help-0

[cedar kraken]

part b is what im struggling on

[lapis panther]

ohhh

part B is on the basis of part A

if u can send a picture of ur work of A, it'd be easier to help u understand

[lapis panther]

so for square x is the side so area is x², for rectangle y is the the width and 4y is the length, so area is 4y² so now A = x²+ 4y²

now here u can get y= (80-4x)/10

put this value of y in the equation A= x²+ 4y²

you'll get the answer

lmk if u didn't understand something

isn't it 5y here instead of 10y?

[cedar kraken]

ummm

[lapis panther]

wait nvm it's 10y, yes

[cedar kraken]

yeah i tried subbing it in and got x^2-1.6x-48=0

the answer says A=1.64x^2 - 25.6x_256

[lapis panther]

can u send the picture of ur work

[cedar kraken]

wait i didnt squre it hold up

[lapis panther]

alr alr

[cedar kraken]

@lapis panther

[lapis panther]

expand the (20-x)² and simplify it more

yep you're getting the write answer if u simplify it more

[cedar kraken]

this right then?

[lapis panther]

you just divided the whole equation with 25

[cedar kraken]

yep then thats the correct answer

far out thats a stupid question

[lapis panther]

but u supposed to divide only 16(20-x)² by 25

[cedar kraken]

.close

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[cedar kraken]

cheers

i get it now pre sure thanks hitman

[lapis panther]

wait did u get the right answer?

[lapis panther]

this

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[gloomy steeple]

Is anyone familiar with kuhn tucker conditions? Need help proving this

[gloomy steeple]

Part of a general optimisation question and I have reached this and need to prove it works

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[mossy current]

What relationship does this graph have? Strong, medium or weak?

[zenith carbon]

Strong correlation

When it’s scattered with no pattern it’s no correlation, strong correlation is when the graph is going slowly up in a strsight fashion and weak correlation is the opposite of strong

What’s medium

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@mossy current Has your question been resolved?

[mossy current]

Well the graph looks medium because its not all going up continuously.

The graph looks like a medium relationship to me. I'll go with your answer, strong though.

but why would it be strong?

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[vivid kayak]

Can I get some help with a question?

[vivid kayak]

nvm .close

.close

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[vestal igloo]

is the following the correct method to answer the above question?

so basically take all the winning/correct paths

calculate their probability branches

and then add them together to get the answer in the square box at the bottom

[split temple]

I mean, you could do it that way I guess.

But it's horribly tedious.

[flat ore]

^

[split temple]

I would recommend an easier solution.

Namely:

[flat ore]

combinatorics

[split temple]

Number the yellow balls from Y1 to Y5, number the red balls from R1 to R3. Find out how many total combinations there are of 3 balls, find out how many total combinations there which contain ONLY R1 or ONLY R2 or ONLY R3.

[vestal igloo]

ok so basically write out at the start the possible winning combinations?

R Y Y

[split temple]

Don't write them out.

[vestal igloo]

Y R Y

[split temple]

Calculate them.

[vestal igloo]

Y Y R

[split temple]

Hmmm, this isn't right.

[flat ore]

Y1 Y2 R1
... so on

[split temple]

These combinations are all equivalent, because order doesn't matter.

[split temple]

This is correct.

[vestal igloo]

yeah that's what the tree diagram shows too

each branch has equal probability

so the answer was basically 3*(branch)

[flat ore]

do know what $nC_r$ is?

god

[ocean seal]

ホタル

[vestal igloo]

no?

[split temple]

You've also goofed up in your solution, come to think of it.

[flat ore]

or this $n \choose x$

[vestal igloo]

have i?

[ocean seal]

ホタル

[split temple]

Indeed.

[vestal igloo]

no I've not seen either method before?

[flat ore]

i think it is correct

[flat ore]

leave it then

[vestal igloo]

if there's a quicker method i would appreciate it

this is a non calculator question and it took me quite a while

[flat ore]

$\frac{3C1 \times 5C2}{8C3}$

[ocean seal]

ホタル

[split temple]

Then I seem to have made a mistake, I think.

Ah, yes, I have. You're right.

I mistakenly considered permutations rather than combinations in my analysis.

[flat ore]

$\frac{3C1 \times 5C2}{8C3} = \frac{5*(54)/2}{(876)/6} = \frac{510}{8*7} = 30/56$

[split temple]

Yup, this is correct.

[ocean seal]

ホタル

[flat ore]

this is the easier method

[vestal igloo]

what does the C stand for?

[split temple]

"Choose".

[vestal igloo]

ok thanks

[flat ore]

nCr means choosing r objects from n objects

[split temple]

It's very convenient notation.

[vestal igloo]

and this is combinatronics?

[split temple]

It is!

Combinatorics, to be precise.

[flat ore]

https://youtu.be/XJnIdRXUi7A

if you'd like to learn more about it

[vestal igloo]

ok thank you for your help both, I will have a read about it

[flat ore]

it's quite helpful

[split temple]

👍

[vestal igloo]

it looks very convenient, thank you

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[undone rivet]

S1 = 2 btw
Does anyone know why the substitution y=2-x gives the roots for alpha + beta and so on...
shouldn't it give the roots for 2-alpha etc.....
https://papers.gceguide.com/A Levels/Mathematics - Further (9231)/2021/9231_s21_qp_13.pdf
this is the paper and the 2nd question
Ping me if you know why please, and thanks in advance

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@undone rivet Has your question been resolved?

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@undone rivet Has your question been resolved?

[undone rivet]

bump

[icy sun]

What's S1 actually equal to?

[zenith carbon]

😐😐😐😐😐😐😐

😐😐😐

[icy sun]

I can read

But there's an expression from which they got 2 from

[undone rivet]

ahh that is sum of the roots

[icy sun]

Mhm

[undone rivet]

like alpha + beta + gamma

so -b/a

b is -4 and a is 2

-(-4)/2 = 2

[icy sun]

yup

[undone rivet]

So do you know why the substitution y=2-x gives alpha+ beta, .... roots?

instead of 2-alpha roots?

[icy sun]

Yup

so it's not y = 2-x it's y = alpha + beta + gamma - x correct?

Does that help?

[undone rivet]

uh

yes I understand that part

oh...

ok I understood ty ty

[icy sun]

Sounds like you got it

👍

[undone rivet]

yes tysm

[icy sun]

Don't forget to close

[undone rivet]

.close

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[undone rivet]

was about to

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[orchid pulsar]

need to find P(X=2) (solving part e). But before even attempting to solve it the hint tells me this is a homogenous system

[orchid pulsar]

how is this a homogenous system? the probabilities are not equal becaise its without replacement

if first key didnt work there is a higher chance my next key is going to work

i did get 1/10 which is correct but sstill dont see how its a homogenous distribution

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[orchid pulsar]

<@&286206848099549185>

[placid zinc]

I may not be familiar with the term. What does "homogenous system" mean here?

X follows a negative hypergeometric distribution, which isn't one I see often haha

[orchid pulsar]

yeah I also though tit follows a hypergeometric distribution. Anyways this is what they mean by homogenous:

ur probably familiar with the formula so I guess u can see what they mean

[tacit arch]

so it means discrete uniform?

[tacit arch]

show the whole problem for context

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[neon berry]

aye

[neon berry]

With numbers 1-7 you can form 5040 seven-digit numbers, if every number is going to contain all seven numbers. Think that you're writing all the numbers by value, beginning with the lowest number. What number will fall on place 2161?

[alpine sable]

with 1 being the first you can make 720 numbers, with 2 being the first you can make also 720 number, 3 also 720 and 720+720+720=2160 so you need the next number which will be 4123567

[neon berry]

So I basically multiply 720 by 2161 in order to get the number

[alpine sable]

?

[keen pasture]

Nice solution :)

[neon berry]

Nevermind that, we know that 2161 begins with number 4, we can now easily find the solution

which should be 4123576

[neon berry]

In the future, are there formulas you could recommend for combinatorics?

[alpine sable]

4123567

[alpine sable]

just do as many problems as you can, better than just looking up and learning formulas

[neon berry]

so the 2160'th is 3765421, since it's the last and largest digit of numbers with 3 being first

[alpine sable]

yeah should be

[neon berry]

thanks for the answer nevertheless

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[undone agate]

Not really sure how to continue, dont think my work is correct either

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[cloud zodiac]

Is this still correct as long as it a multiple of 4?

[quasi scarab]

no

u have 4n^2 twice

[alpine sable]

The thought is correct but the calculation is not

[quasi scarab]

the question is a bit duh though, even a single even number squared would be always divisible by 4

[cloud zodiac]

Oh yea mb I forgot there was 2

Thanks for the help

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[cloud zodiac]

.reopen

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✅

[cloud zodiac]

I have one more question

On the mark scheme they did this. Did I need to do this. Or is it ok? as long as it a multiply of 4.

Mb for asking the same question just want to be sure

[alpine sable]

This might be better because it says for any two even numbers

In the first one ur even numbers are dependent on n

[merry depot]

proving something is a multiple of 4 is (often) showing that it can be written as 4 times something. So yes you should factor out the 4

and also what little said

[fossil cedar]

I mean
(2n)² + (2m)²
= (2n+2m) (2n-2m)
Works too

But it varys on the question

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[cloud zodiac]

Ty for the help

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[oak temple]

Hello

[oak temple]

Is it possible I can get some quick help

[lime matrix]

Send your question

[rare gale]

just post your question

[oak temple]

I was trying to workout

Them

What confused me was how it is 111

[rare gale]

all the edges are cut off of your screenshot...

[open folio]

How is what 111?

[oak temple]

The two radius

[open folio]

What?

[oak temple]

Make it an

Isosceles

Triangle right

[open folio]

Yes

111 + x + x = 180

[low lily]

Look at 60 degree corner

CORNER

Shouldn't there be 90°

[oak temple]

Ok

Yes

55.5 is first one

[low lily]

Not sure is it 90 cant see from that screenshot

[oak temple]

The other is 85.5

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[open folio]

Do you know how to set up a proportion in sector area?

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[shell vault]

mo

no*

[tacit arch]

use
https://www.cuemath.com/measurement/area-of-a-sector/

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[golden sequoia]

What if I have two integrals that have bounds that have variables

[golden sequoia]

Which one do I evaluate first

[gloomy wolf]

which one of these will cause you to end up with a number?

try writing your bounds into the triple integral

and make sure you integrate with respect to a given variable AFTER all variables whose regions depend on it

if you end up with any variables at the end, you chose an order that doesnt make sense

[tacit arch]

so many videos exist about changing the order, but none about how to pick the right one

[golden sequoia]

so by default the order is dxdydz?

[gloomy wolf]

by default i usually go backwards

dy dx in a double, dzdydx in a triple

[golden sequoia]

so to change orders id have to make adjustments on the bounds?

[gloomy wolf]

you would only need to rearrange the bounds so they line up with the variable they correspond to

so yes, you could call that adjustments

[golden sequoia]

Ok thank you. So I don't need to do any of these

[gloomy wolf]

i think basically the idea is that the default actually IS dz dy dx

you would only need to make sure that you adjust your bounds to match the region you are talking about if you want to make any switches from that

if bounds do not depend on any variables, then you can switch freely without worrying about it

*any variables you are integrating with respect to 🙂

[golden sequoia]

ok thank you ill try it right now

[tacit arch]

absolutely integrable

[gloomy wolf]

lmao- i guess i am sweeping stuff like discontinuities and boundedness under the rug when i say that, as should definitely be pointed out 🙂

[tacit arch]

nah i'm just bored. not worth bringing up when it's clearly a first multivariable calc course

your explanations are great

[gloomy wolf]

thank you

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[tough dawn]

I need help with this problem

[gloomy wolf]

A good first step would be to plot those points in the coordinate plane

[tough dawn]

Ok I did that

[gloomy wolf]

ok so you are trying to find the length of the line connecting these two points

draw that line on your paper, and find the right triangle that has that line as its hypotenuse, and draw it into your graph

[tacit arch]

one of dem pythagorean triples

[tough dawn]

Ok I got 6 and 8

[gloomy wolf]

for your side lengths?

[tough dawn]

Yes

[gloomy wolf]

you now have everything you need to find the distance between the two points

how can we find the hypotenuse of a right triangle, given its two other side lengths?

[tough dawn]

I squared the two and added them and it made 100

[tacit arch]

https://courses.lumenlearning.com/waymakercollegealgebra/chapter/distance-in-the-plane/

[gloomy wolf]

there is one more step

[tough dawn]

Oh square it

The answer is 10

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[upbeat peak]

does this converge or diverge

[worn fox]

What tests do you know

[upbeat peak]

Alembert and cauchy criterions

comparaison

[worn fox]

Try comparison with something bigger

[upbeat peak]

$\frac{2n}{2^{n}}$

[ocean seal]

Elyazid

[upbeat peak]

i tried this but it doesnt give me any answer

[last ether]

Have you tried seeing whether this diverges or not

@upbeat peak

[upbeat peak]

how can i

[worn fox]

Ratio

[last ether]

I gotta stop laughing every time I see "ratio"

But yes

Do ratio/d'Alembert test

[upbeat peak]

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[slate kayak]

Anyone know what they did here?

Just remove the x from the 2 and put it under the tan x ?

Hiw is that possible

I know how to get the answer etc the only thing i dont get is how they removed the x

[worn fox]

Divide the top and bottom by x

[slate kayak]

Thx

I get. It

[slate kayak]

Do u know how to write that?

[worn fox]

You can times it by (1/x)/(1/x)

[slate kayak]

Oh yeah

Thx

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[tough ether]

How do I find all intergers n such that n-2 divides 2n

[raven rover]

I'd start by simply experimenting

[tough ether]

so n = 3, 6 seem to be the only ones that work

just by plugginh in small values

but I dont see why

[raven rover]

well you want 2n / (n - 2) = k for some integer k, right?

[tough ether]

yes

[raven rover]

Can you rewrite $\frac{2n}{n-2}$ in some way so it's easier to work with?

[ocean seal]

Shen

[tough ether]

the only thing that comes to mind to to muiltiple it by (n+2)/(n+2)

but that doesnt seem to do much

[raven rover]

Try adding 4 and subtracting 4 on the numerator

And splitting the fraction

[tough ether]

yea i see it now. ty

n = 3,4,6 are the only ones

[raven rover]

👍

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[silent ruin]

how dya do this?

[gloomy wolf]

use rules of exponents, and break it up into two different fractions

[silent ruin]

I'm sorry but I reeeaallly don't get it

[gloomy wolf]

how can you break up $$\frac{a + b}{c}$$ into two fractions

[ocean seal]

JamesH

[silent ruin]

a/c + b/c ?

[gloomy wolf]

right

step one

then theres a rule of exponents that says something about how we can rewrite $$\frac{x^a}{x^b}$$ without a fraction

[ocean seal]

JamesH

[gloomy wolf]

do you know what it is?

[silent ruin]

yeah by minusing it right?

[gloomy wolf]

right

[silent ruin]

so 20--15?

or 15-20?

[gloomy wolf]

20 - (-15) is the same as...

those two things you wrote are not the same

[silent ruin]

I know that

I'm confusing myself I'm sorry

[gloomy wolf]

thats ok

[silent ruin]

35?

[gloomy wolf]

yes

[silent ruin]

25--15 would be 40

[gloomy wolf]

ok great

so now we have a sum of two integers, which are powers of two

[silent ruin]

so add those two?

[gloomy wolf]

don't add them, read what the problem wants us to do

[silent ruin]

oh yeah

actually I kinda don't get it

[gloomy wolf]

what have we done so far?

we have rewritten the fraction, right? as a sum of two integers

what are those integers

[silent ruin]

yep

35 and 40

[gloomy wolf]

those are not the integers

[silent ruin]

oh

2^35 2^40?

[gloomy wolf]

those are the two integers

but what is our expression now

[silent ruin]

the two added?

[gloomy wolf]

yeah

ok

so we are asked to write this as m*2^k, where k is an integer, and m is an odd integer

[silent ruin]

its m*2^n

but yes

[gloomy wolf]

sure

[silent ruin]

yes

[gloomy wolf]

if we know an integer is odd, but dont know what it is specifically, how can we write it down?

it might be better to first talk about, what defines an even number?

how do we know if an integer is even

[silent ruin]

2 4 6 8...

[gloomy wolf]

you have listed some even integers

[silent ruin]

so odd would be 2x + 1

[gloomy wolf]

right

where x is an integer

[silent ruin]

yup

[gloomy wolf]

we want to write 2^35 + 2^40 as a product 2^n* (2x + 1), where x and n are both integers

how do we usually turn sums into products?

it has a name

[silent ruin]

just say it

[gloomy wolf]

i want you to think about it

[silent ruin]

I know but my brain is blank

[gloomy wolf]

how would you write x^5 + x^2 as a product instead of a sum

[silent ruin]

ohh that's what you're asking

(x^5)^2 ryt

or not

[gloomy wolf]

(x^5)^2 is x^10, which is not x^5 + x^2

[silent ruin]

wdym by product

like

[gloomy wolf]

multiplication

[silent ruin]

if its just adding

x^7??

[gloomy wolf]

is that true?

when x = 1, x^5 + x^2 = 1 + 1 = 2

and x^7 = 1

so those are not the same

[silent ruin]

right

[gloomy wolf]

how would you write the polynomial x^2 + 2x + 1 as a product of two polynomials

[silent ruin]

please just say it I really don't know or can't recall rn

[gloomy wolf]

do you know the distributive property?

im not going to hand it to you, you have to think of it yourself

[silent ruin]

ok ok ok

[silent ruin]

and I probably do but I've never heard my teacher say it b4

[gloomy wolf]

the distributive property tells us how to do a(b+c)

[silent ruin]

yeah I do know that

[gloomy wolf]

so what does it say?

[silent ruin]

ab + ac

[gloomy wolf]

ok

so the distributive property take a product , a(b+c)

and turns it into a sum, ab + ac

we want to do that, but backwards

[silent ruin]

oh so taking out a factor?

[gloomy wolf]

YES

thank you

in fact, taking out the greatest common factor

[silent ruin]

yep

[gloomy wolf]

so lets do that to 2^35 + 2^40

[silent ruin]

so it's 5..?

or not..?

it's not..?

[gloomy wolf]

what is 5?

right now we want to rewrite $$2^{35} + 2^{40} = 2^n(1 + 2x)$$

we are trying to identify $n$ and x

[ocean seal]

JamesH

JamesH

[gloomy wolf]

we need to pull out as many twos as we can, so that we are left with a 1 for one of them

[silent ruin]

5 is x?

[gloomy wolf]

you are close but not being very careful

there is a 5 somewhere in the expresion for x

[silent ruin]

it's in my dreams

u mean 2^5?

[gloomy wolf]

that is what i mean

yes

x is 2^5

and what is n?

actually sorry, technically x is 2^4

but you factored right i think

how much did you factor out?

[silent ruin]

?????

[gloomy wolf]

when you factor this

what do you write down

[silent ruin]

idk man

[gloomy wolf]

how did you factor it?

write it down

you factored it right

[silent ruin]

u mean 2^35 + 2^40?

[gloomy wolf]

thats what we started with

we factored stuff out

what do we end up with

[silent ruin]

I just took 5 out of the powers

like

[gloomy wolf]

no words, just write the equation

[silent ruin]

pls

[gloomy wolf]

2^35 + 2^40 = ...

[silent ruin]

bro fr you're killing me

[gloomy wolf]

you are killing yourself bro

[silent ruin]

bro ik bro

it feels like I'm getting held at gunpoint

[gloomy wolf]

that sounds like a you problem

[silent ruin]

I know bro

I just can't think

[gloomy wolf]

relax

do it with something simpler maybe

remember we want to do the distributive property

a(b+c) = ab + ac

but backwards

we have something that looks like ab + ac

and we want to write it as a(b+c)

[silent ruin]

I wrote 5(7+8) bro

[gloomy wolf]

ah i see

our question does not say 35 + 40

[silent ruin]

ik

it's a supposedly simpler way of doing it

[gloomy wolf]

so why are you factoring 35 + 40? factor the thing were talking about instead

[silent ruin]

isn't it 5 anyways

[gloomy wolf]

n o

we are factoring 2^35 + 2^40

what does 2^35 mean?

it means 2 x 2 x 2 x..... x2 35 times

[silent ruin]

a v long number

[gloomy wolf]

2^40 means 2 x 2 x.... x 2 40 times

[silent ruin]

yep

[gloomy wolf]

how did you factor 35 + 40

how did you figure out that it was 5

[silent ruin]

guessed it

[gloomy wolf]

you didnt guess it

you dont guess correct answers

[silent ruin]

BUT I DID BRO

WHAT DO YOU WABT ME TO SAY

[gloomy wolf]

why didnt you guess 3?

or 2?

[silent ruin]

??

why would I guess that

[gloomy wolf]

the same reason you would guess anything else

[silent ruin]

gcf of 40 and 35 is 5

that's literally ir

it

wdym want me to say

[gloomy wolf]

ok so you didnt guess

[silent ruin]

??

[gloomy wolf]

you factored out the gcf

[silent ruin]

yeah

[gloomy wolf]

right

so factor out the gcf of 2^35 and 2^40

[silent ruin]

isn't it too big

it's like

[gloomy wolf]

its big but its sitting right there in front of us

[silent ruin]

yeah I'm fucking blind bro

[gloomy wolf]

no you arent

you might not see this right away but you for sure can

do you agree that 2^40 = 2^35 x 2^5

[silent ruin]

so it's 2^35

[gloomy wolf]

thank you...

[silent ruin]

yeah

you're welcome

[gloomy wolf]

what are we trying to do with that?

[silent ruin]

with this?

can u just TELL ME

its been 1 hour

I'm not getting smarter

[gloomy wolf]

i wont tell you

good bye

[lethal stump]

bruh

so you split the fraction and then just use index laws right

[gloomy wolf]

he has gotten that far

[silent ruin]

yeah

alright its all good I'll figure the rest out by myself

ty for helping me

[gloomy wolf]

no problem

good lukc

[silent ruin]

ty

[lethal stump]

you can write 2^40 as 2^5 * 2^35

basically using the law in reverse

[gloomy wolf]

please dont tell him the answer

he has gotten this far

he needs to put the final pieces together

[lethal stump]

fair enough

[silent ruin]

I'm a her

but yeah

alright again ty

.close

[lone heart]

Channel closed

Closed by @silent ruin

Use .reopen if this was a mistake.

[lethal stump]

np np

[lone heart]

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[gloomy wolf]

how far have you gotten?

[lone heart]

@cinder schooner Has your question been resolved?

[cinder schooner]

im kinda confused about the whole question because if all drivers are gonna work then isnt d the correct answer?

since it can get use 360 tons per day and has the least cement trucks

[gloomy wolf]

that seems to be correct to me

[cinder schooner]

oh thats it rlly?

[gloomy wolf]

actually, im not sure that we can do the whole 360 tons with 7 dump trucks and 2 cement

oh wait nevermind

7 dump trucks x 8 trips per day x 6 tons is kinda laready good enough

[cinder schooner]

ye

[lone heart]

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

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[plush granite]

it got taken

mb

[alpine sable]

u good c;

[naive valley]

to start, first find h(10)

[alpine sable]

√8

is what i got

after doing it

[naive valley]

can you explain how you got that? what did you get for h(10)?

[alpine sable]

ok so

you plug in 10

so

2(10)-8
= 20-8
=12

[naive valley]

yep

[alpine sable]

then you plug in h

for f

so

[naive valley]

then g(h(10)) is just g(12)

[alpine sable]

g(12) = g = √12-4

=

√8

[naive valley]

yep

[alpine sable]

but

[naive valley]

all looks correct

[alpine sable]

im confused

[alpine sable]

it doesnt seem to be an option

[vestal wedge]

Wha..?

[naive valley]

and you're sure those answers go with that question?

[alpine sable]

precisely

[naive valley]

and do you see sqrt(8) in there, perhaps in another form?

[alpine sable]

how could i test

to find that

[naive valley]

can you simplify sqrt(8)

hint: 8 = 4*2

[vestal wedge]

sqrt(8) = sqrt(2×2×2)

[alpine sable]

basically

2x2x2

what Q said

[naive valley]

right

[vestal wedge]

Do you know that sqrt(a^2)=|a|?

[alpine sable]

no sir

i do not

[vestal wedge]

Oh

[naive valley]

more concretely, sqrt(4) = 2?

[alpine sable]

yes

[naive valley]

ok, do you see how you can simplify sqrt(8) = sqrt(4*2) = ...?

[alpine sable]

no

[naive valley]

another hint, in general for positive numbers a and b, you can do sqrt(ab) = sqrt(a)sqrt(b)

[vestal wedge]

Sqrt(ab)=sqrt(a)sqrt(b)

[alpine sable]

what would the value b represent

in this problem

[naive valley]

try a=4 and b=2 or vice versa

[alpine sable]

its like 2.82

oh ye

both are

[alpine sable]

what should i do now?

[naive valley]

well looking back at the options...

do you see one that equals sqrt(8)?

[alpine sable]

yes

2√2

[naive valley]

right

because 2 = sqrt(4)

so 2*sqrt(2) = sqrt(4)*sqrt(2) = sqrt(4*2) = sqrt(8)

[alpine sable]

yes

[alpine sable]

that makes alot more sense ty

I got another quick question regarding another problem

[naive valley]

ok

do you know what $(f \circ g)(0)$ means?

[ocean seal]

OurBelovedBungo

[alpine sable]

yes

f of g

you can write it as

f(g(x))

[naive valley]

yes

[alpine sable]

in this case

[naive valley]

so in this case f(g(0))

[alpine sable]

f(g(0))

yes

[naive valley]

ok

so same drill, first find g(0)

then feed that value to f

[alpine sable]

would g be

2x-2?

oh wait

im dumb

i mean

[naive valley]

you can even do it without trying to find a formula

[alpine sable]

o

[naive valley]

just read off the y value of the g function when x = 0

[alpine sable]

2

right

[naive valley]

yep

now feed that to f

i.e. find f(2)

[alpine sable]

would i look

at the y value again

[naive valley]

yeah, the y value (this time of the f function) when x=2

[alpine sable]

it looks like

its at 1.5

[naive valley]

mmm

x=2 is two ticks to the right of the origin

then looking up to see where that hits the blue function (f)...

doesn't look like 1.5 to me

[alpine sable]

oh

i did it

the opposite way

i did y =2

[naive valley]

ah

[alpine sable]

so im looking for

when the two lines meet

[naive valley]

which two lines

[alpine sable]

f(x) and g(x)

or

im looking for

when x= 2

where that hits the

blue line

[naive valley]

right

[alpine sable]

so thats

3?

[naive valley]

looks right to me

[alpine sable]

is that all i got to do?

[naive valley]

yep

[alpine sable]

oh thanks

i really liked your explanation of that

@vestal wedge thank you aswell

[naive valley]

cool, glad it helped

[alpine sable]

.close

[lone heart]

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Closed by @tropic hinge

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[lone heart]

Please don't occupy multiple help channels.

[ocean spindle]

how did my teacher know this angle was 45%

45 degrees *

is it because triangle is 4- 3 side lenghts

and 3rd is unknown

so its some special triangle

[lone heart]

@ocean spindle Has your question been resolved?

[echo wadi]

It was calculated using the direction they are heading in

[lone heart]

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

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

[ionic bison]

so I have this problem, and I'm trying to find the particular solution.
in problems of this type, where the right side function is an exponential, I've seen answers where the particular solution is of the form Axexp(λx) and some where it's just Aexp(λx)
how do I know which form of guess solution to use?

[slate jolt]

have you heard about the variation of parameters?

[ionic bison]

I have, but I thought I use undetermined coefficients if the right side is exponential

[slate jolt]

oh yeah

do that

[ionic bison]

right

but which guess solution do I use

for this problem

[slate jolt]

in this case its just Aexp(4x)

[ionic bison]

yeah ok

but I'm looking at this other problem

and the guess solution was Axexp(-2x)

what was the giveaway that it was different, or would I just try Aexp(-2x) first and see that it didn't work

[slate jolt]

oh

the x comes from the fact that -2 is a solution to the characteristic equation

x²+x-2 = (x+2)(x-1)

[ionic bison]

hey

that makes perfect sense

thanks!

.close

[lone heart]

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

2ln5x=6

[worthy badger]

ik u do like ln5x^2=6

then u add e to both sides so it's

5x^2=e^6

[slate jolt]

be careful its ln((5x)²)=6

[worthy badger]

?

[alpine sable]

U take the 2 to the power, then it acts as the power of the while argument

Whole*

[slate jolt]

call 5x y

2ln(5x)=2ln(y) =ln(y²) = ln((5x)²)

[worthy badger]

idk what's going on

whats wrong with dividing 5

[alpine sable]

What

[worthy badger]

here lemme send a oic

[alpine sable]

Look at this

[slate jolt]

theres nothing wrong

the mistakle is earlier

[worthy badger]

i dont understand that

[alpine sable]

What is e6💀

[alpine sable]

😞 write it out maybe

[worthy badger]

u add e onto both sides to cancel ln

[alpine sable]

e^ln(5x)^2 = e^6

Then you interchange (5x)^2 with e and you get
((5x)^2)^lne = e^6

So (5x)^2 = e^6

[slate jolt]

the rest seems right

[worthy badger]

so u just added more parentheses

[slate jolt]

no

left side you have 5x²

right side

25x²

[worthy badger]

right side is =6

u put or to seperate the 2

[slate jolt]

no i put or to show the result of (5x)² which is 25x²

[worthy badger]

ok

[alpine sable]

Think of it like this 2(ln5x)
2(ln5 + lnx)
ln5^2+lnx^2
ln25x^2

[worthy badger]

ohh ok

i get it now

.close

[lone heart]

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

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[past sundial]

Hello! I don’t really understand my hw and would like guidance on it.

It takes 9 pumps to fill a 18 inch soccer ball
It takes 30 pumps to fill a 27 inch soccer ball

How many pumps is needed to fill a 71 inch soccer ball?

[slate jolt]

remember the volume for a ball is something like this

V=a*R3

[past sundial]

4/3 pi r ^3

Yep

But I’m not sure about pump volume

Like how much does a pump pump?

[slate jolt]

here we have Volume = 9 pumps

a pump is a unit of volume but we dont know

how much it is

so use the first values to get a

9 pumps = a * 18^3

a is the good constant that relates radiuses in inch and volumes in pumps

so a = (18^3)/9

[past sundial]

Where is pi?

I’m confused

[slate jolt]

ok so i'm starting over, if you have a radius in inch

[slate jolt]

this formula gives the volume in inch cubed

and if we have the diameter d its 1/6 pi d^3

so you could do V in inches cubed which is pi/6*18^3

and you would have Vp in pumps be related to it linearly

so Vp = b * V

with b some constant

[past sundial]

how did 1/6 appear

[slate jolt]

so all in all you have Vp = b . pi/6 . d^3

[slate jolt]

4/3 pi r ^3 = 4/3 pi (d/2) ^3 =4/(3x8) pi d^3

[slate jolt]

so its overall Vp = a*d^3

with a = b*pi/6

and knowing a gives you b also

so you can use the formula for the volume of a ball if u want

but you can also use Vp = a*d^3

if you know what a is numerically

[past sundial]

Wouldn’t you do 3/2 so 8/3?

4/3/2 = 4*2/3 = 8/3

[slate jolt]

4/3 pi r ^3 = 4/3 pi (d/2) ^3 = (4pi/3) (d^3 /2^3) = (4pi/3) (d^3 /8)= (4pi d^3)/(3*8)=pi d^3 /6

w/e it doesnt matter

in the end you have Vp = a * d^3

whatever is a

containing the formula of the sphere and the pump to inch cubed ratio

[past sundial]

So 4/3pi r ^3 / p?

[slate jolt]

whats p?

[past sundial]

Pumps

To make ratio you divide both?

[slate jolt]

you divide by "pumps"?

nvm

[past sundial]

I’m very lost

[slate jolt]

i'm drawing it

2sec

but i started off by just saying Vp = a*d^3

without doing the first calulations

[past sundial]

oh I see you shifted the 2^3 = 8 and under the 4 ah

what does the 4th word say?
B is the (something)

[slate jolt]

ratio

constant or number would work

[past sundial]

ah, and is b base?

nvm I see

[slate jolt]

i dont understand your question

ok then

u can verify that a works also for ball 2

[past sundial]

Where does the part that vp = (b*pi/6)d^3 come from?

[slate jolt]

i substitute

the value of V from the left side in the equation on the right

so what u could do to solve this exercice is to say: i compute for each ball its volume in inch cubed

using the leftside formula

then using the volume in pumps given you find b

[past sundial]

Is “a” area?

[slate jolt]

or as i did, you say Vp = a * d^3 and you find a

a is just a way to call b*pi/6

it could have been anything really

its a constant that we dont know the value of before inputing the values given

[past sundial]

Vp = b((pi/6)*d^3

how come the parenthesis is around b and pi/6 and not like this?

[slate jolt]

well if you have three numbers x y and z

(x y) z = x(y z) = x y z

[past sundial]

yes

[slate jolt]

so i put the parenthesis to put an emphasis on the value of a

they could be removed

its just visual

[past sundial]

what does a mean?

[slate jolt]

well it means the same as 4/3 pi ion the regular formula with radius and same units

like it doesnt have a deep meaning

i just did some maths

and i arrived to Vp = (something)*d^3

i called that something a

[past sundial]

but p and a are 2 different variables

and my teacher once said that 2 variables are unsolvable unless you know the answer to one variable
uhhhhhh

4/3pi r^3 * p = a * d^3

4/3pi r^3 * p = (bpi/6) d^3

[slate jolt]

lets forget about a

i used it as a way of simplifying calculations but i think this way its simpler conceptually

[past sundial]

b = p?

[slate jolt]

yeah

b is p

[past sundial]

but how?

b is ratio using pumps

pumps is pumps

what

[slate jolt]

oh no

i ditn get b= p

p is not a constant

its a unit

25 =/= inch

so b=/=p

since b is a constant

[past sundial]

what is =/=?

[slate jolt]

not equal

[past sundial]

but if they are the same how are they not equal?

[slate jolt]

theyre not the same

[past sundial]

b is p but b isn’t equal to p?

[slate jolt]

i didnt understand what you meant by p

[past sundial]

p is pumps?

vp=bv
So p = b

[slate jolt]

oooh i get it

p in Vp

is not V*p

its just V_p

its a little p attached to V

to say its in pumps

[past sundial]

oh

How do you make the v to be a v little p?

[slate jolt]

would you be ok for a voice call, i think it would be better if i could speak

youd just have to listen

[past sundial]

it’s night right now qwq I’m next to a sleeping person

so I prefer text

[slate jolt]

ok

[slate jolt]

i didnt transform anything

i said that if we call Vp the volume in pumps

and V the volume in inches cubed

they are proportional

so there exists some constant b

such that Vp = b* V

[past sundial]

What is g?

[slate jolt]

its a 9

[past sundial]

And on line 2 is that a 7 as in 78^3?

[slate jolt]

18

as in 18 inches

do i write that badly 😦

[past sundial]

qwq sorry I am bad at reading handwriting

how did 9 appear?

[slate jolt]

9 pumps

Vp = 9

since Vp is volume in pumps

[past sundial]

Why do we need to multiply b when we add pumps?

Is it like a units thingy?

[slate jolt]

well the formula we used on the right

is a formula such that if you put in a distance unit in it it gives you something in that unit cubed

because of the d^3

(18 inch)^3 = 18^3 * inch^3

so V in our case is in inch cubed

but we are given a volume in pumps

that we call Vp

Since pumps and inch cubed measure the same thing

they are proportional

so there exists a ratio (that i called b but could be any name) such that for any volume in inches

if you multiply that volume by b

you get the volume in pumps

the goal of the exercise is to find that relationship between volume in inches cubed and pumps

and then use it to find the volume in pumps for the last ball

[past sundial]

oh so you multiply b because you turn the volume to pump volume

And since you put 9 in pumps then both sides needs pumps so you convert it

[slate jolt]

not sure about the second sentence

but yeah for the first

[past sundial]

What does the 2nd arrow point to mean?

[slate jolt]

it means : so

its an implication

we can deduce what comes after the arrow from what is written before

[past sundial]

how did ratio = 96/pi18^3?

[slate jolt]

you get the jump between the last 2 lines right?