help-17

[ornate ember]

if you're going to troll then at least be clever about it

[vast shale]

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[vast shale]

I don't understand how the equation is formed...

[vast shale]

This was my attempt to solve it but I got 1/3 units²

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@vast shale Has your question been resolved?

[vast shale]

wtf

https://math.libretexts.org/Bookshelves/Calculus/Calculus_(OpenStax)/06%3A_Applications_of_Integration/6.01%3A_Areas_between_Curves

$$ \int_{a}^b (f(x) - g(x)) dx$$

[twin meteor]

Brandon H

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[river lance]

i am not able to do this HW

i am able to find the eigenvalues are 3i and -3i

eigenvectors (1-i ; 2 ) and (1+i ; 2) correspondingly

but i failed to obtain the real form

[granite oxide]

@river lance is “x” in C^2

What is x? A vector?

[river lance]

yes

[river lance]

I don’t het it

Get

[granite oxide]

Never mind, it’s fine

Okay so x is a vector

[granite oxide]

I’m confused, how do you multiply a column vector with a column vector

[river lance]

Hold up give me 5mins

I have a similar question

But with different eigenvalues

Can be used for reference

@granite oxide

To be honest

I don’t even know if this is correct

But that’s all I could trust

<@&286206848099549185>

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@river lance Has your question been resolved?

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[vast shale]

I literally don't know how to solve this question at all.

[patent nymph]

Do you understand the words in the question?

[vast shale]

Yes I do

[patent nymph]

Then do you understand the question?

[vast shale]

I tried solving it

but i kept failing

[patent nymph]

Why

[vast shale]

its basically saying which which table has the equation y=mx+b

[vast shale]

i tried finding slope for each table dead end

and tried plugging in values

[patent nymph]

not a dead end, what did you do

[patent nymph]

also works

[vast shale]

y2-y1/x2-x1

for each table

got a set of points

but they all led to the same slope

of m

the slope would literally be m

[patent nymph]

did you try that for all 3 pairs of points

[patent nymph]

what did you get when you plugged in values

[vast shale]

yes

[vast shale]

i just plugged in values for y and x

and i didnt rlly know where to move on from there

like how would i determine which one is the answer?

[patent nymph]

you are supposed to plug them in and see if they give you something like 1=0, I think

[vast shale]

but how?

i dont get it

why would that mean that is the answer

[patent nymph]

you can try it first and send your results here

plug the values of x and y in the tables into y = mx + c

[vast shale]

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[peak shard]

does a function have to be continuous so that I can take the Improper integral because for Riemann Integration the function doesn't have to be continuous

[peak shard]

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[limber kernel]

For this question

[limber kernel]

Where did I go wrong?

Surely you don’t “have” to use quotient rule

Mark schemes answer

[opal obsidian]

@limber kernel

[limber kernel]

Yo

[opal obsidian]

which one you want to check?

[limber kernel]

My answer

For some reason is not the same as the mark schemes

[opal obsidian]

So the second one is correct?

[limber kernel]

According to the mark scheme

[opal obsidian]

Please give some time

[opal obsidian]

let me check this

[limber kernel]

Any update?

<@&286206848099549185>

[outer warren]

the issue is that the specified point isn't even on the curve

[limber kernel]

Oh I see so I have to use quotient rule there is no other way @outer warren

[outer warren]

no

you don't

the question is inherently flawed

[limber kernel]

Oh I see

Feel sorry for the guys who sat that exam

[outer warren]

if the values presented are consisted,
valid methods lead to the same end result

[limber kernel]

I see thank you so much

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[stiff oriole]

so I got Cos wrong and i got the points on the terminal side wrong. Can someone explain to me what i did wrong please?

[stiff oriole]

like y can’t be anything more than 0 because it won’t be on the line right?

[outer warren]

how are you getting -1/4

[stiff oriole]

for x i got -4 and for r i got 16

wait did i do it backwards

wait no because the terminal side is still wrong

[outer warren]

can you show the full original question

[stiff oriole]

gotchu

the one i did before was 180 degrees and i got that one correct

but since the degrees is 0 that changes the whole thing huh

[outer warren]

your graph is wrong

note that the angle rotation is measured anticlockwise from the positive x-axis

[stiff oriole]

huh

so it’s supposed to be going downward?

[outer warren]

no

[stiff oriole]

kinda?

[outer warren]

no

where would that blue line be when theta is 0

(don't overthink this)

[stiff oriole]

ummmmm 10?

[outer warren]

no

10 isn't a location

focus only on the image i gave you

indicating the terminal side of some angle theta

(which here is around 150° anticlockwise from the positive x-axis / red line)

where would that blue line end up if that angle was instead 0

[stiff oriole]

on the y axis?

[outer warren]

no

you're overthinking

[stiff oriole]

oh god

[outer warren]

you're way overthinking this

starting at the red line

making a rotation of 0°, i.e. pretty much doing nothing at all

you'll be right where you started

[stiff oriole]

fucking hell so it would be the other way around then?

[outer warren]

wdym by other way around

[stiff oriole]

you said it starts at the red line so would i just switch the blue line to the opposite side to where the red line is??

[outer warren]

yes

(what you initially posted was what you would've used for 180°)

[stiff oriole]

i see

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[jagged willow]

367. For which graph point A(-2;9) is for?

[vast shale]

Plug and test which case is correct

[jagged willow]

How to plug

[vast shale]

They tell you the coordinates

(-2, 9)

You should be able to figure out what x and y is

[jagged willow]

So I just plug -2

[vast shale]

Plus -2 and 9

x = -2 and y = 9

Test for each case, which is correct

[jagged willow]

Answer D

[vast shale]

Good job

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[dull lynx]

i found A^B = I but what is next?

[lyric fossil]

what is A^2B?

[dull lynx]

the idenitity matrix

[lyric fossil]

and what was A^2?

[dull lynx]

oh wait

A^2 was identity

[lyric fossil]

ok

[dull lynx]

A^2B =B

[lyric fossil]

then note A^(2022) = (A^2)^1011

[dull lynx]

so I^1011?

[lyric fossil]

yes

[dull lynx]

is that just identity then?

[lyric fossil]

I^1011 = I yes

but what is IB?

[dull lynx]

B

thank you

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[inland gyro]

i tried doing (x-a)(x-b)(x-c)(x) for the quartic function but don't know how to proceed

[lyric fossil]

do you calculus?

[inland gyro]

no it's a pre-calculus problem

[icy spear]

You're not told that 0 is a root

Why do you have a factor of x

[inland gyro]

what should I do then?

[lyric fossil]

i suggest you take cos(x) and find the first three roots greater than 0

[icy spear]

Square one of the factors

For instance you could have (x-a)(x-b)²(x-c)

[lyric fossil]

starting with the quartic is a bad idea

it's easier to fit a quartic to a trig function that the other way around

[inland gyro]

pi/2 + npi

[lyric fossil]

so what are the first three roots

[inland gyro]

so pi/2, 3pi/2, 5pi/2

[lyric fossil]

so use that as three of the zeroes for your quartic

and then just pick any other root (other than x = 0 of course)

[inland gyro]

what do you mean any other root?

[lyric fossil]

like (x-10)

or (x-50)

or (x+150)

whatever you want

[inland gyro]

can i just square one of them instead?

[lyric fossil]

sure

[inland gyro]

oh ok thanks!

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[lyric fossil]

you're not done

@inland gyro

.reopen

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✅

[lyric fossil]

you need to fit the y intercept to be the same as the cosine function

[inland gyro]

oh

let me write down the quartic right now

so do this I would just do x = 0

and then find the product then make it equal to 1?

[lyric fossil]

yes

[inland gyro]

is it like a(x-pi/2)^2(x-3pi/2)(x-5pi/2)

the product lets say = P

then i do aP = 1

[lyric fossil]

yes

[inland gyro]

and make the variable a what adjusts the y int?

okay got it

also why did u ask me to find the roots of cosx = 0?

oh nevermind its for the roots of the quartic right

[lyric fossil]

yes

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[signal kiln]

how i resolve this "If a circle has a radius of 0.5m, can I place 41 circles in a rectangular area of ​​5x8m? Show how without dividing them"?

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@signal kiln Has your question been resolved?

[granite oxide]

What does “place” mean here? What is “dividing them”?

[signal kiln]

Without dividing the circles

and what do you mean places?

just like I didn't understand?

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[brisk oxide]

would this just be

[brisk oxide]

S_j = Sigma i = 1 to j: a_i

[maiden iron]

yes

[brisk oxide]

and this would be yes, even though they aren't necessarily guaranteed to be the same

right?

[maiden iron]

yes there are

[brisk oxide]

got it thank you.

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[cobalt breach]

Could I get help with this equasion

[cobalt breach]

kx +2y =5

y+2 = -3(x +1)

looking for answers for K the give no solutions

[iron flame]

do you know when two lines will have no solution?

[cobalt breach]

when the are the same slope?

[iron flame]

yup

so what is the slope of y+2 = -3(x +1)

[cobalt breach]

-3x

[iron flame]

correct

now create an equation for a line for kx +2y =5 and find the value of k where the two equations will have the same slope

[cobalt breach]

ok, thanks

[iron flame]

np

[cobalt breach]

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[cursive fern]

I wish I didn’t have to come back here again, but I can’t figure this problem out

[cursive fern]

I can’t get lambda to be 1/2

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[vast shale]

I need help with this question.

Probability is want/total. Therefore, the probability would be 3550/80216

To get the total which is the denominator equal to 80216 I did 3550+2500+c=80216

The answer says otherwise however

[flat whale]

Show

[vast shale]

i read it wrong

its out of the lecturers from all countries

i need to pay more attention lol

hahaha

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[sharp birch]

heyo so I was curious whether 9 or 1 was right on this, I assumed that if division and multiplication were together you did multiplication first, but maybe it's the opposite or you do it left from right?

6/2(2+1)
6/2(3)
6/2x3

[rugged vortex]

It's pretty bad notation tbh. I'm reading, due to PEMDAS, as 6 / 2 • 3 which is 9

[vast shale]

The calculator is correct

[rugged vortex]

But regardless it's bad notation

[vast shale]

Your smartphone app was probably made in 10 minutes by some random guy in a WW2 bunker

[sharp birch]

so it's both 1 and 9?

[vast shale]

No the calculator is correct

app != calculator

[boreal remnant]

no the phone is correct according to pemdas

[rugged vortex]

It's just bad notation in general though

I know Ann has a whole jeremiad about these kinds problems

[heavy yoke]

it depends on how you define order of operations. some systems include a special case called "implicit multiplication" in which multiplication due to parentheses has a higher priority than normal. this is mostly to allow calculations involving factorized rational functions (1+4)(6-3)/(7+9)(4-7) to work as "expected"

[boreal remnant]

but pemdas is just a convention

[vast shale]

Parenthesis Exponetiantion multiplication divsison

[boreal remnant]

it's not multiplication then division, it's multiplication AND division on the same step

left to right

[rugged vortex]

The multiplication-division goes left to right regardless of order, just like addition-subtraction

[mystic skiff]

i learned a new word today

[rugged vortex]

It's because division is just multiplication in a diff form

[boreal remnant]

personally i do think that multiplication should take precendence over division though, so we can write things like h bar = h / 2pi without excessive parentheses. you see this a lot in physics

[rugged vortex]

And likewise subtraction is addition in a diff form

[sharp birch]

oh I thought of both of them as calculators xD

when I googled it I think it said that PEMDAS was used in the US and BODMAS in the UK which made it sound like the answer is 9 in USA and 1 in the U.K

[rugged vortex]

Yeah well too bad hpi/2 now

[heavy yoke]

both pemdas and bomdas are mnemonics, not definitions

[mystic skiff]

those are the same thing

brackets are parentheses

[rugged vortex]

Idk what the O would he

[mystic skiff]

orders are exponents

[sharp birch]

aight so is 1 or 9 the answer

[mystic skiff]

or roots

[rugged vortex]

Order oh

[mystic skiff]

it's bad notation. 0 points.

[rugged vortex]

I thought it was British way of spelling exponents to match their accent

9

It's 9

[sharp birch]

bad notation?

[mystic skiff]

yeah

[rugged vortex]

Yeah it's ambiguous

That's why no one uses the division symbol

[sharp birch]

what does 0 points mean

[rugged vortex]

Whether it implies 6/(2(2+1)) or (6/2) • (2+1) is wayyyy too vague

[heavy yoke]

i recommend excessive use of parenthesis when using calculators to avoid this

[rugged vortex]

F- you fail

[mystic skiff]

$\frac{6}{2(2+1)}$ or $\frac{6}{2}\cdot(2+1)$

[twin meteor]

baro | awake

[mystic skiff]

who is to know

[rugged vortex]

We destroyed the man

The man never came back to revile

[sharp birch]

yeah which is why I wonder which one is right or is there no right answer?

[boreal remnant]

the phone is right

[heavy yoke]

the fact that you can follow two accepted order of operations systems and arrive at different answers shows the question is too vague

[sharp birch]

does everyone agree with you or is this something people aren't sure on

[knotty granite]

is what world

the notation looks fine to me

[heavy yoke]

https://www.themathdoctors.org/order-of-operations-implicit-multiplication/

[sharp birch]

If division and multiplication are the only things left do you solve left to right?

[knotty granite]

do people say distributing is not multiplication

[knotty granite]

that and addition and subtraction

[sharp birch]

ye ofc

but so is that the rule then

[knotty granite]

yes

[sharp birch]

okie

and everyone agrees I hope?

[knotty granite]

Parenthesis
Exponents
Multiplication/Division
Addition/Subtraction

wait

fixed

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[fallen hedge]

what law allows this to change?

[fallen hedge]

is that because of the absolute value

[rose raft]

yes

[fallen hedge]

ahh thanks

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[night beacon]

can someone correct me if my answer is right?

[night beacon]

<@&286206848099549185>

[glossy maple]

seems right

[night beacon]

Really?

[glossy maple]

yeah

[night beacon]

alright another one

correct me if i'm wrong please

[solar grail]

The answers are all right

[night beacon]

how about this one?

<@&286206848099549185>?

[solar grail]

Checking rn

[night beacon]

can someone correct me if it wrong or right

[solar grail]

It's all correct

[night beacon]

slr Alright Thank you

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[formal vapor]

can someone help me with question b

thanks

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[wind sequoia]

can anyone give hints? im stuck…

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[soft walrus]

[vast shale]

try to set up a relationship

y = mx + b

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@vast shale Has your question been resolved?

[cursive turret]

richard22 gave you the right hint. maybe you have a question with this hint -> then ask.

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[urban obsidian]

why can't i rewrite this as:

[urban obsidian]

$\int {\frac{1}{\sqrt{3^2 - \left(\frac 32 v\right)^2}}}$

[twin meteor]

hii_

[patent nymph]

!show

[vocal sleet]

Show your work, and if possible, explain where you are stuck.

[urban obsidian]

i am doing so

sorry

anyway why can't i rewrite as that

and then say the integral is arcsin(3/2v / 3) + C

which is just arcsin(1/2 v) + C

@patent nymph

[patent nymph]

I’m checking

[patent nymph]

consider what happens when you use the substitution of v: what happens to dv?

[urban obsidian]

substitution of v with what?

besides i've written it exactly in the form the picture shows

[patent nymph]

or you can differentiate this function and see what you get

[urban obsidian]

why should i care about anything else?

[patent nymph]

you also forgot the dv, which is a part of why directly using the formulas doesn’t work

[urban obsidian]

what

how does that affect anything

i'm confused

i should've written it yeah

$\int {\frac{\text{dv}}{\sqrt{3^2 - \left(\frac 32 v\right)^2}}}$

[twin meteor]

hii_

[patent nymph]

I would suggest that you differentiate what you got initially

[urban obsidian]

yes i know it'll be 4^2 - v^2 inside the surd or something

i need to know how it's different from the picture when it's literally in the same form

[patent nymph]

if you look at the picture you sent out, the thing in the root is u^2, not 9/4u^2

[patent nymph]

there is a slight difference: the form of the variable in the root(it is multiplied by a constant, so you need to do a substitution to make it v^2 again)

[urban obsidian]

i'm expressing it as a difference of squares

i don't get what you mean tbh

u = 3/2v here

hmm okay i see where you're going with that

okay fair

[patent nymph]

the formulas work exactly as written

[urban obsidian]

Okay yes, i understood it @patent nymph

thank you!

any clue how to derive these stuffs?

[patent nymph]

in future cases when you’re confused about the result of an integration, you can try differentiating your result

[urban obsidian]

$\int {\frac{\text{dv}}{\sqrt{a^2 - v^2}}}$

[urban obsidian]

okay yes, that's true

[twin meteor]

hii_

[urban obsidian]

i made the mapping of u to (3/2v) in my head but never substituted it in paper lol

[patent nymph]

for this one, since it’s exactly v^2, you can use that formula you mentioned

[patent nymph]

that’s why showing steps is important

[urban obsidian]

yes i mean but how do you remember these stuffs

you memorize?

[patent nymph]

or you can look at how these formulas were found

[urban obsidian]

true, do you have videos for it?

[patent nymph]

which in this case is by differentiating arcsin x and arctan x

[urban obsidian]

yeah that's true but that seems like a hassle

[patent nymph]

you can check your textbook if you have one, or ask your teacher

[patent nymph]

that may be, but this is the way to understand and learn maths, not to memorize

it is to look at the derivation of every formula and make sure you understand the tricks used

[urban obsidian]

no i'm aware i can differentiate and get the results

but i can't derive it from the very get go

anyway no worries

i'll check something out

thanks

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[crisp yoke]

how do i find that?

[dim quail]

use discriminant

[crisp yoke]

thanks

and for this one why are the solutions for (x,y) infinite?

because and x+y could make 5?

[dim quail]

yes

[crisp yoke]

thanks

[dim quail]

basically the first line of this is (x+y)^2

so yea

infinite sols

[crisp yoke]

thanks

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.reopen

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✅

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[nimble moss]

A, B and C are points with coordinates (6,1,8), (2, -3, 1) and (3, -5, 3).
Angle ABC = θ.

Use a method involving the scalar product to find the exact value of cos θ.

My answer was -10/27.

The mark scheme was 10/27, and it said I need to change the sign. Why is that?

[nimble moss]

Is that the reason?

[bitter copper]

Can I see your original workings please?

[vocal sleet]

@nimble moss Has your question been resolved?

[bitter copper]

Okay

I know what’s happened

So yes there is full reasoning for this

But a quick way to fix it in your workings is when you found AB

What you are doing is trying to find the equation of two lines and finding the angle between them

But because when using the scalar product method all you need is the direction vectors you only need to write down AB and BC

But direction vectors when in the equation are multiplied by a constant

Typically lambda of mew

As a result, they should always be simplified into the smallest form

Or the most natural form

As a result you should have taken out the negatives of AB to turn it into a vector of (4,4,7)

Then when you do the dot product you end up with positive 10

Giving you the answer they had in the mark scheme

[bitter copper]

If you were to more accurately plot the points I believe the angle should be acute, as a result you would be expecting an angle less than 90 or pi/2. As a result you would use 10/27 in that case

I am going to plot them now to double check

[nimble moss]

Isn't it because he wants the inner angle?

[nimble moss]

Like here

Of course I have the angles not drawn accurately

[bitter copper]

Your trying to find the angle between the three points

[nimble moss]

Yeah. Is that why the answer is positive?

[bitter copper]

If you plot the three points onto a 3D graph you can see that the angle between them is acute

So yeah

[nimble moss]

Is the angle between 2 Vectors the smallest angle between them?

[bitter copper]

Basically yes

But to avoid special cases I would just try simplify down the direction vector as much as possible

[nimble moss]

Thank you for your help!

[bitter copper]

All good, I explained it a bit long lmao

Have a good day!

[nimble moss]

You too. Bye!

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[violet flax]

the first part of this question is confusing me

do u not also times the 3 by the inside of the bracket? and then flip it over to make it dy/dx?

nvm omg i keep on sending these and then figuring it out a little later

sorry lol

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[warped chasm]

so uhh

the answers are there

but for b, im questing why I can't use an alternate method

which would be

let w=x^3

so x=cube root(w)

and sub this into the equation

but that doesnt seem to work

e.g:

am I dumb?

<@&286206848099549185>

[vocal sleet]

@warped chasm Has your question been resolved?

[vast shale]

this is correct

[warped chasm]

But this is what I get

Which is off

[warped chasm]

could you explain where I went wrong

No matter how I rearrange this or times it by w^3, I cant get the original

[cold blade]

Had someone with a question earlier similar to this but for a cubic

Try x^2-3x+1=(x-a)(x-b)

Expand the RHS out and setting coefficients equal to each other you get two equations for a and b

Then expand out (x-2a+1)(x-2b+1)

And use those equations you found for a and b to solve for the numerical values of the coefficients of this new polynomial

[vocal sleet]

@warped chasm Has your question been resolved?

[warped chasm]

Let me try

Ummm

That's not

The right one

I wanted the first image

And saying why my method doesn't work

When it works in image no.2

[warped chasm]

essentially

[warped chasm]

Why does this not work

when it works for a different question

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@warped chasm Has your question been resolved?

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@warped chasm Has your question been resolved?

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[dim quail]

for each movement a particle moves 1 cm to the right or 1 cm to the left. Knowing that after 10 moves the particle is back in its original position, calculate the total number of different ways it can perform this sequence of moves.

[dim quail]

if the particle is back in its original position, i know that it got to move the same amount of times to the left and to the right

but idk how to use this information to solve the problem

<@&286206848099549185>

[lean harbor]

we already know that that it moves left 5 times and right 5 times, so what we're looking for is the amount of distinct combinations can be made moving five times left and five times right

so if we visualize this with ten blank spaces

these ten blank spaces represent the order this particle moves left or right in. each dash can either have L for left, or R for right

we can choose any 5 blank spaces to contain R

in doing so, the other five blank spaces will automatically contain L

[dim quail]

but in this way, wouldn't i have to also calculate the permutations of the "L" moves? or it is whatever and doesnt matter

[lean harbor]

no, by counting every way the 5 Rs can go in ten spaces, we automatically also count the L moves because the Ls fill the rest of the spaces

so if i have an R in the first five spaces, i've already accounted for the situation where all the L's are in the last 5 spaces

this is a representation of an n choose k problem

out of 10 (n) spaces, you're trying to choose 5 (k) to have R in them

or L, it doesn't really matter as long as you don't do both

[dim quail]

oh, i got it now

thank you so much! I'll try to solve it again

[lean harbor]

no problemo!

[dim quail]

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[hushed cipher]

What would the behavior be near the non-permissible value and the end behavior for y = (-2)/(x^2-6x+9)?

[soft walrus]

what is the non-permissible value?

just the vertical asymptote?

[hushed cipher]

A value that makes the bottom 0

Makes the denominator 0

So 3 is the NPV

[hushed cipher]

Yes

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[harsh saddle]

Hi, I'm learning a little bit about graph theory and I've come across three primary methods of mathematically representing graphs: Adjacency matrices, incidence matrices, and edge lists. Are there any representations that encode attributes of the vertices though? I haven't found anything. Mainly just curious

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[lime warren]

hi guys, I'm stuck with this question. I know it should be easy enough but I always get stuck when it says "give x as a fraction of y"

[sharp wing]

First, convert years into months.

So Lily would be 2(12) + 4 months old.

Hugo would be 1(12) + 8 months old.

Therefore, Lily is 28 months old, while Hugo is 20 months old.

[lime warren]

i see

[sharp wing]

You're trying to find the fraction of 20 that makes 28.

Divide 28 by 20 to make 28/20.

[lime warren]

1 8/20?

[sharp wing]

This should simplify to 7/5.

[sharp wing]

Yes.

7/5 if you want a fraction, and 1(2/5) if you want a mixed number.

[lime warren]

i understand now. Thanks

good day

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[vast shale]

I am beginning with linear algebra (I'm studying computer science, artificial intelligence more spefically) and I can't seem to find a decent outline anywhere. Is this a good curriculum? Is there something missing? Is there something that's too much/not really needed?

• Vectors and Vector Spaces
  • Definition and properties of vectors
  • Vector operations: addition, subtraction, scalar multiplication
  • Linear combinations, span, and linear independence
  • Vector spaces and subspaces
  • Basis and dimension of a vector space
• Matrices and Matrix Operations
  • Definition and properties of matrices
  • Matrix operations: addition, subtraction, scalar multiplication
  • Matrix multiplication and properties
  • Transpose and inverse of a matrix
  • Systems of linear equations and matrix representation
• Determinants
  • Definition and properties of determinants
  • Calculation of determinants
  • Properties of determinant operations
  • Cramer's rule and applications
• Eigenvalues and Eigenvectors
  • Definition and properties of eigenvalues and eigenvectors
  • Calculation of eigenvalues and eigenvectors
  • Diagonalization of matrices
  • Applications of eigenvalues and eigenvectors
• Linear Transformations
  • Definition and properties of linear transformations
  • Matrix representation of linear transformations
  • Kernel and range of a linear transformation
  • Rank and nullity of a matrix
• Orthogonality and Inner Product Spaces
  • Dot product and properties of inner products
  • Orthogonal vectors and orthogonality
  • Orthogonal complement and projections
  • Gram-Schmidt process and orthogonal bases]]

[vast shale]

Seems pretty normal

Throw in some Optimisation in there at the end too ig maybe

Sorry for the weird format, there are just two levels of bullets 🙂

If I add this to end of the curriculum:

- Optimization
  - Introduction to optimization
  - Objective functions and constraints
  - Linear programming and quadratic programming
  - Convex optimization and non-linear optimization
  - Applications in machine learning and data science

What do you think? @vast shale (btw nice name kk)

Oops, I pinged the wrong one lol @vast shale

Yeah I think it's fine fhen

Imo

About what I did in my introductory linear algebra class

[hard atlas]

why would you have linear transformations that late

[vast shale]

Where would you place it?

Right after vectors?

[hard atlas]

the same time as matrices

they are more or less the same thing

from the start you should see that. translate between them. stuff like that

[vast shale]

Alright! Am I missing something?

[hard atlas]

how much do you know about stuff like set theory

[vast shale]

I am a complete beginner to linear algebra in fact. I only know basic math, like some trigonometry, algebra, etc.

[maiden iron]

Shouldnt you do a bit of abstract algebra ?

learning the basics of groups, rings, etc.

[hard atlas]

rings, ehh. groups and fields a bit might help, yeah

[maiden iron]

and after orthogonality and inner product spaces, you should deal with Hilbert spaces I think

[hard atlas]

thats functional analysis

[vast shale]

Damn, I... really don't know. I'm kind of lost in here. I know it may be too much but I'm trying to self study computer science (I repeat, with a focus on Artificial Intelligence) and I'm trying to organize what I have to study. I'm gathering from different sources and each of them are completely different. I hope that explains my awkward questions x_x

[maiden iron]

optimization needs knowledge of functional analysis too

[hard atlas]

ambitious project

your list is fine

no list will ever be perfect

just start

[vast shale]

Most of the websites and videos I watched mentioned Linear Algebra specifically, and what I do I search wtf does linear algebra include

[vast shale]

indeed

[maiden iron]

good luck dude

[vast shale]

I will need it, thank you very much, but regarding what you said about abstract algebra...

Should I add it?

[maiden iron]

to learn linear algebra, in class we did a bit of abstract algebra before

yea before linear algebra

groups, rings, fields should be enough

it's kinda abstract so take your time

[vast shale]

Thank you and everyone who replied for taking your time 🙂 have a good life 😄

[maiden iron]

no problem, have a good day 👍

[vast shale]

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[radiant harbor]

Help

[radiant harbor]

What do you call this kind of question

And what’s the method

[vast shale]

FOIL

i guess?

[robust island]

you multiply the 2m with m and -2 then multiply 3 with m and -2 then add em all up i guess

[radiant harbor]

Alr

so like foil in algebra

[brittle minnow]

what's that?

[viral copper]

It's a mnemonic to remember what terms to multiply when expanding products like that; first, outer, inner, last

[radiant harbor]

yea

Can I ask another question?

[brittle minnow]

I have never seen that but okay

yeah, sure go ahead

[radiant harbor]

Idk how to do

[merry python]

try to use difference of squares formula

rewrite $\frac{1}{16}x^{2}$ as a perfect square and same for $100y^{4}$

[twin meteor]

B-eard

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[lapis zealot]

Is this a correct way to solve this problem?

[lapis zealot]

Also, I tried another problem earlier and, is this process so far correct aswell?

[patent nymph]

this looks alright, it is very neat

[patent nymph]

No, because you didn’t need to multiply by 1/2

[lapis zealot]

What part did I do wrong then?

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[crisp yoke]

can someone explain this question to me

[jagged gorge]

which part do you not understand

[crisp yoke]

The whole question tbh

what would be the general formula for expressing the parameter of a uniform circular motion?

[open cave]

consider what x and y should be when t = 0

at the start

the starting position will obviously be (5,3)

[crisp yoke]

sinx=3, cosx=5?

i think this is wrong haha

[open cave]

think

you are looking for a parameter such that

at t = 0

x = 5 and y= 3

and

we are told the orbital time is pi

so after pi amount of time

or when t= pi

we should see that it returns to its original position

aka

at t = pi

x = 5 and y = 3

which of these options has these conditions

[crisp yoke]

so the period is 2?

oke, i got it

it's c

thanks

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[bronze wharf]

Hi

[bronze wharf]

I'm a bit confused by this question and was wondering why after using the addition rule to find A intersection B I got a negative number

[vast shale]

you set the venn diagram up wrong

there may be some people who study both literature and math

[bronze wharf]

Ye that's why I put it on the side in a little bubble isntead of the the middle

That's why I had used the addition rule

[vast shale]

maybe it should be 25/31 - 12/31 - 10/31

so you get 3/31

that makes sense to me

[bronze wharf]

But 31 is the total, including those who are not in either

[vast shale]

i know but you discount those outside

31 - 6 = 25

[bronze wharf]

Ye and that gives me a negative for some reason

[vast shale]

hmm yeah that doesnt make sense

this question seems wack

might help to have a second opinion about this

[bronze wharf]

Ye I think it is whack

And it's not like it's 0 so it's mutually exclusive

It gives me a weird negative

[vocal sleet]

@bronze wharf Has your question been resolved?

[vocal sleet]

@bronze wharf Has your question been resolved?

[vocal sleet]

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[solemn remnant]

Hello, Is this the correct approach to solve this question? if so , how do I deal with x and y being different >=0 , < 0 in the injective part?

And is showing surjectivity equivalent to finding the inverse function ?

[hard atlas]

finding a proper inverse function is equivalent to showing the function is bijective

surjective is only equivalent to a right-inverse

injective is equivalent to a left-inverse

your notation during the injective part is absolutely terrible

when you rearrange equalities you need to use <=> and not =

[solemn remnant]

So how would i deal with case3

[cold blade]

I think that case 3 isn't something you need to check

If x>0 and y<0 then x=/=y already

You only needed to check 1 and 2

[solemn remnant]

i just dont get the logic behind that, would x>0 y<0 also mean that f(x) could never be == f(y) ?

[cold blade]

Injective is a property that says x=y implies f(x)=f(y) or vice versa

If you're assuming x>=0 and y<0 then y<x already

Hence that's why it fails

Certainly if you had found f(x)=f(y) in case 3 then this would be a nice way to prove this isn't injective

Also with what someone mentioned earlier, bijectivity iff invertible is true

But also by definition bijectivity means injective + surjective so you're fine checking it this way

[solemn remnant]

i see it now, thank you !

[cold blade]

No problem

But in a less harsh way they are correct use either the inplication symbol

--> or double sided arrows in these things

Equality isn't something you wanna say unless you really mean equal in some way

[hard atlas]

x=y obviously implies f(x)=f(y)

its really the other direction you have to worry about

[cold blade]

Equality implies an equivalence relation

[hard atlas]

the key for case 3 is that f(x) has the same sign as x and f(y) has the same sign as y

[cold blade]

I disagree the key for case 3 is that x isn't equal to y to begin with so it's fine that f(x) isn't f(y)

[hard atlas]

but what if it was equal

that's what you have to check

[cold blade]

Then what I said

[cold blade]

.

[hard atlas]

but they didnt find anything

they tried some stuff and then stopped in the middle of nowhere

[cold blade]

They did

Here lemme outline the proof

He assumed x=y

Then split those possibilities into two cases

The third was erroneous, a better way to go about it though is certainly to say that if x=/=y then we cannot have f(x)=f(y) more directly

But all 3 cases is the same

Regardless

Just change the first statement if you wanna be more exact

[hard atlas]

you cant assume what you want to show

[cold blade]

Yeah sorry where f(x)=f(y)

My bad

Is what he was checking

He checked all cases where f(x)=f(y)

Case 3 doesn't violate injectivity

Like I said because x and y are not equal in case 3

Ahh wait I see what you mean

[solemn remnant]

[hard atlas]

yes thats the whole point. they definitely arent equal but what if they still have f(x)=f(y)

you have to find some sort of contradiction for example

[cold blade]

Yes so you just needed case 3 to show that

Or swapped x and y

In a case 4 scenario of the same as case 3

[solemn remnant]

how could that contradiction look like?

if i would find f(x) = f(y) => x=/y i would prove that the function is not injective while it is ?

[cold blade]

Wait

[hard atlas]

look at the signs of f(x) and f(y)

[cold blade]

Case 3 isn't something we should check

Yeah

Because of that

[hard atlas]

.

[cold blade]

f(x)>f(y)

[solemn remnant]

ahhh

[cold blade]

All you have to state is that when exactly one of x or y is negative you have that f(x) and f(y) cannot be equal

And then you've got everything covered

Statement 3 proves that much

[solemn remnant]

i seee

ok now i got that

@cold blade @hard atlas

thanks again!

.close

[vocal sleet]

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[vocal sleet]

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[half light]

I don’t understand why the gamma function is replaced by a different variable if alpha is the exponent?

In the proof

[mild ledge]

in the first line where they expand Gamma(a)*Gamma(b)?

cause that does look like a typo

Gamma(b) should be $\int_0^\infty y^{b-1} e^{-y} , dy$

[twin meteor]

Stole

[mild ledge]

they fix it in the next lines though

[vocal sleet]

@half light Has your question been resolved?

[half light]

thanks

[vocal sleet]

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[vocal sleet]

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[raw fog]

Stuck on these 2

[rose raft]

cos^2 x is not cos(2x)

[raw fog]

Where

[raw fog]

I have this now

[jade cairn]

2tanx is (2sinx)/(cosx), not (2sinx)/(2cosx)

[nova fulcrum]

You rather change tan =sin/cos in the begining and expand not to get confused

[vocal sleet]

@raw fog Has your question been resolved?

[raw fog]

Ok I fixed that

So what should the next step be

Actually I think I know let me try smt

This is what I have now

It’s close

[raw fog]

<@&286206848099549185>

[jade cairn]

don't convert cos^2(x) to 1-sin^2(x). Rather, get a common denominator of cos^2(x)

Then make one big fraction and use your double angle identity to get the answer

[vocal sleet]

@raw fog Has your question been resolved?

[raw fog]

Can I see what it looks like?

[jade cairn]

I simply multiplied the numerator and denominator of the second term by cos(x) to get that first line

[raw fog]

Thanks

[raw fog]

How did you divide by cos^2x in the first line?

[jade cairn]

What do you mean? I combined the fractions since they have the same denominator

[raw fog]

It was originally 1/cos^2x + 2sinx/cosx

How did 2sinxcosx become 2sinxcosx/cos^2

[jade cairn]

like I said, I multiplied the numerator and denominator of the second term by cos(x)

[raw fog]

Ohhhhhh

I see where I messed up

.close

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[smoky terrace]

One of the projects has the task of reaching the role of Minima Ambasador. Define a strategy in which you can achieve the role of Ambasador by sending messages and making invites (invitations for new project members). How many hours you need to spend to achieve this goal, if one message (10XP) takes 2 minutes and an Invite (100XP) takes 3 minutes. In one hour we write 10 messages, the rest of the time we spend on invites.

[smoky terrace]

I got a result of 7.1 hours. Is that correct?

[vocal sleet]

@smoky terrace Has your question been resolved?

[vocal sleet]

@smoky terrace Has your question been resolved?

[vocal sleet]

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[vocal sleet]

@ripe parrot Has your question been resolved?

[calm minnow]

Well notice that you have 3 of each kind of side correct?

And you only need 1 of each kind of side.

How would you solve this then @ripe parrot?

[floral pike]

I see 5

[calm minnow]

There are 5 squares but 3 kinds of each side

[floral pike]

...you mean 3 sides for each square?

[calm minnow]

Yup!

And you only need 1 side of each square to cover the distance wanted

[vocal sleet]

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[floral pike]

.reopen

[vocal sleet]

✅

[floral pike]

🪄

the 6 km is for the traveled distance, not the straight-line distance, right?

[calm minnow]

Yup

So 6/3=2km

[floral pike]

oh.

I'm not seeing the forest for the trees here

tricky.

[calm minnow]

So each square had 3 sides: 3a+3b+3c+3d+3e = 6

And we want a+b+c+d+e

[floral pike]

do we though?

[calm minnow]

So we divide both sides of the equation by 3 to find that

[floral pike]

at least two (three, I think) are the same

[upbeat chasm]

doesnt matter

[calm minnow]

^

[floral pike]

oh, I see

[calm minnow]

Does this answer your question @ripe parrot?

[ripe parrot]

yess

[vocal sleet]

@ripe parrot Has your question been resolved?

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[wild jolt]

Help

[dreamy geyser]

okay

have you learned trigonometry?

[wild jolt]

Yes

[dreamy geyser]

okay

[wild jolt]

I just want to check my answer

[dreamy geyser]

ohhh

okay

[wild jolt]

My answer is 76 feet

But I don’t know if it is correct

[dreamy geyser]

nvm i dont know how to solve this

i thought it was like

solving for a side or angle

[wild jolt]

Ok

[dreamy geyser]

but the moving 16 feet closer is confusing 😭

[wild jolt]

Yeah

[wraith python]

You can use the tangent of the angle for the slope of an equation.

[wraith python]

Yes, it's approximately 76 feet.

[wild jolt]

Okay thanks

[vocal sleet]

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[wild jolt]

Help

[wild jolt]

I just want to check if my answer is correct.

I got 4.01 miles.

[floral pike]

show us your work

[wild jolt]

.

[wild jolt]

Is it correct?

[wild jolt]

I showed you the work. Now tell me please.

I asked so many people yesterday and none of them could help me.

[floral pike]

gimme a second, I need fresh paper

your hypotenuse is correct

[wild jolt]

Ok

[floral pike]

yes

confirmed

you didn't need to find the hypotenuse

(of the outside triangle)

Actually I take that back. I tried that, but I think I messed something up later (using an incorrect length).

you can get the answer just as quickly finding the outside hypotenuse first

[vocal sleet]

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[wild jolt]

Thank you

[vocal sleet]

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[pliant coyote]

I am supposed to use L'Hopital, but how do I turn that into a fraction?

[silk comet]

use that x = 1/(1/x)

[pliant coyote]

hmm?

nevermind

.close