help-27

[merry sundial]

Wait

Lemme check

[stuck field]

Here

[merry sundial]

Wait a sec lemme try that again real quick

[merry sundial]

What should I do next

[woven radish]

! What the hell am I doing here?

[stuck field]

You can now integrate both sides.

[merry sundial]

Ou

So I'm supposed to write
f(x)dy+d(f(x))y as
d(f(x)y)

Right?

[stuck field]

That's what I did here, yep

[merry sundial]

Ou

I somehow got the same answer💀😭

@stuck field

[stuck field]

How even?

$\int -2x(x^2-5) \dd x$

[woven radish]

! What the hell am I doing here?

[stuck field]

What does this equal?

[merry sundial]

Wait

[stuck field]

How did you even get that integral?

[stuck field]

The first image, is correct.

[merry sundial]

Is this part correct

[stuck field]

That's wrong.

[merry sundial]

Ah shit

[stuck field]

How'd you get (x^2-5)/|x^2-5|??

[merry sundial]

I changed that part bcs

🤦‍♂️ shit I'm stupid

[stuck field]

I specifically pointed it out earlier too. That it's (x^2-5)^2/|x^2-5|

It's not the same.

As what you've written.

[merry sundial]

Quick question

[merry sundial]

I'm supposed to multiply both sides with the integrating factor right?

[stuck field]

You are.

Otherwise there won't be an equality anymore.

[merry sundial]

Then,
I changed the third step
Cuz I thought that I forgot to multiply with the IF on the rhs

In the first image

[stuck field]

But you already did.

You did it twice after that in every other image.

[merry sundial]

There's no absolute value denominator on the rhs right

[stuck field]

Stop.

I have now more than once pointed out the exact error.

If you really wanted help, you'd listen to me.

Here I go once again,

I want you to really listen to me this time.

[merry sundial]

Ok

[stuck field]

There is no absolute value in the denominator because it's \
$-2x \cdot \f{(x^2-5)^2}{|x^2-5|}$, see? It's there.\
This however is the same as,\
$-2x|x^2-5|$, there doesn't have to be an absolute value in the denominator anymore. This surely isn't the first time you're cancelling out fractions, is it?

[woven radish]

! What the hell am I doing here?

[merry sundial]

I'll try it from that step

Wait a sec

There's no square on the rhs even before multiplying with the IF

Cuz I divided with (x^2-5) on both sides

[stuck field]

OH

I actually thought it was presented to you that way already.

[merry sundial]

Nope

[stuck field]

You're going to have to forgive me now.

[merry sundial]

See there

[merry sundial]

It's fine

[stuck field]

My bad, but yeah then you do divide by |x^2-5| twice

Which means you're integrating -2x in the rhs, just that

But of course remember.

(x^2-5) isn't the same as |x^2-5|

[merry sundial]

That was where I was stuck at

[stuck field]

Yeah but there's this fact too.

[merry sundial]

Yup that makes sense

[stuck field]

(x^2-5)/|x^2-5| could be +-1

Either of the two.

So while you do just integrate -2x

You can't simply say you have +x^2 or -x^2 ultimately

So you just let it be.

$-\f{(x^2-5)}{|x^2-5|} \cdot x^2$

[woven radish]

! What the hell am I doing here?

[stuck field]

Is correct so far.

Then you find C

[merry sundial]

Wait

[merry sundial]

Would this be wrong?

[stuck field]

This is a first order differential equation. You'll typically do this in highschool.

There's also an integral, you're integrating -2x

[stuck field]

@merry sundial

Which is -x^2

[merry sundial]

I got that 2x part while differentiating

[stuck field]

What did you differentiate?

Oh wait, |x^2-5|

Why even?

[merry sundial]

Yup

[stuck field]

Let's start over.

It has become a mess.

Here.

[merry sundial]

Just a sec

I generally think of the differential of |x| as the signum function of x

Ah shi...

Nvm

What you said makes sense

I'll continue

[stuck field]

From there, you just have to integrate the rhs

Which is (x^2-5)/|x^2-5| (-2x)

You have to merely integrate -2x

Because what's multiplied with that, is a constant.

It could be 1 or -1, but it's a constant.

Which is it, does not matter. Because ultimately to find y(x) you'll multiply everything by |x^2-5| later on.

[merry sundial]

So, this?

K I got it

[stuck field]

Ah!

You're wrong.

[merry sundial]

😭

[stuck field]

Because you multiply|x^2-5| after the integral

It means you multiply that to +C as well.

Can't believe this is what you were having a problem with for this long.

[merry sundial]

Ah kk silly mistake

[stuck field]

I completely misjudged the problem.

[merry sundial]

Yup

[stuck field]

Fair. But yeah, after this you find C.

[merry sundial]

Just forgot to think that the chain rule doesn't work for this assumption

K

Anyways

[stuck field]

If you mean the sign function

Then that's actually correct.

And you mean derivative not differential

[merry sundial]

Yeah derivative

K thanks

And bye

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[restive river]

i am lost

[restive river]

Can't u go diagonally

well yes

but i am lost i dont know how to solve this at all

there are a total of 50 blocks

thats all i know

U can do A to B diagnolly then B to C vertically then C to D diagnolly

but how many ways tho

there are many ways

the answer is C(15,10) for (i)

Well you can also do A to C then go down to B then diagnolly to D

[restive river]

?

its choosing formula

which equals to 3003

If u don't mind can I stop helping u

so 3003 different ways

[restive river]

ye sure

[restive river]

Hi do u need help?

yes

there are 50 squares total

but the maximum amount of blocks i can do to get from A to D is 15

[restive river]

ii. 100 ways.

whats i tho

i dont understand why 1 3003

Well

It is 5 moves to the right

10 moves up

What is 10+5?

Please tell me u know

15

Correct

i knew there are 15 blocks

So 15 factorial

Now

,w (15!)/(5!10!)

[woven radish]

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Requested by wxec

[restive river]

Does that make sense

what did u do

no

you did C(15,5)

why 5

Remember, 5 is the moves to the right

This can be calculated using the binomial coefficient "15 choose 5" or "15 choose 10"

yes

5 moves to the right

why 5 moves to the right

we can move 10 paths to the right

Look

and 5 up

We are maxed to

15 moves

Our horizontal and vertical movies CANNOT exceed 15

@restive river Following?

yes

So to account for that

We use factorial 5 to cover the moves left to right

and factorial 10 to cover the moves up and down

i see

is it arbitrary?

can we use factorilal 8 and factorial 7?

For?

Ok, let us do ii now

Explain to me how to do ii

In your own words

[restive river]

for I)

[restive river]

No

How does that make sense

You can't go 8 right or left

Nor 7

max i can go right is 10

max i can go up is 5

Please move on to ii.

[restive river]

ok

so

for II

well this one is harder

but ill do compliment

ill let U be all possible paths

which is 3003

and A^c be al possible paths where we dont go by B

Ok

You're doing great, derivative.

the amount of paths where we dont go by B is tough

i am stuck

because we still go Right 10 tmes and up 5 times

[restive river]

Not quite

[restive river]

but i try different possibilities without going through B

and i always get 10 right 5 up

[restive river]

Look closely what the key difference is

Between going from A to D and A B D

from A to D 10 right, 5 up

from A C D, 10 right 5 up

we just dont go on the B square

wait

there are 7 blocks to get to B

which leafs 8 blocks to get to D

Good job @restive river

Idek if it's right I'm just motivating u

im not done

this one is a tough one

[restive river]

Continue

so ignoring the compliment

i didnt choose to go that route

this way is easier i think

i split the map into two destinations

first destination is to get to B

how many blocks: 7

2 up 5 right

so the amount of ways i can do this is either 7 choose 2 or 7 choose 5

now from B to D there are 8 blocks (or 8 possible moves)

which leaves me with 8 choose 3 or 8 choose 5

now, i need to be consistent

either i only do the right moves or the up moves

so i choose C(7,2) times C(8,3)

for my answer to (iii)

i decided not to go by the complement way

is this correct?

@restive river

Perfect

I'm very proud of you

thanks

now for (iii)

so

i do the exact same method

i already know that path B to C is obligatory

so i split the map into 2 sections

each with 7 moves

i get a total of 14 moves

Yes

answeris C(7,5) times C(7,5)

for (iv) the complement rule works.

answer is 3003 - 441

[restive river]

👍

so we have solved it

great problem

these counting problems are tough

[restive river]

Yes now that you know the # of ways for BC you just subtract

Great work

thanks

i tried

but thanks for your help especially

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[restive river]

@restive river Did I help too?

yes

[restive river]

Let's go

Im officially Einstein

yes

[restive river]

Yes

.close

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[peak charm]

When solving a system with 2 equations, If I want to subtract the terms of the first equation from the terms of the second equation, will I get a correct answer?

[peak charm]

For example

it never happend to me to find such thing

or is it just for linear equations?

[winter torrent]

You've identified something important. You've proven that either x = -1 or x = 0, but it's not guaranteed that those are both actually solutions, just that they might be solutions.

With linear systems, you'll only really ever have either one solution or a full space of solutions (with one or more free variables) or no solutions, so this doesn't come up, as you've seen

[peak charm]

Yea i've seen that .However , i always tought that by substracting an equation from the another , the solutions will also be the solutions of the system

[winter torrent]

Well, that's technically not even true for linear systems since if you subtract x + 3 = 4 from itself, you end up with 0 = 0 which has infinitely many solutions

[peak charm]

I've done a year of matrix and determinants and now i feel like my whole life is a lie

Thanks for the help tho!

[winter torrent]

you shouldn't, you seem to understand fairly well. when i saw your problem just now it was somewhat of a surprise for me as well.

[peak charm]

Now all i want is to find proof for every surface and volume area :))).

[peak charm]

I KNOW RIGHT?

[peak charm]

gotta go back to calculus

thanks!

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[verbal cosmos]

I need help finding the fixed points of the system of differential equations: first pic is task, second is solution. i dont understand how i end up with picture number 3. i understand that the triangle delta is the determinant of the jacobian matrix

[cinder bobcat]

that's the trace of the jacobian matrix. if you recall, the trace of a 2x2 matrix is the sum of its eigenvalues, and the determinant is the product of its eigenvalues

[verbal cosmos]

oh ok i will try that out

omg yes ur a blessing

ok will try the next step

[devout snow]

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[strong fiber]

I dont know how to find the limits of integrating?

[restive river]

well it might be easier if you find the area of one petal first

[restive river]

I don't either

because the petals are symmetric and the same size

so you can just multiply by 2 after

anyways i dont think ur graph is particularly correct

[strong fiber]

still what are the limits of one petal

?

[restive river]

your graph is not correct still

so i cant answer that yet

[restive river]

yeah

so like

find the first two instances of theta for which 4sin(3theta) = 0

because the petal starts at 0 and then loops back to the origin

[strong fiber]

Isn't it what's I wrote in grey?

[restive river]

yeah

so

[strong fiber]

0, +- pi/3 ...

[restive river]

0 and pi/3 work

those will be ur bounds

[strong fiber]

How ?

How did u know

[restive river]

well we are considering one petal

and second

like

the green is the area u r finding

and the red rays are like, the values of theta for which the function gets to the origin again

so, effectively, defining the entire area

[strong fiber]

sometimes the integral is from 0 to pi/2, I thought this question was the same. So, how can I make sure to always to get the right limits

[restive river]

ok so like

here is the way i think about it

[strong fiber]

also, are the negative theta values that i found correct? -pi/3 , -2pi/3

[restive river]

do you see that first petal?

[restive river]

sure but u dont need them

[restive river]

you can try drawing it by hand

your lower bound will be where u start from (so theta =0 obviouisly) and ur upper bound is where ur pen hits the origin again (so theta = pi/3)

[strong fiber]

why is my answer still wron

g

oh i forgot sin 3theta

[strong fiber]

thank you

btw did you mean like how i drew it ?

[restive river]

no like

just trace ur drawing

and figure out where ur graph hits the origin

[strong fiber]

I dont think I know what's that

I actually dont understand these graghs I just try to memorise them

do you mind showing me what you mean by tracing ?

[restive river]

hard to explain it without like you know you actually being here

maybe look up some videos about this topic

could be helpful

[strong fiber]

what should I write

[devout snow]

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[strong fiber]

I dont get the highlited part

Why did they do that in blue

[near trout]

simultaneous equations

they're doing elimination

[devout snow]

@strong fiber Has your question been resolved?

[devout snow]

@strong fiber Has your question been resolved?

[strong fiber]

why did they seperately calculated the area instead of subtracting r - r

and what did they do in the 2nd line?

like what does (by symmetry of the petal) mean

are there any helpers?

<@&286206848099549185>

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[half sand]

i cant solve equations, Terms

Equations and formulas

Linear equations and systems of equations

Quadratic equations

funktions

[topaz beacon]

try using khanacademy to help out

[half sand]

i did, really good btw, but i feel like the stuff im learning are pretty obvious and simple, and then i go to do actual exercises and i dont know anything

like i feel like it has things i need and things i dont need and i dont have THAT much time

[topaz beacon]

khan academy is pretty complete

make sure youre going over everything

[winter torrent]

trying to rush through will make it take longer

[topaz beacon]

try to find your weak points

[half sand]

my weak point is math

my ass is so embarassing at it

[topaz beacon]

than spend your time wisely

[half sand]

im too old to be struggling with this level of maths god

well thanks guys ill take it step by step

[topaz beacon]

dont worry about age

what matters is getting it done

[devout snow]

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[dire python]

Where can I go from here? I’m stuck

[lyric hornet]

Are you familiar with u-sub?

[dire python]

Yeah

Give me a sec, lemme try it

Does this look right

[lyric hornet]

Yeah looks right

[dire python]

Thx!

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[frail sun]

I'm pretty awful with word problems, how would I start this?

[urban jungle]

Differentiate both sides with respect to T

[frail sun]

What would both sides be?

[patent marsh]

well be careful. T represents the tip's constant value and it's in terms of hours

everything needs to be in terms of minutes

it's a little t

[urban jungle]

Use differentiation rules…

[shrewd thistle]

The issue is in setting up the equation to start with.

[urban jungle]

The equation is set up though…just use R = T/W

(Sorry if I was unclear on that)

[shrewd thistle]

It's not clear; T and W need to be converted first, and then W will have to be modified to account for the rate of decrease

[urban jungle]

There’s unit conversions, but that should still be the equation to use

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[drifting zinc]

can someone explain why the answer to this is 8y and not 0?
What I did is figure out that z is x + 4y^2 which means the partial derivative of z wrt y is 8y and then the partial derivative of that wrt x is 0.

[winter torrent]

you computed z incorrectly

[drifting zinc]

Isn't the partial derivative of z wrt x 4y^2?

or is there like an easier method to do questions like this without computing the original function?

[supple knot]

Plus a function of x

[drifting zinc]

oh yeah that explains.. but i won't be able to calculate the value of the constant right? so is there another method?

[supple knot]

Yea you use a theorem about mixed partials

[drifting zinc]

young's theorem?

[devout snow]

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[stone gust]

can someone explain why this is always false to em

me

always true*

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[winter torrent]

here's a really silly way you can do this

let f(x) be some multiple of 2^x + 1

[restive river]

Can you give more details

[winter torrent]

assume f(x) = a (2^x + 1) and then figure out what a needs to be

[restive river]

I got it, tysm

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[dire slate]

checking to see if is right, proof by induction

but i also want to check to see if i have the correct understanding of the induction step

to prove p(k+1), we have to show what i wrote in blue for the induction step ends up being the same i wrote in red right?

[devout snow]

@dire slate Has your question been resolved?

[devout snow]

@dire slate Has your question been resolved?

[devout snow]

@dire slate Has your question been resolved?

[covert root]

writing is a bit hard to read

[dire slate]

mb, is there an easier way to write it out?

it would seem a lot to write out the latex

but i can try it out, just give me a few min

[covert root]

You don't you use latex for typing. Just use sigma for sum or / for fraction

[dire slate]

sure thing, making it right now

lets hope this works

[
P(k+1): \frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\dots +\frac{1}{k(k+1)} + \frac{1}{(k+1)(k+2)}=\frac{k}{k+1} + \frac{1}{(k+1)(k+2)}
]
[
= \frac{k}{(k+1)} + \frac{1}{(k+1)(k+2)}
]
[
= \frac{k(k+2)}{(k+1)(k+2)} + \frac{1}{(k+1)(k+2)}
]
[
= \frac{k(k+2)+1}{(k+1)(k+2)}
]
[
= \frac{k^2+2k+1}{(k+1)(k+2)}
]
[
= \frac{(k+1)(k+2)}{(k+1)(k+2)}
]
[
= \frac{(k+1)}{(k+2)}
]

[woven radish]

TheKingPin

[dire slate]

wow

*Trying to show, $\sum_{j=1}^{k+1} \frac{1}{j(j+1)} = \frac{(k+1)}{(k+2)}$

[woven radish]

TheKingPin

[dire slate]

@covert root

you dont have to read all of it

i just want to know, is the idea of my work above to prove that its RHS is equivalent to the * (second equation)

[devout snow]

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[sullen bough]

f(x) = (sin(x^2))^n
lim x->0 f(x)f'(x)/(1-cos(x))^m = 32 sqrt(2)
2m + n =?

[sullen bough]

here's what I have so far: lim x-> (n) (2^(m+1)) (x^(4n-1)) / x^(2m) = 2 ^ (5.5) => 4n - 1 = 2m and (n)(2^(m+1)) = 2^(5.5)
(n)(2^(m+1)) = 2^(5.5), n = (2m + 1)/4 => (2m+1)(2^(m+1)) = 2^(7.5)

so basically I need to solve for m such that (2m+1)(2^(m+1)) = 2^(7.5) and find n based on 4n - 1 = 2m but I don't know how to

looks a lot like lambert w stuff but they haven't taught us that yet so not sure

[twilit comet]

just a question, but what level/grade/course is this?

[sullen bough]

12th grade, derivatives

[twilit comet]

i see

thank you

[devout snow]

@sullen bough Has your question been resolved?

[devout snow]

@sullen bough Has your question been resolved?

[devout snow]

@sullen bough Has your question been resolved?

[sullen bough]

<@&286206848099549185>

[analog karma]

Hi

Could i ask a question?

[faint zinc]

@sullen bough I'm not following your work, what did you arrive at for f'(x)?

!occupied @analog karma

[devout snow]

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).

[sullen bough]

nsin^(n-1)(x^2)(cos(x^2))(2x)?

and x->0 so we get x^(2*(n-1))(1)(2x)

[faint zinc]

sin(x^2) = 0 as x -> 0

So everything goes to 0

You should use l'hopitals

[sullen bough]

sin(x)/x = 1 as x-> 0

[sullen bough]

so you want me to find the second derivative of (sin(x^2))^n

[faint zinc]

You don't have an x in the denominator

[sullen bough]

yes you do

lim x->0 f(x)f'(x)/(1-cos(x))^m = 32 sqrt(2)

(1-cos(x))^n
1-cos(x) = 2sin^2(x/2)

bruh

it's a classic 0/0 limit

[faint zinc]

Where's the bare x in that expression?

[sullen bough]

sinx is congruent to x as x->0

it's a 0/0 limit

I'm like pretty confident of my own work that far

[faint zinc]

And I'm not yet.

[sullen bough]

f(x) = (sin(x^2))^n
f'(x) = n((sin(x^2))^(n-1)) * cos(x^2) * (2x)

[faint zinc]

Right

[sullen bough]

f(x) f'(x) = (sin(x^2))^n * n((sin(x^2))^(n-1)) * cos(x^2) * (2x)

[faint zinc]

So in total you have 2xn sin^(2n-1)(x^2) cos(x^2) in the numerator

[sullen bough]

lim x-> 0 (sin(x^2))^n * n((sin(x^2))^(n-1)) * cos(x^2) * (2x) / 2((sin(x/2))^2m)

now do the denominator as well

replace all the sin(x) with x

(sin(x^2))^n * n((sin(x^2))^(n-1)) * cos(x^2) * (2x) / 2((sin(x/2))^2m) = x^(2n) * n(x^(2*(n-1) * 1 * 2x / 2 (x/2)^2m

[faint zinc]

You wind up with $\frac{2xn (x^2)^{2n-1} \cdot 1)}{2 (x/2)^{2m}}$

[woven radish]

全能の存在

[sullen bough]

(n) (x^(4n-1)) / x^2m

so basically the powers of x need to be the same so it cancels out, otherwise our limit becomes either 0 or infinity

4n-1 = 2m

[faint zinc]

So 4n - 1 = 2m alright

And n = 32 sqrt(2)

Hmm

[sullen bough]

we forgot something in the denominator

(x/2)^2m

[faint zinc]

We seem to have dropped a factor of 2^-2m

Yeah

[sullen bough]

we get a 2^2m in the nominator

yeah

4n - 1 = 2m and (n)(2^(m+1)) = 2^(5.5)

this is what I got

[faint zinc]

Why m+1 there?

Thought it was 2m

[sullen bough]

because there is a 2n

there is an extra 2

in the nominator

[faint zinc]

That 2 in the numerator should have cancelled with the 2 in the denominator

[sullen bough]

should I do it on paper?

and send you the screen shot?

[faint zinc]

(n) (x^(4n-1)) / x^2m

From what you wrote

#help-27 message

[sullen bough]

yeah because I can't write it with all these parenthesis in discord...

I'll just send screen shots because this is impossible lol

just to double check

[faint zinc]

\begin{align*}
\frac{2xn (x^2)^{2n-1} \cdot 1}{2 (x/2)^{2m}} &= \frac{n x^{4n-1}}{(x/2)^{2m}} \
&= \frac{2^{2m} n x^{4n-1}}{x^{2m}} \
\lim_{x \rightarrow 0} &= 2^{2m} n
\end{align*}

[woven radish]

全能の存在

[faint zinc]

Anyway, I need to go afk

Hope this helped

[sullen bough]

it didn't and I just double checked everything

and you're wrong

screenshot on the way

all you had to do was help me solve the system of equations I gave you..

[faint zinc]

Well, I'm sorry it didn't help

[sullen bough]

https://i.ibb.co/m4SRSZw/20240219-200702.jpg

[tender cobalt]

absolutely no defense to using x to mean times

use •

[sullen bough]

fair enough

[tender cobalt]

can confirm that the work you have done so far is correct, but you need to make the size of your steps consistent

this should be better

I managed to find a solution but you'd have to do it through inspection

the first idea is to let n = 2^k, so that the n 2^(m+1) = 32√2 equation can simplify down to k + m = 4.5

the second idea is to combine both equations into one larger one

the way Ive done this, I ended up with 2 * 2^k + k = 5, which by inspection shows that k = 1 is a solution

lmk if you get stuck and Ill send over my work

[sullen bough]

let me check

yeah

8 * 2^(4.5) = 2^7.5

so oh wow

yeah ok makes sense

thanks

.close

[devout snow]

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[tender cobalt]

crackpot option to have this be an equation

you can even see that 2 * 2^k + k is monotonic so youre not going to find any other real solutions

np otherwise

[devout snow]

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[dense lynx]

when is a matrix invertible?

(there's many answers to this, but a specific one will help you solve this)

[devout snow]

@restive river Has your question been resolved?

[restive river]

when the determinant is different than 0

[dense lynx]

yes

then, in this problem, it helps to compute the determinant

[restive river]

i dont understand

how do i use the equation

[dense lynx]

are you familiar with viete's relations?

[restive river]

yes

-b / 2a?

[devout snow]

@restive river Has your question been resolved?

[devout snow]

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[devout snow]

@open ferry Has your question been resolved?

[devout snow]

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[orchid wasp]

9x-25=y^2

[orchid wasp]

How I sqrt left side to find y?

[solar goblet]

you want to solve y in terms of x?

[soft umbra]

I don’t think there’s only one set of (x,y) in this question

[orchid wasp]

Find f(x)

This is the question

@solar goblet @soft umbra

.close

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[noble skiff]

When doing Gaussian elimination why do I have to have it in a form like this and not this

[noble skiff]

I’ll send an image

If you have it in the second situation wouldn’t you just be able to solve for z then use that to solve for y then use that to solve for x

[devout snow]

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[brittle reef]

hey, so we're doing number theory and I'm so lost... I am unsure how to solve these without trial and error?

[restive river]

what's confusing you

you are just asked to find an integer pair

[brittle reef]

yes, but I was just wondering if there was a more efficient method rather than using trial and error, like for the first one, I can guess m is -3 and n is 2 but for the larger numbers, I can't just guess.. idk sorry

<@&286206848099549185>

[devout snow]

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[brittle reef]

.close

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[barren cloak]

is anyone free to check if this is correct?

[devout snow]

@barren cloak Has your question been resolved?

[devout snow]

@barren cloak Has your question been resolved?

[devout snow]

@barren cloak Has your question been resolved?

[faint zinc]

@barren cloak aside from some dodgey notation in the "Evaluate the limit" section, it looks good to me.

[devout snow]

@barren cloak Has your question been resolved?

[barren cloak]

oh ok thanks

[devout snow]

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[faint zinc]

Yw ❤️

[devout snow]

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[modern badge]

what this can be?

[modern badge]

i tried different values but neither of them made sense

[eager lodge]

I think it's area

[soft umbra]

Wtf is this question?

[eager lodge]

Let triangle have area of x

[modern badge]

it's like a + bc = 80

shapes represents some numbers

[eager lodge]

Square = 2x

Pentagon = 3x etc

[modern badge]

what is upside down triangle for example

[eager lodge]

Since square can be divided into two triangles etc

Just a distraction i think

Oh wait

Triangle = 8 so it's the same upside down

[modern badge]

8?

what you said about area can be a path

but upside down triangle makes it confusing

[eager lodge]

Treat the triangles as the same for now

[modern badge]

then 2x + x^2 = 80

x = 8

[eager lodge]

I think that's why the triangle is upside down

Since 8 is the same upside down

[modern badge]

but you don't even need to find the x value if you gonna calculate shapes below that way

so you get 1/3?

i swear to god whoever made this a problem has mental health issues

[eager lodge]

I got 1/2

[modern badge]

how?

[eager lodge]

Pentagon = 3 triangles

Hexagon = 4 triangles

(4∆-3∆)/2∆ =1/2

[modern badge]

for some reason i took hexagon as 6 triangles

which makes no sense

so if we gonna calculate it this way

that equation above is useless

x variable cancels itself out anyway

let me check what's the answer

answer is 2

i'm gonna skip this one

.close

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[devout snow]

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[solar goblet]

$f(x)$ is continuous on $\mathbb{R}$

[woven radish]

FungusDesu

[river solar]

yes

indeed it is 😋

[solar goblet]

$\int_0^{\frac{\pi}4} \tan(x)f(\cos^2x)dx = 2$

[woven radish]

FungusDesu

[solar goblet]

$\int_e^{e^2} \frac{f(\ln^2x)}{x\ln x}dx = 2$

[woven radish]

FungusDesu

[solar goblet]

Evaluate $\int_{\frac14}^2 \frac{f(2x)}xdx$

[woven radish]

FungusDesu

[solar goblet]

not sure if i can find f(x)

[river solar]

wow so its a question about functional equations

eek

[lavish sigil]

was there a purpose to doing this

$\int_0^1 \frac{f(x)}{x}dx = 4$

[woven radish]

jewels!

[lavish sigil]

Because I was able to simplify it to this

(and you can just go f(x) = 4x if you want to)

[river solar]

can you?

[lavish sigil]

the only condition is that its continuous

surprise surprise it is

[solar goblet]

how?

[lavish sigil]

oh wait no the limits

grr

cos x = u

$-\int_1^{1/\sqrt 2} \frac{f(x^2)}{x} dx$

[river solar]

shouldn't there be a minus sign? am i being silly?

[woven radish]

jewels!

[lavish sigil]

yeah my bad

[river solar]

ah okay

okay now i see how you got it

i was also confused about the limits

[lavish sigil]

$\int_{1/\sqrt 2}^1 \frac{f(x^2)}{x} dx$

[woven radish]

jewels!

[solar goblet]

i would prefer you give me the hint and i do the work myself please

[lavish sigil]

yeah that would be better 😅

can you guess what the next sub would be

[lavish sigil]

if we wanna find this

[solar goblet]

ahhh

ywah i got it, let me do some calculations

$\int_{\frac{\sqrt2}2}^1 \frac{f(u^2)}udu = \int_1^2 \frac{f(v^2)}vdv$

[woven radish]

FungusDesu

[solar goblet]

for reference, u = cosx, v = lnx

does it make sense to take derivative of both side?

<@&286206848099549185>

[river solar]

im not sure where you are making lnx

[uneven notch]

wdym?

to take derivative of both side

[solar goblet]

i mean what i said

differentiating both sides

and thats my question too, does it make sense?

[uneven notch]

I dont think so

they are definite integrals

with both constant on intervals

[devout snow]

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[devout snow]

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[ivory saffron]

how am I meant to do 15b?

[ivory saffron]

if it helps, answer to 15a was (3, 1.5)

<@&286206848099549185>

this must be easy I'm just stupid

[ivory saffron]

omg you just sub in the X and y of the point of intersection cuz it's in both lines

FUCK

Holy I'm failing

@eager narwhal ty

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[eager narwhal]

alg

[devout snow]

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[inner ibex]

hello

[inner ibex]

i dont get dis

for the first one

wtf do they mean

a quadrilateral with four congruent sides* is a square

is this true or false how do i know

[onyx rune]

start with telling me what a quadrilateral mean

[inner ibex]

quadrilateral have 4 sides

[onyx rune]

yeah and what does congruent angles mean

[inner ibex]

same angles

sry i didnt mean a gle

[onyx rune]

ok now can you tell me figures that have those?

[inner ibex]

i meant sides

square

[onyx rune]

are there others?

[inner ibex]

idk

i only know square

[onyx rune]

thats correct

ok now a rectangle have 4 sides and 4 same angles? (thats not a square)

[inner ibex]

srry i meant the question says four congruent sides and a quadrilateral

not angles

[onyx rune]

ok mb

so its cannot be a rectangle

[inner ibex]

yea

[onyx rune]

but do you know a rhombus?

[inner ibex]

whats a romb

yeah is it like a kite

can it be like a parallelogram

[onyx rune]

sorry its in my native language xd

ok then, does rhombus have 4 sides?

and does have 4 equal sides

[inner ibex]

yes

yes

[onyx rune]

so that means it can be square but it can be rhombus

so it doesnt mean its square for sure

[inner ibex]

ohh

wait

so i have to just think of that

during the test

..

and i only have 30 minutes to complete the whole thing

[onyx rune]

for a) you just need to give a counterexample i think

[inner ibex]

ok let me try part ii

so one five sided shape i know is a

pentagon?

what idk how to do this

[onyx rune]

ok for the ii) the aanswer is true i think

[inner ibex]

what the flipp

what are some examples of 5 sided

shapes

[onyx rune]

five sided regular polygon is called a pentagon

[inner ibex]

bru

How am i supposed to know

its obtuse

[onyx rune]

do you know whats obtuse means?

[inner ibex]

more than 90 degrees

[onyx rune]

there is a formula for the sum of angles (n-2)*180 degrees

where n is the number of sides

use that

[inner ibex]

what the flippp

[onyx rune]

and check if the single angle is obtuse

[inner ibex]

😭😭 why they expect me to know this if ive never learnt in my life

[onyx rune]

now you know xd

[inner ibex]

converse just means opposite right?

[onyx rune]

yeah

[inner ibex]

so if angle B<90 then angle A <=90

like if P => Q then the converse would be Q=> P

[onyx rune]

i dont think that is what it asking 😭

[inner ibex]

oh...

bruh THIS IS SO HARD

[onyx rune]

i think that means P => Q then the converse would be ~P => ~Q

but im not sure

[haughty quest]

≤ ≥ use this

[inner ibex]

what is the squiggly line

[inner ibex]

im in my phone

[onyx rune]

means not

[haughty quest]

sane

[onyx rune]

not P => not Q

[haughty quest]

same*

[inner ibex]

iphone

[haughty quest]

ok

[inner ibex]

i thougbt thats

counterpositive?

[onyx rune]

im not sane 😭

[inner ibex]

counter what

[haughty quest]

I was writing same

[inner ibex]

contra

positie

contrapositive?

i thougjt contrapositive was if P => Q then not Q => not P

[onyx rune]

this is correct sorry

im not native sorry

[inner ibex]

yeah no work

[inner ibex]

its ok

[onyx rune]

i checked in google

[inner ibex]

is it corect

[onyx rune]

so i think youre right

[inner ibex]

can u ols check this?

srry am asking so much my test is tomorrow 😅😅

.close

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[clear gust]

Can anyone help me understand this quadratic function

[clear gust]

L is y=1over2x^2

[restive river]

This quadratic equation is of the type: y=ax

[clear gust]

Yea A on the x thingy is 4

[restive river]

Are you saying that the x-coordinate of point A is 4?

[clear gust]

Ya

[restive river]

Hmmm, so a=1/2

Because 8=1/2*4^2

Therefore, P= 1/2*(-3)^2=4.5

[clear gust]

Wait this is mb

Let the x coordinate of point p be a, and the y coordinate be b.

When the range of values ​​of a is -3≦a≦2, the range of values ​​of b is ?≦b≦?

[devout snow]

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[desert cloak]

yall smart aint even gon lie

[desert cloak]

whats the difference between a ray and a line segement?

[dense lynx]

a ray ends a point and stretches on forever

a line segment ends at two points

(looking this up online might give you a more specific answer)

[lost laurel]

A ray is a line that extends to infinity in one direction, a line segment is a part of a line

[glossy dew]

A ray is like o-----------> and it keeps on going on one end
A line segment is like o--------------o and it has two ends

[devout snow]

@desert cloak Has your question been resolved?

[devout snow]

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[restive river]

Can I multiply (3 x 3) with (3 x 1) in matrix?

[eager lodge]

matrix multiplication is defined for mxn matrix with nxk matrix

so yes

you get mxk matrix as result

[restive river]

Thanks

.close

[devout snow]

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[devout snow]

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• Show your work, and if possible, explain where you are stuck.
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[ionic prism]

I'm trying to understand better recursive sequence formulas.
Let's say I have $f_{n}=3\cdot(f_{n-1})^2$ with the base case of $f_1=a$
How do I find the simplified formulas? The one without recorsion

[woven radish]

horizon2.0

[sand dove]

there are a few ways you could do it

have you heard of AG progression (AGP)?

arithmetico-geometric progression

[ionic prism]

Nope

But, might have in a different language

[sand dove]

$u_n = au_{n-1}+b$

[woven radish]

rafilou2003

[sand dove]

you know how to solve this?

[ionic prism]

Not really

[ionic prism]

I can try to like I did with my example but haven't really learned it

But the issue is I'm not sure that my way of doing this is right

[sand dove]

what have you tried? Show your work

[devout snow]

@ionic prism Has your question been resolved?

[ionic prism]

Not really organized but I hope it's understandable

[sand dove]

I get $f_n = 3^{2^{n-1}-1}f_1^{2^{n-1}}$

[woven radish]

rafilou2003

[sand dove]

if you didn't get that then maybe there is something wrong with your approach

[ionic prism]

Can you explain how you got to it exactly?

[sand dove]

Which progressions do you know immediately how to solve

like you have a well nice formula for it?

[ionic prism]

I got really close probably just a problem counting since it's my first time doing something like that

[ionic prism]

We don't really have anything like that

[sand dove]

really

like if $u_n = u_{n-1} + d$ you have no formula?

[woven radish]

rafilou2003

[ionic prism]

It's a course in discrete math/combinatorics so we got almost no formulas, we derive them ourselves

[ionic prism]

This is an arithmetic series

I know the basic ones, but we don't use those

Like we were once asked to find the formula for the Fibonacci sequence using induction

And I'm completely lost as to how to do these kinds of questions

[devout snow]

@ionic prism Has your question been resolved?

[sand dove]

so

I will give you two steps as to reliably solve your question

step 0 : know that if you have a geometric progression $u_n= au_{n-1}$ the solution is $u_n = a^{n-1}u_1$ (you should have seen this)

[woven radish]

rafilou2003

[sand dove]

step 1 : solve for an arithmetico-geometric progression $u_n = au_{n-1}+b$

[woven radish]

rafilou2003

[sand dove]

this has a few parts :

• find l such that l = al+b
• show that v_n = u_n - l is a geometric progression
• use step 0 to get the formula for v_n in terms of v_1
• add l on both sides to get the formula for u_n

And finally, step 2

the one that ties it all together

[sand dove]

find a recurrence formula that $\ln(f_n)$ verifies

[woven radish]

rafilou2003

[ionic prism]

Alright I'll try it and update

[ionic prism]

So I created a new function called g such that $g_n=\ln{f_n}$ and $g_{n-1}=\ln{f_{n-1}}$

[woven radish]

horizon2.0

[ionic prism]

And got to the formula $g_n=2^n\cdot(g_1)+(2^n-1)\cdot\ln{3}$

[woven radish]

horizon2.0

[ionic prism]

So I get that $f_{n+1}=a^{2^n}\cdot3^{2^n-1}$

Assuming a is $f_1$

[woven radish]

horizon2.0

[ionic prism]

I got exactly 1 more power of 2 than you I just see

Wait no

[woven radish]

horizon2.0

[ionic prism]

I corrected this

Thank you I got it

And how can I do so for more complicated sequences?

For example I got this one:
$f_{n+1}=f_n\cdot f_{n-1}$

[woven radish]

horizon2.0

[sand dove]

similar trick

what does ln(f_n) verify

[ionic prism]

It's all expressed with f tho

[sand dove]

yes but

now if you let u_n = ln(f_n)

u_(n+1) = u_n + u_(n-1)

tada, fibonacci

so you would solve identically to fibonacci

and then revert back to f_n = e^(u_n)

[ionic prism]

Ah wow

[ionic prism]

Thanks

.close

[devout snow]

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[restive river]

Where did i go wrong in my calculation? How is my sign different?

My professor said this:

^ thats why i flipped the sign

but when i look at the textbook

the rule is a bit different

and seems like it wouldnt apply here?

not sure

can anybody help me