help-39

[pseudo oxide]

very good

12x + 8 is the derivative of f(x)

at any x

the x u need to find the derivative at

is -3

so just plug it in

[midnight coral]

oh

right

-28

how do i do itoc

the second one

[pseudo oxide]

that is literally the same thing

[midnight coral]

yeah

that’s was i was confused at

that means the slope of the line tangent is the same thing as iroc?

is this right for c?

,rotate

[midnight coral]

pls help 😔

<@&286206848099549185>

[versed remnant]

at x=-3 , plug that in ur f(x) to find y coordinates , and then use that

[midnight coral]

and not use use anything for y?

[versed remnant]

f(x) = y = 6x^2 + 8x

y = 6(-3)^2 + 8(-3)

[midnight coral]

-28

OHH

bro

my brains gone

.close

[pearl pond]

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[midnight coral]

.close

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[dark ingot]

If m_i and 1/m_i , m_i>0 i=1,2,3,4 are four distinct points on a circle then prove that m_1 ×m_2×m_3×m_4=1

[cyan bronze]

Opposite angles in a cyclic quadrilateral add up to 180 might help

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[delicate dock]

How do I solve for the** minimum value of F** to get a quadratic with roots that are complex conjugates

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[teal hazel]

how

why is it (n-1)! for only the engineers

[tall flint]

It shouldn't be

I don't see why it wouldn't be 3!

[teal hazel]

yes

but for any circle

isnt there n-1

!

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[wheat vale]

how would i attempt to do part b. I can't seem to get to the answer

[wheat vale]

its a mechanics question

[pearl pond]

@wheat vale Has your question been resolved?

[wheat vale]

<@&286206848099549185>

thank you 🙂

i got 40 seconds but as the time for deceleration of N

i didnt know what to do after that

yh

the mark scheme graph looks like this

?

sorry i didnt understand the seperating the speeds

yh

yh that makes sense

wait so how did u construct the N equation? cause i saw that on the ms but didnt know how to get to that

is it the area?

hmm im sorry but i still dont think i understand this 😦

yh split it into triangle and rectangle or area of the trapzium it will equal the displacement 975m

yes

idk

im stuck on that do i use M or N

yh but idk how i got to 40

like i used s=ut+1/2at squared

and then made u = -30/t

sorry i meant a

u know what dw

its fine

ill try to get through the ms

its fine ds

dw

ok thanks for the help

[pearl pond]

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[drowsy flare]

why isnt the result 0? i dont understand, on the denom theres x and on the top theres x-x so why not 0

[compact veldt]

Wait what is the limit

-1?

[random ermine]

denom is inf * 0 so already non trivial

i would analyze the limit of the denominator first separately

[drowsy flare]

yes

[drowsy flare]

the x is a stronger inf and wins

so the denom is x

[drowsy flare]

how

[random ermine]

the denominator goes to zero according to desmos

[drowsy flare]

wtf

now im confused^2

[drowsy flare]

can u explain bro

[random ermine]

l'hopital

[pearl pond]

@drowsy flare Has your question been resolved?

[pearl pond]

@drowsy flare Has your question been resolved?

[drowsy flare]

Cant do without?

[random ermine]

idk?

[drowsy flare]

can u show how u found with hopital

[random ermine]

i didnt do it

[drowsy flare]

do u know how to solve?

<@&286206848099549185>

[slow grove]

define F(x) = (sqrt(x^2 - 2) - x) / (x arctan(1/x^2)) = (sqrt(1 - 2/x^2) - 1) / arctan(1/x^2)
now let f1(x) = sqrt(1 - 2x) - 1, f2(x) = arctan(x), g(x) = 1/x^2
we find that f1'(x) = -1/sqrt(1 - 2x) and f2'(x) = 1/(1 + x^2)
then F(x) = f1(g(x))/f2(g(x))

using l'hopitals lim x -> inf F(x) = lim x -> inf [f1(g(x))]' / [f2(g(x))]'
= lim x -> inf f1'(g(x)) g'(x) / [f2'(g(x)) g'(x)]
= lim x -> inf f1'(g(x)) / f2'(g(x))
= lim x -> inf [-1/sqrt(1 - 2/x^2)] / [1/(1 + 1/x^2)]
= -1/1 = -1

[plush vale]

shouldnt f1 be sqrt(1-2/x^2)-1?

[slow grove]

you could define it that way and then define f2 as arctan(1/x^2)

but the way i defined it always you to skip a bit of computation, admittedly not much

as you get the g'(x) to cancel (from chain rule)

[plush vale]

i see

thanlks a bunch

[slow grove]

nws

[drowsy flare]

bro r u able to latex it

i dont understand it

[slow grove]

just as in its hard to read?

or you dont understand a step

[slow grove]

\begin{align*}
F\left(x\right) &= \frac{\sqrt{x^2 - 2} - x}{x \cdot \mathrm{arctan}\left(\frac{1}{x^2}\right)} \
&= \frac{\frac{1}{x} \left[\sqrt{x^2 - 2} - x\right]}{\mathrm{arctan}\left(\frac{1}{x^2}\right)} \
&= \frac{\sqrt{\frac{1}{x^2} \left(x^2 - 2\right)} - 1}{\mathrm{arctan}\left(\frac{1}{x^2}\right)} \
&= \frac{\sqrt{1 - \frac{2}{x^2}} - 1}{\mathrm{arctan}\left(\frac{1}{x^2}\right)} \
\text{Define } f_1\left(x\right) = \sqrt{1 - 2x}& - 1,, f_2\left(x\right) = \mathrm{arctan}\left({x}\right),, g\left(x\right) = \frac{1}{x^2} \
\text{then } f_1'\left(x\right) = \frac{-1}{\sqrt{1-2x}} &\text{ and } f_2'\left(x\right) = \frac{1}{1 + x^2} \
\text{and } F\left(x\right) &= \frac{f_1\left(g\left(x\right)\right)}{f_2\left(g\left(x\right)\right)} \
\
\text{Note } \lim \limits_{x \to \infty} f_1\left(g\left(x\right)\right) = 0 &\text{ and } \lim \limits_{x \to \infty} f_2\left(g\left(x\right)\right) = 0 \
\text{hence (by L'Hopital) } \lim \limits_{x \to \infty} F\left(x\right) &= \lim \limits_{x \to \inf} \frac{\left[f_1\left(g\left(x\right)\right)\right]'}{\left[f_2\left(g\left(x\right)\right)\right]'} \
&= \lim \limits_{x \to \infty} \frac{f_1'\left(g\left(x\right)\right)g'\left(x\right)}{f_2'\left(g\left(x\right)\right)g'\left(x\right)} \
&= \lim \limits_{x \to \infty} \frac{f_1'\left(g\left(x\right)\right)}{f_2'\left(g\left(x\right)\right)} \
&= \lim \limits_{x \to \infty} \frac{\frac{-1}{\sqrt{1 - \frac{2}{x^2}}}}{\frac{1}{1 + \frac{1}{x^2}}} \
&= \frac{\frac{-1}{\sqrt{1 - 0}}}{\frac{1}{1 + 0}} = -1
\end{align*}

@drowsy flare

[drowsy flare]

Where did x² go in first step?

[slow grove]

hows this

[jolly parrot]

Acman

[drowsy flare]

Thx

[drowsy flare]

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[hollow pike]

i need help with the circled parts as i did 3/1500 does not equal 15/1500 * 45/1500

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[errant flax]

can someone tell me the formula

[errant flax]

on x and y -axis on rotation

on the rotation volume on a body

[thin sigil]

Uhhh... can you be more pacific?

[errant flax]

Let Ω denote the area in the first quadrant between the line y =
16 and the curve y = x^4. When the curve Ω is rotated one revolution around y-
the axis forms a body of rotation. Calculate the volume of the body.

like this

[late venture]

for x=axis its $2\pi \int ^a _b \sqrt{1-\frac{df}{dx}}dx$

[cinder flower]

be more atlantic

[thin sigil]

Right, you need the integral

[late venture]

i think

wait no im stoopid

[errant flax]

whats that ?

[late venture]

$2\pi \int ^b _a xf(x)dx$

[thin sigil]

"Let Ω denote the area" and then "When the curve Ω"

[thin sigil]

What is it, an area or a curve

[cinder flower]

i was just poking fun at aki's message haha dw about it

[errant flax]

my professor have done like this

[late venture]

volume of a rotated body

[errant flax]

but i wanna know the formula and the difference between when its a volume and the rotation area both on x and y

[thin sigil]

Yeah but the questions defines Ω as an area and then refers to it as a curve

[late venture]

im stoopid again

$V = \pi \int _a ^b f^2(x)dx$

[jolly parrot]

spookyspaghetti

[late venture]

that for around x-axis

[errant flax]

and for y?

[late venture]

for y-axis its $V = \pi \int _{f(a)} ^{f(b)} (f^{-1}(x))^2 dx$

[jolly parrot]

spookyspaghetti

[late venture]

with $f^{-1}(x)$ the inverse function of $f(x)$

[jolly parrot]

spookyspaghetti

[errant flax]

so why have my proffesor have 2pi?

[errant flax]

here

[late venture]

not a clue

[errant flax]

this is the rule he uses

okay nw i will just use it xD without understanding why

thats why i am confused

the rule when i search it up its only pi and not 2 pi

[late venture]

if i can see the function they use maybe i can see

[late venture]

yeah no idea

the one with just pi is the one i learned and it works fine

wikipedia says the same as me so idk

[errant flax]

okay nw

i gonna ask it again here

maybe someone else

know this weird rule

.close

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[stoic imp]

nnmnmbn

[thin sigil]

mmmnmmnmnmnmbbbnnnbb

[unkempt yacht]

what have you tried?

[stoic imp]

nothing

but i know there is a solution if it passes through 0

[thin sigil]

well

you have e^x * (x^2 - x - 5) = 1/3

right?

[stoic imp]

sure

[thin sigil]

If we subtract 1/3 from both sides

[stoic imp]

okay

[thin sigil]

e^x * (x^2 - x - 5) - 1/3 = 0

we get this

Is this more managable now?

[stoic imp]

right

no it isnt is same thing

[thin sigil]

We can no subtract that

and see what happens

also factor the quadratic

[stoic imp]

,w factor x^2 -x-5

[thin sigil]

Ayo don't cheat

[stoic imp]

i git the roots now what

got

[thin sigil]

Ew them roots ugly

but correct

[rough forge]

is this even solvable analytically?

[stoic imp]

yes, its homework by teacher

[thin sigil]

The question asks how many solutions it has

not what the solutions are

[rough forge]

ah ok then

[thin sigil]

So the important thing here is that you figured out this quadratic has 2 real roots

keep that in mind

[stoic imp]

okay

[thin sigil]

So now we got this

[stoic imp]

?

[thin sigil]

When you factor the quadratic

I didn't bother writing in the numbers

they're x_1 and x_2

[stoic imp]

i got lost here

[thin sigil]

yeah you have that - 1/3

[stoic imp]

where is the 1/3 you put here?

okay

okay ok

a moment

okay

[thin sigil]

So when will this be 0?

[stoic imp]

When 3ex….. is 1

[thin sigil]

Basically

more specifically when this holds

[stoic imp]

ye

[thin sigil]

Aaaaannnnddd I'm honestly not sure what to do next

I thought it'd work out into a nicer form

[wraith hare]

what I would do is prob try sketching the graph by finding the roots (which you've done) and the turning point and then determine the shape of the graph

then see how many times y=1/3 intersects the graph

[stoic imp]

exactly

how to do i do that tho

[thin sigil]

Oooohhh

[stoic imp]

inflection points?

[thin sigil]

I approached this wrong

completely

[wraith hare]

do you know how to find turning points?

[thin sigil]

this is about extremes again

[stoic imp]

first derivative test

[wraith hare]

yh

[stoic imp]

i think

[thin sigil]

Remember Rolle's theorem

it comes in handy here

Do the same analysis we did before

find the extremes

[stoic imp]

,w differentiate (e^x)(x^2-x-5)

[jolly parrot]

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[stoic imp]

,w solve (e^x)(x^2 + x -6) = 0

[jolly parrot]

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[thin sigil]

Okay now calculate their values

And then check if their max or mins

[stoic imp]

Nmm

we got two critican points, one maxima one minima

critical point 2 evaluated makes a maxima

[thin sigil]

Well to check maxima or minima you have to take the second derivative

remember

[wraith hare]

you don't have to do that here

[stoic imp]

Whhat

ohh

[wraith hare]

observe aki

you have to observe and think about the numbers you got

[stoic imp]

what do they mean, the numbers

[thin sigil]

I mean yeah, you could figure it out from these numbers but like, it's not certain in general

[wraith hare]

think about the two numbers 7/e^3 and -3e^2. Is there really a need to do second derivative in this case

[thin sigil]

You can have situations like these for example

I know what you're getting at

but it's bad practice

in my book

[wraith hare]

it's good practise though imo to logically think and never always rely on doing more maths. Like isn't it better to think actually think about it than always do maths routinely. The function here is cts so that situation is mitigated

[thin sigil]

It's not when you're new to something because all of that can confuse you. It's best for op to learn the methodology first and then figure out tricks later

[wraith hare]

fair point

I did assume @stoic imp knows enough basic cuz he's undergrad

[stoic imp]

,w differentiate (e^x)(x^2 + x -6)

[thin sigil]

Anyway, we got our extremes

[jolly parrot]

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[thin sigil]

Just plug in the x values and check the signs

[stoic imp]

Nn

[thin sigil]

Ye

so the left one is a ?

the one at -3

I mean

[stoic imp]

wdym left

ahh maxima

[thin sigil]

ye

and the other one is?

[stoic imp]

minima

[thin sigil]

Good

[stoic imp]

now wat

[thin sigil]

Well what we wanna check now is how the function behaves left of -3

and right of 2

[stoic imp]

checking -3.5 and 2.5?

or do you mean lateral limits

[thin sigil]

Best to check limits

[stoic imp]

okay

how to do that lol

[thin sigil]

This is a sketch of what we know so far btw

No not to 2

to inf

and -inf

[stoic imp]

ohh

[thin sigil]

we wanna see if there are any horizontal asymptotes

just in case

[stoic imp]

horizontal asymptotes

[thin sigil]

yeah

[stoic imp]

what do i do here

its + inf

?

literally

,w lim n to +infinity (e^x)(x^2+3x-5)

[jolly parrot]

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[thin sigil]

wolfram is drunk

[stoic imp]

????

yeahhh

[thin sigil]

,w lim x to +infinity (e^x)(x^2+3x-5)

[jolly parrot]

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[thin sigil]

it's inf

[stoic imp]

Ahhh

[thin sigil]

inf times inf is inf

[stoic imp]

okay

[thin sigil]

now check for -inf

[stoic imp]

0

[thin sigil]

good

so, now nowing this we can graph our function

like so

[stoic imp]

,w lim x to -infinity (e^x)(x^2+3x-5)

[jolly parrot]

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[stoic imp]

how 😭

[thin sigil]

Don't mind my inaccurate graphing

Omg my pc had a seizure

Well we knot the max and mins right

and we know that it goes to 0 on the left

to inf on the right

and we also know there are not extremes between -3 and 2 right

so we can also add

This

Right?

[stoic imp]

?

oh yeah because only teo critical points

two

[thin sigil]

yes

so the function must always decrease from x=-3 to x=2

at least in this case of a differentiable and continuous function

[stoic imp]

How?

[thin sigil]

because if at any point it started increasing

it would have an additional minimum

we know it doesn't

so it can't

Remember Rolle

[stoic imp]

Can you do the visual justification

lowkey i studied rolles but forgot!

what does it say?

[thin sigil]

See how purple

in order to go up

had to have a bottom point

we know that it doesn't have those

anywhere else

so it can't do any ups or downs anywhere else

based on our analysis the function goes from -inf to -3 increasingly, then it goes from -3 to 2 decreasingly and then it goes from 2 to inf increasingly

which is what I sketched in blue

It can't make any other changes in up or down anywhere because it doesn't have extremes to do so

Basically

[stoic imp]

what is the number of solutions, 2?

[thin sigil]

Now that's the cool part

On our sketch we will now draw the line of y = 1/3

now let's first check

is 1/3 greater or lesser than 7/e^3

[stoic imp]

lesser

[thin sigil]

good

so I'll sketch it as suc

such

Now do you see how many solutions there are?

[stoic imp]

3, why?

[thin sigil]

yes

[stoic imp]

what is this 1/3 number

horizontal line test haha

wtf

[thin sigil]

Look at what the problem is asking for

The number of solutions for f(x) = 1/3

will be the number of times the graph of f(x) crosses the line y=1/3

[stoic imp]

shiit

awesome

[thin sigil]

yes

so here's a bit more justification

in order to go from 0 at -inf to 7/e^3 at -3 the function must pass the point 1/3 once and exactly once

it can't pass it more than once since it has no extremes inbetween

and it can't pass it less than once since it can't do jumps

(at least this one can't)

[stoic imp]

due to continuity differentiability?

[thin sigil]

then going from 7/e^3 at -3 to -3e^2 at 2 the function must cross 1/3 again for the same reasons

and only once at that

[thin sigil]

ye basically

and then again to go from -3e^2 at 2 to inf at inf it must pass 1/3 once and only once again

so in total it must pass 1/3 three and exactly three times

meaning the equation f(x) = 1/3 must have 3 and exactly 3 solutions

So to summarize, we checked the extreme points to get the places where the function "turns" so to speak, then we checked how it behaves left and right of the extremes to see to which value it goes and then we sketched the graph and the line y = 1/3 making sure to position it accurately under 7/e^3

this gave us the answer

[stoic imp]

Lowkey hard

[thin sigil]

The reason we had to check the left and right limits

is because the function could've behaved like this

for example

and then it wouldn't cross 1/3

left and right

in light blue

had the right limit been 0

then it wouldn't cross 1/3 on the interval of 2 to inf

which is why we needed to check those

in order to accurately draw our sketch

[stoic imp]

how can i check my ans

though

[thin sigil]

By using a graphing calculator

lol

If you did everything right

the way I told you, then you must be correct

[stoic imp]

,w plot (e^x)(x^2 -x-5)

[jolly parrot]

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[thin sigil]

Imma give you another problem to practice this if you want

[stoic imp]

sure

[thin sigil]

Here's the graph

in scale

See how it resembles our sketch

[stoic imp]

yeah

[thin sigil]

đ

Here's your practice problem

see how many solutions that upper function has

for each of these y values

individually

[stoic imp]

okay

ty for the help

[thin sigil]

Np, good luck

[stoic imp]

imma be closing

.close

[pearl pond]

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[pearl pond]

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[wild fable]

I’m having trouble showing the formulas highlighted in blue are equal, I expanded but then I get 2x*mu which I don’t know what to do with

[warm current]

!original

[pearl pond]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[wild fable]

That is the original

hiiiiii @merry carbon

The question says to prove this identity starting with the LHS and expanding the brackets

[warm current]

Try $f(x)=\mu=1$ and see that it is false.

[jolly parrot]

SWR

[merry carbon]

I'm guessing the mu is supposed to be the integral of x [w/ limits] divided by (b - a) or something?

Or wait, what is mu supposed to be? Looks statsy to me

[wild fable]

It’s cut off a bit but this is the question

[thin sigil]

Define mu

[warm current]

Given your context so far, it's false

[thin sigil]

proceeds to mooo like a cow

[wild fable]

Expected value mean

[thin sigil]

Oh this is some statistics shenanigans

[wild fable]

yes

[thin sigil]

no, that's too evil for me

[wild fable]

grrrrr

[warm current]

People often miss this bit

[thin sigil]

LMAOOOOOOOO

[wild fable]

idk I thought mew mu was a giveaway it was stats

[wild fable]

FMTB

[rough forge]

https://tenor.com/view/caught-in4k-4k-camera-caught-sus-gif-20846600

[thin sigil]

WHAT IS THIIIIIIIISSSSSS????

[warm current]

https://tenor.com/view/ross-friends-im-not-a-mind-reader-gif-14904319

[wild fable]

mind read me

[wild fable]

it’s FMTB

[thin sigil]

I always assume it's not stats because stats are evil and I see no evil

[warm current]

https://tenor.com/view/benjammins-no-nope-not-doing-it-never-gif-16758419287806165606

[wild fable]

https://tenor.com/view/forget-it-flash-men-in-black-will-smith-gif-19235596

[cinder flower]

what are we talking about it this lovely channel

[warm current]

stats

[thin sigil]

evil

[warm current]

https://tenor.com/view/veeunus-spongebob-spongebob-squarepants-squidward-done-gif-24674884

[wild fable]

stats should not exist

[rough forge]

maybe it's some ibp stuff

[thin sigil]

statistically some of most staticians have a chance to agree with that

[warm current]

Something either happens or it doesn't. If you ask me, it's all 50/50

[merry carbon]

[the $\mu$ is then basically $\int_a^b x f(x) \dd x$ then, right?]

[jolly parrot]

@merry carbon

[wild fable]

Yes my favourite statistician @last summit would definitely agree

[merry carbon]

(allow me, I don't know stats stuff )

[thin sigil]

Imagine having a whole ass integral replaced by a letter

and calling it math

[wild fable]

I think so at least that’s how E(x) is defined

[merry carbon]

And f is such that its integral between a and b is 1, yea?

[rough forge]

I would never thought that you would swear

[wild fable]

Stats will make anyone swear even the purest of hearts

[thin sigil]

Stats bring out the evil in me

[warm current]

Volume, charge, moment of inertia, electric/magnetic fields, current, ...

[wild fable]

ye I think so

[thin sigil]

That's different, those are well established, liked and funny

[merry carbon]

In which case then, expand the brackets at the bottom and use that

[cinder flower]

i don’t like those

[thin sigil]

Moment of inertia never hurt my expectations

[rough forge]

too much love i will die

[cinder flower]

too little love i will die

[rough forge]

AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA

[wild fable]

Expand like get more integrals?

[cinder flower]

that’s me

[merry carbon]

Basically yea, split them up

[wild fable]

wot about the 2xmew do I convert mew into an integral

[thin sigil]

https://tenor.com/view/mew-cat-kitten-kitty-cute-gif-17624380

[merry carbon]

mu is a constant, you can take it out, and don't forget the f

[wild fable]

https://tenor.com/view/cat-silly-silly-cat-silly-cats-mewing-gif-14812443053695024113

[merry carbon]

(does that at least help a bit @wild fable? )

[wild fable]

I think so

[merry carbon]

Oh also turns out I should've known this

How's it going?

[thin sigil]

Not your fault, at all, blame stats

"pRoBaBiLitY DeNsItY FuNcTiON"

[merry carbon]

-wait is today becoming stats day for me

Oh no, the end is coming 🕊️

[pearl pond]

@wild fable Has your question been resolved?

[pearl pond]

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[midnight haven]

broda are you doing jee questions

[sharp smelt]

yeah, just for fun

couldn't write advanced due to my centre being far away

[midnight haven]

i thought I had seen the last one before

[sharp smelt]

so I'm going though the paper now

[midnight haven]

Ah

[sharp smelt]

this is the 2024 paper

[midnight haven]

mmm

[sharp smelt]

so Q would match to 3, right?

the differential at 0 would be 0

,w is x|x|sin(1/x) differentiable at 0

[jolly parrot]

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[sharp smelt]

no

wonderful

just wonderful

,w differentiate x^2 sin(1/x)

[jolly parrot]

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[sharp smelt]

ah

makes sense

[midnight haven]

f(h) - f(0)/h , h tends to 0
so we have f(h)/h, h tends to 0

[sharp smelt]

I was trying to do it mentally

got it

I'll close this for now

.close

[pearl pond]

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[high geyser]

Im failing my final someone PLEASE help

[high geyser]

Can anyone explain why theta is in between pi/2 and -pi/2 for the boundary circle???

(This is double integrals with polar equations)

[marble sigil]

think of arrows pointing from the origin to the rim of the shape

it goes between up (pi/2) to down (-pi/2)

[high geyser]

OHHHHH

cuz i thought the arrows started at the origin 😭

thank you

[pearl pond]

@high geyser Has your question been resolved?

[high geyser]

.close

[pearl pond]

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[wraith hawk]

could someone check if my answers to these 2 questions are right? we got them for homework but the solutions wont be discussed in class

[midnight haven]

jee ?

[worldly glacier]

the food?

[midnight haven]

entrance exam

[worldly glacier]

for next time you can check wih wolfram alpha and symbolab

[wraith hawk]

yeah

[midnight haven]

i don't know where else do they give such horrendous questions

[worldly glacier]

oh doesn't look too bad

must be a chill uni

[wraith hawk]

i tried but i couldn't get an answer 😭

im in highschool lol

[worldly glacier]

lmao why not just plug in the value for sin(π/2)

[worldly glacier]

yeah yeah we have calc bc in america

for high schoolers

also symbolab spit me an answer

I can say ur final answer is right for the first q

[wraith hawk]

oh wait 😭 i thought we were supposed to do that after simplifying the derivative

[wraith hawk]

okok ty

[midnight haven]

because it's (π/2)x ?

[worldly glacier]

ohh where are the paranthesis

but 2nd question looks good

lemme redo one

[midnight haven]

it's right

just that you can't directly plug in

not until after you derivate

[worldly glacier]

yeah yeah I didn't see the paranthesis tho

[wraith hawk]

yeah my bad

[worldly glacier]

interesting

wait is this the thingy for IIT?

[wraith hawk]

yup

[worldly glacier]

I thought it would be a lot harder ngl

but maybe the harder questions are scattered

with some easier ones

[wraith hawk]

this is a v basic q

yeah

the actual jee adv questions are something else

[midnight haven]

it's not an exam question, this is for practice

[worldly glacier]

yeah I just think the JEE and the other testing systems are kinda stupid

but I guess when you have that population

what to do

anyway good luck @wraith hawk

and have a good morning!

[midnight haven]

good morning

[wraith hawk]

thank u sm

<3

https://www.jeeadv.ac.in/documents/JEEAdv2024_Paper1_English.pdf

heres the adv paper if u wanna see the questions in the exam btw

[worldly glacier]

it seems p chill until I got to the physics part 😭

[wraith hawk]

yeah phys is tough

anyways ty

.close

[pearl pond]

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[mild grotto]

is this just if water is 1g/cm^3

the rectangles volume is 1 x 2 x 2 = 4cm^3

and the mass is 3

so mass/volume = density ( 3/4 = 0.75)

so it floats?

[sharp vigil]

yes

[pearl pond]

@mild grotto Has your question been resolved?

[pearl pond]

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[wicked sluice]

Can someone help me on how to find the domain and range for this graph as well as graphing it by hand

[wanton burrow]

So

what are your two vertical asymtpotes

@wicked sluice

and what is your horiz asymptote

[jovial oar]

Domain

[wanton burrow]

so (-3/2, 1)

Just doesn’t exist

!?

[wicked sluice]

-3/2 and 1

yepp

[wanton burrow]

x is equal to that

but

what’s ur horiz

[wicked sluice]

not sure tbh

[wanton burrow]

Asymptote

[wicked sluice]

is it just 0

[wanton burrow]

y=0

bc it’s bottom heavy

[wicked sluice]

could u quickly explain why that is pls

[wanton burrow]

degree on denominator

Is larger

[wicked sluice]

ohhh alright

so whenever denominator is larger do i assume y= 0

[jovial oar]

Range

[wanton burrow]

What

That’s wrong

it’s just y cannot equal 0

Is your range

[jovial oar]

why tho

[jovial oar]

this mean except 0 and from 0 to inf

[wanton burrow]

why cannot it not equal -.31

Either or

Both are acceptable

For the range

[jovial oar]

table it

idk

[wanton burrow]

I’m like 99 percent sure

[jovial oar]

exract the bottom

[wicked sluice]

these r the solutions

im not too sure how there is like a second graph below the x axis

[wanton burrow]

oh wait ur right

about range

[jovial oar]

smallest value of the equation is -3.125 at x= -1/4

[wanton burrow]

Domain ur still wrong

but

[jovial oar]

fr?

[wanton burrow]

u forgot (-3/2,1)

but that’s all

[wicked sluice]

so that quadratic at the bottom

[jovial oar]

wdym by that

[wicked sluice]

is that the (x-1)(2x+3)

[wanton burrow]

like

[jovial oar]

yess

[wanton burrow]

.

[wicked sluice]

ohh how come its flipped

[jovial oar]

i did all of that on my caculator

wonder where i got it wrong

[wanton burrow]

idk that’s why set builder

notation

is easier for these

[jovial oar]

ig im not wrong with those range and domain tho

[wanton burrow]

ur right for range

but for domain it’s just x cannot equal -3/2 and 1

it can still equal 0

There is a y intercept

on this graph

and ur domain implies there isn’t

[jovial oar]

oh

Domain

[jovial oar]

what flipped tho

[wicked sluice]

its a negative

like it has a minimum value

nvm i got it

thanks

[jovial oar]

nvm

[pearl pond]

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[lone nebula]

Hi, is this even correct

[summer gorge]

No

What is the question about

[lone nebula]

If its 12sinx -5cosx=13 find sinx and cosx

[summer gorge]

There is an error in ur expansion of (13+5b)²

[lone nebula]

How

[summer gorge]

it is 169+130b+25(b)^2

[lone nebula]

Aaa

Okay

[summer gorge]

after that use the quadradtic eq to slove for cos x

[lone nebula]

Okay

[summer gorge]

take the vale that belongs to[-1,1]

*value

[lone nebula]

Okay

Wait

[summer gorge]

its (13b+5)^2 this time

srry my bad

ur right i didnt see properly

now u can plug it the eq 12a-5b=13 to get a

[pearl pond]

@lone nebula Has your question been resolved?

[pearl pond]

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[vernal grove]

For a) this is what I did

I found the prob of finding no defective bulb which was

(2 (48 (50
0) 5) / 5)

And then subtracted that by 1

so 1 - 0.8081 = 0.19...

but for part b

Im not sure how to do it

[tall flint]

$\frac{\binom 20 \binom{48}{5}}{\binom{50}{5}}$

[jolly parrot]

Steakanator

[tall flint]

that?

[vernal grove]

yuhh tysm

i dint know how to latex that

I subtracted that by 1 to find the prob of finding atleast one defectrive bulb

thats correct right?

[tall flint]

seems fine

[vernal grove]

but now for part b

could you give me a hint

[tall flint]

start by generalizing part a to any number of bulbs chosen, not specifically 5

[vernal grove]

how would I do that

do I just sub in a variable for 5 ?

which can be anything?

lol im not sure

[tall flint]

go through the same thought process that got you to your first answer

if the inspector chooses n light bulbs at random without replacement, what is the probability he finds at least one defect?

[vernal grove]

$1 - \frac{\binom 20 \binom{48}{n}}{\binom{50}{n}}$

[jolly parrot]

Calc III Victim

[vernal grove]

this?

[tall flint]

alright

the question asks you to find n such that ^ >= 1/2

[vernal grove]

so I just keep trying random values?

[tall flint]

seems inefficient

do algebra, not guesswork

[vernal grove]

ohhhh

$1 - \frac{\frac{2!}{1(2)!} \cdot \frac{48!}{n!(48-n)!}}{\frac{50!}{n!(50-n)!}} > \frac{1}{2}$

and then solve for n?

[tall flint]

i take issue with this term

[vernal grove]

0! = 1 no?

oh im dumb

[jolly parrot]

Calc III Victim

[tall flint]

better

[vernal grove]

alr gimme a fe wmins imma try solving and then get back to u

[tall flint]

[vernal grove]

$- \frac{48!}{n!(48-n)!} \cdot \frac{n!(50-n)!}{50!} > -\frac{1}{2}$

[jolly parrot]

Calc III Victim

[vernal grove]

one sec imma latex the whole thing bcz my handwriting is shittttt

$= - \frac{48!(50-n)!}{50!(48-n)!} > -\frac{1}{2}$

[jolly parrot]

Calc III Victim

[vernal grove]

$=-\frac{(50-n)(49-n)}{50 \cdot 49} > -\frac{1}{2}$

[jolly parrot]

Calc III Victim

[vernal grove]

$= -\frac{2450 - 99n + n^2}{2450} > -\frac{1}{2}$

[jolly parrot]

Calc III Victim

[vernal grove]

$=\frac{-2450 + 99n - n^2}{2450} > -frac{1}{2}$

[jolly parrot]

Calc III Victim

[vernal grove]

$= -1 + \frac{n(99 - n)}{2450} > -\frac{1}{2}$

[jolly parrot]

Calc III Victim

[vernal grove]

$\frac{n(99 - n)}{2450} > \frac{1}{2}$

[jolly parrot]

Calc III Victim

[vernal grove]

$n(99 - n) > \frac{2450}{2}$

[jolly parrot]

Calc III Victim

[vernal grove]

now what

😭

@tall flint

[tall flint]

Now you solve the quadratic

[vernal grove]

How does that go again

I forgot

been so long

[tall flint]

Well if you keep the algebra going you end up with $$-n^2 + 99n - 1225 >0$$

[jolly parrot]

Steakanator

[tall flint]

So it's more or less just finding the roots

[vernal grove]

ohhhh

tysmmmmm

bless u

.close

[pearl pond]

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[vernal grove]

.close

.reopen

[pearl pond]

✅

[vernal grove]

But then I will get 2 diff values

how will I know which one is the solution

[warm current]

only positive

[vernal grove]

Im getting 14.496 and 84.503

[warm current]

oh

[vernal grove]

both are positive

so it could be both?

[warm current]

hm. no clue

[vernal grove]

oh wait

im actually an idiot

ofc it cant be 84

we have 50 bulbs to choose from

smhhh 🤦

[tall flint]

Indeed

https://tenor.com/view/its-shrimple-shrimp-funny-haha-meme-gif-9312733212572166512

[vernal grove]

one more issue for part a in my book this is what the solution says

this wont be the same as mine right

[tall flint]

It is the same

[vernal grove]

oh how

[tall flint]

Your solution just includes an extra $\binom 20$

[vernal grove]

oh wait

[jolly parrot]

Steakanator

[vernal grove]

(2 0) = 1 im dumb

tysm

.close

[pearl pond]

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[pearl pond]

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[wild fable]

I’m trying to calculate Var(x) for f(x) but I don’t know what went wrong cuz it doesn’t give me the right answer

[rough forge]

Isn't Var(X) = E(X²) - E(X)²

[wild fable]

i was using this formula

$Var(X) = \int_{a}^{b} x^2 f(x) , dx - \mu^2$

[jolly parrot]

water beam

[wild fable]

whats E(X^2) how do i find that?

[rough forge]

Oh it's actually this lmao

[wild fable]

yeah

[rough forge]

They are the same

I think

You subtracted twice μ²

[wild fable]

dont i have to calculate the Var(X) for each of the parts in the piecewise separately and add them up thats why theres to mews sqwuared

[rough forge]

,,Var(X) = \left ( \int_0^2 \frac{x^2}{8} \dd x + \int_2^5 \frac{x^2}{4} \dd x \right ) - \mu^2

[jolly parrot]

𝔸dωn𝓲²s

[rough forge]

I dont even think thats really true

It should be R instead of a to b

[wild fable]

ohh ok

i got the right answer with this

[rough forge]

I would stick with this then

[jolly parrot]

𝔸dωn𝓲²s

[wild fable]

whats the difference with fsubX of x? and just f(x)?

[rough forge]

Notice for R\[0,2]U[2,5] the integral is 0

[scenic aurora]

just a notation to denote that this is density from random variable $X$

[jolly parrot]

print("NAME")

[wild fable]

o

ok

btw this was just a precursor to the actual question i wanted to ask 💀 i just had to complete this before asking the other one

wait

shit

[scenic aurora]

ok...

[wild fable]

question 6 i just completed wasnt exercise 16A

it was exercise 16C

💀

my bad chat

[scenic aurora]

at least you prepare for your exam with that one extra question

[wild fable]

yeah lol

anyways i dont even have exercise 16A on this paper 💀

.close

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[teal hazel]

what do i do?

[summer gorge]

!status

[pearl pond]

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 was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[teal hazel]

1

[teal hazel]

1

[late venture]

whats your first thought in tryuing to solve this ?

@teal hazel

[summer gorge]

k

[teal hazel]

100C1+100c1+100C!

1

[late venture]

and what is C1 ?

[teal hazel]

combination 1

[late venture]

and how did u think of this equation, just trying to understand your thought procces

[teal hazel]

I thought you get one mark from each

[late venture]

you recieve an integer mark from 1-100 folr each category

so all the natural numbers in the interval [1,100]

[teal hazel]

I dont understand

[elfin patio]

mhmm let me see

to score a total mark of 200, the marks for design, technique, and originality must add up to 200, with each category scoring between between 0 and 100.

[teal hazel]

yh

[elfin patio]

lets use a generating functions approach to solve this problem. we can set up the generating function for each other category as follows

[teal hazel]

what is a generating function??

[elfin patio]

a generating function is a formal power series used to encode a sequence of numbers.

[teal hazel]

one question

does this use combinations?

[elfin patio]

yes

[teal hazel]

How do u do it?

[elfin patio]

wait i will tell you

[surreal smelt]

@teal hazel

[teal hazel]

Woahh

Thanks a lot for the effort

[surreal smelt]

don't thank me chatGPT gave me the idea 😂

[summer gorge]

nice solution regard less

[teal hazel]

The power of ai ❤️

Thanks aswell Zaakoo

.close

[pearl pond]

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[pearl pond]

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[kindred flame]

Can someone explain what part I did wrong ?

[surreal smelt]

looks correct to me

[kindred flame]

i put it into the answer book like this and it said it was incorrect

ill send ss

the above was incorrect when i did the "check" so im not sure what part i missed...

[surreal smelt]

maybe the ± needs to be next to 2sqrt(2)

[kindred flame]

ohh ur right

that was a silly little mistake, thank you for noticing 🙂

.closed

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[pearl pond]

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[pearl pond]

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[quasi turtle]

So if I can scale each vector such that the sum is zero, its independent

[midnight haven]

you need the concept?

[quasi turtle]

I'm asking if I understand whats been written or not

Nm i get it

.close

[pearl pond]

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[pearl pond]

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• Show your work, and if possible, explain where you are stuck.
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[cerulean smelt]

Determine the magnitude of an earthquake with approximately twice the intensity of Tremor
measuring 3.5 on Richter scale?
can someone help me with this?

[cerulean smelt]

M = log(I/I0)

this is the eq i know

is the equation like

3.5*2 = log(I/I0)

wait 3.5 is intensity not magnitude

so M = log(7/I0)

?

im not sure

<@&286206848099549185>

[sacred bough]

m

R=log I/I0

​

I0 = 3.5

I = 2 x 3.5 = 7

[cerulean smelt]

ok I get this part

what do I do next?

M = log(2)