help-0

[craggy stream]

$\frac{4x^{-1}}{1+x^{-2}}$

[ocean seal]

AKS

[craggy stream]

now as x tends to inf u get 0/1

which is 0

[ivory gulch]

i rewrote it

$e^{\frac{4x}{x^2+1}}$

[ocean seal]

ski

[craggy stream]

so basically its of the form e^f(x) where f(x) tends to 0

[ivory gulch]

so its 1

is the answer

[craggy stream]

since e^x is continuous then u can say lim e^f(x) = e^( lim f(x) ) as x tends to inf

so yeah its 1

[ivory gulch]

ok for this one

i just need to basically do what i did

but then set to zero and solve for a and b

right

[craggy stream]

are a and b supposed to be numbers of functions of x

[ivory gulch]

that's all the info i have

[craggy stream]

it seems like theyre numbers

[ivory gulch]

yeah

they are

[craggy stream]

then a/x^2 tends to inf so u need to somehow control that for lim to be 0

so u should first group it all together as 1 fraction

[ivory gulch]

yeah i did that

i have my first derivative

$\frac{2tan\left(2x\right)+x+bx^2sin\left(bx\right)+2xsin\left(bx\right)}{3x^2}$

[ocean seal]

ski

[ivory gulch]

couldn't i set this = 0

and solve for b

[craggy stream]

thats wrong

[ivory gulch]

how so

not disagreeing

[craggy stream]

i think u didnt diff the tan

[ivory gulch]

damn

[craggy stream]

and the sin and the ax

[ivory gulch]

ur right again

i did to the ax

[craggy stream]

that x should be an a

[ivory gulch]

yeah

just did it wrong

ok lemme redo

$\frac{2sec^2\left(2x\right)+a+bx^2cos\left(bx\right)+2xsin\left(bx\right)}{3x^2}$

[ocean seal]

ski

[craggy stream]

yeah thats right

[ivory gulch]

set to 0?

[craggy stream]

then u get something/0

[ivory gulch]

oh so i have to diff again

and i'll need to again after that

[craggy stream]

u should diff only if the lim is of the form 0/0 or inf/inf

which it isnt

[ivory gulch]

oh ok

[stone minnow]

can someone help me with surds dm me if you can

[ivory gulch]

so set to 0 or what

[craggy stream]

so what happens to the individual terms when x tends to

[ivory gulch]

well denom is 0

but for numerator

[eager fern]

How do I find when I don’t know what a(t) is

[craggy stream]

just look at the terms separately

[ivory gulch]

ok so for sec^2 2x

that's the same as 1/cos^2 (2x)

right

[ivory gulch]

#❓how-to-get-help

[ivory gulch]

#❓how-to-get-help

[craggy stream]

yeah

[ivory gulch]

1/pi^2

[craggy stream]

no

[ivory gulch]

typo

[ivory gulch]

#❓how-to-get-help

jesus

4 ppl back to back

[barren belfry]

No one was typing 🤷‍♂️ sorry

[ivory gulch]

1/pi^2

[craggy stream]

it shouod be 1

cos(0) is 1

[ivory gulch]

cos(2pi/2)

wait i was plugging in 2pi

[craggy stream]

lol

[ivory gulch]

i meant to say

wait what

am i dumb

at 0,1 (cos(0), sin(0))

its pi/2

no?

oh wait

am dumb

yeah ok its 1

alr i will figure it out later, i have to go to class now. can i @ u in like 45 mins? @craggy stream if you're not on it's cool

i appreciate the help thus far

[craggy stream]

im about to sleep lol

[ivory gulch]

it's fine, just read the rules next time

[ivory gulch]

oh ok np. well gn, and thanks again 😄

[barren belfry]

Ok, is this idle now?

[eager fern]

No

How do I do c

My questions too hard

Is it that all math teachers give homework on stuff they haven’t taught

[bleak ridge]

$p(t)=(e^{0.05t} - cos(0.35t))(6-\frac{t}{5})$

[ocean seal]

PapaBread

[bleak ridge]

Is this right for the initial function?

[amber iron]

Hi, is the channel open?

[eager fern]

It’s .05

But everything else is right

e^.05t

I just replaced t with 7 since I don’t know any other way

[bleak ridge]

Oh I see

Do you know your derivative rules

$p(t)=(e^{0.05t} - cos(0.35t))(6-\frac{t}{5})$

[ocean seal]

PapaBread

[eager fern]

Yes

[bleak ridge]

So uhh

Try just starting with the first term

The e^0.05t-cos(0.35t)

[spring mountain]

hi does anybody know here how to solve weighted average mean? T_T

[bleak ridge]

No

Pi is irrational

[ocean seal]

shoury

[bleak ridge]

Aka it cant be expressed by a ratio

Its a bit bigger I think

[raw shard]

smaller

[ocean seal]

shoury

[raw shard]

no

just remember the first few digits

i think 355/113 might be a good approximation

i might remember that wrong though

[queen wasp]

Is there a formula for (1/2 + 1/4 + ... + 1/n)?

[raw shard]

what you sent looks like 1/2^n

and yes, that equals 1

[queen wasp]

for a finite n?

[raw shard]

are you talking about what i sent?

because what you sent doesn’t look like it, although the two first terms do

[queen wasp]

I'm talking about $\sum_{i = 0}^{\log_2 n} \frac{1}{2^i}$

[ocean seal]

JellyRekt

[bleak ridge]

Youd prob want i=1

[queen wasp]

How does that change it?

[glass lichen]

i=0 is fine

Note it's a geometric series

[bleak ridge]

Thatd be the difference between (1/2 + 1/4 + ... + 1/n) and (1 + 1/2 + 1/4 + ... + 1/n)

Idk which he meant

[glass lichen]

Oh true

[queen wasp]

For my purposes, I think I could use either, so I can do whatever is going to be simpler

[bleak ridge]

1 iteration simpler ftw

[eager fern]

How do I find a

[queen wasp]

If I start from 0, it comes out to $1 + 1/2 + (1/2)^2 + ... + (1/2)^{n - 1}$ which there is a formula for, right?

[ocean seal]

JellyRekt

[cloud summit]

((2^n)-1) / (2^n)

[eager fern]

How I find a?

[cloud summit]

this looks better :)

[strong wing]

is anyone good at physics?

Yo I need some help with this, I need to find T, N0 and Lambda sign

using the exponential function and the formula

[untold warren]

is someone able to give me guidance on this?

not sure how to prove/disprove it

im assuming this doesnt hold since it says if and only if

[tight locust]

i think a simple counterexample would suffice

[untold warren]

Like x=3?

x=3 isnt rational but x^3 is

[tight locust]

consider sqrt2

[untold warren]

ahhh

[tight locust]

x=sqrt2 is irrational
x^2 = 2 is rational

[untold warren]

wait what am i saying lol 3 is def rational

[tight locust]

the converse(?) statement is true though. if x is rational then x^2 must be rational

[untold warren]

i did proof by contradiction

said that since x^2 = 2 is rational, then x should be rational, and so on

[mighty dune]

Can anyone help with this problem? I thought the answer was 135 but it is saying wrong

[wary stream]

Did you need to include the units?

[mighty dune]

I also put the units and it’s saying wrong also

[pseudo pebble]

Can someone look over this differential equations assignment? Not sure if I did them correctly, mainly 4 and 5? 4 felt too short?

[spare birch]

https://www.blooket.com/play?id=223566 can u join?

lol thats my question

LOL

[pseudo pebble]

?

[scarlet quest]

anyone who could help me with dif. eq? 😄

is not the same question I asked yesterday

[pseudo pebble]

@scarlet quest #❓how-to-get-help

[scarlet quest]

so I can receive help here

[pseudo pebble]

read it

[scarlet quest]

The function y = f (x) is the solution to the differential equation. dy/dx=(x-1)*(y+1). One of the integral curves has tangents l and m at the two tangent points (0,0) and (2,0).
a) Solve the differential equation by the separation method. Remember to pay attention to where in the coordinate system you solve - ie. what reservations you may take (for example, you may not divide by 0)

[pseudo pebble]

2. Before asking a question, make sure that the channel you are using is not currently in use. An occupied channel would generally have an ongoing discussion or a trailing unanswered question. Likewise, when you are done using a channel, make this clear so that it's open for others to use.

[scarlet quest]

Yea, I know - thanks. 🙂

[pseudo pebble]

<@&286206848099549185> (Differential Equations) Can someone look over #4 and #5 in the picture I posted? I don't feel like I did #4 right and #5 I just want to make sure.

[merry umbra]

Lecture this for me

All that you wrote

[pseudo pebble]

?

[merry umbra]

This hold on

[pseudo pebble]

You want me to explain #4 ?

[merry umbra]

Explain in clear definition on what that means

In all that

[tired drum]

How many ways can ABCD be shuffled such that non of them are in the same place? Can someone help me with this?

[merry umbra]

Idk

Math takes a lot of critical thinking

[pseudo pebble]

I'm solving homogeneous linear equations with constant coefficients, in the distinct root case. I use the auxiliary equation f(m)=0, finding its roots and using them to a write a solution for y.

[merry umbra]

In which in America we don't get

[pseudo pebble]

With roots such as a,b,c, the solution is y=C_1e^(ax) + C_2e^(bx) + C_3e^(cx) for non-repeated roots. For repeated roots such as a, a, a, b, c, the solution would be of the form y=C_1e^(ax) + C_2xe^(ax) + C_3x^(2)e^(ax) + C_4e^(bx) + C_5e^(cx).

u there?

guess i'll just find out if i did them right if they're marked wrong

.close

[heavy otter]

hippity hoppity this channel is my property

[lone heart]

Available help channel!

Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[kind sonnet]

What might fit in the Part B is x times bigger than Part A?

[kind sonnet]

I’m hoping my current work is okay 😬

[tardy timber]

Seems fine

[kind sonnet]

I was stuck because I’m going from a smaller Part B to a bigger Part A

What might I put into the red square?

[tardy timber]

If Part A is 5/2 as large, then Part B is 2/5 as large

[kind sonnet]

Oh LOL well that makes sense for some reason I was thinking of a word

Thank you so much!!

[tardy timber]

Np :)

.close

Oh you need to do it

Cool system

[heavy otter]

imagine not having any of the 5 roles that can do it

[tardy timber]

Sadge

[lone heart]

@kind sonnet Has your question been resolved?

[lone heart]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[unkempt vector]

https://gyazo.com/5ad0e8aca373e2ca8ede33dba8ac355c can someone help me solve this? I don't know where to start

[unkempt vector]

wait nvm im dumb

.close

[lone heart]

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• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[misty sphinx]

@upbeat gorge thank you! i’m also confused on what the reason for 4 be and i think i know 5 is it SAS?

[polar barn]

is it for

[lone heart]

@misty sphinx Has your question been resolved?

[misty sphinx]

it’s for this. i already got 2 done. i’m confused on 4 tho and i think i got 5

[polar barn]

oh sorry that was the messag i wrote like days ago and idk why i press enter now

[misty sphinx]

it’s okay😭😭

[lone heart]

@misty sphinx Has your question been resolved?

[slender girder]

can i ask a question real quick?

[bleak ridge]

Try an available channel like #help-6

[lone heart]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[tame quiver]

A runner increases their speed from 2.5 m/s to 4.3 m/s in 3.5 seconds what was the acceleration of the runner?

[covert cairn]

The answer is 0.51 ms^-2

[wary stream]

Don't give answers

[raw shard]

(4.3-2.5)/3.5

[covert cairn]

Ok

[tame quiver]

am i allowed to ask another question

[raw shard]

yes

[tame quiver]

a person is sledding down a hill begins with a velocity of 0.5 m/s and accelerates down the 40 m hill at a rate of 1.8 m/s^2. what was their velocity at the bottom of the hill

[raw shard]

do you have any idea of what to do

[tame quiver]

do you i dont

my science teacher is a bit lazy

[raw shard]

divide 40 by 1.8

[tame quiver]

22.2 repeating

what do i do with that

[raw shard]

i just realized i’m not sure i know how to do this

repost the question and wait for someone else to come

[tame quiver]

a person is sledding down a hill begins with a velocity of 0.5 m/s and accelerates down the 40 m hill at a rate of 1.8 m/s^2. what was their velocity at the bottom of the hil

[pulsar yarrow]

${v_f}^2 = {v_0}^2 + 2a ({x_f- x_i})$

[ocean seal]

Yone

[tame quiver]

what goes where

[pulsar yarrow]

Lol? Just put the numbers everything's given w description

v_0 = initial velocity, v_f = final velocity, a = acceleration, x_f = final position, x_i = initial position so (x_f - x_i) = distance of hill

[tame quiver]

ik that

i need help plugging them in

the numbers

[pulsar yarrow]

What's so hard plugging the numbers in? I mean it literally says "begins with a velocity of...." "accelerates down the 40m hill" "rate of 1.8ms^-2"

"what is the final velocity at the bottom"

[tame quiver]

uh

ill try

[olive arrow]

A sailboat is sailing toward a 15m tall lighthouse. The sailor drops anchor 80m from the base of the lighthouse. From the point at which the anchor was dropped, what is the angle of elevation from the boat to the top of the lighthouse?

[pulsar yarrow]

The channel's occupied unless mewtwo closes it

[tame quiver]

i am still using this

[tame quiver]

is 1.8 finall position

[pulsar yarrow]

what are the values you used?

V_0 = ? , (x_f - x_i) = ?, and a = ?

Oh wait you were asking me what the final position is

No it clearly says acceleration is 1.8 and the length of the hill is 40

[tame quiver]

so is it 0

[pulsar yarrow]

? what is 0

[tame quiver]

final position cuase

[pulsar yarrow]

if your final position is 0 then what's your initial position?

[tame quiver]

you right

what i have writtn rn is

vf^2 = 0.5 m/s + 2(40)(x

[pulsar yarrow]

v_f^2 = 0.5m/s + 2(40) (a) *

x_f - x_i = 40 because the final position is 40 and the initial position is 0 since he's starting at the very top

you could also consider the final position as 0 like you said but then the initial should be -40

[tame quiver]

that makes sense now

i was thinkin too much

[tame quiver]

om rocking with vf^2 = 0.5 + 80a now

[pulsar yarrow]

put in the value of a?

[tame quiver]

yea was i not supposed to

wait

[pulsar yarrow]

what

[tame quiver]

i still think its 80

so the final answer would be v_f^2 = 80.5

[lone heart]

@tame quiver Has your question been resolved?

[obsidian trail]

well the distance would be 80m, height would be 15 metres we could calculate the angle with
$arctan(15/80)$

[ocean seal]

KyrohJc

[lone heart]

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[pulsar yarrow]

(v_f)^2 = 0.5^2 + 2(40)(1.8)

,w \sqrt{0.5^2 + 2(40)(1.8) }

[ocean seal]

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[lone heart]

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[minor berry]

I always watch it cuz im bad at math so if anyone good at math so DM me

Thx u

[tough hatch]

please read #❓how-to-get-help

we're not allowed to DM anyone here

here

[alpine sable]

everyone

2 + 2 = 5

hello

[wary stream]

Don't troll

[alpine sable]

bruh

it's a joke

[wary stream]

Help channel is not the place to do that

[alpine sable]

ok

[lone heart]

@minor berry Has your question been resolved?

[lone heart]

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[past sundial]

Random question but is -6^2 = -(6^2) = -36 or is it -6^2 = (-6^2) = 36?

[charred flint]

The convention is to do the exponent first, so -36

[austere urchin]

“PEMDAS”, so exponents come before multiplication

[past sundial]

Oh okie thanks

.close

[lone heart]

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[ocean seal]

marc

[cunning obsidian]

What is the range of p

Also isn’t like 2^2-1 a prime number already ksdhfivij idk

[austere urchin]

I feel like we’re missing some context. Can you rephrase the question possibly?

[ocean seal]

marc

[austere urchin]

Ah, that might be important

[cunning obsidian]

I thought the q was “prove if 2^k-1 is prime then k is also prime”

xd

[lone heart]

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[umbral kindle]

can someone help

i didnt done question like this before

[minor berry]

Sorry

[lone heart]

@umbral kindle Has your question been resolved?

[rich basin]

anyone know any good visualisers for complex numbers, trying to udnerstand how z is changed with arg(z-1) in the complex plne

[wild wagon]

Think of writing a new function g(x) = 3x-2 and use chain rule to calculate the new derivative

[umbral kindle]

ok

.close

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[rich basin]

Can anyone help me undesrstand the graphing of this complex number?

[rich basin]

these are the solutions, which i don't seem to understand

[lone heart]

@rich basin Has your question been resolved?

[rich basin]

<@&286206848099549185>

[lone heart]

@rich basin Has your question been resolved?

[lone heart]

@rich basin Has your question been resolved?

[devout summit]

What didn't you understand exactly? Which line

[rich basin]

I managed to figure it out 20 minutes ago ,but it is kind of odd how this was really difficult to kind of work out

[devout summit]

Then, type .close when you are done @rich basin

[rich basin]

@devout summit and also is it fine if i ask a series question which I don't really quite get?

[devout summit]

Yes you can (I may or may not be able to solve though)

[rich basin]

So i got theta = sin^-1( (n+1)!)

but i'm not sure

so i deduce that for all numbers that the subscript of theta can be, it should always be 1

[vale wigeon]

sin^-1( (n+1)! )

are you out of your mind?

[rich basin]

Yeah, this might be insane

[vale wigeon]

(n+1)! is almost never in [-1,1]

[rich basin]

Sorry

@vale wigeon i'm confused how can i reduce this down into a plausible number?

[vale wigeon]

can you post your question again?

[vale wigeon]

this looks incomprehensible to me

[vale wigeon]

you just posted the exact same thing -_-

[rich basin]

what do you not get in this thing, i can try to expaln what it said?

[vale wigeon]

from the beginning

what is alpha_n?

it just looks like an overcomplicated way to say alpha_n = 1 for all n

[rich basin]

alpha n is = (Beta n) to the power of 4n

[vale wigeon]

and what's beta_n?

[rich basin]

$B_{n}\ =\ -1^{\frac{1}{2n}}$

[ocean seal]

BorutoEyePower

[vale wigeon]

\beta exists, you know.

[rich basin]

that is what it equals to

[vale wigeon]

also parentheses

did you mean $\beta_n = (-1)^{\frac{1}{2n}}$?

[ocean seal]

Ann

[rich basin]

yeah

[vale wigeon]

so shouldn't $\beta_n^{4n}$ evaluate to $(-1)^{\frac{1}{2n} \cdot 4n} = (-1)^2 = 1$?

[ocean seal]

Ann

[rich basin]

so that means alpha would basically be equal to 1

[vale wigeon]

so i repeat

why go through all this trouble?

why not say $\alpha_n = 1$ if you mean $\alpha_n = 1$?

[ocean seal]

Ann

[rich basin]

well that is part of the question though?

part of the trick i guess?

[vale wigeon]

............

okay, where did you get this question from?

idk about you but i do not appreciate nonsensical fluff like this

[rich basin]

I got it from a professor

[vale wigeon]

what professor

[rich basin]

A uni professor

[vale wigeon]

seriously, this all looks like nonsense

[devout summit]

They think it is a fun question?

[vale wigeon]

for some extremely questionable definition of "fun"...

[rich basin]

So how can i reduce down the sin^-1 ( (n+1)!)

[vale wigeon]

where did you get that from in the first place

[rich basin]

so what i did was

so you know that alpha to the subscript of anything would be 1 right

[vale wigeon]

yes, all the alphas are 1

[rich basin]

i immediately used that and reduce everything

[vale wigeon]

as i have said multiple times

$\gamma_z = \sin(\theta)$ by the looks of it

[ocean seal]

Ann

[rich basin]

and then i calculated the geometric mean of psi

[vale wigeon]

and the definition of psi_n(n) is weird bc on the one hand it's named as a set and on the other hand it's written to be a member of {1, 2, ..., n} so like, wtf

[rich basin]

which would be (n!)

[vale wigeon]

what

[rich basin]

because it is S subscript n to the power of n

[vale wigeon]

S_n^n = n! sure

$\frac{n!}{(n+1)!} = \frac{1}{n+1}$

[ocean seal]

Ann

[rich basin]

but then three is a sin(theta) as well

which i reduced from the epsilon

[vale wigeon]

epsilon?

[rich basin]

i mean the counter clockwise alpha

[vale wigeon]

that's gamma

you need to learn your greek alphabet

[rich basin]

yeah

[vale wigeon]

what you have is $\frac{\sin(\theta) \times n!}{(n+1)!} = \frac{1}{n+1}$

[ocean seal]

Ann

[vale wigeon]

the n! is not inside the sine

[rich basin]

yeah

[vale wigeon]

and it never was

and YOU shoved it in there

and YOU ignored the (n+1)! in the denominator

for whatever reason

[rich basin]

that is arcsin

[vale wigeon]

what arcsin

we never brought up arcsin

[rich basin]

arcsin = sin^-1()

[vale wigeon]

fckhnirfngsdffaLJDSGJDLSAGF

[vale wigeon]

do you understand this Y/N

[rich basin]

yeah

[vale wigeon]

ok so???

do you understand that n!/(n+1)! = 1/(n+1)

Y/ N

[rich basin]

yeah

[vale wigeon]

okay great

so you have sin(θ)/(n+1) = 1/(n+1)

do you understand this

[rich basin]

yeah

[vale wigeon]

THEN WHY ARE YOU SHOVING N FACTORIAL INTO AN ARCSIN

this is just sin(θ) = 1

nothing more nothing less

no factorials to be shoved into arcsin

[rich basin]

okay thanks

[harsh wharf]

The body moves at an acceleration of 5 m / s², subjected to a force of 35 N. Calculate body weight?

what's the answer

[rich basin]

!close

.close

[lone heart]

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[kind raven]

can someone help me with 6ii

<@&286206848099549185>

[minor berry]

Too hard

[kind raven]

@minor berry r u a helper???

[alpine sable]

I'd start by moving the y terms to one side and the x terms to another, then integrating both sides

[minor berry]

No but is kinda hard

[alpine sable]

Which class

[kind raven]

college?

[alpine sable]

Like 10th 11th 12th ?

is this calc 2 or differential equations

[kind raven]

differential

[alpine sable]

Oh that's a college question. Sorry

@kind raven what have you tried so far

[kind raven]

differntial

but i think i did wrongly

😦

[alpine sable]

hmm ok

what was your integral

[lone heart]

@kind raven Has your question been resolved?

[glacial cave]

help :I

<@&286206848099549185>

[lone heart]

From

#❓how-to-get-help

Rule 4

If your question has not been answered for a minimum of 15 minutes, you may use the Helpers tag once. Please do not try to bump your question using this ping unnecessarily. Do not abuse this ping. Do not individually ping users with the Helpers tag without their express permission.

[polar barn]

lemme try this

[glacial cave]

thx

[polar barn]

@glacial cave alright

so

im going to explain to you something

[glacial cave]

ok

[polar barn]

and then we apply that to the number

[glacial cave]

aight

[polar barn]

do you what an explaination or the answer directly?

[glacial cave]

explanation will be better

answering directly is nothing

[polar barn]

good

i like that

lets call $1^2+2^2+3^2+...+n^2$ to be A

[ocean seal]

alephcomputer

[glacial cave]

ok

[polar barn]

so

[glacial cave]

so...

[polar barn]

A = $1^2+2^2+3^2+...+n^2$

[ocean seal]

alephcomputer

[polar barn]

so we can say that

[glacial cave]

yeah

[polar barn]

A = $1+2(1+1)+3(2+1)+...+n(n-1+1)$

[ocean seal]

alephcomputer

[polar barn]

right?

and then

[glacial cave]

i dont undestand the n(n-1+1)

part

[polar barn]

n(n-1+1) is basically n . n

im putting it there for it to follow the pattern

[glacial cave]

oh ok

[polar barn]

$A = 1 + 1.2 + 2 + 2.3 + 3 +.......+ (n-1).n + n$

[ocean seal]

alephcomputer

[polar barn]

you take one out of the factor and then expand it

[glacial cave]

ok

[surreal meadow]

is . times?

[polar barn]

ye

[surreal meadow]

$\cdot$

[ocean seal]

maximo

[polar barn]

oh

sorry haha

im not really good with this bot

alright thanks

[surreal meadow]

you can also do $*$ or $\times$

[ocean seal]

maximo

[polar barn]

@glacial cave now what you do is that you rearrange the numbers

[lone heart]

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[glacial cave]

hmmm

[polar barn]

to have A = $(1+2+3+...+n) + (1$\times$2+2$\times$3+3$\times$4+...+(n-1)n)$

[ocean seal]

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

[lone heart]

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[forest quest]

I'm looking for a function of some sort.

Given a slope or direction (0-359, represented in red here), how could I calculate the isometric projected perpendicular angle (the blue lines)?

The output could be a slope or a 0-359 angle

[royal meadow]

so first we transform the angle from this viewpoint into the angle that we would get looking vertically downwards

so 270 would go to 270, 90 would go to 90; 30 would go to 45, 210 would go to 225

so what is this function

0 would remain at 0

and 180 at 180

wild guess but it might be f(t) = arcsin(sqrt(sin(t)))

or something like that

obviously it would depend on which quadrant you were in, but that's just some if statements

then you find the perpendicular angle, just add 90

so that's g(t)

and then you transform back, so do f^-1(t), which is like... arcsin(sin(t)^2)

so do f^-1(g(f(t)))

[forest quest]

I'll give that a try. Thank you so much!

[gusty oyster]

F(x) =−3x+ 1. The largest open interval for which |f(x)−f(−1)|<1 is given by the inequality?

Pls help

I got |-3x-3|<1

But the ans is |x+1|<1/3

[lone heart]

@forest quest Has your question been resolved?

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[gusty oyster]

F(x) =−3x+ 1. The largest open interval for which |f(x)−f(−1)|<1 is given by the inequality?
I got |-3x-3|<1

But the ans is |x+1|<1/3
Pls help

[glass lichen]

those are equivalent

$\abs{-3}\abs{x+1}<1$

[ocean seal]

Mosh

[gusty oyster]

Oh

Thank you so muchh!!

.close

[lone heart]

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[magic warren]

how do i do this

[hardy gulch]

[ 3072x^4 - 2880x^3 + 840x^2 - 90x + 3 = 3072(x - \alpha) ( x - r\alpha) (x - r^2 \alpha) (x - r^3\alpha) ]

[ocean seal]

ricey

[hardy gulch]

expand and compare the coefficients

[magic warren]

oh yeah @hardy gulch im so dumb how did i not think of that

[lone heart]

@magic warren Has your question been resolved?

[lone heart]

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[tough hatch]

don't forget to .close so others may be able to use the channel faster

[lone heart]

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[bronze flint]

help

how do i get the answers?

[magic root]

its pretty simple, so you know how a full circle is 360degree right?

[tough hatch]

have you learned anything about the different kinds of angles formed from intersecting a transversal with two parallel lines?

[bronze flint]

ye

[bronze flint]

what does that have to do with this?

[magic root]

that will help you figure it out, because you can subtract the angle you already have from 360, and then you will find the circle. One sec let me just show u

[bronze flint]

ok

[magic root]

this circle here is 360, so since we already have 1 angle, which is 113. then we know that the sum of that circle would be...

360 - 113 = 247

So the other 3 angles around the circle have to be under 247

[tough hatch]

or you can say that the sum of the measures of the four angles formed by intersecting two lines is equal to 360 degrees.

[bronze flint]

huh

ok

so

how do I know what e is?

[tough hatch]

you literally didn't answer my question, nice

[bronze flint]

oh

ye

those stuff

.close

[lone heart]

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[royal meadow]

nope

is this the original question

so it wants you to solve for n?

just... that?

what do you actually want to show

like

[manic nova]

what is 999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999+99999999999999999999999999999999999999999999999999999999999999999999999

[royal meadow]

fuck off

this is a question from a test or a paper or something?

ok

riiight

so you want to find the limit

so are you trying to show that

for all k, there exists n such that the left side is larger than k

ok

so this is equal to 2n - 2.5 + 1.5/n - 4/n^2

agreed?

why not

(a+b)/d = a/d + b/d

yes

ok so

for positive n:

2n - 2.5 + 1.5/n - 4/n^2 is less than 2n - 2.5 + 1.5/n

now notice that 1.5/n < 1.5 for n > 1

so 2n - 2.5 + 1.5/n < 2n - 2.5 + 1.5

ok wait no this is the wrong way round

we want to show that it's larger

starting over

for positive n:

i mean sure that works

i mean it basically works

something something ratio test

imma just keep doing it my way

for the sake of completion

we start with: 2n - 2.5 + 1.5/n - 4/n^2

notice that 1.5/n < 1.5 for n > 1
so 2n - 2.5 + 1.5/n - 4/n^2 > 2n - 2.5 - 4/n^2 for positive n

notice that 4/n^2 < 4 for n > 1
so 2n - 2.5 - 4/n^2 > 2n - 2.5 - 4

therefore 2n - 2.5 + 1.5/n - 4/n^2 > 2n - 6.5 for all n > 1
but as n goes to infinity, clearly 2n - 6.5 goes to infinity, and the LHS is greater so it also goes to infinity

[tawny fable]

Hey!
Can someone please explain to me 5.b and 5.c?

[royal meadow]

oh i hate life

#❓how-to-get-help

just read my argument

[vale sapphire]

broski

there is an entire system set up so that people don't ask questions in occupied channels

#❓how-to-get-help

[lethal dock]

Oh sorry? This is occupied?

[vale sapphire]

look at the category in which the channel is located

[lethal dock]

Oh-

[bleak ridge]

Use an open channel like #help-1

Why am I blue

[vale sapphire]

u be active

i should claim back that role sometime

[vale sapphire]

????

or... use one of the channels marked "available"???

this is the whole point

[lone heart]

@lucid birch Has your question been resolved?

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[raw nexus]

hello anyone good at working out effiency? i have a question but i think its more of a science question?

[raw nexus]

i got 0.2121 kg/s (4dp) which is correct

i got for this 25.29 kW which is correct

i got 43.3 degrees C for this which is correct
just confused about this equation below?

im not sure how to work this out out the correct answer is 86.1% but not sure how to work it out, please can someone help?

im not sure where to begin on this question

<@&286206848099549185>

[lone heart]

@raw nexus Has your question been resolved?

[west fjord]

yo @raw nexus

[raw nexus]

Yo

[west fjord]

i need help with Polynomials

[raw nexus]

I'm dead ass at polynomials, sorry I can't help you

[west fjord]

all good

[lone heart]

@raw nexus Has your question been resolved?

[raw nexus]

i dont want to hog the channell

.close

[lone heart]

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[frail echo]

dammit

[strange raven]

Hi

[lone heart]

@frail echo Has your question been resolved?

[orchid python]

hi

[lone heart]

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[turbid anchor]

a question regarding abstract algebra/group theory:

Let $G$ be a finite group and $H\leq G$. For $a\in G$, let $f(a)$ be the smallest positive integer $d$ such that $a^d\in H$. Prove that $f(a)\mid|a|$.
[Hint: Consider the cyclic group $\langle a\rangle$].

[ocean seal]

Snodlop

[turbid anchor]

from considering the hint, i know that the fundamental theorem of cyclic groups states that every subgroup of <a> is cyclic, and each has a different order k_i, meaning for each subgroup H of G=<a>, f(a)=k_i

i'm not sure if my logic is completely sound for this case, and even if it it, i'm confused as to how i can expand this notion for non-cyclic groups

on second thought, i have an additional question i'd like to ask towards the solution to this problem: does this have to do with lagrange's theorem? as in, given |G|, we know that for each H<=G that |H| divides |G|, so are we able to use this information to find f(a)?

[gilded smelt]

Im trying to figure if this is convergent or not.
Could u please help me

[turbid anchor]

bro

[gilded smelt]

Yessir

[turbid anchor]

this channel is occupied

[gilded smelt]

Oh okay my bad
Sorry about that

Seen it was inactive for 10min

[turbid anchor]

yeah i'm waiting 15 to ping someone

it's been around 15 since i first asked the question, so i'll drop the <@&286206848099549185> now, thank you in advance :)

[charred flint]

@turbid anchor yeah lagranges theorem will solve it

[turbid anchor]

how would it change depending on the order of a tho

i.e. say a^k is in H but b^k isn't, how do you make the distinction

[charred flint]

so langranges theorem is about subgroup orders dividing into group orders

f(a) | |a| means that you want the argument to use the group <a>, and a subgroup with order f(a) or with f(a) many left cosets

[turbid anchor]

ok i get that

that's definitely a better proof than what i came up with regarding the fundamental theorem of cyclic groups

... if G isn't cyclic, then are some elements only contained in the subgroup H=G?

meaning for b in that subset of elements, f(b) must equal 1?

[charred flint]

"some elements" of what? a^n's?

[turbid anchor]

some elements of the group

since G isn't cyclic the elements don't necessarily have to correspond with a^n do they?

[charred flint]

right

a^n won't hit all of G usually

you're really only going to have to look at powers of a though

try the cyclic group of a^f(a)

[turbid anchor]

so by lagrange's theorem, every subgroup of G with |G|=n has order n/d for some d, meaning if an element in G is not of this order, it is trivially delegated to the subgroup H=G, meaning we only need to focus on those elements whose order divides n?

[charred flint]

umm the order of an element in G always divides |G|

[turbid anchor]

right sorry i meant if it equals n rather than is less than n but still divides n

i'm really frazzled today my bad

[charred flint]

I don't quite understand but let me restate what you have:
H<G
there's a, which by Lagrange |<a>| divides |G|
there's a^(f(a)), which is in H

[turbid anchor]

oh shit i think it just clicked

for some reason i was circling around the idea of "every possible subgroup" or "every possible element"

but i don't need to focus on that at all

ok never mind i've gone one step back: if f(a) is the smallest integer such that a^(f(a)) is in H, doesn't that make it the generator?

[charred flint]

right, so you can get a cyclic group from it

[turbid anchor]

so wouldn't we trivially have f(a)=1 then? since that would be the smallest integer that puts a in <a>

[charred flint]

it's smallest integer for a^f(a) in H

so like <a^f(a)> is a subgroup of H

[turbid anchor]

oh got it alright

we're not concerned with G anymore

we know that the order of every subgroup H divides the order of G

[charred flint]

yup

[turbid anchor]

so we use f(a) to create a subgroup of H which divides H

so no matter what, f(a) must divide |a|

fuckin genius man

thank you

i'll close the chat now, but thank you again for all the help

.close

[lone heart]

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[static saffron]

i am stuck, i need to extract the C_l from here. The P_l are legendre polynomials. They are an orthogonal base of L^2 [-1,1] which i am currently working with.

[peak lance]

Can some one help me plz

[static saffron]

occupied.

[peak lance]

With number 13 (b)

Sorry

[static saffron]

just delete the message and use another

[charred flint]

you're asking how to get C_l's from the top expression equaling the third expression?

[static saffron]

yeah

[charred flint]

gotcha you match terms, like P_2 involves the l=2 term

[static saffron]

yeah. That i do know. But i need to find an actual equation for C_l since i will need to use the C_l for an electric potential

[charred flint]

yeah so only c_0 and c_2 are nonzero

[static saffron]

i know that all uneven C_l's = 0
i know that C_0 = R^2/epsilon_0 * 2*sigma_0
i know that C_2 = R^2/epsilon_0 * 3/25 sigma_0
and i know that for l in (4,6,8...) C_l = 0 too

[charred flint]

right, not sure what else you want

[static saffron]

let me show you^^

i will need to insert the C_l in these series after

would i really only just use the 0ths and 2nd term and be done?

[charred flint]

yup

[static saffron]

i really am stupid sometimes.. and here i am studying physics -.-

[charred flint]

emag is cursed don't worry

[static saffron]

it really is.

thank you very much. Cheers.

.close

[lone heart]

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[ancient shale]

Hi guys. Im worried about how I can multiply 2 vectors with different numbers of components?
(4) (8)
example: (2) * (5)
(6) (3)
(1)

[charred flint]

you can't

[ancient shale]

ok thank you

do you know how the numbers in a vector are named? Components?

[charred flint]

elements or components yeah

[ancient shale]

ok thx

.close

[lone heart]

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[lone heart]

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[alpine sable]

if a and b are not divisible by 3 prove that a^2+b^2+1 is divisible by 3

[alpine sable]

any hingts please

[charred flint]

numbers can be either 3n, 3n+1, or 3n+2

try all the possibilities of those forms with a and b, and you'll see a^2+b^2+1 is divisible by 3

[alpine sable]

as in (3n)^2 || (3n+1)^2?

and a and b cannot be 3n

[charred flint]

not sure what the || is for

[alpine sable]

or

[charred flint]

yeah a and b can either have a remainder of 1 or 2

[alpine sable]

ah yeah

I get 3n(...)+3q(...)+3 = 3k

which is true

thx

[lone heart]

@alpine sable Has your question been resolved?

[lone heart]

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[south pivot]

I disregarded everything that was less than six. Then split the shape left over into a triangle and a trapezoid but i couldnt remeber how to find the area of a trapizoid lol so i ended up splitting it into two triangles and a rectangle then found the area of all three of those shapes and added it together, i double checked it another time and got a different answer.

[glass lichen]

why did you disregard everything less than 6?

[fringe turret]

hey i forgot about decimals and stuff but what will i do in this ?

[wary stream]

Don't ask in an occupied channel. Read #❓how-to-get-help

[south pivot]

I get it now. I looked at everything other than what I was suppose to.

[glass lichen]

yes

[south pivot]

Thank you

[lone heart]

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[magic root]

What could this be?

[magic root]

are the options wrong?

[wary stream]

No, the options are right

What did you get?

[magic root]

shouldn't it be something like -3x - 15x^2

:///

I think I did it wrong or something

[wary stream]

Check your math

How did you get that?

[magic root]

Well you see...

-3x (1-5x)
The one becomes a -3x [-3 x 1]
The 5 becomes -15 [-3 x 5]

Okay I can't explain thru typing lol

its kinda hard like this

😅

[wary stream]

First don't use x as times

[magic root]

oh

[wary stream]

So you understand that, if given $$a(b + c)$$ that equals to $$ab + ac$$

[ocean seal]

dldh06

[wary stream]

Right?

Distribution

[magic root]

Oh Yeah ik that.

thats what I tried to do i think

[wary stream]

So you do $$-3x \cdot 1$$

[ocean seal]

dldh06

[wary stream]

First right?

[magic root]

oh, how would i take care of the "x" variable? do I just do -3 x 1, and then add the variable to it after.

[wary stream]

It's all multiplication

[magic root]

Oh

[wary stream]

In words, negative 3 times x times 1

Right?

[magic root]

yeah

[wary stream]

That's the ab part

So what do you get?

[magic root]

Oh okay, I think you will get -3

-3 x 1 = -3

[wary stream]

You're missing the variable

[magic root]

oh yeah.. and then...

-3x

right?

[wary stream]

Yes

[magic root]

Okay.

[wary stream]

Now the ac part

That's $$(-3x) \cdot (-5x)$$

Right?

[ocean seal]

dldh06

[magic root]

yeahh

like that

[wary stream]

So what should you get?

[magic root]

so it would be 15x?

right

[wary stream]

Close

What's x times x?

[magic root]

ohh x^2

15x^2

[wary stream]

Almost

There you go

[magic root]

I forgot about the x part.

[wary stream]

The main issue, you forgot was negative times negative equals positive

[magic root]

Yeah probably

[wary stream]

That's why you got -15

[magic root]

yeah

[wary stream]

Make sense?

[magic root]

btw in this picture, there is A and C,

A is those 2 nums minus
and C is those 2 nums plus.

Would I do plus because there is no operator left?

thats all, and then I would be finished.

[wary stream]

Well, $$-3x(1 - 5x) = -3x \cdot 1 + (-3x) \cdot (-5x)$$

Right?

[ocean seal]

dldh06

[magic root]

yeah

[wary stream]

So you get -3x + 15x^2

Which choice matches that?

[magic root]

Option C.

So option c is right! :)))

Thank you so much!

,close

oops

how do i do it lol

[wary stream]

.close

[magic root]

ohh

thanks

.close

[lone heart]

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[finite kite]

Terminology question:
How do I calculate for instance:
8C1 x 7C1 + 8C1 x 7C2 + 8C1 x 7C3 + 8C1 x 7C4 + 8C1 x 7C5 + 8C1 x 7C6 + 8C1 x 7C7

[finite kite]

Is there an efficient way?

[wary stream]

Honestly, a calculator

[finite kite]

Yeah next step is to do that for 32 instead of 8

I'm looking to calculate every way to arrange 32 elements into 3 groupings, such that only one of the groupings can be empty, and the other two must always contain at least one element

[glass lichen]

factor out 8C1 then simplify all the 7Cwhatevers

[finite kite]

Ordering of the elements in the groupings not being an issue

[finite kite]

Oh goodness how did I not think of that

well that should solve it

Just need to find a way to code it up

.clsoe

.close

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[stone minnow]

can someone help me with surds?

[rapid verge]

what is the question mate?

[stone minnow]

im new to surds

[vague coral]

you can start to get rid of the square root on the denominator

I'll let you guess how

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[lone heart]

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[lapis citrus]

Vanessa has a coupon for 5$ off 2 off dinner for 4 nights she spends a total of $100 on dinner for all 4 nights

[lone heart]

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[nocturne oyster]

Hi, I was currently a little confused about how to do this question

[nocturne oyster]

I thought those would be my c values but I think I am approaching this question completely wrong

[empty ridge]

Hi @nocturne oyster, why do you think this is wrong? Isn't what you did a representation of the MVT?

[nocturne oyster]

I plugged it in and the answer came out as wrong

like these are hw questions that we get assigned on a weekly bases that only have 3 attempts

this is what i mean

[maiden urchin]

Why were you giving three values when it asked for two?

[nocturne oyster]

I gave 2 values

3 (c)

[maiden urchin]

oh that's a . not a ,

[alpine sable]

[nocturne oyster]

yea lol

[empty ridge]

I think 6 and 2.5 are bad estimates... (this question is weird) You need "closer" values

the derivative at 6 is closer to 0... and that's why it seems wrong

the same at 2.5

[nocturne oyster]

right so i should try something like 2.7 and 6.2 ?

[empty ridge]

Nope, the derivative at 2.7 is negative, right?

[nocturne oyster]

yea i just realized 2.3

would be better off

[empty ridge]

Right.

Done?

[nocturne oyster]

Yea i think im good, i just wanted to check something rq tho

if thats fine

i need like 2min

I just feel like if I entered 6 and 2.5 it would have still taken it as an answer

i dont think the answer would be that strict

[empty ridge]

@nocturne oyster I really don't know. This question is a bad one for testing the MVT.

[nocturne oyster]

alright well thank you, ill try figuring out

.close

[lone heart]

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[queen citrus]

Hi, could anyone here help talk me through a problem concerning partial derivatives in multivariable calculus?

[vale sapphire]

just ask the question

[solar pebble]

^

[queen citrus]

I think I'm supposed to explain what the partial derivative represents

but I don't really understand it in terms of price and quantity

Would 6 represent the rate that grapefruit sales are increasing when their price is 3 and oranges are 4?

[lone heart]

@queen citrus Has your question been resolved?

[wispy hull]

hello

hello

[lone heart]

@queen citrus Has your question been resolved?

[solar pebble]

in kg yes

[queen citrus]

ok cool

what wouldthe unit of time be though?

cause it would be kilograms per some unit of time right?

[glass lichen]

the units are Q/r

so kg/dollar

[lone heart]

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[hard ether]

How do you do this?

[ornate timber]

@hard ether , which step are you struggling with

[hard ether]

Pretty much everything

like i dont know where to go after square root of 81

[ornate timber]

Were you able to draw a diagram of the problem?

[hard ether]

well yes i think

[ornate timber]

Great, can you attach it to this thread so I can make sure you're on the right track?

[hard ether]

Ok

Is it correct

[novel spruce]

yes although they probably want you to write 9 instead of sqrt(81) @hard ether

[lone heart]

@hard ether Has your question been resolved?