is that?

[eternal wagon]

It's the chain rule

[bright lagoon]

I'm not seeing a more nuanced answer than this so far:

[potent lotus]

Melvin Eugene Punymier

[bright lagoon]

Disregard this ^

[quaint sphinx]

use such property:

[bright lagoon]

that's what I was looking for

[hybrid flint]

Ok wait let me read through everything

[quaint sphinx]

bro

do u know that is wrong

[bright lagoon]

Yes, I see I was mistaken

[hybrid flint]

Why was that wrong

[quaint sphinx]

bcz see this

[bright lagoon]

$\frac{d^2 y}{dx^2} = {\frac{d}{dx} \left(\frac{dy}{dx}\right)} = \frac{d}{dt}\frac{dt}{dx}\left(\frac{dy}{dx}\right) = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}$

[potent lotus]

Melvin Eugene Punymier

[bright lagoon]

there we go.

that's the "simple" part I horribly misremembered

[hybrid flint]

Bruh I thought dx^2 means d(x^2) 💀

[bright lagoon]

no, the notation means different things

I haven't lost that yet 👀

[hybrid flint]

what's the difference between dx^2 and d^2x?

[bright lagoon]

I can't speak authoritatively on that

or rather, get the precise meaning

in context, d^2 y/dx^2 is the second derivative of y with respect to x

oh, lol

just look at the first two expressions

first derivative is just dy/dx

The derivative operator is d/dx

so we do d/dx (dy/dx)

and that's the wishy-washy part in my early calculus schooling

[hybrid flint]

ok I kinda get that

[bright lagoon]

a professor told me "we can basically multiply these but not really"

but that is effectively what happens!

we have two factors of d in the top, so d^2

[hybrid flint]

so is d^2y/dx^2 = (d/dx)(dy/dx)

so the second derivative?

[bright lagoon]

and two factors of (dx) in the bottom, notated dx^2

[bright lagoon]

yes

[hybrid flint]

What does x(t) mean 💀

[bright lagoon]

"x is a function of t"

[hybrid flint]

What is the last two parts of the equalities for then

[bright lagoon]

the penultimate expression shows that d/dx can be re-expressed as d/dt * dt/dx. The 'dx' parts cancel and we only have dt on the bottom

this is Chain Rule

the last part changes the way the penultimate expression is written. Instead of putting dt/dx on the top, we divide by dx/dt, which is equivalent.

this comes from what we already know about dividing or mulitplying by fractions

"dividing by a fraction is the same as multiplying by the recipricol of that fraction"

[hybrid flint]

Isn’t there a dt/dx term tho

[bright lagoon]

(in this case, we are doing the reverse direction of inference)

[hybrid flint]

Yea I got that

[hybrid flint]

^

[bright lagoon]

? All the terms are there

Ok, did I confuse you by making it sound like we actually cancel something? We don't cancel. We need those terms there to get it in the form we want to end up with.

Effectively, it's the same as the factor we started with , d/dt

d/dx = (d/dt)(dt/dx)

the dx parts WOULD cancel

these expressions are equivalent

[hybrid flint]

Isn’t d/dx*dx/dt=d/dt and not d/dx?

[bright lagoon]

yep

oops

one sec

fixed

[hybrid flint]

Ok so I get the steps but why would you do them?

[bright lagoon]

I just fixed it up there too ^

[bright lagoon]

to get dx/dt on the bottom

[bright lagoon]

this part was always written correctly

[hybrid flint]

Yea I followed it

[hybrid flint]

What do you do with dx/dt?

[bright lagoon]

$\frac{dt}{dx} = \frac{1}{\frac{dx}{dt}}$

[potent lotus]

Melvin Eugene Punymier

[hybrid flint]

I get that it’s in the denominator but what do you do with it

[bright lagoon]

we are able differentiate your x(t) with respect to t, so we can use this form to get the second derivative of y(t)/x(t)

[hybrid flint]

What does respect to t mean?

[bright lagoon]

have you ever differentiated an expression before?

because you don't seem familiar with function notation that's a distant precursor to calculus

[hybrid flint]

Maybe not with proper notation

[bright lagoon]

t is the variable we are doing calculus with. we are taking the limit of the function as changes in t become infinitesimally small.

although we are just leaning on some rules we memorized at this point, not actually taking limits

(though we could)

[near lily]

is this chain rule

[hybrid flint]

Ok I get that

[near lily]

do you use f'(x) in functions

as derivative notation

[bright lagoon]

that is a derivative notation

I have been useing Leibniz notation

let's solve your problem

[hybrid flint]

I’ve never actually done any calc questions, I’ve just seen people do it and I’m like ok the derivative of this is this

[hybrid flint]

Ok yea

[hybrid flint]

So what’s the step after this

Oh wait

So is dx/dt the derivative of x(t)

@bright lagoon

[bright lagoon]

yes

here I am just using chain rule

you can see in the last step that we just need the second derivatve of y with respect to t
and the the square of the first derivative of x with respect to t

or, in other notation:

$\frac{y''(t)}{(x'(t))^2}$

[potent lotus]

Melvin Eugene Punymier

[bright lagoon]

...which we can easily do

[hybrid flint]

Ok I got it thanks!

Why is d^2y/dt^2 the second derivative of y(t)

Nvm

It’s d/dt *dy/dt

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[restive dove]

little confused here

[dull wagon]

which part

[restive dove]

All of it

[dull wagon]

first find the derivative, applying the relevant rules

[restive dove]

Wouldnt it just be (43+31)’=0?

[dull wagon]

don't worry about the x=0 or the table

[restive dove]

Oh

[dull wagon]

differentiate first then plug

otherwise you're essentially implying that the derivative evaluated anywhere for any function is 0

[restive dove]

wait so what am i differentiating

the 4f(x) + 3g(x)?

[dull wagon]

yes

[restive dove]

im

dumb or smth

whats f(x) here

[dull wagon]

f(x) is f(x)

what's the expression for its derivative

[restive dove]

to me i see (4(3) + 3(1))'

is this product rule stuff?

got it

just had to use my whiteboard lol

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[shrewd arch]

How do you eval this limit?

[calm coral]

@shrewd arch Has your question been resolved?

[calm coral]

@shrewd arch Has your question been resolved?

[shrewd arch]

<@&286206848099549185>

[regal geyser]

(?)

hm interesting limit ah

I mean I would guess it goes to 0

as a guess

lol

it goes to 1

I am not sure how you would evaluate this

I think there might be a way

hm

oh nvm I tought you could solve it with polylogarithms but nvm

[calm coral]

@shrewd arch Has your question been resolved?

[kindred drum]

Riemann sums

I think you take it in university

[calm coral]

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[shrewd arch]

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[surreal wedge]

use riemann the limit equals to : integral 0 to 1 (x*e^x)dx

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[hallow linden]

I think it might be 42 cuz this is the only approach from which i could get a possibly definite answer

as since it's 8 so it's 16, then 7 so it should be 14 and 6 so 12 and so 16 + 14 + 12 = 42

sry i'm dumb

[dull orchid]

Sounds like sound logic to me :), do you know why you can do that though?

[hallow linden]

no

that's the reason i opened this

[dull orchid]

Cool, have you learnt about similar triangles?

[hallow linden]

yeah. a bit

[dull orchid]

Right, so you know how if the triangles have the same angles and triangle "A" has a side of size "8" and triangle "B" has a side of length "4", then all the rest of the sides of "B" would be half of the sides of "A"?

[hallow linden]

yea

[dull orchid]

Well, this is just the same thing, but the triangles are inside each other

[hallow linden]

https://tenor.com/view/loading-thinking-shiba-inu-gif-25291442

Ohhhh ye

I got it

[dull orchid]

So if the length of AD is 8 and AE is 16 then AE is twice AD, so AC would be twice AD + DF + FB :)

Because ADE and ABC are similar

[hallow linden]

ok so basically, I'll just consider ABC an enlarged version of ADE, and calculate the the values accordingly

[hallow linden]

yea I got it

tysm

[dull orchid]

Np :)

[hallow linden]

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[lucid junco]

2018^2018/20

[lucid junco]

Whats the remainder

[main musk]

ok first simplify 2018, whats 2018 (mod 20)?

[lucid junco]

What's a mod

[main musk]

wait are you not doing number theory?

[calm coral]

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[rigid iris]

Can someone correct my answers please

[rigid iris]

i. 11/18/2020
ii.83.33
iii. I didn't know how to get it

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[calm coral]

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[blazing coyote]

$\arctan\left(\frac{2x}{1-x^2}\right)+\operatorname{arccot}\left(\frac{\left(1-x^2\right)}{2x}\right)=\frac{\pi}{3}-1$

[potent lotus]

ƒ(Why am. I here)=I don't Know

[blazing coyote]

If the sum of the solutions of this equation is

$\alpha - \frac{4}{\sqrt3}$

[potent lotus]

ƒ(Why am. I here)=I don't Know

[blazing coyote]

find $\alpha$

[potent lotus]

ƒ(Why am. I here)=I don't Know

[blazing coyote]

so to start

[craggy holly]

hmm

[blazing coyote]

$2\arctan\left(\frac{2x}{1-x^2}\right)=\frac{\pi}{3}-1$

[potent lotus]

ƒ(Why am. I here)=I don't Know

[blazing coyote]

$\arctan\left(\frac{2x}{1-x^2}\right)=\frac{\pi}{6}-\frac{1}{2}$

[potent lotus]

ƒ(Why am. I here)=I don't Know

[blazing coyote]

now let $x=tan(u)$

[formal tulip]

Just make sure to note the domain when you invert it

[potent lotus]

ƒ(Why am. I here)=I don't Know

[blazing coyote]

so If I do this, I only solve in $(0, \frac{\pi}{2})$

right?

[formal tulip]

yeah

x > 0

[potent lotus]

ƒ(Why am. I here)=I don't Know

[formal tulip]

or whatever the argument is

[blazing coyote]

so $2u= \frac{\pi}{6}-\frac{1}{2}$?

[potent lotus]

ƒ(Why am. I here)=I don't Know

[craggy holly]

Yes

[blazing coyote]

hmm

$u=\frac{\pi}{12}- \frac{1}{4}$

[potent lotus]

ƒ(Why am. I here)=I don't Know

[blazing coyote]

so this is one of the solutions

how do I find the other solutions now?

[formal tulip]

when you inverted the arccot

theres the other case

[blazing coyote]

I got rid of some solutions, I know

but

uh

[blazing coyote]

so here,what do I do exactly

x=tan(u)?

[formal tulip]

$\arccot \frac{1}{x} = \pi + \arctan x$ for x < 0 I believe

[potent lotus]

jewels!

[blazing coyote]

right

[formal tulip]

You should do that where you made the case to invert it

[blazing coyote]

so $2\operatorname{arctan\left(\frac{2x}{1-x^2}\right)+\pi=\frac{\pi}{3}+1$

?

[remote mural]

this has got to be some form of art dude

[formal tulip]

arctan

also what is that operatorname stuff

just do \arccot lol

[craggy holly]

Yep

[potent lotus]

ƒ(Why am. I here)=I don't Know
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[blazing coyote]

give me a minute

[formal tulip]

$2\arctan \left ( \frac{2x}{1 - x^2} \right ) + \pi = \frac{\pi}{3} + 1$

[potent lotus]

jewels!

[blazing coyote]

thanks!

[formal tulip]

remember now that 2x/(1 - x^2) < 0

[blazing coyote]

yes

$2\arctan\left(\frac{2x}{1-x^2}\right)=1-\frac{2\pi}{3}$

[potent lotus]

ƒ(Why am. I here)=I don't Know

[blazing coyote]

so $\arctan\left(\frac{2x}{1-x^2}\right)=\frac{1}{2}-\frac{\pi}{3}$

[potent lotus]

ƒ(Why am. I here)=I don't Know

[blazing coyote]

hmm

let $x=tan(t)$

[potent lotus]

ƒ(Why am. I here)=I don't Know

[blazing coyote]

wdym

[formal tulip]

you should have done that with the original one as well

would've helped massively 😛

[blazing coyote]

didn't I

lol

[formal tulip]

I meant you just left it at this

[blazing coyote]

ah

so $2t=\frac{1}{2}-\frac{\pi}{3}$

[potent lotus]

ƒ(Why am. I here)=I don't Know

[formal tulip]

slow down

uh

so when you defined $x = \tan t$, with our domain consideration $\arctan \tan 2t \neq 2t$

[potent lotus]

jewels!

[blazing coyote]

oo

right

[formal tulip]

since $t \in \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right )$

[potent lotus]

jewels!

[formal tulip]

it would open up as pi - 2t yes?

[blazing coyote]

yeah

so ?

$2t-\pi=\frac{1}{4}-\frac{\pi}{6}$

[formal tulip]

uh no

tan (pi - 2t) = -tan 2t

so it would open as 2t - pi

[blazing coyote]

ah

ok

[potent lotus]

ƒ(Why am. I here)=I don't Know

[blazing coyote]

so

[formal tulip]

this is the positive case or the negative case

[blazing coyote]

negative, right?

[formal tulip]

I think pi - 2t isnt right either wait we need to figure this out properly

[remote mural]

arc means inverse right ?

[blazing coyote]

yes

[formal tulip]

alright

so 2t lies in the interval (-pi, 0) in that case

pi + 2t shifts it to (0, pi) but that still isn't in the invertible range for us

this is why I hate inverse trig

[blazing coyote]

sorry

[formal tulip]

I think what you should do is

Break the interval here

And use phase shifts then

[blazing coyote]

wdym by phase shifts

[formal tulip]

shifting the argumnet to get it in range?

[blazing coyote]

ok

[formal tulip]

the stuff we've been doing so far

• pi, -pi etc

[craggy holly]

where did you even get the question tho

[blazing coyote]

a past JEE paper

[blazing coyote]

what kind of intervals ?

[formal tulip]

probably (-pi, -pi/2) and (-pi/2, pi)

[blazing coyote]

hmm

I'm really confused

[blazing coyote]

staring from here, again

[formal tulip]

im sorry for switching up gears again but I think I know what's best

since the invertible range of arccot is (0, pi) it would be better to do it with arccot in mind

so just converting the arctan to arccot and substituting tan 2t will be enough

[blazing coyote]

hmm

ok

[blazing coyote]

so

$\left(2\operatorname{arccot}\left(\frac{\left(1-x^2\right)}{2x}\right)\right)=\frac{\pi}{3}-1$

[potent lotus]

ƒ(Why am. I here)=I don't Know

[blazing coyote]

$\left(\operatorname{arccot}\left(\frac{\left(1-x^2\right)}{2x}\right)\right)=\frac{\pi}{6}-\frac{1}{2}$

[potent lotus]

ƒ(Why am. I here)=I don't Know

[blazing coyote]

now what

[formal tulip]

x = tan u

why do you have parenthesis around the arccot

also is this the negative or positive case

[blazing coyote]

Negative

Right

So 2t=π/6 -1/2

But this isn't close to the form of the og question

[formal tulip]

wheres the -pi then

[blazing coyote]

Uh just a minute

-π +2x=π/6-1/2?

[formal tulip]

uh

$-\pi + 4x = \frac{\pi}{3} - 1$

[blazing coyote]

Why 4x?

[potent lotus]

jewels!

[formal tulip]

2arccot(cot(2t))

[blazing coyote]

Got it

[formal tulip]

i'll take your leave now sorry

this has taken much more time than i initially thought it would

[blazing coyote]

Thanks a lot for the help

[formal tulip]

😅

just work on this you're probably close

[blazing coyote]

I will

Thanks

[calm coral]

@blazing coyote Has your question been resolved?

[blazing coyote]

so $4x=7 \pi/3 -1$?

[potent lotus]

ƒ(Why am. I here)=I don't Know

[blazing coyote]

this feels very sus

<@&286206848099549185>

#help-42 message OG problem

nvm, I'll try this tomorrow

.clos

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[dense falcon]

The question is sus too

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[copper notch]

hi

[copper notch]

how can i decompose this matrix into sum of symetric and antisymetric matrix

$\begin{pmatrix}1&0&0\1&0&-2\0&0&1\end{pmatrix}$

[glass heart]

lets call the matrix A

[potent lotus]

Slowaq

[glass heart]

we want A=B+C where B is symmetric and C is antisymmetric

[copper notch]

yes

[glass heart]

what happens if you transpose both sides

[copper notch]

wdym by both sides?

[glass heart]

A=B+C

[copper notch]

ah

i dont remember buuuuut A^T=B^T+C^T?

[glass heart]

yes

can you simplify the right side?

[copper notch]

(B+C)^T?

[glass heart]

no that would have been a step actually before obtaining A^T=B^T+C^T

[copper notch]

ah true

but sadly then i dont know

[glass heart]

you know that B is symmetric and C is antisymmetric

[copper notch]

ah

B^T+C^T=B-C?

[glass heart]

yes

so we know A=B+C and A^T=B-C

[copper notch]

yes that should help

give me a second

$\begin{pmatrix}1&0&0\1&0&-2\0&0&1\end{pmatrix}=\begin{pmatrix}1&0.5&0\0.5&0&-1\0&-1&1\end{pmatrix}+\begin{pmatrix}0&-0.5&0\0.5&0&-1\0&1&0\end{pmatrix}$

[potent lotus]

Slowaq

[copper notch]

@glass heart thx for help

.close

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[glass heart]

yw

[calm coral]

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[buoyant mirage]

someone help

[buoyant mirage]

with b

p is 12.18

12,18

i cant get visual of the triangle

also they use trapezium formula in ms should i jump

.close

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[split meadow]

Can someone help me with this

[blazing coyote]

is this. a test?

[fast magnet]

are these the eassesments

they were fr annoying

just remember the probabilities always add to 1

[split meadow]

This is a past paper

I've solved questions like this

But for this one they give almost no information

[fast magnet]

so for the not shortest its 3/5 because 2/5+3/5 is 5/5

there is no other way other than these two so the top parts need to add together to get the bottom part

[calm coral]

@split meadow Has your question been resolved?

[split meadow]

So to get not lost
I need to do 2/5 + 7/10?

[calm coral]

@split meadow Has your question been resolved?

[calm coral]

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[keen yarrow]

heyy so I have this fraction $\frac{(\frac{y^2-x^2}{y})}{y^2}$ and I want to simplfy it

[potent lotus]

Specific Joe

[keen yarrow]

$\frac{(\frac{y^2-x^2}{y})}{y^2}$

[potent lotus]

Specific Joe

[blazing coyote]

well, what have you tried?

[keen yarrow]

I tried multiplying the whole fraction by $y^2$ to cancel out the $y^2$ at the denominator $(\frac{y^2-x^2}{y})*y^2$

[potent lotus]

Specific Joe

[small shuttle]

to be clear you have $\frac{(y^2 - x^2)(\frac 1 y)}{y^2}$

[potent lotus]

b

[keen yarrow]

yes

[small shuttle]

do you know $\frac 1 {y^2} \cdot \frac 1 y$?

[potent lotus]

b

[keen yarrow]

no ?

1/y^3 ?

[small shuttle]

that's it yeah

[keen yarrow]

wait

so its gonna be $\frac{y^2-x^2}{y^3}$ ?

[potent lotus]

Specific Joe

[small shuttle]

👍

[keen yarrow]

I see

[small shuttle]

at which point you can split that into two fractions

[keen yarrow]

yeah that's a good idea

okay I think I got it

thanks

.close

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[remote mural]

Geometry Question (formulas are provided aswell as calculator- this is district assesment prep)

[remote mural]

heres what ive done so far

ive found the volume of the fertilizer required, which is 576

ive set up the equation of 576= (volume of cylinder)+(volume of cone)

im thinking of substituting 8 and 14 in but im not sure because i dont think im supposed to do that

Would love some help on this, and suggestions if i come across problems like this

[quiet jay]

I think the equation should be set up so that h_1 +h_2 = 14

and r= 4

[remote mural]

so how would i do that

[quiet jay]

If you expand the equation you have for volume

you will have a h_1 and h_2 naturally

then use substitution

h_1 = 14-h_2

that will get you h_2

then sub that back in and you will have h_1

[remote mural]

OH

wait

[quiet jay]

I havent done any calcs on it

but i think that apporach should be solid

[remote mural]

hm wait

so i sub r=4 into both the cylinder and cone volumes right?

[quiet jay]

ya

[remote mural]

w

so 16πh= 14- 1/3(16πh)

[quiet jay]

this for h1 or h2

[remote mural]

h1

solving for height of cylinder

[quiet jay]

okayy

[remote mural]

ya

[quiet jay]

I do think this question is a bit bad tho

[remote mural]

fr my curriculm sucks

we never went over 3d shapes yet we doing it

[quiet jay]

because the height doesnt need to be 14

its only max

[remote mural]

we never went over it in algebra either

yea thats weird

[quiet jay]

so u could say the cone hieght is 0

and get maximal gains with the cylinder

[remote mural]

thats the problem i had

the boundaries arent specific

[quiet jay]

bad question in my op

[remote mural]

not even my tutor could help me 😭

this is my last resort

[quiet jay]

thats kinda wild

i wanna get into tutoring

at that point in maths where i should learn how to teach

joined this today for that

[remote mural]

ohhh i didnt see u joined rn

😮

also im at the point where 16πh-14=-1/3(16πh)

[quiet jay]

rearrange for h at this point

it will be squared

take the pos value

negative heights are booky

[remote mural]

yea

okay so 16πh+1/3(16πh)=14 i rearranged how i was solving

[quiet jay]

ahh its a quadratic equation

times by the whole thing by 16pih

then do you know the quadratic formula

[remote mural]

yea

if i wasnt i wouldnt be here xd

[quiet jay]

true true

still gona ask i was trying to help a guy with a triangle problem and i assumed he knew what cos and sin was

i was wrong

[remote mural]

LOL

NO WAY

[quiet jay]

yeah kinda his courses fault tho

[remote mural]

oh damn

im learning this in 8th rn

hehe

14 and doing WHATEVER THE HECK THIS IS

[quiet jay]

i dont do america i have no idea what that means

[remote mural]

8th grade

[quiet jay]

ahhh

[remote mural]

which is between 13-14 years

old

but since im a grade ahead

[quiet jay]

yehh that would be 2nd year high school for me

[remote mural]

what the

farkas

[quiet jay]

scotland

[remote mural]

that would be 1st year highschool for me

anyway

256π²h²+ 1/3(256π²h²)=224πh

[quiet jay]

ohh wait im silly

[remote mural]

x=h

[quiet jay]

i tought the middle part

[remote mural]

https://tenor.com/view/yeojinvevo-roblox-gif-24225969

[quiet jay]

h on the bottom

its way simplier

no need for quadratic

my bad

[remote mural]

what

what

[quiet jay]

^

you can pull h out here

and divide by the other side

[remote mural]

so i divide both sides by h

[quiet jay]

i thought i was written as 1/(3*16pih)

[remote mural]

nah

[quiet jay]

oka oka

[remote mural]

thats ok

[remote mural]

do that

?

[quiet jay]

so pull out h then divide by the thing h is multyping

[remote mural]

wdym pull out h

[quiet jay]

h(16pi +16pi/3)=14

[remote mural]

OH

so THIS is h1 right

[quiet jay]

yuh yuh

then the question is pretty much joever

[remote mural]

so

14 over 16pi+16pi/3

[quiet jay]

yeas

all over

[remote mural]

just add 16pi and 16pi/3

and its just 14 over than

*that

[quiet jay]

yupp

i would leave it in pies and that

dont know what ur curriculum would want

[remote mural]

i think my curriculum wants me to cut my main arteries cause i never learned any of this

[quiet jay]

dam cuz

[remote mural]

we never went over any of this, they just gave the formulas

[quiet jay]

i mean thats kinda cool actually

it may suck but you will improve

[remote mural]

fr

[quiet jay]

because you gotta find out the details of what these formula mean

[remote mural]

yea

[quiet jay]

in uni atm its really nice its all derived from fundamental theoroms

VERY DIFFICULT i will say

but cool

[remote mural]

i only know this cause its something ive studied online

[quiet jay]

i recommend 3blue 1 brown

[remote mural]

im asking this question of the behalf of like 20 kids who dont know how to do this LOL

[quiet jay]

Amazing math youtuber

ahahahah

welp u can show of the answer

enjoy!

[remote mural]

ya

TY

U SAVED MY LIFE AND DEPRESSION WITH THIS ONE

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[hybrid flint]

Ok so I manipulated the result to PQ = PB*PC/(PB+PC)

By Ptolemy I got AP = PB+AC so (PQ)(AP) = (PB)(PC)

Just some things I found in case I need it:
By power of point, (BQ)(QC) = (AQ)(AP)

So I also rewrote the result as QP^2 = (BP)(PC) - (BQ)(QC)

I tried stewart's theorem on BPC because angle BPA and CPA are 60 degrees

and got (BC)(QP) = (BP)(QC)

tried to manipulate it with some of the earlier things but hasn't worked out

not sure how to isolate BP*PC

[calm coral]

@hybrid flint Has your question been resolved?

[hybrid flint]

<@&286206848099549185>

[hybrid flint]

I think I'm close

Ok so I applied cosine law with bpq bcq and bcq and manipulated them

.close

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[remote mural]

is this solution correct

[remote mural]

i found this somewhere and i attempted it in the same way

[tawdry garden]

Yes

[remote mural]

thank you

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[remote mural]

i’m wondering how my teacher got 0.999 (:

[calm coral]

@remote mural Has your question been resolved?

[calm coral]

@remote mural Has your question been resolved?

[calm coral]

@remote mural Has your question been resolved?

[remote mural]

.close

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[calm coral]

@slow birch Has your question been resolved?

[upper sparrow]

(do you have an idea of how to do these?)

[slow birch]

ik it does but its wrong

[upper sparrow]

how do you mean it's wrong?

[lyric ravine]

As in the solution says that the vectors dont span R^2

[storm sail]

To span $\mathbb{R}^2$ you need two linearly independent vectors

[potent lotus]

fwa

[slow birch]

it does span R^2 no?

[lyric ravine]

It does

[storm sail]

It should

[slow birch]

oh well

.close

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[eager forge]

How do i do this

[fringe reef]

recall that concavity is determined by the second derivative

[eager forge]

yea

[fringe reef]

can you write down g''(x)?

[fringe reef]

you are given g'(x) so g''(x) is the derivative of g'(x)

[eager forge]

using ftoc

[fringe reef]

hint: fundamental theorem of calculus

yes

[eager forge]

g'' = e^-x^2

[fringe reef]

yes

[fringe reef]

is this function positive on (0, 2)?

[eager forge]

should be

exponential is always positive right

[fringe reef]

yep

so your function must be concave up

[eager forge]

so i know its concave up

how do i find inc / dec?

[fringe reef]

I'll get to that

[fringe reef]

note that you can also eliminate E

[eager forge]

eb d

[fringe reef]

the only way to get g'(x) = 0 is to set x=0, but x>0 hence no point of inflection

[fringe reef]

try estimating the integral, roughly

between 0 and 2

[eager forge]

ik it would be positive

Right?

[fringe reef]

yes

[eager forge]

but how do i do it mathematically

[fringe reef]

try plotting e^(-x^2)

[eager forge]

got it

Wait no like

how do i integrate it

Without a calc

[fringe reef]

you can't

[eager forge]

ibp?

[fringe reef]

nope, you cannot integrate this function

,w integral e^{-x^2}

[potent lotus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by arte_m_etra

[fringe reef]

yeah

[eager forge]

so on these kinds of problems do i just estimate?

[fringe reef]

but notice that as you do integral frol zero to x, and increase x, you "gain" more area hence the integral is always positive

in general if $f(x)$ is positive on some interval $(a, b)$ then $\int_{a}^{b} f(x) dx > 0$

[potent lotus]

artemetra

[eager forge]

if the normal function?

[fringe reef]

wdym

by normal function

[fringe reef]

sorta, you write what i wrote

[eager forge]

I see

[eager forge]

like say

f(x) = x^5 and intabf(x)

from 0 to inf since normal function is positive, would that also mean int ( 0 --> inf ) is also positive

[fringe reef]

yes

well it will be infinite but it will be +infinity

[eager forge]

makes sense

thanks

[calm coral]

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[glacial raptor]

Stats and Probability, Least Squares Regression

[glacial raptor]

how do I find the mass of the object at 27 minutes

[hasty fiber]

plug 27 into your line

[glacial raptor]

yeah like substitute it into the x right?

[hasty fiber]

that's what it means to plug in

[glacial raptor]

nvm I realized what I did wrong

mis type

thanks for the help

.close

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[swift tide]

$\int_{1}^{x^4} -\sin{(xt)}dt + \frac{4\cos{(x^5)}}{x}$

[potent lotus]

Jasmine<3

[swift tide]

That's what I'm getting

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[fleet zodiac]

$x^3 + 12x^2 + 39x + 28$

[potent lotus]

Alex

[fleet zodiac]

i need to factor this withour RRT

any ideas?

[blazing coyote]

RRT?

[prime skiff]

rational root theorem

[plush vapor]

Why have I never heard of this thing

Does it make factorisation easy

[blazing coyote]

Vitaes' formula

[blazing coyote]

you'll get 3 equations in 3 variable from this, solve them to get the roots

[plush vapor]

Lmao what a troll

[blazing coyote]

what do you mean, troll?

[fleet zodiac]

I dont understand that

[plush vapor]

It'll result in the same cubic

When you solve that system

You'll return to the same cubic eventually

[blazing coyote]

yeah, but you'll get alpha, beta and gamma

[fleet verge]

Complete the cube, it is very easy tbh

[plush vapor]

I don't get what you mean, we'll arrive at the same cubic equation in Alpha or beta or gamma

When we attempt to solve the equation

[blazing coyote]

We want to factorise it

so for that finding alpha, beta, gamma are sufficient

say we get alpha=1, beta=2, gamma=3

(x-1)(x-2)(x-3) will be the factorisation

[plush vapor]

My point is, to get those 3 roots you'll end with the same equation to solve

[blazing coyote]

didn't realise, sorry

[plush vapor]

Go ahead and try to solve the system you make

It'll just lead to the original equation

[fleet verge]

Just this

[plush vapor]

I'd recommend noticing the coefficients and realising -1 just satisfies this thing since 39+1=40 and 12+28 also equals 40

Then dividing the whole thing by x+1

To get it into the form (x-1)(quadratic)

And factorise the quadratic

[fleet zodiac]

i know to complete the square but not the cube

[fleet zodiac]

isnt that using rrt?

[plush vapor]

What is rrt even

I didn't use that lmao

[fiery python]

rational root theorem

[blazing coyote]

The rational root theorem in algebra states that a polynomial equation with integer coefficients can only have rational number
roots if the leading coefficient is divisible
by the denominator of the fraction.

[plush vapor]

I don't see how what I recommend uses that?

[fiery python]

those questions in the cube square format are pretty common for rrt

[blazing coyote]

I don't either, just saying what it is

[plush vapor]

Regardless, competing the cube won't be a problem either, break 28 into 27+1 and continue from there

[fiery python]

think its a way of factorising using synthetic division

[fleet verge]

It is the same, u can realize the (x+a)^3, makes the x^2 coefficient = 12 when a = (find this)

[fleet zodiac]

4?

$(x + 4)^3 - 9x - 36$

[potent lotus]

Alex

[fleet zodiac]

ok i got it thx

.close

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[native hollow]

i'm stuck at B and C.

i've tried getting the 90th percentile to get the value in B, but i'm still not sure about my answer.

i thought, since above 10% of the distribution was required, getting the 90th percentile would be the right choice (since if i would get the 10th percentile, it would either be equal to 10% or less than 10%,)

same goes with C. i don't know what to apply since B and C sounds like they're looking for the same thing

[calm coral]

@native hollow Has your question been resolved?

[native hollow]

<@&286206848099549185>

[calm coral]

@native hollow Has your question been resolved?

[calm coral]

@native hollow Has your question been resolved?

[unkempt drift]

Yes, so you need the score of the 50 * 9/10 = 45th student
There are 50 students overall so count backwards by 5, (50, 49) and (48, 47, 46)
So the category is actually 185-189

I'm not sure what the difference between B and C is

[calm coral]

@native hollow Has your question been resolved?

[native hollow]

ohhhhh i see

thanks a lot!

.close

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[coarse minnow]

I want to prove by induction that

[coarse minnow]

$P(\bigcap_{i=1}^nA_i)=P(A_1)\cdot \prod_{k=2}^nP(A_k|\bigcap_{i=1}^{k-1}A_i)$

[potent lotus]

Bob Goldham

[coarse minnow]

to do this

I did something like

first I show that it holds for n=1

then

assuming it holds for n, we can

$P(\bigcap_{i=1}^nA_i)=P(A_1)\cdot \prod_{k=2}^nP(A_k|\bigcap_{i=1}^{k-1}A_i)$

[potent lotus]

Bob Goldham

[coarse minnow]

$\Leftrightarrow P(\bigcap_{i=1}^nA_i) + P(A_{n+1}) - P(A_{n+1} \cup \bigcap_{i=1}^nA_i) =P(A_1)\cdot \prod_{k=2}^nP(A_k|\bigcap_{i=1}^{k-1}A_i) + P(A_{n+1}) - P(A_{n+1} \cup \bigcap_{i=1}^nA_i)$

[potent lotus]

Bob Goldham

[coarse minnow]

adding the same term to both sides

$\Leftrightarrow P(\bigcap_{i=1}^{n+1}A_i) =P(A_1)\cdot \prod_{k=2}^nP(A_k|\bigcap_{i=1}^{k-1}A_i) + P(A_{n+1}) - P(A_{n+1} \cup \bigcap_{i=1}^nA_i)$

[potent lotus]

Bob Goldham

[coarse minnow]

LHS is just the intersection for n+1 since P(A intersect B)=P(A)+P(B)-P(A union B)

now what? How do I proceed from here?

that RHS looks a bit... messy

[coarse minnow]

can't use the same trick as here, since that'd indirectly use the statement I'm trying to prove about conditional probabilities

I could also put it into a form like this, though I'm not sure it's helpful:

$\Leftrightarrow P(\bigcap_{i=1}^{n+1}A_i) = P(A_1)\cdot \prod_{k=2}^n
\frac{P(\bigcap_{i=1}^{k}A_i)}{P(\bigcap_{i=1}^{k-1}A_i)}
+ P(A_{n+1}) - P(A_{n+1} \cup \bigcap_{i=1}^nA_i)$

[potent lotus]

Bob Goldham

[coarse minnow]

<@&286206848099549185>

[calm coral]

@coarse minnow Has your question been resolved?

[sacred valley]

questions like these is what makes math interesting for me
like, I dont understand any of that stuff but I wanna know

[coarse minnow]

it's probabilities

[coarse minnow]

baye's theorem yields this, which is the same thing again? I'm out of ideas here

[sacred valley]

yeah, I recognise the P lmao

[coarse minnow]

I could also do this, but resolving the fraction in the product requires me to assume the statement I'm trying to prove

[calm coral]

@coarse minnow Has your question been resolved?

[calm coral]

@coarse minnow Has your question been resolved?

[calm coral]

@coarse minnow Has your question been resolved?

[calm coral]

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[remote mural]

whats the next step to this

tryna divide the fraction

[wise ridge]

?

did u see what i said in #help-35

[remote mural]

thank u

sophiaaaa

but

how do i remove thr fraction

[wise ridge]

you dont have to

10/1 is simply 10

it simplified itself

[calm coral]

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[calm coral]

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[calm coral]

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[velvet wasp]

is this correct so far?

If the derivative at any point DNE then the whole derivative DNE right?

[cerulean cradle]

non diff fn

[velvet wasp]

wdym

the derivative at f(3) does not exist right?

And for the last question

@cerulean cradle

Look good?

[calm coral]

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[buoyant mirage]

what does a period mean

i dont get it

[calm coral]

@buoyant mirage Has your question been resolved?

[lucid umbra]

Period is the distance it takes for a trigonometric function repeats itself, i.e. complete one cycle

[blazing hinge]

ehhh not really it's more the length of a cycle

[buoyant mirage]

that made sense

tysm

tysm

.close

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[austere moth]

How do i determine the area (green) ?

[austere moth]

I understand that i'm supposed to make a integral with set boundaries where it goes from (0,0) to (a,0)

[vital raven]

it looks like a polar coordinates integral

idk

[austere moth]

That might be right, i haven't touched polar coordinate integral yet however

How do i create a double integral equation to solve the area A?

I can't seem to understand why i would want a double integral for this. Isn't this just "area" meaning a single variable or a single integral

[vital raven]

yes you can solve it using a single integral

[austere moth]

In that case the bounds would be from (0,0) to (a,0)

[vital raven]

im trying to figure it out im still not sure

[austere moth]

$\int^a_0 \frac{x^2}{a^2} + \frac{y^2}{b^2} -1 <= $

However i have to somehow fit the second criteria that 0 <= y <= x

[hollow lion]

Is there any more info in the question?

[austere moth]

Yes

[vital raven]

ah i know

oh so they want you to use double integrals

[austere moth]

"Determine the area A and the coordinates(ux,uy) to the center of gravity of the ellipse sector D if..."

Well my first job would be to find the area A

Which to me seems like i just want to integrate the given function?

[vital raven]

you can seperate the area into a triangle plus an integral

[austere moth]

$\int^{c}_0 \frac{x^2}{a^2} + \frac{y^2}{b^2} -1 da$

This would be the area B?

[vital raven]

ux is the just the center of the region

doesnt give us the area

[austere moth]

oh woops i assumed it was the point on the curve of the ellipse

[vital raven]

x = B is the point where the ellipse meets the line y = x

[austere moth]

Can we make it c instead

[vital raven]

okok

the integrand isnt right tho

what are you integrating with respect to ?

[potent lotus]

Merineth

[vital raven]

now im confused lol

[austere moth]

i'm integrating wrt a?

[hollow lion]

Why would you split the area into a triangle and the rest?

[austere moth]

idk he said so

[hollow lion]

Yeah I’m asking him

[austere moth]

oh sorry

[vital raven]

makes it easier to calculate imo

$R_2$ is just a triangle. $R_1$ is an integral

[potent lotus]

Obotron

[vital raven]

if we rearrange ellipse equation into the form y = f(x) then $R_1 = \int_{c}^{a}f(x) dx$

[austere moth]

Since i'm given restrictions and i want to maximize a given function, wouldn't i be able to use Lagrange?

[potent lotus]

Obotron

[vital raven]

which function are we maximizing ?

[austere moth]

We are maximizing the function for the ellipse and the restriction would be the 0 <= y <=x

So i take the gradient of both and make a function

$\nabla f = \lambda \nabla g$

[potent lotus]

Merineth

[austere moth]

where f is the given function for the ellipse and g is the restriction

[vital raven]

i thought we are finding the area under the ellipse where 0 <= y <= x

[austere moth]

yes

Isn't that was lagrange does?

[vital raven]

lagrange finds the maximum/minimum of a function

[austere moth]

@upper sparrow

[austere moth]

Oh right

[vital raven]

Equation of ellipse is: $\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$ which can be written as: $y = \frac{b}{a} \cdot \sqrt{a^2-x^2}.$

[austere moth]

It kind of makes sense to divie it into two integrals and sum them up

since as you pointed out, one could just be a triangle

[potent lotus]

Obotron

[vital raven]

yes i think its the easiest method

you can find point c by equating the equation of the ellipse equal to x

$\frac{b}{a} \cdot \sqrt{a^2-x^2} = x$

[potent lotus]

Obotron

[vital raven]

idk if im making sense lol

i have it figured out in my head but cant explain

[calm coral]

@austere moth Has your question been resolved?

[vital raven]

Ok here is a summary of my method:

1. Write equation of ellipse is: $\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$ as: $y = \frac{b}{a} \cdot \sqrt{a^2-x^2}.$

[potent lotus]

Obotron

[vital raven]

2. split the area into 2 regions as such

3. Calculate c by equating the ellipse equation to the line y = x: i.e $\frac{b}{a} \sqrt{a^2-c^2} = c$

[potent lotus]

Obotron

[vital raven]

4. Now R2 is just a simple triangle with height and width equal to c. $R_2 = 0.5c^2$

[potent lotus]

Obotron

[vital raven]

5. R1 is a fairly simple integral: $R_1 = \int_{c}^{a}f(x) dx$

[potent lotus]

Obotron

[vital raven]

@austere moth im trying :(

[calm coral]

@austere moth Has your question been resolved?

[desert edge]

Can I get help?

@calm coral

[calm coral]

@austere moth Has your question been resolved?

[calm coral]

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[onyx quiver]

is a correct?

Im new to vectors

this is how the teacher did ti

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[cedar quartz]

Yo

[cedar quartz]

Any idea here

Btw it | Un - 2 | its just a mistake

Um

<@&286206848099549185> AAAAH

......

[chrome robin]

Help pls this ez but idk how:

[cedar quartz]

I can give more details its just because that all it needs

[cedar quartz]

What is the questiom