help-23

[sinful iris]

ohhh

then i use the corresponding trig identities?

[rigid inlet]

rewriting them in terms of a, b, c will doubtlessly be helpful

[sinful iris]

like this?

damn the quality

[rigid inlet]

something like that

it depends on how exactly you labelled your triangle

[sinful iris]

a hipotenuse

c adyacent

[rigid inlet]

sure

now starting with this term

here's where you use the thing you were asked to use

[sinful iris]

a^2 = b^2 + c^2

?

[rigid inlet]

that's the thing you were asked to use

[sinful iris]

yeah

uhh

so am i supposed to find b and a

in the given formula

[rigid inlet]

you can't

[sinful iris]

nono

in a^2

= b^2 + c^2

[rigid inlet]

it'd be helpful to rewrite this as b^2/a^2

[sinful iris]

just to know

i dont know what im doing here

do i have to prove

this is real

[rigid inlet]

that's what you're doing right now

[sinful iris]

okok

[sinful iris]

so now that ive done this

i rewrite a^2 as

b^2 + c^2?

[rigid inlet]

which leaves you with?

[sinful iris]

b^2 / (b^2 + c^2)

[rigid inlet]

b^2 / (b^2 + c^2)

alright then

what do you think comes next?

[sinful iris]

do that with the other sides probably

[rigid inlet]

what's "that"

[sinful iris]

rewrite them using this

[rigid inlet]

well there is no a^2 on the other side

[sinful iris]

would it be helpful to use c^2 = a^2 - b^2

[rigid inlet]

not really

[sinful iris]

then idk how to proceed

[rigid inlet]

this is your current right side

[sinful iris]

yes

[rigid inlet]

this is your current left side

$\frac{b^2}{b^2+c^2} \overset{?}{=} \frac{ \frac{b^2}{c^2}}{1 + \frac{b^2}{c^2}}$

[sinful iris]

yeah

ive distributed the square in the parenthesis

[flat frigate]

Steakanator

[sinful iris]

for the rhs

[rigid inlet]

let's make it a little clearer

now what can you do to the left side so that it looks the exact same as the right side?

[sinful iris]

divide it

by c^2?

[rigid inlet]

what's "it"

[sinful iris]

numerator and denominator

[rigid inlet]

precisely

[sinful iris]

im not sure tho

alright so if i do that

the only left to prove is the far right one?

[rigid inlet]

seems like it

[sinful iris]

do i

multiply

denominator and numerator

far right side

by b^2/c^2

[rigid inlet]

that works

[sinful iris]

then its done?

[rigid inlet]

do you get the same thing everywhere?

[sinful iris]

yup

[rigid inlet]

then what else is there to do?

[sinful iris]

nothing

[rigid inlet]

nothing

[sinful iris]

alright

ty

.solved

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[quiet plume]

Try 00

[scarlet comet]

S->UTU->0U0TU->00TU->00U->00, ?

Omg

S->UTU->TU->U->0U0->00

[scarlet comet]

Thank you bro

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[restive laurel]

ive been lost on this for a while and not sure what i did wrong

[restive laurel]

i understood its a linear differential equation

got the Integrating factor as x+1-ln(x+1)

but tryna integrate

e^-x * the IF

is a bit annoyin

and hwen i checked with an online calc

it did some wierd sutf

[quiet plume]

Can you show your work for the integrating factor?

If you put the ODE in the form $\frac{dy}{dx} + p(x)y = f(x)$, you should get the integrating factor as $e^{\int p(x) dx}$, so something went wrong trying to get it.

[flat frigate]

Azyrashacorki

[safe radish]

@restive laurel Has your question been resolved?

[royal kiln]

@restive laurel The best explanation of linear differential equations on the internet:
https://tutorial.math.lamar.edu/classes/de/linear.aspx

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[inland flax]

Where did I make a mistake?

[inland flax]

I just found the answer and i was wronggg

[rough storm]

it may help NOT to expand immediately

[inland flax]

yeah that's why i factored again 😭

[rough storm]

i wouldn't expand the denominator

so in this step

[inland flax]

oh dang

[rough storm]

just expand out the numerator and note you'll have an extra factor of 'h' that will cancel

:3

[inland flax]

i just noticed that if i expand the numerator

yeah i got it!

thanks!!!

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[quiet plume]

Because (0.75 + 0.25*0.75) != 1.
If I understand, you would want to compute a quarter of 35.36 to add that onto the total. The issue is that the actual value of Q is not 35.36, but 47.15.
Thus, a quarter of 35.36 does not correspond to a quarter of 47.15.

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@sick urchin Has your question been resolved?

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[zenith hare]

Anyone know what is the name of this function? or at least what can I call it? I need to call it with something

f(1) = 1
f(2) = 1+2 = 3
f(3) = 1+2+3 = 6
f(4) = 1+2+3+4 = 10
f(5) = 1+2+3+4+5 = 15
f(6) = 1+2+3+4+5+6 = 21

[finite igloo]

uhh

summation function?

[flat frigate]

:D

[zenith hare]

oh!

thx

[lean otter]

I need help with quadratic function. Its really a basic concept question but why is y=x^2 a parabola shap in the graph? Like why do we also get the negative side part as well? Why does it have to be symmetrical in the graph.

[finite igloo]

what

[finite igloo]

the shape is because of the function

wdym

the symmetry is cuz even function

or f(-x) = f(x)

[lean otter]

but why do we need to do f(-x)?

cant we just only draw

f(x)

[finite igloo]

ok if (-x) = f(x)

then the function is the same for both left and right of the y-axis

[zenith hare]

you only need to draw one of them since they basically the same

as such drawing f(x) also mean you have drawn f(-x)

I'm closing this

.close

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[wicked nova]

I'm having some problems with this proof. I get this far then I don't know what to do with this extra 1

[rotund crest]

any log with 1 equals 0

cause of the index law

x^0=1

[wicked nova]

So ln(1)=0

[rotund crest]

ye

[wicked nova]

Thanks

[rotund crest]

np

[wicked nova]

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[thorn bough]

so i have this expression that im stuck on:
64^4 :(2^2 * 4^2) - [(12^5 : 3^5) * 4]:16^2
i did some passages:
64^4:(8^2)^3 - [(4^5)*4]:16^2=
64^4 : 8^6 - 4^6 : 16^2
the problem is that im stuck here bc i dont understand what to do after,the solution of this should be 48(its writed on my book) but i dont know what to do

[thorn bough]

like this is what i did for now

[fathom adder]

4^5 * 4 = 4^6

64 = 8^2

16= 4^2

[thorn bough]

aaa yes!ur right sorry but i have adhd and i sometimes do wrongs on this

thank u now i understood

[fathom adder]

Keep going, its all good otherwise

[thorn bough]

thank u sm!

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[burnt notch]

You speak English very well, considering your age

[thorn bough]

i’ve learned it since i was like 6,im abt to turn 16 and i have like a C1 level based on a test of cambridge that i did 2 years ago in my school haha

[burnt notch]

Wow nice!

[storm maple]

Che classe fai?

[thorn bough]

fai conto che sono stata rimandata in matematica quest’anno ma sarei in 2 superiore

[storm maple]

Ma tu sei il mod nel gruppo telegram help matematica/fisica/chimica?

[burnt notch]

Che superiori?

[burnt notch]

Sì haha

[thorn bough]

allora tecnico grafico,ma cambio scuola e a settembre vado a fare un professionale socio sanitario

[storm maple]

Grande

[burnt notch]

Ah ok ok

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[rough igloo]

can some pls help😭😭

[flat jolt]

are the cordinates for the midpoint of PR given?

wait nvm

we can derive it easily

[fathom adder]

Its (1,3)

Could it lead somewhere with vector dot product

[flat jolt]

i dont think so

[fathom adder]

Like PR • QI

[flat jolt]

maybe properties of traingle will apply

[fathom adder]

Pythagore ?

[flat jolt]

nah

maybe the midpoint of PR is the circumcentre of triangle PQR?

nah mb it wont help

wait

i got it

find the slope of PR

and QS is perpendicular to PR

so u can find the slope of QS too

from there i think we can get the value of q through equation of straight line

@rough igloo

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@rough igloo Has your question been resolved?

[fathom adder]

Original

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[west hedge]

What is binomial variable

[west hedge]

Is it related to binomial theorem

[light shoal]

binomial random variable? or something else

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[high aspen]

Hey guys, I'm in a discrete math class and I'm being asked this and I have no idea how to solve this or figure out how to. Could anyone help me?

[tardy mango]

Are you familar w/ the process of re-indexing a sum?

That's what this question is asking you to do

[high aspen]

The lectures didn't cover how to do that

[tardy mango]

o

[high aspen]

So I'm not sure how to

[tardy mango]

The idea is that $$\sum^{k}{n=1} a_n=\sum^{k+r}{n=r+1} a_{n-r}$$

[flat frigate]

Civil Service Pigeon

[tardy mango]

ex. let's take k=2 and r=2

Then, the sum on the left is $a_1+a_2$

[flat frigate]

Civil Service Pigeon

[tardy mango]

and the sum on the right is $\sum^{4}{n=3} a{n-3}=a_{3-2}+a_{4-2}=a_1 +a_2$

[flat frigate]

Civil Service Pigeon

[tardy mango]

In essence, the idea is that you're just "shifting" all the terms by some amount

and adjusting the bounds of the sum accordingly

[high aspen]

Ohhhhhkay

[tardy mango]

(btw, r can be positive or negative here, as long as the bounds still make sense)

ex. we could've taken r to be -1 as well

[high aspen]

Thank you, that helps a lot

I was so confused

[tardy mango]

[high aspen]

.close

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[candid shoal]

Could anyone help me with this Integration by Substitution problem? It has something to do with filling in the integral as f(u)du, but I can't figure out how to handle it when the actual du isn't the same as the du actually there in the function

[candid shoal]

in the top left is the initial problem, and in the top right is the rule I'm trying to apply to it

[tardy mango]

I'd just sub $u=\frac{1}{\theta}$

[flat frigate]

Civil Service Pigeon

[tardy mango]

It helps to build intuition for this if you rewrite the fractions using negative exponents

[candid shoal]

I did do that, with integral $theta^(-2)cos(theta^(-1))$

I did not format that right

[tardy mango]

ik what you meant

dw

[candid shoal]

alright, thank you

but so yeah, doing that, I saw that the thing outside the cos was the derivative of the thing inside the cos

so then I figured cos theta ^-1 was my u

I think part of why I'm struggling is because a lot of the stuff here is kind of arbitrary. Like I can't really think of a mathematically provable reason as to why I should pick what as my f(u) and du for any given problem, and then I really don't know what to do after I actually calculate the du from my f(u)

By making u just theta^-1 I was able to make this. I still don't know what to actually do next though

[tardy mango]

I gotta head off soon, but recall that $$\frac{d}{dx} \sin(x)=\cos(x)$$

[flat frigate]

Civil Service Pigeon

[candid shoal]

right, then doing it again, cos becomes -sin

[tardy mango]

you forgot a negative sign on here tho

[candid shoal]

you're right, thank you for that

[tardy mango]

👍

imma head off now

anything else?

[candid shoal]

I think I need to make what I have in the integral into f(u)du, so that then I can take whatever if my f(u) in that statement and integrate it. But right now the u is stuck in that cosine, and I don't know how to get it out

or if that's even when I need to do next

[tardy mango]

find the integral in terms of u and then substitute $u=\frac{1}{\theta}$ back in

[flat frigate]

Civil Service Pigeon

[tardy mango]

In general, when you integrate by substitution, find the integral in terms of the substituted variable and then replace the substituted variable with the original definition right at the end

ex. if you have an integral that's originally in terms of x and you then substitute a variable u, you would find the integral in terms of u and then substitute the definition for u in terms of x back in

[candid shoal]

so wait, does that mean that I can just integrate what I have now and then undo the substitutes?

even though it's not exactly f(u)du?

[tardy mango]

Rewrite it as $-\int \cos(u) \dd{u}$ if that makes you happy

[flat frigate]

Civil Service Pigeon

[candid shoal]

the contents of du is negative, does that mean I can jsut pull the negative sign out like that?

Is it still substitution if I'm picking pieces of the substitution out of it?

[tardy mango]

You can factor constants out of an integral

Take that as you will

[candid shoal]

can the constants be additives or do they have to be multiples? Can you even get an additive inside an integral?

[tardy mango]

??

Wdym by “additives”

[candid shoal]

Like

Extracting a constant multiple is taking a number like 4 and pulling it out of the integral, but that's when the 4 is being multiplied with the rest of the function

What if the 4 was a constant being added inside the integral instead?

[candid shoal]

what if instead of kf(x)dx, you had (k+f(x))dx
could it still be factored out then?

probably not, I guess

[tardy mango]

You could do $k \int \left[1+\frac{f(x)}{k} \right] \dd{x}$ ig

[flat frigate]

Civil Service Pigeon

[tardy mango]

alright I rlly gotta go now

[candid shoal]

alright, sorry for keeping you so long

thank you for your help, I really appreciate it

.close

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[spiral saddle]

Ping me

[terse lichen]

integration is the reverse of differentiation

[craggy sedge]

Ren integrate f''(x) then use the fact that f'(0)=2 to find the value of the constant (c) and repeat it again to find f(x) and use the fact that f(0) =2 to find the constant then you are done

[spiral saddle]

can someone check my stuff, where did i mess up

[craggy sedge]

You did right lol

[spiral saddle]

For real?

[craggy sedge]

yes this is the 4th option

[spiral saddle]

Ty

.close

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✅

[spiral saddle]

ping me

[flat frigate]

938c2cc0dcc05f2b68c4287040cfcf71

[safe radish]

@spiral saddle Has your question been resolved?

[drifting knot]

&8!_ the intersection points

You draw the curves

You will understand

[spiral saddle]

I already found the intersection points

0, 2, 4

[drifting knot]

Draw the curve

[spiral saddle]

how

[drifting knot]

e^-2.e^x

[spiral saddle]

how do I draw the curves, I only have 1-2 minutes to solve this

[drifting knot]

See u know curve of e^x

You need the curves

Cz it will be ez to see

Which one is the answer

You know

e^x

[spiral saddle]

im dont know

[drifting knot]

Ok see

Then check which curve is greater in that interval

The intersection point interval

U need higher-lower

[spiral saddle]

,w plot e^x

[flat frigate]

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[drifting knot]

,w plot e^x/2

[flat frigate]

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[spiral saddle]

how do I prove it

[drifting knot]

Na

See

[spiral saddle]

how to prove it

[drifting knot]

w,plot e^(x/2)

,w plot e^(x/2)

[flat frigate]

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[drifting knot]

Now can u see it

,w plot e^(x-2)

[flat frigate]

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[drifting knot]

Can u see b/w intersection points which point is above or below

That is b/w 2 and 4

Check which curve is higher and which curve is lower b/w intersection points

@

[spiral saddle]

i did a little table from 0 to 2 to 4

[drifting knot]

Ok so e^x/2 is greater

Ok

[spiral saddle]

but this is not rigurous proof that g(x) > f(x)

[drifting knot]

Oh you need this

You should have told me

See let h(x)=g(x)-f(x)

See u know integral of e^(x/2) and e^(x-2) ?

[spiral saddle]

mmm

unsure

[drifting knot]

See the idea is that

We will let

h(x)=integral of e^(x/2)-integral of e^(x-2)

[spiral saddle]

let what

[drifting knot]

Then we diff. H(x)

[spiral saddle]

definite integral or indefinite

[drifting knot]

definite

[spiral saddle]

you are not indicating the upper and lower bounds

how did you know that g(x) > f(x) btw

[drifting knot]

I mean indefinite ofc*

I will tell

Wait wait let me recall

Yes got it

See let h(x)=g(x)-f(x)

Then diff w.r.t x

[spiral saddle]

why g(x) - f(x)

[drifting knot]

And apply condition for increasing function that is h'(x)>0

I will explain

Apply

[spiral saddle]

fair

[drifting knot]

did u do ut

Tell

The

[spiral saddle]

,, h(x) = \int e^{\frac{x}{2}}dx - \int e^{x-2} dx

[drifting knot]

See

Can u tell texit functions

I will explain ezily

Tell me*

See I'm explain

[spiral saddle]

what do u want to know, how to compile latex?

[flat frigate]

938c2cc0dcc05f2b68c4287040cfcf71

[drifting knot]

,w d/dx e^x

[flat frigate]

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[drifting knot]

Ok got it

[spiral saddle]

fair

what now?

[drifting knot]

,w d/dx[h(x)=e^(x/2)-e^(x-2)]

[flat frigate]

Results provided by WolframAlpha

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[drifting knot]

Na bruh

,w d/dx (e^(x/2)-e^(x-2))

[flat frigate]

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[drifting knot]

Yes now see

this is h'(x)

[spiral saddle]

see what

[drifting knot]

We took h(x) as e^x/2 - e^(x-2)

I took its derivative so it is h'(x) now if apply increasing function that is h'(x)>0 and find interval in which it is increasing

Find

w,

w,help

[spiral saddle]

why do we care if its inscreasing

both f(x) and g(x) are increasing

but that doesnt prove that g(x) > f(x) 😭

[drifting knot]

See if h'(x)>0

We have h(x) as increasing in that interval

[hardy narwhal]

te esta diciendo que como la derivada de la resta es positiva implica que la funcion es creciente

la h(x)

[drifting knot]

No bro

We can't understand

[hardy narwhal]

.-.

[drifting knot]

U know that if x1>x2
And function is increasing

Then f(x1)>f(x2)

We checking the same thing

We gonna find when h'(x)>0

And that the x where h(x) is equal to 0

We gonna compare

[spiral saddle]

ahh

okay thats smart, thank u

so h(x) = g(x) - f(x)

and h'(x) > 0

mmm

[hardy narwhal]

para no molestarlos mucho, que area se supone que esta pidiendo?

[spiral saddle]

,, h(x) = g(x) - f(x) \ h'(x) = \frac{e^{\frac{x}{2}}}{2} - e^{x-2}

[flat frigate]

938c2cc0dcc05f2b68c4287040cfcf71

[spiral saddle]

can someone help I am still hardstuck

[hardy narwhal]

can i try it

is a question

[spiral saddle]

https://media.discordapp.net/attachments/903481106819612714/1264449662559981578/1187916384341078070.png?ex=669dea03&is=669c9883&hm=09319cb97691c960676b8db7a8c7f92c3c62efbc0240821b2074fa4aeeaf8302&

[hardy narwhal]

I drew the three graphs in geogebra and I want to know which area is important

blue is g(x) and green is f(x)

[spiral saddle]

Unsure

[hardy narwhal]

.-.

[spiral saddle]

I just know the intersection points are 0 2 4

[hardy narwhal]

in the drawing if we take the function h(x) = g(x) - f(x) in the interval [0,4] that we are taking and then we integrate, we are calculating this area

do you agree?

Suppose we are asked for this area

[spiral saddle]

Fair fair

[hardy narwhal]

How do we remove the bottom area would be the question

[spiral saddle]

1 to 6

[hardy narwhal]

I didn't understand

could u make a draw

and send

[spiral saddle]

Idk what i am doing with my life

[hardy narwhal]

I also don't know what I'm doing trying to help when my English is terrible and I have to use the translator

but i think que right answer is the a)

[spiral saddle]

okay im ready lets do this

depression wore off

im ready for tackling this

[hardy narwhal]

I made this drawing

[spiral saddle]

no

first definite integral is between 0 and 2 doe

[hardy narwhal]

whats mean doe

[spiral saddle]

that part aswell no?

ohh

im so stupid

im stupid

im stupid

[hardy narwhal]

if we see splits the first integral

[spiral saddle]

your drawing is correct and mine is stupid

[hardy narwhal]

We can see it more safely

[spiral saddle]

is it possible without drowing

drawing the curves*

[hardy narwhal]

Yes, this is just to gain intuition and continue with formal work

[spiral saddle]

fair

how to proceed

[hardy narwhal]

i have no idea

do u need prove something?

[spiral saddle]

no

how did u found a

[hardy narwhal]

only with the drawings

[safe radish]

@spiral saddle Has your question been resolved?

[spiral saddle]

mmm

[jade mountain]

wait what’s ur question

i can try to help

[spiral saddle]

is about finding area between three curves

[rocky quarry]

try sketching the functions if you haven't already, that would help you out alot

[craggy sedge]

!occupied

[safe radish]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[safe radish]

@spiral saddle Has your question been resolved?

[safe radish]

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[unborn kayak]

hello everyone, im a 9th grader and currently studying exponent, i need someone to explain to me about the answer and how to get to that answer. thank you :)

[violet stratus]

so 2^(n+3) = 2^n * 2^3 right?

[unborn kayak]

ah yes

[drifting knot]

Yes

[violet stratus]

then you have 8*2^n + 2^n at the numerator
and you have 4*2^n - 2^n at the denominator right?

[unborn kayak]

wait im confused, this is the b equation righht?

[violet stratus]

yes

[unborn kayak]

cuz theres 2 equatopn

aighht

[unborn kayak]

'ah okay alr go on

[violet stratus]

so numerator is2^n(8+1)
and denominator is 2^n(4-1)
I guess you can do the rest

[unborn kayak]

okay thank you

can u do thhe c equation aswell?

[cyan vale]

what have u done with part c

$\frac{3^{2024} - 3^{2022} + 16}{3^{2022}+2}$

[unborn kayak]

just like

wait

[flat frigate]

Newt

[cyan vale]

hint :$$3^{2024} - 3^{2022} = 3^{2022}(3^2-1)$$

[unborn kayak]

thhis is confusing

[flat frigate]

Newt

[cyan vale]

why?

$\frac{3^{2022}(3^2-1) + 16}{3^{2022}+2}$

[flat frigate]

Newt

[unborn kayak]

wheres the -1 comes from

[cyan vale]

remeber distribution?

$ac-ab = a(c-b)$

[flat frigate]

Newt

[unborn kayak]

oh yes

[cyan vale]

so in this case a = 3^2022

[unborn kayak]

ohhhh

wait lemme take it for a sec

[cyan vale]

All easy from here

[unborn kayak]

can u jus do this or

so its jus this now?

sorry bad writing

[cyan vale]

no wait

[unborn kayak]

?

[cyan vale]

$\frac{3^{2022} \times 8 + 16}{3^{2022}+2}$

agree?

[flat frigate]

Newt

[unborn kayak]

yep

then

[cyan vale]

don't do calculation if it not simplified

[unborn kayak]

ohhh

[cyan vale]

$\frac{8 \times (3^{2022} + 2)}{3^{2022}+2}$

[flat frigate]

Newt

[unborn kayak]

so thhe answer its jus 8?

[cyan vale]

yeah

[unborn kayak]

what is the process called in this

[cyan vale]

$3^{2022} \times 8 + 16= 8 \times (3^{2022} + 2)$

[cyan vale]

distribution

[unborn kayak]

oh again

[cyan vale]

the $ab+ac=a(b+c)

[unborn kayak]

i rlly dont understand distribution lol

[flat frigate]

Newt

[cyan vale]

did you get this one?

[unborn kayak]

i get it

[drifting knot]

,w factors of 16

[flat frigate]

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[cyan vale]

?

[drifting knot]

16=8x2

8 is taken common

[unborn kayak]

uh

im still kinda confused on the b equation bcs i dont understand fractions to muc

kinda confusing for me

[drifting knot]

See

8x2+8x4 can't u take 8 common?

[unborn kayak]

huhh

@cyan vale is thhere a some kind of trick for doing distribution?

uh so where do i go from here?

please help me anyone

[violet stratus]

maybe replace 2^n with x?

[unborn kayak]

so

[violet stratus]

order of operations

[unborn kayak]

mb

omg

im so dumb

:(

this?

[violet stratus]

yes

[unborn kayak]

then what do i do next

x = 2^n

[violet stratus]

I mean if you have 8 apples and if you add one more apple what do you get

[unborn kayak]

do i convert it back again

[unborn kayak]

9x

[violet stratus]

so

[unborn kayak]

9x/3x?

[violet stratus]

yes

[unborn kayak]

so 3x

3 (2^n)

[violet stratus]

recheck

[unborn kayak]

uhh

wdym

[violet stratus]

do this again

9x/3x is true but the answer is not

[unborn kayak]

im confused

[unborn kayak]

so i convert x into 2^n back again?

[violet stratus]

you don't have to

what if you do 9x / 3x

[unborn kayak]

3x?

or im wrong

[unborn kayak]

so the answers just 3?

[violet stratus]

yes

[unborn kayak]

im mega confused

dam.

why did you jus divide the x

but not the entire thing?

[violet stratus]

just simplified the fraction

[unborn kayak]

is that some kind of trick or smth

dam

[violet stratus]

same thing is done when you do $\frac{3}{6} = \frac{1}{2}$

[flat frigate]

MetuMortis

[unborn kayak]

oh

u divide both by x

now i geet it

sorry for not understanding

.close

[safe radish]

Channel closed

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Use .reopen if this was a mistake.

[unborn kayak]

thank you for hhelping me all

[violet stratus]

no problem

[unborn kayak]

.close

[safe radish]

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[steep magnet]

I mentions a step 2 but I dont think its important for this question

What does it mean by 'independent of'? It means like if we change the value of p we get a different answer for the volume of the body B right?

I know that no matter what value of p we pick the volume of body B will always be the same, but does that mean its independent? Or not independent?

[obtuse jackal]

independent -> does not depend on

i.e. iff B(p) is a constant function

[steep magnet]

Ok so it is independent?

[obtuse jackal]

if you found so

[steep magnet]

Hmm it said its not independent

[steep magnet]

What was wrong with my reasoning?

I know I wasnt wrong in my calculations, no matter what value I put for p I always get 2

Oh wait

I see what i did wrong

.close

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[bold phoenix]

Help me with 9th grade maths

[drifting knot]

Here to help

[safe radish]

@bold phoenix Has your question been resolved?

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[green mica]

Hi

[green mica]

My teacher gave me both graphs of a function and its derivative but i am confused if I do the A to D for both of them or just the original function and the derivative is there just to help

[safe radish]

@green mica Has your question been resolved?

[green mica]

<@&286206848099549185>

[marsh walrus]

for the original function

the derivative is just helpful

[green mica]

Thanks

So are my answers correct

based on the original function i got these

if you could just verify

[marsh walrus]

no

[green mica]

oh

[marsh walrus]

you are to state the intervals of increase and decrease

you only have one

there is both a local maximum and local minimum

D looks right

[green mica]

What abt c

[marsh walrus]

wait

D isnt right

C is fine

[green mica]

Ok let me redo this and come back

[marsh walrus]

for concativity you need the derivative of the derivative right

but its just a line

[green mica]

Idk really

[marsh walrus]

sure you do

f'' > 0 means concave up

f'' < 0 means concave down

so whats the derivative of a parabola look like

its a line

whats the line look like in this case?

youre given the equation of the parabola

[green mica]

Ok i will come back and answer this

[marsh walrus]

okay

if you could ping

just in case

or i should say, feel free to ping

[green mica]

I need help with a

@marsh walrus

I think i figured out the rest

[marsh walrus]

sure yea

but A is still wrong

[green mica]

I know

i need help on A

[marsh walrus]

alright

so f' > 0 means f is increasing

and f' < 0 means f is decreasing

[green mica]

yes

[marsh walrus]

theyve given you a picture, though

we can just read off these intervals

where is the parabola greater than 0?

[green mica]

Positive infinity

[marsh walrus]

to be clear which intervals

the answer should be in interval form

[green mica]

x < 0

I dont know interval form

[marsh walrus]

I guess your form is fine

x < 0 is one of the places, yea

which others?

0 < x < 2 is where the function is decreasing

that leaves what else?

[green mica]

Wait i got it x < 0 and x > 2 is also increasing

[marsh walrus]

yea

so we'd usually write it like $( - \infty, 0) \cup (2, \infty)$

[flat frigate]

jan Niku

[marsh walrus]

at least, this is interval notation my teacher would like us to use

but maybe your class its not that important

you got the right answer either way

[green mica]

Ok I will see that is the rest right

[marsh walrus]

yea

i think you got it

[green mica]

Thank you

[marsh walrus]

np

[green mica]

.close

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[icy crystal]

what do I need to do to find the equation of a graph?

[red delta]

When x is 0 what is y from the graph

[icy crystal]

okay so it's always based on when x is zero?

[red delta]

No

But in this case that's all you need

[icy crystal]

okay so what will I need in other occasions?

[delicate sphinx]

The easiest method since you have answer choices and the graph, plug in a point that isn't an intercept and check it

[red delta]

Intercepts, gradients ,area it makes with x or y axis

[icy crystal]

okay

[red delta]

...

Depends on the question

[delicate sphinx]

So for example, using the graph, you know that if you did x = 3, then you should result in 32, so plug in 3 into all the equations and see which results in 32

[icy crystal]

ah okay

but the goal is to always find the graph where x equals zero?

[delicate sphinx]

Not exactly

[icy crystal]

okay so what should I attempt to solve for when it pertains to these questions?

[delicate sphinx]

You can use the y intercept to eliminate choices

[delicate sphinx]

.

[icy crystal]

ah okay so to put the y intercept in an equation?

[delicate sphinx]

.

As I stated, you can use the y intercept to eliminate options but it's not going to necessarily give you the final answer

[icy crystal]

okay thanks!

[delicate sphinx]

Because the y intercept of that graph is (0, 4) so you can have choices like y = 4(2)^x or y = 4(4)^x or y = 4(8)^x which all has the same y intercept so would need to use another point along that graph to check

[safe radish]

@icy crystal Has your question been resolved?

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[bleak cove]

I don't know where to start, I tried using derivatives but didn't really end up getting anything

I have no idea what to do now

[quiet plume]

Maybe same denominator -> l'Hopital would work? Have you tried something of the sort?

[bleak cove]

We're not supposed to use l'Hopital here

[safe radish]

@bleak cove Has your question been resolved?

[bleak cove]

<@&286206848099549185>

[balmy hearth]

?

[sleek lava]

these may be helpful for solving this

[bleak cove]

@balmy hearth

[balmy hearth]

yea what do you need help in this?

[bleak cove]

Solving it, there's an indeterminate form and i have no idea how to remove it

[balmy hearth]

do i evaluate the limit?

[bleak cove]

the limit is 1/2 btw but i have no idea how you get to that

[balmy hearth]

i cant solve it entirely mate the server has rules

i can explain it

[bleak cove]

Yeah sure please do

[safe radish]

@bleak cove Has your question been resolved?

[sleek lava]

howdy! i can help

if still needed

@bleak cove

[bleak cove]

Yes please

This is killing me

How would you approach the limit ?

[sleek lava]

i would first use the limit law sheet i posted earlier to split each fraction into a collection of limits

[bleak cove]

yeah but you still get indeterminate forms

[sleek lava]

yep! but we can continue to split it down from there

i would start by combining the fractions

if you havent yet

[bleak cove]

the worst part is, with Hopital its really easy but without isnt

[bleak cove]

i tried doing that but it doesn't seem to work

[sleek lava]

hopital is veryy nice

may i see your work with combining them?

[bleak cove]

Yeah but im not supposed to use it in this exercise unfortunately

[bleak cove]

Sure one sec

$test$

it gives $\frac{2^x - 1-x \ln(2)}{x \ln(2) (2^x -1)}$

[flat frigate]

crampter (prof univ AC)

[bleak cove]

Then i cant simplify anything

[sleek lava]

that shouldnt be multiplying on the bottom

wait nvm it should

next i would do the limits on the fraction

[bleak cove]

Yeah but they give 0/0

[sleek lava]

hmmm

i would normally just use l'hoptial here

😭

[bleak cove]

Same

Im sick of this i wanna stab my computer

.close

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[wind kite]

how do i rationalise this denominator

[crisp mortar]

What did you tried

[safe radish]

@wind kite Has your question been resolved?

[humble blade]

hi basically in rationalizing those kinds of fractions with radicals at the bottom u js have to multiply the numerator and denominator with that radical

[safe radish]

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[candid shoal]

I'm trying to find the function P based on its given derivative function, but what I got out doesn't match the answer, and for the second part, my answer is off by exactly 1000. What did I do wrong?

[candid shoal]

this is the original problem

[meager crow]

I think the c value is off

ln(1) = 0

So 1000 = 12000*0 + c. So c should be 1000

[candid shoal]

right, I figured out what I did wrong. When I was taking the 12000 out of my function when solving for C, I didn't divide C by 12000 at all. I figured I didn't have do, since C was an arbitrary value at the time, still not actually meaning anything

that was wrong though

[past cosmos]

p(3) = 1500t^2 + 1000 = 14500

This is my opinion

[candid shoal]

whatever it is, my answers match now, and I'm going to bed cause my brain is fried

goodnight all

.close

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[valid umbra]

Where did I mess up, the real asnwer is 1/6

[valid umbra]

I have not shown a lotta steps but I think you make still make out what I did

[pulsar pecan]

if you ever run into this problem in an exam, take each of the lines and put them into your calculator but replace every “x” by something like 0.0001

[valid umbra]

We aren't really allowed to use calculator in exams lol

But I'll try that right now because I am not in an exam

[pulsar pecan]

oh

[valid umbra]

Are all my steps correct? I used a lotta standard limits

Like lim x-> 0 sinx/x = 1 and lim x->0 (cosx-1)/x^2 = 1/2

Do you see where I made the mistake?

[golden verge]

I really can't tbh

Isn't this from vikas Gupta

[valid umbra]

I dont' know who is that

This is a jee mains problem

[golden verge]

Oh nvm

Let me think

[valid umbra]

k

[old chasm]

I don't understand a bit in the last part, your limit should be diverge given what you wrote

[valid umbra]

Which line?

[old chasm]

like, where did 1/4x^2 go?

[golden verge]

He combined it

[valid umbra]

I took lim x-> 0 common

[old chasm]

oh ok

[valid umbra]

and put the - sign inside to make it cosx-1

[golden verge]

Also limit is 1-cosx not cosx -1

Like the format

But still

Only changes the sign

[old chasm]

and why is sin(sin(x))cos(x) = cos(x)

[valid umbra]

oh yeah sorry I forgot

[golden verge]

He did a lot of intermediate steps

Divide and multiply by sin x

U should see why

[old chasm]

I don't think that can be simplify to that

[valid umbra]

Yeah that was a bit rusehd on my part so I did not show the steps

[golden verge]

It can

sin(sinx)/sinx

Is 1

Similarly sinx/x is 1

So residue is only cosx/x²

[old chasm]

sin(sin(x))cos(x) < cos(x)

[golden verge]

Yeah

[old chasm]

amplitude that is

[golden verge]

It's a limit

Wait

Let me show u

[valid umbra]

⬆️

[golden verge]

Like this

[old chasm]

but isn't cos(x)/4x^2 diverge

[golden verge]

So if x tends to 0 sinx also tends to 0

But he combined it

[old chasm]

like... I thought lim f(x)g(x) = lim f(x) lim g(x) only if both are finite

here one is not

[golden verge]

Oh right

True

[old chasm]

that's why I say how did you get there

[golden verge]

Agreed

The thing is

The way I'd solve this is by using Taylor series

That will give the answer

[old chasm]

I would also use taylor if I see this

[golden verge]

Yeah

[valid umbra]

so the simplification cannot be done there right?

[old chasm]

yes... you have to have finite limit on both

because one will dominate anyway

if you have one that is infinite

[valid umbra]

why did I even combine then, I should have just let that one approach 0 lol

1/4x^2 approches 0 right?

as x->0

[old chasm]

no

it goes to infinity

[valid umbra]

oh yeah

so it DNE

alright got it

my bad

I'll be more careful next time distributing limits

[golden verge]

Use Taylor series

Till 4th power

[valid umbra]

The teacher used cos(a) - cos(b) identity, then applied LH rule 2 times and then got the answer by simplifying

[valid umbra]

k

.close

[safe radish]

Channel closed

Closed by @valid umbra

Use .reopen if this was a mistake.

[safe radish]

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[rapid pendant]

hello

does anyone understand prefix tables and how to sum between two indexes

like if we turn abc into a prefix array [0, 1, 3, 6]

because a = 1, b = 2, c = 3, and prefix[1] = a, prefix[2] = a + b, prefix[3] = a + b + c

and we want to find the sum btween prefix 2 and 3

why TF are a lot of sources using prefix[3] - prefix[1] to do that

[timber night]

Wait could you give one of the sources?

[rapid pendant]

yea

[rapid pendant]

https://akd3257.medium.com/understanding-the-concept-of-prefix-sum-array-70fb333169a4

hold on that might be a bad example

i think this is just making the prefix array

i assume if left prefix index is 3, and right prefix index is 3, it should always be 0

why would we use the index of L-1

[terse lichen]

so you want to find prefix[2] + prefix[3] is the question?

[rapid pendant]

the sum between them

the sum between two prefixes

all sources i find online in regard to this are saying we subtract 1 from the left prefix index

[terse lichen]

so what im asking right?

sum between prefix 2 and 3

[rapid pendant]

i would imagine, but online sources seem to be implying that we would structure that like prefix[3] - prefix[1]

sum between index L and index R

but it's L-1

like im so confused rn

wait

im seeing this now

it's conflicting

but this makes more logical sense

im going to trust my gut and assume that a lot of online sources are incorrect

and likely relying on AI

because AI says the same incorrect thing every time

[terse lichen]

hopefully this makes sense

[rapid pendant]

i dont understand

[terse lichen]

so you want to find sum between 2 and 3

which in this case L = 2 and R = 3, their indexes

its the same as taking the sum of the whole array, subtracted by the sum of the first two elements

sum of first two elements is the L-1 index of the prefix array

and sum of whole array is R index of prefix array

[rapid pendant]

wouldnt prefix 2 be 3 and prefix 3 be 6

making the difference 3

[terse lichen]

prefix 2 = 3?

[rapid pendant]

no?

if L is 2 and L-1 is 1

L, or prefix 2, is 3 right

[terse lichen]

oh yeah

but you'll be taking L-1 index of the prefix array

so in the original array, your L is 2

thus L-1 is 1, taking the [1] element of the prefix array

[rapid pendant]

this would be L = 3 and R = 4 right

[terse lichen]

arrays count start from 0

[rapid pendant]

index-wise

oh yeah

my bad

i dont understand this

why are we doing L-1

[terse lichen]

ok maybe instead of using arrays we'll use equations

[rapid pendant]

ok

[terse lichen]

so you want to find 2 + 3

that is the same as (0 + 1 + 2 + 3) - (0 + 1) right?

[rapid pendant]

wouldnt it be (0 + 1 + 2 + 3) - (0 + 1 + 2)

[terse lichen]

that will be just equal to 3

but you want to find 2 + 3

[rapid pendant]

oh

i just want to find the difference between two prefixes

or the sum between them

which i think is the same thing

[terse lichen]

is the prefix here meaning the prefix array, or the original array

because prefix array is made from another array

[rapid pendant]

hmm

the question is posed as "difference between prefix l and prefix r"