help-0

[carmine oar]

like no offense

[alpine sable]

ok

[carmine oar]

but please consider why I asked the question

here

[alpine sable]

ig most people know parabolas

and quads

by pre college

right?

it doesn't even require calculus

[carmine oar]

it doesn't even require equations tbh

[alpine sable]

YOUR ASSUMPTION THAT IT'S A SQUARE IS NOT MAKING ANY SENSE TO ME

[carmine oar]

a good way to learn is to prove that a square always has the largest area for quadrilaterals

[carmine oar]

then I just leave

if you're so good

just keep it to yourself

it's not an assumption for me
just like this is a result can be proved

[alpine sable]

what square are u talking about?

[carmine oar]

just I think my english is good enough for brutal to understand

if you don't understand then a pity for you ig

he understands my idea and he knows how to do it then it's good for us all

[alpine sable]

ok ya now I get what ur talking about

[carmine oar]

I just think that just keep it simple

like in elementary
we don't use calculus to proof the area of a circle

we just use it

without knowing why

[alpine sable]

u meant that the area of square is greater than the area of a rectangle with same perimeter

[carmine oar]

yeah

[alpine sable]

ya I see

[carmine oar]

I think you just think too general

try fitting in the context of these types of questions

[alpine sable]

finding the maximum value of a quad doesn't require calculus ig

[alpine sable]

ye I overlooked that fact

[carmine oar]

I'm just using this as an example

[lone heart]

@random haven Has your question been resolved?

[random haven]

I understand it now guys i am just gonna do the square thing, thank you both

[lone heart]

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[slender sparrow]

hi! I'm trying to solve this for x in R. I already know the answer but I'm not sure how to get there.. I've tried asking before but so far no success. Could someone help me ?

[slender sparrow]

this is the answer

[novel night]

I only have a couple minutes, but I'd start with some trig identities... for example the identity $\sin(a+b) = \cos(a)\sin(b) + \sin(a)\cos(b)$ ought to come in quite handy.

[ocean seal]

daveamayombo

[novel night]

Or $\cos(a-b) = \sin(a)\sin(b) + \cos(a)\cos(b)$ would also be a good starting point.

[ocean seal]

daveamayombo

[lone heart]

@slender sparrow Has your question been resolved?

[slender sparrow]

thank you and you're right. I divided everything by 5 and obtained sin(arcsin(3/5))sin(2x) + cos(arccos(4/5))cos(2x)

so i can now use this identity. I will report when I'm done.. sorry for answering a bit late

now I have the problem that arcsin(3/5) and arccos(4/5) are different. However I was told that arccos(4/5) = arcsin(3/5) because
$(4/5)^2+(3/5)^2 = 1$

[ocean seal]

Nini

[slender sparrow]

but I don't understand why that is

[balmy warren]

do you know the Rcos(x+a) or Rsin(x+a) method

[slender sparrow]

i do not

[balmy warren]

you can rewrite the left hand side as a transformation of sin or cosine

you know compound angle formulae?

[slender sparrow]

yes

[balmy warren]

ok so lets try writing it as rsin(x+a)

can you expand this identity out?

Rsin(x+a) = ?

[slender sparrow]

what is R supposed to be ?

a real number ?

[balmy warren]

Yes

I will get to that in a sec

[slender sparrow]

Rsin(x+a) = R(sin(x)cos(a) + cos(x)sin(a))
= Rsin(x)cos(a) + Rcos(x)sin(a)

?

[balmy warren]

Yes

Can you see how this compares to your lhs of your question

[slender sparrow]

ah right! i managed to do this a bit earlier, is that what you meant?

this is what I've done so far:

arccos(3/5) = arcsin(4/5) is what someone from my class told me we used, but I don't know how we obtained this, hence the question marks

[worn fox]

Try drawing a 3,4,5 right triangle

[lone heart]

@slender sparrow Has your question been resolved?

[lone heart]

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[vague blaze]

Hey

[last ether]

Cancel low order terms and use your constant rule

Also <@&268886789983436800> #help-0 message

Guy with cat pfp sure loves being a dick

Anyways you wanna cancel low order terms (end behavior) and use your constant rule when evaluating the limits

[vague blaze]

Hey Im quite stupid what does this mean

;c

[last ether]

End behavior

[vague blaze]

So I would cancel the 1+2 in the bottom?

[last ether]

So let's say you have $$\lim_{x\to\infty}\frac{30}{1+2^{3-x}}$$

We can rewrite that as:

$$30\lim_{x\to\infty}\frac{1}{1}$$

[vague blaze]

I see

What can we do with that?

[last ether]

Just evaluate the limit

[sly mantle]

@alpine sable pls dont be obnoxious with pings. muted for 24h

[last ether]

Actually wait

Hold on

Sorry my bad

[vague blaze]

No it's okay

how could I evaluate the limit tho

[ocean seal]

Umbraleviathan

[vague blaze]

where did the x go

[gray ingot]

2^-inf is basically 0

[last ether]

That's because $\lim_{x\to\infty}\frac{1}{2^{3-x}} = \infty$ (basically 1 will be bigger than $2^{3-x}$ as x keeps on getting bigger)

[ocean seal]

Umbraleviathan

[vague blaze]

I see

[last ether]

So I cancelled the low order term

Which was the x term

[vague blaze]

Yea i think I get it

[lone heart]

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[lone heart]

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[woven hollow]

(a^-5 b^2)/(ab^-4)

[woven hollow]

I don’t know how to solve this at all

And would like to learn how to

This is what it looks like normally

(I’m not cheating, it’s a practice and There are 6 problems I just need to know how to solve)

[mortal trellis]

do you know any rules to simplify $\frac{a^n}{a^m}$ ?

[ocean seal]

Denascite

[woven hollow]

Yes

[mortal trellis]

and do you know $a^{-n} = \frac{1}{a^n}$ ?

[ocean seal]

Denascite

[woven hollow]

Yes

👀

[mortal trellis]

write the fraction as $\frac{a^{-5}}{a}\cdot \frac{b^2}{b^{-4}}$ and treat each of those separately

[ocean seal]

Denascite

[woven hollow]

Ok

Found my pen

I’m somewhat starting to get it

Crap I still can’t solve it

Wait

Wait

I have to use the properties

So how do I use the quotient of powers property on it

[lone heart]

@woven hollow Has your question been resolved?

[woven hollow]

<@&286206848099549185>

[placid zinc]

So we're back to a question asked earlier

do you know any rules to simplify $\frac{a^n}{a^m}$ ?

[ocean seal]

Kaynex

[woven hollow]

Yes but not when the exponent is negative

[placid zinc]

Same rule still applies

Don't let the negative throw you off

[woven hollow]

So it would be a^-6 • b^6

Right?

[lone heart]

@woven hollow Has your question been resolved?

[last ether]

They have the same base

$$\frac{a^b}{a^c} = a^{b-c}$$

[ocean seal]

Umbraleviathan

[lone heart]

@woven hollow Has your question been resolved?

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[woven hollow]

I’m just not gonna understand 👁

[lone heart]

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[north adder]

okay and?

[rigid ice]

I should solve this indefinite integral using the part integration method

[north adder]

alright

what do you think you should differentiate and integrate

[rigid ice]

i tried to integrate but i ended up in a loop

[north adder]

thats fine

show me where you are

wait

what did you integrate and differentiate

you should be able to get the value after 4 ibps

[rigid ice]

I integrate 2x^3

[balmy warren]

2x^3 + x yeah?

[north adder]

no

[north adder]

why would you integrate that

have you heard of LIATE

[rigid ice]

I divided it into two integrals

no

[balmy warren]

you have 2 functions of x

[north adder]

@balmy warren u wanna take over

[balmy warren]

yes plz

[north adder]

oki

[rigid ice]

cause xe^(-2x^2) was an immediate integral

[balmy warren]

oh so you expanded the brackets out

ok that's fine

[rigid ice]

yes

[balmy warren]

so you are integrating 2x^3e^-2x^2

why did you integrate 2x^3?

[north adder]

nvm

my bad didnt read the notation

pls put brackets 🥺

[balmy warren]

I think differentiating 2x^3 and integrating e^(-2x^2) would be easier

[rigid ice]

why

[balmy warren]

ok lets go with your method for now

so you integrate 2x^3

and differentiate e^(-2x^2)

[rigid ice]

yes

[alpine sable]

u want to solve

$\int (2x^3+x)e^{-2x^2}dx$

[ocean seal]

∆y/∆x=πy+π^2x

[north adder]

because when you differentiate an algebraic equation over and over it will become 0 eventually

obv when the power is an integer

[balmy warren]

what is your result for by parts

[rigid ice]

one moment sorry

[alpine sable]

ig u will only require integration by parts for

$\int x e ^{-2x^2} dx$?

[ocean seal]

∆y/∆x=πy+π^2x

[alpine sable]

the other part can be done by u sub?

[alpine sable]

actually

intgeration by parts

is not even needed

u can use u sub for everything

[rigid ice]

for exercise I have to do it like this

[alpine sable]

oh ok

I see

[languid bolt]

,rotate

[rigid ice]

$\int (2x^3+x)e^{-2x^2}dx$

[ocean seal]

blackhawk0206

[rigid ice]

how can i integrate e?

[minor needle]

expand to get 2 integrals, then IBP

⬆️

[rigid ice]

already done

[north adder]

bruh why are 5 differnet people helping

[lime bobcat]

XD

[rigid ice]

is what I am doing correct?

[languid bolt]

other channels are pretty unresponsive.. that's probably why

[north adder]

@balmy warren

[eternal shale]

i have the same problem

[north adder]

@balmy warren dude you here?

[rigid ice]

<@&286206848099549185>

[mortal trellis]

it would probably be a good idea to split up the sum and do two integrals. also use a substitution u=-2x^2. then for one of the integrals you need to use some IBP

[mortal trellis]

you shouldn't in the second step of IBP integrate the function which you differentiated the first time. that just undoes the first step and leads you nowhere

[north adder]

@rigid ice

btw as a general principle

this is what is used

it works like 80% of the time

[rigid ice]

ok i'm trying thx

[mortal trellis]

well it's a good rule but not if you have stuff like exp(-2x^2). should first do that substitution

[rigid ice]

.close

[lone heart]

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[north adder]

lol

gg wp

i g

[north adder]

mhm hence the 80%

@rigid ice btw for future reference a thank you comes a long way

[mortal trellis]

they said thx

[lone heart]

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[rigid ice]

sorry thx means thank you

[north adder]

im illiterate my b

.close

[lone heart]

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[lost estuary]

hello!
i've got some questions about solving indefinite integrand excercises

[minor needle]

integrate each term separately

[lost estuary]

first of all: what we'd need to do here is separate the first division into two chunks:
$\frac{{sin x / 4}}{{3 \cdot sqrt} / 4}$

[ocean seal]

sla-ppy

[lost estuary]

whoops

[minor needle]

$$\frac{\sin x + 3\sqrt{x}}{4}=\frac{\sin x}{4}+\frac{3}{4}\sqrt{x}$$

[ocean seal]

Modus

[lost estuary]

but ye, even though latex is wrong, the point gets across i hope :)

[minor needle]

u can also factor out constants when integrate

[lost estuary]

yes, okay so anyways

what do you do with sin x / 4

do you carry out the division?

wait

do you just factor out 1/4?

and leave sin x for integrating

[minor needle]

yes

[lost estuary]

awesomeeee

i always get shocked when i see this, and think its some super complicated thing :`)

thanks mate

.close

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[lost estuary]

.reopen

[lone heart]

✅

[lost estuary]

any clue

what that huge dot might mean?

or perhaps its just a formatting error?

o yea, inputting it to wolfram re-formatted it, just a huge cdot lol

.close

[lone heart]

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[fluid zealot]

Can someone help me with part b?

[alpine sable]

geometrically, what does y=kx look like?

[fluid zealot]

Y= mx +c?

A line passing through the origin?

[alpine sable]

yup

If k didn't have to be an integer, where would it intersect the triangle for k to be the largest?

[fluid zealot]

A

[alpine sable]

yup, and what would k be if it went through A?

[fluid zealot]

I'm not sure

[alpine sable]

what is the x,y coordinate of A

[fluid zealot]

I only know the y coordinate 5

[alpine sable]

A is on the line 7x+5y=35, so can you solve for x with y=5?

[fluid zealot]

Ooh yeah

X = 10/7?

[alpine sable]

yup

so what would the slope of the line through (0,0) and (10/7,5) be?

[fluid zealot]

3.5

[alpine sable]

yup

so k is the largest integer smaller than 3.5

[fluid zealot]

Why does it have to be smaller than 3.5?

[alpine sable]

because 3.5 is the largest non integer slope that goes through the triangle

so any other possible slope needs to be smaller

[fluid zealot]

Oh okay so 3

Thank you so much!

[alpine sable]

yw :)

[fluid zealot]

. close

.close

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[vale wigeon]

there is no such thing as matrix division

i mean you could try multiplying the second matrix by the inverse of the first matrix (in that order), and in this case the inverse exists, but that is not always going to be possible

no

you did as i instructed

the idea is that $E_1 = (E_1A)A^{-1}$ if $A^{-1}$ exists

[ocean seal]

Ann

[vale wigeon]

is wolfram alpha giving you some kind of bullshit result?

[lone heart]

@acoustic plover Has your question been resolved?

[hazy atlas]

you can find it yourself ig
inverse of matrix exist when matrix is non singular, use the formula A^-1 = Adj A/|A|. where A is square matrix of any order

is wolfram smoking weeds or something lol (idk is the answer wolfram giving wrong?)

wait let me look up for examples

yours

are you familiar with the concept of finding cofactor matrix from a given matrix?

Adjoint of a matrix is transpose of the cofactor matrix

and "two pipes" is determinant of that matrix

try solving it by yourself ig

(if the answer by wolfram is wrong cuz idk)

transpose of this matrix is Adj matrix

and determinant i hope you know

try solving it by yourself smh

i am not quite familiar with wolfram

[lone heart]

@acoustic plover Has your question been resolved?

[lone heart]

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[tacit arch]

whatever topics were covered in the book you got that problem from

[dry ridge]

Algebra: Solving equations, exponents, log.

[lone heart]

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[ocean seal]

The

[fathom field]

bc according to my prof its the first but according to the internet its the later one

and he exclaims its 1/p multiple times too

on lecture slides

[lone heart]

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[fathom field]

what

no

.closwe

[tacit arch]

.reopen

[fathom field]

ohhh

[tacit arch]

oh that only works in timeout

you'll have to open a new channel

[fathom field]

.reopen

fuckk

[tacit arch]

don't delete your first message next time

[fallen verge]

I think they deleted the texing

[fathom field]

yeah wont clicked on delete source didnt know this happens

[lone heart]

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[cyan mountain]

Is this graph odd,even, or neither and how can I tell

[balmy warren]

do you know what odd and even functions are?

[cyan mountain]

Yes

[balmy warren]

tell me

[cyan mountain]

Odd is when the exponent goes to an odd number

And the changes are off

Odd

And even is when exponent goes to an even number and the changes in the graph are odd

I believe

[balmy warren]

no

[cyan mountain]

Oh

[balmy warren]

I suggest you research what they are

[cyan mountain]

Ok

.end

[fallen verge]

its close

[cyan mountain]

I see it’s neither

.close

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[lone heart]

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[untold garden]

N = number of samples | I know G aswell. Not sure how to proceed with the big E. Where can i learn about this: what is this kind of math called? I am calculating gravity from a series of pictures

[heady pollen]

the big E is a sum

[untold garden]

i figured that out but i dont get the context

[heady pollen]

so

basically

you have n samples

[untold garden]

the sum of what 😛

[heady pollen]

each with a certain value

[untold garden]

yeah 5 of em or so

[heady pollen]

if you want to get the mean value of say you have 5 samples

than you would logically add the five together and divide by 5 right?

that is exactly what this means

[untold garden]

thats what i got. What variable should i the sum of

hmm 😛

[heady pollen]

this sum $\sum{i=1}{n}$ means, that you add the element after it together

oops my latex is wrong

[untold garden]

could you give me a hint based on that?

i really dont get it lol

[heady pollen]

wait let me write it up real quick

[untold garden]

❤️

i will learn thank you (from your help )

[heady pollen]

the i is called an index

it "runs" from 1 to n

and you add up the different g_i

[untold garden]

yeah i get the i part

[heady pollen]

that means that the enumarator equals g_1+g_2+g_3+...+g_n

so you add all the g_i together and divide by the number of them

[untold garden]

thats almost jibberish for me haha what tutorial should i watch xD

[heady pollen]

so in your example you would add up
9.87+11.6+13.0+12.7+12.9

and divide by 4

[untold garden]

oh that makes a bit more sense to me

😛

oh g i(1)

[heady pollen]

basically add everything together, divide by number of things you added

[untold garden]

so Eni = 1

should be

the sum of 9.87+11.6+13.0+12.7+12.9 = 1(gi)

[heady pollen]

what do you mean with Eni?

[untold garden]

The first symbol

[heady pollen]

ahh ok

[untold garden]

sum of symbol the n and the i

😛

like how should i write in the calc lol

[heady pollen]

how do i put this

[untold garden]

9.87+11.6+13.0+12.7+12.9 = 1(gi)

guess thats kinda right hmm

9.87+11.6+13.0+12.7+12.9 = 1(gi) / n

[heady pollen]

do you understand programming language?

[untold garden]

a little

g = √ 9.87+11.6+13.0+12.7+12.9 = 1(gi) / n

https://cdn.discordapp.com/attachments/490557019623915520/989622789101219942/unknown.png

is that the same?

[heady pollen]

maybe this helps

your g_i is not a number

[untold garden]

it should thanks ;D

[heady pollen]

g_i is representing more like an idea

you run i from 1 to n

so i becomes that number

[untold garden]

that sounded funny

[heady pollen]

so there is no g_i

but there are g_1 g_2 and so forth

[untold garden]

gotta sit with this a bit dude thanks xd

[heady pollen]

https://www.youtube.com/watch?v=XJkIaw2e1Pw

this might help

it is long but you dont have to watch the whole video

[untold garden]

ok 🙂

[heady pollen]

but spoken words are always easier to follow than written words

[untold garden]

thanks a lot! ( i kinda see u did everything up there 🙂 )

[heady pollen]

your welcome

[lone heart]

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[pseudo spear]

$\sum_{i=1}^{i=m} \sum_{j=1}^{j=n} C_i_j \text{ where } C= (AB)\circ W \text{ and W is symmetric}$

[ocean seal]

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

[pseudo spear]

I'm looking for an effecient way to compute the sum

W has a specific pattern to it. In practice m and n are both 2017 and everything is square

[lone heart]

@pseudo spear Has your question been resolved?

[pseudo spear]

Each element of W only has one of 7 elements

I'm thinking there may be a way to compute 7 partial sums and then multiply them by each of the unique values

[lone heart]

@pseudo spear Has your question been resolved?

[tacit arch]

try asking in a computer science server

[pseudo spear]

Do you know of any software that would solve the equation for a given W?

as in, simplify it automatically by looking at all the different terms that arise

[tacit arch]

if you write out an example on paper and upload a picture maybe

but what you're asking is too vague

[pseudo spear]

There's my work so far

if you open the original the resolution should be enough

[tacit arch]

What are you even solving for?

Separate your givens from what you're looking for

[pseudo spear]

sum of all elements of the matrix

its in the latex i wrote at the start

[tacit arch]

you're calculating the double sum?

[pseudo spear]

no doubles, just every element of C summed

so the line (a + d + g)(j + k + l) is where I found you could get AB via row sums, but I'm not sure how to get the weights in there

[tacit arch]

what are your matrices A, B, and W here?

[pseudo spear]

at the top of the page abc def ghi would be A, jkl mno pqr would be B. the table in between them would be C

And I've scribbled some weights in for W in the table as well

just a pattern 2 2 3, 2 3 2, 3 2 2

and then at the bottom I go off and try to calculate that to see if W falls away some how

but I didn't get anywhere

I'd probably have to write out a much bigger example to see if a pattern emerges

[tacit arch]

you're probably better off finding an algorithm that does hadamard products efficiently

https://static.hlt.bme.hu/semantics/external/pages/LSTM/en.wikipedia.org/wiki/Hadamard_product_(matrices).html

then sum the elements at the end

[pseudo spear]

the problem is not the hadamard so much as AB

[tacit arch]

Hadamard multiplication is built into certain programming languages under various names. In MATLAB, GNU Octave, GAUSS and HP Prime, it is known as array multiplication, or in Julia broadcast multiplication, with the symbol .*

matrix multiplication is already efficient in most programming languages

unless potentially you add more symmetry/conditions

[pseudo spear]

it's n^3, there's a way to get this with something like 2n

[tacit arch]

nope

[pseudo spear]

There is, I know someone who's done it

[tacit arch]

https://en.wikipedia.org/wiki/Computational_complexity_of_matrix_multiplication

go ahead and update the wiki

[pseudo spear]

Yeah the thing is because we're finding the sum of elements, the full matrix multiplication is not necessary

I'm not looking for C

[tacit arch]

The current best bound on ω is ω < 2.3728596,

[pseudo spear]

As in my latex i'm looking for the sum of the elements of C.

[tacit arch]

then go ask them?

[pseudo spear]

It's a competition, they're not going to tell me

[tacit arch]

.

[pseudo spear]

The line showing (a + d + g)(j + k + l) shows you can get the sum of AB with 2n by summing the rows and columns of A and B

but the issue is how do you factor the weights in

I doubt they'd have anything, it's not really a comp sci question

[pseudo spear]

If you wanted to simplify a large expression which software would you use?

[pseudo spear]

.close

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[sinful tapir]

Wind is blowing a kite horizontally at a speed of 4 ft/s. At what rate is the angle between the string and the ground
changing when the kite is 14 ft above the ground and 50 ft of string has been let out?

[sinful tapir]

im not sure how do to this

can someone plz help? thx so much

[placid zinc]

Let's first get an equation for that angle

[sinful tapir]

tangent?

tan14/50

[placid zinc]

What's the angle of the string, in terms of the x-distance?

I think tan, yeah. That is:
tanθ = x/14

[sly mantle]

@sinful tapir pls dont multipost or post in others channels

[sinful tapir]

ok sorry

what are u defining x as

[placid zinc]

I'm letting x be the horizontal distance

I know it's 14 high, and x far.

[sinful tapir]

isnt that 50?

[placid zinc]

It will be 50 at some point

[sinful tapir]

oh ok

so we got tan theta = 14/x

[placid zinc]

For now, it's good to know how θ changes with x, so we leave x as a variable

Then we can finish it up by taking the derivative of both sides in terms of t:
sec²θ (dθ/dt) = 1/14 (dx/dt)

[sinful tapir]

and then we plug in dx/d

dt

right

[placid zinc]

Yeye. We also want x = 50 like you said

[sinful tapir]

so thats 4

[placid zinc]

Can solve for dθ/dt this way

[sinful tapir]

wait

but u cant plug in x

anywhere

[placid zinc]

Oh haha

No x to plug in, good point

[sinful tapir]

then how do we find it

[placid zinc]

We do want to plug in θ

[sinful tapir]

which one

arc tan 14/50?

[placid zinc]

We want dθ/dt. That's solving our problem

[sinful tapir]

arctan(14) / 50 =
1.71828766 degrees

?

[placid zinc]

Yeah, that's right. θ = arctan(14/50)

[sinful tapir]

ok

1.00089992627

thats equal to sec^2 (1.72)

right

[placid zinc]

I'll take your word for it, haha

[sinful tapir]

it seems low

[sinful tapir]

and what are the units?

2/7 deg/se

sec

thats what i got

[placid zinc]

,w 4 / (14sec(arctan(14/50))^2)

[sinful tapir]

why are u multiplyngby 50

?

[placid zinc]

Can't work in degrees, doesn't agree with calculus

Oh mb

[sinful tapir]

why by 4

im confised

i just did

[placid zinc]

0.308 rad/s?

[sinful tapir]

sec^2(1.72) x = 4/14

and solved for x

[placid zinc]

Wow, I'm not in it.

[ocean seal]

Results provided by WolframAlpha

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

Requested by Kaynex#7520

[placid zinc]

Okay that looking a little better? Haha

[sinful tapir]

shouldnt it be

(4/14) / that

[placid zinc]

That's the same

[sinful tapir]

oh ok

is that in degrees

[placid zinc]

I mean (4/14) / u = 4/(14u)

That's in rad/s

[sinful tapir]

ok

[sinful tapir]

i think its in deg

[lone heart]

@sinful tapir Has your question been resolved?

[lone heart]

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[spare canopy]

how can can 16/9 + 16 get into the root of 160/9?

[weary wyvern]

$16 = \frac{16\cdot 9}{9}$

[ocean seal]

giannis_money

[spare canopy]

i dont get it

why is there a sudden 9 above

[fallen verge]

You have 16 wholes which are 16*(9/9) since 9/9is one whole

[weary wyvern]

wdym

so you think $16 = \frac{16}{9}$?

[ocean seal]

giannis_money

[spare canopy]

no i was confused, i got it now ty

👍

.close

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[oak fiber]

let f(x) = x^2 +ax +b , where a and b are constants. v(-4,-5) is the vertex of the graph y = f(x)

a) use the method of completing square, find a and b

b) given p(p,-1), q(q,-1) are two distinct points lying on the graph, where p>q. find the distance between p and q

[oak fiber]

help me pls

[tawny schooner]

What do they mean by "vertex of the graph"?

In the first qn

[fallen verge]

Critical point

[tender dew]

Vx-q = p-Vx

[tender dew]

we know this because function is symetrical

so if for 2 different arguments function has same value, then those points are same distance away from vertex

Vx = -a/2

a = 8

b = 11

[oak fiber]

ty

.close

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[gusty pendant]

I have a circle with the origin (0, 0)
How would I find the required radius of the circle to intercept any point along a 2-D plane?
Note: This is not for a homework assignment or anything, just curiousity after messing around with desmos.
Also lmk if question doesn't make sense lol

[last ether]

Do you mean that the circle is centered at the origin?

[gusty pendant]

Yes

[last ether]

Well

You can have the point (a, b) that's on the circle

[weary wyvern]

Do you want the circle to contain every point in the plane?

[last ether]

Then $x^2 + y^2 = a^2 + b^2$

[ocean seal]

Umbraleviathan

[last ether]

Assuming that (a, b) is a point on the circle

Is that what you're looking for?

[gusty pendant]

Oh, that's actuall pretty simple. Thanks!

[last ether]

Yeah it's just distance formula squared

[gusty pendant]

Yup :)

[last ether]

Which is just gonna be it's argument lol

[gusty pendant]

.close

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[edgy flare]

Complex Variables😅:

[edgy flare]

I’m not sure how to even start this one.

[lone heart]

@edgy flare Has your question been resolved?

[edgy flare]

<@&286206848099549185>

[edgy flare]

I’ll ask tomorrow.

.close

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[final oriole]

hello. How do I shift this graph such that it has an origin of (0,0)? It's okay if the shape of the graph changes a little but I still want it such that it increases from negative infinity reaches a peak and asymptotes to kinda like pictured above (so same behaviors).

[median oar]

what do you mean it has an origin of 0,0

[vale wigeon]

do you mean that you want the graph to pass through (0,0)?

you could just subtract 4 from it

[final oriole]

oop. my bad at explaining. I basically want a horizontal shift of some sort. not a vertical one

so that the current x intercept is at the origin now

nvm i figured it outtttt. tyyy

.close

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[lament basalt]

Help ima explain pls wait

[lament basalt]

This is about the graph of Circular functions
Specifically the graph of tan function

So i wrote an example of function where it has 2 for the amplitude (distance of center to a peak, or d of center to a trough)

Since it is tan (y/x)
And circle has 30-45-60 triangles before going to 90 degrees

I multiplied each values of the 30-45-60 by 2

And I got this conclusion
Is this correct?
Thanks

[vale wigeon]

is your goal to graph y = 2tan(x) for 0 < x < 90°?

[lament basalt]

No

I just graphed it until 90 degrees to find out if what I did is correct

[vale wigeon]

then what is your goal, if not to graph y = 2tan(x)?

[lament basalt]

I just wanna find out if what I did to get the values of 30, 45, & 60 degrees are correct

I multiplied each of their original values by 2 and rounded them off

Ima ss the original values pls wait

[vale wigeon]

your y-axis is unlabeled

[lament basalt]

Oh sry ima write it

[vale wigeon]

also jeez what kind of textbook is this that you're being fed decimal approximations to all these values

[vale wigeon]

no, 2tan(30°) isn't 1 and 2tan(60°) isn't 3...

your awfully coarse rounding explains why your graph looks so wonky

[lament basalt]

oh sry

[vale wigeon]

not to mention that like. the x axis scale is off the shits

if you're making graphs then why the devil are you not making them on graph paper?

[lament basalt]

Wait sry I forgot to see the message above the, "you not making them... paper."

I don't have graph paper and pardon if it looks terrible

What if I didn't round them off?

[fallen verge]

Your x axis spacing is still kind of off

[fallen verge]

30 should not be halfway in between 0 and 45

[lament basalt]

Meaning, 1.5?

[lament basalt]

I thought more than 1 but less than 1.5

[fallen verge]

No, where you put the line for 30

It shoudld be more to the right

It shouldnt be halfway

30 is 2/3 of the way to 45

[lament basalt]

Pardon

Okay

Ohhh I get it

But how about the values of y?

May I ask if it's correct?

[fallen verge]

They should be fine i think

[lament basalt]

I multiplied each values of tan which is seen here

By 2

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[weak seal]

$\sum _{n=1}^{\infty :}\frac{1}{4n^2-1}$

[ocean seal]

Senoune

[weak seal]

is this a harmonic serie?

[tacit arch]

try telescoping

[weak seal]

hmm idk if we've seen that and I dont feel like I've missed a class

lemme check just to be sure

[tacit arch]

$\frac{1}{a^2 - 1} = \frac{1}{(a-1)(a+1)}$

[ocean seal]

riemann

[hazy atlas]

use the formula a²-b² = (a+b)(a-b) in denominator, now subtract those two in numerator and make it 1 by multiplying/dividing by that number which comes so as to make numerator 1, now you will see that all the terms but the first and last gets cancelled out.

[weak seal]

ok this is smart

[hazy atlas]

is this method called telescoping?
im not quite familiar with the names of concepts/theories lol

oh yeah it is just Googled

anyway this works @weak seal good luck

[fallen verge]

Partial fraction decomposition from here

[weak seal]

.close

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[vale wigeon]

do you mean the set of all matrices of this form, where a, b, c ∈ R? @hallow light

are addition and scaling defined in the usual way or are some extra definitions given?

so that's yes to both

in that case...

what's your zero vector?

you didn't answer my question

and also what you did say is wrong

the neutral element wrt matrix addition is the matrix of all zeros

what?

"isn't it the case that when you take any matrix and multiply it by -1, the result is the matrix of all zeros?"

the zero matrix is not in your set

this question doesn't make sense

the 4×4 zero matrix EXISTS

it just isn't a matrix of this form.

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[lone heart]

@feral quiver Has your question been resolved?

[lone heart]

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[sharp moth]

Why is this written this way?

[sharp moth]

Instead of this

Like the +/- overrides the denominator's sign?

In the case of the first term I think I understand that the two negatives cancel each other out.

[vale wigeon]

$\mp$ is only really used in conjunction with $\pm$ when there are multiple plus-minus signs and you need to clarify which combinations of plus and minus are to be considered

[ocean seal]

Ann

[sharp moth]

There is a difference between the two signs?

[vale wigeon]

yes

when you see a $\pm$ and $\mp$ in the same experssion it usually means that they are to be taken as opposite signs

[ocean seal]

Ann

[vale wigeon]

thats about it

[sharp moth]

What do they both mean

they clarify which of the root is being taken

[hollow shale]

When the first sign is positive, the second is negative, and vice versa.

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[long smelt]

How do you do C? Can you walk me through the rules you're using? I'm having a hard time trying to learn them

[vale wigeon]

do you know the rule $p \log_a(x) = \log_a(x^p)$?

[ocean seal]

Ann

[vale wigeon]

use that & the sum of logs rule

[long smelt]

So x^4*y inside the log, right?

[vale wigeon]

exactly

[long smelt]

Awesome, not sure why that didn't come to me. Appreciate the help!

.close

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[paper stirrup]

Hi can someone help me on polynomials

[paper stirrup]

How draw this graph?

Question 2

[high rapids]

Where did (x+1) come from?

Oh nvmd

Didn't see -1

Well since the coefficient of x^3 is 2 what can you conclude from this?

[paper stirrup]

uhh

cant really say anything

[high rapids]

Do you know what happens what the sign of the coefficient of x^3 says about the curve?

[paper stirrup]

it is wavy

but im not sure in which direction it goes

there are 2 ways

[high rapids]

What do y=x^3 and y=-x^3 look like?

[paper stirrup]

first one should be above the x axis

or it is increasing

yeah

[high rapids]

Yes

Now can you find the y-intercept?

[paper stirrup]

how

i sub in values?

for x

[high rapids]

Yes

What is x in the y intercept?

[paper stirrup]

the 3 values

i sub in something random like 0.2?

[high rapids]

No

On the y-axis what is the value of x?

[paper stirrup]

it depends

ohh

its 0

so i sub x with 0?

[high rapids]

Yes

[high rapids]

Shouldn't it be [x-(2+√5)]?

[lone heart]

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[rocky condor]

what si the firmula

[paper stirrup]

.reopen

[paper stirrup]

This?

[languid bolt]

@paper stirrup this help is already occupied btw

[rocky condor]

help him this poor guy

[paper stirrup]

ik, it closed by itself huhu

wanted the guy to reply me

[languid bolt]

CB * BP = PA^2 ?

tangent secant theorem

[rocky condor]

[languid bolt]

?

[rocky condor]

thanks for the name

[languid bolt]

np

[rocky condor]

do you know this name

[languid bolt]

hmm

i dont

[rocky condor]

i'm in the darkweb of math

[languid bolt]

wdym

[rocky condor]

I found a video about circle theorems with the name you mentioned above tysm

.close

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[languid bolt]

@rocky condor power theorems you mean?

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[slim fern]

I have 5 microcomputers to be connected to 3 printers. how many connections are necessary between the computers and printers in order to be certain that whenver any three computers each require a printer, the printers are available

[vale crag]

is that something you're trying to do for real ? Just checking

[slim fern]

No lol

[vale crag]

Okok

Availability sounds a bit vague here tbh

whenver any three computers each require a printer, the printers are available

Let's imagine you got 3 printers spamming a lot of print jobs

Regardless of the way the printers are connected, I wouldn't consider the printers available anymore

Is there a certain way the computers behave though?

Do they wait for their current job to be finished to send a new one or whatever ?

[mortal trellis]

I assume in terms of graph theory the problem is as follows:

Let $G=({1,2,3}\cup {1,2,3,4,5}, E)$ be a bipartite graph such that for every selection of three vertices $A={a,b,c}\subseteq {1,2,3,4,5}$ there exists a perfect matching in the subgraph $G_A = ({1,2,3}\cup A, E_A)$. What is a lower bound on $|E|$?

(the union here is supposed to be as multisets with two ones etc)

[ocean seal]

Denascite

[vale crag]

Ah yeah I guess I was seeing the problem in a queuing theory style for some reason

[vale crag]

That's probably a more pertinent interpretation here ^^

[mortal trellis]

if it is the correct interpretation I honestly have no clue how to tackle it

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[slim fern]

Thank you!

[mortal trellis]

that helped?

[slim fern]

I figured it out

Myself

[mortal trellis]

what's the answer

[slim fern]

12

[mortal trellis]

and how did you get it

don't leave us hanging

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[slim fern]

actually wait, I've changed it to 9

[slim fern]

C1 -> P1
C2 -> P2
C3 -> P3
C4 -> P1, P2, P3
C5 -> P1, P2, P3

.close

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[slim fern]

@mortal trellis

[stuck grail]

But how did you prove that 9 is minimal?

[slim fern]

Idk how to type it up and send it rn, its in my book, but thats the solution

I'll send it after I'm done with this worksheet

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[olive juniper]

i dont get why

[olive juniper]

,, x*-\frac{20}{x}

[ocean seal]

minto7

[olive juniper]

is -20

[vale wigeon]

is this a "i think it should be something other than -20" or "the entire expression reads like gibberish to me and i am endlessly intimidated"?

[olive juniper]

latter

im afraid of fractions

[gilded vessel]

well if you divide -20 by any number say 5 and then multiply it by 5 you will of course get -20

x just represents any number

[olive juniper]

ooh i get it

[vale wigeon]

yes

dividing by x and then multiplying by x brings you back to where you started

in this case the "where you started" is -20

[olive juniper]

.close

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[real flicker]

how do I solve these with log

[lament glen]

for the first one, you can factor a $3^{2x - 2}$

[ocean seal]

Doggo

[lament glen]

then divide both sides by something

and take log base 3 of both sides

then do some algebra

I think you don't need a calculator

[real flicker]

may I ask what $ is

[mortal trellis]

you can also just guess the first one using the "law of nice numbers"

[lament glen]

it's for latex

[lament glen]

this is the actual message

[real flicker]

o

[lament glen]

then scribble something for proof by intimidation

[real flicker]

but how do I factor it I'm new to this topic so I'm not so sure

[lament glen]

factoring isn't related with logarithms

anyways

$3^{2x - 1} = 3^{2x - 2} \cdot 3$

[ocean seal]

Doggo

[real flicker]

oh i see what u mean now

so this is how I'm supposed to write it

[lament glen]

and now you can factor a $3^{2x - 2}$

[ocean seal]

Doggo

[lament glen]

$ka + kb = k(a + b)$ this thing

[ocean seal]

Doggo

[real flicker]

then does it become 6^2x-2

[lament glen]

it doesn't become 6 but yeah this step is correct

then you can divide both sides by 2 to leave the $3^{2x - 2}$ alone right?

[ocean seal]

Doggo

[real flicker]

oh my God ur so smart

[ornate grotto]

3^2x-2=27
(Take log base 3)
2x-2=3
2x=5
X=5/2

[real flicker]

the log step is correct right

[lament glen]

yeah

[ornate grotto]

Yes good job

[real flicker]

oh my

tysm

I'll try the second one on my own

🙏

[ornate grotto]

Question, can you do it without factoring?

[lament glen]

not sure but I don't think so

[real flicker]

can we leave this channel open for a while so u could check my second one

[lament glen]

the bot sometimes sends a message, as long as you say the question isn't answered when it does the channel will stay open

[real flicker]

alr ty

can I remove the log if both sides have the same log with same base

referring to this step

[lament glen]

log refers to base 10 or base e?

[real flicker]

I meant base 10

or base any number

[lament glen]

yeah removing log from both sides is just raising both sides to whatever base power

[real flicker]

alright

thx for ur help

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[mental socket]

I just need the answer to know if I was right

[mental socket]

I got 5 times square root 10

[mortal trellis]

,w sqrt(5)(sqrt(8)+sqrt(18))

[ocean seal]

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[mortal trellis]

yes

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[ionic bear]

I have to find the limit of a_n using squeeze theorem but I think I did something wrong because the limits of those 2 sequences are 0.

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@ionic bear Has your question been resolved?

[ionic bear]

<@&286206848099549185>

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@ionic bear Has your question been resolved?

[harsh girder]

so the limit of an is 0

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@ionic bear Has your question been resolved?

[ionic bear]

I don't think is possible to be 0

[tacit arch]

why

[harsh girder]

,w Limit[Sum[(n + k)/(n^3 + k), {k, 1, n}], n -> Infinity]

[ocean seal]

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[ionic bear]

Wouldn't a_1 be (1+1)/(1^2+1)=1 which is >0?

[tacit arch]

so?

The sequence 1, 1/2, 1/3, 1/4, ... begins with 1 and is all positive but it converges to 0

[ionic bear]

Oh so I did this exercise right

Thank you

.close

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[slender sparrow]

I am trying to calculate this

[slender sparrow]

for pi/2 < phi < 2pi

I already know that cos(phi) = 4/5 and sin(phi) = 3/5

So I have this :

[alpine sable]

Nini

Is that thing in the middle an o?

[slender sparrow]

yes

[alpine sable]

Ok hmmm.

[slender sparrow]

okay hmm i made a mistake

[lament glen]

why is 4/5 in cos

[tacit arch]

you substituded $\cos(4/5)$ instead of 4/5

[alpine sable]

Yeah

[ocean seal]

riemann

[slender sparrow]

i just realised haha my bad ... thank you!

Though I do have another question: I don't know when the condition pi/2 < phi < 2pi comes into play

[weary wyvern]

Are you given cos phi is 4/5?

[slender sparrow]

no
I am given sin(phi) = 3/5 and I found cos(phi) using pythagorus

[weary wyvern]

cos phi is not 4/5 then

Think about which quadrant phi is in

[slender sparrow]

the first quadrant