help-10

[fallow cove]

idk how to start. Do it just make this equation in terms of y and then integrate?

[fallow cove]

ping if help thx

[timid silo]

polar coordinates

[fallow cove]

Haven't learn

[timid silo]

so basicly, let subsititute
x = rcos(theta)
y = rsin(theta)
and solve

[fallow cove]

what is r

[timid silo]

radius

[fallow cove]

ok

but first

[fallow cove]

why we sub this?

pls give explanation

[rich tapir]

yah correct, but remember we have 2 solutions for y

wrong message

oh nvm

[fallow cove]

huh

[timid silo]

so basicly, its the same as solving for y=y(x) then intigrating

but when u intigrate ud have to do a trig subsitituion

[fallow cove]

oh ok

[rich tapir]

what are you confused about

[fallow cove]

your messages

I thought you were helping another person

and also

[rich tapir]

Do it just make this equation in terms of y and then integrate?

[fallow cove]

this is 9(pi) right

[rich tapir]

i was just answering your question

[timid silo]

or what you can do is integrate 1 over the normal domain

[fallow cove]

after making the equation in terms of y I got y=(6-3x)/2

so what are the bounds when we integrate?

[timid silo]

actually i think this is easier

[fallow cove]

ok but what are the bounds?

is it 9 and 4

[timid silo]

$\int_{-\sqrt{9(1-\frac{x^2}{4}}}^{+\sqrt{9(1-\frac{x^2}{4}}} 4 dy$

[fallow cove]

what

[rich tapir]

[fallow cove]

I haven't learn multivar calculus yet

I'm just calc 2

[rich tapir]

we still answering the original question?

about the ellipse?

[fallow cove]

I guess?

I'm just calc 2 man

[rich tapir]

yah that's what i assumed, Jester is getting a bit ahead

[fallow cove]

yeah

what

can you just express this thing in dx only

[rich tapir]

well anyways, like you said, did you solve for y?

[fallow cove]

without the dy

[fallow cove]

yeah

[rich tapir]

what did you get

[fallow cove]

here

[timid silo]

lets goo

[fallow cove]

what

[rich tapir]

show your work, doesnt look correct

[timid silo]

so basicly this thing

[fallow cove]

,rccw

[fallow cove]

also seems incorrect to me

[fallow cove]

give me the video link'

[timid silo]

https://www.youtube.com/watch?v=Mw3MeDLL8d4

[fallow cove]

ok

thx

[fallow cove]

and also this

[rich tapir]

$\sqrt{36-9x} \neq \sqrt{36} - \sqrt{9x}$

[warm shale]

alihsaas

[rich tapir]

critical mistake

[fallow cove]

oh shit

no wonder

[fallow cove]

and also this

[rustic bough]

i mean ARea of ellipse = pi * a * b

[rich tapir]

so you know now how to solve the first one correctly?

[rustic bough]

im pretty sure area is = 6pi

[fallow cove]

yeah

the radius is 3 right

[rich tapir]

what?

[fallow cove]

huh

[rich tapir]

what's radius 3, x moves from -2 to +2 in original question

[obtuse pebble]

@fallow cove Has your question been resolved?

[rich tapir]

ok back

show your work @fallow cove

[obtuse pebble]

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[lapis spruce]

29b

[obtuse pebble]

@lapis spruce Has your question been resolved?

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[willow narwhal]

What's next? Should i multiply 1/2 and 25 to sin0 and cos0? And what do i need to look up when i want to solve for theta?

[brisk matrix]

use a trig identity

do you know another way to write 2sin(x)cos(x)

[willow narwhal]

Sadly no

[brisk matrix]

i suggest you look at the trig identities you’ve been taught

[willow narwhal]

I remember some but not all of them

[brisk matrix]

yes that’s why i said you should look at the ones you’ve been taught in the past (or just look up “double angle” laws)

[willow narwhal]

After that what's next?

[brisk matrix]

you integrate

the substitution will make it clear

[willow narwhal]

Is the way i integrated wrong?

[brisk matrix]

how did you go from the second to last, to the last line?

also you suddenly picked up an extra 5 after you square rooted

and the substitution should be x-5 = 5sin(theta), not x^2 - 5^2 = 5sin(theta)

[willow narwhal]

Wait i'm writing this down

Isn't that what i did in the third line?

[brisk matrix]

sure but not on the right

where you were writing the substitution

[willow narwhal]

Ohhhh i messed that up

How should i write that then?

[brisk matrix]

.

[willow narwhal]

I looked up how to integrate cos^2 theta and found a simplified answer 1/2 (sin0 * cos0) + c

.close

[obtuse pebble]

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[brisk matrix]

oh i misunderstood your question

[brisk matrix]

all you'd do is say x-5 = 5sin(theta)
so theta = arcsin((x-5)/5)

[obtuse pebble]

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[lost harness]

I dont know how to start with this question

[kind hawk]

it might help to separate the LHS into two different series

[lost harness]

how?

what does it mean when it says converges?

it goes to a point?

[kind hawk]

yes

[lost harness]

how would we split them?

slit the x and 2x?

[kind hawk]

that could be a strategy. split the terms without the 2 in front from those with the 2

[lost harness]

so would that be

x + x^3 + x^5 + x^7 + ...

-2x^2 - 2x^4 - 2x^6 ...

[wary jewel]

yes

then take out the common factors

[lost harness]

x(1+x^2+x^4)

omg wow

and then we get

-2x^2(1+x^2+x^4)

[wary jewel]

yes, don't forget the + ... tho ahaha

[lost harness]

yes that too

[wary jewel]

but yes you get it, you know how to solve further?

[lost harness]

i dont know

[wary jewel]

okay

[lost harness]

but i can see they both have 1 + x^2 + x^4

[wary jewel]

yes

[lost harness]

but idk what to do with them

[wary jewel]

x( 1 + x^2 + x^4 + ...) - 2x^2( 1 + x^2 + x^4 + ...) indeed both have ( 1 + x^2 + x^4 + ...), so take that out of both terms

[lost harness]

so now we have x- 2x^2 ( 1 + x^2 + x^4 + ...)

[wary jewel]

(x- 2x^2) ( 1 + x^2 + x^4 + ...) what you mean right

[lost harness]

yes with the bracket

[wary jewel]

w8 i'll check if i'm correct

[lost harness]

ok

[wary jewel]

okay so

I was kind of wrong, but it's no problem

have you learnt the formula for infinite sum of geometric series?

s= first term / (1-r)

[lost harness]

yes

[wary jewel]

ok

we need to go back in time though

x + x^3 + x^5 + x^7 + ...

-2x^2 - 2x^4 - 2x^6 ...

lets take the first series: x + x^3 + x^5 + x^7 + ...

you see how this is a geometric series?

[lost harness]

wait

when we factored out x-2x^2

we have (x- 2x^2) ( 1 + x^2 + x^4 + ...)

so if we take 1 out and factor out x^2 from that we r left with (1 + x^2)( 1 + x^2 + x^4 + ...)

could we not make (1 + x^2)( 1 + x^2 + x^4 + ...) equal the same thing
so if we let A = ( 1 + x^2 + x^4 + ...)
(1 + x^2)A = A
and then solve for A?

or is that not allowed 💀

i get 1/(1-x^2)

[wary jewel]

let me make it simpler

because i don't get you lmao

[wary jewel]

oh yes that is indeed correct

[lost harness]

oh is it!?

[wary jewel]

i had another explanation but the same results

so

and your solution might be a little easier but it's really no difference

[lost harness]

how did ur solution work?

[wary jewel]

(x - 2x^2)( 1 + x^2 + x^4 + ...) you see how the 2nd part is a geometric series

so if you use the formula for the sum it is 1/(1-x^2) since r=x^2

which is the same as your A

[lost harness]

the gemoetric series is a(1-r^n)/(1-r)

right

[wary jewel]

yes as we know that the series converges, we can conclude that x is smaller than 1

since otherwise 1+x^2+x^4+... would diverge

[lost harness]

yes

[wary jewel]

so

(r is x^2, x^2n would be the r^n here, but x^∞ = 0 for x smaller than 1)

so you just have a1/(1-r)

[lost harness]

so thats just infinity

sum to infinty

[wary jewel]

w8

so

sum of a geometric series for example sum of the first three terms would be

a(1-r^3)/(1-r)

so the infinite sum would be a(1-r^∞)/(1-r)

and r is smaller than 1 here

so that term becomes zero

[lost harness]

ok

[wary jewel]

so

for the infinite sum the formula is a/(1-r)

a=1 here

[lost harness]

oh yes that makes sense

[wary jewel]

but like it really doesn't matter, since your method is fine too

[lost harness]

thank you tho

[wary jewel]

np

[lost harness]

how do you do maths very fast??

[wary jewel]

wdym?

[lost harness]

i always run out of time in exams

[wary jewel]

oh

[lost harness]

even if i was able to do this question i wouldnt be able to do it cuz of the time

i would run out of time

[wary jewel]

idk I'm also still in high school, but I'd say just know what formulas to use for every possible exercise on the exam

and make sure you understand every sort of exercise your teacher could possibly ask

and don't lose too much time on smaller exercises, i always make the ones with the most points on it the first

[lost harness]

you do the questions with the most marks first??

[wary jewel]

yes

[lost harness]

and leave the lower marks out till last?

[wary jewel]

yes, but i always have enough time to make everything

[lost harness]

my exam is tomorrow

[wary jewel]

damn

mine is friday

i have 3 math exams

[lost harness]

same

pure 1, pure 2 stats and mechanics

ive already failed stats

but i got a shot at pure

[wary jewel]

mine are integrals, stats and conic sections

stats are hard yeah

[lost harness]

ah we do everything

can i send another question?

[wary jewel]

you're in your last year?

sure

[lost harness]

yes

[wary jewel]

what are you going to do next year in uni?

also you can send the question whenever you want

[lost harness]

computer science

[wary jewel]

nice

[lost harness]

hbu

[wary jewel]

civil engineer

[lost harness]

cool

[lapis spruce]

do quotient rule

[wary jewel]

yeah

quotient rule does the trick

[lost harness]

i kind of dont understand the fundementals of maths so i messed up in the middle of it

so thats (vu' - uv')/v^2

[wary jewel]

yes

[lapis spruce]

if u are more comfortable with product rule, u can bring the denominator up like this:
(x-4)(2+root(x))^(-1)

ur choice

[lapis spruce]

yes

[lost harness]

is this correct

[lapis spruce]

u have to get it in the form they want

basically find A

[wary jewel]

you could also just notice that the numerator is (sqrt(x)-2)(sqrt(x)+2)

[lost harness]

i was just thinking this

[wary jewel]

and then scrap (sqrt(x)+2)

[lost harness]

and its the the sqrtx - 2

[wary jewel]

yes

and then take the derivative of that

[lost harness]

and thats 1/2sqrt x

[wary jewel]

exactly

[lost harness]

thats disgusting

i would over complicate this

[wary jewel]

,w derivative of (x-4)/(2+sqrt(x))

[warm shale]

Results provided by WolframAlpha

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

Requested by ! Ilya#2861

[wary jewel]

yep

lmao

you'd get there without though

[lost harness]

i hate the questions they ask

[wary jewel]

but lets not

[lost harness]

they dont tell us what to do

we have to figure it out 💀

[wary jewel]

oof

[lost harness]

cuz u can technaiqllly do quotent rule

but the difference square option is way better

[wary jewel]

yes

[lost harness]

thank you everyone

[wary jewel]

np good luck tmr

[lost harness]

thank you 💪

[lapis spruce]

but thats the idea for any exam....

[lost harness]

yh 💀

[lapis spruce]

is this AS level??

[lost harness]

no

A-Level

what should i do on the day of the exam tomorrow?

should i do more questions or just chill?

[lapis spruce]

honestly if ur done u can chill,dont stress out too much

[lost harness]

im kinda not done

and im kinda not great at maths

[lapis spruce]

sleep early and all

[lost harness]

yh

[lapis spruce]

revise concepts and gl

u will do fine

[lost harness]

thank u

bye everyone

[lapis spruce]

is this your last year in school??

[lost harness]

yes lol

[lapis spruce]

ah ok

i did A levels last year as well

[lost harness]

what did u get

was it hard?

[lapis spruce]

my paper was actually easy but it might be becuz i like maths

[lost harness]

ah

im only ok at maths

but oh well

[lapis spruce]

dw u will be fine

[lost harness]

thank u

[lapis spruce]

do the latest year papers of A levels cuz they can be the same format like your paper and they are tougher than past years

[lost harness]

2022??

[lapis spruce]

yup

[lost harness]

i think ill do one now

before bed

and then chill tomorrow

last time i got sick from stress

[lapis spruce]

whens your exam

[lost harness]

tomorrow

💀

1:45 pm 💀

[lapis spruce]

i wouldnt advise to do a past paper one day before

[lost harness]

so today?

right now ? 💀

[lapis spruce]

yea it will stress u more i did that mistake

just revise full text book and see xample qs

[lost harness]

ok ill do that

and on exam day?

do more qustions?

or just look at it

[lapis spruce]

look

absolutely dont do past papers tomorrow

[lost harness]

yh ok

.close

[obtuse pebble]

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[trail jewel]

does anyone know how i can find f39(0) without taking the derivative 39 times

[versed flume]

try maybe the first 2 or 3 derivatives and find the pattern, then proof the pattern (if you have to)

[trail jewel]

but because the point is centered at 0 when we plug in x = 0 wouldnt that make the entire equation 0?

[north quarry]

sum of 0 is 0

[hot sonnet]

That doesn't mean the derivative is 0 🤔

[trail jewel]

oh right..

wait but if we took the derivative of x it would be nx^(n-1)

so the derivative would be 0?

[hot sonnet]

yes

no

[trail jewel]

but youre multiplying n*0^(n-1)

[hot sonnet]

x^2 becomes constant after 2nd derivative, right?

[trail jewel]

yes

[hot sonnet]

when you have x^k, each differentiation takes 1 from the exponent

[trail jewel]

oh i see so the answer would be constant and i just need to find the coefficient of c39?

[hot sonnet]

mmm I don't know where that coefficient comes from 😵‍💫

[hot sonnet]

only one term will be constant, which one will it be?

[trail jewel]

(-1)^n/(39*8^n)?

[hot sonnet]

no I mean

[trail jewel]

sorry im kinda having trouble with taylor series

[hot sonnet]

what term becomes constant after 39 derivatives

[trail jewel]

the x becomes constant

[hot sonnet]

the terms of the sum are n=1, n=2, ...

there's only one that after 39 derivatives won't depend on x

[trail jewel]

the one at n = 39?

[hot sonnet]

yes 👍🏽

the rest are 0 after 39 derivatives

or become 0 cause x=0

so you have to take the 39th derivative of that single term

[trail jewel]

ohh okay i think i understand

thanks for the help!

.close

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[plain grove]

lets say N(t), t ∈ [0, ∞), and its a Poisson process with rate λ. How would you find Cov(N(t1), N(t2)), assuming t1, t2 ∈ [0, ∞).

[plain grove]

i get as far as

E(N(t1) x N(t2)) - E(N(t1)) x E(N(t2))

i dont know how to find E(N(t1) x N(t2)) though

[obtuse pebble]

@plain grove Has your question been resolved?

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[fluid snow]

How would you prove Newton's inverse square law using Kepler's second and third law?

[obtuse pebble]

@fluid snow Has your question been resolved?

[obtuse pebble]

@fluid snow Has your question been resolved?

[obtuse pebble]

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[dusky mauve]

How do I find the surface area?

And ignore the 15 I wrote

[wooden cipher]

do you know the length of the other dimension?

[dusky mauve]

Nope!

[wooden cipher]

then you cant find the surface area

[dusky mauve]

That's why I was asking I was so confused lol

Alr thanks

[wooden cipher]

ask your teacher when you get the chance

[dusky mauve]

Yep

Tyy

.close

[obtuse pebble]

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[wooden cipher]

youre welcome

[obtuse pebble]

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[lean plover]

Does anyone have any videos, where i can learn the fundamentals of algebra and where to position/locate it in problems

[crimson torrent]

helo

i've been reading a book for that

it has fundamentals of algebra and very good explanations as well

[grizzled shore]

just algebra in general?

[lean plover]

Just algebra in general

[grizzled shore]

what level of algebra are we talking

[lean plover]

the basics, never really learnt it in highscohol and now need to take a uni course on it. Im good with numbers just struggle identifying what goes where

[crimson torrent]

ah

mathematics for practical man, the one im reading rn

u can get a good grasp at fundamentals with this

it has fundamentals till calculus

[lean plover]

Where could i find this book + any videos i can get a start on?

[grizzled shore]

https://www.youtube.com/watch?v=hURKivs0GIs&list=PL4C9296DF81B9EF13&index=7

this guy is pretty good

[lean plover]

Thank you

[grizzled shore]

maybe you'll have some luck with this

[crimson torrent]

https://www.aproged.pt/biblioteca/mathematicsforthehowe.pdf

[lean plover]

Thanks guys

[crimson torrent]

hmm do u also know any books for practicing algebra ? like after finishing the fundamentals

[grizzled shore]

i honestly just used my school textbook, would be the cambridge stage 4 book i think that one has algebra?

[crimson torrent]

Oeh i see

thank u

[obtuse pebble]

@lean plover Has your question been resolved?

[obtuse pebble]

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[jolly radish]

A population consist of three numbers (2,4,6) Consider all possible samples of size 2 which can be drawn without replacement from the population

[jolly radish]

Find the following
a.) Population Mean
b.) Population Variance
c.) Population Standard Deviation

[obtuse pebble]

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[wild estuary]

How are the “axis angles” helpful in graphing the parent graphs for all six trig functions?
Thank you so much!

[fathom flicker]

what do you mean by "axis angles"?

[wild estuary]

Im not really sure tbh

[dreamy forum]

Where is this question coming from, meaning what is the source of the problem you are currently facing

[wild estuary]

my math teacher made it up

[dreamy forum]

Ah wonderful

I would really like to know what they meant by axis angles

Because as far as my very short research goes the axis angle representation "parameterizes a rotation in a three-dimensional Euclidean space by two quantities"

[timber island]

Maybe the angles 30, 45, 60, 90, 120, 180?

Or simply the angle formed by the two axis

[wild estuary]

yea i was reading that too

[dreamy forum]

This may make sense

[wild estuary]

but it was under trig unit

so i think it is unit circle related

something like that

but with a unit circle

[dreamy forum]

Yeah rotation is a different burden

But with the unit circle there is some sort of cheat sheet for remembering some useful sin and cos values if I'm not mistaken

[wild estuary]

yea

[dreamy forum]

This one

[wild estuary]

so i think my teacher maybe wants me to use stuff on the unit circle to find cos sin tan

for different trig graphs

[dreamy forum]

Probably

[wild estuary]

alr ty bro

[dreamy forum]

Its the only thing that would make sense imo

[wild estuary]

actually i thin it is coterminal

[obtuse pebble]

@wild estuary Has your question been resolved?

[obtuse pebble]

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[pure sun]

In a four card hand from a standard deck, how many different hands have 2 pairs?

I’m not sure how they get 13C2 * 4C2 * 2, could someone explain to me please?

[zenith raft]

who got that?

[pure sun]

The solution uses that

[zenith raft]

that seems very small

[royal basin]

what solution

are you sure it isn't 13C2 * (4C2)^2 btw

[pure sun]

Oh wait yeah it is squared xd

This one actually has an answer key

[royal basin]

does that dispel your doubt

[pure sun]

Nope sorry I’m still kind of confused

I think I’m mostly confused by how they get 4C2

Cause I feel like it should be 4C1 so that’s what I’m confused by

[timber island]

Yo

[timber island]

2 pairs of what?

[pure sun]

Two pairs of any cards

Like a hand of 2,2,3,3

[obtuse pebble]

@pure sun Has your question been resolved?

[obtuse pebble]

@pure sun Has your question been resolved?

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[neon lintel]

listed in my lecture notes is this proof that i believe is examinable (i.e. they will ask you to recall the proof) however the proof relies on a general case as a last statement and from what i can see it doesn't look like a proof by induction and i'm not sure why they used 2 examples in the early part of the proof

[neon lintel]

would it be safe to omit the first two examples? or would i need to keep either m!+2 or both?

[obtuse pebble]

@neon lintel Has your question been resolved?

[neon lintel]

.close

[obtuse pebble]

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[dense drift]

"all white numbers are odd" "all black numbers are even" how do i determine the color of number 3?

[obtuse pebble]

Please don't occupy multiple help channels.

[leaden ginkgo]

is 3 odd or even?

[candid yarrow]

are all odd numbers white?

[dense drift]

3 is odd

[dense drift]

that's what im trying to find out

[leaden ginkgo]

huh

you just said all white numbers are odd

[dense drift]

if all white numbers are odd that doesn't necessarily mean that all odd numbers are white

[leaden ginkgo]

oh damn

that is true

[high lily]

so insufficient info

[leaden ginkgo]

hmmm

[high lily]

there could be other colours

[dense drift]

the answers are : A) 3 is a white number B) 3 is an odd number C) it's impossible to determine the color D) being odd is condition necessary to be black E) all even numbers are black

I thought that it wasn't possible to determine the color but im not sure

[candid yarrow]

is this a test on logic?

[dense drift]

yes

it could be D aswell i think

[candid yarrow]

can you explain why

[dense drift]

no my bad it says black not white

so it can't be D

[candid yarrow]

can you send the full question?

[dense drift]

"all white numbers are odd" "all black numbers are even" : A) 3 is a white number B) 3 is a black number C) it's impossible to determine the color D) being odd is condition necessary to be black E) all even numbers are black

[candid yarrow]

As you said, 3 is odd, so one of them is clearly correct

[dense drift]

no im so sorry but i made a mistake in the answers, it says that 3 is a black number

i think it's impossible to determine its color

[royal basin]

you're wrong about that

well, ok, maybe colorless numbers exist, we don't know

[dense drift]

we can't determine whether its black or white, because all white numbers are odd doesn't necesarily mean that all odd numbers are white

[royal basin]

ok wait hold on

do we or do we not know that every number is assigned a color

that is the real question here imo

[dense drift]

i don't think every number has a color, but white and black have assigned numbers

idk if that makes sense

[royal basin]

ok so yeah some numbers could be colorless

[dense drift]

we only know that white numbers are odd and black are even. that doesn't give us enough info to determine the color of 3

[dense drift]

so do you think that it's impossible to determine its color?

[royal basin]

yes

[dense drift]

great, thank you!!!!

.close

[obtuse pebble]

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[regal jewel]

a swimming pool is 2.5 m deep, and has room for 693.75 m^3 water. How far away is the water away from the swimming pools surface, when 656.25 m^3 get poured in?

[candid yarrow]

!show

[obtuse pebble]

Show your work, and if possible, explain where you are stuck.

[candid yarrow]

!status

[royal basin]

!status

[obtuse pebble]

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

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

[royal basin]

lmao sniped

[regal jewel]

3

so what i thought was, divide height of swimming pool by the amount of water

so we get how much water there is per cm(i converted the unit)

[candid yarrow]

what answer did you get

[regal jewel]

something along the lines of 6 cm

[candid yarrow]

Can you show your expression

[royal basin]

not the other way around?

[regal jewel]

tiny detail mistake

$frac{693.75}{250}$

nah

[candid yarrow]

$$ \frac{693.73}{250} $$

[warm shale]

(a + b)^2 = a^2 + b^2

[regal jewel]

is this some kind of update?

never used double "$"

and then

$$ 656.25l/(\frac{693.73l}{250m})$$

no

wait

[candid yarrow]

at this point it is useful to put in units

[regal jewel]

i dont know man

[warm shale]

紅卫兵，周

[regal jewel]

just show me how to solve this

my head cant wrap around this

[candid yarrow]

this should be right

oh wait no

this is the height of the water, not the distance of the water from the surface

[regal jewel]

but then

$$ 693.75l-(656.25l/(\frac{693.73l}{250m}))$$

[warm shale]

紅卫兵，周

[regal jewel]

$$ 250cm*(693.75l/(656.25l/(\frac{693.73l}{250cm})))$$

bro i dont know how to proceedd

i just made a mess

[candid yarrow]

the units of this are meters

in the right parentheses you’re subtracting liters and meters

[warm shale]

紅卫兵，周

[regal jewel]

bro

[candid yarrow]

shouldn’t it be 2.5 m - (2.5 m*656.25L/693.73L)

[regal jewel]

i dont know man

my brain doesent work

its actually incorrect

that was actually what i got on one of my attempts

in the book it say 10 cm

[obtuse pebble]

@regal jewel Has your question been resolved?

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[obtuse pebble]

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[tired sierra]

how do you integrate this?

[candid yarrow]

!show

[obtuse pebble]

Show your work, and if possible, explain where you are stuck.

[candid yarrow]

and where’s the dx

[tired sierra]

huh good point

anyways no working out because I actually have no idea how to integrate this

I initally thought it was a normal trig function but then I realised sec^2x has an x in it

[candid yarrow]

consider d/dx tan x

[tired sierra]

so that was rulled out

d/dx tanx

[candid yarrow]

can you evaluate that?

[tired sierra]

sec^2 x?

[candid yarrow]

yes

do you know how to do u sub

[unkempt bramble]

Put tanx=t

Differentiate both sides

[tired sierra]

can I really only do this with u sub?

[candid yarrow]

you can

[unkempt bramble]

Sub is straightforward

Maybe there's an odder way

[tired sierra]

I know but I'm doing a test where they don't expect that so

oh btw this isn't a test, I am going to do one, this is just a practice paper

don't want to make it sound like I'm cheating lol

[viral blade]

If u-sub is too advanced I really don't know what else they'd expect

any other technique would be much more advanced

[unkempt bramble]

Try making everything sin cos lol

[dreamy acorn]

thats painful and i dont think it work lol

[unkempt bramble]

Try cubing haha

[viral blade]

,w integral of sec^2(x)tan^3(x)

[warm shale]

Results provided by WolframAlpha

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

Requested by thewizardofOU#0205

[tired sierra]

you forgot the 1

[unkempt bramble]

Just tell em u found it more intuitive

[viral blade]

still u-sub tho

[tired sierra]

wait a minute

tanx-1 differentiates into sec^2 x

[viral blade]

I didn't forget I was just testing one term of the expanded binomial

[tired sierra]

does that mean I could do a reverse chain rule

[candid yarrow]

you can

[tired sierra]

now would you look at that

no u sub

:^)

[unkempt bramble]

Like...

[dreamy acorn]

you can take away sec^2 x and put dx to be d(tanx-1)

its same as u sub

[viral blade]

That's what u sub is

Its always reverse chain rule

[tired sierra]

I see

welp

[unkempt bramble]

Reverse chain works

[tired sierra]

I just did it so uh thanks for the help lads

[viral blade]

usually people just learn one notation scheme for chain rule (newton stuff) and the opposite for u-sub

[tired sierra]

ah I see

[viral blade]

but chain rule is also dy/dx = dy/du * du/dx

[unkempt bramble]

Now this removes the cheating suspicion element doesn't it

[tired sierra]

since f'x = fx

therefore its (1+tan)^4/4

evidence??

[unkempt bramble]

Integral(f'x)=fx

[tired sierra]

I literally showed it to you

[unkempt bramble]

Yea just think what shd i differentiate via chain rule to get the the stuff inside the integral

Thats f(x)

So start with finding f

[tired sierra]

nibberge I understand where your going

[unkempt bramble]

Intuitively

[tired sierra]

but I have already gotten the solution for mine

thank you tho

[unkempt bramble]

Alr

[tired sierra]

your effort has not gone underappriciated

.close

[obtuse pebble]

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[candid yarrow]

!show

[obtuse pebble]

Show your work, and if possible, explain where you are stuck.

[candid yarrow]

!status

[obtuse pebble]

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

[opal trail]

3

i got 1701

[proper kelp]

U prolly didn’t find the first time they were collinear

Wait both runners r

Same speed right

[opal trail]

@proper kelp yes

i think i am doing something wrong

[proper kelp]

Ok I’ll try it ig

What was your way of proving colinear

[obtuse pebble]

@opal trail Has your question been resolved?

[opal trail]

i just took 2 circles

@proper kelp i am blatantly wrong as i figured out a major mistake

[proper kelp]

Which was

[opal trail]

nvm my answer becomes 1

i dont know where to begin

[proper kelp]

I think a way is just to let A be origin and draw two circles

Find equation of each circle

And y =mx

And find point which satisfies all equations

I actually have no idea either

[opal trail]

can we make a line across point a intersecting the 2 circles

this s where they might be collinear at some point

take the theta

[proper kelp]

I mean they will be colinear at that point at some point I’m pre sure but it’s most likely not going to be the first point of collinearity

[opal trail]

no we take it for some theta and then think that 1 does k rotations

[proper kelp]

Right

[opal trail]

and equate the distance covered

[proper kelp]

Yeah this question is too hard sorry 😦

[obtuse pebble]

@opal trail Has your question been resolved?

[opal trail]

<@&286206848099549185>

[high lava]

hi

what problem do you have?

[proper kelp]

https://cdn.discordapp.com/attachments/903430332324405288/1115607760105644123/IMG-20230606-WA0001.jpg

[opal trail]

anyone

[high lava]

😮

[obtuse pebble]

@opal trail Has your question been resolved?

[obtuse pebble]

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[north quarry]

the < has to flip to >

well if u plug in -6

its not true right

yep

its a property of inequalities

5<7

in the main equation u plug in

the thing ur lookin for is flipped

not the equation

here its 8<7

u dont flip the base equation

u flip when u divide/multiply by a negative

well now u do

11-2x>3

u sure ?

whats the 1st thing u do ?

ye

yes now try plugging x=0 for example in the main equation

since 0<4

yea

yep

now try x=5

this disproves the found inequality

so it should disprove the main one

no no

u dont make it fit

u have

x<4

this is all the values of x that respect 11-2x>3

why 3<x

u just pluggedx=0

and it worked

[tardy helm]

off topic, but why did you both have to pick bracket names?? This is so confusing to read, lol!

[north quarry]

u already solved that

ok lets restart then

solve 3x+9<18

i also see invisible

why u flip ?

[north quarry]

^

by a negative

yep

now verify it with the base equation

u solved that x needs to be smaller than 3 to fit the base one

yes

so 2<3

and it works

but wb now u try to plug somethig that isnt smaller than 3

ur not respecting the condition

4<3

right

yes

here u didnt respect the condition

and therefore the base equation doesnt work

bc 21 isnt < 18

this

the problem ur solving for

the 1st thing

this is the meaning

is 21 smaller than 18 ?

then it doesnt work

well u solved it

no, u didnt multiply/divide by a negative

it has been proven

u solved for it

to show u a value that makes it not work

yes

if u dont respect the solution u found, x<3

the 3x+9<18 wont work

YES GOOD JOB

ok one more

5x-21<9

yes

perfect

the only thing u forgot when u asked the question was to flip when u divide/multiply by a negative value

so back to the problem u sent

how would u do it now

always test, ull see that it doesnt work and u forgot something

now verify

ok ok

there u go

[obtuse pebble]

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[bleak skiff]

hello, i need to find this sum

using riemann sum properties

[tardy epoch]

Divide top and bottom by n^2

[tardy epoch]

Stop spamming your messages everywhere

[bleak skiff]

ok

[obtuse pebble]

@bleak skiff Has your question been resolved?

[bleak skiff]

@tardy epoch actually i still don't know what to do after that

the formula i know is this one:

<@&286206848099549185>

@tardy epoch

.close

[obtuse pebble]

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[timid silo]

Hey i need help on geometry homework

[dark stirrup]

!status

[obtuse pebble]

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

@warm epoch Has your question been resolved?

[obtuse pebble]

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[trail jewel]

hey guys i need some help determining if this series is convergent or divergent

i understand that 1/ln(3n) is larger for 1/n so we can evaluate the series 1/n (harmonic series) by comparison test

however 1/ln(3n) is larger for 1/n when n is greater then 1 so can we still use the comparison test?

[north quarry]

u could use LCT i think

or u can separate the ln

[trail jewel]

if i use lct would my bn = 1/n?

[north quarry]

doable

[trail jewel]

alright ill give it a try

[obtuse pebble]

@trail jewel Has your question been resolved?

[trail jewel]

alright thanks!

.close

[obtuse pebble]

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[coral stone]

I understand everything here but the actual answer

[coral stone]

probability of dry alone is .5

while probability of windless is .1 I think

or is it .6?

.6 i think

how would you get the actual answer to the question

<@&286206848099549185>

[odd urchin]

have you tried venn diagrams?

those are helpful

[coral stone]

yh look its at the bottom right

[odd urchin]

will

well*

the probability of dry and windless is where dry and not wet are the same

you can see that in the venn diagram somewhere

you just have to think about it a little

[coral stone]

🤔

[odd urchin]

windy* i said wet earlier i meant windy

there’s a place in the venn diagram where it is both dry and not windy

[coral stone]

hmm

[odd urchin]

sorry, where it’s dry and not windy

that’s where it’s dry and windy

because that’s where they overlap

[coral stone]

oh shit

[odd urchin]

we’re looking for the place where it’s in dry

[coral stone]

lmfao

its .5

[odd urchin]

yes

[coral stone]

idk why i didnt notice that

[odd urchin]

lmao i’ve done the same thing😭

.close

[obtuse pebble]

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[obtuse pebble]

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[maiden whale]

5𝑥2 + 105 𝑦2 = 25

[robust sleet]

!status

[obtuse pebble]

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

[robust sleet]

what even is the question

[maiden whale]

Find the ellipse, center, a and b, and focus

[obtuse pebble]

@maiden whale Has your question been resolved?

[normal oracle]

!show

[obtuse pebble]

Show your work, and if possible, explain where you are stuck.

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Show your work, and if possible, explain where you are stuck.

@weary idol Has your question been resolved?

[brisk matrix]

show your work please @weary idol

why did you swap the order

@weary idol

if the canvas is originally 8.5 inches x 11 inches

and you want the number of pixels on a inches x b inches of the canvas

then you need to do (2550 * a / 8.5) * (3300 * b / 11)

[obtuse pebble]

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[obtuse pebble]

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[tacit wing]

I need some help, translation is "Show that the equation (equation here) has exactly one solution"

[brisk matrix]

i suggest you square both sides

[tacit wing]

so 3x^2+1 = 1 and then from there intermediate value theorem?

[brisk matrix]

oh is this an analysis question

i thought it was like an algebra question haha

[tacit wing]

yep, hard to translate these questions that are already very vague, sorry

[brisk matrix]

hm how are you defining sqrt(x)? some would say the above has no solution

[tacit wing]

Im not even sure exactly what is asked, this is an old test question from a derivatives test and all I know is its something to do with bolzanos theorem

[sullen flume]

That's what I'm saying, that has no solutions (as far as I can see)

[tacit wing]

It could be asking for "show this has exactly one zero point"

but I am not sure

actually that cant be the case

im really lost

[obtuse pebble]

@tacit wing Has your question been resolved?

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[lucid marsh]

Hey, I was wondering if someone can explain how to go through this?

[lucid marsh]

<@&286206848099549185>

<@&286206848099549185>

[somber thistle]

it's a circular cylinder

[lucid marsh]

How did you get that?

[somber thistle]

it's a circle

and y has nothing to do with it so it's constant for same values of x and z

[lucid marsh]

Ooh okay, so how do I start an equation?

[somber thistle]

it's the same for all values of y

x^2 + z^2 = 6

[lucid marsh]

It says cross-section so all I have to do is plug it in?

Like the first one says, “ y=-6” using x and z”, am i replacing the values?

[somber thistle]

enter the equation i sent

[lucid marsh]

Into the cross section for the first one>

*?

[somber thistle]

yes

[lucid marsh]

Along with the second like x^2+z^2=0

?

[somber thistle]

no

[lucid marsh]

So then what is it?

[somber thistle]

it's always x^2 + z^2 = 6

[lucid marsh]

So I enter the same one for each of the boxes?

[somber thistle]

i'd assume so

[lucid marsh]

Ok

<@&286206848099549185>

[somber thistle]

what

[lucid marsh]

Is it suppose to be different rings or one ring?

[somber thistle]

probably same ring

[lucid marsh]

Ok

.close

[obtuse pebble]

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[hard minnow]

Can someone help me with this? My

[dapper silo]

what did you try?

[obtuse pebble]

@hard minnow Has your question been resolved?

[desert sinew]

There should be a standard formula for such volume. Plz try check your notes or textbook first. 🙂

[obtuse pebble]

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[covert lantern]

consider the following equation 100000000000000000000×((150+(A×25)+(B×10)+(C×5))÷100)×((1.1+(W×0.01)+(X×0.0075)+(Y×0.005)+(Z×0.0025)+(D×0.0015)+(E×0.001)+(F×0.0005))^200

[covert lantern]

the sum of W,X,Y,Z cannot be greater than 1, the sum of A,B,C,D,E,F cannot be geater than 9 if the sum of W,X,Y,Z is 1, if the sum of W,X,Y,Z is 0, the sum of A,B,C,D,E,F must not exceed 12

is there a easy way to find the smallest solution to this thats bigger than 10^30

the best one i have so far is

,w 100000000000000000000×((150+(2×25)+(0×10)+(1×5))÷100)×((1.1+(1×0.01)+(0×0.0075)+(0×0.005)+(0×0.0025)+(4×0.0015)+(2×0.001)+(0×0.0005))^200

[warm shale]

Results provided by WolframAlpha

Click here to refine your query online
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Requested by OttLight#1782

[covert lantern]

im not sure if thats the best one and idk how to check other than trying every possible combination

[obtuse pebble]

@covert lantern Has your question been resolved?

[obtuse pebble]

@covert lantern Has your question been resolved?

[somber thistle]

i believe you can put in letters and conditions

[covert lantern]

into what program?

the conditions would be

[somber thistle]

well

you can simplify

so simplify as much as possible

[covert lantern]

{A,B,C,D,E,F,W,X,Y,Z} are natural, W+Z+Y+Z≤1, A+B+C+D+E+F≤12-3(W+X+Y+Z)

i think tahts the most simple i can get

[somber thistle]

you can divide out the 10^whatever

[covert lantern]

what program would i input the terms into

[obtuse pebble]

@covert lantern Has your question been resolved?

[obtuse pebble]

@covert lantern Has your question been resolved?

[covert lantern]

<@&286206848099549185>

[prime ravine]

؟

[hidden garnet]

...........oh my god this is a real problem?

[covert lantern]

yes.

i formatted the messy equation to this

((150+25A+10B+5C)*0.01)*(1.1+0.01W+0.0075X+0.005Y+0.0025Z+0.0015D+0.001E+0.0005F)^200

all the conditions are the same

btw natural includes 0 in this case

[covert lantern]

<@&286206848099549185> consider the following expression:
((150+25A+10B+5C)*0.01)*(1.1+0.01X+0.0075W+0.005Y+0.0025Z+0.0015D+0.001E+0.0005F)^200

find the smallest solution that is larger than 10^10 where: {A,B,C,D,E,F,W,X,Y,Z} are natural, W+X+Y+Z≤1, A+B+C+D+E+F≤12-3(W+X+Y+Z)

brute forcing seems impossible...

and i dont really know where to approach this from

[timid silo]

Can you write the equation in LaTeX?

[covert lantern]

i can try

havent learned how to use latex yet

[timid silo]

Ask chatgpt to do it

[covert lantern]

oh ill try that

[((150+25A+10B+5C)\times0.01)\times((1.1+0.01X+0.0075W+0.005Y+0.0025Z+0.0015D+0.001E+0.0005F)^{200})]

[warm shale]

OttLight

[covert lantern]

uh it doesnt fit

[timid silo]

Wtf, it's that big??

[covert lantern]

is there a way to make it smaller?

[warm shale]

OttLight

[covert lantern]

how do i make it smaller

imma be honest that might be harder to read