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[pliant dune]

then you need properties about cubics being divisible by primes

[thick lynx]

Is there any easy way or is this the easiest?

the problem is

[pliant dune]

there might be another way, this is just the way I know

[pliant dune]

to be more specific if prime p divides a^3 then p^3 divides a^3 too

[thick lynx]

Why is this?

[pliant dune]

(assuming a is an integer) if you factor a into primes, say $a = p_1 \cdot p_2 \cdot ... \cdot p_n$ then $a^3 = p_1^3 \cdot p_2^3 \cdot ... \cdot p_n^3$ so if p is one of the factors of $a^3$ it must be one of the $p_i$ and then you have $p_i^3$ in the product

[ocean seal]

Tom

[thick lynx]

if p is one of the factors of a³ it must be one of the p_i
Oh, for some reason I was thinking something along the lines of "cubing a could create a new prime factor" but that can't be, since it just cubes all the existing ones and factors with the exponent 0 will stay 1 since the exponent stays 0

thx

$$\frac{4}{5} = \frac{a^3}{b^3}$$
$$4 * b^3 = 5 * a^3$$

[ocean seal]

Kepe

Kepe

[thick lynx]

But now what?
4 * b³ is not a prime and 5 * a³ is only a prime if a = 1, no?

[pliant dune]

(using | to mean divides) if you have 5 | 4*b^3 then either 5 | 4 or 5 | b^3

obv it's not 5 | 4 so you get 5 | b^3

[alpine sable]

(usually the notation is the other way round a|b if a divides b)

[pliant dune]

(mb I haven't used it in a while)

(spoilers if you want to work it out yourself) then there's another small step that ||once you have 125 | 4b^3 you get 125 | 5a^3|| and then you've shown ||a^3 and b^3 have 5 as a common divisor so a and b have 5 as a common divisor||

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[vale wigeon]

apply linearity

$T(4v_2 + 6v_3) = 4T(v_2) + 6T(v_3)$

[ocean seal]

Ann

[vale wigeon]

linearity means the definition of a linear transformation.

this is a concept so basic that it should be burned into your head during a linear algebra course.

like

it cannot be stressed enough just how important the concept of a linear transformation is

literally ALL of linear algebra centers around it

i wrote out the application of linearity for your problem up there.

the rest is arithmetic. and i will preemptively say that no, you do not need wolfram alpha to do it.

it is basic arithmetic

well then why don't you have numpy do this for you too

read the problem!

Define the linear transformation T: R^3 -> R^3 by T(x) = Ax.

you are asked to compute T(u) and T(v), where u and v are given to you.

[lone heart]

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[oblique pier]

What do you guy think

[vale wigeon]

,w evaluate 1/(2a+b) + 1/(2b+c) + 1/(2c+a) for a = 5, b = 5, c = 1/25

[ocean seal]

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[fallen verge]

I would multiply by the product of the three denominators and see what cancels

[vale wigeon]

this is false @oblique pier

[warm zephyr]

bring to a common denominator first

[oblique pier]

[thick star]

,w evaluate 1/(2a+b) + 1/(2b+c) + 1/(2c+a) for a = 2, b = 2, c = 1/4

[ocean seal]

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[oblique pier]

try when a=b=c=1

[vale wigeon]

sure you might get 1 in that particular case so what

[oblique pier]

Yeah that what i need

[vale wigeon]

your problem states: Given that a, b and c are arbitrary real numbers with abc = 1, prove that (your expression) = 1.

[fallen verge]

So its not prove, its solve

[vale wigeon]

and this statement is just false.

[thick star]

that isnt what your expression said.

[vale wigeon]

but it is NOT true that abc = 1 implies (your thing) = 1.

simple as that.

[oblique pier]

kk

[vale wigeon]

what'd you need this for anwyay

anyway*

[fallen verge]

if a=b=c=1 it is true tho lol

[oblique pier]

Im solving an inequality

"=" happends when a=b=c=1

[vale wigeon]

show us the inequality.

http://xyproblem.info/

[fallen verge]

Oh that is very different

[oblique pier]

wait

Find S maximum value

[fallen verge]

,rotate

[vale wigeon]

so in fact you want to maximize S subject to abc = 1 (and presumably a, b, c > 0)? @oblique pier

[oblique pier]

yeah

[vale wigeon]

ok

[oblique pier]

since it said a,b,c are real positive number it said

Im almost done

[oblique pier]

i just need this

[vale wigeon]

well this is FALSE!

[oblique pier]

[vale wigeon]

what is this supposed to mean.

is this supposed to mean "haha you're funny but i refuse to actually believe you"

[oblique pier]

i think im wrong somewhere along the way

sorry

let me check a again

[vale wigeon]

yeah you're wrong in thinking that your original statement is true despite now getting hit with TWO counterexamples to it

[oblique pier]

oh

Sorry

[vale wigeon]

does anything interesting happen if we just trudge through the grueling algebra of adding these three fractions together

,w simplify 1/((a+1)^2+b^2+1) + 1/((b+1)^2+c^2+1) + 1/((c+1)^2 + a^2 + 1)

[high rapids]

Have you tried solving LHS?

[vale wigeon]

why are you still indulging OP in attempting to prove this FALSE FUCKING STATEMENT

ok so the short answer is nothing interesting is going to happen if we do it the way i had WA do just now.

[thick star]

i can't lie. some of the ignorance over here at the math server gets to my head.

[oblique pier]

alright

[vale wigeon]

i kind of want to try and do some geometry here, though admittedly i'm not sure how much good thatll do us either.

convinced theres some stupid trick that either isnt obvious until you know it already or some CRAZY geometry with like a hundred points on the diagram that youre just expected to magically know to do

[ocean seal]

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[vale wigeon]

just found a typo in my WA input

not that it changed anything

but still

let u = (a+1, b+1, c+1), p = (0, 1, c), q = (a, 0, 1), r = (1, b, 0). then S = 1/|u-p|^2 + 1/|u-q|^2 + 1/|u-r|^2.

and also p+q+r = u.

not sure how much that helps at all.

it probably doesn't.

not obvious yet how abc = 1 ties into it either.

@oblique pier how much punishment are you facing for failing to solve this

[oblique pier]

Im self studying

So yeah free style

[vale wigeon]

right

does the book give anything resembling an answer key or is it just "lmao figure this shit out we're gonna tell you fuck all"

because honestly im curious how they expect this to be solved

[oblique pier]

Ann, Do you think we can use Cauchy Inq here?

[vale wigeon]

if you can find a way to apply it then i don't see why not.

but tell me how you're going to apply it.

@oblique pier i'm not sure if you meant "do the Math Gods™️ allow the application of cauchy's inequality here" or "is there a good way to apply cauchy's inequality here"

...actually, what do you mean when you say cauchy's inequality? do you mean AM-GM or cauchy-schwarz?

[oblique pier]

i meant AM-GM

[vale wigeon]

but tell me how you're going to apply it.

[lone heart]

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[lone heart]

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[alpine sable]

is my approach correct and what is the next step !

[real gazelle]

why not just treat this like single variable calc

hm wait

okay the thing I would do next is multiply the numerator and denominator by xy to clear the fractions

that will kill the nested fractions

so $$-\frac{xy^2e^{xy}-y}{x^2ye^{xy}-x}$$

[ocean seal]

Eric Tao (he/him)

[real gazelle]

and then I'm not sure you can do anything to simplify it after that

or hm there's gotta be some way to simplify this

$$-\frac{y}{x}\frac{xye^{xy}-1}{xye^{xy}-1} = -\frac{y}{x}$$

[ocean seal]

Eric Tao (he/him)

[real gazelle]

yeah you can simplify it down to -y/x

[alpine sable]

hmmm

got it

thank you

[real gazelle]

no problem!

[alpine sable]

.close

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[livid raven]

Im calculating the surface between this 2 lines and I just wanted to know if I’m on the right path

Before I insert the answer, and also can I get rid of the natural logs since all the answers have them, or is that only in a function ?

[tardy rapids]

1. You skipped the whole integration part
2. No you cannot remove the logartihms just like that

$\int ln(x) dx = (ln(x) - 1)x + c$

[ocean seal]

5lttrz

[livid raven]

Oh true, I did do that : / thanks for noticing, but does it look right up till then?

[tardy rapids]

Yes its fine until there

(Also small pet peeve please write the dx)

[rich cargo]

is ln(x+6) the blue curv?

[livid raven]

Yeah

[livid raven]

Im kinda self leaning integrals, that goes after the ln(7x+5) right?

[rich cargo]

$\int (toBeIntegrated) dx$

[ocean seal]

Aranea

[rich cargo]

if you talk about the dx

[livid raven]

Ah ok, thank you, so much to learn haha

.close

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[kindred zinc]

Hey gyus, could someone help me find x?

[rich cargo]

there will be a lot of x

[devout lantern]

you sure you wrote it correctly ?

[kindred zinc]

yeah

that whats on the paper

but I have no idea even where to start on this one

[weary wyvern]

try to turn the cos into sin

[devout lantern]

I dont think you can solve it without using computer

[kindred zinc]

yeah but not even that helps enough

[languid portal]

You could graph it to find some points I guess

[weary wyvern]

i think you can

[slender gull]

Convert cos into sin of sin into cos.

For starters.

Squaring creates extraneous solutions.

cos(A) = sin(pi/2 - A)

[livid raven]

Oh ok

[kindred zinc]

So i simply try andfind X from that formula?

[slender gull]

What formula?

[kindred zinc]

.

[slender gull]

Sure.

So let's convert everything into cos.

For the sake of simplicity.

Do it.

[kindred zinc]

Ok then

thank you

[slender gull]

You're welcome.

[kindred zinc]

.close

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[real field]

hey

[real field]

can someone help me with this

[fresh parcel]

both of them?

well the "alternating" part is (-1)^n

[real field]

yea, i am sort of behind on my work

[fresh parcel]

so everything thats not the alternating part is a_n

so what would that be?

[real field]

1/n^9

?

[fresh parcel]

correct

[real field]

then the limit of that would be 0?

[fresh parcel]

yes

and we can recall that the alternating series test has one more condition

which is that a_n must be a decreasing sequence from [n, inf)

[real field]

so less than or equal to?

[fresh parcel]

not equal to

strictly less than

[fresh parcel]

huh

yeah im stupid

sorry

[real field]

dude you are good

thanks

[fresh parcel]

each term must be less than the previous one

[real field]

and the series is convergent i assume

[fresh parcel]

yeah

its convergent

[real field]

.close

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[brazen summit]

why does this not work in a graphing calculator as it is supossed to [x,x,x,y]=[2,6,4,12]{[4,4,13,2]<[y,y,y,x]<[12,12,14,6]}

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@brazen summit Has your question been resolved?

[vale crag]

what are you trying to do here?

[lone heart]

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[dapper python]

Can someone explain why it's B?

[fresh parcel]

it is not necessary to know f(a) and g(a)

its necessary to know f(x) as x approaches a and g(x) as x approaches a

[placid zinc]

In fact, it isn't enough to know f(a) and g(a), unless you know that f and g are continuous (at a)

[dapper python]

though us knowing f(x) as it goes to a isnt it f(x) - f(a) / x - a? in that case dont we still need f(a)

like dont we need to know what we dividing on i.e. g(a)?

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[alpine sable]

this is the correct question

<@&286206848099549185> pls

[lone heart]

@alpine sable Has your question been resolved?

[alpine sable]

nobody can help me

[lone heart]

@alpine sable Has your question been resolved?

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@alpine sable Has your question been resolved?

[true pulsar]

I recommend finding the parametric form of the line you're given, and finding a line parallel to that which passes through the given point. It's easier to conceptualize

[real gazelle]

do you still need help

[lone heart]

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[rich cargo]

How can I search for my name on this server. I want to see the help i got yday and when i search for my name it just shows that nothing was ever written by me x(

[normal ingot]

the messages may not be saved depending on how long ago that was

[rich cargo]

yeterday

*yester

nvm now it works

probably then i just had bad internet

.close

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[molten lotus]

Is this talking about a right triangle or a non right triangle

I also have no idea how to solve this

[fickle kelp]

Begin by drawing a line of 19 cm

[molten lotus]

Ok

[fickle kelp]

Then we have a line of 12 cm on one side of the line, and another of 15 cm on the other

There is one point where those can meet

[molten lotus]

Ok so it isnt a right triangle lo

[fickle kelp]

No

You can plot that point by drawing a circle with radius 12 at one tip of the line, and another of radius 15 on the other tip

Now, where those two intersect, is the third point on your triangle

[molten lotus]

Can you show me visually?

[fickle kelp]

Yeah

Like this

[molten lotus]

Bruh

Okay

[fickle kelp]

Where AB=19 cm

[molten lotus]

Ive never seen that before

[fickle kelp]

Should I explain why I'm doing this?

[molten lotus]

Yeah

[fickle kelp]

We have three givens: the lengths of the three sides of a triangle

So what we want to figure out is where the points of that triangle lie given those lengths

The first two don't really matter, as long as the distance between them is one of the sidelengths, in this case I chose 19 cm

[molten lotus]

Ok

[fickle kelp]

Now, how do we figure out where the third point lies with respect to the first two?

First let's call our first two established points point A and point B

[molten lotus]

Yeah

And thats 19cm

[fickle kelp]

Yeh

We know two things, our third point (let's call it C) has a distance of 12 cm to A, and a distance of 15 cm to B

Agree?

[molten lotus]

Yeah

[real gazelle]

do you know the law of cosines

if so then just use that

[molten lotus]

Ohh

Like the sin a / a = sin b/b

[fickle kelp]

I don't really know what they're asking for in this occasion

[molten lotus]

Theyre asking for what side has the smallest degree

[real gazelle]

no that's law of sines

law of cosines is like

c^2 = a^2 + b^2 - 2abcos C

or something like that

[molten lotus]

Yeahh I was gonna put that

Okay so I can just plug all that in

[real gazelle]

yeah all you have to do is use that formula

and solve for C

[fickle kelp]

Ow wait

Yeh

That makes sense

Oops

[molten lotus]

Okay

Wait so is drawing the circle necessary?

[fickle kelp]

Didn't know if they wanted an algebraic approach or a measured sketch

[fickle kelp]

Forget drawing everything

You can just solve it algebraically using the formula

[molten lotus]

Okay and then I can use the law of sines to solve for the angles?

[real gazelle]

law of cosines

[real gazelle]

yeah this thing ^

[molten lotus]

Im so confused cant I use this formula instead

The one where they solve for cos A

[fickle kelp]

Yeah that's the one

It's the same

[molten lotus]

Okay

So is a 19?

and b is 15

And c is 12?

[fickle kelp]

Yes

[molten lotus]

Ok

[fickle kelp]

If you then take the arccosine on both sides you get one of the angles of the triangle

[molten lotus]

The arcsine?

[fickle kelp]

Basically cos^-1

That one

[molten lotus]

Ohh okay

Umm I got 0.02222

I dont think thats right

[real gazelle]

make sure you are in degrees mode on your calculator

[fickle kelp]

Then if you shuffle around the values for a b and c (so for example a=15, b=12 and c=19) you get another angle of the triangle

[molten lotus]

Nvm I got 88.726655

Ohh okay

[fickle kelp]

Then you compare them and pick the smallest one

[molten lotus]

Ok I see

[real gazelle]

yeah you need the smallest angle

[fickle kelp]

Round it to the nearest integer and there you have your answer

[molten lotus]

Ok thanks!

.close

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[wintry vortex]

So I followed this until the last 2 steps

What I got down to was 8x^3/2+c

I dont get how they ended up with a srt

[real gazelle]

x^{1/2} is sqrt(x)

[wintry vortex]

shouldnt it be 3/2 though

wait, no fml i see what i did wrong

even now though shouldnt it be sqrtx over 1/2+c

.close

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[fleet peak]

I have written out functions already for transforming between (i,j) to (r, psi).

I know the 1/r comes from abs(dt/psi)

I simply don't really understand how I'm supposed to use the chain rule here.

[tight locust]

So i and j are the Cartesian basis here presumably

What do r and psi mean

[fleet peak]

Indeed, and r and psi for polar coordinates

[tight locust]

Weird

Use theta

[fleet peak]

Even in my study, they mix psi and theta between different subjects, very annoying.

[tight locust]

ok let's be absolutely correct here: that is not a psi, that's a phi

and the reason for sometimes seeing phi instead of theta for the angle from the x axis is because physicists are dumb

[fleet peak]

You're right it's a phi, I kept using psi in Quantum Mechanics so that's the confusion.
Anyways, I don't have an idea how to proceed.

[lone heart]

@fleet peak Has your question been resolved?

[tight locust]

https://math.stackexchange.com/questions/586848/how-to-obtain-the-gradient-in-polar-coordinates

this is a really helpful derivation

[fleet peak]

helps a lot, albeit quite complicated

[tight locust]

the trickiest thing to understand is what r and theta mean as unit vectors

[fleet peak]

Well assuming it's a ball on a cord rotating around the Origin, I would assume that the Theta is is the current momentum (cut the rope and the ball flies in that direction) , and the (negative) r is the force of the rope keeping the ball in a circle

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[alpine sable]

Hello, I need to ask something, typing it out

[raven rover]

Next time you can just type it out first and then send that as the first message

[alpine sable]

what if someone books the room while im typing 😛

So I am factoring out polynomials, finding Greatest Common Factor, GCF, and I don't understand why they didn't carry the x variable into the GCF like they did before.

Here they carried it, 2x²

But then

Why didn't carry the x here, to make it 5x?

oh wait, its staring at me in the face. the 15 doesnt have it

LOL

It has to be -all- participants right?

[raven rover]

Then you move to a different one

[raven rover]

Yes, all terms

[alpine sable]

ahh all terms, dang, i need to remember that, okay thanks

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[tawdry swallow]

I got question rq

[tawdry swallow]

I put this as the answer but I wanna figure why i couldnt just do (1m^2/1*10^3mm)

and not have it squared outside the parentheses

or is it that the mm wouldnt cancel out with the mm^2

[raven rover]

1 m^2 isn't equal to 1 * 10^3 mm

[tawdry swallow]

oh that's why

thanks

.close

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[rich cargo]

can someone explain me difference from Latex to tex?

[normal ingot]

https://tex.stackexchange.com/questions/49/what-is-the-difference-between-tex-and-latex

[rich cargo]

ty

!

.close

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[daring egret]

How do I simplify the fraction?

[daring egret]

Making it a mixed number doesn't help at all

[rapid sundial]

Convert the 1 to 3/3 and apply something similar to the denominator

Make it an improper fraction not a mixed number

[raven rover]

@daring egret ^^

[alpine sable]

A pile of cards facing down has 5 spades, 6 hearts and 2 diamonds in it. You randomly pick up 3 cards. What is the
probability that the three you choose include at least one spade?

[normal ingot]

#❓how-to-get-help

[daring egret]

Thank you

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[alpine sable]

Last 2 rows confuse me

Why can we say that a product of a nonzero vector with a matrix in a lower dimension is equal to zero

[lone heart]

@alpine sable Has your question been resolved?

[placid zinc]

@alpine sable
Row 2 → Row 3
Is just the distributive property

Common factor v.

[thick star]

youre basically applying the first two rows to the third row. like kaynex said, it's just algebra

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[carmine bronze]

I'm very confused

[elfin snow]

[last ether]

They technically did

It's just the questions

[elfin snow]

sorry the imagine didn't load for me

[last ether]

Discord moment

[carmine bronze]

Oh the relation I did on the side I don't think even means anything since this is using trig

just realized now

[last ether]

Yeah

[lone heart]

@carmine bronze Has your question been resolved?

[carmine bronze]

<@&286206848099549185>

[last ether]

Note how the ladder side can be two triangles

[carmine bronze]

yes

[carmine bronze]

yes

the angle at the bottom left there should be the same as the other angle

[last ether]

Let the ladder be split by sectors A and B:

$$\sin{(θ)} = \frac{24}{A}$$
$$\cos{(θ)} = \frac{6}{B}$$
$$L(θ) = A + B$$

[ocean seal]

Umbraleviathan

[last ether]

Solve for A and B

Plug in

@carmine bronze

[carmine bronze]

Trig wasn't exactly my highest grade

[last ether]

This is just rearranging variables

[carmine bronze]

oh

oh

[carmine bronze]

just like this then

[last ether]

Yeah

But then you'll eventually have to take the derivative of make your life easier

[carmine bronze]

Yeah I keep forgetting them

[last ether]

$$L(θ) = 24\csc{(θ)} + 6\sec{(θ)}$$

[ocean seal]

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

Umbraleviathan

[lone heart]

@carmine bronze Has your question been resolved?

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[stable night]

hi, can someone explain this?

[stable night]

im not entirely sure about multisets as well

[lone heart]

@stable night Has your question been resolved?

[stable night]

<@&286206848099549185>

[lone heart]

@stable night Has your question been resolved?

[stable night]

<@&286206848099549185>

[prime badge]

so the store has 4 kinds of snacks you want and you want to buy exactly 6 for some reason

then you have 84 ways to buy 6

[stable night]

hmm can we briefly talk about multisets first?

is it right if i say a multiset is just a set with duplicates

like {1,1,2,4,4}

[prime badge]

yes

if a normal set is like (1 A, 1 B, 0 C, 0 D, 1 E), a multiset can have more than one

the order is not a part of it

[stable night]

ah i see

so (1A, 1B,0C, 1A)

is a multiset

[prime badge]

no i meant 2A

[stable night]

oh like 2 As'

[prime badge]

that's just something broken and not a multiset

yes

[stable night]

ohhh lol

okay i geddit

okay so the formula above is implying that

i have a multiset and i want to take items from it with replacement

[prime badge]

no, it's talking about making a multiset

the amounts don't matter "in the source", you can;t run out, it's just a set

[stable night]

ohhh i see

okay so lets use ur example

the supposedly infinite shop has 4 kinds of snacks and i want to buy exactly 6

[prime badge]

yes

[stable night]

so n = 4 and r = 6

[prime badge]

yes

[stable night]

then by putting in the formula will give me 84

okay i get how the formulas used

but intuitively i dont get

why is it r + n - 1 thou

[prime badge]

like if you get 3 of B and 3 of D it maps to this string |***||***

0, then 3, then 0, then 3

any combinations would be like that, it would be 3 "bars" and 6 stars mixed together

so you have 9! / 6! / 3! choices

[stable night]

ohhhh

makes sense

[stable night]

so this is the intuition but when i solve questions i can just use the formula

[prime badge]

the wording can go in the opposite direction "you have 4 charities you like and you want to donate 6 oil tankers"

you're giving not buying, but it's the same calculation

[stable night]

i see

i get

tysm!

.close

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[native cloud]

What's the question

[hard remnant]

which one is opposite and the hypotenuse

[native cloud]

Hypotenuse is the longest side

[hard remnant]

so it's c

[native cloud]

Yes

[hard remnant]

and it's the same as opposite right?

[native cloud]

Adjacent is the shortest side

The last side is your opposite

[alpine sable]

Aren't adjacent and opposite sides with reference to some angle
Adjacent can be larger than opposite

[native cloud]

Really?

[hard remnant]

so a is opposite
b is adjacent
and c is hypotenuse?

[native cloud]

I wasn't taught like that

[alpine sable]

Okay nvm angle isn't given here, go with what airtoastie says

But the question is a bit incomplete

[native cloud]

side 'a' seems shorter

[alpine sable]

Like if you take angle B, adjacent would then be a
But if you take angle A, adjacent would be b

Trigonometry basically

[hard remnant]

@alpine sable the angle I'm talking about is C

[alpine sable]

So your reference angle is C?

In the question?

[native cloud]

Go with whatever your textbook/teacher says

[hard remnant]

yea

[alpine sable]

Then longest is hypotenuse, c
Opposite side to angle C is c

And next time please post the full question

[hard remnant]

that explains why the sin of 90° is always 1

[hard remnant]

sorry about that

@native cloud you could find a triangle where the opposite is smaller than the adjacent

[native cloud]

Then those sides changes

[alpine sable]

No

You declare those sides with reference to some angle

[hard remnant]

if B was 60° and A was 30° so the opposite of 30° will be smaller than the adjacent

[alpine sable]

Exactly

[native cloud]

So sin 60 is opposite /hypotenuse?

[hard remnant]

yea

[alpine sable]

sin(theta) in general is opposite/hypotenuse

sin(60) is √3/2

[native cloud]

Hmm okay, thanks for the recap

[alpine sable]

You may close this help channel if your doubt is cleared @hard remnant

[hard remnant]

when adjacent > opposite then the cos > sin, and cos 30 is sqrt(3)/2 and sin30 is 0.5

may I ask another question

[hard remnant]

is doesn't seem that if we devide b by c we will get 0 if C is the angle, like we know that cos 90 =0

[alpine sable]

For that

Picture a triangle

With angle theta

[alpine sable]

So ABC triangle, A = theta, C = 90°

Okay so

So picture

[hard remnant]

ok

[alpine sable]

Theta slowly approaching 90°

Side C wil come closer and closer to A

And when it's 90°

Side AC disappears

Point A becomes point C

Implies AC length becomes 0

So cos(theta) as theta approaches 90°, approaches 0
It becomes 0/hypotenuse at 90°

[sharp sail]

its better with the unit circle method

[alpine sable]

Yes

But idrk if they know the unit circle

[hard remnant]

what's unit circle?

[sharp sail]

oh

[alpine sable]

Well

You'll learn it soon

[hard remnant]

isn't that circle that has some famous angles and their sin and cos?

like 30 45 60 90 180

[sharp sail]

no lol

it basically defines the sine and cosine functions

[alpine sable]

It's a circle that introduces relation between angle and arclength, a new unit of angle, radians, on which you define those trig functions

[hard remnant]

so what's this circle then

[alpine sable]

Yes that's the unit circle

[sharp sail]

yea this is the unit circle but dont think this is a kind of chart drawn to remember values of trig functions

[hard remnant]

even if I do I can't

[sharp sail]

its just a circle centered at origin

With radius of 1 unit

and triangle of angle theta is inscribed inside the circle

[hard remnant]

oh I understand

.close

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[olive juniper]

$\frac{-5x}{-5} < \frac{9}{-5} = x > \frac{-9}{5}$

[ocean seal]

minto7

[olive juniper]

can someone tell me why the answer is \frac{-9}{5} instead of \frac{9}{-5}?

[gray isle]

full context please?

[olive juniper]

i was doing this

$-9x-4<-4x+5$

multistep inequalities

[ocean seal]

minto7

[gray isle]

what was the step after that

[olive juniper]

$-5x<9

\frac{-5}{-5}<\frac{9}{-5}

x>\frac{9}{-5} or \frac{-9}{5}?$

the hell was that texit

[gray isle]

-5x < 9
is valid
you made an error when attempting to divide both sides by -5

[olive juniper]

its -1.8 rite?

so it doesnt matter if its -9/5 or 9/-5?

[gray ingot]

this is true

[gray isle]

$-\frac ab = \frac{-a}{b} = -\frac{a}{-b}$

[gray ingot]

last one isnt true

$-\frac ab = \frac{-a}{b} = \frac{a}{-b}$

[ocean seal]

Swas.py

[gray isle]

fk sry typo

[olive juniper]

big thanks guys

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[violet panther]

Hi how can I solve this???

[clear stump]

MATH

[gray isle]

try expressing that equation in terms of a single trig function'

[thick star]

can you divide both sides by a common trig function to yield tan x?

[violet panther]

I tried until here

But I don’t know the √3

[thick star]

$\sqrt{3} \tan x = 1
\
\implies \tan x = \frac{1}{\sqrt{3}}$

[ocean seal]

Yash

[violet panther]

Ohhhh

I got it

Thank you so much @thick star

.close

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[alpine sable]

Would someone please tell me, how x is defined in the equation of y + z = 3?

[alpine sable]

.close

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[alpine sable]

Hi guys

[alpine sable]

someone can help me with this?

6. Find the following double angles
   a) Sin = − 𝟏/2

b) Cos = − 𝝅/4

c) Tan = - 𝟐/3

i don't understand

[vale wigeon]

neither do we

can you post a picture of the problem?

[alpine sable]

yeah

give me a moment

i need help with this

[vale wigeon]

...

[alpine sable]

i don't understand it

[vale wigeon]

wtf lmao

[alpine sable]

you don't understand spanish?

[vale wigeon]

no it's not that

i understand enough Spanish to read the words

[alpine sable]

oh

[vale wigeon]

but the notation makes no sense still

[alpine sable]

why?

[vale wigeon]

"sin=-1/2" is nonsense

[alpine sable]

why?

Lmfao

[vale wigeon]

sin is a trigonometric function. what needs to come after it is an angle, and here you have nothing

it's sin(something), not just sin

[alpine sable]

oh ok

thanks

[lone heart]

@alpine sable Has your question been resolved?

[alpine sable]

hey

@vale wigeon you can help me with something more?

[alpine sable]

the next excercise

[vale wigeon]

next one doesn't make any sense either 🤡

"Calculate the following angles..."

[last ether]

Wtf

Lmao the left hand side makes no sense

The trig functions have no argument

[lone heart]

@alpine sable Has your question been resolved?

[lone heart]

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[frank plover]

I can't see how the yellow follows from the former, can anyone flesh it out a bit more 4 me?

[frank plover]

Its a proof on the transpose of a product matrix equalling the transpose of the multiplicands in reverse order (A.B)^T = B^T . A^T

[alpine sable]

the transpose of $(c_{i,j})$ is simply $(r_{j,i})$ where $r_{j,i}=c_{i,j}$

[ocean seal]

AimaneSN

[frank plover]

Oh thanks alot! for some reason i kept seeing the indices of a and b in the sum switched alsp

thats where my confusion came from

thanks !!

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[gilded gazelle]

I have a formula to calculate levels by the current xp of an user.
I do that with a while loop, which is slow...
I want to try to improve it, so it's one line, self updating and therefore faster.

The formula:

# This is the standard xp for level 1. It's used for the next calculations
level = 1
xp = 75

# Level 2 needs 250 xp
# 75 is the max xp of the current level (which is now 1)
# 2 describes the next level to calculate the xp of
# level_2 now holds the value 250, which is the xp needed to get level 2
# (50 + ( max_xp_of_level / level )) * next_level

level_2 = (50 + ( 75 / 1 )) * 2 = 250
level_3 = (50 + ( 250 / 2 )) * 3 = 525
level_4 = (50 + ( 525 / 3 )) * 4 = 900
...

Now I don't want to have a loop, as this formula only calculates the next level and starts from level 1...
This is very slow, because with level 1000 it should take a lot of time to complete...
Is there a way to make it into one line, where the max_xp_of_level and level are getting "updated" automatically?
I tried to make multiple loops in this formula itself, but I think that's the wrong way...

[mortal trellis]

exp for level n = 25(2n^2+n)

does that help you?

I'm not yet sure how you are calculating stuff

[gilded gazelle]

You did understand it actually very well.
Thanks a lot, that works perfectly 👍

I will close it in 2-3 minutes

.close

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[slate kayak]

In a vase there are: 3red, 6blue, 7white marbles. A girl picks 3 marbles.
A) Calculate the chances of her picking 0 blue, 1 blue, 2 blue, 3 blue.

B) Sum up the 4 chances together. Why can you know the results beforehand?

[slate kayak]

So i have already done question A

I only have trouble with answering question B

I summed up all the 4 chances together and i got the number one (1)

We can know the results beforehand because?
Idk the answer my guess is because every chance together equals 100%?

[real gazelle]

Yes

[slate kayak]

Thanks, have a nice day!

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[neon girder]

I couldn't reach the answer.

I tried to use double-angle identities but it just got more complicated.

[real gazelle]

wait what's the question

is it the integral at the bottom

[neon girder]

yup

[real gazelle]

oh alright

[neon girder]

sorry I thought the arrow was obvious

[real gazelle]

you're all good

try substituting 1/(2x) = sin theta instead

2x is the hypotenuse of the triangle

1 is one of the legs

and then the other leg is sqrt(4x^2 - 1)

[neon girder]

yeah but I don't think I can do that before integrating

[real gazelle]

what do you mean

[neon girder]

i should not substitute instead of any angle before integrating

or can i ?

[neon girder]

ok you know what let me try it

[real gazelle]

yeah you should substitute before taking the antiderivative

[neon girder]

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[tacit ferry]

hello, i'm currently making a videogame, and in it i want to make it so that you have a gun that shoots lasers in random directions, but confined in a circle, so it can only go inside the circle, how could i calculate what the maximum angle (as a 3d vector) of a line should be based on the diameter of the circle?

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[gusty heart]

I need help linearising my data

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@gusty heart Has your question been resolved?

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@gusty heart Has your question been resolved?

[real gazelle]

do you know what sort of curve should fit this data?

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[gusty heart]

A parabola seems to fit

[real gazelle]

okay then try square rooting the y value

[gusty heart]

okay

[gusty heart]

I have this now

does that count as linearised?

[real gazelle]

that doesn't really look linear to me

[gusty heart]

same

[real gazelle]

where did this data come from?

[gusty heart]

I was on a physics site and here's what I got:

[gusty heart]

from a physics experiment

[real gazelle]

what kind of experiment

what pattern would you expect the data to follow?

[gusty heart]

I'm expecting a linear pattern

Negative correlation

[real gazelle]

wait but your original data are clearly not linear

[gusty heart]

yh

Ig I'll just use what I have and in my evaluation explain that something is likely to have gone wrong

[potent garnet]

Eric Tao in all the help channels at once...

[real gazelle]

yeah if you're expecting the original data to be linear and it's not, then there's no point to linearizing it

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@gusty heart Has your question been resolved?

[lone heart]

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[spare canopy]

do you need to apply chain rule here bcs2e = 2 x e

[real gazelle]

are you trying to take the derivative of h(x)?

[spare canopy]

ye

but i used the chain rule but apparently you dont need to

[real gazelle]

oh okay

no, here you can just use the fact that derivative of 2e^x is the same as 2 derivative of e^x

sometimes people call that the multiplication by a constant rule

[spare canopy]

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[spare canopy]

.reopen

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✅

[spare canopy]

.close

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[humble barn]

If I know b and L, I can find K no problem.

But I don't know how to write an expression for K given b and L.

[lone heart]

@humble barn Has your question been resolved?

[real gazelle]

can you explain how to find K given b and L without using specific values for b and L

like in general what is your process for finding K

it's just the sum you wrote right?

oh I see you've written the sum like a for loop lol

are you a programmer

[humble barn]

Yes, so the example I made up in the question was for specific values of b and L. I don't know how to do it without specific values of b and L.

[real gazelle]

okay awesome

[humble barn]

I can write basic programs, but not professionally no.

[real gazelle]

okay so first of all we want to write the sum without the x+=L bit

which is pretty simple

so if you expanded this out it would be 0^2 L + L^2 L + (2L)^2 L + ... + (b-L)^2 L

so the way to write this sum would be $$\sum_{n=0}^{(b/L) - 1} (nL)^2 L$$

does that make sense

wait small typo

[ocean seal]

Eric Tao (he/him)

[real gazelle]

okay cool does that make sense ^

I basically rewrote your sum without the x += L bit

[humble barn]

n increment by 1 each time right?

[real gazelle]

yep

[humble barn]

I see. How do I evaluate this sum?

Is there a way to evaluate it?

[real gazelle]

and then next we can simplify a bit to get $$\sum_{n=0}^{(b/L) - 1} n L^3$$

[ocean seal]

Eric Tao (he/him)

[real gazelle]

the L^3 part doesn't vary so we can take that out

and then next we can simplify a bit to get $$L^3\sum_{n=0}^{(b/L) - 1} n$$

[ocean seal]

Eric Tao (he/him)

[real gazelle]

is that good

like does that make sense

[humble barn]

Yes this makes sense so far.

[real gazelle]

cool

now are you familiar with arithmetic series

[humble barn]

If I know what b is, can I evaluate your expression without using the summation symbol?

[real gazelle]

yes

is the expression n(n+1)/2 familiar to you?

[humble barn]

Yes, that's the sum of f(x) = x.

[real gazelle]

yup

so now we apply it here

[humble barn]

From google, so it's the bottom one?

Wait, it's the top one.

[real gazelle]

$$L^3 \frac{(b/L - 1)(b/L)}2$$

[ocean seal]

Eric Tao (he/him)

[humble barn]

Because L is fixed.

[real gazelle]

yeah it's the top one

and then if you want to get rid of the nested fractions you can simplify a bit

[humble barn]

Nice, now I can divide your answer by 1/3 * b^3 which was the exact integral.

[real gazelle]

$$\frac{bL(b-L)}2$$

[ocean seal]

Eric Tao (he/him)

[real gazelle]

wait hm

lemme check if I made a mistake somewhere this seems wrong

oh oops

it should be the sum of n^2 not n

[real gazelle]

which means it's the middle formula here

[real gazelle]

mb I simplified this expression wrong

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[humble barn]

.reopen

[lone heart]

✅

[humble barn]

Oh ok I see.

[real gazelle]

but so that means it's $$L^3 \frac{(b/L - 1)(b/L)(2b/L - 1)}{6}$$

[ocean seal]

Eric Tao (he/him)

[real gazelle]

and then simplifying that is pretty sttaightforward

$$\frac{b(b-L)(2b-L)}6$$

[ocean seal]

Eric Tao (he/him)

[real gazelle]

that's correct now ^

[humble barn]

And after dividing by 1/3 * b^3 I get

$$\frac{(b-L)(2b-L)}{2b^2}$$

[ocean seal]

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

[humble barn]

If L == 0 then answer = 1, which makes sense.

If L > 0, answer is less than 1/3 * b^3 which makes sense.

Wow you're a genius sir, thank you very much!

I'll now go and apply this to other functions to practice.

.close

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[tranquil yoke]

I have ruled one out but to be honest i have no idea

[raven dagger]

Using the formula
1+ tan²x = sec²x
You can figure out if d is true or not

[tranquil yoke]

ty

[real gazelle]

tbh the easiest way is just to plug in values of x and see if you get a false statement

for example plug in x = pi/4 to each one

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[alpine sable]

can somebody check if this irght

[ocean oyster]

The first step should be last

Have u tried following the steps to draw the bisector of A?

[alpine sable]

yes i have

[ocean oyster]

Wait no

I meant put the first step last

And keep the others in the same order

[ocean oyster]

Alr how’d it go?

[alpine sable]

im confused

it didn;t work

do you mean like this

[ocean oyster]

Let’s work off of the first pic

[alpine sable]

k

[ocean oyster]

Yes

Great

Have u tried following that order?

[alpine sable]

wait a sec

ill do it

yup it worked

thx

[ocean oyster]

Ur welcome!

[alpine sable]

how do i unoccupie

.close

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[peak silo]

If I was trying to solve for dy/dx for the top equation, would this be how I do it or have I made a mistake

[waxen flame]

That last line should be 1/(2y), not (1/2)y.

[peak silo]

Ah, ok

Thank you

.close

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[wooden raven]

I’m trying to figure out a

Since the bottom triangle shares the same point as the top triangle wouldn’t the angles for the bottom be 30 45 and 115?

[lone heart]

@wooden raven Has your question been resolved?

[alpine sable]

yo

show me ur work

ur in p3.18 right?

[thorn vortex]

what angles are you specifically talking about?

[wooden raven]

Ye

I’m just trying to figure the coordinate at the top

[alpine sable]

u need to go step by step brad

first we start with the angle

[wooden raven]

So rn I’m trying to find cos 30

[alpine sable]

why u trying to find cos 30?

[wooden raven]

Should I just focus on 45?

I figured cause there’s another angle it’s important

[alpine sable]

u want to draw 30?

manually?

we start from the buttom

[wooden raven]

Whatever allows me to solve the problem lol

[alpine sable]

lets start from buttom

[wooden raven]

Ok

I get for cos 30 5sqrt3/2

Then for sin 5/2

[alpine sable]

why

u try to find

what cos 30 is

[wooden raven]

Don’t do that?

[alpine sable]

dont

i know why

but no need

trust me

[wooden raven]

Ok what should I do the 45 degree?

[alpine sable]

get a white sheet

ill tell u what to do

[wooden raven]

Ok

I know cos is adj/h

That and sin is opp/h

[alpine sable]

draw x and y

[wooden raven]

Ye I drew the graph

Basically just focus on 45 and ignore 30?

[alpine sable]

so

listen

we need to start with 30

how to draw 30?

draw a triangle which their side are all equal

ull get 60

right?

[wooden raven]

Yeah cause a triangle is 180