help-39

[hollow cobalt]

I never said that proves it

[fast zealot]

have you tried playing around with it ?

[summer torrent]

yes

[hollow cobalt]

Try simplifying some stuff

[summer torrent]

btw what grade is this is it highschool or uni

[fast zealot]

earlyish highschool

(for my UK experience)

[hasty eagle]

6th grade or something?

[summer torrent]

to prove it 💀

[fast zealot]

you didn't play around with it enough if you don't see how that proves it :D

[hasty eagle]

you don't need complicated math to square a number tho

[summer torrent]

i cant just calculate it i need to prove it

[hollow cobalt]

We are aware of that

[fast zealot]

lol

[hasty eagle]

ok how about this. you got yourself an expression you don't understand. all we're saying is you can manipulate that expression into a form that you DO understand

[fast zealot]

just use the properties of exponents you know, and play around with the lhs, ((5/2)^(5/2))^2

and see if that gets you anywhere

[summer torrent]

yall are not helping f this

[fast zealot]

getting used to doing work yourself might be hard :D

good luck tho

:)

[hasty eagle]

well if you refuse to listen to reason, then best of luck i guess

[summer torrent]

speak swedish plz

math in english is hard i dont udnerstand

[viscid sierra]

somebody call Ann

[summer torrent]

is she swedish

[hasty eagle]

,w what's the squaring in swedish

[jolly parrot]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by toomanycooks_

[hasty eagle]

nvm

[summer torrent]

roten ur or upphöjt

[fast zealot]

((5/2)^(5/2))^2 < 10^2 is all you need

lol

[summer torrent]

why speak swidhs

[fast zealot]

vad blir vänster sida till nu?

[summer torrent]

idk

[fast zealot]

applying (a^x)^(y)

[summer torrent]

ok this is to hard bye

[fast zealot]

bye

:D

[upper shoal]

https://tenor.com/view/cat-flashbang-gif-25002731

do .close

[pearl pond]

@summer torrent Has your question been resolved?

[pearl pond]

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[tulip cradle]

Most confused on the first one

[tulip cradle]

For the second one is it just 3A^-1 x A on its own and 4B^-1 x B on its own so u get 3A + 4B

And for the third one would it be A^-1 B C^-1

[timid spindle]

X*X^(-1) is the identity matrix for any invertible matrix X

First do (3A^(-1) + 4B^(-1))*B and then multiply the result on the left by A

[tulip cradle]

Ohhh

Yeah ok thanks its 3B+4A

[grizzled dust]

just use the fact that matrix multiplication is distributive over matrix addition

i think that's the point you're supposed to understand with this exercise

[tulip cradle]

Oh I got the first one now

[tulip cradle]

Is this correct tho

[timid spindle]

No taking the inverse of a product reverses the order

[tulip cradle]

Ah ok

[timid spindle]

Aswell as inverting each matrix

You can check if you're right by timesing the original by the inverse you calculate and seeing if you get the identity

[tulip cradle]

Makes sense

Thanks

.close

[pearl pond]

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[ebon dagger]

Can anyone help me with where i went wrong in using the chinese remainder theorem

im being told the answer is 136

but i got nothing like that

[limpid lily]

You got 105m + 31, which gives 136 when m is 1.

If m is 0 or 2, you're outside the range.

[ebon dagger]

ohh i thought you always sub in 0 regardless

i see now

thanks

.close

[pearl pond]

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[limpid lily]

No problem.

[pearl pond]

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[ember tiger]

Hey

[regal herald]

ello

[ember tiger]

After i got the derivative how do i get the second?

[regal herald]

,rccw

[regal herald]

get the second derivatives of y and x with respect to t and do the same thing ig

[ember tiger]

I dont think so

I remember half the way

Which is

But how do i use this up

[regal herald]

oh yeah
its $\frac{d}{dt}\left(\frac{dy}{dx}\right)\div \frac{dx}{dt}$

[jolly parrot]

AℤØ

[regal herald]

mb

[ember tiger]

Am i correct?

[regal herald]

,rccw

[regal herald]

i follow you to here

wait

one moment

yeah

[regal herald]

i think this is okay

hm

[ember tiger]

Its correct right?

[regal herald]

the only thing thats bothering me a bit, is with the first derivative
you cancelled the t's or the 1/ts however you put it
which somewhat changes the function

because now its defined at t=0 when it shouldnt have been

i might be being pedantic

[ember tiger]

No i am sure of the first derivative tbh

I used ai to prove it lmao

Idk if that is reliable

But my application cant do second derivative

[regal herald]

actually, you may get away with it since t=0 isnt in the domain of the x or y anyway

just feels slightly meh

[regal herald]

but if we allow it, this is fine

[ember tiger]

Thanksss

[pearl pond]

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[long meteor]

Pls help

[robust sentinel]

where are you stuck

[long meteor]

Idk where to start

[robust sentinel]

okay

two lines are perpendicular if the product of their slopes is -1

and you are given the slope of L1

finding m of L2 should be straightforward

[long meteor]

is it 4x + something

[robust sentinel]

-1/-4 is not 4

[long meteor]

so -1/-4 + something?

[robust sentinel]

yes

[long meteor]

what is the + thing

[robust sentinel]

let's simplify to 1/4

ok so you are given a point

(4, 6)

and you know the slope of L2

[long meteor]

+5 is it?

[robust sentinel]

yes

[long meteor]

I just put the 4 and 1/4 into y=mx+c then figured it out

Can you help me on 4 more pls?

[robust sentinel]

i'll do what i can

[long meteor]

Ive been struggling on these for ages but no one helped me yesterday

same with the day before 😭

[long meteor]

Do i just send all of them?

[robust sentinel]

nah let's do one by one

im assuming gradient = slope?

not familiar with that term

[long meteor]

Yes

[robust sentinel]

okay good

[long meteor]

or gradient is the m bit of y=mx+c

[robust sentinel]

so both lines have the same slopes

since they're parallel

[long meteor]

Idk if this is it or not but is it (change in y) / (change in x)

[robust sentinel]

okay so it's the same

so same thing as before, try to find y=mx+c for both lines

you know the slope is 3 and a point for each line

[long meteor]

Where did you get 3 from??

[robust sentinel]

from the question

[long meteor]

oh im blind 💀

Sorry

[robust sentinel]

happens

[long meteor]

Do i just put it into the equation

so 7=2x3+c

so c=1

Is for a its (0,1) and b (0,2)

[robust sentinel]

right so that's like G

[robust sentinel]

you're right

soo all good?

[long meteor]

I have 3 more

2 more on the lines 1 on bounds

the bounds one i might be able to do

[robust sentinel]

okay

[robust sentinel]

well you're doing good from what i'm seeing

maybe try those by yourself and tell me if you get stuck?

[long meteor]

If you get it wrong twice it changes the question

thats why im struggling loads

Ok

[robust sentinel]

i see

[long meteor]

5x-8?

@robust sentinel

[robust sentinel]

seems right

[long meteor]

next one

This one is confusing me im worse at the perpendicular

its 1/3x - something

[robust sentinel]

remember two lines are perpendicular if the product of their slopes is -1

yes and you can find x at y = 0 for line D

so that you get a point to find the equation of E

[long meteor]

its 7,0 for the cordinates on the y=0 line

but 7x1/3 is like 2.333333 and lots more 3's

[robust sentinel]

you can keep it as a fraction

[long meteor]

is 2.333333... equal to 7/3

[robust sentinel]

indeed

[long meteor]

So this is the equation?

[robust sentinel]

seems good

didn't really check

[long meteor]

it said its incorrect

[robust sentinel]

okay

[long meteor]

it changed it 😭

[robust sentinel]

well it's the same thing

[long meteor]

Can you just tell me the anwer

cause you can probably answer it really quickly

is it 1/3 - 5/3

1/3x - 5/3

[robust sentinel]

oh i see where you went wrong last time

when finding P

you set x = 0 and not y

im guessing

[long meteor]

Thats what i did

[robust sentinel]

it's asking y = mx+c

and you just kept the P

yup

[long meteor]

Could you just tell me the answer to this one cause i gotta go in like 3 mins sorry

I can do it tomorrow morning

I got it

1/3x - 5/9

1 last one

[robust sentinel]

yup

[long meteor]

I might be able to do the bounds one but i accidently like type in the wrong decimal then i have to fully redo it

73.5+6.25 for the lower and 74.5+6.35 for the upper?

[robust sentinel]

im not sure what the question is asking for tbh

oh right

[long meteor]

i think you have to find what the lowest number could be before rounding

so if you had 73.5 you would round it to 74 i think

[robust sentinel]

oh okay

[long meteor]

I read it wrong again its decreassed

73.5-6.35 for higher 74.5-6.25 for lower i think

Is that it?

[robust sentinel]

seems right

[long meteor]

68.25 (Upper)
67.15 (Lower)

I got it right

finnally done with it 😭

[robust sentinel]

good job!

[long meteor]

Tysm for all the help

[robust sentinel]

all you dude haha

gotta trust yourself

[long meteor]

Ik but i make like tiny mistakes then have to redo it all fully

[robust sentinel]

yeah i understand

you can close the channel if you're done

have a good day!

[long meteor]

You two

Can i friend you so next time i get stuck i can ask you

otherwise ill have to wait like 2 days again for someone to respond

[robust sentinel]

yeah, might not be available always tho

but sure

[long meteor]

Its fine

I gtg now ill cya next time i need help maybe

.close

[pearl pond]

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[spark tinsel]

Hey

[spark tinsel]

Just needing some help

With a question

In regards to sectors.

Anyone willing to help

Wolverines, you willing to help?

?

[calm wing]

just ask

send the problem

[spark tinsel]

Ok

Sending now

The first one

@calm wing

[calm wing]

ok

[spark tinsel]

So we have a sector of a circle

[calm wing]

yes

[spark tinsel]

The arc length is 15

Angle is 41

I need the radius

[calm wing]

yes

[spark tinsel]

You think you fan help?

Can*

[calm wing]

what's the formula for the arc length

[spark tinsel]

I've got the arc length

It's 15 look at the thing I send

You need to press on it there is a question above not the one below

You there ?

[calm wing]

bruh

ik

what's the formula for the arc length

you need to use the formula for the arc length to find the radius

[spark tinsel]

41/360 x pi x rx2

Rx2 is basically diameter

Ay dawg you left me in the dust

@calm wing

[calm wing]

you don't have to tag me evrey minute

[spark tinsel]

41/360 x pi x d

That's the formula

[calm wing]

yes

[spark tinsel]

We need the d

[calm wing]

yes

[spark tinsel]

We Want it and we need it.

[calm wing]

so, equate this formula to the arc length you have

15

[spark tinsel]

After d we will divide it by 2

Yes

Arc length is 15

[calm wing]

yes

equate the formula and the 15

[spark tinsel]

41/360 is what we gotta do first right?

[calm wing]

$\frac{41}{360}\cdot\pi\cdot 2\cdot r = 15$

[jolly parrot]

artemetra

[calm wing]

solve for r

that's it

[spark tinsel]

How do I solve

[calm wing]

do you know how to solve

uh

equations

[spark tinsel]

Yes

[calm wing]

in general

[spark tinsel]

Yeah

[calm wing]

well, what's the issue here

so i can explain

[spark tinsel]

Alr so 41/360 is 0.3916 recurring

Then we x by pi

[calm wing]

you don't need that info but sure yeah

[spark tinsel]

Wait no

0.1138 recurring

[calm wing]

it doesn't matter

[spark tinsel]

X pi

[calm wing]

you don't need decimals rn

[spark tinsel]

Oh

Ok so after all of that we have 0.3578

[calm wing]

also in a test you might be asked to find an exact solution

without decimals

so you better learn that

[spark tinsel]

So basically

[calm wing]

for r?

[spark tinsel]

No

That's 41/360 x pi

Then what do I gotta do?

[calm wing]

let me show you a general way

[spark tinsel]

Ok

[calm wing]

so, you want an expression for r

but you have all this messy stuff in front of it

$\frac{41}{360}\cdot\pi\cdot 2$

[jolly parrot]

artemetra

[calm wing]

wouldn't it be nice if we got rid of it?

[spark tinsel]

Yeah

[calm wing]

what you can do is divide by it both sides

[spark tinsel]

0.7156 is the answer to that

[calm wing]

to what

[spark tinsel]

41/360 x pi x 2

Ok if ubwere to solve it how would u do it

[calm wing]

sure

[spark tinsel]

Can you tell me a direct answer and tell me where you got it from using what formula

[calm wing]

by dividing both sides we get

$r=\frac{15}{\frac{41}{360}\cdot\pi\cdot 2}$

[jolly parrot]

artemetra

[calm wing]

this can be (and should be) simplified though

[spark tinsel]

Oh

Ah so this was it after all

Ah so basically

Pliz correct me on this

But

Arc length divided by the angle/ 360 x pi x 2

[calm wing]

yes

[spark tinsel]

Is the radius 20.96

I appreciate you alot.

[calm wing]

,w 41/360 * pi * 2 * 20.96

[jolly parrot]

Results provided by WolframAlpha

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

Requested by arte_m_etra

[calm wing]

yes

[spark tinsel]

Wait

[calm wing]

20.96

you got it

[spark tinsel]

Why it say 14.9987

[calm wing]

pi has infinite digits

[spark tinsel]

Ahhh yes

[calm wing]

your calculator has finite precision

[spark tinsel]

Brother

If you do not mind

I will ask you some questions about some other topics

Oh shit

Yo

I got 20.978 on the calculator same question

[calm wing]

better accuracy

yk what

put 21

as your answer

[spark tinsel]

Oh nvm

I forgot to include all digits

Anyways it was very good learning from you

Wait

[calm wing]

np

[spark tinsel]

Also

Also

I got another one

[calm wing]

ok

[spark tinsel]

Last one

Need to find radius

It it just the same equation

[calm wing]

60 cm^2 is area?

[spark tinsel]

Looks like i5

It

[calm wing]

use the formula for area

and do the exact same thing lol

i gtg now, sorry

[spark tinsel]

Nooo

[calm wing]

there will be other people to help you

[spark tinsel]

Last question man

No you are the one

Cmon dang

For me

Ok atleas say if I'm on the right tracks

131/360 x pi x r squared

.Close talk to me tomorrow bro

.close

[pearl pond]

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[limber pivot]

"How many numbers in the closed interval [1, 500] are divisible by 3 or 5 but not by 7?"
I found all the total of numbers that are divisible by 3 or 5 or 7 by principle of inclusion-exclusion which gave me the number 271, then subtracted the numbers divisible only by 7 (71), leaving me with 200 numbers. Is this correct?

[vestal tapir]

200 is right

i wish i understood what you did

[shrewd basin]

They just found the number of numbers that are divisible by 3,5,7 and then number of numbers that are divisible by 15,21,35,105 now you can find how many numbers are divisible by 3 or 5 or 7 out of which 71 are divisible by 7.

Tho…

[pearl pond]

@limber pivot Has your question been resolved?

[pearl pond]

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[valid owl]

Q = ρ ∗ V ∗ cp ∗ ∆T
I am given this formula with an array for x which is final tempature and y is Q.

[valid owl]

I used linear regression on this array to get an a0 and a1 and now i need to solve for L and Ti
all other varibles are known,
i just need help setting this up for how to get L and Ti using my known varibles plus a0 and a1

so I have Q = a0 + a1* Tf

and Q = ρ ∗ V ∗ cp ∗ ∆T

I was able to get L = a1 / (p * area * cp)

but am confused on how to get Ti

Ti = tempature initial and Tf = tempature final with P A and Cp all being known values

I am unble to use Q and Tf as a known value since those are my x and y values of the array

.close

[pearl pond]

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[west condor]

I got this, but photomath says it's wrong

[warm current]

,rotate

[west condor]

.close

[pearl pond]

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[pearl pond]

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[compact token]

Can someone please help me with this problem?

[compact token]

<@&286206848099549185>

[thorn hamlet]

yes

[compact token]

so am I supposed to use moment generating function here?

is that the only way?

[thorn hamlet]

Yes

[compact token]

is there no other way to do it tho?

[thorn hamlet]

nope

[compact token]

So the thing is that I have not given much importance to moment generating functions, could you tell me if yk any example or similar problem like this?

[thorn hamlet]

what grade are you in

[compact token]

I am in uni

[pearl pond]

@compact token Has your question been resolved?

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[outer hare]

How would I solve this inverse laplace?

[west sapphire]

have you tried factoring the denominator and using partial fractions?

[outer hare]

Idk what factors to use

[west sapphire]

it's a quadratic in s^2

if all else fails, you can use the quadratic formula

[outer hare]

Yeah let me go and try that

@west sapphire so i solved it but for some reason I cannot see my mistake in this problem

[west sapphire]

can you show what you did?

[outer hare]

the answer is

@west sapphire i got hte orginial y(s) wrong so i redid it

and it came out to

Im wondering if this is still correct just a different solution to the same ivp

or is that not possible?

[west sapphire]

this doesn't seem to be the same as the one you were asking about above

[outer hare]

It is, I just made a mistake on the original y(s)

[west sapphire]

i see

[outer hare]

its the same problem just fixed a mistake prior to the inverse laplace

but it still is incorrect and I really am having trouple figuring out where I went wrong

[west sapphire]

so when you factor (s^4 + 5s^2 + 4), did you get (s^2 + 1)(s^2 + 4)?

[outer hare]

Yes

[west sapphire]

and what did you get for your partial fractions

you seem to have skipped a few steps

[outer hare]

I put this one through wolfram because i just want to see if my other steps were correct

but here:

its just doing partial fractions of s^2+1 and s^2+4

but im more curious on where I could have gone wrong with the problem, my prof solved it using the elemination method which is why I cant check my work as I am using laplace transforms

nvm i figured it out

I forgot to distribute one of the multiples

thank you for taking a look @west sapphire

.close

[pearl pond]

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[midnight haven]

I need help on algebra 1.

[midnight haven]

Can I get a review on Scatter Plots & Trend Lines?

[plush bramble]

[pearl pond]

@midnight haven Has your question been resolved?

[pearl pond]

@midnight haven Has your question been resolved?

[pearl pond]

@midnight haven Has your question been resolved?

[pearl pond]

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[inland lantern]

what is "the slope equation"

y = mx+b is the standard form for a line though

[pallid dust]

The slope formula is a general formula used to calculate slope between two points, m = (y2-y1)/(x2-x1)

and as @inland lantern said above y =mx+c is the standard form for a line

[pearl pond]

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[green arrow]

Does anyone knows how to solve this shenanigan? Thank You, I've run out of ideas, and I'm dumb, thanks to all of you and good luck

[green arrow]

This equals 2^9

And it must become that

[old marsh]

Cbrt (2^27 * (2 + 2^3)/10)

= 2^9 * cbrt(1)

= 2^9

[green arrow]

what

Could you make the bot run that?

[old marsh]

Sure

[green arrow]

thank you

[old marsh]

$(2^{27} \cdot (\frac {2 + 2^3}{10}))^\frac 13$

[green arrow]

Oh

shit I see it now

[old marsh]

Omg

[jolly parrot]

Stephen

[old marsh]

There we go

Do u see

[green arrow]

ya ya, now I do

fuck I would havr never get that

[old marsh]

It’s just a matter of seeing that u can factor out a big term out of that

With an exponent that is a multiple of 3

[green arrow]

Ya, but, it is late, I'm having a big fail in the test

oh well

thanks

[old marsh]

Np

[green arrow]

&close

$close

#close

[old marsh]

Use a period

[green arrow]

a what?

[old marsh]

.

[green arrow]

/close

oh

.close

[pearl pond]

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[green arrow]

damn that simple

[old marsh]

Lol

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[round trail]

Hello! I am kind of stuck on a proof. Can anyone help get me get started?

[round trail]

this is the Exercise 70 the problem refers to:

[pearl pond]

@round trail Has your question been resolved?

[humble lintel]

That page is a bit too short to really understand what the notation being used is without more mental gymnastics than I am willing to use.

If you've taken a course like probability theory, you can borrow a generating function approach by noticing that if you define M(x) = 1 + e^x + e^2x + .... then M'(0) is your left hand side of what you want to prove for k = 1; M''(0) is the left hand side for k =2, etc. You can then use a Taylor Expansion of M(x) and grind out the details directly without doing any induction.

[round trail]

i have not learned that 😭

is there any more information I can provide that might be helpful?

[humble lintel]

I don't know what exactly the a_n are or what the context of this course even is. If you want help it would help if you applied the hint in the question and posted your working because that path looks very tedious compared to the strategy I outlined. I'm definitely too lazy to write it down on paper but maybe someone else is more patient

[round trail]

oh
introduction to discrete structures
the chapter is on counting and the section the problem is from is about the binomial theorem
in the second image i think a_n can be anything idk

[round trail]

ok ill do that

i will close for now xd

.close

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[vale grail]

I'm thinking this is AAA am I right?

[sweet bone]

having 3 angles the same does not guarantee congruence, only similarity

you also need the fact that you have shared side lengths as well

[vale grail]

Oh

So theres not enough info then?

To determine it

[sweet bone]

no, they are congruent

the dashes on the sides mean they are the same length

[vale grail]

Oh

I'm just dum lol

[sweet bone]

you can work out the third angle on the right-side triangle then use a different congruence rule

[vale grail]

Gotcha

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[wild fable]

Not quite sure what to do from here

[edgy stone]

u substitution looks like it made everything worse

try just doing partial fractions on the first line

@wild fable

[wild fable]

I can’t

[edgy stone]

$\frac{1}{y(y+1)}$

[jolly parrot]

Kaisheng21

[wild fable]

The only integration techniques I’ve learnt are u-sub and the ln integration stuff.

[edgy stone]

can you do partial fractions on this?

[sweet bone]

try u = sqrt(x) + 1

[edgy stone]

no, don't do that

actually maybe do that idk

but yeah the point is like

[wild fable]

I don’t know partial fractions

I only know u-sub and the natural log integration

[edgy stone]

$\frac{1}{y(y+1)} = \frac{1}{y} - \frac{1}{y+1}$ \
\
$ \frac{1}{ {\sqrt{x} \left( \sqrt{x} + 1 \right) }} = ?$ \

[wild fable]

okay

[edgy stone]

suddenly i hate latex

ah i get it

[jolly parrot]

Kaisheng21

[wild fable]

That looks a lot more complicated than what I just did ngl

[edgy stone]

it isn't

[sweet bone]

normally partial fractions is nicer, but the u-substitution in this case makes the integral 2/u

[wild fable]

Well I mean I already told you guys I can’t do partial fractions

Because I don’t know partial fractions

[wild fable]

Is that what happens if you let u = sqrt(x) + 1

[sweet bone]

should be; dx = 2sqrt(x) du, so you have integral of (2*sqrt(x))/(u*sqrt(x)) = 2/u

[wild fable]

Yeah I got it

Then I just have to change the bounds and then yeah

That was a lot easier

.close

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[glad wing]

What in the world is three less than half the total number 💀 are they trying to make us solve a riddle in another language????

Need help to make equation out of this non sense

[glad wing]

I can do it out of normal story problems but this is just awful

Don’t know where to even start

[last summit]

If y equals 3 less than half of x

y=x/2-3

[glad wing]

X/2 is a fraction right

And then -3

[last summit]

Yes

[glad wing]

In other words can x/2 be like 1/2x or something

Or 2/1

[last summit]

It’s half of whatever x is

[glad wing]

So it can be 1/2 x ?

[last summit]

That is not the same

Unless you mean

(1/2) * x

[glad wing]

Oh

Yes

Yes yes

That’s what I meant

Ok Tysm!!!

Wait

What would be the inequality sign

@last summit

Helloooooo

[glad wing]

How to figure out what the inequality sign is????

For that equation

Pls

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[vale grail]

How can I prove these are/are not congruent? Could I just place the left triangle over the right and call it a day?

[pallid dust]

Do you think they are congruent?

[vale grail]

Yeah

[calm wing]

they are congruent

[vale grail]

Is there a formula for this?

[pallid dust]

not sure what type of detail you need but you can prove SSS here with the distance formula if needs

[vale grail]

Yeah I do

Gotcha

[pearl pond]

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[severe bolt]

how would you approach the problem?

[glass meadow]

1. write the equation for the tangent at P
2. find Q
3. 2Qx = Px
4. ???
5. plug in 2, equate to 1

[pearl pond]

@severe bolt Has your question been resolved?

[severe bolt]

.close

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[river sun]

is this correct

[pearl pond]

@river sun Has your question been resolved?

[shrewd crescent]

looks right

[river sun]

wording too?

[shrewd crescent]

yeah think so, you got the the three stages and the conclusion

[pearl pond]

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[rancid crow]

Not too sure whether this statement is true or false

[rancid crow]

"For all real numbers x there exists a real number y such that if x < y, x^2 < y^2"

I think it is true

[sacred crest]

Well Z is the set of integers

[rancid crow]

ohhh

still think its true

[sacred crest]

It is but why

Its not as obviously true as it might seem (nvm actually it is, but there’s still an interesting property here)

[rancid crow]

because if you have a negative number as x

you can make y have a larger modulus value

and if x is positive

[rancid crow]

still this

is my logic correct

[sacred crest]

I think so yes

How would you prove this?

[rancid crow]

I have no idea tbh

base cases?

[sacred crest]

Well one approach would be to say “let x be some integer” and then come up with some y for which the conditional is true

Now the fun part: When is a conditional true?

[rancid crow]

if that first statement is true and the second statement is true

and if the first statement is false

[sacred crest]

P ⇒ Q is true if ¬P and/or Q

Can you see why this makes the proof very simple?

[rancid crow]

hmm

not sure sorry

[sacred crest]

Well, all we need to do is show that for any x, there exists some y such that ¬P

What’s ¬P here?

[rancid crow]

x >= y?

[sacred crest]

Yes

[rancid crow]

ohhhhh

so a proof would be let y be >= x?

[sacred crest]

Yeah, just say y=x

In that case, x<y is false, making the conditional true

[rancid crow]

that makes sense!

thx ❤️

[sacred crest]

Glad I could help!

And just in case you want a more “satisfying” proof, you could choose $$y=|x|+1$$ for which both $y>x$ and $y^2>x^2$ are true

[jolly parrot]

FirstNameLastName

[pearl pond]

@rancid crow Has your question been resolved?

[pearl pond]

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[arctic breach]

Let A(4,−4) and B(9,6) point on the parabola,
y^2=4x. Let C
be chosen the arc AOB
of the parabola, where O
is the origin ,such that the area of △ACBis maximum.Then, the area (in sq.units) of △ACB is:

[sudden pumice]

have u been taught the parametric cordinates of parabola

[arctic breach]

yes

I got till the area formula

[sudden pumice]

then whats the problem

[arctic breach]

but i don't know what does maximum area mean

I don't have 't' for c point

[sudden pumice]

can u show how u solved

[arctic breach]

ok wait

avsec

@sudden pumice

[sudden pumice]

since point c lies on arc ab, the range of t3 would be (t1,t2)

[arctic breach]

(-2, 3)

but what about the

maximum area

[sudden pumice]

t belongs to (-2,3) so( t-1/2)^2 would belong to (0,25/4) then ....

[arctic breach]

didn't understand

[sudden pumice]

uhh

is this your homework question

[arctic breach]

no

I mean

[sudden pumice]

practice?

[arctic breach]

it's a pyq

in solution it's directly give as t3 = 1/2

[sudden pumice]

u are preparing for jee?

[arctic breach]

yes

[sudden pumice]

same here

[arctic breach]

oo

2024?

[sudden pumice]

no 2025

[arctic breach]

ohh

[sudden pumice]

i am in 11

[arctic breach]

good

do you know this?

[sudden pumice]

yes

i solved

it

[arctic breach]

then help

pls

[sudden pumice]

from where did u not understand

[arctic breach]

why is t= 1/2

for maximum area

[sudden pumice]

when did i say that

[arctic breach]

in solution

it's given

[undone quail]

wassup

[sudden pumice]

how would i know the about solution u are reading

[arctic breach]

what answer did you get

[sudden pumice]

the answer is 31.25

[arctic breach]

ya

how did you do

can you show

[sudden pumice]

wait

u got area= 5|(t-1/2)^2 -25/4|

right?

[arctic breach]

yup

[sudden pumice]

the the range of t is [-2,3]

[arctic breach]

true

[sudden pumice]

the range of t-1/2 is [-2-1/2,3-1/2]=[-2.5,2.5]

[arctic breach]

ya

true

[sudden pumice]

then the range of (t-1/2)^2 is [0,(2.5)^2]

[arctic breach]

yaaaa

ya

makes sense

[sudden pumice]

then the range of (t-1/2)^2-25/4or6.25 would be [-6.25,0]

[arctic breach]

ya

[sudden pumice]

then take its mod which would be [0,6.25]

[arctic breach]

ya

[sudden pumice]

then multiply by 5 and u got the range of the area

from there take maximum value

[arctic breach]

oh

ya

wow

yaa

[sudden pumice]

bye and best of luck

[arctic breach]

thanks man

.close

[pearl pond]

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[pearl pond]

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[jolly parrot]

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

[pearl pond]

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[midnight haven]

Trying to solve this problem with hyperbolic trig sub $$\int x^3 \sqrt{x^2 - 100} dx$$

[jolly parrot]

Several people

[midnight haven]

My first step is $ x = 10 cosh(\theta)$ which makes the integral $$10000 \int cosh^3(\theta) \sqrt{100}\sqrt{cosh^2(\theta) - 1} sinh(\theta) d \theta$$

[jolly parrot]

Several people

[midnight haven]

simplified
$100000\int cosh^3(\theta) sinh^2(\theta) d \theta$

[jolly parrot]

Several people

[midnight haven]

Have i made a mistake yet?

then next step is $100000\int cosh(\theta) (sinh^2(\theta) - 1) sinh^2(\theta) d \theta)$

[jolly parrot]

Several people

[midnight haven]

then u-sub $u = sinh(\theta)$ $$du = cosh(\theta)$$

[jolly parrot]

Several people

[midnight haven]

$100000\int u^4-u^2 du$

[jolly parrot]

Several people

[midnight haven]

Right?

[near haven]

where did the 10000 come from?

oh wait no

i see it

[near haven]

$\cosh\left(x\right)\left(\sinh^{2}\left(x\right)+1\right)\sinh^{2}\left(x\right)$

[jolly parrot]

CH3

[near haven]

fourth step

[midnight haven]

oh thank you

.close

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[outer drift]

new question:
what about this

[outer drift]

i'm assuming I

end up with 7x^3/7x^4

but idk where to go from there

[thin lynx]

plug in infinity

why did u get rid of e^-x

[outer drift]

oh

[outer drift]

how would you plug in infinty?

[thin lynx]

Well u think of it as magnitude

so like 7(infinity)^3 would be really really big

[outer drift]

but like

[outer drift]

I can't really plug infinity in can I?

[thin lynx]

Well that is why you think of it as magnitude

Or u can actually just plug into your calculator a huge number

[outer drift]

magnitude?

[thin lynx]

99999

[outer drift]

so I just do 7(999999)^3

and solve?

[outer drift]

yes but in math

i understand

the word

[thin lynx]

Magnitude is the

size of a number

abs value of it

like the girthiness

[outer drift]

lol

but that still makes no sense

you can't just plug in 999999

in for x

there has to be better way of doing this

[thin lynx]

no that was an example

U plug in finity

[outer drift]

bro

idk what that means

[thin lynx]

U think of it

[outer drift]

im sory

[thin lynx]

as big number

over small number

[gloomy hare]

7x^4e^-x as x->infinity = 0
so no, you won't be left with 7x^3/7x^4

[thin lynx]

= infinity

[outer drift]

okay

[thin lynx]

so for example

[outer drift]

lets resest

i have 7x^3/7x^4e^-x

right now

[gloomy hare]

no

[outer drift]

no?

[thin lynx]

no

[outer drift]

idk how I'm supposed to plug in an imaginary humber

[gloomy hare]

one second i'll write something

[thin lynx]

No get past this

[gloomy hare]

$\frac{7x^{3}+14x^{2}-5x+4}{x^{3}\left(7xe^{-x}+10+\frac{90}{x}-\frac{15}{x^{2}}+\frac{7}{x^{3}}\right)}$

[jolly parrot]

Combustion

[gloomy hare]

$\frac{7+\frac{14}{x}-\frac{5}{x^{2}}+\frac{4}{x^{3}}}{7xe^{-x}+10+\frac{90}{x}-\frac{15}{x^{2}}+\frac{7}{x^{3}}}$

[jolly parrot]

Combustion

[gloomy hare]

now you should be ready

[thin lynx]

That was good

[outer drift]

that

still doesn't really make sense

[gloomy hare]

which part

[outer drift]

i'm trying to plug in infinity?

why did you do that btw

[gloomy hare]

well not "plug in"

[thin lynx]

infinity in denominator turns it to 0

^ hint

[gloomy hare]

^

[outer drift]

do I just remove all the x

[thin lynx]

Think about that

[outer drift]

and the numbers associated with it

[gloomy hare]

no

are you studying precalc?

[outer drift]

this isw aht ti's showin gme

take higher power numbers

[gloomy hare]

sure you could do that

but in this case

won't work

[outer drift]

which in my case is 7x^3/7x^3x^-x

oh

why not

[gloomy hare]

$lim_{\left(x\to\infty\right)}\frac{7x^{3}}{7x^{4}e^{-x}}=\frac{1}{0}$

[jolly parrot]

Combustion

[outer drift]

why did you put it in this form

[outer drift]

so the limit doesn't exist

[gloomy hare]

it does

simplifying it incorrectly doesn't mean that it doesn't exist

[gloomy hare]

what's 1/x as x->infinity

[thin lynx]

would 1/99999 be close to infinity, -infinity, or 0

[outer drift]

okay

so first

I'm simplfying

yes

and when simplygin, i get the equation taht you got

[gloomy hare]

can you answer the question?

[outer drift]

which is this

[outer drift]

0?

[gloomy hare]

yeah

well

[outer drift]

it's infinty

[gloomy hare]

yeah

[outer drift]

x can be anything

[gloomy hare]

alright

[outer drift]

so 1/x is trying to get to 0

[gloomy hare]

so

[outer drift]

but it never will

it'll just get smaller

infintieyl smaller

1/9999999

[thin lynx]

basically 0

[outer drift]

yes

[thin lynx]

good

[gloomy hare]

basically simplified it this way so everything becomes ->>> some constant + 0 + 0 + 0 divided by some constant + 0 + 0
because 1/x as x->infinity is 0

[outer drift]

which is what you got

the form

[gloomy hare]

yep

so

now just evaluate it

[outer drift]

exactly but

how

like

do I put it in my calc

[gloomy hare]

no

[thin lynx]

U dont

[outer drift]

you see that's my confusion

idk what you guys mean

by evaluating

[thin lynx]

this is not calculations

this is manipulation

[gloomy hare]

okay just "plug" in infinity

[outer drift]

infinty

just 0

oh

do I just remove all x numbes

[thin lynx]

so everything with x in denom goes away

[gloomy hare]

^

[outer drift]

so 7/10

[gloomy hare]

yeah

[outer drift]

well

[thin lynx]

that is why we put that into the form

[outer drift]

i could have just done 7x^3/10x^3

and then gotten rid of x^3

7/10

but the answer is 0

so that doesn'tmake sense

[gloomy hare]

the answer is 0?

[outer drift]

7/10 = 0.7

[gloomy hare]

,w limit as x-> infinity \frac{7+\frac{14}{x}-\frac{5}{x^{2}}+\frac{4}{x^{3}}}{7xe^{-x}+10+\frac{90}{x}-\frac{15}{x^{2}}+\frac{7}{x^{3}}}

sin(x)/x lmao

[thin lynx]

Huh

Thats wild

[outer drift]

soooo

is it 0.7 or 0

[gloomy hare]

let me check with wolfram

[outer drift]

alrighty

[feral leaf]

You realize you typed -infty here, right?

[gloomy hare]

oh there you go

[outer drift]

oh you genius

[outer drift]

ty

so limit exists at 0.7

btw how would oyu do this

[gloomy hare]

the limit is equal to 0.7

[gloomy hare]

power rule

[outer drift]

I get 1

but i don't think that's right

[gloomy hare]

$x\sqrt{x}=x^{\frac{3}{2}}$

[jolly parrot]

Combustion

[thin lynx]

combine your x

[outer drift]

how'd it turn into 3/2