help-19

[zenith tartan]

🤨

[prisma cairn]

I was explaining the horizontal forces

[teal adder]

I know

But @zenith tartan called the horizontal equilibrium the sum of Fy

That’s how he notated it in his explaination

[zenith tartan]

oh yea

switch them

[teal adder]

Omfg

[prisma cairn]

💀

[teal adder]

Ur kidding

Ur kidding bro

Ur genuinely kidding

This kid should never get a helpful role

15 minutes of explaining cuz this kid notated wrong

[prisma cairn]

Bruh how old are you

[zenith tartan]

don't call me a kid

[teal adder]

16

[zenith tartan]

im 97 years old

my eyes r getting weak

things happen

[teal adder]

Bro

This guy is not qualified

Yeah explanation makes sense

Thanks jits

Predicate you @prisma cairn

[zenith tartan]

jits?

[teal adder]

Not @zenith tartan though

[teal adder]

U called me dawg in every single text

[zenith tartan]

ok that's enough

[teal adder]

And u question jit lmao

Bros offended

[zenith tartan]

mf i left my study for you, how can you say that

[teal adder]

Mf it took me 45 mins to understand this

I couldve explained this in 5 minutes

[prisma cairn]

Imma go sleep i Gotta wake up at 6 tmr

Lmao

Is 3:25

[zenith tartan]

bro it's only 3 hours lfet for 6

[teal adder]

.close

[odd edge]

Channel closed

Closed by @teal adder

Use .reopen if this was a mistake.

[teal adder]

Stop yapping guys

[zenith tartan]

im laughing so hard

mb tho

[teal adder]

I know

It is your bad

In never using this server again

[prisma cairn]

Wow

[zenith tartan]

😂😂😂😂😂😂😂😂😂😂😂😂😂😂

[teal adder]

I’m kidding @zenith tartan

[cold urchin]

aight u cant say that to my apprentice lil bro

[zenith tartan]

yea ik

[teal adder]

Bro

This guy

[zenith tartan]

yes

[teal adder]

Took 45 minutes to explain

@cold urchin new appreciate

[zenith tartan]

i didnt

that was ur

dumb ass

[teal adder]

Apprentice

[odd edge]

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[teal adder]

Stfu @zenith tartan

[cold urchin]

.close

[odd edge]

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[zenith tartan]

lmao

[teal adder]

You switched the sum of y and x

Confused me so much

I was baffled

Then u have the audacity to insult my intelligence

[cold urchin]

bro its a new help channel

[zenith tartan]

that was common sense tho, you could have figured that out earlier

[prisma cairn]

Z

[teal adder]

.reopen

[odd edge]

✅

[cold urchin]

omd

[teal adder]

.close

[odd edge]

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[teal adder]

@zenith tartan

yes

what

[zenith tartan]

ur msgs getting deleted

[teal adder]

Lmao

By who

My intelligence quotient is higher than yours

There we go

[zenith tartan]

idk, let's just stop here

[teal adder]

Have a good day friend

Don’t help anymore ppl tgo

[zenith tartan]

dont start again

[teal adder]

🙏

[prisma cairn]

My intelligence remainder is zero after this

[teal adder]

Okay

.close

[white basin]

bruh 💀

we needa get @zenith tartan helpful asap

[odd edge]

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[slender patrol]

I asked something about the same question before but now I'm actually trying to do it and it's not working out... here's the question

[slender patrol]

part a, I've differentiated and set that equal to 2

then tried to solve for x but got a quartic that I don't think i can solve with the level of maths I've covered

[nimble blaze]

solving for x isn't needed,
you should be able to manipulate what you have to the equation requested

can you show your derivative?

[slender patrol]

ooh wait it's alright, i just messed up basic arithmetic

this question is also proving to be an issue

im unsure of a nice way to prove part c and therefore part d

i initially went based off the fact that the sum of the roots of unity are equal to 0 and we showed that w⁵=1 in part a

but I'm not sure if I'm allowed to assume that fact, but it is only 2 marks

.close

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[mystic saffron]

Determine if the series converges. Having trouble starting problem, tests to be used comparison test, limit comparison test, ratio test, and integral test

[river lily]

I would try a comparison test here

wait

lemme work it out

ratios is a better choice

[mystic saffron]

okay let me try that

give me a sec

[pulsar elbow]

can't you compare this to the sum $$\sum_{n=1}^\infty \frac{8}{e^{2n}}$$, which converges by the comparison test again $$\sum_{n=1}^\infty \frac{8}{n^2}$$ which converges

[river lily]

can you do that?

[mystic saffron]

i dont think you can? becase e^2x and x^2 dont act the same way when going to infinity?

[river lily]

also missing the sqrt terms

unless im wrong

lemme google

[pulsar elbow]

sure, but 8/n^2 > 8/e^2n > 8/e^2nsqrt(n^3+1)

[river lily]

ahhhh

[clever fjord]

smay

[pulsar elbow]

i should stop editing that message

[river lily]

and since 8/n^2 is larger than the an and it converges we can argue that an will also converge

i see

i agree

[pulsar elbow]

there is a heirarchy of functions here that you should think about, e^x > any polynomial in the limit, so if you have it in the denominator and nothing super interesting or fast growing in the numerator, then you should recognize comparison as an option

[odd edge]

@mystic saffron Has your question been resolved?

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[wheat apex]

i dont get this

[wheat apex]

what does it mean by decimal expansion?

[cold urchin]

i think it just means the same decimal representation

[pulsar elbow]

yeah this is strange wording and ambiguous, real numbers don't "have a decimal expansion", you may as well say (x,y) such that x = y

[odd edge]

@wheat apex Has your question been resolved?

[wheat apex]

ima be so sad if this is assigned on anything then lol

[wheat apex]

answers were 0.1 2.1 4.1

[pulsar elbow]

well, that is not a decimal expansion, if you want a better name it’s called the “fractional part”

you will usually see it in any resource about the floor function

[wheat apex]

ah i see

then how is 2.1 and 0.1 elements of 4.10?

[pulsar elbow]

well they have the same fractional part. 4.10 is 4 + 0.1

so the fractional part would be 0.1

and the answer choices that have that fractional part are 0.1, 2.1, and 4.1

you can think of it as “the stuff after the decimal point”

[wheat apex]

i see

[odd edge]

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[warm rapids]

What’s good y’all, I tried getting help on this from the physics disc but wasn't able to get any hits, so Im trying my luck here. The question is basically: At how many points along the line connecting A and B is the interference constructive? So far, Ive tried to count the amount of points N of constructive interference existing between the the two antenna by using the equation r1-r2=mλ for constructive interference.

Ive used the fact that r1+r2=40m (where r1 and r2 are the path lengths of EM waves coming from antenna A and B respectively) and also the fact that r1-r2 should equal an integer multiple of 7 which I then used to cycle through different integer values of m. I got an answer of about 5 points of constructive interference by dividing the total length 40m by the wavelength 7m and rounding it, which still wasnt right.

[odd edge]

@warm rapids Has your question been resolved?

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@warm rapids Has your question been resolved?

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@warm rapids Has your question been resolved?

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[astral flare]

<@&286206848099549185>

[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.

[lament walrus]

stop

[astral flare]

how do you round a decimal that has 3 digits

[mint mirage]

Google it

[lament walrus]

you are using channel and u already opened one

[mint mirage]

If you don't have a specific problem you are stuck on, please stop opening channels like this

[odd edge]

@astral flare Has your question been resolved?

[odd edge]

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[hoary marsh]

Hello

[hoary marsh]

Why does dx = 2cos(θ)dθ?

[potent nexus]

wdym

just differentiate x=2sintheta

rearrange for dx

[cedar geyser]

D/dx 2sin(theta)

[hoary marsh]

But where did the dθ come from?

Implicit differentiation?

[cedar geyser]

Oh my bad I meant d/dtheta

Differentiate w respect to theta

[hoary marsh]

So dx/dθ?

[cedar geyser]

Yea

[hoary marsh]

Ahh I see

Thank you

.close

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[south anchor]

This is a graph that represents a relationship. Would the relation be not transitive because A is related to B and B is related to A but A is not related to A. Do the vertexes have to be distinct?

[crisp wadi]

The relation is not transitive, yeah

Do the vertices have to be distinct for what to happen though?

For that to be a graph?

[south anchor]

oh I meant the elements x,y,z in the definition of transitvity

thank you

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[crisp wadi]

No, they may be identical

[odd edge]

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[tough bridge]

I'm not being able to figure it out, could someone help me with it?

Noo😭

[rocky juniper]

fcck sry

there are three other channels!!

[tough bridge]

Wrong timing again🥲

[rocky juniper]

😦

[tough bridge]

It's cool de

Dw*

[rocky juniper]

Hi, I'm sorry but it's the first time in linear algebra that I'm completely lost and don't know at all how to do the exercise. Can you help me?

I've started with the formulas for the left and right pseudoinverses for the specific matrices with full column or row rank but don't know if it's right or what to do next

[pastel dew]

calm down guys

[rocky juniper]

didn't do anything

[odd edge]

@rocky juniper Has your question been resolved?

[odd edge]

@rocky juniper Has your question been resolved?

[odd edge]

@rocky juniper Has your question been resolved?

[odd edge]

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[winter quest]

Does anyone, who also knows some biology, know the probability of independant segregation of chromosomes as well as the probability of crossing over? I know independent assortment/segregation is 2^23. Assume crossing over is the same probability, what is the total number of combinations for both?

[winter quest]

but they can also randomly cross over

so when you combine these probabilites do you add or mutiply

[odd edge]

@winter quest Has your question been resolved?

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[winter quest]

.REOPEN

.reopen

[odd edge]

✅

[winter quest]

.close

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[green ledge]

is there any way to solve for θ or Φ, given the values of α and β

[green ledge]

couldn't get further than this

[weary pelican]

you could apply sin on both sides

[weary pelican]

if you managed to get this, then $\sin(\beta) =...$ and $\sin(\alpha) =...$

[clever fjord]

rafilou2003

[green ledge]

oh right

i somehow missed that

$phi = \arccos(\frac{sin(\beta)}{sin(\alpha)})$

[clever fjord]

inf1425

[green ledge]

i guess

is that right?

.close

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[dry star]

Can someone confirm that the answer is in fact -3x/1-x

[dry star]

If it isn't can you check my thought process on how to solve for errors

[flint heart]

How do I check your thought process

[dry star]

I guess I will walk you through how I got that answer

Is it incorrect?

[south plume]

you are not correct

[dry star]

Okay

So I started by setting x = y+1/y-3

[south plume]

mhm

[dry star]

I canceled out the denominator on the bottom through multiplication

[flint heart]

Ok

[south plume]

okay

[dry star]

I set it equal to (y-3)x = y + 1

[south plume]

yes

[dry star]

substituted in the x for both the y and the 3

So it would be yx - 3x = y+1

[south plume]

oh

distributed

[sullen ferry]

distributed?

[cold urchin]

Ah lol

[flint heart]

Ok

[south plume]

anyway continue

[sullen ferry]

i believe you mean distributed instead of substituted, just for the future

[dry star]

Okay thank you lol

I do appreciate that, still learning

subtracted yx from both sides

now: -3x = y + 1 -yx

distributed the 1 in for y

[flint heart]

Huh

1 is not in bracket

[dry star]

I beleive I messed up doing that so it would be -3x = y ( 1 - x)

[flint heart]

Where did the 1 go

[south plume]

yeah those are not the same

[south plume]

RHS here is y-yx, but before it was y+1-yx

[dry star]

okay, so maybe (-1+x)?

[south plume]

missed the one

[south plume]

still no

[south plume]

move +1 to the other side

[dry star]

okay so it would now be -3x + 1 = y - yx

[flint heart]

Bruhhhhh

[dry star]

lmao

[south plume]

nope

-1

on the lhs

-3x-1

[dry star]

okay lmao, simple mistake

-3x-1 = y - yx

[south plume]

yep

now continue as you did before

factor y out

[dry star]

I'm guessing that it will be -3x-1 = (1 - x)

or just x

do the y's cancel?

[south plume]

y-yx = y(1-x)

there will be just one y

it won't cancel with anything

-3x-1 = y(1-x)

[dry star]

okay

divide on the right side

[south plume]

yes

[dry star]

to get -3x-1/1-x = y

[south plume]

yes

which is your answer

[dry star]

okay thank you, I think I just struggle with remembering that tiny park

part

I appreciate all the help though!

.close

[odd edge]

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[pseudo crag]

I did these absolute values, in the first it says that x belongs from - infinity to 0 and from 2 to plus infinity, but it also includes them and the second one says that when x is negative then x belongs from zero to 2. For the second one, when x is less than or equal to 3 and second that x>3. How should I set the conditions and resolve?

[odd edge]

@pseudo crag Has your question been resolved?

[flint heart]

U can have first condition that x less than or equal to 0 x<=0 then x is in btw 0,2 then x is in btw 2,3 then x is greater than 3

There are 4 conditions in total

[pseudo crag]

[odd edge]

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[winged turret]

Greetings, can someone tell me what is outcome of this? ^^'

[restive delta]

Hmm?

[royal herald]

Let her post the q LOL

[restive delta]

Oh it's loading?

Kk

[winged turret]

Ye

[restive delta]

The derivative of that, is it?

[short temple]

is that a derivative ?

[winged turret]

Yeah

[restive delta]

Nice

[royal herald]

What have u tried

Phony

[winged turret]

Its actually just a part of full exercise

Should i sent it

[restive delta]

Ok ig

[winged turret]

Idk how to continue

[royal herald]

,rotate

[warped glacier]

d/dx (2x^2 - 5x + 5) = 4x - 5

[winged turret]

That's it?

[restive delta]

That's it.

[winged turret]

Thanks 👍

[warped glacier]

yep

[winged turret]

.close

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[fiery ermine]

Can an unit circle become a mobius strip if we apply the relation $(\cos(x), \sin(x) )\sim (\cos(x+\pi), \sin(x+\pi))$

[clever fjord]

bigpufik

[hallow fractal]

yes

[fiery ermine]

What if I relate an arbitrary ring that is bounded by a circle with radius r>1, and the unit circle

[hallow fractal]

no

[fiery ermine]

why?

[odd edge]

@fiery ermine Has your question been resolved?

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[supple rose]

Other than the obvious, 8 x 10^x = 80000 and 6 x 10 ^y = 600 and 2 x 10^z = 20

is there another legitimate way to solve this?

at the end x = 4 , y = 2 and z = 1

[sharp oak]

There's infinite solutions. Choose any y, z. Solve for x.

[supple rose]

no there cant

because it = 80620

[mystic saffron]

yeah

theres always a number

lol

[hybrid carbon]

But there are three unknowns

[mystic saffron]

if you make 2 constant

[supple rose]

yea but ur saying x = infinity

however it cant

[mystic saffron]

no

were saying theres infinitely many solutions

not x=inf

[supple rose]

really?

then give me some of them

[mystic saffron]

pick a y and a z

and put it into a computer

[supple rose]

?

[mystic saffron]

im not about to find them lmfao

[hybrid carbon]

Your solution is fine.

It seems like the question is about place value.

[supple rose]

there is only 1 solution to all of them

[sharp oak]

Like, choose y = z = 0.

Then you want a solution to:
8(10^x) = 80620
x ≈ 4.00335

[fair prism]

1 integer solution

[supple rose]

wait what?

[fair prism]

There are many (infinite) if they can be real numbers

[supple rose]

but it says its = 80 620

so how are there infinite?

[sharp oak]

Can you elaborate?

[supple rose]

refer to the question

[fair prism]

Lets do an example. Let y=1, z=1. We can now solve for x using subtraction, division, then logs

[sharp oak]

Like, choose y = z = 0.

Then you want a solution to:
8(10^x) = 80620
x ≈ 4.00335

That's already another solution

[supple rose]

no, where does it say that?

[sharp oak]

Where does what say what?

[supple rose]

how are there infinite?

[sharp oak]

x = 4.00335, y = 0, z = 0 is another

[fair prism]

Think about this and you will realize why

Just spend a minute thinking instead of insta responding

[supple rose]

it doesnt make sense

80619.4789584 is what 8 x 10^4.00335 gives

[sharp oak]

Right. If I included more decimal places it would be more accurate

[supple rose]

and then is y = 0, 6 x 10^ 0 = 6

[supple rose]

it never would

because u add 6 and 2

[sharp oak]

Oh whoops, you're right. Let me fix that

But I hope it makes sense that it can be fixed by lowering x a little bit

[supple rose]

also how are you finding these alternate values?

[sharp oak]

So if I let y = 0, z = 0:
8(10^x) + 6 + 2 = 80620

8(10^x) = 80612

10^x = 80612/8

x = log(80612/8)

[supple rose]

so log finds the base power?

we havent learnt this yet

[sharp oak]

Yeah logarithms help you manipulate exponents like that

[supple rose]

huh

[sharp oak]

x ≈ 4.0033097092284

[supple rose]

and if i let y = 1 and z = 1 it would find another value?

[sharp oak]

Yep, just repeat with the other values

[supple rose]

woah

thats kind of cool

[sharp oak]

,calc 8(10^4.0033097092284) + 6 + 2

[clever fjord]

Result:

80620.000000008

[supple rose]

so there are infinite solutions

[sharp oak]

And if we had a more accurate value, we could get even closer

[supple rose]

Ok this is a revelation

lol

i got x = 4 y = 2 and z = 1

Thanks!

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[ember shell]

Problem 5 :D

[quasi sparrow]

Write two equations with two variables

[ember shell]

Oh ok

Uhhh

Scratch thst I have no clue how

😭😭

Uhm ok

And then r plus n is like 4 dollars?

Ah ok..

Uhh

Ok...

Sure

No 😭

Kk

Uhm

Slightly

Uhm ok

[mint mirage]

The overall question is asking how many kilograms for raisins and nuts. Using the info above, the equation that was given is R + N = 20 kg, where R and N are weights. You want to use the price per kilogram some how, so if you take the price per kilogram for raisins which is 3.5 dollars per kilogram times by the R, you get the dollar amount

[quasi sparrow]

If you don't understand something just ask

[gentle abyss]

a+b* = twenty three quarter snickles - james

[ember shell]

Ah ic

[quasi sparrow]

ac do you know why that equation is correct

This one

[ember shell]

Uhh slightly

[quasi sparrow]

Why not fully

[ember shell]

Uhh I understand why you need to add the two rates on top but idk abt dividing them by the total weight

Oh yeah I understand the second thing

Wait nvm I misread it...

Like ok

[quasi sparrow]

No i did something wrong

[ember shell]

I understand thst you need to know the two amounts of the raisins and nuts

However I'm not exactly sure how to do that???

Since both spaces are empty

[quasi sparrow]

Two equations

[ember shell]

For the amounts

[quasi sparrow]

They gave you one equation for R and N already

Read the rest of your problem to figure out the other equation

Use the hints people gave you

[ember shell]

Ye

[quasi sparrow]

Try writing some facts down

That's different from what people already gave you

[ember shell]

Uh ok

Yeah but it's useless if I don't understand why it works 😭😭

Uh sure but I won't be able to talk

[quasi sparrow]

You should write down something

[ember shell]

Uh ok

[quasi sparrow]

You really need to write down something

[ember shell]

Yeah I've been writing stuff down

[quasi sparrow]

What's the second equation then

This is the first

[ember shell]

Ye

[quasi sparrow]

@ember shell this question is for you to answer

[quasi sparrow]

Can you explain that equation in words

What's x, y

[ember shell]

Isn't it like uhh since ur trying to find the total mixture u like multiply both the raisins and the nits costs and kg

Bur since you don't know the kg you just make it like x and y

Which equals to the total mix?

And then uhhh

The first equation instead of being like r and n its x and y

I think

Which is like a system equation that you solve?

Or smrh

😭

I will try to find the answer then 😭

Ye

So the answer for y

Or the nuts is 8?

Tysm lol I'm dying bruh

[odd edge]

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[mystic saffron]

how do (b)

how do i find A from cof(A)

because that is the essential step

i mean i know the long way

but i am way too lazy to do it

so long

need to solve algebra

i just want quick and ez way

because my way is setting up the matrix like so:
$\begin{bmatrix}
2&1&-1 \
a_21 & a_22 & a_23 \
a_31 & a_32 & a_33 \
\end{bmatrix}$

[clever fjord]

Derivative

[mystic saffron]

then you would try to find the 6 variables

which is long af

[echo ginkgo]

yeah don't solve for A you don't have to

[mystic saffron]

ah what do i need to do

[echo ginkgo]

remember

what are the entries in the cofactor matrix ?

what do they mean ?

[mystic saffron]

they are the individual determinants

[echo ginkgo]

yeah they're the determinant when you remove a row and column right

[mystic saffron]

yes

[echo ginkgo]

and you're given a row of A to start with

can't you use that row combined w/ some entries of the cofactor matrix to get the determinant of A ?

spoilers || laplace expansion || give it a bit of thought first

[mystic saffron]

i didnt learn laplace expansion

[echo ginkgo]

ah

how do learn about cofactor matrices without knowing laplace expansion ?

that's a bit weird

[mystic saffron]

ahhh we call it cofactor expansion

[echo ginkgo]

ah fine

so many names

[mystic saffron]

cofactor expansion along the ith row: $det(A) = a_{i1}C_{i1} + a_{i2}C_{i2} + a_{i3}C_{i3} + ... + a_{in}C_{in}$

[clever fjord]

Derivative

[mystic saffron]

so i have all my cofactors

AHHAHAH

i got it

i just need to do cofactor expansion along the first row!!!

and they gave me it!

[echo ginkgo]

yea

[mystic saffron]

yessss

thanks chief

much appreciated

peace ✌️

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[sleek falcon]

how does principle component analysis related to svd

[quasi sparrow]

https://stats.stackexchange.com/questions/134282/relationship-between-svd-and-pca-how-to-use-svd-to-perform-pca

https://jonathan-hui.medium.com/machine-learning-singular-value-decomposition-svd-principal-component-analysis-pca-1d45e885e491

[sleek falcon]

thanks!

also in low rank approximation, you are given a matrix a with some rank (in full rank format), which eventually becomes rank r format right

or compressed svd

[odd edge]

@sleek falcon Has your question been resolved?

[sleek falcon]

is the difference between low rank approximation and rank r format SVD that for low rank approximation, you keep k singular values, and for rank r SVD you keep all singular values that are positive and NOT 0 ? I might be confused about the difference since they seem relatively similar.

[sleek falcon]

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[spare hornet]

Teacher was telling us how to do these but gave up. Mid way

Pls help 😭

And this one too

[latent scaffold]

When is sin(x) zero?

For the first image, a

For b) it's mostly the same I think, what you erased seems good.

[spare hornet]

I got those ones but the next picture

Can you check if I did that one right and how I would do the next one

[odd edge]

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[spare hornet]

Noo

[latent scaffold]

Just open another I think or write again once it is free again

[spare hornet]

K

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[raw acorn]

Is this right?

[raw acorn]

<@&286206848099549185>

<@&286206848099549185>

This question too

[odd edge]

@raw acorn Has your question been resolved?

[raw acorn]

no

<@&286206848099549185>

[odd edge]

@raw acorn Has your question been resolved?

[raw acorn]

<@&286206848099549185>

[odd edge]

@raw acorn Has your question been resolved?

[raw acorn]

<@&286206848099549185>

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@raw acorn Has your question been resolved?

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[small siren]

I have gotten pi/3 and 2pi/3

The answer says 5pi/3 and 4pi/3 are also answers within the domain

But I thought the new dominant is -pi/2 </x-pi/2</3pi/2

Domain*

[mystic saffron]

dont

its porn

[small siren]

Bruh

[mystic saffron]

dumbasses be dumbassing

[tardy lagoon]

<@&268886789983436800> porn?

[mystic saffron]

okay someone help kookie

[restive delta]

cos(2x- pi)= 1/2

= -cos(2x)

[small siren]

The original domain is [0,2pi]

[restive delta]

Ye ye mb

I didn't see

[small siren]

Um could you further explain please

[restive delta]

Yeah one sec

Domain is within set of values that x can take

[small siren]

Y

[restive delta]

Wdym

[small siren]

Yes

[restive delta]

Yeah

So why would it change

[small siren]

So it says solve for [0,2pi] but wouldn't the domain change when solving the question?

[restive delta]

No

[restive delta]

This Is what we get

x is within [0,2pi]
So 2x is within [0,4pi]

You mean like that?

[small siren]

Yes

[restive delta]

Ah

But both are the same tho

[small siren]

Oh, thank you

[restive delta]

You would still get those 2 other answers

[small siren]

Thank you for your help

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[muted sierra]

Can someone solve this please

I have the first part just the second parts a bit confusing

It’s almost times three of a but I’m stuck

[bronze epoch]

When you have the equation of a parabola in standard form: y = a(x+b) + c, the vertex (minimum/maximum) is (-b,c)

[muted sierra]

So do I do simultaneous equation with my answer to part a and b?

[nimble blaze]

consider what you could do to a) to give b)

[muted sierra]

Times 3 and and add 4

[nimble blaze]

yeh

min value will be transformed in the same way

[muted sierra]

Ah I’m kind of getting it

Wait

So

I get minimum value of a and just times 3 and add 4?

[nimble blaze]

yeh, pretty much

[muted sierra]

Ah I see Thankyou very much👍🏽

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[tepid badge]

i just need some help confirming my answer

[tepid badge]

is it 2x+y=1?

[noble forge]

yes

[tepid badge]

ty chad

[eternal ermine]

does it need to be = 0

2x+y-1=0

[tepid badge]

idk i just rearranged my y=mx+c equation

[eternal ermine]

y=mx+c is slope intercept form

general form is Ax+By+C=0

[tepid badge]

idk what i was thought was that general is ax+by=c

and i think thats what my teacher wants but idk

[eternal ermine]

welLLL

give what ur teacher wants

[tepid badge]

ill probably get incinerated if i dont lol

[eternal ermine]

yea im wrong

u r correct

mbmb

[tepid badge]

allg

ty anyway

[mystic saffron]

we can use the point-slope form. The equation is y - y₁ = m(x - x₁), where (x₁, y₁) is the given point and m is the slope. Plugging in the values, we get y - 5 = -2(x + 2). The equation of the line is y = -2x - 1. Since the general form is Ax + By = C, it's going to be 2x + y = -1

[tepid badge]

Ok thanks

Anyway while this is still open

I have the sneaking suspicion I did this wrong

Is there something I missed

SOMOENE PLS

[brittle beacon]

Last step

[tepid badge]

wha

r

[odd edge]

@tepid badge Has your question been resolved?

[odd edge]

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[pallid tinsel]

have u drawn a picture?

[ashen lily]

idk how

i don’t understand

[pallid tinsel]

u r on the ground at a certain location. u have to look up thru an angle of 30deg to see the top of a cliff

then u walk 210 meters towards the cliff from that location

from that new location u now have to look up thru an angle of 45deg to see the top of the same cliff

[ashen lily]

oh

smth like this?

[pallid tinsel]

well the angles have to swap

[ashen lily]

u said

[ashen lily]

this

[pallid tinsel]

yes but u arrive at the new location by walking towards the cliff

[ashen lily]

oh

ok what to do after

[pallid tinsel]

its trigonometry

[ashen lily]

i know

but

[pallid tinsel]

call the height x or something

[ashen lily]

tan theta =perpendicular/base what value do i put in here

tan 45=

perp/210

[pallid tinsel]

how is it 210

[ashen lily]

base is 210

[pallid tinsel]

can u redraw this correctly and send? and what exactly is the length u have marked as 210 here?

[ashen lily]

wont 30 be at C

[pallid tinsel]

u marked 210 wrong

210 is the distance from A to the new location

everything else is marked correctly

[ashen lily]

oh ye

so we take triangle BDC tan 45=BD/BC 1=BD/BC……BD=BC

now?

the stick figure is d

can someone helpp me 😭😭😭💔

.close

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[marble forge]

I would apreeciate help with matrix (adjoint and inverse).
I have an exam next week and I struggle with this particular concept

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@marble forge Has your question been resolved?

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[austere rover]

Can someone help me recall what this bolded thing means. I wrote it down but forgot 😅
Let $A \in M_{m,n} ( \mathbb{R} )$ and $r = \rho (A)$. Then
$dim C(A) = dim R(A) \leq$ This -> min{m,n} <- This

[clever fjord]

LeGM

[austere rover]

nvm I remember

.close

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[ashen lily]

find the quadrant in which the terminal side of the angle likes

lies

i think 3

quadrant

[zenith tartan]

that's right

[ashen lily]

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[mellow axle]

,rotate

[mellow axle]

Is there an efficient/short way to solve this?

Or do i have to go through the 8262526226 billion steps?

I feel like there must be something im not seeing

[knotty oasis]

You can use the quadratic formula, but it might get messy

[royal herald]

Does i4 mean i^4

[mellow axle]

No

ix4

[royal herald]

Oh ok

[pulsar current]

take the 7 to the other side
then the entire left side is a real number, equate that with its conjugate

[mellow axle]

This is how im doing it right now

But i see its going to be hell

[mellow axle]

This is what i did basicly right?

[pulsar current]

in other words simply conjugate both sides and add

[pulsar current]

i dont think so, no

i didnt read what u sent but it definitely doesnt look like what i suggested

[mellow axle]

I dont really know what you mean

Conjugate of what?

Oooh wait

Leave the channel open

[pulsar current]

this whole complex number is equal to 0

[mellow axle]

Should i put z as x+iy?

[pulsar current]

so its conjugate should also

be equal to 0

[pulsar current]

no it will become too clumsy

we are reducing it first

to a simpler equation

[pulsar current]

write the equation for this and then add

[mellow axle]

So z=7?

The solution is supposed to be 1-2i and -2+3i

[pulsar current]

what in the actual world is that?

[mellow axle]

What does the conjugate look like

[mellow axle]

I have no idea 😭

[finite hollow]

group imaginary

[fervent hound]

i suggest multiplying by -i first

[finite hollow]

and real

then since it is equal to zero

you can equate ur

real and

imaginary

components

to zero

and solve

for z

[mellow axle]

Like thisv

?

[finite hollow]

no

[fervent hound]

how'd u know that z is real or imaginary??

[finite hollow]

it should be

[finite hollow]

what

it usually is real

like the question says

solve for z

such that z is a real component blah blah

[mellow axle]

Z cant be real here

I dont think

[finite hollow]

ok mb

[mellow axle]

The answer is an imaginary number

[fervent hound]

where are u reading that from ?

[finite hollow]

but u should get (z-7)+(z^2+z+4)i=0

[fervent hound]

there's no question (just equation)

[finite hollow]

so theres

[mellow axle]

,rotate

[mellow axle]

^answer (the one in the middle)

[finite hollow]

wattttt

[mellow axle]

Yup :(

[finite hollow]

bo

bro

i got

z=7

and z= -1/2+- root15/2i

[fervent hound]

multiply the whole equation by -i so that the coefficient of z² becomes 1
then apply quadratic formula

[fervent hound]

how did u get that ??

[finite hollow]

i just put

th real components

and solved for zero

and the imag

and solved for zero

ive never

[fervent hound]

what are the real components ??

[finite hollow]

sen this

question

in my life

[fervent hound]

how did u know that z is imag or real ???

[finite hollow]

(z-7)+(z^2+z+4)i=0

[fervent hound]

so any number multiplied by "i" is an imag number ??

[finite hollow]

i have no clue

i think so

[fervent hound]

what if z = 4i for example

iz = -4

[finite hollow]

-4

omg

real

[fervent hound]

that's what i'm saying, u can't use that method cause u don't know if z is pure imag/real or complex

[finite hollow]

ok u know for part c

z^2=i

i mean z^2+i=0

[finite hollow]

cant u just say that

[mellow axle]

:(

[finite hollow]

cant u say that

z^2=i

then solve it

normally

[mellow axle]

Didnt work :(

[fervent hound]

this is correct

now just apply quadratic formula

u can't use that method cause u don't know if z is real or imag

[finite hollow]

thats

what i

was

thinking

but

other way around

🤔

[fervent hound]

$z^2+(1-i)z+4+7i=0$

[clever fjord]

Adam Chebil

[mellow axle]

Is quadratic formula the pq formulav

?

[fervent hound]

calculate b²-4ac

[finite hollow]

nah what

this shit

is so

confusing

lol

[mellow axle]

Yep

Every single question is like this ;-;

[finite hollow]

nah

the ones above

are easy

this one

is

asscheeks

[fervent hound]

(1-i)² = -2i

[mellow axle]

Ye working on it

Stuck again

[fervent hound]

find the square roots of b² -4ac which is -16 - 30i first

[mellow axle]

,rotate

[fervent hound]

how did u get that ??

[fervent hound]

it's -28i not +28i

[fervent hound]

correct

If you don't want to simplify it the solutions are just: $\ z_1 = \frac{i-1+\sqrt{-30i-16}}{2}$ $\ z_2 = \frac{i-1-\sqrt{-30i-16}}{2}$

[mellow axle]

How would you simplify this?

I feel like thats not possible

[clever fjord]

Adam Chebil

[mellow axle]

I get what you mean

[fervent hound]

by getting rid of sqrt

[mellow axle]

How?

Squaring both sides wont help?

[fervent hound]

no, by calculating the square roots of the inside

[fervent hound]

but if your teacher wrties comlplex numbers inside sqrt no need to do that

[mellow axle]

I dont think she does

How would i do it?

[fervent hound]

well, u can caculate square root of complex numbers by solving this system of equations : $\ \sqrt{z}=x+iy \ x^2+y^2=|z| \ x^2-y^2=Real(z) \ 2xy=Im(z)$

[clever fjord]

Adam Chebil

[fervent hound]

or write it directly: (if u're allowed to do so)

[mellow axle]

Jeez

Well

Thanks for the help!

[fervent hound]

np!

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[snow solstice]

hii

[snow solstice]

need to calculate radius of convergence for this one

the answer is one

but no idea why

[odd edge]

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[odd edge]

@snow solstice Has your question been resolved?

[odd edge]

@snow solstice Has your question been resolved?

[worldly rune]

Wow what in the world is that

[snow solstice]

no idea

[worldly rune]

Yeah ig you just have to use some common convergence test to find a radius maybe

You can use the ratio test

Ratio test says that the series converges absolutely if $\lim_{n\to \infty}|\frac{a_{n+1}}{a_n}| < 1$

Then you can see what values of z makes it that this holds true

$a_n$is from $\sum_{n=0}^\infty a_n$ btw

[clever fjord]

edwardborn

edwardborn

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[still axle]

it doesnt factor

so how do i find critical pts

[shy smelt]

The question is a bit tricky, the cost of making x items is x^3 - 22x^2 + 20000x

reading it again, I think they've made a mistake

they probably want you to calculate the minimum average cost of a single item, not the minimum average cost of x items

[still axle]

okay but how would i proceed with that

[brave flint]

For critical points u can use quadratic formula

[still axle]

@brave flint undefined

on both + and -

not sure what to do

<@&286206848099549185>

[odd edge]

@still axle Has your question been resolved?

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[vital flicker]

I need help

[vital flicker]

This is what I have

[odd edge]

@vital flicker Has your question been resolved?

[vital flicker]

No

@rancid moat

@tidal glacier

[odd edge]

@vital flicker Has your question been resolved?

[kind cradle]

what the hell is this

[vital flicker]

Differential equations 😭