help-23

[untold topaz]

How does one solve this without using matrix?
We've only learnt cross/dot product like that

[past birch]

substitutions

represent an unknown as an operation of other unknowns

[untold topaz]

Aight lemme try

[patent grove]

yey

typing

u figured it out

[untold topaz]

So if I let z = t, whr t (e) R ||idk how to type the symbol :P||
Then
x = - 13 t + 54
y = - 5t + 21
z = t

[patent grove]

haha?

hah?***

[untold topaz]

Wait... is it not?

[patent grove]

why need t

dont need t

xyz is enough

$$x-3y-2z=-9$$$$2x-5y+z=3$$$$-3x+6y+2z=8$$

[flat frigate]

Question mark

[untold topaz]

x = 2
y = 1
z = 4
?

[patent grove]

uhm

just plug the value of xyz into these three eq and check will do

[untold topaz]

Yep the coordinates are correct

[patent grove]

then

its correct

[untold topaz]

Does this just mean 'find the point of intersection'?

[patent grove]

?

what?

oh, solve with matric then i d k

[untold topaz]

Issokay, idk too 😂

.close

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[hasty hamlet]

im doing this limit, and i want to use a equivalences, but im not sure if i cant do log(1+(an)), understanding an as x^3-2x^4

[brittle mesa]

sen = sin

[hasty hamlet]

log(1 + an) ∼ an. with this equivalence

[lime dust]

Give me a moment

You can transform it into this.

$\lim_{x\to{0}}{\dfrac{\tan^{2}\left(x\right)-\operatorname{sen}^{2}\left(x\right)}{x,\left(x^{3}-2,x^{4}\right),\ln\left(\left(-2,x^{4}+x^{3}+1\right)^{\frac{1}{x^{3}-2,x^{4}}}\right)}}$

Wait, the latex xd

[flat frigate]

Samuel

[brittle mesa]

we'll wait

thx samuel

[hasty hamlet]

thx

[lime dust]

Have in mind this $\lim_{x\to{\infty}}{(1+\frac{1}{x})^{x}}=e$

[flat frigate]

Samuel

[brittle mesa]

thx bro

[hasty hamlet]

i would like doing it with equivalences

like this

sinx~~x

[lime dust]

I hate red, I am not able to read that

[brittle mesa]

I agree with that, but bro doesent have the resources needed for a new pen right now, he is using my laptop in order to use discord

[hasty hamlet]

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[ionic oar]

2x^2

X = 5

[ionic oar]

Is it 10^2?

[quasi bison]

no

$2x^2$ means $2 \cdot (x^2)$ --- exponents have higher priority than multiplication.

[flat frigate]

Ann

[quasi bison]

if they wanted 10^2 with this, the expression would have been (2x)^2.

[ionic oar]

Kk

[ionic oar]

So it’s 2 * 5^2?

[quasi bison]

yes

[ionic oar]

ok!

and it’s 50

Whenever is odd number when multiplied by minus number is always -, but when multiplied with even number is always +?

Like

4 * -3?

@quasi bison

[quasi bison]

i think you are confusing multiplication and exponents...

or just crashing into a language barrier.

[ionic oar]

Okok

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[ionic oar]

4145

1982 was Jónas x years old. How old

a) was he 5 years ago

[ionic oar]

@quasi bison

How to do this

[quasi bison]

!noping

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[ionic oar]

x-5=1982?

[quasi bison]

,rccw

[quasi bison]

does this mean "Jónas was x years old in 1982"?

as in he was x years old in the year 1982?

[ionic oar]

I’m not quite sure

In 1982 Jonas was x years old

[quasi bison]

if you aren't sure then i can't help you because i do not speak icelandic

[ionic oar]

And it asks how old was he 5 years earlier

[ionic oar]

Yes

[quasi bison]

if you're going to use google translate, you have to use entire sentences.

otherwise it can produce weird results because the syntaxes of the source and target language do not always line up.

anyway ok

jónas was x years old in 1982 and we are asked how old he was 5 years prior.

that is just x - 5. that's the answer. equating it to 1982 is wrong.

do not ping me in the future, by the way.

i don't like it.

[ionic oar]

will 5 - x be also correct?

B) he will be 15 years later

So it’s x + 15

<@&286206848099549185>

Hi??

-_-

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[mild lake]

for IVT/Intermediate Value Theorem, when the question states

using IVT, prove that there is a root in the interval (x,y)

that root refers to 0 right

[void cosmos]

yes

IVT is an existence theorem so we have to show that a root exists. To prove this the plug the endpoints into the function if ur given one.

If one of the endpoints is negative and one is positive

we can then ensure that f(x) will take on every value from a to b

therefore by IVT we can prove that there exists a root c on the intervals that is an element of (a, b) assuming b is larger.

Note f(a) cannot equal f(b)

[a,b] however must be continous.

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[winter sable]

Can someone help me come up worth an idea of a pulley with 1 horizontal output, 1 vertical output, and 1 rotational output using a pull string as an input

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[mint berry]

I have a question about algebraic structures.
Specifically, if
(Z, ∗) where * is defined as x ∗ y = x + y − 1
is a group.

I've already determined that it is a magma. (It's fairly obvious that the result of * will always be an integer, so it's closed)
I've then determined that it is a semigroup, as x ∗ y = x + y − 1 is associative.
Further, I've determined that it's also a monoid. (the neutral element is 1)

I am now trying to find out if (Z, ∗) is a group.
So far I've found that
for x = 0 the inverse is 2
1 is it's own inverse
for x = 2 the inverse is 0
for x > 2 the inverse is (x-2)
and for any negative x, the inverse is (-(-(x+2)))

[mint berry]

I'm not entirely sure if I'm right though. It seems unintuitive that the inverse of 0 would be 2.
And that 1 is it's own inverse.

[fast gazelle]

the inverse of 0 is 2, and 1 is its own inverse

given that the identity element is 1, the identity element has to be its own inverse

you're wrong about x > 2 and about negative x

maybe just think of it as, try solving the equation x + y - 1 = 1 for y

[mint berry]

Ah, I think I've got it.
2-x

It's correct for x=0, x=1, negative x (which would make it 2-(-x) ), and any positive x

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[wooden oyster]

would this here be correct

[empty gyro]

I do not think this is correct

[wooden oyster]

what is wrong

[empty gyro]

How did you get lambda+2?

[wooden oyster]

lambda - -2

= lambda +2

one sec ill tru to fix it

this should work

[final halo]

yeah i think you were confusing with lambda*I - A in which case you would have had to flip the signs of the off diagonals

[empty gyro]

yeah that looks better

[wooden oyster]

well it was right

or i can the exact same answer at least

this way is better anyways, then you avoid trouble with $$ \pm $$

[flat frigate]

// mav

[wooden oyster]

How do i find v_1 and v_2

[empty gyro]

Your eigenvectors come from your eigenvalues. Do you remember how to find eigenvectors normally?

[wooden oyster]

by normally you mean normalized?

[empty gyro]

No I mean in general do you know how to find an eigenvector?

[wooden oyster]

by inserting the value?

[empty gyro]

yeah that's the start

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[fast pike]

Hi, below will be images from a practice exam on Gateway math. These questions posed an issue to me and I found myself completely lost on how I should solve them. Not a test / not graded, just need to prepare for the legitimate test. On the question I got wrong, I’ll attach my work as well for it. Any help at all is greatly appreciated. I know it’s a lot so please do not feel pressured to help with each question unless you feel inclined to. Thank you! (I do not have work for first few questions as I genuinely just do notnknow what to do)

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[opal falcon]

Hi I was able to solve the first half of my homework problem I’m not sure how to set up the second half of the problem to find the height of the building the window is in

[opal falcon]

I figured the previous question out this question I am having trouble sketching it out

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[south sparrow]

$\overline{A}B$ and $\overline{AB}$ seem tricky, because on the Venn diagram AB is the middle part, and not AB are A xor B. In terms of two events that can both happen, $\overline{AB}$ means ø.

[flat frigate]

kytsu1

[rough storm]

$\overline{A}B$

is read as, if I'm understanding correctly, not in A and/(but still) in B

[flat frigate]

🫎Mοοsey🫎

[rough storm]

and you can see precisely where this occurs in the venn diagram

[south sparrow]

This is what I think, but I will try to show where I am confused.

The original task is about two conveyor belts, one red, one blue, and A, B are events, where there are products on them. A, there is a product on red, B, on blue.

[rough storm]

looks good, the rectangle is what we would call universal set not A and not B would not be empty set in that context. I think it would be better for you to post your original question

[south sparrow]

Express in English language different given states, for example, $\overline{AB}$ and $\overline{A}B$.

[flat frigate]

kytsu1

[south sparrow]

When A means the event, when there is a product on the red conveyor belt, and B means there is a product on the blue conveyor belt.

I don't know, if the complement of the intersection of AB, (AB)' includes A-B and B-A, or is it the unviersal set, which is in this case ø.

[south sparrow]

Maybe because A-B ⊂ (AB)' ^ B-A ⊂ (AB)' ^ ((A-B)^(B-A)=ø) -> (AB)' = ø

AB is expressed as there are products on both conveyor belts, so $\overline{AB}$ is the event, where AB is not taking place. In other words, $\overline{AB}$ is when there is a product on only one of the belts, or on non of the belts.

[flat frigate]

kytsu1

[south sparrow]

Mathematics is mind twisting, I like it. I thought (AB)' meant there are no products on either.

Haha, I don't know what is true anymore.

These are De Morgan rules

[south sparrow]

I love it! It seems that this image shows the correct solution.

Except it does not express it in English.

$\overline{A}$ means there is no product on red, B means, there is a product on blue. Their product, or intersection in event algebra, means that both of those statements are true. When $\overline{A}B$, there isn't a product on red, and there is a product on blue!

[flat frigate]

kytsu1

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@south sparrow Has your question been resolved?

[south sparrow]

Q = AB ∪ A' ∪ A-B ->
A-B ⊂ (AB)' ^ B-A ⊂ (AB)' ^
((A-B)^(B-A)) = ø, but
A ⊂ (AB)' ^ B ⊂ (AB)' ^
AB ∉ (AB)' ->
(AB)' = { A, B, ~(AB),

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@south sparrow Has your question been resolved?

[south sparrow]

I wonder if (AB, A', and A-B is the complete event space) is true for event spaces with non mutually exclusive events.

[south sparrow]

$\overline{A+B}$

[flat frigate]

kytsu1

[south sparrow]

A+B means there is a product at least on one of the conveyor belts, so $\overline{A+B}$ is when there are no products on either conveyor belts.

[flat frigate]

kytsu1

[south sparrow]

https://tenor.com/view/cat-gif-6997856904020830598

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[elfin acorn]

Nitrogen gas can be prepared from the following reaction
2NH3 + 3CuO → N2 + 3Cu + 3H2O
If 20.0g of NH3 reacts with 92g CuO, what mass of N2 can be produced?

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[celest ferry]

Ok close channel

[plucky elk]

it will. just takes time. go open a new one

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[quartz wasp]

a pyramids base is a isosceles triangle the base of the triangle is 6 the edges of the pyramid is all 13 and the hight of the pyramid is 9 need the volume of the pyramid

[indigo birch]

drawing a diagram would help

[quartz wasp]

Its a bad drwaing tho

<@&286206848099549185>

[humble sandal]

Mayvbe Pythagorean theorem?

[quartz wasp]

how?

[humble sandal]

You have to find the Hypotenuse

[quartz wasp]

how?

[humble sandal]

Bruh I can't do the root

Btw:
√ C² + c² = √ 13²+6²

@quartz wasp

And you should find the missing side.

Remember: do 13² and 6² before doing the square root

Or it won't be right

[quartz wasp]

Can you try doing it and tell me what you get i cant do it

[humble sandal]

Ok

I made it just now and it comes a comma number

Oh wait

[humble sandal]

Can u say the problem's data? Cause I can't understand from the drawing, sorry

[quartz wasp]

a pyramids base is a isosceles triangle the base of the triangle is 6 the edges of the pyramid is all 13 and the hight of the pyramid is 9 need the volume of the pyramid

[humble sandal]

So all of the edges are 13, also the ones without 13 written near them?

Talking about this two

Cause if it is, you alr have everything, if not, I think you must apply Pythagorean theorem

[quartz wasp]

no

[humble sandal]

Ok so

[quartz wasp]

that is the base of the pyramid ik 1 side is 6

[humble sandal]

So, you must do 6:2 which makes 3, you found the half of the base, which you need to find the hypotenuse, then you must find the C, you must make 9:2, equals 4.5, then you must do √4.5 × 4.5 + 3 × 3 which equals 5.4083269132 (You can round it up if you want) and you found the two remaining edges of the pyramid (I think)

[quartz wasp]

.close

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[cedar storm]

when we have $ \begin{cases}
\mu \neq 1 \
\mu \neq 0
\end{cases} $
can i just say that $\mu \in \mathbb{R}$ ?

[flat frigate]

bazylenius27

[median vigil]

if (\mu \in \mathbb{R}) then barring any other restrictions (\mu = 0) and (\mu = 1) would be valid values for (\mu)

[flat frigate]

cloud

[cedar storm]

sorry i thin k i messed sth up

the problem is as follows

"prove that for every real value of μ the equation (μ-1)x+μy+μ²=0 makes up a straight line

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[drowsy dagger]

Can anyone recommend a free online book on combinatorics that is easy to understand

[drowsy dagger]

<@&286206848099549185>

[jagged badger]

try #book-recommendations

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[sour hill]

I’m completely lost on this trig question

[sour hill]

i don’t know how to solve for QS

And google is not helping me out

[golden forge]

you can use the sine law

[simple oak]

You don’t need sine law just sine ratio

Isn’t this right angle triangle

[sour hill]

yes

[simple oak]

We don’t need law yet

Do you know the primary trigonometric ratios

[sour hill]

no

[simple oak]

Do you know sine cosine tangent

They are ratios

Easy way to remember is SOH CAH TOA

[sour hill]

yes

[simple oak]

sinx=o/h cosx=a/h tanx=o/a

[sour hill]

Oh

[sour hill]

Yes I know these

[simple oak]

Ok

[sour hill]

My teacher never used this wording

My bad

[simple oak]

So what side is QS

a o or h

[sour hill]

H

[simple oak]

Ok

And what information do we have

[sour hill]

QR = 13

[simple oak]

Yes

[sour hill]

S = 42
R = 90
Q = 48

I think

[simple oak]

Ok

[sour hill]

Angles

[simple oak]

We only need S

[sour hill]

Ok

[simple oak]

Okay what kind of side is QR

a o or h

[sour hill]

a?

I think

[simple oak]

A is the adjacent side

Is that side adjacent to angle 42

[sour hill]

oh

no

my fault

[simple oak]

It is opposite

[sour hill]

yes

[simple oak]

Okay so we can use sine ratio

sinx=o/h

[sour hill]

soh

[simple oak]

sin42=13/h

How to proceed

Do you know how we can isolate h

[sour hill]

Try nd isolate

Yes

X 13

Both sides

or / sorry

[simple oak]

Why do we do that

[sour hill]

😔

[sour hill]

to get just h on one side

[simple oak]

But if you multiple both sides it becomes

13sin42=169/h

So here’s what we can do

sin42=13/h

Multiply both sides by h

[sour hill]

smart guy

[simple oak]

hsin42=13

Then divide both sides by sin42

h=13/sin42

H has been isolated

[sour hill]

yes

so now do the math

Sin42

[simple oak]

Yes

[sour hill]

Thank u bro

U better than my teacher

[simple oak]

[sour hill]

is it cool if I friend u

[simple oak]

Uhhhhhuh

I don’t accept friend but if u want help just msg I check

[sour hill]

fs

.close

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[bronze jasper]

#proofs-and-logic message

[bronze jasper]

I’m asking help here since I have no answers in the channel

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[bronze jasper]

<@&286206848099549185>

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[daring crescent]

?

[golden forge]

elaborate

[simple oak]

[empty gyro]

!original

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Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

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[high gate]

Needing help with this question thats as far as I got

[solemn vault]

U sure with this?

[high gate]

Meant to cross out (3+x)

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@high gate Has your question been resolved?

[sturdy thunder]

oop i think theres a mistake on your first line of working

[sturdy thunder]

the third part of the first line you put x+3, when the equation is x-3

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[ripe nymph]

I don't know how to proceed

[desert pasture]

find b1, b2, b3. See if there's a pattern

try applying king's rule too

you'll get $2I=\ \int_0^{\frac{\pi}{2}}\frac{\cos^2\left(nx\right)}{\sin x}+\frac{\sin^2\left(nx\right)}{\cos\left(x\right)}$, see if that helps

[flat frigate]

Why am. I here

[desert pasture]

does this make sense ?

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[ripe nymph]

.reopen

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✅

[ripe nymph]

But how do I get the ans

I tried this

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@ripe nymph Has your question been resolved?

[ripe nymph]

No

<@&286206848099549185>

[crystal ridge]

Evaluate the integral first.
Here you have to apply that infamous property ∫f(x) dx from a to b = ∫f(a+b-x)dx
Add them

yea it ain't working out

wait lemme take a different approach

[crystal ridge]

Try evaluating (bn+1)-(bn)

cause that is being asked in the question

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[crystal ridge]

The correct answer is option (B) @ripe nymph and @desert pasture

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[pseudo carbon]

Find a degree-three polynomial function of one variable f(×) such that f(1)= f(4)=f(-2)=0 and f(2)=-16.

[pseudo carbon]

what is function of one variable in this case ?

[umbral mantle]

just a basic function that is in terms of x

so f(x) = <an expression that only uses x>, so no y, z, etc.

[pseudo carbon]

so it means i have to find only one coefficients in this function right ?

[umbral mantle]

i mean, it's a polynomial so you'll have to find the coefficients of x, x^2, x^3, x^4, etc. however many you need

as a hint i would start in factored form, e.g. (x-x_1)(x-x_2)(x-x_3) etc.

[pseudo carbon]

so i just add an unknown value as a. Then, f(x) = a(x−1)(x−4)(x+2) right ?

then f(2) = -16 so on

[umbral mantle]

yeap

just solve

[pseudo carbon]

ahh alright thx

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[hardy lion]

Just leave pi as pi

[solar cradle]

What u mean

[hardy lion]

Dont change pi into 3.141

[solar cradle]

But then how will I get a figure

[hardy lion]

What

[desert pasture]

your answer will be in terms of π (the options are in terms of π)

[solar cradle]

Omg….

I thought they were tiny M’S!!

Gosh I’m dumb

.close

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[lean otter]

Cantor-Schroeder Bernstein, before you come up with injective functions, do you always have to show that they are not surjective?

[clear blade]

not sure I understand the question

can you maybe rephrase

an injective function can be surjective but that doesn’t change the fact that it’s injective

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[lime grove]

do u have an example of a function like this

[past birch]

what is l

?

[lime grove]

l and a are real numbers

[past birch]

oh

[final halo]

sure, f(x) = l for all x

[lime grove]

no i meant that has a limit of that form

[final halo]

i dont know what you mean, the function i just gave has a limit of that form

[lime grove]

how all x that doesn't verify this

because there's no a in Df

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[zenith igloo]

this is what I wrote so far

[hasty wagon]

the last form z=... where k=0,1,2,3,4
is nice except a few notation mistakes
but i feel weird about the cos(2π/5)=135°

[zenith igloo]

forgot the z_k

oh

2pi/5 is 72

my bad

then it's in quadrant I

since all functions are positive in the 1st quadrant

how do I write this in the rectangular form though, wait actually

well our reference angle for 2pi/5 is already 72

there is no need for a reference angle

only problem is I don't know how to write them in algebraic form now

[zenith igloo]

what should be my next step

<@&286206848099549185>

[rose holly]

Yep

[hasty wagon]

sorry, i had no signal just now 😔

[hasty wagon]

oh, you need to write them out without sine and cosine?

[lean otter]

hi

[zenith igloo]

yup, sadly

[zenith igloo]

no worries xD

I have the solution given by my professor but

I got 0 idea how he solved it

so I tried seeing if this can give me the same solution

apparently what I got so far is right

I got everything the same as he did

just that he didn't use the polar form at all

he wrote the numbers in a geometric progression

then factored them

and then got a 2nd degree equation

[hasty wagon]

i see what your prof did there

[zenith igloo]

could you explain

[hasty wagon]

sure

[zenith igloo]

he did this 1year ago with another guys

and thank you

[lean otter]

There are a number of chickens and rabbits, the chicken has two legs, the rabbit has four legs, from the top there are 243 heads, from the bottom there are 1218 legs, ask how many chickens, how many rabbits?

Do you know?

[hasty wagon]

firstly, do you know that
z^n-1=(z-1)(z^(n-1)+z^(n-2)+...+z^2+z+1)

[hasty wagon]

please read #❓how-to-get-help

[zenith igloo]

this looks like a derivate from the sum of all nth roots of unity

but no I didn't

[hasty wagon]

nah, it's just factoring out the (z-1) out of z^n-1

[lean otter]

what's wrong

[hasty wagon]

!occupied

[safe radish]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[hasty wagon]

it's okay, for now, just take it as a fact, in case you wanna know more, do go search online for "factor x^n-1"

[zenith igloo]

alright

let me write it here

[hasty wagon]

now that we have
z⁵-1=0
so we factored it into
(z-1)(z⁴+z³+z²+z+1)=0

since we know that z=1 is a root, we will solve the remaining roots by solving z⁴+z³+z²+z+1=0

[zenith igloo]

okay

[hasty wagon]

since we know that z≠0 we divided the equation by z²
we get
z^2+z^1+1+z^(-1)+z^(-2)=0

now what your prof did is very useful

[zenith igloo]

ohhhh

one problem is that

when he gave us this homework

he mentioned no archives

[hasty wagon]

consider this form is so "symmetrical" we can consider let
w=z+1/z

[zenith igloo]

so basically I shouldn't do exactly what he did to solve it

[hasty wagon]

oh...

[zenith igloo]

but since I got no other idea on how to solve it

XDDD

[hasty wagon]

maybe we can first understand what he did

[zenith igloo]

agreed here

[hasty wagon]

now we rewrite the modified equation of z in term of w

recall:
w=z+1/z
then
w²=(z+1/z)²

after expanding we get
w²=z²+1/z²+2

[hasty wagon]

and can you construct the new equation?

[zenith igloo]

yup

it's

w^2 + w = 0

[hasty wagon]

careful

[zenith igloo]

oh

-1

we only have a 1

w^2 + w - 1 = 0

alright

now we can solve this

obvs

[hasty wagon]

now we can solve this Quadratic

[zenith igloo]

now we can swap back in the original one

since we have w1, w2

original one aka w = 1 + 1/z

[hasty wagon]

and what your prof did is using the fact that conj(z)=1/z for these z

[zenith igloo]

oh

so it's

left side = z + conj(z)

damn

and if we do (a + bi + a - bi)

[hasty wagon]

this is because these are roots of unity, so |z|=1 and z*conj(z)=|z|²
so 1/z = conj(z)

[zenith igloo]

we will be left with 2a

so we get our a after solving this

[hasty wagon]

we have 2(Re(z))

[zenith igloo]

I got no idea what Re() is

[hasty wagon]

real part of

[zenith igloo]

oh

alright

yeah well

we have 2a = .../2

so our a is

-1 +- sqrt(5)

shouldn't there only be one a?

[hasty wagon]

careful

[zenith igloo]

NVM

it's real

square root can't be negativ4

so it's a = -1 + sqrt(5)

I mean it isn't anyway

[hasty wagon]

2a = (-1±√5)/2
a=(-1±√5)/4

[zenith igloo]

oh forgot the 2

oopsie

alright but how can a have two values?

if z = a + bi

is one value of z

and another value of conj(z)

[hasty wagon]

note that we have 4 z's remaining

and clearly they are in pairs

so we have 2pairs
a±bi

and we have 2 a's

[zenith igloo]

oh alright

okay so I wrote everything done

we got the a = (-1 += sqrt(5)) / 4

what we also got

[hasty wagon]

and using Pythagorean identity

[zenith igloo]

if I write this in the polar form

is the cos

[hasty wagon]

we can find sin

[zenith igloo]

alright

I guess I have to take it as the formula

(-1 + sqrt(5)) ^2

(a + b)^2

not just each element squared

[hasty wagon]

well, you'll need some careful calculations

you can split it into + and - case before calculating

[zenith igloo]

ignore the first +1

[hasty wagon]

in case you wanna keep the ±, just remember that

$(-1)(\pm a)=\mp a$

[zenith igloo]

oh

[flat frigate]

Biscuity

[zenith igloo]

I thought they're the same

then I'll swap them around

-+

• a

[hasty wagon]

yea

[zenith igloo]

well

knowing

we will multiply by (-1)

I guess I can re-swap them around, right?

oh wait

this isa fter multiplying by (-1)

alright

[hasty wagon]

so we have
a=(-1±√5)/4
b=±(√(10±2√5))/4

[zenith igloo]

oops I forgot a 1

[hasty wagon]

nope

[zenith igloo]

wait, I should swap -+ into +- again?

donezo

alright we got cos and sin

now since I have z

how do I replace them into z actually

[hasty wagon]

so, we have
z=a±b
for both sets of (a,b)

[zenith igloo]

oh

so one z will be

a + b

conj(z) should be

a - b?

alright

let me write them down

oops I forgot the sqrt around the sin

[hasty wagon]

hmmm

[zenith igloo]

nvm I FORGOT nothing

[hasty wagon]

something wrong in the middle

[zenith igloo]

my brain is fried, is ok

[hasty wagon]

i found it

[zenith igloo]

ohhhhhh

[hasty wagon]

to be honest it's really difficult to catch up with all the - and +

i really suggest to split cases

[zenith igloo]

true

tough for me but probably easier for others

but will do that next time

[hasty wagon]

anyways this is correct

[zenith igloo]

i just wrote the 16 as 4

and left the sqrt on top

[hasty wagon]

oh yeah

[hasty wagon]

except that

[zenith igloo]

so we got Z and conj(Z)

[zenith igloo]

forgot an i there

[hasty wagon]

should be -+

[zenith igloo]

oh alright

[hasty wagon]

as i have said something wrong in the middle

[zenith igloo]

I'll revert what I wrote above

[hasty wagon]

it's hard to keep track with all +-

[zenith igloo]

these should be, right?

oh wait

wait

• • • down

WTF

[zenith igloo]

wait wait

why are we only taking one variant

we had +- for both a and b

[hasty wagon]

a_1+b_1 i
a_1-b_1 i

[zenith igloo]

yes

[hasty wagon]

a_2+b_2 i
a_2-b_2 i

[zenith igloo]

OH

FOR A PAIR

[hasty wagon]

2 pairs

[zenith igloo]

YES

[hasty wagon]

so finally we have the final answer

[zenith igloo]

thanks for the help

[hasty wagon]

do tell if you need another method for this Q

[zenith igloo]

no, it's fine

I want to understand it more

than write another version

the truth is that

[hasty wagon]

good luck!

[zenith igloo]

no way I would've wonder to make it into an ecuation like that lol

thank you!

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[versed wave]

question 30

rough translation, given A B C with following coordinates, find the orthocenter H of the triangle ABC

i dont know where to start, in 2d i can just draw it out, but in 3d it becomes much more sophisticated

[west hedge]

Fungusdesuuuuuuuuuu

I’m Humandesu

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[versed wave]

<@&286206848099549185>

[stable kettle]

hello

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[keen notch]

To solve for matrix X in the equation
PAX = i
AX = P^-1
X = A^-1 P^-1

[keen notch]

Would that be the correct solution?

All matrices are 3x3

[vague hare]

ori-motek

[umbral swan]

assuming they are invertible yes

[keen notch]

As in they have an inverse?

[shadow sparrow]

yes

[keen notch]

Thanks

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[digital anvil]

why my answer vs teacher answer different

[umbral mantle]

uh

[digital anvil]

what me doing wrong

[patent grove]

question is what

[digital anvil]

its whether its true or false that root of -3 times root of -3 is 3

in my case its true

while in his case its false

[umbral mantle]

well

[tiny wraith]

sqrt(a) * sqrt(b) = sqrt(ab) only holds for positive a and b

[digital anvil]

hmm so you are saying that a and b must be positive

then only we can multiply them?

[patent grove]

,calc sqrt(-3)*sqrt(-3)

[flat frigate]

Result:

-3

[patent grove]

ok

-3

not 3

[tiny wraith]

Actually, it also holds if just one of them is positive

[patent grove]

o

,calc sqrt(3)*sqrt(-3)

[digital anvil]

hmm so one of them should be positive if we want to multiply roots?

[flat frigate]

Result:

3i

[digital anvil]

LMao

[patent grove]

ohh

[tiny wraith]

So sqrt(-3) = sqrt(3) * sqrt(-1) = sqrt3 * i

[digital anvil]

ooooh

i get u now

[digital anvil]

how is it 3i here tho

wait

[flat frigate]

Result:

-1.2247448713916 + 1.2247448713916i

[digital anvil]

huh?

root of alr root/

[patent grove]

?

[digital anvil]

u said sqrt of sqrt of -1

[flat frigate]

Result:

-0.70710678118655 + 0.70710678118655i

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[lean otter]

what's the best tips for self studying maths

[lean otter]

To stop thinking and start doing

lol

[lean otter]

where do i begin

do i learn it from yt? or wat

wasn't it u, who asked about similar things yesterday

what/where/how do i begin

[lean otter]

nah

[lean otter]

whichever is easy for u

[quasi bison]

what level of education are you at

[lean otter]

a level

[quasi bison]

as in A-levels?

[lean otter]

yh

[quasi bison]

so like... that must be highschool ish

[lean otter]

do u know it>?

[quasi bison]

right

[lean otter]

check out khan academy, it's good

[quasi bison]

i have some passing familiarity with it from work.

[lean otter]

oh right

[lean otter]

k

[quasi bison]

i would say khan is good for really "from nothing" basics

and up to & including earlyish university

[lean otter]

no, it's good either way

[quasi bison]

no im just saying what its scope is

[lean otter]

some topics are covered by 3b1b, if u know him

[quasi bison]

to the best of my memory

3b1b is not that beginner-friendly

i'd say you'd need to have some mathematical maturity to really appreciate his stuff

[lean otter]

yup

on khan academy, he starts from basics, wahtever he teaches. He has lectures for basic level. Ofc it's a bit advanced on his own channel, but not on khan academy

[final halo]

(A level is post high school pre uni)

[lean otter]

ok i will try khan academy

it's a suggestion, try whichever works for u

some people like solving textbook problems, some like yt

it depends on the person

I just shared my opinion as I have tried khan academy a few years ago

[quasi bison]

i mean tbh

you can't get far without solving problems in one form or another

[lean otter]

What method did u use

[rustic goblet]

"math is not a spectator sport" is a common saying

watching videos can help a lot, but you have to be willing to do tons of problems to learn the math properly

[quasi bison]

what method did i use for what

[lean otter]

To study maths

[quasi bison]

oh if only that could be summarized in one method.

i did not use any predefined study strategy (or any study strategy at all) if that's what you're asking about.

[lean otter]

right

[lean otter]

Yh thanks for ur opinion

[lean otter]

So where can I find these questions?

Like more advanced than school questions

[quasi bison]

uhhhh

god idk

i know MIT has something called open courseware

with course materials published online for some (idk which) of their courses

so perhaps that

but like generally you'd want to look for stuff in more specific subjects

like analysis, or linear algebra, or group theory etc.

[lean otter]

Thanks

.close

[safe radish]

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[raven sky]

<@&286206848099549185>

[raven sky]

Huhuhu help mee

[safe radish]

Please don't occupy multiple help channels.

[severe pond]

stop pinging helpers

u only ping after 15 minutes

if u don’t get an answer

and u didn’t even post a question

[raven sky]

im sorry mine was removed

so i pinged again

mb

i already did but it was removed

[severe pond]

no u didn’t

[raven sky]

sorry

[severe pond]

u didn’t give a specific question

u asked for help without a question

[raven sky]

as i was saying, can you give me capitalized costs problems with solutions?

[severe pond]

and pinged helpers

immediately

[raven sky]

ok mb

<@&286206848099549185>

[red delta]

stop

[safe radish]

@raven sky Has your question been resolved?

[safe radish]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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

[covert berry]

how to show this is a group?

[covert berry]

i am used to a*b but i dont see it here

[marsh walrus]

theres not much to do other than confirm that it satisfies the necessary properties right

[covert berry]

Yea

[marsh walrus]

and it appears that the operation is polynomial addition

[covert berry]

i think addition already gud

does that mean poly is auto good?

[marsh walrus]

have you checked associativity?

[marsh walrus]

theres no reason to approach the problem this way i dont think

well, it seems like this will be pretty easy just doing it directly

[covert berry]

C, A, Id, Inv

those are the req rite

[marsh walrus]

im working off the set on wikipedia lol

[covert berry]

lol

[marsh walrus]

i dont have my old algebra notes here

[covert berry]

it also ahs to be closed i think

[marsh walrus]

sure

[covert berry]

but i think thats obvious cuz 2 addition of polynomial degree 2 always ghives polys degree 2

?

[marsh walrus]

maybe i can find a better collection

lol mathisfun has a page on groups

[covert berry]

Nice

[marsh walrus]

okay so closure

[covert berry]

I loved that page its simpler

[marsh walrus]

i mean i guess, which property are you struggling with

assuming youve done one of these proofs before

[covert berry]

i just want confirmation

from someone who actually knows 100%

so my reasoning for closure is it wilol be in the same set cuz adding 2 polynomials of the form also gets a polynomial of the same form?

[marsh walrus]

this is a place i think you actually use your old work

to argue that the reals are closed under +

[covert berry]

i think our prof said we can just say that

so all gud

[marsh walrus]

alright

[covert berry]

and it kinda makes sense

[marsh walrus]

then yea, why is your argument failing to convince you

[covert berry]

for associative

I tested the thing they said and i got 3ax^2+3bx+3c for both

[marsh walrus]

you shouldnt get a number

[covert berry]

cuz this is math

forever doubt

if no answekey in textbook

oh yeah it was bad wording

i wanted to make the associative property equal

which is basically saying i got the same val for both

[marsh walrus]

idk what youre writing but it might help to slow down and be clear

we now show the operation is associative. That is, we show that given some elements .... that the property ... holds.

[covert berry]

Agreed

So the arbitrary elements are ax^2+bx+c right?

Oh fuck

Does that mean I have to make more variables

[marsh walrus]

they took the subscript so im trying to think of how to name them

maybe well use p q and s

$p = a_1 x^2 + b_1 x + c_1$ and $q=a_2x^2+\dots$

[flat frigate]

jan Niku

[covert berry]

ohhhh

so this is wrong

[marsh walrus]

and desire to show that $p+(q+s) = (p+q)+s$

[flat frigate]

jan Niku

[covert berry]

cuz i used the same variable name for both right?

[marsh walrus]

i mean theyre not very general

right

these should be any polynomials

[covert berry]

cuz i set a and b to = ax^2+bx+c

so the only fix i need for this one is to change the variables for b and c?

[marsh walrus]

you need these to be any polynomials

[covert berry]

yea same thing

i think*

[marsh walrus]

you are reusing a for so many things here

[covert berry]

cuz if i keep the variable then it is limiting it

[marsh walrus]

its also nonsensiical because youre using a for both a polynomial and a real

[covert berry]

oh yeah ur right

[covert berry]

How does this one look

[marsh walrus]

youve just started right

but yea id do something like this

[covert berry]

this is my entire associativity

and then i just say they're equal

and thats done i think

[marsh walrus]

👀

[covert berry]

For the identity element, is it 0?

all these symbols are removing me intuition but i think this is the idea

its just the thing that when the operation is done to the other thing result is the other thing

[marsh walrus]

i dont know that you checking this way is like the best way to approach the problem

you should generate an idea of what you think the identity element is

then see if you can prove it

then if you have issues with your logic you know thats addressable

[covert berry]

Say A is a rando polynomial ax^2+bx+c. let e = 0 be identity element. then 0+ax^2+bx+c=ax^2+bx+c=A

it is also the same the other way around if you add 0 from the right

so thus, 0 is identity elkelemtn

[marsh walrus]

hmm? 0 isnt in P_2

[covert berry]

how do u know

[marsh walrus]

$P_2$ is the set of polynomials $ax^2+bx+c$ where $a,b,c \in \mathbb R$

[covert berry]

is 0 not a real number

[flat frigate]

jan Niku

[marsh walrus]

im just bein sure

0 doesnt look like ax^2+bx+c

[covert berry]

true

but it seems like the abc in R allows it

maybe it defines P2 differently than the usual?

i see the requirement right there

abc have to be in the reals and im pretty sure 0 is in the reals

and otherwise they'd exclude it i think

[covert berry]

for inverse, do i just say it's the negative of the given polynomial?

[marsh walrus]

i just gave you feedback on this lol

[marsh walrus]

.

[marsh walrus]

these dont look careful enough tbh

you need to show that the identity commutes

also youre still using 0 as just a number

you should be explicit

[covert berry]

0x^2+0x+0 ?

[marsh walrus]

yes

[covert berry]

wdym by commute?

commute to work?

[marsh walrus]

youve written it there

the commutative property

here, lets say for natural numbers

[covert berry]

1,2,3...

[marsh walrus]

yea lets do integers lol

[covert berry]

...-1,0,1,2,3...

[marsh walrus]

An identity element must satisfy $e \cdot b = b \cdot e = b$ for all $b \in \mathbb Z$.

[flat frigate]

jan Niku

[covert berry]

Agreed

Oh shit so i just say that instead of implying it

[marsh walrus]

The identity element is 0. We must check that 0 commutes with $b$, e.g. that $b\cdot e$ = $e \cdot b$

[flat frigate]

jan Niku

[marsh walrus]

Clearly, $b+0 = 0 +b$ by commutativity provided by the integers.

[covert berry]

the wordiness makes me want to jump off a roof

[flat frigate]

jan Niku

[marsh walrus]

well you arent being clear

[covert berry]

That's one word I agree with

It's clear

[marsh walrus]

you cant prove something by obviousness

math is littered with problems that seem obvious and clearly almost certainly true but remain unproven

[covert berry]

So basiclaly my thing makes sense? I just need to be more formal?

[marsh walrus]

im just saying its not about having an answer

thats the beginning of the work

not the end