help-38

[rapid horizon]

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[wide ravine]

Quick question

[wide ravine]

Is 1 and half :2:3 the simplest form of this ratio

Is it 3/2:2:3

Or 3/4/6?

Which is considered simpler

[stoic iron]

Depends who you ask

[dapper swift]

I would expect 3:4:6 tho

[stoic iron]

Generally ratios written without fractions I would consider easier to interpret

[wide ravine]

Ok

Simplest form tho

I jus dk which one to go with lol

U do write 1.5 as 3/2 tho correct

[dapper swift]

Yes

[wide ravine]

Thanks so much

[dapper swift]

np

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[rotund spear]

Anyone have any idea on how to find the nth derivative of a trig function without doing successive differentiation

[gritty harness]

You need to write the first few to find a pattern, no other way

[rotund spear]

I see

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[real pebble]

Have you tried plugging in

[wraith hinge]

zero?

[young arch]

i forgot to mention

i already have the answer

its 0 but my teacher is saying its 1

i have no idea why

[wraith hinge]

is he crazy

[real pebble]

FR

[young arch]

nvm they saying dne

not one

[real pebble]

Look at a graph of the function

You'll see that it's clearly 0

[young arch]

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[wraith hinge]

Can i get help with these

[past river]

What have you tried

[lapis star]

!status

[trim jolt]

What step are you on?
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2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
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[wraith hinge]

2

[past river]

Can you show your work please

[wraith hinge]

so for 3c i got the answer as PI and that its convergent

yea i can hol on

[past river]

It does converge, but it‘s not exactly pi. You‘re close though

[wraith hinge]

its rly all over the place

but for 3a and 3c i got it converges

i have no idea how to do 3b

[past river]

3b is the gamma function if you‘ve heard of it. If not, integration by parts should work

Wait it won‘t nvm

[wraith hinge]

nah i dont know gammma function

[past river]

3a is integration by parts, as for 3b, let me think about it

[wraith hinge]

its all improper integrals i just dk bout 3b

[past river]

Yeah. I‘m not sure how you‘d do that without using the Gamma function. It does converge tho, that‘s for sure

[wraith hinge]

yea our teacher hasnt taught us that

[rough goblet]

how is 3B gamma?

[wraith hinge]

but he makes the questions harder then what we do in class normally

[rough goblet]

3B is defo not gamma

gamma is from 0 to inf

[past river]

Dude

[rough goblet]

you forget

[past river]

Sub

[rough goblet]

it's gamma and incomplete gamma mixed

[past river]

Yeah exactly

[rough goblet]

that's my point lol

ur not entirely wrong

[past river]

But if they haven‘t gamma no point in mentionning that

[rough goblet]

just adding on

agreed

fyi IBP doesn't work for 3B

idt

[past river]

Yes I did realize that

And said it

[rough goblet]

ah mb

didn't see

,w plot e^-t * t^(2/3)

king's property...? no

[past river]

All good. If you know how to do that without gamma function then feel free to take over this help channel cause I know I don‘t. Only thing I can think of is maybe get this to look like a Gaussian integral but even then, would be meh to solve

[wraith hinge]

im guessing the teacher mustve spiked the difficulty

[pseudo grotto]

what class is this for

[wraith hinge]

this is for a calc 2 question

[pseudo grotto]

looks like contour integration

ah nvm

[wraith hinge]

yea this is improper integrals in calc 2

[rough goblet]

huh

first and last are easy enough but i can't think of what to do for 3B

[pseudo grotto]

maybe rewrite e^-t as a power series, divide by t^2/3 and interchange limits

[rough goblet]

huh

didn't think of that

so fubini and taylor basically

[pseudo grotto]

idk what else you can do in a calc2 setting

[rough goblet]

erm

[pseudo grotto]

[rough goblet]

one sec

$$\int_0^1 t^{-\frac 23} \sum_{n=0}^{\infty} (-t)^n \text{ d}t$$

[solid kiln]

ren

[rough goblet]

@pseudo grotto

look good?

uhhh
tbh not entirely sure what to do now

$$\int_0^1 \sum_{n=0}^{\infty} \frac{(-t)^n t^{-\frac 23}}{n!}\text{ d}t$$
$$\sum_{n=0}^{\infty} \frac{(-1)^n}{n!} \int_0^1 t^{n -\frac 23} \text { d}t$$

tf????

huh

@pseudo grotto

better?

@wraith hinge u following this?

[wraith hinge]

how could i plug in to find convergence/divergence from there?

[rough goblet]

not really? expected

[rough goblet]

i mean

idk abt convergence/divergence

it isn't really necessary

just... solve the integral?

then plug in the resulting value

and u get the form e^{something}

[wraith hinge]

i need it for the question prompt

[rough goblet]

solve the integral here, then figure out the taylor series for e^{whatever}

[wraith hinge]

ah

[rough goblet]

"compute each of the following integrals"

[pseudo grotto]

you're missing an n! in the denom but yea

[rough goblet]

is what i see

[rough goblet]

MB LMAO

[wraith hinge]

i think it means i have to show converge or diverge too.

[solid kiln]

ren

$$\int_0^1 \sum_{n=0}^{\infty} \frac{(-t)^n t^{-\frac 23}}{n!}\text{ d}t$$
$$\sum_{n=0}^{\infty} \frac{(-1)^n}{n!} \int_0^1 t^{n -\frac 23} \text { d}t$$

[wraith hinge]

This is what chat gpt showed what do yall think?

or is it just wrong asl

[rough goblet]

!nogpt

[trim jolt]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[rough goblet]

tf is "e^-t x"

[pseudo grotto]

then the integrals become $\left.\frac{t^{n + 1/3}}{n+1/3}\right|_0^1$

[rough goblet]

so it turns into...

hm

ah

uh-

[solid kiln]

chebyshev's infinite pee norm

[rough goblet]

what happened to eval bar

yea

which is just 1 / (n + 1/3)

lemme latex that rq

[pseudo grotto]

which is just $\frac{1}{n+1/3}$

[solid kiln]

chebyshev's infinite pee norm

[pseudo grotto]

then the sum becomes $\sum_{n=0}^\infty \frac{(-1)^n}{n!}\frac{3}{3n+1}$

[solid kiln]

chebyshev's infinite pee norm

ren

[rough goblet]

tf??

@wraith hinge

and to be honest, no idea how to find sum convergence for that

uhhh

*mumbles*

[wraith hinge]

yea i think my teacher just dragged it

[rough goblet]

lmao

[wraith hinge]

doesnt exactly seem like a standard improper integral problem for calc 2

[pseudo grotto]

any divergence test prob works

[rough goblet]

it isn't

[rough goblet]

i may or may not have forgotten all of them

let's see

we had...

root test

r- something test

limit comparision

those are all i can remember

[wraith hinge]

Ratio, root, alternating series, integral test

ratio test might work here

[pseudo grotto]

wait uh

[rough goblet]

yea probably ratio

[pseudo grotto]

you're meant to actually find the value of convergence

[primal steppe]

i think question 3b is a mistake

[pseudo grotto]

not just show it converges

[rough goblet]

ahahaha

oops

erm

[primal steppe]

the answer for 3b is not elementary

[wraith hinge]

imma just take is as that cause no way bruh expects me to figure this out. If he puts it on a test imma just hold that

[rough goblet]

u sure fam

,w int from 0 to 1 of e^-t * t^(2/3)

yep

like i said

incomplete gamma

so nonelem

[wraith hinge]

There is 12 questions and each have 2-3 parts. imma just move on so i dont get cooked

[rough goblet]

alr

[wraith hinge]

professor tweakin

/stop

!stop

!end

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[vague pier]

$4x dy - y dx = x^2 dy$

[solid kiln]

rainy

[vague pier]

How would you solve this diff. eq.?
I first have to know what form it is

So I was thinking, first divide all by dy

then im left with:

$\frac{dx}{dy} = \frac{x^2}{4x-y}$

But now idk what to do

[kindred pier]

I would divide by dx instead of dy. Seeing dy/dx is much more common

[kindred pier]

Also, you made a slight algebra error here. Better double check

[vague pier]

Okay, I just did dy because then the right side cancels out and i just get x^2

$4x\frac{dy}{dx}-y=x^2\frac{dy}{dx}$

[solid kiln]

rainy

[vague pier]

think i found something

$y=\dfrac{C,\sqrt[{4}]{x}}{\sqrt[{4}]{x-4}}$

[solid kiln]

rainy

[vague pier]

Appears to be correct

thx anyway

c.lose

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[open kelp]

i have to find the nth term

i dont quite get that

[wraith hinge]

well start with the numerator

what progression do u notice there

assume ur n starts from 1

[open kelp]

i got the nominator

its the denominator

it has +2 in it

i dont get the +2 part

[wraith hinge]

well remember u r starting from n = 1

so like

realise that they are powers of 3 of course

but also realise that they are starting from 27 (so no 3, 9 for example)

and u know 3^3 = 27

AND you also know it just carries on the powers of 3 normally from there, so 3^3, 3^4, 3^5, ...

so ur index starts from 1, ur first term should start with exponent 3, and the successive terms should increment that value

does that help you piece it up in your mind better

[open kelp]

ne

see what i did was 27*3 ^n-1

thats what i did

[wraith hinge]

ok sure but thats the same thing

,, 27\23^{n-1} = 3^3 \23^{n-1} = 3^{n-1+3} = 3^{n+2}

[open kelp]

ohhhhhhhhhhhhhhhhhhhh

so im write

even if dats written i the marking scheme

[wraith hinge]

ye

they're equivalent

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[drowsy shell]

So far I've got ((n-3)!/(4(n-8!)))((n-2)+2(n-3)+3(n-4)+...+(n-4)3+(n-3)2+(n-2))
I think I should be able to write the (n-2)+2(n-3)+3(n-4)+...+(n-4)3+(n-3)2+(n-2) as a summation but I haven't studied summation in school yet and I'm quite bad with it so I'd need help with that
Also I think I can just simplify the first part like this: (n-3)!/(4(n-8)!) = (1/4)(n-3)(n-4)(n-5)(n-6)(n-7) but I'm not competly sure if it's actually correct

[drowsy shell]

The formula I got comes from the combinations of everything but the yellow and white pawns (the fraction with the factorials) times the different ways the yellow and white pawns can arrange (first possible white pawn fixed position times ways yellow pawns can dispose+ second possible white pawn fixed position times ways yellow pawns can dispose ect.)

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[drowsy shell]

<@&286206848099549185>

[timid obsidian]

Hello, how can I help you?

[drowsy shell]

Hi I was trying to write a formula has a summation and needed help since I'm not good at summation

the formula is this one ((n-3)!/(4(n-8!)))((n-2)+2(n-3)+3(n-4)+...+(n-4)3+(n-3)2+(n-2))

[timid obsidian]

Ok…

[wraith hinge]

I cant help srry.

I aint advance in math

[timid obsidian]

:/

[drowsy shell]

I think that it's jut about the part starting from (n-2) cause then I should just be able to multiply the first part by the summation right?

so I have to write as a summation a series of addends that are the product of 2 elements, one goes from 1 to n-2 and the other from n-2 to 1

[timid obsidian]

Yep

[timid obsidian]

Yep

[drowsy shell]

alright

[timid obsidian]

Try using the X5-9 method

Gtg bye

[drowsy shell]

ok thanks

wait I might have got it is this right?

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[gaunt mist]

help

[gaunt mist]

oh i did it right

question 6

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2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
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7. None of the above

[gaunt mist]

1

[alpine raptor]

what is the formula for the space of a prism?

Hbside?

[gaunt mist]

i have no idea im only in the eighth grade

[alpine raptor]

Surface area right?

[gaunt mist]

yes

[alpine raptor]

okay then, so you know that the two small triangles are 8cm^2 each

[gaunt mist]

does ^ mean multiply?

[alpine raptor]

power

doesn't really mean anything, just the measuring unit for area

you have the area of the two triangles known to be 8cm^2 each

all the rectangles are of length 15, as it shown in the picture

and 2 of them are of width 4, and 1 is of width 5.7

area of rectangle = hxw:
area 1: 4x15 = 60
area 2: 4x15 = 60
area 3: 5.7x15 = 85.5
are of triangles:
area 1 = area 2 = 8
Combining all of them: 60+60+85.5+8 +8 = 221.5 cm^2

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[flat wing]

for these kinds of question

how am i supposed to know which direction v2 is going in?

[craggy sage]

U know alpha and angle between walls is 75° there u can find the third angle then using that u can work out like done in the first collision

[flat wing]

i know how to find B

but the question

doesnt original tell me the direction thno

[craggy sage]

I am having difficulty to understand ur point here

If u know beta doesn't that tells u which direction ball is going

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[thick hull]

i guess you draw a diagram and you can sort of tell which direction the ball is going to go

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[high ridge]

why was an X added in the first part of this question

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[elder saffron]

Have you learned the method of separation of variables?

[clever cave]

separating y terms with dy and x terms with dx

right

[elder saffron]

Yup

[clever cave]

i did that but my teachers answer key is different than my answer

i got y=sqrt(16x-16)

but my teachers key says

[elder saffron]

Ah so what did you get on both sides before you integrated

[clever cave]

dy-y = 8dx

[elder saffron]

Looks like you tried to multiply both sides by dx first

You will get dy=(y+8)dx

now you must divide both sides by (y+8) as a whole

[clever cave]

oh wait what

[elder saffron]

so you will integrate 1/(y+8) dy and 1dx separately

[clever cave]

i can’t just subtract the y in the (y-8)

?

[elder saffron]

Nope

[clever cave]

oh

[elder saffron]

it functions similiarly to the usual operations

Just think of dy/dx as a fraction

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[clever cave]

but how did my teacher get e

i’m still lost abt that

[elder saffron]

After you integrate 1/(y+8) you get ln|y+8|

thats where the e comes from

[clever cave]

I am so cooked for the ap exam

😔

i’m so close to an A in calc too, i’m at an 89.96 😭

[elder saffron]

are you taking BC or AB

[clever cave]

AB

i’m taking BC next year tho

[elder saffron]

Ah, don't fret then

Id say if you are just learning differential eq, they are very tricky

but once you get used to implicit differentiation, they become more intuitive

[clever cave]

here’s the thing, i get the concepts but after the unit passes and we move onto a new topic that doesn’t necessarily use the same concepts

i forget it entirely

like we learned implicit differentiation a while ago, but i forgot it

[elder saffron]

Hmm yes

I'd say practice more then

[elder saffron]

But yes differential equations is a big jump from implicit differentiation

Theres alot of integrals in the middle. When I was learning it, I forgot alot of differentiation as well!

[clever cave]

You got any study resource recommendations for the ap exam?

[elder saffron]

I really recommend this guy on yt called redpenblackpen

He talks about many calculus 1&2 topics

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@clever cave Has your question been resolved?

[clever cave]

@elder saffron i am currently at y+8=e^(x*ln|12|)

would it be right to simplify the right side into

12x

oh wait

12e^x

ohhhh

im stupid i figured it out

.close

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[wraith hinge]

Can anyone provide me the proof for this one?

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[edgy willow]

z=r*e^(i*theta) but the e part has magnitude 1, it just rotates around

[kindred pier]

Because $e^{i\theta} = \cos{\theta}+i\sin{\theta}$

[solid kiln]

SWR

[kindred pier]

So take magnitude of RHS and see that it is 1

any

[solid kiln]

snow

snow

[kindred pier]

correct

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[split epoch]

helpp please i feel so dumb

[vapid grove]

the area under the curve is distance traveled

[split epoch]

yesyes

but idt im doing the strips right or something??

[wraith arch]

does it want LRAM, RRAM, or MRAM

[vapid grove]

middle >.>

[split epoch]

ehat r these😭😭

[wraith arch]

left, right, and middle endpoint rectangular approximation

basically what height are we using for each strip

like the y value at the very left of the interval, the middle, or the very right

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@split epoch Has your question been resolved?

[split epoch]

ohh

yeah idek

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@split epoch Has your question been resolved?

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@split epoch Has your question been resolved?

[robust wagon]

i’m gonna assume they want middle riemann sum which would look something like this for the first 2 rectangles

since you want 5 strips of equal width, each width is 2

depending on left, right, or middle approximation, the height you use for the rectangles will be different

[split epoch]

ohh okay thank you

[knotty garnet]

@split epoch close it

[split epoch]

oh my bad soery

.close

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[knotty garnet]

Its alright

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[regal hatch]

how do I prove this?

[regal hatch]

im not sure how to make the left side = right side

[crisp lantern]

try shuffling things around

[nimble stone]

cos^4=(1-sin^2)^2

maybe could be of use

[regal hatch]

I was trying to find this one, but I only know cos^2

[crisp lantern]

its pretty simple once you shuffle things the correct way

[nimble stone]

ive just uses sin^2+cos^2=1

[nimble stone]

wdym

[crisp lantern]

try proving cos^2-sin^2=cos^4-sin^4

[nimble stone]

you cant do that

[crisp lantern]

this follows immediately

[crisp lantern]

why not

[regal hatch]

but that's cos^2 not cos^4

[crisp lantern]

ofc you can

[nimble stone]

if theyre proving an identity, there cant be any operations across the sides

[crisp lantern]

what

why not

this is completely equivalent to my statement

[nimble stone]

because then youre assuming the identity rather than proving the identity

[crisp lantern]

no

shuffling things around doesnt do that

[regal hatch]

@nimble stone can you explain this please?

[crisp lantern]

that only happens if you apply something non injective on both sides

like squaring or something like that

adding or subtracting something is completely fine

[regal hatch]

I now have
sin^2x + 1 - sin^2x - sin^2x + sin^4x

[crisp lantern]

did you try proving this instead?

[regal hatch]

where did this come from?

[crisp lantern]

this is the same as your statement with things shuffled around

[regal hatch]

when do I shuffle around equations like this?

instead of just turning the LS to RS

or vice versa

[crisp lantern]

i mean that depends on the question really

if it makes things easier

[regal hatch]

do I just expand:

(cos^2x + sin^2x)(cos^2-sin^2x)

[crisp lantern]

why would you expand this

you dont have to

[regal hatch]

what do I do then?

[crisp lantern]

you've proved it already

since sin^2+cos^2=1

[regal hatch]

ohh

[regal hatch]

may I ask how did you think that shuffling the equation around like this would make it easier

[crisp lantern]

as i said there's no hard written rule as such

you do it when it makes things easier

[regal hatch]

ah ok.. thanks 👍

.close

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[wraith hinge]

Hello! Can someone explain how the formula in b) works? Thanks!

[wraith hinge]

(Meaning why you do those calculations)

And I know a is 13 spades here, but what value does a usually stand for?

[crisp lantern]

usually you need to find the probability in each scenario to find the expected value
so the probability to get 0 spades 1 spade 2 spade and so on

then the expected value is 0*p(0 spades)+1p(1 spade)....

but there might be some trick here that i'm missing idk

ok there's one trick that i can see

the expected value of all four suits is the same so 4*E[x]=8

and you get E[x]=2

[wraith hinge]

Hmm, I'm confused as to what a stands for, like in general, not specific to this question

In this formula^

[crisp lantern]

here they seem to have taken a different approach

it's basically saying the expected number of spades is 8*(probability each card is a spade)

so a is the total number of spade cards in the deck

dont get too bogged down with the notation here i'd say

its completely unnecessary here

[wraith hinge]

Alright!

Thank you.

.close

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[wraith hinge]

.reopen

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✅

[wraith hinge]

Question: A geologist mixes 10 rocks with fossils in a bag with 90 other similarly sized rocks that do not contain fossils. She then removes 20 rocks from the bag at random.

b) What is the probability she will get less than 3 fossils?

c) What is the probability she will get at least 4 fossils?

Correct answer:

For b) I tried to do C(10,7)×C(90,13)÷C(100,20)

But the answer isn't correct, lol

I'm confused as to what exact calculations they did

[crisp lantern]

what are you trying to do here?

[crisp lantern]

isnt this the probability to get 7 fossils?

[wraith hinge]

Math : D I was trying to find the probability of 7 so I can minus it by 1 and get the probability of 3

Something indirect method something

But apparently not correct

[crisp lantern]

that gives you the probability to get any number of fossils but 7

[wraith hinge]

oh

[crisp lantern]

so 1,2,3,...6,8,9,10

[wraith hinge]

Applied the wrong formula ahahaha

I don't understand the formula that they did use

[crisp lantern]

the probability to get more than or equal to 4 fossils is 1 - probability to get less than 4 fossils

[wraith hinge]

So how do I get the probability to get less than 4

[crisp lantern]

you have to get probability of 3 in the first part first

less than or equal to 3 i mean

which is the sum of probability to get 0,1,2 or 3

oh wait it's just less than 3

so only 0,1 or 2

which they've done in the b) part

[wraith hinge]

https://tenor.com/view/pikachu-pokemon-cali7-nodding-nod-gif-15699416

[crisp lantern]

so what do you not understand here?

[wraith hinge]

The red line

I don't understand what calculation they're doing

[crisp lantern]

you have to do it like you did it here

for 0,1 and 2

instead of 7

[wraith hinge]

So, C(10,0)×C(90,20)÷C(100,20)

For 0

[crisp lantern]

yea

[wraith hinge]

Ok!

I got that part

[crisp lantern]

now i dont know if you can use the calculator here

to get the decimal part
you'll need stirling here ig

[wraith hinge]

I have it right here, doing calculations as we go

[crisp lantern]

ok cool

[wraith hinge]

Let me send a pic of my calculator

One sec

[crisp lantern]

but how will calculate these factorials

i dont think you can simply do that

[wraith hinge]

Factorial is n!, no?

[crisp lantern]

yeah i mean

it's a factorial

just do it and you'll see what i mean

i dont think it would be able to handle this

[wraith hinge]

We're trying to figure out P(x=3) and P(x<=2)

Right

[crisp lantern]

yes

[wraith hinge]

It says we use the complementary values we already know, what would those be?

[crisp lantern]

these

to get P(x>=4)

[wraith hinge]

Ok I see how we know P(x<=2)

But where does P(x=3) come from

[crisp lantern]

what do you mean

[crisp lantern]

because 3 is less than 4?

[wraith hinge]

Do we do C(10,3)×C(90,17)÷C(100,20)

[crisp lantern]

yes

[wraith hinge]

OK

I get it now

Thank you

Again

[crisp lantern]

did you calculate these tho?

[wraith hinge]

Yes

[crisp lantern]

can your calculator even do this?

[crisp lantern]

did it work?

[wraith hinge]

Yes

Yup yup

I did the calculations they showed

And got the same answer

[crisp lantern]

ok cool then

[wraith hinge]

Have a good evening!

[crisp lantern]

you too

[wraith hinge]

.close

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[delicate lance]

was my way of solving for x right?

[austere cedar]

You're using a lot of brackets for multiplication, which is going to throw you off someday

[austere cedar]

This line sus

[delicate lance]

Oh ok

[delicate lance]

Sus is crazy

[austere cedar]

You'll want this identity:
3^(5x) = e^(ln(3)5x)

Then by taking the ln of both sides, you can turn this into a linear equation for x

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[unborn estuary]

HELP MEEE

[maiden zinc]

Do you know log rules

Your first step is wrong

Can you see why?

Well first of all, set domain restrictions as well

@unborn estuary

[unborn estuary]

@empty yacht

[maiden zinc]

..

bro I’m literally helping you 😭😭😭

[empty yacht]

yes

me

i’m the one

asking

my other acc

that’s my phone

[maiden zinc]

oh

[empty yacht]

ok so

[maiden zinc]

You confused your log rules

[empty yacht]

oh

[maiden zinc]

But first of all

What is the restrictions on x calues

Values

[empty yacht]

uhm

[maiden zinc]

x must be larger than what

[empty yacht]

0!

v

[maiden zinc]

And

[empty yacht]

and

e

[maiden zinc]

No

[empty yacht]

aait

0 and

uhm

[maiden zinc]

argument must be larger than 0

Check first term

[empty yacht]

1/2

[maiden zinc]

ln (x-1/2)

Ye

[empty yacht]

okok

[maiden zinc]

So

[empty yacht]

so

[maiden zinc]

When you have an expression in your argument with subtraction or addition

It can’t be broken up

[empty yacht]

ok ok

so

i can’t do

[maiden zinc]

If you have an expression in your argument that is division or mitolicafikn it can

[maiden zinc]

Ye

[empty yacht]

okay

got it

[maiden zinc]

But

[empty yacht]

but

[maiden zinc]

You are adding two logs of the same base

You can mutiply their arguments

[empty yacht]

yes

so we turn ln to

log?

[maiden zinc]

no

you can keep ln

[empty yacht]

oh

ok

[maiden zinc]

You have ln(x-1/2)+ ln(2)

[empty yacht]

yes

[maiden zinc]

Recalling these rules

[empty yacht]

yeah

rule 2

[maiden zinc]

you have the top one in play

no

[empty yacht]

oh

[maiden zinc]

Its addition

[empty yacht]

oops

yes one

[maiden zinc]

so

Rewrite the left side

Into one ln expression

[empty yacht]

so it’ll be

ln (x-1/2) times (2)

[maiden zinc]

which is

[empty yacht]

which is

uh

give me a sec

[maiden zinc]

not times two

Only the argument

[empty yacht]

oh

[maiden zinc]

Is times two

[empty yacht]

uhm wait

[maiden zinc]

If you have ln(x), x is ur argument

if u didn’t know

[empty yacht]

Ln ((x-1/2) (2))

er what

[maiden zinc]

which is

That’s right

[empty yacht]

which is Ln (2x-1

[maiden zinc]

Just simplify

ye

[empty yacht]

ok then

[maiden zinc]

So you have ln(2x-1)=2ln(x)

Rewrite the right side

[empty yacht]

the right side

[maiden zinc]

Using rule 3

[empty yacht]

x^2

[maiden zinc]

ye

So

[empty yacht]

so

we solve

for x

[maiden zinc]

So you have ln(2x-1)=ln(x^2)

What’s the next step

You think

[empty yacht]

you

log

no

yes

[maiden zinc]

no

[empty yacht]

no

we subtract

[maiden zinc]

no

You could

But don’t

Easier way

[empty yacht]

we

[maiden zinc]

exponentitate

[empty yacht]

right

how

[maiden zinc]

do you know what that means

[empty yacht]

yes

power

[maiden zinc]

ok

so ln

Is log base e

[empty yacht]

yeah

[maiden zinc]

So you would do e to the power

Of both expression

So

[empty yacht]

ohhhh

[maiden zinc]

2x-1=x^2

[empty yacht]

ohhh

[maiden zinc]

Or you could just rember that bc they are the same on both sides

[empty yacht]

ok

[maiden zinc]

2x-1=x^2

But that’s where it comes from in essence

So now just move everything over

0=x^2-2x+1

And factor

[empty yacht]

you get

1

[maiden zinc]

Ye

[empty yacht]

OK

[maiden zinc]

And we said x>1/2

[empty yacht]

ONE MKRE

IS THIS RIFJT

[empty yacht]

oh

yeah

[maiden zinc]

so it checks out

[empty yacht]

OK

and

sorry this

one

[maiden zinc]

Very end

You factored wrong

[empty yacht]

oh

OH

wait

x=1

[maiden zinc]

I would always try to keep x^2 positive

[empty yacht]

x=-2

[maiden zinc]

So you don’t mess up

Ye

That’s right

[empty yacht]

ok ok

THANKS

[maiden zinc]

👍

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@unborn estuary Has your question been resolved?

[empty yacht]

i’m bacckkckck

what’s going on here

oh nvm

ooops

e^5?

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[strong horizon]

can someone help me with this problem?

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[serene crane]

Why isnt this correct?

for ratio test?

My first answer is wrong, which would lead to the limit possibly being wrong

[rugged latch]

it wants u to simplify probably

[serene crane]

holy hell

it did

@rugged latch one more question

I made bvn+1 = 1/(n+1)!, and once I put in 6 it was smaller than .0005.

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[molten warren]

this confused me for a second because I know I don't use my name for my username fellow Gabe

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[serene crane]

hazzah fellow gabe

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[wraith hinge]

how do i prove 1/1+x + 2x/1+x^2 + 4x^3/1 + x^4..................... to infinity = 1/1-x

[zinc dove]

Uh I think you could say

Let’s set the series to S

And then multiply both sides by x

[wraith hinge]

oh

[zinc dove]

Wait no that doesn’t work as nicely as I thought in my head

[wraith hinge]

can i not do this using differentiation

i am studying it under

differentiation

chapter

[zinc dove]

Hmm

[wraith hinge]

so i think there must be some way

[wraith hinge]

also this would work if there was some geometric

proggresion

here is none

[zinc dove]

I didn’t see the numerators were being multiplied

[wraith hinge]

yeah

[zinc dove]

Maybe integrate?

Looks like it would cancel nicely

[wraith hinge]

i know integration due to physics

but

it dosent make sense to use it here

as i am learning it

the differential calculus first

then the integral

[zinc dove]

Same thing tho basically

[wraith hinge]

so i wanna do it clearly using idea of differentiation

[zinc dove]

If you have a series ln(1+x) + ln(1+x^2) +.. so on the derivative is the lhs

[wraith hinge]

yes but then why would i even do this if know some stuff beyond it actually the courses taught in schools in my country are illogical they make you apply math in physics chemistry you dont have any intuition about

[wraith hinge]

yes

i figured that

[zinc dove]

Same everywhere probably

[wraith hinge]

hmm

[zinc dove]

If you know how to take derivative you can find integrals

[wraith hinge]

i know how to find integrals

[zinc dove]

It’s not like there’s any new integration techniques used here

[wraith hinge]

hmm

what does it mean to integrate a sum

[zinc dove]

Well derivative of a sum is just sum of derivatives

So same thing with integration

[zinc dove]

Idk what to do from here though because if you combine into one ln it becomes a crazy product

[wraith hinge]

so you mean i integrate the general term t(n) then apply limit n tends to infinity

[zinc dove]

Yeah

[wraith hinge]

but can i say something about what that product will be equal to

i saw something like this in complex numbers where we have things like z^n -1 = ( z - 1)(z-a)(z-a^2)......(z^-a^n-1)

the roots of unity thing

i got it i guess product is just (1-x^2n-1)(1 +x^2n-1)

.close

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[zinc dove]

Yoo I got u

Wait lemme write this thing out

[zinc dove]

Come back

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[narrow prawn]

Didn't get that modulo 8 part...

[vague citrus]

odd power of 3 is in the form 2k+1
now, 3^(2k+1)=3*(9^k) which can easily be proved to be 3 mod 8

same way for even

[gritty harness]

,rotate

[narrow prawn]

No not that part ..but how $3^n +3^n =0 (mod 8)$ ?

[solid kiln]

ideal_37

[amber python]

because 3^m + 3^n + 1 is a square mod 8

and it's also odd

the only odd square mod 8 is 1

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[narrow prawn]

.reopen

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[narrow prawn]

Didn't get it like WLOG let 2b+1>2a, $$3^{2a}+3^{2b+1}=3^{2a}\cdot (1+ 3^{2b+1-2a}) $$ here 1+3^{2b-2a+1} is even$ ? Am I doing something wrong??

[solid kiln]

ideal_37
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[narrow prawn]

:/

Ohh well never mind got it lol

.close

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[fading pebble]

Not sure how to find a_2 and w, but I know that a_1 is the slope in a linear regression of the last points since -a2cb+a2cv is stable and negligible at the end

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[polar wigeon]

what terms are a_2 and w coefficients of in the solution equation

solving coupled ODE takes time

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@fading pebble Has your question been resolved?

[fading pebble]

Hmm not sure what you mean

[fading pebble]

I have found the general solution

But it’s extremely long and painful to look at

[polar wigeon]

i mean the data gives you loads of boundary conditions

but some boundary conditions will be better than others

its 2nd order so the slope and value at t=0 should be enough though

[fading pebble]

How do you see it’s second order?

[fading pebble]

How do I find the slope at t=0 here?

[polar wigeon]

you could probably do a crude (y2 - y1) / dt

[polar wigeon]

you have to take a derivative of one equation and substitute in the other equation to solve one

[polar wigeon]

although it also says "given the last observations in the table, "

maybe you're meant to use final value boundary condition and gradient?

boundary conditions are really important and can sometimes require more information about a process

i know nothing about blood concentration

[fading pebble]

I was told I could just do a linear regression on the final 5 observations

And then my slope, A, would be a_1

[fading pebble]

I guess so, but again I’d only get information about the first diff equation

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[slim mountain]

Why does it have e^f(x) in front ?

[slim mountain]

Isn’t it supposed to just be xe^f(x) times f’(x)

[burnt mulch]

Answered in another server ^

[slim mountain]

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[quiet remnant]

How to solve first quartile, we use the formula but the answer is not included in the choices 😭😭

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[carmine junco]

whats open and closed interval

[carmine junco]

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[quiet remnant]

.reopen

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[bleak basalt]

"How many multiples of 3 less than 1000 can be formed from the digits 2, 5, and 9?"
I'm learning fundamental counting principles with problems such as: "how many even 2-digit integers less than 50 are there" This were never too hard to solve; however, that first above stumped me. How do even I solve this not manually?

[stark sandal]

0?

if something is a multiple of 3, then it also has to be divisible by 3

and the condition for divisibilty by 3 is if the sum of digits is divisible by 3, then the number itself is divisible by 3

but any number formed with the digits 2,5, and 9 will have sum of digits = 16

so nay number formed by 2,5, and 9 cannot be divisible by 3

[bleak basalt]

I think I'm beginning to understand, thanks
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[limpid fjord]

I need help with: "Prove that 1/1.4 + 1/4.7 + .... + 1/((3n-2)(3n+1)) = n/(3n+1)" My first idea was to try and find the limit of the sum of the series, but before that I set out to find the limit of 1/((3n-2)(3n+1)) and using that the lim of 1/n when n-> infinity = 0, i think that the limit is 0. So the sum is 1/4 + 1/28 + 1/70 and so on until the numbers we add on almost reach zero and they are each smaller than 1/4. So I guess it would make sense that the value of the sum should be smaller than 1/4 + 1/4 = 1/2 so smaller than 1/2. Then i checked the limit of the right side: n/(3n+1), again using the same law, i think it is 1/3. And that is indeed smaller than 1/2. But I have no idea now how to find the limit of the sum of the series so that it equals this expression.

[regal bough]

u prove this through induction

not through limits

[lyric hawk]

or you can use telescoping

[limpid fjord]

how so? I am not sure what induction is in my language 😅

[regal bough]

https://www.youtube.com/watch?v=tHNVX3e9zd0

it takes a specific case and generalizes it over a domain

[limpid fjord]

ooh i see

ook imma gonna give this a try.

[lyric hawk]

alternatively, you can note that $$\frac{1}{1\cdot4} + \frac{1}{4\cdot7} + \ldots + \frac{1}{(3n-2)(3n+1)} = \frac{1}{3}\left(1-\frac{1}{4} + \frac{1}{4} - \frac{1}{7} + \ldots - \frac{1}{3n+1}\right).$$

[solid kiln]

lpieleanu

[lyric hawk]

so a lot cancels

[limpid fjord]

oh nice

perfect then that approach makes more sense no idea why i started with limits

thanks!

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[deft moth]

what exactly is the question asking for

?

[thorny gazelle]

Equivalence means. It's Reflexive, symmetric, and transitive. So you have to prove these to show its equivalence.

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@deft moth Has your question been resolved?

[deft moth]

hm

ik the defintion of equivalence

a relation on N would be a subset of NxN

[deft moth]

these are not subsets of [5]x[5]

so what exactly does it mean "write down the correspondng equivalence relation"

<@&286206848099549185>

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[tight prism]

I need help with Combination

Inside a box there are 7 red balls 5 white balls and 3 green balls, how many ways can you select 3 balls that are

• all red?

[tight prism]

so apprarently what our teacher said is that the total number of objects are 15 since 7+5+3

then the selected objects r 3

but Idk what to do next

when I solved it what comes out is 455

[runic dew]

There will be only 1 combination if the balls are all red

[tight prism]

how

[runic dew]

What is the original question

This comes directly from the definition of combination

combination is what something consist of

[tight prism]

Idrk wait let me reconstruct my question:
A BOX CONTAINS 7 RED BALLS, 5 WHITE BALLS, AND 3 GREEN BALLS. IN HOW MANY WAYS CAN WE SELECT 3 BALLS SUCH THAT THEY ARE ALL RED?​

[runic dew]

If you want to regards the balls as individually different, doing so we would get $(765)$/3! (the 3! factorial accounts for permutations)

[solid kiln]

team132

[tight prism]

oh

wait

Why is it 7 6 5?

whered it come from

[runic dew]

Think about it

[tight prism]

my head hurts 🥲

[runic dew]

I think the actual answer they want to go for is $765$ not the 3 factorial

[solid kiln]

team132

[tight prism]

how did u come up with 765?

[runic dew]

Take your time. We are applying the rule of product axiom (refer to wikipedia if need be). An axiom is statement that is true an cannot be proven,

[tight prism]

okay

[tight prism]

ur actually right haha

but is 1 rlly the final answer though?