help-10

[limber torrent]

how do I make a linear function with these points?

the points in question are (0,210) and (2,68)

[scarlet locust]

can you find the slope between those points?

[limber torrent]

how would I do that?

[scarlet locust]

if the points are (x1, y1) and (x2, y2) then the slope is
$ m = \frac{y_2-y_1}{x_2-x_1}$

[warm shale]

Swas.py

[limber torrent]

I got 145/-2

[scarlet locust]

alright now you need to find where the line crosses the y axis

or in other words the y-intercept

can you do that?

[limber torrent]

yeah it's 210

[scarlet locust]

alright now you can use those 2 in the equation
y = mx + c, where m is the slope and c is the y-intercept

and that gives you the function

wait I think you calculated the slope wrong

its 142/-2 not 145/-2

[limber torrent]

ohhh

[scarlet locust]

so this would be your final equation

[limber torrent]

how would I solve this?

[scarlet locust]

nope there is no need to solve anything

that's your function

y = -71x + 210

where y is the the room temperature and x is the cooling time

[limber torrent]

oh

thank you!

[limber torrent]

wait

why is it 142?

[scarlet locust]

210-68 is 142

[limber torrent]

ah wait I accidentally put 65 instead of 68

on my calc

thank you for the help!

[obtuse pebble]

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[ripe island]

Hello, I just want to verify if this is how you would solve the given problem written on the yellow paper. Thanks

[fierce lagoon]

Not quite

Find the derivative of $x^2 +2$

[warm shale]

Umbraleviathan

[ripe island]

Ohh I see, how do I do that though? The examples we were given didnt explain how to T^T

[fierce lagoon]

Well, two ways

You can use your power rule

Or use the limit definition

I would use the power rule though

@ripe island does this make sense to you?

[ripe island]

I understand how I would use the second one buut the first one idk since Idk what h represents in the example given

[fierce lagoon]

h is a change in x

Essentially what the first one is saying is that "as the distance between x and h get really small (almost instantaneous), what is the slope between the two points?"

Where the distance between x and h is "infinitely small"

So h is a small change in x

But regardless, what would $F'(x)$ be

[warm shale]

Umbraleviathan

[ripe island]

Ohhhh I see I see, Thanksskyuuu

I'll go re-do my answers, starting by finding the derivative~ thankskyuuuu sooo much

[fierce lagoon]

Np

[ripe island]

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[fierce crater]

Hello

[fierce crater]

I have a question

I am counting intergals right now and have been wondering

I have a following result:

cos(y) = C / sin(x)

Can I put the trigonometric function of cos(y) inside its inverse function [cyclometric arccos()] in order to simply have y?

Like that:

y = arccos[C/sin(x)]

I thought it could be possible, since if f = g^-1 => f[g(x)] = x and g[f(y)] = y

[obtuse pebble]

@fierce crater Has your question been resolved?

[fierce crater]

<@&286206848099549185>

[obtuse pebble]

@fierce crater Has your question been resolved?

[fierce crater]

<@&286206848099549185>

[obtuse pebble]

@fierce crater Has your question been resolved?

[obtuse pebble]

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[wise oak]

A bag contains 4 red marbles, 9 blue marbles, and 5 green marbles. what is the probability of choosing a blue marble when one marble is drawn?

[wise oak]

9 blue marbles, 18 total.

oh

i really overthought that

lol thanks

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[vagrant compass]

for part b i get 2 equations, but it asks for THE equation (the mark scheme also only shows one equation)

on the left is the Matrix M (worked out from part a), and my 2 equations (only the red one is in the mark scheme)

heres the mark scheme

[obtuse pebble]

@vagrant compass Has your question been resolved?

[thick fog]

Not sure where you are getting that green line from. You can follow the linear transformation visually and see it isn't a solution

[vagrant compass]

this is my working

[thick fog]

I don't really have the patience to go through the work to see where your substitution is creating an extraneous solution. One of your steps along the way to creating the m^2 has made your calculation no longer equivalent.

If you set A = PQ, then Ax = x is equivalent to finding eigenvectors of A corresponding to eigenvalue 1. This only has one linearly independent solution, so you can't get two solutions like you did.

[vagrant compass]

no idea what any of that means

[thick fog]

You can see the green line is not a solution, take any point in Q3, Q reflects that point about the x-axis

then P rotates that point by pi/3

You don't end up anywhere near the same spot

[vagrant compass]

what is the reason for that?

[thick fog]

If you start on the red line, every point you start at, if you reflect about x and then rotate you end up in the same place

[vagrant compass]

ive used this method in the past, and its always worked

i usually have 2 invariant lines

why does this one only have 1

[thick fog]

It depends heavily on the P and Q

[vagrant compass]

should i use a different method?

or did i make a mistake?

wait a second i think i figured it out

i had one solution to be -(sqrt3)/3

but ti should of been

(sqrt3)/3

which would mean i have a c value

which means it wouldnt be linear anymore?

since the origin has moved

[thick fog]

The whole way you look at the problem is just not quite correct.

You are asked to find invariants of QP. Which means transforming any vector [x,y] by QP does not change [x,y] at all

So you have the equation QP[x,y] = [x,y]. Your notation from the start is already a bit wrong here because you are multiplying by x * y * on one side and x prime y prime on the other.

When you multiply the matrix on the left by that vector, it gives you two equations that must be simultaneously true, x = -1/2 x - sqrt(3)/2 y etc. You are instead equating -1/2 x - sqrt(3)/2 y to something that doesn't really make sense. I am honestly surprised you are getting an sensible answer at all

[vagrant compass]

its not x* and y*, i put x' and y' on accident and crossed it out

[obtuse pebble]

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[south inlet]

Can someone explain why they differentiated y and x? usually they only differentiate y. This is implicite differentiation

[short spruce]

uhh because it's implicit differentiation

you're differentiating with respect to x, treating y as a function of x

[south inlet]

i don't get it

is that not the same as keeping y? and only differentiating y

[short spruce]

i don't know what you mean by that

[south inlet]

because I saw videos on implicite differentiation, and they only multiply (dy/dx) by the terms that have y in them

[short spruce]

you're treating y as a function of x, you'll have a dy/dx whenever you need to take the derivative of y

[south inlet]

so

y becomes y(x)

[short spruce]

yes, you are treating y as a function of x

[south inlet]

ok give me a second let me try to solve it

i got the right answer

but I don't get why we should think of y = y(x)

[short spruce]

because you're differentiating with respect to x

[south inlet]

is it always that case?

[short spruce]

depends on what you're differentiating with respect to, if i understand your question

you want to find the derivative of y with respect to x, therefore you are treating y as a function of x

y=f(x)

[south inlet]

oh, so the question has to include which variable we want to differentiate?

[short spruce]

yeah

you want dy/dx, the derivative of y with respect to x

if it asked for dx/dy, it'd be the derivative of x with respect to y, treating x as a function of y [x=f(y)]

[south inlet]

ok yeah, i get it now

thx

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[inland thistle]

Hi, I understand how to do the criss cross method and then get the delta for x1 and x2

What I don't understand is supposedly the first step to this problem

[teal prawn]

x cant be 1.25, because of it is, you get an undefined expression (division by zero)

[inland thistle]

That makes sense, but why did we pick 2x-2,5 out of the equation, not the 4x-4 or 2x?

[teal prawn]

because 2x - 2.5 is in the denominator

4x - 5 evaluating to 0 is perfectly fine

[inland thistle]

What if we had an X in both denominators?

[teal prawn]

then you would have to check for both

i see an equation in the ques... not a func im not sure why they are finding domain, perhaps there is more to this?

[inland thistle]

The question is prove the solution's a prime number

It's in polish, so ignore the left

Like I understand how and that 7 is the solution, I just have issues with defining stuff at the beginning

Like I don't understand what R{1.25} means

I guess it means belongs to, but then I'd have no idea what it'd look like it belonged to more than one number

[teal prawn]

it says x belongs to R except 1.25

R \ {1.25}

[inland thistle]

Oh so that's why x1 can't work?

[teal prawn]

yeah

[inland thistle]

Makes sense ty

No more qs

.close

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[latent panther]

Can someone walk me through this?

[latent panther]

<@&286206848099549185>

[versed turret]

Do you know the integral of a power of t?

[latent panther]

I do not

[versed turret]

In general, $\int x^n \ \mathrm{d} x = \frac{x^{n+1}}{n+1}$

[warm shale]

castroploiin

[versed turret]

In your question, x is t

[latent panther]

So should I rewrite this but replace the x's with t's?

[versed turret]

I'm guessing you know how to substitute in limits?

[versed turret]

Apply this rule for in your question

[latent panther]

Yes I am pretty sure

[versed turret]

Ignore the 4 in your question, just apply the above rule on t^3

$\int af(x) \ \mathrm{d}x = a \int f(x) \ \mathrm{d} x$

[warm shale]

castroploiin

[versed turret]

Yep! Now does anything cancel?

[latent panther]

If I put the 4 back in it would cancel right?

[versed turret]

Yes.

And then you just sub in your limits

[latent panther]

so I just have t^4

[versed turret]

Right

[latent panther]

So what should I sub in again?

[versed turret]

The upper bound is x, the lower is 1

[latent panther]

Oh so do both of them right?

[versed turret]

You need to use both of them, if that's what you mean

[latent panther]

so if I sub in 1 I get 1 and x is x^4

so 2 answers?

[versed turret]

You sub in the upper limit and then subtract from that the same expression but subbed in with the lower bound

[latent panther]

wait what?

[versed turret]

So, $t^4 |^{t=x}_{t=1} = (x^4) - (1^4) = x^4 - 1$

The LHS is just me saying I need to sub those values in

[warm shale]

castroploiin

[latent panther]

So thats the answer?

[versed turret]

Yes.

You take the upper bound, sub that in

Then you take the lower bound, sub that in

And subtract lower bound answer from the upper bound answer

[latent panther]

Alright

That makes sense

Ill have to do more practice problems

[versed turret]

It's easier to think in terms of area

[latent panther]

Do you know of any resources I could use to get practice problems like that?

[versed turret]

Old exam papers should have plenty

https://www.physicsandmathstutor.com/maths-revision/a-level-aqa/integration/

[latent panther]

Alright sweet

Thank you for your help

I appreciate it

[obtuse pebble]

@latent panther Has your question been resolved?

[obtuse pebble]

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[onyx dust]

did my professor make a mistake here?

[drifting wraith]

on the last line yeah

[onyx dust]

trying to find the critical points for an optimization problem

but i dont think x^2 - 8x + 48 is factorable is it?

so im not sure how i would solve the problem

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[timid silo]

how would I solve this with this perfect square method

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

<@&286206848099549185>

[timid silo]

nobody knows?

[steel marsh]

You could take the derivative and solve for max value(s)

[obtuse pebble]

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[slim leaf]

Try to write it in vertex form by completing the square

[brittle blaze]

First derivative and set equal to 0

Or that

[drifting wraith]

also it should be half the root

physically, t when it hits the ground is double t at the max nvm it's not from ground level

[brittle blaze]

Very true. Id personally do derivative bc its only a polynomial

[drifting wraith]

but it says estimate instead of find

so i;ve no idea lol

[waxen flame]

Ok

@brittle blaze @drifting wraith

What do I do

[drifting wraith]

can you find first derivative?

[waxen flame]

1761/16?

What

[drifting wraith]

i've no clue personally

i mean i can't guess what it would expect you to do

[waxen flame]

Oh

I think I might be right

Nvm

[drifting wraith]

you can solve -16t² + 82t = 0

[brittle blaze]

?

[waxen flame]

41/8

[brittle blaze]

@waxen flame do you know how to differentiate

[waxen flame]

No

I wasn't taught that

[drifting wraith]

so that's when it returns to 5 meters

so take half that

[waxen flame]

20.5/8

[drifting wraith]

and find h(t) from that

that's the answer

[waxen flame]

How tho

[hushed comet]

@waxen flame do u know of any equations that help you find the x coordinate of where the y lies?

if not then here you go -b/2a

[waxen flame]

That's what it was

Now what

[hushed comet]

the answer to that needs to be plugged back into the function.

Because -b/2a gives you an x value. Specifically the x value on which the height can be found.

-b/2a gave x=2.5

make sense?

[waxen flame]

Yeah

So is 2.5 the answer

[drifting wraith]

it's 2.6

they rounded wrong

no that's t

plug this t in your function

[hushed comet]

No, your question is asking for the Max Value. Not where the Max Value occurs (2.5).

[waxen flame]

Ok

[drifting wraith]

2.6

[waxen flame]

110.04 I got

I plugged 2.6 in thr equation

[drifting wraith]

that's the answer

[waxen flame]

Ok thanks

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[waxen flame]

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[sterile pewter]

@ripe crag I tried it with three variables and it's wrong

fx = z, fy = 2y, fz = x

<9, 10, 1> dot product <2, -1, 2> = 10

[ripe crag]

hmmm
perhaps they do want the unit vector after all?

<9, 10, 1> • <2/3,-1/3,2/3> might work?

[sterile pewter]

Oh yes I just tried that and it works

thanks again!!!

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.reopen

[obtuse pebble]

✅

[sterile pewter]

you really saved me so much time thank you!

[ripe crag]

ah great!
yeah i think the convention is to take the unit vector, even though the direction will be the same and you'll just get some scalar multiple of the answer

[sterile pewter]

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[thorn ruin]

Anyone understand 47 b

[thorn ruin]

<@&286206848099549185>

[brittle blaze]

Do you know how to find normal force here?

[thorn ruin]

Yes i did 47a

[brittle blaze]

A is supposed to be friction

Thats parallel

[thorn ruin]

I am not a physician explain in normal people terms

[brittle blaze]

😂

Hol up

[brittle blaze]

so basically the perpendicular force you want is mg cos

and the parallel force is mg sin

so to make it not move you need mgsin in the other direction too

thats a

[obtuse pebble]

@thorn ruin Has your question been resolved?

[obtuse pebble]

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[charred latch]

hello

so i understand divergence theorem

i just need to know how to get the bounds for the triple integral

i tried doing it in spherical coordinates

heres my work

the volume is a half sphere with radius a

hold on is the bound for phi pi/2 and -pi/2

i am so stupid

lemme try that

that ddnt work

[obtuse pebble]

@charred latch Has your question been resolved?

[obtuse pebble]

@charred latch Has your question been resolved?

[charred latch]

wait

its 0 to pi/2

ok i got it

.close

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[vague condor]

I have this horrendous thing for homework and have no clue where to start. I know how to do the first one

its just

$\sum_{n=0}^{\infty} x^{n}$ i think

[warm shale]

beanbeanjuice

[vague condor]

but i have no clue how to differentiate both sides

[hybrid gull]

Power rule!

[vague condor]

hmm

uh

so

im guessing that is right?

im not sure

anymore

im lost

[brittle blaze]

This is correct

Its a known fact you shouldn’t have to derive

[obtuse pebble]

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[vague condor]

i give up 😭

[obtuse pebble]

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[brittle blaze]

I was gonna help my guy💀

[vague condor]

dw i dont think i'll ever get it

[obtuse pebble]

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[onyx dust]

why is there only one inflection point?

shouldnt there be multiple since f'' is changing concavity?

[civic zealot]

-2 and 8 are points of discontinuity

[onyx dust]

ohhh okay

gotcha

thanks!

.close

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[timid silo]

anyone has pdf with all trigonometric formulas with proof?

[timid silo]

https://lmgtfy.app/?q=All+trigonometric+functions+with+proof

loser

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[obtuse pebble]

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[final sky]

Is there a faster way of doing this?

[teal prawn]

(x + 2)(x - 2) = (x^2 - 2^2) = (x^2 - 4)
which would've reduced the number of factors to open

.. perhaps you could also factor out (x + 2) instead

[final sky]

So I shouldn't have factored that huh

[final sky]

?

[teal prawn]

this is good enough

[final sky]

Yeah but

This Sure will take my time in an exam

[teal prawn]

yeah.. well

with more practice, you'll get faster

[final sky]

Thanks ^^

.close

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[onyx turtle]

plz help

[onyx turtle]

i have a test in 10 minutes

[versed turret]

A. Looks like you could just use a binomial distribution

[onyx turtle]

i remember learning this but i forgot so im freeking out

wait so how do i set it up again omfg

im freaking out

[versed turret]

n is 9

p is 7/25

[onyx turtle]

wha

what would x be

ok i did a

how do i do b

quick i have 5 minutesplzzzz

[obsidian isle]

7/25 × 9

Linearity

[onyx turtle]

oh ok thx

[obsidian isle]

,calc 7/25*9

[warm shale]

Result:

2.52

[obtuse pebble]

@onyx turtle Has your question been resolved?

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[autumn veldt]

This is determinant from my prof, arnt the signs off for the 20e^t term and the 4. The denominator should be e^0 - (-3e^0) = 1 + 3 = 4?

[swift marsh]

It’s right

[gloomy valve]

you could cancel the minus signs of both numerator and denominator

[swift marsh]

They just multiplied both the numerator and denominator by -1, making every term negative instead of positive, for whatever reason

[autumn veldt]

The prof goes on to use -4 for the denominator tho

I see how u1' as a whole might be equivalent, but the determinant of the denominator is not equal to -4 right

[nocturne minnow]

Because your prof multiplied everything by -1

[obtuse pebble]

@autumn veldt Has your question been resolved?

[autumn veldt]

.close

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[timid silo]

I got a probability quiz and I need helpo

[obtuse pebble]

Please don't occupy multiple help channels.

[teal prawn]

why did you open another channel? #help-4 just ask it here

[trail wharf]

My understanding of Asymptotes is rudimentary. y=1/x will get you the following: y=1, x=1. y=2, x=1/2. y=3, x=1/3. Etc. Essentially, the denominator for x will incriment by 1 as y does. I need to know the formula that will cause the denominator for x to increase exponentially as y incriments. For example, y=1, x=1. y=2, x=1/2. y=3, x=1/4. y=4, x=1/8. y=5, x=1/16. Etc. I'm sure the answer is right in front of me but I'm struggling.

I think I ended up posting in Jupy's duplicate. Should I try and make my own again?

[teal prawn]

yeah

[obtuse pebble]

@grave crypt Has your question been resolved?

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[timid hare]

How do the solutions to these 2 questions differ? My professor said they have different valued slns??
1)I have two children. Given that at least one of them is a boy, what is the probability that I have two
boys?

2 I have two children. Given that the older one is a boy born on Tuesday, what is the probability that I
have two boys?

[hoary cargo]

@timid hare In both case, the sample space for respective experiments are different. That is they have different probability.

[timid hare]

I got 1/3 for the first question by doing prob = (1/4)/(3/4)

can you tell me what is the answer to 2?

[hoary cargo]

@timid hare i don't think you are right in first case. Can you show the sample space created for both cases?

[novel knoll]

Its a common “paradox”

Worded poorly so multiple answers make sense

Read Boy or Girl paradox on wiki

[obtuse pebble]

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[timid silo]

how do I simplify from top line to bottom? 0_0

[worthy comet]

its just wrong, you can't simplify the top to get the bottom, there is an error somewhere. Eg if $z_{n+1}=1$ and $z_k=0$ otherwise, then the top sum is 1 and the bottom sum is 0.

[warm shale]

Alexander42

[timid silo]

.close

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[small oak]

Hello, I just need a quick guide on how to get into trigonometry, as in do i start by learning the formulas first or do i need some other knowledge before?

[brave bramble]

You'll want to know how to solve linear equations first

Other than that, go ahead!

[small oak]

those are no problem actually, i've had a year's pause after my highschool and now im looking to get into a uni, so im trying to remember stuff

[brave bramble]

Then relearning trig shouldn't be an issue

[small oak]

anyways thanks

.closed

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[timid silo]

solve for y

[timid silo]

x is 60 degrees

[obtuse pebble]

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[nova violet]

what are the special traits of a square?

[nova violet]

is there anything more?

[woeful cove]

Diagonals are equal

The diagonal bisecting forms 2 similar isosceles triangles

And diagonal=✓2×side

But yeah that's all I know @nova violet

[nova violet]

yeah I saved that for the next slide

not sure anything else? That’s different from what I have

[woeful cove]

Nothing more, you could talk about its area

[nova violet]

hmm

[woeful cove]

And cube too

[nova violet]

.close

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[lavish arch]

I'm confused on what is happening

[lavish arch]

actually nvm

.close

[obtuse pebble]

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[vague fable]

no clue on how to solve this - how do i relate the distance to middle to the length of the chord?

[vague fable]

<@&286206848099549185>

[long sinew]

Have you drawn it yet?

[vague fable]

have not, will do rn

just did, do not really understand what to gain from it

[tardy epoch]

share the pic

[vague fable]

where do i start counting the distance from the chord?

[vague fable]

was a lil confused on how to correctly graph it

[obtuse pebble]

@vague fable Has your question been resolved?

[tardy epoch]

can you label 6 and 3 lengths on your picture?

[vague fable]

i did a little bit, but i cannot really figure out where to label it

[obtuse pebble]

@vague fable Has your question been resolved?

[vague fable]

i'm genuinely still very confused

could someone pls help

[obtuse pebble]

@vague fable Has your question been resolved?

[obtuse pebble]

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[jaunty schooner]

Could someone help me with this question? I don't know if I got it right and the next part I don't understand

[obtuse pebble]

@jaunty schooner Has your question been resolved?

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[sudden otter]

I have no idea how to find the distance on a 3d plane.

[sudden otter]

plz ping me

[woeful cove]

I'm not sure but do xaxis, yaxis then z axis

Please ping me too if you solve it

But distance I know

[nocturne minnow]

It's pretty much Pythagorean theorem

[woeful cove]

Yes

[sudden otter]

ohh triangle

but how do I apply that

[nocturne minnow]

Literally what I said

Don't give it to them

You teach

It's an application of Pythagorean theorem

sqrt(x^2 + y^2) is for distance formula 2D, what could you do to make it for 3D?

[sudden otter]

ah ok

um

^3?

[nocturne minnow]

No

(x, y) is 2D, (x, y, z) is 3D

What could you do to the distance formula, that makes it for 3D?

[sudden otter]

Just add the z

[nocturne minnow]

Not just z

[sudden otter]

so sqrt(x^2 + y^2+z^2)

um

[nocturne minnow]

Yes

That

[woeful cove]

Yes

[sudden otter]

ah ok

[nocturne minnow]

I was saying, it's not just z, it's z^2

[sudden otter]

so just do that for both

[nocturne minnow]

Well, it's distance formula

Also, I made a mistake. It's $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ for the distance formula

For 2D

[sudden otter]

ahhh

[warm shale]

dldh06

[nocturne minnow]

Now, what should you do to make it 3D?

[sudden otter]

sqrt(x2-x1)^2+(y2-y1)^2+(z2-z1)^2

[nocturne minnow]

Yes

That's it

[sudden otter]

so like this?

[nocturne minnow]

Yes

[sudden otter]

sqrt19

thank you!

.close

[obtuse pebble]

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[pale wedge]

is the Y always the dependent clause? for example if there were two terms: one being time since thrown from machine and the other being distance from ground, would the second one be Y?

[fierce lagoon]

Well

I'd assume so

Because essentially, time relates to your position

So there are two scenarios

"After some time, this is your position after being yeeted"

"This is the time after you've been yeeted because of your position"

Which one makes more sense?

Essentially, if you think about any projectile, the distance will always be related to the time thrown since

So yes, "y" would be the second 'term'

[pale wedge]

tysmmm i really needed this

.close

[obtuse pebble]

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[twilit bobcat]

what does a^infinity equals?

[twilit bobcat]

a = constant

[hybrid gull]

Depends

[zenith orbit]

CAN u pls help me im desperate

i cant find any1 else

i need help with 6 probability algebra 2 questions

[hybrid gull]

@zenith orbit Read #❓how-to-get-help

Not hard.

[zenith orbit]

i did one guy offered and then left

pls im desperate

[hybrid gull]

Read what I said. Or I won't try helping you

That's up to you.

[zenith orbit]

i just need the answers no explanation cuz the answers are already listed

ok fine

[hybrid gull]

You're joking right?

[zenith orbit]

no?

wanna see?

pls i rlly need help

its jusy 6 six multiple choice questions

[hybrid gull]

i just need the answers

the answers are already listed

Lmao

[zenith orbit]

no like its multiple choice

[hybrid gull]

Then make another channel lmao

[zenith orbit]

thats what i meant

i literally joined here for help

[hybrid gull]

Poor guy can't read the entire channel :/

[daring rock]

Dont try to take over someone else's help channel. And don't expect anyone to just give you answers. We're here to help you learn, not to do your homework for you.

[zenith orbit]

ok thx

i wa sjust confused cuz ppl made help servers for me and they all left

i thought you guys wre supposed to help but it doesnt look. like that

bunch of lying fuckers

[hybrid gull]

XD

[zenith orbit]

xd

[woeful cove]

Help you learn

[zenith orbit]

yes i need help

im going crazy fuck math

[hybrid gull]

#❓how-to-get-help

[woeful cove]

I'll help you if you're gonna do it yourself

[hybrid gull]

This guy's something

[zenith orbit]

fuck you shen piece of motherfuckin goshit eating cow asshole go eat ur mommas shit. every1 else thx for being kind

[hybrid gull]

Lol

I was willing to help - and ya didn't want it

[woeful cove]

What's the matter

[crimson berry]

Good lord

[hybrid gull]

Scroll up if you'd like to see the convo - but man apparently couldn't read #❓how-to-get-help

[crimson berry]

Hold up, this isn't even you channel wtf

[woeful cove]

😭

Ain't no way bruh

[hybrid gull]

¯_(ツ)_/¯

[woeful cove]

@zenith orbit I can still help if you occupy an empty channel

[zenith orbit]

nah im good.i found ppl who actually care baout me

[warm shale]

Edwin25

[obtuse pebble]

@twilit bobcat Has your question been resolved?

[hybrid gull]

Is this a bannable offense lmao

It's a pretty creative insult tho

I wouldn't expect much from the guy regardless - so I wasn't surprised

<@&268886789983436800> Thoughts?

[obtuse pebble]

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[strange kelp]

hi im struggling to understand this one. what formula would i use to go about solving this?

[strange kelp]

also colorful previous dude

[hybrid gull]

Yes - I would agree lmao

[hybrid gull]

There isn't an explicit formula for this, but you can reason this out

[timid silo]

Power of Jordan is 1/2
Power of sisters is 1/3

Can you finish the rest?

[strange kelp]

if you can give me a second i dont have paper cuz im on computer

[timid silo]

(1/2+1/3)t=1

[hybrid gull]

Keep in mind a few things:

1. They both work for the same amount of time.

2. The time they each take to complete a birdhouse will tell you how fast they work

[timid silo]

Solve for T

[strange kelp]

5/6=t?

[timid silo]

(1/2)t1=1
(1/3)t2=1
One stands for one house
(1/2+1/3)t=1
t=6/5

[strange kelp]

feel like im missing something here

imma be honest

[timid silo]

1/2 + 1/3 = 5/6

[strange kelp]

yes

[timid silo]

(5/6)t=1

[strange kelp]

okay

[timid silo]

multiply both sides by 6/5

[strange kelp]

ah

[timid silo]

…

How old are you?

:D

[strange kelp]

15

[timid silo]

This from grade 3

Or grade 4

[strange kelp]

im homeschooled i have no people to ask lol

[timid silo]

Oooh

Happens :(

So yeah
Power adds

[timid silo]

This is enough

(1/2)t1=1
(1/3)t2=1
One stands for one house
(1/2+1/3)t=1
(5/6)t=1
Multiply by 6
(5)t=6
Divide by 5
t=6/5

Each step👍

[strange kelp]

11/5=1 bird house?

[timid silo]

Why 11/5

6/5 hours

For one house built together

[timid silo]

For this problem

Together they build faster

[strange kelp]

alright so if i convert this to time

as in minutes

[timid silo]

boy:
P1 times 2=1
P1=1/2
girl:
P2 times 3=1
P2=1/3
One stands for one house
(p1+p2)*t=1
(1/2+1/3)t=1
(5/6)t=1
Multiply by 6
(5)t=6
Divide by 5
t=6/5 hours
Which is 1 hour 12 minutes

[strange kelp]

yeah i think i understand its literally more than an hour

yup

got the same thing

[timid silo]

👍🇰🇿

[strange kelp]

thanks amigo for helping me out. its a struggle.

have a good one

[timid silo]

De nada mi amigo
Disfruta te

[strange kelp]

.close

[obtuse pebble]

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[compact shadow]

You need to ask a specific question
Anyway enumerative combinatorics written by Richard Stanley talks a lot about them, especially chapter 7, representation fo symmetric groups

https://en.jp1lib.org/book/439958/8d4e69

https://en.jp1lib.org/book/439959/3e07c5

[obtuse pebble]

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[dreamy current]

hello

[dreamy current]

https://cdn.discordapp.com/attachments/873510218540142622/966930937462005831/unknown.png

heres my best understanding of this question

this would be segment 1 and segment 2 added up right?

but how would i find theta?

area of a segment = (1/2)(theta-sin(theta))(r)^2

COSINE RULE OH

.close

[obtuse pebble]

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[obtuse pebble]

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[glossy sparrow]

i got derivative of sine, which is cos

f prime (0) = cos(0)

=1

thats the slope

since x equals zero, what's y?

[burnt scroll]

The original function was f(x) = sin(x) right

so y = sin(0)

[devout lantern]

Can someone help plz

[glossy sparrow]

bro

i took this

[devout lantern]

Do you know how to solve it ?

[raven spire]

yes. and the point should be (0, sin 0)

[raven spire]

y = sin(0) right? cause it's a graph for the sine function

[obtuse pebble]

@glossy sparrow Has your question been resolved?

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[hot iris]

how are they saying ab = c^p

or b^2 - ac = 0

how does that work if cube root of p is irrational

[supple pumice]

equation 3 is just a linear combination of the basis vectors for $F(\sqrt[3]{p})$

[warm shale]

please request a new nickname

[supple pumice]

assuming this is about Field extensions

[hot iris]

i have no idea what is a field extension

[supple pumice]

can you tell what class this is

so i can tailor my answer a bit

[hot iris]

high school level

[supple pumice]

ohh

[hot iris]

10th 11th

[supple pumice]

ohhhhhhhh

[hot iris]

yeah

[supple pumice]

im so sorry

[hot iris]

so how does that happen

[supple pumice]

so basically there are no multiples of sqrt(p) on the right hand side right

[hot iris]

yeah

i mean, its just 0

[supple pumice]

yeah so any multiple of sqrt(p) on the lhs must be zero as well

[hot iris]

wait....

what

tf

[compact shadow]

If x+yp^(1/3)=0, then py^3=-x^3. If y doesn’t equal 0, then p=-(x/y)^3 contradiction

[hot iris]

$If x+yp^{(1/3)}=0, then py^3=-x^3. If y doesn’t equal 0, then p=-(x/y)^3 contradiction$

[warm shale]

azorfus

[hot iris]

makes it better to look at

[compact shadow]

Thanks

[hot iris]

how does this work

[compact shadow]

“p is not a perfect cube”

Written in your first line

[hot iris]

yeah

[compact shadow]

So contradiction

y has to be 0, so is x

[hot iris]

this is what u meant ryt

oh i see

[compact shadow]

Now x =ab-c^2p and y=b^2-ac

This answers your question

[hot iris]

oh

Nice

idk how i'm going to get used to proofs like this

ok

now

how do i prove whether if something is an integer

[compact shadow]

Induction of course

[hot iris]

oh

ok then the Sn+1

[compact shadow]

Since It satisfies that recursive formula

[hot iris]

uhhh

[hot iris]

yeah so i use induction and then get to Sn+1

where now idk how to prove this statement that it is an integer

like

$S_{n+1} = (3 + \sqrt{5})^{n+1} + (3 - \sqrt{5})^{n+1}$

[warm shale]

azorfus

[compact shadow]

The n-2 and n-1 th term being integers can give you n-th term being integer

[hot iris]

oh

[compact shadow]

You just prove s_1 and s_2 are integers

[hot iris]

so that is just valid for all terms?

no need to write a big proof for that?

[compact shadow]

Yeah

[hot iris]

nice

[compact shadow]

Just induction

[hot iris]

bro

lemme write this wait

so the base case is $S_1 = 6$

[warm shale]

azorfus

[hot iris]

now i have to use this to prove the n+1 term is an integer\

which is S_2

wtf am i speaking

uh

[compact shadow]

Just directly show that s_2 is an integer

[hot iris]

ok what if it is?

what about S_3

isn't that a rule in induction

[compact shadow]

For n>=3 use induction

[hot iris]

oh

[compact shadow]

You never used induction before?

[hot iris]

i have

this case is different

[compact shadow]

The sum of two integers is an integer

[hot iris]

oh

i'll write it down

wait

Base case:$$ S_1 = 6$$
$$.$$assuming S_k is true for some integer k, we now prove S_{k+1}

this is how it usually goes right

[warm shale]

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

[hot iris]

uh nvm this

[compact shadow]

The other half is also proved by induction: $2^{n-2}$ divides $s_{n-2}$ and $2^{n-1}$ divides $s_{n-1}$ so $2^{n}$ divides both $6s_{n-1}$ and $4s_{n-2}$, so $2^{n}$ divides the sum of them which is $s_{n}$
And show $2|s_{1},4|s_{2}$

[woeful cove]

Stole my pfp

[hot iris]

[hot iris]

oh

Alright

[warm shale]

Cogwheels of the mind

[hot iris]

Nice

bro

i think i'm misunderstanding induction

can u do an example by proving $\sum^n_{k=1} k = \frac{n(n+1)}{2}$

[warm shale]

azorfus

[hot iris]

how i do it is

for n = 1

s = 1, 1(1+1)/2 = 1

so it works for 1, so assuming it works for k which is some integer which can also be 1

we try for k + 1, or 1 + 1

then

the lhs = rhs

thats how i usually do it

but here

[compact shadow]

First when n=1, we have 1=1(2)/2 so it’s true for 1
If it’s true for n-1 so the sum of n-1 terms =(n-1)n/2, then the sum of n terms =the sum of n-1 terms +n=(n-1)n/2+n=(n^2-n+2n)/2=n(n+1)/2 so it’s true for n

[hot iris]

i'll do the text thing

wait

[compact shadow]

Therefore it’s true for any positive integer n

[hot iris]

ok not needed ig

[hot iris]

yeah okay

so now in this question

i'll just try it on paper

first

$S_{n} = S_{n-1} + ?$

[warm shale]

azorfus

[hot iris]

what do i add in this case

[compact shadow]

?
$s_{n}=6s_{n-1}-4s_{n-2}$

[warm shale]

Cogwheels of the mind

[hot iris]

how do i prove this

same by induction

ah

[compact shadow]

No Directly show that

[hot iris]

wdym, do i just substitute values?

[compact shadow]

?You have the expression of s_n

Just directly show that s_n satisfies that

[hot iris]

oh

so i expand it

and solve

is that what u mean

[compact shadow]

You just show this

[hot iris]

yeah okay

so if RHS = LHS at some point

we prove it

[compact shadow]

Yeah

[hot iris]

alright

next question

wait no

[compact shadow]

Actually

[hot iris]

?

[compact shadow]

You show that $(3+\sqrt{5})^{n}=6(3+\sqrt{5})^{n-1}-4(3+\sqrt{5})^{n-2}$ and $(3-\sqrt{5})^{n}=6(3-\sqrt{5})^{n-1}-4(3-\sqrt{5})^{n-2}$ respectively

[warm shale]

Cogwheels of the mind

[compact shadow]

It’s same as n=2

[hot iris]

wait what

[compact shadow]

?

[hot iris]

so its not this?

[compact shadow]

It becomes showing that $(3+\sqrt{5})^{2}=6(3+\sqrt{5})-4$ and $(3-\sqrt{5})^{2}=6(3-\sqrt{5})-4$ respectively

[warm shale]

Cogwheels of the mind

[compact shadow]

It is

[hot iris]

oh

[compact shadow]

I am just telling you how to show that by showing the following two things

[hot iris]

oh

alright

nice

another doubt

$(3+\sqrt{5})^{n}=6(3+\sqrt{5})^{n-1}-4(3+\sqrt{5})^{n-2}$

[warm shale]

azorfus

[hot iris]

can u do this for other things in algebra?

[compact shadow]

Like I said

It’s the same as n=2

Because you can divide that thing to the power of n-2 both sides

[hot iris]

oh

[compact shadow]

Here

[hot iris]

ok nvm

i forgot that we expand

ok nice

nice

this

i don't even know how to approach this

should i prove it like usually by substituting q = 1 for a base case

and then use normal induction

like the n(n+1)/2 example

[compact shadow]

Use induction in terms of q

[hot iris]

alright

thanks mate

[compact shadow]

Np

[hot iris]

.close

[obtuse pebble]

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[wild marsh]

which series divergence test proves that this function diverges?

[obtuse pebble]

@wild marsh Has your question been resolved?

[wild marsh]

<@&286206848099549185>

[eternal bloom]

The limit of the partial sums as n goes to infinity does not exist

[wild marsh]

could you elaborate?

I imagined that the series goes something like (-1)+(1)+... which would eventually equal zero? or is that incorrect

[eternal bloom]

S_n is the partial sum.
S_ 1= a_1
S2 = a2 + a1
S3 = a3 + a2 + a1

You're asked for the sum of a_n, not the sum of S_n.

[wild marsh]

I see, thank you very much.

.close

[obtuse pebble]

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[weary horizon]

#15 first

[sweet cobalt]

Do you understand what similar means in this context, and the implications of them being similar

[weary horizon]

not rlly

[sweet cobalt]

So basically if two shapes are similar

One is a scaled up version of the other

so a similar shape to the rectangle on the right would be a rectangle with side lengths 4 and 10

Because you have just halfed each side length

Does that make what similar means any more clear?

[weary horizon]

oh

ye

[sweet cobalt]

cool

[weary horizon]

so would it be 2 and 5

[sweet cobalt]

Yes

Did you do that algebraically or intuition/trial and error? It may be useful for other questions to be able to do it with algebra

[weary horizon]

algebraically

[sweet cobalt]

Ok cool

Do the other parts make sense then?

[weary horizon]

hmm, 1m

[sweet cobalt]

yeah np

[weary horizon]

ehh, not rlly :/

[sweet cobalt]

Sure no worries

For sake of example, if we consider a rectangle with side lengths 1 and 2

This is similar to a rectangle with sides 2 and 4

[weary horizon]

ye

[sweet cobalt]

We notice as we just scale up each time

that the ratio 1:2

is the same as the ration 2:4

[weary horizon]

ye

[sweet cobalt]

i.e, when we have similar shapes, the ratio of the sides is the same

So looking at our shape

8:20

8 x 2.5 =20

So if the short side is length x

the long side is 2.5x

so the area is 2.5x^2