help-19

[teal grove]

I need help verifying my work, I’m a bit confused as to what I’m doing (especially for d)

[velvet heart]

notice e^x starts at n = 0, but that sum starts at n = 1

so your answer is off by 1 (3^0/0!)

[teal grove]

I did see that, but I didn't know how to fix it.

Would the correct answer be (e^3) - 1?

[velvet heart]

yeah

you could just adapt that sum by adding and subtracting 1

so 3 + 3²/2! + ... = (1 + 3 + 3²/2! + ...) - 1 = e^3 - 1

[teal grove]

Got it, thank you!

Does everything else look okay?

[velvet heart]

yep

[teal grove]

Thank you!

.close

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[raw gazelle]

Knief help

[raw gazelle]

@amber schooner

[amber schooner]

send it

the image

[amber schooner]

$D = R_0\cos(\varphi_2) \pm \sqrt{R_0^2 \cos^2(\varphi_2) - R_0^2 + R_i^2} = R_0\cos(\varphi_2) \pm \sqrt{R_0^2(\cos^2(\varphi_2) - 1) + R_i^2}$

[clever fjord]

knief

[amber schooner]

what

ok well you can still read it

no idea why it did that

anyways

[raw gazelle]

Ohh I get it

Is that the most simplified form?

[amber schooner]

$\sin^2(x) + \cos^2(x) = 1 \implies \sin^2(x) = 1-\cos^2(x)$

[clever fjord]

knief

[amber schooner]

$D = R_0\cos(\varphi_2) \pm \sqrt{R_i^2 - R_0^2\sin^2(\varphi_2)}$

[clever fjord]

knief

[raw gazelle]

I’m still processing it hold on

[amber schooner]

from this if you multiply both sides by -1

you’ll get -sin^2 = cos^2 - 1

[raw gazelle]

I still don’t get it 💀

[amber schooner]

do you get sin^2 + cos^2 = 1?

[raw gazelle]

Yes

Pythagorean identity

[amber schooner]

ok now subtract 1 from both sides

sin^2 + cos^2 - 1 = 0

[raw gazelle]

Yep

[amber schooner]

now subtract sin^2 from both sides

cos^2 - 1 = -sin^2

[raw gazelle]

So now it’s negative tho right?

Ye

[amber schooner]

from here we have cos^2-1

so i just wrote it as -sin^2

[raw gazelle]

Oh I think I get it

[amber schooner]

$R_0^2\cos^2(\varphi_2) - R_0^2 = R_0^2(\cos^2(\varphi_2) - 1) = R_0^2(-\sin^2(\varphi_2))$

[clever fjord]

knief

[amber schooner]

yes?

[raw gazelle]

Ye

[amber schooner]

and we can just write that like this ^

[raw gazelle]

Alr I got it

Thx knief 👍

[amber schooner]

you’re welcome sir

[raw gazelle]

Oh and

Get some sleep bro 💀

[amber schooner]

it’s 930 wdym

i’m going to workout now

😭

[raw gazelle]

I be catching you lurking on the math channel at night bruh

Aight cya

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[silver mulch]

was this retracted because a sub n is 0 according the nth term test?

[silver mulch]

so it would be like 0 / 6 or something, so the limit comparison test doesnt apply?

[forest sky]

yes, also the new series clearly diverges by the nth term test

[silver mulch]

thanks

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[summer kite]

A boy is flying a kite from the top of a building. He sees the kite at the end of the stretched thread at
an angle of elevation of 500
.Another boy standing at the bottom of the same building sees the kite
an angle of elevation 650
,if the length of the thread is 100m.Find the height of the building.

[mystic saffron]

LMAO

What is 500? Radians, degrees, what?

[plush meteor]

hoping it was 50.0 💀

[feral halo]

😭

wtf is 500

[velvet heart]

maybe the 0 is a °

[feral halo]

ooh tru

but who tf does that

so fucking weird

if ur too lazy to copy paste deg

50* works

or 50 deg

[odd edge]

@summer kite Has your question been resolved?

[summer kite]

I'm sorry when I copy pasted it , the computer switched ° to 0

Idk why

50° and 65°

[feral halo]

what grade r u in btw

[summer kite]

The answe I got was 214

[summer kite]

Hell no

I don't wanna get bullied

[feral halo]

bro what 😭

[summer kite]

I'm not gonna explain why

Wait I got the answe 137 m

Wait shit this only figures how high the kite is

Give me a min

[feral halo]

idk man i get 61m 😭

[summer kite]

Ohk so I got 60m

[feral halo]

i've just been sitting here for like 10 min

[summer kite]

I'm gonna trust you and beleive it's 60

[feral halo]

thinking it looks wrong

[feral halo]

61

sine rule

[summer kite]

Oh yeah

[feral halo]

pretty sure
x/sin15 = 100/sin25

[summer kite]

Hmmmm

I'm gonna close this then

I guess I got the answe

Thank you

.close

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[mystic saffron]

hey

[mystic saffron]

hey guys in Art. 10

it says there that ax^4 has 5 dimension or 5 degree

but when i asked chatgpt it tells me it has 4 dimensions and the degree is not applicable perhaps what's the modern terminology or what?

I'm comfortable with this book so please don't ask me to switch out because of it's old terminology 😭

[sharp lotus]

To me it depends on what you consider a and x to be. Generally, when you introduce an x it is a variable or something that you want to identify. So when you think about all the functions f(x) = ...x^4 + ...x^3 + ...x^2 + ...x + ... (all the functions that you can get replacing the dots with numbers), it can be usefull to replace the dots with a, b, c etc to generalize (f(x) = ax^4 + bx^3 + cx^2 + dx + e). But in this example the values replacing the dots are known, for example, one function among all the one in that form could be 1x^4 + 2x^3 + 3x^2 + 4x + 5). So in that case, we say that the function is a polynomial of degree 4 because, the biggest power of an unknown quantity is 4, a was equal to one but as it is known, it "doesn't count" as part of the degree (in your example, they said that 8a^2b^5 is of degree 7 for example). As it is relatively common to name it this way, I think that that's what chat GPT meant here. So to sum up, if a is known, your expression is of degree 4 and else it is of degree 5.

[mystic saffron]

idk functions..

[odd edge]

@mystic saffron Has your question been resolved?

[sharp lotus]

Ah

So just remember that chat gpt was thinking that there was a known number behind your 'a' so if that's the case, it's degree 4 and else degree 5

[mystic saffron]

ah i see

[odd edge]

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[slow kindle]

Prove that n^6-n^2 is divisible by 60 using mathematical induction and without using modular arithmetic. I get to the expression k^6+6k^5+15k^4+20k^3+15k^2+6^k+1-k2-2k-1 and I can't figure out how to use the induction assumption here.

[warped glacier]

well by the induction hypothesis, k^6 - k^2 will be divisible by 60

that means you should try proving that (k + 1)^6 - (k + 1)^2 - (k^6 - k^2) is divisible by 60

,w (k + 1)^6 - (k + 1)^2 - (k^6 - k^2) simplify

[clever fjord]

Results provided by WolframAlpha

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Requested by sooouth.

[warped glacier]

yikes doing this without modular arithmetic

[mental lotus]

This factorisation is nice tho

[steel spire]

it feels easier to me to just try to "reinvent" modular arithmetic for the problem by factoring the original thing a bit and trying to argue that

[warped glacier]

the expanded form is $4 k + 15 k^2 + 20 k^3 + 15 k^4 + 6 k^5$ btw

[clever fjord]

southlander!

[mental lotus]

I mean we can directly see 3 and 4 are factors

[steel spire]

n^6-n^2 = n^2(n^2+1)(n+1)(n-1) has 3 consecutive numbers so it's divisible by 3, work to prove it's divisible by 20 by the factors etc

[warped glacier]

basically you want to show that $k^5$ leaves the same remainders as $k$ mod 5

[clever fjord]

southlander!

[warped glacier]

cheating I know but

[warped glacier]

ah right that directly

I think yeah that's the neatest factorisation

the one I did doesn't look neat

[mental lotus]

There is term of n² and n²-1 so one of them is divisible by 4

[steel spire]

yep

only have to break on the cases n odd and n even to get that too

as opposed to doing the full set of residues 0,1,2,3 mod 4

we're doing broke version of CRT now lmao

[mental lotus]

I can't see how we get 5 as factor without moular arithmetic tho, or at least splitting into cases

[steel spire]

I think I see a decent-ish way if you have binomial theorem

n^6-n^2 = n(n^5-n) and I guess induction step focusing on n^5-n will force (n+1)^5 - (n+1) and expanding the (n+1)^5 will throw a lot of 5s in idk

haven't realy thought it through tbh lol

in my mind I'm rederiving fermat's little theorem I guess sorta

I'm going to bed that's what I'm really doing right now lmao

[odd edge]

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[remote token]

Four circular disks of radius 1 are placed in a box
as shown. Some of the circular discs are partially covered
other circular disks. What is the total area of ​​the
hidden parts of the circular discs?

[remote token]

i need to find 4x this area

[swift lake]

wdym by hidden parts?

[remote token]

so the circles overlap

[swift lake]

ow

[remote token]

yeah

oh and this is no aid btw

so just pen and paper

[swift lake]

the are of it it just 2 times of the ( 1/4 circle - a 1 x 1 triangle

wdym no aid

[remote token]

i can only use pen and paper

so cant look up formulas or use a calculator etc

[swift lake]

this is not a test right?

[remote token]

not right now no

there are prep questions

for a test

[swift lake]

to find the are of red just use the area of 1/2 cirle radius of 1 - square 1 x 1.

[remote token]

why

[swift lake]

imagine you overlap 1/4 circle on top of each other

[remote token]

yuh

[swift lake]

the remove the 1 x 1 square

bro i can't draw RIp

[remote token]

ok so what i thought to do:

i saw that this shaded area fills up a bit less than a quarter of the square

i estimaed 1/6 of a square

since r=1, A=pi

so I said that area of red is pi/6

we have 4 of em so 4pi/6 or 2pi/3

since i dont have a calculator ill say pi is 3 and a bit more

so that means 2pi/3 would be 4/3=1.5

so i just said pi=3 and then added a bit to the end result

so ill go through the options like this:

a) 2pi-4=6-4=2.3

b) 3.14

c)12.5-12=0.5

d) 1

e=pi^2/4=10/4=2.5

so here a seems to be a tiny bit closer

and i dont see how we'd ever get pi^2/4 while doing this calculation

so it should be safe to assume a is correct?

.close

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[oblique heath]

i need to find a and b here,i put it under one denominator , force n in front but i have n^2(....)+n(....) the insides are different , the denominator has n not n^2

[oblique heath]

a friend said that n^2(..) should be 0 but why???

[supple thunder]

wdym by n^2(..)

u mean n^2 * ..?

[oblique heath]

yes

sorry

[mystic saffron]

change xn into a single fracrion

[oblique heath]

thats what i did

[mystic saffron]

i dont see it

[oblique heath]

check what i typed above the pictures

[mystic saffron]

divide both the numerator and denominator by n

[oblique heath]

we don t type * normally sorry

[oblique heath]

i did and one n still remains

[supple thunder]

nono its okay i was just confused i thought u meant like ^2(..) for a sec

[mystic saffron]

that means the n² term on the numeraror should be 0

because xn is 2

[oblique heath]

but why thats what i don t get

why does it happen

[mystic saffron]

because xn is not infinity

[oblique heath]

oh , so like for example if i had n^2 on the denominator the n term would be 0?

[mystic saffron]

depends on the question

[oblique heath]

wdym

like yeah the case still the same xn is not infinity

[mystic saffron]

if the denom's power is bigger than the numerator then the limit is 0

[oblique heath]

if the numerator would contain both n^2 and n and the denominator would contain n^2 then the n term from the numerator would be 0?

xn is equal to smth not infinity

[mystic saffron]

the n term alone would basically be 0 yes

but not the numerator

[oblique heath]

yes yes

n term i meant

thank you i got it now

.close

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[oblique heath]

.reopen

[odd edge]

✅

[oblique heath]

wait how do i make it available?

[steep mantle]

do ".close", it will close on its own

[oblique heath]

.close

[elfin zodiac]

.close

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[mystic saffron]

.sendhere

https://discord.com/channels/268882317391429632/1309876644625780799

[odd edge]

@mystic saffron Has your question been resolved?

[mystic saffron]

.close

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[mystic saffron]

I think i'll assume my issue is too complicated to get a timely answer.

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[robust wyvern]

seriously

[crisp wadi]

<@&268886789983436800>

[mortal trench]

hahaha

[robust wyvern]

question was so hard to solve

[pulsar elbow]

what was that site

[robust wyvern]

moderators had to interfere

[hollow ginkgo]

Porn site

[robust wyvern]

some porn site

[pulsar elbow]

.close

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[pastel dew]

https://tenor.com/view/porn-communitychannel-natalie-tran-phone-customer-support-gif-4911930

[odd edge]

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[plain apex]

I got up to 11/16 on my own

I then just did 11/16 * 1/2(the probability that it doesn't catch the 5th fish) to get 11/32 as my answer

why is this reasoning wrong and why do we need to find the complement

[odd edge]

@plain apex Has your question been resolved?

[grand mural]

Have you solved it?

[plain apex]

.closed

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[silent comet]

Hi so i did everything right and i found R=A+B but somehow its wrong

how did R become 9

[quasi sparrow]

Your Bx and By are swapped

The angle given is with respect to the y-axis

But the trig function should be evaluated at the angle with the x axis

You can fix it by just using 90-30=60 degrees

[silent comet]

yeah i mean i did sin30 for the Bx

Dont matter if i do Cos60

right?

[quasi sparrow]

This is wrong

Bx = B * sin(angle between x axis and vector B)

You did
Bx = B * sin(angle between y axis and vector B)

[silent comet]

Bro sin30=cos60

I just didnt subtract 90 and used it on y instead

how is that wrong

Its like i did the pythag triangle using y vector as x

[quasi sparrow]

Oh I see

Your Ax and Ay are swapped, not Bx and By

[silent comet]

oh

[quasi sparrow]

Your axes aren't labeled

Since I thought your A components was correct, that meant your B components was wrong

But it's the reverse

[silent comet]

thanks bro im dumb asf

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[tulip aspen]

pic 1 is q, pic 2 is answer, im confused on how we can derive that 96 = 50 + K? wouldnt we need to find x to confirm that?

[limber glade]

compare the sides

cuz that part does not involve x, we don't need to find x

since 2x²-20x are on both sides, the rest must be equal too

[tulip aspen]

OHH

so could we cancel that out?

[limber glade]

or the equal sign will make no sense

yeah

if that suits you

[tulip aspen]

ohh okay

thank youu

[limber glade]

welcome

if you're done then close the channel

[tulip aspen]

.close

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[fast ivy]

wat da heck is this word problem

how do i even set this up

like i know 11.2% is hydrogen then

[velvet heart]

mass hydrogen = 11.2% * total mass

mass water = 88.8% * total mass

you are given the hydrogen mass, you can find mass water

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@fast ivy Has your question been resolved?

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[mystic saffron]

I have recently found out by chance that (oops)

[mystic saffron]

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[sturdy magnet]

id like to see if my answer for part e is correct.
My answer is 15.6?

[sturdy magnet]

(a clearer pic)

[signal trail]

what have you tried

[sturdy magnet]

ill take a pic of my working out-

[signal trail]

ok

[sturdy magnet]

i used cosine rule to find angle ANB

and did 180 minus that angle to find APB

and did sine rule to find the length of AP

[signal trail]

gimme a second

[sturdy magnet]

yup nws

[signal trail]

Law of cosines?

[signal trail]

yeah ik

[sturdy magnet]

yupp thats what i used

[signal trail]

looks good!

[sturdy magnet]

kk ty!!

.close

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[proper maple]

$A \in \mathbb{R}^{5x5}$ which verifies $A^5=2AA^T$ Find all possible values for det(A)

[clever fjord]

ransik (gmdn)

[proper maple]

I think I may have taken a wrong path here

because that final det i arrived at is kind of contradictory... unless det(A) is 0?

or wait no

that's the 5th root by the way not square root

i think i might be completely deranged here

[slow sphinx]

solve it like you would solve a polynomial

[proper maple]

yeah i was thinking i might do that

it just seems

i don't know

[slow sphinx]

you have a^5 = 2^5 a^2, now just simplify insead of taking the 5th root

[proper maple]

i think that might be right

[slow sphinx]

how did it become a cuberoot

[proper maple]

you had 5th root of det(A) ^2

[slow sphinx]

yeah

not 6th

[proper maple]

ohh i'm an idiot

i subtracted them

sorry

[proper maple]

so remember how you told me not to take the 5th root

oh wait nevermind

i did as you told me i think

yeah anyways i think that's right now

[odd edge]

@proper maple Has your question been resolved?

[slow sphinx]

it was 32x^2 not x^3

just correct that and the steps are right

[odd edge]

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[odd edge]

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[lofty patio]

Here is the question:

[lofty patio]

I started by multiplying both sides by 1+ sin(theta), and got to this point: $$r+r\sin{\theta}=1$$

[clever fjord]

KXLI

[lofty patio]

i am not sure how to continue

[ember oak]

Do you recall the equations for converting between polar and rectangular coordinates?

[lofty patio]

yeah, x = r cos theta and y = r sin theta

so i can see that i can replace r sin theta with y and get r+y=1

or is that not the right way

[ember oak]

yes you are correct

now you just need to replace r

[lofty patio]

ty

.close

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[somber umbra]

I have a math project I am completely stumped on

[somber umbra]

i filled out the first graph (question 1) and i did the second question but i do not think it's right at all to be honest.

[odd edge]

Please post images (such as PNGs or JPGs) of the question rather than other filetypes such as PDFs which have to be downloaded. Non-image downloads can potentially contain viruses or other security risks.

[mossy tinsel]

yes thats filled correctly

[somber umbra]

thanks!! i have a few more pics to send in

[mossy tinsel]

wait sorry

where u asking about q1 or q2

the diagram or the table

[somber umbra]

i kinda need help with the whole project 😭

[mossy tinsel]

Your Q1 is correct, the numbers in the table are correct

[somber umbra]

thanks! i made a graph for Q2 but i think it's like way off. i can send a pic

[mossy tinsel]

idk abt Q2 but I can help you with Q3

Since the only restriction is that you have to start and finish at SCF, you'll have to start by listing out each of the possible permutations of this occurring. There should be a total of 4! ways this can happen.
I can start it off for you:

SCF → UF → FSU → UCF → USF → SCF
SCF → UF → FSU → USF → UCF → SCF
SCF → UF → UCF → FSU → USF → SCF
SCF → UF → UCF → USF → FSU → SCF
SCF → UF → USF → FSU → UCF → SCF
SCF → UF → USF → UCF → FSU → SCF

there would be 18 more, continue by changing the second stop from UF to either FSU or UCF

then you would have to sum up all of these distances to find the total distance

is this by any chance a programming assignment? seems tedious to do manually

For Q4:

1. Start at SCF-Bradenton (SCF). (in this case it would be USF which is only 54 miles)
2. From USF , travel to the nearest unvisited location (college) based on the distance table
3. Repeat this process until all locations have been visited.
4. Return to SCF-Bradenton.

sum up all the distances and this is ur answer

[somber umbra]

you'd be surprised it is not lol. just a college math course

[somber umbra]

so i put those into the graph?

[mossy tinsel]

into the table

yes

but that is only 1/4 of the routes

theres are 18 more

[somber umbra]

like this? and then get the total

[mossy tinsel]

correct

and then whichever one has the smallest total distance is ur answer to q4

and the answer to q5 is 250,000 * the total distance of the route you chose in Q4

[somber umbra]

wait is there actually 18 more since it says "assuming you visit each college only once"? cause i filled it out and it filled in the table

[mossy tinsel]

yes in the other 18 alternatives, you also only visit each once

but in a different order

for example

the 7th one could be

SCF → FSU → UF → UCF → USF → SCF

still only visit each once but the second stop is FSU instead of UF

[odd edge]

@somber umbra Has your question been resolved?

[somber umbra]

gotcha okay

i think i got everything else. the last question is asking how much would have to be invested each month for the railroad. i got $192,900,000 for question 5

[mossy tinsel]

what did u get for q4 and q3

like which route was shortest, how many km was it and how much does it cost

ehh it doesnt matter

[somber umbra]

i'm gonna add onto the graph i just wanna finish everything else first cause it's due in like an hour lol

[mossy tinsel]

unfortunately u still only have 1/4 of the permuations so theres a good chance you missed the actual shortest distance.
but since u dont have the time to complete it now we'll just consider the shortest out of the ones you've chosen

Q4 would be correct
Q5 would be ur Q4 answer * 250,000 so that is also correct

i think youve made a few silly mistakes here and there because i calcauted the distance of that route and got something different

[somber umbra]

the distance of the first one? damn :(

this project is 100 points lmao.

[mossy tinsel]

also for that last problem in ur word document, at the very bottom, u sub the figures straight into the formula
A would be ur answer for Q5
r would be be the interest rate 3.3% converted into a decimal
n would be 12 (months in a year)
t would be 10 (we're told that the sinking fund is establised over a period of 10yrs)

[mossy tinsel]

u really should have started earlier

[somber umbra]

p=A(0.033/12)/(1+0.033/12)^12(10)-1)

[somber umbra]

i knowww i had the flu tho

[somber umbra]

if that's the equation i cannot figure out the answer

:(

[mossy tinsel]

why not

[mossy tinsel]

@somber umbra

[odd edge]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[mossy tinsel]

my bad boss

oh wait i cant show that either

just sub it in ig

[mossy tinsel]

also ur gonna see certain routes have the same distance because of the symmetry of the matrix e.g. the same route but backwards but i would still leave it in the table because it says show all possible travel options

gl

[odd edge]

@somber umbra Has your question been resolved?

[odd edge]

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[lilac horizon]

Hi guys, I'm trying to write a paper about finding out which polynomial approximation methods are the most appropriate, looking at accuracy and computational efficiency, for different types of functions and I wanted to check if there was a way to find out the inaccuracy of the approximation as a whole rather than just at a certain interval, I have tried using lagrange error bound but it seems to be limited to a certain interval? The polynomial approximation methods that I'm looking at include, Taylor Series, Pade Approximants and Lagrange Interpolation.

[boreal crag]

What do you mean as a whole?

[lilac horizon]

Instead of an interval like [0,1], I was wondering whether I would be able to find out the inaccuracy of the approximation from (-inf,inf)

[boreal crag]

I would generally expect infinite inaccuracy, though I still don't know exactly how you're measuring inaccuracy

[lilac horizon]

Or will that just make the inaccuracy tend to inaccuracy

[lilac horizon]

Originally, I was looking at Lagrange Error Bound but that seems to only calculate the maximum inaccuracy for an interval

[boreal crag]

Something like e^x for example, every polynomial will eventually end up arbitrarily far away

[lilac horizon]

It gives the maximum possible inaccuracy rather than the definite inaccuracy

[boreal crag]

You'll have a hard time finding functions where the error doesn't tend to infinity

[lilac horizon]

Oh yea, thats true.

Then do you have any tips on how to justify choosing a certain interval? Because I'm looking at ln(x), sin(x) and tan(x)

For sin(x) and tan(x) I assume it'll be quite easy since they're periodic and I can just define the interval as one period, but I'm abit lost for defining a certain interval for ln(x)

[boreal crag]

Not really, just depends on what your purpose is

[low locust]

any power series you define for ln(x) has finite radius of convergence anyway

because lnx diverges at x=0

so the classic approach is to just approximate it in the interval (0,2)

due to log properties you can approximate any value ln(x) from that anyway

[lilac horizon]

Oh yea that's true

But, how do I calculate the inaccuracy if I use pade approximants/lagrange interpolation instead of taylor series

[dawn tiger]

so long as you're comparing different polynomial approximations on the same interval for the same function, you'll be able to compare their relative errors, no?

[low locust]

you can always integrate over the difference

or rather, absolute value of difference or square of difference

or in general |diff|^p

thats called the L^p error

[lilac horizon]

So I can just do something like | integral of [f(x) - p(x)] along the interval | / | Integral of f(x) along the interval | x 100%

Or is that flawed?

[low locust]

you need to make the error positive

otherwise the positive and negative errors can cancel each other out

[dawn tiger]

most reasonable thing is probably just look at (f(x) - p(x))^2

[low locust]

the scaling factor is up to you, I wouldnt bother honestly

[dawn tiger]

and just square root the value of the integral after? i'm not sure if that's necessary since you're only comparing relative errors

[lilac horizon]

Is finding the intersections with the x-axis the only way to make sure the errors don't cancel out?

[low locust]

are you trying to do this by hand?

otherwise, whats the issue with finding the intersection

if you do the difference squared then you dont have to worry about that intersection

squaring is much nicer anyway

[lilac horizon]

Uhhhh, I'm thinking of doing a graph from the 1st degree of taylor approximation up to 100th degree of taylor approximation

[lilac horizon]

So I would integrate (f(x) - p(x))^2 along the interval?

[low locust]

yes

[lilac horizon]

Okay thank you guys for all your help!!!

Thank you for being really patient with me

.close

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[oblique heath]

any ideas i need to find the limit of an when n tends to infinity

[south turtle]

as n gets bigger and bigger we can express ln(1+e^5n) as something simpler

[oblique heath]

?

[south turtle]

Just think for a sec

think about 1+e^5n

n becomes very big

what can we ignore

[oblique heath]

e itself?

[south turtle]

the 1

Do you see why?

[oblique heath]

why?

why 1

bcz of ln?

[south turtle]

if n becomes very large

e^5n is going to be much much larger than the 1

so we can just forget about it

so ln(1+e^5n) is approximated by ln(e^5n) for large values of n

[lethal kettle]

Use series expansion method with Taylor series

Ig that should work

Thentake common and cancel out

Or use l hospital rule

Differentiate up and down

Gimmie a min : ans 5/2

[odd edge]

@oblique heath Has your question been resolved?

[oblique heath]

thx

[south turtle]

You can just do

then 5n/2n=5/2

[odd edge]

@oblique heath Has your question been resolved?

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[snow breach]

-4(5- 2x)(x+3)

[snow breach]

how do i expand and simplify this

i got up to

(-20x+8x)(x+13)

now what do i do :(

[feral halo]

i would expand the second and third term first

not the coefficient

have u heard of F.O.I.L

(a+b)(c+d) = ac + bc + ad + bd

[snow breach]

yeah

[snow breach]

wait wjat

what

[feral halo]

like

just expand (5-2x)(x+3) first

[snow breach]

with FOIL?

[feral halo]

ac + bc + ad + bd
(5x) + (-2x^2) + (15) + (-6x)

[snow breach]

then expand with coefficient

[feral halo]

= -2x^2 -x + 15

then multiply that by -4

[snow breach]

im confused

i can still just expand with coefficient then use foil with

[snow breach]

that

right?

[rare roost]

@snow breach if ur struggling with expanding just make a table

[lethal kettle]

It will work in this case yes , but in other cases it might fail so I used the proper method so that he can know how to solve it if something like this comes up but this trick doesn't work

[rare roost]

Want me to show u

[snow breach]

the window method?

[snow breach]

??????

[feral halo]

replying to someone else

[snow breach]

ohh

[proven harbor]

Disappointed 💀

[rare roost]

Oops i made a mistake

[snow breach]

SHUSH

GEORGE

[rare roost]

It’s -2x^2

[proven harbor]

Lol

[snow breach]

im slow ok 😭

[rare roost]

Then u add all the blue sections

@snow breach

[snow breach]

uhh

[proven harbor]

!nosols

[odd edge]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[rare roost]

Oh thank you

[lethal kettle]

Btw @snow breach split one of the equations (a+b)(c+d) = a(c+d) + b(c+d)

[snow breach]

wait wait awit

[rare roost]

@snow breach do u understand ??

[snow breach]

let me try foil

[rare roost]

@snow breach ill let the other helpers help you, but if u dont get it still dm me and i can 100% teach u step by step ill help u ace ur test lol

[snow breach]

ok

ty

anyways

so

-4(5- 2x)(x+3)

this is what i started with

i expanded the first half

(-20x+8x)(x+13) then got this

now do i use FOIL on this?

[lethal kettle]

Don't do that

[proven harbor]

Solar is right

[lethal kettle]

Leave the coefficient

[snow breach]

oh ok

[lethal kettle]

Put it in multiply later

[snow breach]

ok so then

[lethal kettle]

Because it will get complicated

[snow breach]

(5- 2x)(x+3)

we got this

what do we do with this

[rare roost]

Just a tip leave coefficient at end its easier

[lethal kettle]

.

[snow breach]

ok so

it would be

5 (x+3) -2x(x+3)?

[lethal kettle]

Yes

Now expand each bracket

And then multiply whole thing by 4

Then simplify by adding cutting etc

[snow breach]

5x+15 -2x^2-6x ??

wait

there

is that it?

[lethal kettle]

Yes

Now

Multiply 4

Thatukept aside to each

Before that simplify the term

Make -5x + 6x into -x

[snow breach]

ty

.close

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[tidal hollow]

Help please.I'm trying to solve this exercises, im not sure that the 3. is correct or not. And how can i prove the 4.

[odd edge]

@tidal hollow Has your question been resolved?

[tidal hollow]

<@&286206848099549185>

Help 🥺

[odd edge]

@tidal hollow Has your question been resolved?

[odd edge]

@tidal hollow Has your question been resolved?

[tidal hollow]

.close

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[rough iron]

In a coin toss game, a square checkered board is provided as shown in the picture. How to play it is the player is asked to throw a coin onto the board. If the coin hits the side of the square box, the player is declared the loser. If the bounce between the coin and the board makes the coin fly out of the board, the player is allowed to throw the coin back onto the board. If the coin manages to enter completely into one of the boxes without touching any side of the box, the player is declared the winner.
If the ratio between the radius a of the coin and the side of the square box b is known to be 1: k, the value of k and the following possible chances of winning are ....
Answer:
A. When k = 2, P = 10%
B. When k =, P = 16%
C. When k = 4, P = 20%
D. When k = 5, P = 36%
E. When k = 8, P = 56.25%

figure A is the checkerboard, figure b is the relative position of the coin that landed in one of the checker

[velvet heart]

so what are your thoughts on this?

[rough iron]

honestly, i'm so lost and my brain's fried LOL

been doing math practice papers since 9 pm and it's 12 am

now

<@&286206848099549185>

[celest acorn]

looks intersting

[celest acorn]

whar is K for B?

[rough iron]

k is b but in ratio for a?

[celest acorn]

I mean option b

[celest acorn]

here

[rough iron]

10-3

10/3

man this is a olympic level question

[celest acorn]

so what i would do is look at the possible locations of the center of the coin

[rough iron]

someone js told me

[celest acorn]

no

he is lying

[rough iron]

what

[celest acorn]

i look at the center of the coin

if it is within a of side of the square we lose

[rough iron]

yeah

[celest acorn]

so we win in (b-2a)^2 areas out of b^2 areas

yes or no?

[rough iron]

where is 2a from?

oh circle's radius

rightv

[rough iron]

what

[celest acorn]

?

[rough iron]

oh so basically the inner square is the point where the center of the circle liesv

?

okay i get it

[celest acorn]

so chances of winning is (b-2a)^2 / b^2

= (1- 2a/b)^2

=(1-2/k)^2

when k = 2 chances of winning is 0

when k = 10/3

chance of winning is 16%

when k = 4
chance of winning is 25%

when k = 5
chance of winning is 36%

and when k = 8
chance of winning is 56.25%

so option BD and E are correct

[rough iron]

you did something i couldn't jave done in 1 day

😭🙏

[celest acorn]

dw, u just have to practice a bit

[rough iron]

i don't understand the 2a^2 part though

[celest acorn]

?

[rough iron]

the 2a

[celest acorn]

so you get this diagram?

[rough iron]

i dont understand how is that 2a

[celest acorn]

the edge of the inner square is b-2a

not the black

[rough iron]

yeah how is it 2a?

OH NEVERMIND

i am so slow

🤦‍♂️🤦‍♂️🤦‍♂️🤦‍♂️

i understand it mow

thanks for the help 😁

.close

[odd edge]

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[celest acorn]

welcome

[odd edge]

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[teal swift]

Question: "How far is the distance between Venus and Earth when the planets are positioned as shown in the figure?

The diameter of Venus' orbit is 216,400,000 km and Earth's is 299,200,000 km.
Answer with an appropriate power of ten."

[teal swift]

i solved it

.close

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[upper onyx]

i am not sure about the meaning of some symbols. could you retype it here?

[edgy dirge]

I think that symbol is a proportional sign. but can you clarify it?

[warped urchin]

its the fish

[odd edge]

@quaint lodge Has your question been resolved?

[odd edge]

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[flat talon]

hello

[odd edge]

Please don't occupy multiple help channels.

[ember oak]

.close

[odd edge]

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Use .reopen if this was a mistake.

[flat talon]

where is general chat

[ember oak]

Stay in #help-12

[ember oak]

#discussion or #serious-discussion or #math-discussion

[flat talon]

I want to contact multiple members for help

[ember oak]

or #chill

[ember oak]

don't

[flat talon]

without cost

[ember oak]

#❓how-to-get-help

[flat talon]

I want to like provide help

[velvet heart]

then answer the questions people post, don't open a channel

[ember oak]

.reopen

[odd edge]

✅

[ember oak]

If you want to offer help, then browse the open help channels or forum posts, and provide guidance there. Do not share solutions and do not help people on quizzes or tests

[winter scaffold]

can someone helpp

with just one kenken puzzle haha

[ember oak]

#❓how-to-get-help

[odd edge]

@flat talon Has your question been resolved?

[odd edge]

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[odd edge]

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[glossy sleet]

Can anyone see how they got the yellow part

[glossy sleet]

It’s based on the table on the top right

[odd edge]

@glossy sleet Has your question been resolved?

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@glossy sleet Has your question been resolved?

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[oblique rapids]

Have I done this right

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@oblique rapids Has your question been resolved?

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[foggy zephyr]

how would i find the domain here. i get how x cant be 4 and x has to be greater than or equal to plus or minus 2. then why is it not [2,4) U (4, infinity). Note i didnt use test points im not sure if im supposed to use them and the answer key shows a different answer than mine.

[foggy zephyr]

<@&286206848099549185>

[dim sandal]

!15min

[odd edge]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[latent scaffold]

hint ||(-3)^2 - 4 = 9 - 4 = 5 >= 0||

[dim sandal]

What does "greater than or equal to plus or minus 2" even mean, conceptually?

[foggy zephyr]

its plus or minus 2 because i took the square root

[foggy zephyr]

where are you getting these numbers from

[latent scaffold]

It's just an example to show you that you're missing a bunch of negative values that make the inside of the square root non negative as well

If you have $x^2 \ge 4$, you can't conclude that $x \ge \pm 2$.

[clever fjord]

Azyrashacorki

[dim sandal]

Inequalities can be very tricky

[foggy zephyr]

what would you do instead then

[latent scaffold]

But you can conclude that $|x| \ge 2$

[clever fjord]

Azyrashacorki

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[thick fiber]

!status

[odd edge]

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

[thick fiber]

2

Ill show

I tried taking common factor after third line but I got nowhere

[elfin zodiac]

Do you know that $5^{logx} = x^{log5}$ ?

[clever fjord]

Samuel

[elfin zodiac]

@thick fiber

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@thick fiber Has your question been resolved?

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[thick fiber]

.reopen

[odd edge]

✅

[thick fiber]

sorry im back

[thick fiber]

i didnt know

how would that help

[elfin zodiac]

$5^{\log x} = \exp(\log(5^{\log x})) = \exp((\log x)(\log 5))$

[clever fjord]

Samuel

[elfin zodiac]

$x^{\log 5} = \exp(\log(x^{\log 5})) = \exp((\log 5)(\log x))$

[clever fjord]

Samuel

[elfin zodiac]

This makes your exercise really trivial

You can do the same with the other

So we have the same base (x)

[thick fiber]

what does exp mean

[elfin zodiac]

The exponential function

If your log is neperian it would be e^ln

If it is in base 10 it would be 10^log

[thick fiber]

oh yeah ok i get it

[thick fiber]

you did this just to prove right?

[elfin zodiac]

Yes

[thick fiber]

ok

[elfin zodiac]

Just in case you didnt know why they are equal

[thick fiber]

so do i continue with that from where im at

or from the beginning

[elfin zodiac]

It is the same, you did nothing illegal as far as i saw

So u can continue from there

But it will be easier for u

To divide instead of multiplying by 5^(-1) or 3^(-1)

So you can eliminate the fractions by finding the common factor

[thick fiber]

yeah ur right

so i combine the fractions for each side?

[elfin zodiac]

The have different exponent

Just multiply the whole equation by 15

To eliminate fractions

[thick fiber]

i did this

i got

18 * x^log5 = 50 * x^log3

😭

[elfin zodiac]

That is correct

[thick fiber]

what is next

do i

i dont know if it'd be convenient to divide by any of the x's

[elfin zodiac]

You know x^log3 can never be 0

So divide both sides by x^log3

[thick fiber]

x^(log5-log3) = 25/9

[elfin zodiac]

And log5-log3 is

[thick fiber]

log(5/3)

ok i can see 25/9 is (5/3)^2

can i do something with that

[elfin zodiac]

You can try to down that log(5/3)

[thick fiber]

ok

[elfin zodiac]

Take log both sides

U get logxlog(5/3)=log(25/9)

[thick fiber]

yes

oh

[elfin zodiac]

Now divide both sides by log(5/3)

[thick fiber]

wait i think i got it

[elfin zodiac]

Nice

[thick fiber]

yes

(log(5/3)^2)/log(5/3)

(2*log(5/3))/log(5/3)

so 2 in the rhs

log x = 2

x = 100

[elfin zodiac]

Now check

5^log(100)-3^(log(100)-1) = 22

[thick fiber]

yes

its correct

[elfin zodiac]

Perfect

[thick fiber]

thanks

.solved

[odd edge]

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[odd edge]

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[frosty mirage]

Hey guys, what's the link between the fourier transform and the fourier series? If there is one at all. I kinda what they each are on their own, not sure how they relate though. Thanks 🙂

[odd edge]

@frosty mirage Has your question been resolved?

[frosty mirage]

<@&286206848099549185>

[floral stratus]

They both share the same idea, but the Fourier transform essentially applies to non-period functions or to functions where T -> infinity

The Fourier series is only valid for signals that repeat periodically over a specific time

[frosty mirage]

my understanding of the fourier transform was that it shows you which frequencies are present in the function / signal

[floral stratus]

Yes that’s correct

[frosty mirage]

dont frequencies only make sense with periodic functions?

i guess my question is, can i use the fourier transform to get to the fourier series? if so how?

[frosty mirage]

does this mean that you can construct a fourier series over the entire real line?

[floral stratus]

Yes. Exactly

And for non-period functions you would use the Fourier transform bla bla bla

So wait

[frosty mirage]

sorry i meant can't

[floral stratus]

You can set this up by computing the Fourier transform of the period function, then identifying the discrete frequencies. The values at those frequencies then give you the Fourier series coeffs. c sub n

[frosty mirage]

would it give me non-integer discrete values for frequencies?

[floral stratus]

But you can construct a Fourier series over the real line. It extends the periodic function by repeating it

[frosty mirage]

right. also, by periodic function do you mean the f(t) being evaluated?

[floral stratus]

No you can only get integer discrete values for frequencies because otherwise there’d be destructive interference when summing over all periods. So it’d all cancel out

[floral stratus]

Yes 🙂‍↕️

[frosty mirage]

waitt so you can ONLY represent periodic functions with a fourier series?

[floral stratus]

I don’t see why you’d use the Fourier transform, since it’s used in cases where you can’t use the Fourier series

Periodic functions with the Fourier transform lead to a discrete frequency spectrum, which is why the Fourier series is used for periodic functions

If that answers your question

[frosty mirage]

sorry, im very new to this. we havent covered it in class im just doing some extra work

[frosty mirage]

okay

but how about with non periodic functions?

[floral stratus]

No I find it really impressive when students really want to cover a topic with full interest. Your curiosity is something you should be proud of

[frosty mirage]

thank you lol

[floral stratus]

Only the Fourier transform

[frosty mirage]

and because it will outout not all discrete integer in the frequency domain with non-periodic functions you cannot construct a fourier series?

[floral stratus]

Sorry could you maybe rephrase that question lol

[frosty mirage]

sorry my bad.
when the function you put into the fourier transform to generate the frequency domain, if the function is non-periodic, then the output will have not discrete integer frequencies in the output. since the fourier series requires those discrete integer frequencies, it is not possible to make a fourier series for a non-periodic function?

[floral stratus]

You’ve got it bro

[frosty mirage]

okay lol, but then, the saw tooth wave is constructed piecewise, and its linear in those segments right? why is there a fourier series for that? the piece isnt periodic?

[floral stratus]

The sawtooth wave is periodic as a whole. The Fourier series doesn’t care about how it’s constructed piecewise, rather just about if the function is periodic

The linear ramp isn’t periodic by itself, but the sawtooth wave is still periodic because of the repetition of that segment

[frosty mirage]

so if you got the fourier transform of a segment say from 0-1 in the time domain you would get integer frequencies which then help you get fourier series?

[floral stratus]

No not quite because to construct the Fourier series you need periodicity. The segment wouldn’t be periodic. The Fourier transform of the segment from 0-1, as you say, would result to a continuous spectrum in the frequency domain. Not just discrete frequencies

This happens because the Fourier transform assumes that the function is defined over the entire real line, and the single segment is interpreted as non-periodic even though the sawtooth wave IS periodic

[frosty mirage]

woah thats weird.... so if i did the fourier tranform on the snippet of a sin wave id get a continuous spectrum?

[floral stratus]

Yes!! If you take snippets that can be interpreted as non-periodic, then you’d get a continuous spectrum as a result if you apply the Fourier transform

[frosty mirage]

wow that makes me uncomfortable LOLL

[floral stratus]

Just make sure to take the whole function then nothing malicious or dubious will happen 😭

[frosty mirage]

so theoretically, i can have a periodic function thats made up of parabolaes or soomething i dont know. find its fourier series by doing the integral etc and then boom ive got a fourier series?

do you mind if i summarise everything here and you can check if my understanding is ok?

[floral stratus]

I was about to go to bed but go ahead 🙂‍↕️ I’d be glad to help

[frosty mirage]

oh dont worry about it then. i can ask someone else. sleep is important

[floral stratus]

If no one answers then send me a DM and I’ll go even more in depth tomorrow then

[frosty mirage]

alright, ill just leave you a dm then. you can answer it in the mornin if thats alr?

[floral stratus]

Good luck and good night, your intuition is really neat

[floral stratus]

Yes I definitely will

[frosty mirage]

thank you so much, i really appreciate it. Sleep well !

close

[odd edge]

@frosty mirage Has your question been resolved?

[odd edge]

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[ocean geode]

hello i need help using the complemention rule (stats) to answer a true or false question

[odd edge]

@ocean geode Has your question been resolved?

[ocean geode]

@mystic saffron

[mystic saffron]

Ok

Restate the rule of complements for me

[ocean geode]

for any event​ E, ​P(E)=1-​P(not ​E).

is this it

[mystic saffron]

Yeah P(E)=1-P(E^c)

Now for the first question

Notice that at least one question can be written like this

P(X>=1)=P(X=1)+P(X=2)+…+P(X=12)

Do you agree?

[ocean geode]

yes? Im not sure i dont quite understnad this entire concept

can you explain why youre plugging that in

[mystic saffron]

Ok yup

So X is a random variable giving the number of questions guessed correctly right?

[ocean geode]

okay

[mystic saffron]

P(X=1) is the probability that 1 question is guessed correctly, and P(X=6) would be 6 questions right?

[ocean geode]

yes

[mystic saffron]

Now the question (a) asks what is probability that at least 1 is correct.

That is the same as P(X>=1)

[ocean geode]

so 1/12?

[mystic saffron]

Nope not exactly

Because lets make it a simpler example

Say we have 3 questions

True-false. So this means each question has 50% chance

If we look at the probability we answer 1 correctly and the other two are wrong that is 50% for first question and 50% for the other two right?

[ocean geode]

yes

[mystic saffron]

But that could also be 50% second question was right. And 50% the first and last are wrong

[ocean geode]

yes

[mystic saffron]

If you learned in class what I am doing here is looking at all the outcomes where we get 1 question correct and the other 2 wrong

[ocean geode]

is this like combinations or permutations

[mystic saffron]

Which can be represented as a combination

Yup!

[ocean geode]

okay, how should i plug it in

[mystic saffron]

Well for the example of 3 questions

True and false

There are 2^3 outcomes right?

[ocean geode]

yes

[mystic saffron]

Because for each question we can get true or false. And using the multiplication principle we multiply the outcomes for each question

Which gives us 2^3 in total

But we also want to count the number of ways we can get 1 question right and 2 wrong (to count number of ways to get exactly 1 correct)

[ocean geode]

okay

[mystic saffron]

Thats the same as C(3,1)

[ocean geode]

i dont understand quite,,, dont we do 2^12

oh

im sorry im looking back at my question

[mystic saffron]

Ok good

Lets go back to your question

[ocean geode]

i understand now! so you haev 3 wuestions 2 possibilites thats why u squred it

[mystic saffron]

Did you learn about binomial random variables?

[ocean geode]

yesss but its not quiote in my head