help-23

[lean otter]

I see

[fiery merlin]

But that wouldn't work if you need a cutoff point.

[lean otter]

my thing is the teacher said there would be a restriction on domain

[fiery merlin]

1/x goes between y = 0 and 1 forever after that.

[lean otter]

so if the domain is X>0 the trig wouldn’t work?

[fiery merlin]

If you're limited to the basic trigonometric functions, yes, I think so.

Maybe one of the hyperbolic functions like sinh might work, but I haven't really learned them well.

Also, that might not count as trigonometric.

[lean otter]

Yeah i guess I just have to hope that’s not the domain then

Also I was thinking of drawing a linear function as kind of a guide for my functions or would that not be a very good idea

[fiery merlin]

If it was, I would think more about it and see if I was wrong. If I still thought I was right, I'd just say why I thought it was impossible.

Why do you think that could help?

[lean otter]

actually im not too sure

Nevermind

Okay though I understand

thank you very much for your help and your time

[fiery merlin]

You're welcome.

[lean otter]

.close

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[coarse rain]

what is the best way to do f?

[coarse rain]

in the least confusing way

without using integrals

[hardy monolith]

graph it

and count the squares under the curve

[coarse rain]

nooo

i mwna

acc

can you js tell me what im doing wrong

im starting from 3

and since im going forward for 3 seconds

or the velocity is positivie i mean

since the firstr derivative test tells us we are positive for until 3 and after 5

and negative in between

i have a line drawn from 3m to s(3) = 57

and then I have a line drawn from 57 backwards to 53 which is s(5), since in that interval the velocity is negative

which means the particle moved back 4

so then why is it not just 54 + 4 = 58

but answer is 112m somehow

oh wait

i forgot to move forward 3 seconds again

i thought id add the times lmao

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[obtuse jetty]

Can someone explain how to find side length CD

[hardy monolith]

!status

[safe radish]

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

[obtuse jetty]

1

[hardy monolith]

do you know sohcahtoa?

[obtuse jetty]

yes

[hardy monolith]

do you know how to apply that here?

[obtuse jetty]

not really i dont know how to find any lengths unless i made a right triangle using the 3m side as the hypotenuse

but i dont think any of those would help

do i need to make a triangle using side 3 and 5?

[west crow]

'

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[wraith breach]

Can someone help me with a couple of problems?

[fallen heart]

Assuming the television is a rectangle, you can construct a right triangle and then solve for the third side (the width)

[wraith breach]

huh?

[neat kiln]

Draw and label the tv and so the math

[plucky elk]

is there even enough information

[fallen heart]

Should be

[plucky elk]

oh rectangle yes

$a^2 + b^2 = c^2$ for right triangles with hypotenuse length c

[flat frigate]

riemann

[plucky elk]

and side lengths a, b

[wraith breach]

what???

[fallen heart]

Do you know the Pythagorean theorem?

[wraith breach]

yes

[plucky elk]

you know c and one of a or b

solve for the missing side

[fallen heart]

The rectangular TV is secretely just two right triangles

[wraith breach]

yeah

[fallen heart]

Since you have the diagonal length and one of the side lengths, you can use PT to solve for the last side length

Which is what you need to answer your problem

[wraith breach]

what is PT?

[fallen heart]

Pythagorean Theorem

[wraith breach]

oh ok

[fallen heart]

Pythagorean is hard to spell

[wraith breach]

it is

wait but we have height and length, and we need to find width so is it not a rectangluar prism?

[fallen heart]

It's just asking for two dimensions

Width and height

[neat kiln]

It says diagonal height given

Length*

Do you know what is diagonal of a rectangle?

[wraith breach]

no

[fallen heart]

Ah

AD is the diagonal line

[wraith breach]

ok

[fallen heart]

You see how it divides the rectangle into two right angles?

[wraith breach]

yes

[fallen heart]

You have the length of AD and the length of AC

You can rearrange $a^2 + b^2 = c^2$ to solve for CD

[flat frigate]

eyzk

[wraith breach]

so 68^2-45^2?

[plucky elk]

that gives you the square of CD

[wraith breach]

ok

soo i got 2599

[fallen heart]

Find the square root of that

[wraith breach]

51?

[plucky elk]

,calc 51^2 + 45^2 - 68^2

[flat frigate]

Result:

2

[plucky elk]

close enough i guess

[fallen heart]

Interesting

[fallen heart]

Since it asked you to round, I think that answer would suffice

Or 51.0 if you wanna be picky

[wraith breach]

ok

i still have more though, is that ok?

[fallen heart]

Of course

[wraith breach]

i could not understand this

[fallen heart]

I don't like how this problem is worded

[wraith breach]

my teacher isnt the best at wording

[fallen heart]

But this is also stuff with Pythagorean theorem

[wraith breach]

oh ok

[fallen heart]

I think drawing out each problem would help

[wraith breach]

ok except i have no clue how this even makes sense

[fallen heart]

I think it's meant to mean that you're leaning the ladder against the building such that the top of the ladder touches the top of the building

Like

Like this beautiful drawing

[wraith breach]

ohh

but then what about the rest of the problem? like the bottom of the ladder and the bottom of the building

[fallen heart]

That unknown distance is from the bottom of the ladder to the bottom of the building, and you have to again solve for it by rearranging the Pythagorean theorem

[wraith breach]

ok that makes more sense

28^2-15^2?

[fallen heart]

Assuming that the building meets the ground at a 90 degree angle

[fallen heart]

Yess

[wraith breach]

ok

559?

[fallen heart]

Yep now find the square root of that and round to the nearest tenth

[wraith breach]

23.6

[fallen heart]

Yep

[wraith breach]

ok

ok, i have 3 to 4 more problems

im gonna try this one on my own

[fallen heart]

Alrighty good luck

It's very similar to the previous TV one

[wraith breach]

16.5?

[fallen heart]

Yep

[wraith breach]

this one is also similar

i dont get this one that much though

[fallen heart]

Uhmmm

Me neither the phrasing is throwing me off

[wraith breach]

oh

can i see if another person can help?

<@&286206848099549185>

[green granite]

Think abb a right angle triangle

Longest length is 50

[wraith breach]

yeah

hello?

<@&286206848099549185>

[empty gyro]

Your initial ladder setup is like this

[wraith breach]

oh okkk

so 14?

but then after she moves it another 6 feet

do i just add 6 to 14?

[empty gyro]

and the problem wants you find how high the ladder reaches now (the purple line)

[wraith breach]

ok so i do add 6 to x

?

[empty gyro]

you need to find AB first

[wraith breach]

which is 14

[empty gyro]

So now you know what AB and DE are. Use that to find EA

[wraith breach]

20 is AD and then 50 for DE, so 50^2-20^2?

which is 2100 then i root it?

which is 45.8? or would i do 46

hello?

.close

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[sonic bramble]

Can someone check if I did c) correctly?

[woven hound]

what did you get for m

[sonic bramble]

-1

[woven hound]

uh that is not what I got

[sonic bramble]

you have to take the derivative of

$g(x) = xf(x)$

[flat frigate]

Gen

[sonic bramble]

right?

$g'(x) = x*f'(x) + f(x)*1$

[flat frigate]

Gen

[sonic bramble]

$g'(2) = 2*f'(2) + f(2)*1$

[flat frigate]

Gen

[sonic bramble]

$g'(2) = 2(-1) + 6$

[flat frigate]

Gen

[sonic bramble]

$g'(2) = -2 + 6$

[flat frigate]

Gen

[sonic bramble]

$g'(2) = 4$

[flat frigate]

Gen

[sonic bramble]

is that what you got?

@woven hound

[woven hound]

yes

[sonic bramble]

oh ok

I think I tried integrating instead

thanks

.close

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[candid ocean]

How do you do c)

dont u just plug in y as 1500

bc i keep getting it wrong lol

[woven hound]

yes

[candid ocean]

bruhh why am i getting the wrong answer

I get 0.278 = t

answer key says 1.12

[woven hound]

,wolf 1500 = (2100 + 8t)/(1+0.5*t) + 150

[flat frigate]

Results provided by WolframAlpha

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

Requested by woshiari

[woven hound]

you probably just made a mistake

[candid ocean]

@woven hound do u see where 😭

[woven hound]

1350 * 0.5 is 675

you're doubling instead of halving

[candid ocean]

oh

oops

yikes

ugh thank u

[woven hound]

[candid ocean]

.close

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[safe radish]

Please don't occupy multiple help channels.

[quartz jasper]

how do i solve this problem?

[plain sky]

for better ease, convert one hour and 50 minutes to 110 minutes

110 minutes is 1/6 of the journey

so multiply that by 6 to get the full trip

[quartz jasper]

okay

so it takes him 660 minutes

[plain sky]

convert back to hours and then see which of the choices results in 3:30 pm as an end time

[quartz jasper]

ohohhhhhhhhhhh

so its 4:30 am?

[plain sky]

bada bing bada boom ya got it

[quartz jasper]

thanks zyk03

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[desert niche]

this might be a dumb question but... whats the area of this trapezoid?
my work:
A = (4/2)(15+4)
A = 38

[empty gyro]

[desert niche]

oh...

thanks for the help!!

.close

[empty gyro]

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[desert niche]

i might be blind

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[lavish basin]

So I have this inequality and I get the solution ±sqrt(2)

[lavish basin]

But I don't understand why this is the correct solution

[autumn kite]

show yr work

[autumn kite]

how do uk if the inequality wont change after u hv squared both the sides?

cuz -2>-3
but if u square
2<9

and uk the thing inside the root is always positive

and since its given like
root(...) >= x^2 + 2x-3
the Right Hand Side might be negative

[lavish basin]

Sorry, what is +ve and RHS?

[final loom]

Start with 4x^3 + 2x² - 12x + 5 ≥ 0

[lavish basin]

English is not my first language

[lavish basin]

Ok, thx

[final loom]

Hint: is equivalent to (2x - 1)(2x² + 2x - 5) ≥ 0

[autumn kite]

how do u figure it out soo fast?

[final loom]

._> this gave one root = (-1 - √11)/2 so one factor is (2x² + 2x - 5)

alternatively p(1/2) = 0 so, 2x - 1 is factor

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[fast sierra]

heya, got a quick check i want someone to confirm

[fast sierra]

i have a formula as follows:

h is the final height, h0 is height at 0 seconds)

so can it be wrote as ( sqrt (delta h))

?

[autumn kite]

no

sqrt4 - sqrt1 = 2-1 =1
sqrt (4-1) = sqrt(3)

it can be written as delta (sqrt(h))

[fast sierra]

gotcha, thanks

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[zealous dirge]

i'm confused how to solve this problem
this is what i have already try i don't found any number can divide with 156
7 * 1^2 +56 * 1+91 = 156
3 5 7 9 11 13 17 19 23 29

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@zealous dirge Has your question been resolved?

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@zealous dirge Has your question been resolved?

[final loom]

A rigorous mental math-y question
7n² + 56n + 91 = 7(n² + 8n + 13) = 7[(n + 4)² - 3]
Now the question asks you to find a set of "p"s for which there exist a positive integer n such that p | 7n² + 56n + 91

That is, there exist a positive integer m = n + 4, such that p | 7[m² - 3]

Clearly 7 is included in that set, so we can update that condition to, p | m² - 3

So, we need to find a set of p's for which there exists positive integer solution to m² - 3 = 0 (mod p)
or m² = 3 (mod p)

It is easy to check that n = 2, 3, 7, 11, 13, 23 are first 6 members of this set

Can check for m from 9 to 15 to get four more members 37, 39, 47, 59. Now all that's left is to check if the left out primes: 5, 17, 19, 29, 31, 41, 43, 53 have no soln. m² - 3 = 0 (mod p)

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[final loom]

.reopen

[safe radish]

✅

[final loom]

._>

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[split moss]

How were the canonical vector images determined?

i.e it seems to me that they take the first colum = e1 , second column = e2 etc

[hardy monolith]

e1=(1,0,0,0)

e2=(0,1,0,0)

etc

[split moss]

yes

therefore the matrix associated with the map f with respect to the canonical basis is equal to the matrix

mmm

look

h = 0

h = 0

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[dim tusk]

Any idea how to solve this?

[dim tusk]

Not an assignment just wanted to see if there was a proof for this

Do you think it does?

[fathom adder]

Avoid, he was some kind of bot

[dim tusk]

Bro what do you mean?

[fathom adder]

Oh i thought you was asking to the one that proposes money for assignement just right before

Mb if not

[dim tusk]

Oh okay

This problem just came to my head when playing a game so I thought I'd see if anyone has any idea

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[main rain]

does anyone understand fouriertransforms. I get the concept but solving it without integrating it especially it its in the other direction. (Im german so excuse my bad language)

[limber scarab]

i am doing a math quiz the question is 100247356%99538x29016497= what is the answer

[heady widget]

When does maths get this hard?

[main rain]

mechanical engineering 2nd cear im new to this channel is that too hard for this channel ?

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[tacit whale]

Have any of you guys had any experience with the yt vid: calculus in 12 hours? And if so, would you say it covered Calculus AB

[plucky elk]

help channels are for specific math questions

ask in #math-discussion or #discussion

[tacit whale]

Oh sorry. will do

[severe pond]

why would you watch a 12 hour video on youtube

just use khan academy

[tacit whale]

i dont watch it all at once

i dont think khan has calc

[severe pond]

check again

it’s ap calc course is well developed

[tacit whale]

ok thanks bro

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[west crow]

Hi

[west crow]

Guys

I need

[final loom]

help?

[west crow]

Yes

Do u speak

Frensh

[safe radish]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

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

[west crow]

I need somone who expain to me integral dépend d'un paramètre

I need somone who expain to me integral dépend d'un paramètre

[final loom]

[safe radish]

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

[final loom]

Do you want someone to teach you the entirety of Integration?

[west crow]

What's entirety

[rugged arch]

[final loom]

mmmm to the rescue

[final loom]

intégralité

[rugged arch]

no 😭

[west crow]

No

[rugged arch]

that sounds like torture

[west crow]

I need

[rugged arch]

teaching someone the entirety of integration on text

[west crow]

A spesefic

[final loom]

Do you perhaps have a question on mind?

[west crow]

Integrals

[dire fjord]

maybe you are more suitable since you know french 😏

[rugged arch]

[safe radish]

@west crow Has your question been resolved?

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@west crow Has your question been resolved?

[main rain]

.reopen

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[glad dragon]

Is this key wrong? I got 2xe^x

I plugged in the upper bound and took the derivative as well

[plucky elk]

!show

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[icy lance]

you... havent asked a question (in the mathematical sense)

[bronze gust]

What is the question sir/madam

[quiet juniper]

Don't open multiple help channels

[bronze gust]

Yes you may

Alright question answered

[icy lance]

🫡

[severe pond]

lol

[bronze gust]

No clue frfr

[severe pond]

#help-4 message

swear you are the same person

yes

this is your alt

you made this account today

lmao

[quiet juniper]

do u know what an euler path is

[severe pond]

did you get banned?

[quiet juniper]

It's a path that goes through all edges exactly once

[severe pond]

what does that mean?

you were muted?

[quiet juniper]

So just look for the path that does that

Only one of A B C does that

why do you think it's B

it's basically a line connecting two points in those visual graphs

The instructions say it

"The edges are labeled 1,2,3,etc. And so on In the order in which they are travelled"

[pure agate]

The numbering on the edges show the order in which you travel across the graph.

The sequence of edges must be connected. Edge 1 must be connected to Edge2, Edge 2 needs to be connected to Edge 3, and so on.

For C, is edge 8 connected to edge 9?

For three of the four possible answers, there is no edge between some of the points.

Only one of the routes is valid.

I know what I think, what do you think?

If it were a no, I would have given a thumbs down.

Do you have any thoughts on an answer?

Why not take a screen capture of your screen instead of taking photos?

And which answer do you think it is?

[safe radish]

@fierce quail Has your question been resolved?

[pure agate]

Yes.

[safe radish]

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[zealous dirge]

all good @fierce quail ?

[zealous dirge]

.close

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[feral spoke]

If I have a 4% to win and I tried to win 10 times what’s the chance that I will win?

[drowsy moss]

that you'll win at least once? or are you going to stop playing once you win?

[feral spoke]

I will win once

Lemme rephrase it

If there was a game where where the prize I wanted was 4% for one spin and I spun the wheel 10 times all at once what is the chance I will win the prize at least once

[shadow glade]

it's easiest to calculate the probability that you "win at least once": you can calculate the probability that you lose 10 times in a row, which is simple: (0.96)^10 then subtract that from 1
so you end up with 1 - 0.96^10

0.335167364
so you have about 33.5% chance to win at least once

[feral spoke]

Oh

Ok hank you

.close

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[lean tapir]

This is what i can do so far

Ik i gotta find a derivative for y=ate^-kt to show the the maximum at x=1/k

<@&286206848099549185>

[young nexus]

if 1/k =0.25 shouldnt k be 4?

[lean tapir]

i cant even lie im hella lost

my tutor asked me to do this question for a third time now

I still got it wrong

[young nexus]

i dont know what you mean but 1/k = 0.25 is a simple calculation.

[lean tapir]

thats what my tutor said too lmao

i think he might be cooked

ok wait but is the derivative: ae^-kt (1-kt)

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[hushed forum]

I still remain with 3 variables (bottom)

[hushed forum]

System of 3 equations

[young nexus]

and your question is?

[hushed forum]

I made it up

I want to solve

I forgot what XYZ were

Help

[safe radish]

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[native robin]

i dont get how to do this

like arent the force by each charge the same no matter what R is

[light shoal]

no, if you make R bigger then the test point is at a greater distance from the two other charges

[native robin]

wouldnt the forces cancel out due to them being in opposite direction

[light shoal]

if the test point is on the same axis as the other charges then yes

otherwise only the component along that direction cancels

there will be a component orthogonal to that which will not cancel

(i.e. a component parallel to the plane)

think about what happens if R is very large, then it's almost as though the two q charges are a single 2q charge, and the test point is at distance R from it

[native robin]

Ok i got $F_{net}=2\frac{kqQ}{a^2+R^2}\sin\theta$

[flat frigate]

Jash

[native robin]

where theta is the angle from the horizontal to the test charge

i need to get thetea in terms of R tho

[light shoal]

yea

you can use sohcahtoa

get sin(theta) in terms of a and R

[native robin]

ohh u right i was trying to get it in terms of F and q

[native robin]

$F=2kqQ\frac{R}{(a^2+R^2)^{3/2}}$

[flat frigate]

Jash

[light shoal]

yea that looks right

so now you need to find R that maximizes this

[flat frigate]

Jash

[native robin]

do i have to do quotient rule

[light shoal]

notice that if R is 0 then F is 0 as you would expect

since then the test point is aligned with the two charges

yea quotient rule seems most straightforward here, unless there's a shortcut i'm not noticing

[native robin]

,w d/dR(2kqQR/(a^2+R^2)^1.5)=0

[flat frigate]

Results provided by WolframAlpha

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

Requested by jashgg

[light shoal]

something seems wrong, the answer shouldn't be negative

[native robin]

the answer is a/sqrt2 idk why there is negative sign

[light shoal]

yea i mean you get R^2 = a^2/2, i guess wolfram for some reason chose the negative square root

but only the positive root makes physical sense

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[dark vigil]

what number comes next?

[dark vigil]

help me

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@dark vigil Has your question been resolved?

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[lean otter]

yo

[lean otter]

can i get

help

i want notes on like basic statistics

and

visualization

[safe radish]

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[urban shadow]

i dont get part b of this question i just looked at the answer and went with it

[safe radish]

@urban shadow Has your question been resolved?

[mellow cypress]

What did you get for a?

[urban shadow]

6a/a+2

[mellow cypress]

And how did you get that?

[urban shadow]

well area of the rectangle is LxW

the x point is a

and the y is f(a)

so xy = af(a)

so i multiply a with the function when i sub in a

[mellow cypress]

So it’s dependent on f, which has a domain and range of what?

[urban shadow]

[0,6]

[mellow cypress]

And the range?

[urban shadow]

[3/4, 3]

?

[mellow cypress]

Yeah

So what do you think the implied domain and range would be now?

[urban shadow]

would it be [0,6]

[mellow cypress]

Which means the implied range is?

[urban shadow]

[3/4, 3]?

[mellow cypress]

Yes

[urban shadow]

but the answer says its (0,6]

[mellow cypress]

Ah right, slight caveat

You can’t have an area of 0

So the x can’t go down to 0

Does that make sense?

[urban shadow]

ohhhok ok

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[tight bear]

i need help

[strange spruce]

what is the problem?

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[strange spruce]

$ /int(/frac{cosx + secx}{cosx}dx)$

[violet drum]

Backslash

Also use {

Not brackets

[strange spruce]

im trying to undestand that

[thin bridge]

$\int \frac{\cos x + \sec x}{\cos x}\dd{x}$

[flat frigate]

ℝαμOmeganato5

[strange spruce]

thx

[bronze gust]

Split it up into cosx/cosx + sec x/cos x

At least I think that should work

[strange spruce]

$\int 1+ \frac{1}{\cos^2x}\dd{x}$

[violet drum]

It does

[flat frigate]

GreenV_rand

[violet drum]

Whats 1/cos2x

[bronze gust]

Then should be able to continue from here

[violet drum]

Im on ipad its long as hell to type symbols

[strange spruce]

np bro

[violet drum]

Di yk what to do?

[strange spruce]

i dont remember the integral o sec²x

[violet drum]

Its a simple one

[strange spruce]

i would like to prove that integral like derivates

[violet drum]

Do u know the derivatives of all trigonometric functions

[strange spruce]

yea

[bronze gust]

What's the anti derivative of sec^2 x

[violet drum]

Well sec2x is a derivative of a common one

[strange spruce]

tgx

sure, that is ez, but another ones are really hard

[violet drum]

Yeah there are many hard integrals

[strange spruce]

the answer will be x + tgx + C

[bronze gust]

if tg x is tan x then yea

[strange spruce]

i'll like if have another way to prove some integrals

[bronze gust]

There is a pretty intuitive way to derive the anti derivative of sec^2 x

[violet drum]

Wdym

[bronze gust]

So sec^2 x= 1/cos^2 x

So we can imagine this to be the derivative of some function taken with the quotient rule

so we know the denominator of this function must be cos x

So
(cos x*f'(x)+f'(x)*sin x)/cos^2 x = 1/cos^2 x

[strange spruce]

i know how to demonstrate every derivate formula

[bronze gust]

And we're tryna find f(x) such that this is true

And by inspection this is clearly sin(x)

So the antiderivative of 1/cos^2 x is sin x/cos x

[strange spruce]

make a lot of sense

an example: $\int \frac {1}{x^2 - a^2} \dd{x}$

[flat frigate]

GreenV_rand

[strange spruce]

finally

how can i get the answer without knowing it?

[rugged arch]

wdym knowing it?

the first best thing could naturally be to use partial fraction on it

[strange spruce]

i just cant prove that

[rugged arch]

can't prove what?

[rugged arch]

^

[strange spruce]

that integral = 1/(2a). ln((x-a)/(x+a)) + C

[violet drum]

I assume since its a common antiderivative

[strange spruce]

how do i get this?

[rugged arch]

well like mentioned before, try partial fractions

[strange spruce]

i dont know it yet

[violet drum]

Separate it into x+a x-a

The bottom

Then u want to separate that into 2 fractions

[rugged arch]

,, \frac{1}{x^2 - a^2} = \frac{\frac{1}{2a}}{x - a} - \frac{\frac{1}{2a}}{x + a}

[flat frigate]

mmmm7

[rugged arch]

how come?

it's not an "integration" technique

so it's assumed that you know this before calculus ig

[strange spruce]

wait a sec

what are u talking about?

parcial derivates?

[rugged arch]

no

[violet drum]

Partial decomposition

Partial fractions

[rugged arch]

,, \frac{1}{3x(x-1)} = \frac{1}{3(x-1)} - \frac{1}{3x}

[flat frigate]

mmmm7

[rugged arch]

the act of "splitting" the product in the denominator

is partial fraction decomposition

u learn this in algebra 2-ish

[strange spruce]

ok, i get this

[rugged arch]

so you know how to do this?

[strange spruce]

sorry bro, i dont know with that name

[rugged arch]

yeah and that's what i meant then

,, (x^2 - a^2) = (x - a)(x+a)

[flat frigate]

mmmm7

[violet drum]

Then do u know how to integrate this

[rugged arch]

,, \frac{1}{(x-a)(x+a)}= \frac{A}{(x-a)} + \frac{B}{x +a}

use partial fractions on this

[flat frigate]

mmmm7

[rugged arch]

where A and B are for you to find

[strange spruce]

lemme try

[rugged arch]

the answer is here but okay u try yourself and show why what i wrote is correct

[strange spruce]

,, \frac{1}{(x-a)(x+a)}= \frac {x(A+B) + a(A - B)}{(x-a)(x+a)}

[flat frigate]

GreenV_rand

[strange spruce]

after that?

[rugged arch]

yeah smth like that

now match polynomials

,, 1 = x(A + B) + a(A - B)

[flat frigate]

mmmm7

[rugged arch]

the coefficient of x on the left hand side is 0

so what should A + B be?

[strange spruce]

A = -B

[rugged arch]

yeah

[strange spruce]

ao A+B = 0

[rugged arch]

and now match the constant terms

[strange spruce]

but a(A-B) = 1

[rugged arch]

yeah

A - B = ?

[strange spruce]

a = 1/2A

or A = 1/2a

and B = -1/2a

interesting

[strange spruce]

then?

[rugged arch]

yeah plug it in to get this

i'm guessing you know how to integrate something like $\frac{3}{x - 5}$

[flat frigate]

mmmm7

mmmm7

[strange spruce]

ln(x-5)+C?

[rugged arch]

missing the 3

[strange spruce]

true

[rugged arch]

anyway so you know how to integrate this then

1/2a is a constant just like 3 was on the above problem

and yeah

[strange spruce]

ok

ok

i1ll try that

$\frac {1}{2a}ln|\frac{x-a}{x+a}| + C$

[flat frigate]

GreenV_rand

[strange spruce]

that's correct?

[rugged arch]

yeah

[strange spruce]

uhhhh

bro, thank u so much for helping me

[rugged arch]

sure np

[strange spruce]

teach me the anti deravite of 3 funtions

[rugged arch]

,, \int u'vw = uvw - \int uv'w - \int uvw'

[flat frigate]

mmmm7

[strange spruce]

forget it, my english isn't good

[rugged arch]

that?

[strange spruce]

sorry, i speak that wrong

[rugged arch]

yeah i thought so

use translator

or chatgpt to translate your question

[strange spruce]

yeah, i think that's better

i'm trying to calculate $\int lnx \dd{x}$

[flat frigate]

GreenV_rand

[rugged arch]

familiar with that?

[strange spruce]

so so

[rugged arch]

use it

ln(x) = ln(x) * 1

that's a product of two functions

[strange spruce]

yeah

bro $\int u.v'\dd{x} = u.v - \int v.u'\dd{x}$?

[flat frigate]

GreenV_rand

[strange spruce]

correct?

[strange spruce]

are the same formula?

[rugged arch]

yeah

[strange spruce]

im comfused

[rugged arch]

use this if you like

[rugged arch]

about?

[strange spruce]

how they are the same?

[rugged arch]

,, v \mapsto v'

[flat frigate]

mmmm7

[strange spruce]

kskssksksksks, thanks

ohhhhh x(ln x - 1) + C

[safe radish]

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[final schooner]

help with fourier transform

[final schooner]

i found the first part but im stuck with second part

how should i begin

[safe radish]

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[safe radish]

@final schooner Has your question been resolved?

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[slim sphinx]

Can someone help me solve the equation in induction please?

[final loom]

expand the LHS binom

[slim sphinx]

LHS stands for?

[final loom]

left hand side

[slim sphinx]

I tried I get stuck with $n\choose{2}$

[flat frigate]

Horsi135

[slim sphinx]

or $n\choose{k-1}$

[flat frigate]

Horsi135

[final loom]

LHS = $\sum_{k = 4}^n\frac{n!}{(n-k)!(k - 3)! 3!}$

[flat frigate]

Arya

[final loom]

@slim sphinx does anything click with this?

[slim sphinx]

currently solving for n+1

[final loom]

$= \frac{n(n-1)(n-2)}{3!} \left[\sum_{i=1}^{n-3} \binom{n-3}{i}\right]$

[flat frigate]

Arya

[final loom]

mathematical induction?

oh (@_@;) You wanted to do it using induction only

[slim sphinx]

yes

[final loom]

right, can you show your work?

Let me know if you're stuck somewhere

[slim sphinx]

I'm sending my work now

It's a mess

[final loom]

No you did amazing. I lost the trail after this, but I think seeing from what you did so far, and that you concluded.. you prolly did it right

[slim sphinx]

The math is correct but I couldn't make it equal in both sides

[safe radish]

@slim sphinx Has your question been resolved?

[slim sphinx]

<@&286206848099549185>

[final loom]

m sorry if it comes off as irritating but do you absolutely compulsorily have to do it using induction?

if so, I'll have to do it myself once before guiding you ahead ;-;

[slim sphinx]

I already proved it with combinatorics and now I need to prove it with induction

<@&286206848099549185>

[velvet marsh]

I'll try to help but what's this

?

[slim sphinx]

3

[velvet marsh]

Ok

Im Really Sorry but i have no clue

[final loom]

:p

[velvet marsh]

I wish you good luck at solving this

[final loom]

@slim sphinx Are you free for a couple mins? It'll take a bit to type the latex but you'd have to be patient

[slim sphinx]

I'm free for about an hour and a half

[flat frigate]

Arya

Arya

[slim sphinx]

yes

[flat frigate]

Arya

[final loom]

Then $P(m+1) = \sum_{k=4}^{m+1} \binom{m+1}{k}\binom{k}{3}$

[flat frigate]

Arya

[final loom]

$= \binom{m+1}{3} + \left[\sum_{k=4}^{m} \binom{m}{k}\binom{k}{3} + \sum_{k=4}^{m} \binom{m}{k-1}\binom{k}{3}\right]$

[flat frigate]

Arya

[final loom]

right?

[slim sphinx]

yes

[final loom]

$= \binom{m}{3}(2^{m-3} - 1) + \binom{m+1}{3} + \ \sum_{k=4}^{m} \left[ \binom{m}{k-1}\binom{k-1}{3} + \binom{m}{k-1}\binom{k-1}{2} \right]$

[flat frigate]

Arya

[final loom]

right?

[slim sphinx]

yes

[final loom]

$= \binom{m}{3}(2^{m-3} - 1) + \binom{m+1}{3} + \ \sum_{k=3}^{m-1} \binom{m}{k }\binom{k}{3} + \sum_{k=3}^{m-1} \binom{m}{k}\binom{k}{2}$

right?

[slim sphinx]

I think in the last one it should also be k and not k-1

[final loom]

which one, and where

[flat frigate]

Arya

[final loom]

okay now?

[slim sphinx]

yes

[final loom]

now notice that $$\sum_{k=3}^{m-1} \binom{m}{k }\binom{k}{3} = P(m) = \sum_{k=4}^{m} \binom{m}{k}\binom{k}{3}$$

[flat frigate]

Arya

[slim sphinx]

yes

[final loom]

so this equals: $$\binom{m}{3}(2^{m-2} - 1) + \binom{m}{2} + \sum_{k=3}^{m-1} \binom{m}{k}\binom{k}{2}$$

[flat frigate]

Arya

[final loom]

$= \binom{m}{3}(2^{m-2} - 1) + \sum_{k=2}^{m-1} \binom{m}{k}\binom{k}{2}$

[flat frigate]

Arya

[final loom]

right?

[slim sphinx]

the one above

[final loom]

hmm?

[slim sphinx]

.

[final loom]

Yes I skipped some manipulations

but you can check that for yourself

are you confused somewhere?

[slim sphinx]

I got $m\choose{3}$ not $m\choose{2}$

[flat frigate]

Horsi135

[slim sphinx]

in the middle

[final loom]

Hmm, (m+1 C 3) - (m C 3) = (m C 2)

the -(mC3) comes from 2*(2^(m-3) - 1) = (2^(m-2) - 1) - 1

[slim sphinx]

oh okay

[final loom]

Got it?

[slim sphinx]

yes

[final loom]

so we reached here

now we show that: $\ \sum_{k=2}^{m-1} \binom{m}{k} \binom{k}{2} = \binom{m}{2} \sum_{k=2}^{m-1} \left(\binom{m-2}{k-2}\right) = \binom{m}{2} (2^{m-2} - 1)$

$= \binom{m}{3}(2^{m-2} - 1) +\binom{m}{2}(2^{m-2} - 1)= \binom{m+1}{3}(2^{m-2} - 1)$

[flat frigate]

Arya

[final loom]

[slim sphinx]

how did you prove the second one

[final loom]

which one?

the second equality?

[slim sphinx]

yes

[flat frigate]

Arya

[slim sphinx]

actually both

[final loom]

you can show mC2 * (m-2)C(k-2) = mCk * kC2 by expanding either side

[slim sphinx]

ok

and the last one?

[final loom]

last one is just (m-2)C0 + (m-2)C1 + ... + (m-2)C(m-3) + (m-2)C(m-2) - (m-2)C(m-2) = (1+1)^{m-2} - 1

but if your teacher is against this kind of computation, you can derive this from P(m+1) as well.
P(m+1) = 4(m+1)C4 + Sum_{n=3}^{m} (m+1)C(k+1).(k+1)C3 = 4(m+1)C4 + (m+1) Sum_{n=3}^{m} mCk.kC2

[slim sphinx]

ok

thank you very much

[final loom]

in any case, with that, we showed P(m+1) is true as well

[slim sphinx]

I don't know how to thank you enough

[final loom]

I'll hit you with an easier proof whenever it clicks with me ^^" You can rest easy for now tho

[safe radish]

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[lavish hound]

okay this is confusing and my professor did a terrible job explaining what she did

[lavish hound]

what is going on in the red bubble

you integrate x and e^2x separately?

why is there u = x
U = 2x
du = 2dx

[quiet plume]

It's just scratchwork to compute the antiderivative of e^{2x} for integration by parts

[lavish hound]

i am aware

[quiet plume]

And the U=2x is just their substitution to compute it

[lavish hound]

im trying to understand every step of it though

so you integrate both separately

[quiet plume]

You integrate dv = e^{2x} dx

[lavish hound]

okay so following this i choose my u and dv

[quiet plume]

And you differentiate u = x

[lavish hound]

applying that here my dv is e^2x

and u is x

[quiet plume]

Yes

[lavish hound]

so..1?

[quiet plume]

You need du and v, which you get by differentiating u and integrating dv respectively

You can sort of see in the red circle you made that the red writing was added

Initially they just had $dv = e^{2x} dx$, and $u=x$

[flat frigate]

Azyrashacorki

[lavish hound]

okay so
u = x
dv = e^2x
du = 1
v = 1/2*e^2x

[quiet plume]

YEs

[lavish hound]

so we have everything now we just plug it in?

[quiet plume]

Well actually it would be du = dx

But yeah after that you just plug it all in

[lavish hound]

okay whats going on in the black text

i got xe^2x - 1/2 integral e^2x * e^2x

why is there a 1/2 outside, inside, as well as another 2

[quiet plume]

They multiply and divide by 2 to get the dU from their substitution in

But you could also just say that dx = dU/2 and plug it into the original integral

[lavish hound]

ah okay i got it so far

ty

.close

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[blazing cloak]

how do i solve B/

no clue where to start

[shadow verge]

csc(60+45)

[blazing cloak]

Oh

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[blazing cloak]

.close

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[arctic wadi]

hey guys, i could need some help with computing the integral showed in the picture, the density function for Xi = Psi_alpha follows

[arctic wadi]

there is the density function for Psi (ignore that I marked something there)

So as i have to put this density function in for xi, I have an integral over a series I believe

assuming i can change order of series and integral, i would have something like

series of integral of exp(2 * pi * i) * exp(-pi * (y-k/alpha)^2)

i just dont have a clue how to get this to what he claims exp(-pi * alpha^2)

[safe radish]

@arctic wadi Has your question been resolved?

[arctic wadi]

<@&286206848099549185>

[harsh smelt]

YE?

[arctic wadi]

hey whats up m8

got some problems (above two messages)

was hoping someone can help me with it

[harsh smelt]

sorry idk that if u want helps with algebra or geometry u can dm me im not above avarage

[arctic wadi]

alright np, ty for trying

[harsh smelt]

np mate

[arctic wadi]

.close

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[arctic wadi]

.reopen

[safe radish]

✅

[safe radish]

@arctic wadi Has your question been resolved?

[hollow rock]

Anyone German?

[fathom adder]

You have to graph the derivative, give an interval where the slope is positive or 0 and in which points there is max and min ? Right ?

[hollow rock]

Yeaaa

[fathom adder]

Im not too rusty in german wunderbar kekw

[hollow rock]

Hahahaha niceeeee

[fathom adder]

So f going down mean negative derivative and f going up mean positive derivative and when there is an horizontal line then detivative is zero

[hollow rock]

Okay I understand. How do I note that?

Like okey with plus minus and zero?

Only

[fathom adder]

I mean if you have to draw it you would do more

[hollow rock]

Okay okay

[fathom adder]

Would be something like this

[safe radish]

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[proud briar]

stuck on this question

[honest perch]

let O be the middle of the polygon

OBE is isosceles

angle EOB = 178

[proud briar]

wait how do you know that

[honest perch]

= 3*angle EOD so angle EOD = 178/3 = 360/n

there is no such n

[honest perch]

what is angle OBE

[proud briar]

178

[honest perch]

why

[proud briar]

isosceles?

[honest perch]

ok and

[proud briar]

idk

@honest perch

[honest perch]

no

[proud briar]

?

[cedar owl]

you can do it by solving for the angles in a few smaller shapes like octogons nonagons and decagons and trying to find a pattern in the decrease (not very rigorous though)

[proud briar]

ok thanks

/close

.close

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[late oracle]

i know its easy, i just dont understnad what to do. please someone quickly help and tell me, or even explain

[plucky elk]

Do you know sohcahtoa

,tex .sohcahtoa

[flat frigate]

riemann

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[sharp owl]

my question is about finding the maximum/amplitude of this function without using inverse trigonometric functions

[sharp owl]

the answer in the textbook is this right here, but i wasn't sure how to arrive at this without using arctangent. When i try to take the derivative of teh function, i get that it is equal to 0 at arctan(-1/50). how do i get thius irrational expression below from that?

[desert juniper]

are you asked to obtain the value, or just a bound?

[sharp owl]

what is meant by bound?

[desert juniper]

a bound would be the range of values it could take

[sharp owl]

basically, this was part of a larger problem where I had to solve an ODE and as it approcahes infinity it starts oscillating with a certain amplitude

[sharp owl]

just the amplitude i think

[sharp owl]

this is part of the solution to the ode which doesnt go to 0 as x->inf