help-27

[copper mango]

yes, and the furthest from the center you can be (while being on/in the planet of course) is?

[wet fox]

1

?

[copper mango]

.

[wet fox]

R

[copper mango]

yes

the furthest you can be from the center is R

so r in [0,R]

[wet fox]

Ohh so upper limit is R and lower 0

Alright I see

[copper mango]

yes

[wet fox]

Thank u

[copper mango]

the 4 and 1 explain stuff about the density

[wet fox]

So it’s not relate to the bounds on the integral

Right got u

[copper mango]

nope

[wet fox]

Thanks 👍👍

.close

[devout snow]

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[copper mango]

and then theta and phi are standard

[wet fox]

Yep

[devout snow]

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[true oyster]

how would you draw a graph without local max/min but with critical points

[true oyster]

undec

undef

[true oyster]

ok

im not sure

i dont know

ok

i just need 10

it says f is continuous

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[obsidian violet]

In Minecraft, you can barter with Piglins to receive different items. These items have different probabilities of occurring with the important item for speedrunners being Ender Pearls with approximately an 2.18% chance of being dropped per trade. You make 3 trades in a row.

⦁ What is the probability getting Ender Pearls all three times? (2 marks)

[obsidian violet]

hy

<@&286206848099549185> can i get help please

[polar bolt]

Use definition of independence

[obsidian violet]

whats that

[polar bolt]

What you need to use

[obsidian violet]

and how do i do that

[polar bolt]

Check your book what it is

[obsidian violet]

independence
/ɪndɪˈpɛnd(ə)ns/
Learn to pronounce
noun
the fact or state of being independent.

[polar bolt]

Your math book is a dictionary?

Interesting

[obsidian violet]

sorry

im such an idiot ha

i dont have a maths book

we werent given one

[obsidian violet]

i searched this on google will this be it n probability, we say two events are independent if knowing one event occurred doesn't change the probability of the other event. For example, the probability that a fair coin shows "heads" after being flipped is 1 / 2 1/2 1/2 .

[loud gust]

bruh wait theres no way a question had minecraft in it lmao

0.0218^3 if im not mistaken tho

[polar bolt]

Don’t just give answers

[loud gust]

havent done probability in like 2 years so wouldnt take my word for it

lol my bad

[obsidian violet]

yeah my teacher was playing minecraft lol

[obsidian violet]

scapeprof so what do i do next

[polar bolt]

Doubt. Use Wikipedia or smth then…

And look under definition

[obsidian violet]

alright il brb

[obsidian violet]

its says two events

[polar bolt]

Continue reading then if you have no idea how it works for 3 (or n)

[obsidian violet]

where do i read two events?

[polar bolt]

Section: Definition - subsection: more than two events?

[obsidian violet]

i dont understand this at al

[polar bolt]

This is where it is defined for more than 2 events on wiki

Surely you can read that yourself?

[obsidian violet]

its alright i got the answer can you help me with another question

.close

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[past robin]

Question 18

[past robin]

, rotate

[devout snow]

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@past robin Has your question been resolved?

[devout snow]

@past robin Has your question been resolved?

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@past robin Has your question been resolved?

[junior karma]

You still there?

[devout snow]

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[devout snow]

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[restive river]

Huh 😭😟

[frozen elm]

That looks like just difference of cubes

[restive river]

How do figure that out then

[fresh smelt]

i would divide the bracket by the answer using a long division method

[restive river]

Ok thank you

.close

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[alpine vine]

might be a bit of a trivial question, but how do i formally show that the showcased set $M$ is a subgroup of $(\mathbb{R}^2, +)$

[woven radish]

lewis

[devout snow]

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[restive river]

Pls help me

[devout snow]

@ivory sapphire Has your question been resolved?

[inland seal]

@ivory sapphire Since they share sweets in the radio of 5:6, you could say that Sarah has 5x sweets, and Henry has 6x sweets (x is an unknown number)

Try using this, and the info they give you, to find what x is.

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[night oxide]

This is a question from pre calculus, Im really confused on how to answer this question, to initially get alpha and beta you need to use the graph but Im not even sure where to start graphing

[restive river]

its do with double angle formula and and addition formulas

[night oxide]

yes I know that

[restive river]

for the first graph they want u to just draw the tan graph i guess and go across from 2 on the y axis and look at what angle ur at on the x axis

and then similar thing for the second graph

i guess they just want u to know how to find the solutions of trig equations from graphs and not rely on calculator

[night oxide]

we arent allowed to use calculators

[night oxide]

this is wrong

[restive river]

okay

[night oxide]

this is how it is supposed to look

this isnt alphas graph but thats how it is drawn

[devout snow]

@night oxide Has your question been resolved?

[devout snow]

@night oxide Has your question been resolved?

[fleet mauve]

Use trigonometric equations

Basic ones

[night oxide]

That helps nothing

I already said that I Know I need to use formulas

[night oxide]

<@&286206848099549185>

[devout snow]

@night oxide Has your question been resolved?

[devout snow]

@night oxide Has your question been resolved?

[devout snow]

@night oxide Has your question been resolved?

[fleet mauve]

You have tan alpha and sec beta, so u can find, sec alpha and tan beta using
Sec² A- tan²A=2
If you have sec u can get cos.
If u got cos, u can get sin by putting vali in tan
So u will get all terms u need for finding cos(beta-alpha) and second part

[pseudo basin]

Sec² A- tan²A=2

1, not 2.

[night oxide]

Can you show me what you mean on a piece of paper If possible I don't really understand what you mean through the text w

[fleet mauve]

I have solved first part

Similarily u can put values for tan(2beta)

In formula

[devout snow]

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[midnight solstice]

Hello, can someone help me with this question

[night oxide]

you didnt use the graph like I said you would need too

.reopen

[fleet mauve]

U wont use graph here

[night oxide]

yes you would

[fleet mauve]

That won't be right

[night oxide]

yes it would I even gave the part for b

beta

see^

U can wrk it out with a graph

[fleet mauve]

U can do it with just equations too

[night oxide]

but I specifically said that I need to graph it

cause we cant use calculators

[fleet mauve]

U don't need calculators in eqns too

Those r basic calculations

[night oxide]

BUT I SAID I SPECIFICALLY SAID I HAVE TO DO IT THAT WAY

like bruh

[fleet mauve]

Well I didn't read that

Earlier

[night oxide]

also said it just now

[fleet mauve]

When I was solving it.

[night oxide]

😐

[devout snow]

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[midnight solstice]

.close

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[potent topaz]

i need help with this, is this simply as this screenshot, or is there a better answer?

[potent topaz]

https://cdn.discordapp.com/attachments/758895084082429993/918445699148972052/unknown.png

ty

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[steady gale]

help

[restive river]

You here mate?

[steady gale]

?yea

[restive river]

Okay so where are you having problemo?

[steady gale]

okay so put the values into the equation but the answer im getting is 1.6 days whereas the answer behind the book says its 24.6 days

[restive river]

What did you put as the value of R(t)?

[steady gale]

isn’t this correct?

[restive river]

Just a sec

Something is wrong with my Discord or network.

Images aren't showing up.

[steady gale]

oh okay no worries

[restive river]

Yeah looks good.

,w 87.5 = 100(1/2)^{t/8.2}

[woven radish]

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[restive river]

Yep, your book is wrong.

@steady gale you here mate?

[steady gale]

oh are u sure?

[restive river]

Yeah.

See if you replace t with uh

24 days in the given equation.

You get

[shy wing]

i like how there’s a complex solution

[restive river]

,w 100(1/2)^{24/8.2}

[woven radish]

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[restive river]

Actually nvm we both dum

[steady gale]

lmao

wdymm

[restive river]

The equation is given for "amount of radioactivity present after t days".

So if 87.5 percent is gone.

We are left with only 12.5 Percent, no?

[restive river]

@steady gale Got it or shall I rephrase it?

[steady gale]

yeah got it so instead of 85.7, i put 12.5 right?

[restive river]

Yes.

[steady gale]

ahh i see okay ty!!

[restive river]

Did you get answer as 24?

[steady gale]

yes i got 24.6

[restive river]

Good.

.close the channel if you don't have any more questions.

And cheers because both of us are dum dum lol.

[steady gale]

hahah ive got another question

19 pls!!

[restive river]

Alright I'm here.

I can only help with a and b.

[restive river]

First equation is incorrect.

[steady gale]

yea i just need help with a.. idk if my equations are correct

[steady gale]

can u pls explain

[restive river]

Do you know about compound interest?

[steady gale]

kinda ig

[restive river]

Essentially, it says 7% every year so if your salary first year is 37k, you will get a raise of 7% of 37k and next year you will get 7% raise of that so it will be
7% of 7% of 37k = 37k(1+7%)^2 after 2 years

Next year, it will be 7% of that so

7% of 7% of 7% of 37k so

37k(1+7%)^3.

After t years, it will become 37k(1+0.07)^t.

[steady gale]

ohhh

[restive river]

0.07 because 7% = 7/100 = 0.07.

[steady gale]

ah okok i think i got it but then how is my second equation correct?

i mean like why bcs isnt it the same thing

[restive river]

Because it does not deal with any sort of percentages you see.

Percentage is quite a tricky thing.

Because it changes every year.

Let's think about it logically as

0th year you get salary = 45k
1st year you get a raise = 45k+2.5k
2nd year = 45k + 2.5k + 2.5k = 45k + 2(2.5k)
...
Year t you get = 45k + t(2.5k)

Makes sense?

In the second equation, they are like giving you some fixed amount of raise.

[steady gale]

yup i think i get it

thx sm !!

[restive river]

But in the first year, they are like indirectly referencing that they'll give you raise depending on your previous salary so if for some reason your previous salary increases, hence your new raise.

[restive river]

:)

[steady gale]

ohhhh i see

now to solve b i need to put those equations equal to each other right?

[restive river]

Yep.

[restive river]

Wait

[steady gale]

can i add 1+0.07?

[restive river]

B as the b in your question?

[restive river]

Yes.

[steady gale]

oh no i meant a sorry

[restive river]

Yeah.

[steady gale]

okay let me try now

[restive river]

1+7% is not addable so yeah you have to either convert 1 into 100% and then add or convert 7% into normal number.

1 = 100/100 = 100%

0.07 = 7/100 = 7%

[steady gale]

idk how to solve it further

like idek what to next..

do **

[devout snow]

@steady gale Has your question been resolved?

[restive river]

@steady gale You should ping me lol.

[steady gale]

mb lol can u help

[restive river]

The answer should be either never or generally it is between the first to 3rd year.

So you have to manually calculate lol.

[steady gale]

book says 8.7

oh u mean by trial & error?

[restive river]

Ah dAmN, I hate those books.

[restive river]

Welp, not anymore lol.

What I meant before was for time as some integer years.

[steady gale]

lol im so confused

oh

[restive river]

t = 8.7 is a very unrealistic time you see.

Well anyway, let me try to figure out how to solve.

Because logs and algebra is quite messy.

,w evaluate 45000+2500x=37000(1.07)^x

,calc 45000+2500*8.7

[woven radish]

Result:

66750

[restive river]

,calc 37000(1.07)^8.7

[woven radish]

Result:

66656.205919889

[steady gale]

oh alright i’ll be waiting bcs i genuinely have no clue at this point

[restive river]

@steady gale Which grade are you in?

[steady gale]

12

its advanced functions

[restive river]

Country?

Do you know about Lambert W function?

[steady gale]

no

[restive river]

Computers allowed?

[steady gale]

i mean my teacher didnt teach that

no

[restive river]

Calculators?

[restive river]

And the question is something which has been taught by your teacher?

@steady gale Don't leave me here alone ;-;

[steady gale]

not in class its a hw question lol

[restive river]

^

[steady gale]

calculators yes sry just saw this

[restive river]

Graphing calculators?

[steady gale]

yes

[restive river]

Ahhhhhh

You can ignore the long MeThOd NoW lol.

[restive river]

45k+t(2.5k) = 37k(1.07)^t basically means that

If you plot both the 2 equations

Their intersection point

Is the solution.

[restive river]

So just plot these 2 equations individually and see where they intersect.

Is that allowed?

[steady gale]

thats allowed yeah okay got it thank u!

[restive river]

,w plot 45000+2500t, 37000(1.07)^t

[woven radish]

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Requested by TeamSeas.org#7284

[restive river]

Let's ignore the negative years.

I was trying to replace the exponentiation with logs.

Like 1.07^t = y.

And 2500 t as some 2500 * log(y)/log(1.07).

Hoping to get a better chance.

[steady gale]

oh okay i get it now

thx! im gnna close this now

.close

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[devout snow]

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[errant cove]

Hello, i am stuck at this exponential equation, couldn't find any specific form, if someone knows how to solve it, i would be happy to hear it

[devout snow]

@errant cove Has your question been resolved?

[timber pebble]

oh

hrm

can you get there through some mix of 4 and 6 uhh

it looks like you have those mixes of factors

sqrt 2 and 3

i wonder if this factors nicely

like $( \frac 1 4 ) ^{x+1} = \frac{1}{4 ^{x+1}} = \frac{1}{\sqrt 2 ^{4(x+1)}} = \frac{1}{ \sqrt 2 ^{4x+4}}$

[woven radish]

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

[solar veldt]

Can anybody hlp me with 1th grade math?

[toxic mango]

@solar veldt go to any of the help channels like #help-18 #help-17 #help-28 and ask ur question

[timber pebble]

so

Let $p = \sqrt 2$ and $q =3$ then you have

[woven radish]

jan Niku

[timber pebble]

$(\frac 1 p ) ^{4x+4} + ( \frac{1}{p^2} ) ^x (\frac 1 q ) ^x - p ( \frac 1 p ) ^x ( \frac 1 q ) ^x = 1$

[woven radish]

jan Niku

[timber pebble]

wow that looks horrible

$p ^{-{4x-4}} + p^{-2x}q^{-x} - p^{1-x}q^{-x}=1$

@errant cove i have a feeling this may factor way nicer

not sure immediately

[woven radish]

jan Niku

[errant cove]

Well, yeah, i can go to this point, but the real question is what i do afterwords

There should be some easier way to do it probably

I was thinking to bring it to this form

And from here it's easy to solve

[devout snow]

@errant cove Has your question been resolved?

[devout snow]

@errant cove Has your question been resolved?

[errant cove]

Bruh

[devout snow]

@errant cove Has your question been resolved?

[devout snow]

@errant cove Has your question been resolved?

[long fog]

Steel need help?

[errant cove]

If you have some idea, yeah, i still need some help, thanks

[devout snow]

@errant cove Has your question been resolved?

[worldly wagon]

@errant cove so, I've guessed the answer

it's ||x=-1||

but how does one arrive to that answer without guessing or plotting that function of x...

[worldly wagon]

alr ping me if needed

[devout snow]

@errant cove Has your question been resolved?

[errant cove]

I will try it again in a few hours, i will ping you if i still don't find the answear

[devout snow]

@errant cove Has your question been resolved?

[shy wing]

3 days

[icy onyx]

[errant cove]

[stray olive]

$-2^{(\frac{1}{2} - \frac{x}{2})} \cdot 3^{-x}+4^{-x-1}+6^{-x} = 1$

[woven radish]

Master Butler

[stray olive]

yes this somehow will help u!1

this question is so hard that wolfram Alpha doesn't have the steps to solution

[shy wing]

alepk has wolfram premium?

[stray olive]

no I wish

[shy wing]

4 days

[icy onyx]

wow

[finite shuttle]

<@&286206848099549185> Can anybody close this channel ?

[inland seal]

we're helpers, not moderators

[finite shuttle]

aaaaaa sorry, i thought you could help so, sorry for the ping

[untold hamlet]

.close

[devout snow]

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[shy wing]

oh yeah woog

[devout snow]

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@past steppe Has your question been resolved?

[devout snow]

@past steppe Has your question been resolved?

[devout snow]

@past steppe Has your question been resolved?

[shy wing]

@past steppe are you talking about the slope or the gradient vector?

because if you mean slope, a multi variable function doesn’t really have a single slope

i can probably help, just need a minute to refresh on this

@past steppe ok so what’s confusing you?

sorry if i’m pinging too much, i just respond a bit late and want to make sure you don’t take an hour to see this

[devout snow]

@past steppe Has your question been resolved?

[shy wing]

luckily for you i’m still on lol

you just treat one variable as a constant

i don’t really get what’s confusing you

missing something

remember it requires the chain rule

nope

what’s the derivative of the inside function with respect to x

it’s not 1

what’s $\pdv{x}(xy)$?

[woven radish]

quantum

[shy wing]

oops

yeah

that’s the derivative of the inside function

you just seem to be messing up basic chain rule stuff

it’s y*cos(xy)

how did you even get just y inside cosine

the inside function never changes

so can you do $\pdv{y}(\sin(xy))$?

ok

[woven radish]

quantum

[shy wing]

you’re wrong

what’s $\pdv{y}(xy)$

[woven radish]

quantum

[shy wing]

yes, so…

it’s x*cos(xy)

yeah lol

this is just the normal chain rule

are you just getting confused with all the xs and ys

$\nabla$

[woven radish]

quantum

TeshiMeshi

[shy wing]

$\nabla f = \langle{y \cdot \cos(xy), x \cdot \cos(xy)}\rangle$

[woven radish]

quantum

[shy wing]

spend too much time on that lol

anyways

you said <3,4> earlier though

which is it?

ok

so we need this to be a unit vector

[woven radish]

TeshiMeshi

[shy wing]

$\vec{u} = \langle{\frac{3}{\sqrt{13}}, \frac{2}{\sqrt{13}}}\rangle$

[woven radish]

TeshiMeshi

[shy wing]

oh darn

i have messed up

unnecessary latex moment

anyways

[woven radish]

quantum

[shy wing]

so we would do

[woven radish]

TeshiMeshi

[shy wing]

$\nabla f(\frac{\pi}{4}, \frac{\pi}{4}).\vec{u}$

[woven radish]

quantum

[shy wing]

yeah

wow that’s ugly

well whatever

you already have all the values, so now it’s just the dot product

no

[woven radish]

TeshiMeshi

TeshiMeshi

TeshiMeshi

[shy wing]

,w (pi/4)cos(pi^2/16)(3/sqrt(13))+(pi/4)cos(pi^2/16)(2/sqrt(13))

[woven radish]

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[shy wing]

not sure why you subtracted

yes

,w dot product of <a,b> and <c,d>

[woven radish]

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Requested by quantum#6320

[shy wing]

it’s the answer

fortunately there’s an exact form

so use that

what do you mean

this isn’t a point

this is a number

it’s going to be a number

[woven radish]

TeshiMeshi

[shy wing]

nope, it’s the dot product of the gradient vector at the point and the unit vector in the desired direction

you can’t

your problem doesn’t ask for a coordinate

i mean you already have the coordinate, (pi/4,pi/4)

doesn’t ask for a coordinate answer

i’m not sure i’m confident enough in my knowledge of this to decide this for you

but i don’t see how it could be a coordinate

you can send it

,w 3/sqrt(13)

[woven radish]

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[shy wing]

,w 2/sqrt(13)

[woven radish]

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[shy wing]

i’m pretty confident that i didn’t do anything wrong

the dot product is addition

look at this if you want https://tutorial.math.lamar.edu/classes/calciii/DirectionalDeriv.aspx

,w dot product of <(pi/4)cos(pi^2/16), (pi/4)cos(pi^2/16)> and <3/sqrt(13),2/sqrt(13)>

[woven radish]

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[shy wing]

same answer

no clue

sure

still pretty sure the answers they gave are wrong though

ok

that’s not the partial derivative

it’s the gradient vector

and it would be the other way around

you have already shown that you know how

[woven radish]

TeshiMeshi

[shy wing]

that’s the gradient vector

but yes

no

that’s the gradient vector

the gradient vector is made up of partial derivatives

do you even know what a partial derivative is

what’s the question they asked @past steppe

did they ask for the gradient vector or the partial derivative

i’m confident at this point that the answer to both of these questions is none of the above

you have my permission to call your teacher an idiot if you get either of these wrong

because the answers relative to the questions are nonsense

there are two partial derivatives

one with respect to x, one with respect to y

[shy wing]

also i meant to say no to this

not sure why i said it was the gradient vector

i mean i guess

but the formatting or whatever it’s called of these questions is terrible

it should say \\$\pdv{f}{x}$ = something \\ $\pdv{f}{y}$ = something

[woven radish]

quantum

[shy wing]

also what do you mean?

literal trash

these questions actually suck

please call your teacher an idiot for me

nope

the answer will be a single number

as shown by the first directional derivative we did

considering in all the answers, the coordinates are both divided by the same number, i feel like those are actually the unit vectors

actually no

they definitely aren’t lol

[long fog]

That's why we have none of the above

[devout snow]

@past steppe Has your question been resolved?

[raven leaf]

what on earth $\pdv{f}{x}$

[woven radish]

Xetrov

[raven leaf]

I never knew there was a command for this

[devout snow]

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[ripe schooner]

help me plss

[devout snow]

@ripe schooner Has your question been resolved?

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@ripe schooner Has your question been resolved?

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[restive river]

how do i slove for n?

[copper mango]

you dont

[sonic smelt]

I think you meant to say something similar to "How do I get a closed form for S?"

[devout snow]

@restive river Has your question been resolved?

[devout snow]

@restive river Has your question been resolved?

[devout snow]

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[restive river]

linear algebra I think https://i.gyazo.com/551bdfa198b38eae0017424ad402afe5.png based on https://cses.fi/problemset/task/1083 (not homework or anything, I am not a student! Check profile for proof in link in bio, I just want to understand the question) is the correct way to work this out, I can just calculate the power of then multiply right?

[restive river]

so would this be like 10 * 10 * 10 * 10 * 10 * 2?

[copper mango]

not linear algebra, but tbh idk what the question is suppose to be

you have an already solved inequality

[restive river]

its more the 2 ⋅ 10^5

part I am struggling with

[copper mango]

what about it....?

[restive river]

does it work out like

10 * 10 * 10 * 10 * 10 * 2?

[copper mango]

yes.

[restive river]

thanks!

!close

!terminate

.close

[devout snow]

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[stark mason]

We have to find C

[jagged schooner]

start by getting rid of that braket in the sine

[stark mason]

The textbook turns the second term to cos^2

Thats where Im stuck

[polar wharf]

Sin^2(x) = 1 - cos^2(x)

[jagged schooner]

^^

[polar wharf]

From the pythagorean identity

Sin^2(x) + cos^2(x) = 1

[stark mason]

Yeah but how do u turn the second term to cos so u can apply the pythagorean identity

[polar wharf]

Yes

Wherever sin^2(x) is

You can instantly rewrite as

1 - cos^2(x)

That is using the identity

The other option is to try something like

Sin(x) = cos(x - pi/2)

Replace as cosine first

Or pi/2 - x

[stark mason]

I will try this out

Im still confused

This is how it went

I dont know how we got from line 1 to line 2

I think it just clicked

Finally

I just had to use -cosx=sin( 3pi/2-x)

Thank you for trying to help

.close

[devout snow]

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[devout snow]

Channel closed

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[analog nest]

I need help with 8th grade math and the question is for my school work so don't tell me the answer just tell me how to do it:

6c + 14 = -5c + 4 + 9c

c = ?
only tell me how to solve it please (don't tell me the answer)

[copper mango]

move all the terms w/ c to 1 side, and all the constants to the other

[analog nest]

can you say that a little bit more simpler

(I'm not that smart)

[copper mango]

not really

all the terms with c, move them to 1 side of the = sign... all the constant terms, move them to the other side

[analog nest]

so if i'm getting this right(which i'm 90% i'm wrong)

move every number with c at the end of it to the side and add it or just add it all together

?

[copper mango]

collect all the c terms on one side, and all the constants on the other.

[devout snow]

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[analog nest]

so like this:

6 + 14 + (-5) + 4 + 9 on one side

and

c + c + c on the other side

[copper mango]

no.....

[analog nest]

heres what i'm learning from:

https://www.khanacademy.org/math/cc-eighth-grade-math/cc-8th-solving-equations/variables-on-both-sides/e/linear_equations_3?modal=1

[copper mango]

yeah, I doubt Khan ever did what you just did

[analog nest]

I told you i'm not that smart

what steps do you use to solve these kind of problems

[cyan rain]

id say group the terms that are the same, like all the terms with c together and all the constant terms together

6c + 14 = -5c + 4 + 9c

first group them like this -4 + 14 = -5c + 9c -6c

and then solve them, thats how I would do it

[analog nest]

thanks I'll try that.

PS. I'll let you know if it works or not

[cyan rain]

Cool no problem👍

[analog nest]

alright it working for some of my problems thanks I just got figure out the other ones which are the same... just subtraction.

[cyan rain]

just remember when you move the terms around they change signs

Glad I could help

[analog nest]

My last equation was this: 8−4s = s+13

and I typed 1

but I was so close because it said

-1

so I'll keep trying

[cyan rain]

You must have forgotten the minus sign then right?

[analog nest]

yes

was that you

[cyan rain]

what do you mean?

[analog nest]

did you just go to a voice channel

nevermind

[cyan rain]

anyway good luck with your questions

[analog nest]

a - 15 = 4a - 3

[formal haven]

$-15 = 3a - 3$

[woven radish]

sills

[formal haven]

$-12 = 3a$

[woven radish]

sills

[formal haven]

$\frac{-12}{3}=\frac{3a}{3}$

[woven radish]

sills

sills

[formal haven]

$(-4) - 15 = 4(-4) - 3\-19 = -16 -3\-19 = -19$

[woven radish]

sills

[analog nest]

-3 + 5 + 6g = 11 - 3g

[woven radish]

sills

[formal haven]

$2 + 6g = 11 - 3g$

[woven radish]

sills

[formal haven]

$6g = 9 - 3g$

[woven radish]

sills

[formal haven]

$9g = 9$

[woven radish]

sills

[formal haven]

$g = 1$

[woven radish]

sills

[formal haven]

$6x + 2 + 2x + 4$

[woven radish]

sills

[formal haven]

@twilit fog If you're wondering we're in vc

[twilit fog]

oh thank

statistic question
Defind X have Mean = 100 Mean = 25

(1) -200 (2) 100 (3) 50 (4) 10

question 1
Set of Data Y = -2X then what it Y OF standard deviation = ?

any calculator to find answer?

[formal haven]

@twilit fog Don't ask a question here, open your own channel where someone can help you, #❓how-to-get-help

[twilit fog]

ah ok

[formal haven]

Please don't be rude to others in the help channels

[analog nest]

7b - 15 = 5b - 3

[formal haven]

$7b-15=5b-3$

[woven radish]

sills

[formal haven]

$7b-12=5b$

[woven radish]

sills

[analog nest]

$$7b - 12 = 5b $$

[woven radish]

Yasuke

[analog nest]

$$12 = 2b

or

12b = 12$$

[woven radish]

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

[devout snow]

@analog nest Has your question been resolved?

[devout snow]

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[devout snow]

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[main gull]

Don't open multiple channels

[ashen ravine]

I didnt know that going into other channels while asking for help closed them

thats what happened

[main gull]

None of the ones you opened got closr

[ashen ravine]

oh, for me they did

It wont let me back in tem

them*

they stopped showing in Occupied.

[main gull]

Did you accidentally hide the channel?

[ashen ravine]

yes I did

lemme close them

done

anyways...

[spare crypt]

you didn't ask a question

[ashen ravine]

its a true or false question

.close

[devout snow]

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[spare wigeon]

Determine the domain of values for the function $$f\left(x\right)=4\tan \left(\frac{2x}{3}\right)$$

[woven radish]

Theophania

[spare wigeon]

So I reasoned that $$\frac{2x}{3}\ne 90+180n$$
$$⇒x\ne 135+270n$$
$$D=\left{x\in \mathbb{R}:::x\ne 135+270n\right}$$

[woven radish]

Theophania

[spare wigeon]

is this correct?

[sonic smelt]

Yes

[devout snow]

@spare wigeon Has your question been resolved?

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[runic abyss]

need help

dont know what to do

[primal lantern]

puzzling with triangles

[devout snow]

@runic abyss Has your question been resolved?

[formal haven]

@runic abyss Use trigonometry

[primal lantern]

divide triangles in 2

so you get right corners

[devout snow]

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[slate drum]

i think there are some math tricks here but I probably don't know them: a^5+a^5b^5+b^5=2^10 and a,b>=0 and a,b are integers

[primal lantern]

$a^5+a^5b^5+b^5=2^{10}$

[woven radish]

Ruffy🎄

[primal lantern]

you need 2

equations

to solve it exactly

[slate drum]

what do you mean?

[primal lantern]

so you get
$$a^5 = \frac{2^{10}-b^5}{1+b^5}$$
$$b^5 = \frac{2^{10}-a^5}{a^5+1}$$

[woven radish]

Ruffy🎄

[primal lantern]

from reforming the equation

nothing more you can do I think

[slate drum]

wolfram says that a = 4 and b = 0

or b = 4 and a = 0

[primal lantern]

well yea

those are the integer solutions

there are plenty out there

[slate drum]

yes

a,b>=0 and a,b are integers

Solve the equation in a set of non-negative integers

[primal lantern]

you're not going to exactly calculate them without a 2nd equation

unless you're just trying out some random values

what they did

is just make one of them equal to 0

and then a^5 = 2^10

this is b

not 0

or 4

[slate drum]

so i just have to try random integers in this type of equations?

[primal lantern]

what wolframalpha did

was the only possible way you could get 2 integers

put a or b equal to 0

so you have one variable left

instead of 2

2 variables means you need 2 different equations to exactly solve it

what they are asking is just to find one of infinite possible solutions

[slate drum]

so i guess one of my problems is solved

could you help me with another one?

or i have to close this and create another?

[primal lantern]

although if both need to be positive integers then yea those 2 wolfram gave you are the only ones possible

you can ask here

[slate drum]

show that the number k is natural

only first one

second and third are easy

[primal lantern]

what is the middle value

0.4?

[slate drum]

4/9

so 2/3

[primal lantern]

oh

weird

[slate drum]

0,(1) = 1/9

0,(2) = 2/9

etc.

[primal lantern]

I see

[primal lantern]

what do you mean

isn't there only 1

[slate drum]

i mean middle root

and last fraction

btw i use translate for mathematical terms

so you can correct me if i type something wrong

[primal lantern]

one sec I'm trying it myself

basically

playing with roots

and working away the root of the denominator

[devout snow]

@slate drum Has your question been resolved?

[devout snow]

@slate drum Has your question been resolved?

[wraith vine]

i have a question

[devout snow]

@slate drum Has your question been resolved?

[left palm]

a nice fact is

$\frac{\sqrt{a}}{b} = \sqrt{\frac{a}{b^2}}$

i'd bring the denominator under (square the trinomial) and you should be able to reduce from there

[woven radish]

citrusmunch

[left palm]

(btw this is only ok since b ≥ 0)

[mighty edge]

b cant be 0

[left palm]

i debated clarifying that too, but yes, b can't be 0 since it's in the denom, but mostly i care about

$b = \sqrt{b^2}$

[woven radish]

citrusmunch

[copper mango]

if b is positive, that's true

[left palm]

(btw this is only ok since b ≥ 0)

[frozen elm]

Imagine being >= 0

[shy wing]

[devout snow]

@slate drum Has your question been resolved?

[devout snow]

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[fringe wind]

.close

[devout snow]

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[stray flower]

How does question b work

It's weird

Simple probability question

[crystal rune]

I'm thinking this is conditional probability

[languid ocean]

well, the other ball can be any of the other 9 balls

[crystal rune]

He hides the first ball and drops the second

[stray flower]

So even though you know it's red

You don't count it ?

Wait nvm

[crystal rune]

P(First ball is red | Second ball is red)

[stray flower]

I was confused because of the answer

It's not that for some reason

I just need an explanation

[languid ocean]

second ball is red, so you only have 9 balls that can be the first ball

in which there's 1 red ball and 8 non-red balls

[sand thorn]

The dropping business doesn’t matter. It’s the same as if he took out a red ball from the bag, throws it away, and picks a new ball from the bag. We’re asking what the chance is that the new ball is red

Right?

Idk

[stray flower]

So I thought it would simply be the probability of picking a red out of 9 balls

Btw this is from an old test I'm redoing

[sand thorn]

Yeah me too. Bag has 4 red, 3 black, 2 white. So the final answer, the chance of picking red, is 4/9 right?

[stray flower]

Also this is the answer

[sand thorn]

Wait that’s the answer?

That’s not a number

[stray flower]

I know

But it's how you get the answer

[sand thorn]

But it simplifies to the number right

[stray flower]

Yeah

[sand thorn]

Yeah

Ok

[stray flower]

I just don't get why

<@&286206848099549185>

[sand thorn]

Yeah P(A | B) = P(A and B)/P(B)

P(second red | first red)
= P(two balls red)/P(at least one red)

@stray flower

[stray flower]

I don't understand that

[sand thorn]

Do you know what P(A | B) means

Intuitively

[stray flower]

Probably of a/b?

Probability

[sand thorn]

Nah nah

It’s the probability that A occurs given that B occurred

So in this case, the probability that the second ball is red, given that the first one (dropped) is red

[stray flower]

Ohh is that a formula?

[sand thorn]

Well yes but don’t think of it like that

[stray flower]

Why

[sand thorn]

Here

Trash photo

[stray flower]

It looks fine

[sand thorn]

Do you know what $A \cap B$ means

[woven radish]

abs_0

[stray flower]

Yeah

[sand thorn]

Cool

[stray flower]

The middle part

[sand thorn]

So knowing that B occurs, what is the probability of A occuring?

Well

[stray flower]

That would be

[sand thorn]

We’re already restricted to elements of B — since it’s given that B occurred

[stray flower]

Mhm

[sand thorn]

So it’ll be something/(# elements in B)

[stray flower]

Myes

Yes

[sand thorn]

Now, the only way for A to occur given that B occurs is for it to be inside $A \cap B$

[woven radish]

abs_0

[sand thorn]

Does that make sense?

[stray flower]

Yep

[sand thorn]

It can only be outcomes from A that are also from B, since B occurred

Ok

[stray flower]

Mhm

[sand thorn]

So then $$P(A \mid B) = \frac{#(A \cap B)}{#B}$$

[woven radish]

abs_0

[stray flower]

Ohhh

[sand thorn]

Right

Yeah

[stray flower]

Yes

Back to the original question

[sand thorn]

Yeah

So P(A | B) = P(A and B)/P(B)

P(second red | first red)
= P(two balls red)/P(at least one red)

The question is asking what’s the chance that the second ball is red, given that the first ball (dropped) is red?

So you can literally plug it into that equation

P(first red AND second red) is the same as saying P(two balls red)

P(first red) is the same as saying P(at least one red)

Hence your answer

[stray flower]

So is there something going on with 'at least'

Coz I thought it meant that it has to be red

Or can you not assume that

Or what

[sand thorn]

Well

The guy chooses two red balls without replacement

If we know the first ball is red, and the second is unknown, then AT LEAST one ball chosen is red

[stray flower]

Oh yeah

[sand thorn]

When we write it this way, we’re avoiding “first” and “second” notions of time, and whatnot

[stray flower]

Ah

[sand thorn]

Those fumble the problem and make it difficult to do the math

[stray flower]

Ohh

[sand thorn]

So we’re translating it in a sense

[stray flower]

Thats complicated wow

[sand thorn]

Into a completely unambiguous, void context where we can operate on stuff

It seems crazy but you learn to do it naturally

[stray flower]

I thought so

That's cool

[sand thorn]

Now I’m curious how to get from here to the actual answer

[stray flower]

That isn't too hard

Ok thank you so much I'll have to go process this in my mind now

[sand thorn]

Yeah dude, good luck

I gotta sleep lol

[stray flower]

Ty

[sand thorn]

See ya

[stray flower]

Bye

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[whole totem]

what is the formula in the end

[devout snow]

@whole totem Has your question been resolved?

[devout snow]

@whole totem Has your question been resolved?

[devout snow]

@whole totem Has your question been resolved?

[wanton pecan]

It's not really a formula, it's just a logical conclusion

[whole totem]

nvm

i found it

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[umbral bolt]

So i have a function f(x) = sin x and I need to find a g(x) function that would allow the graph to show both a vertical and horizontal asymptote. Im stuck on finding an equation that would also give the horizontal asymptote

[devout snow]

@umbral bolt Has your question been resolved?

[umbral bolt]

<@&286206848099549185>

[wanton pecan]

You want to find a g(x) such that f(x)/g(x) had a horizontal asymptote too?

[umbral bolt]

okay so

i need to find a second function

that can either add/subtract/multiply/divide with f(x) = sin(x)

that the graph would have both a vertical and horizontal asymptot

wait..

i think i got it

im pre sure if i do somthing where i would have to add something to the sin functiona and both are being divided

hmm

maybe something like 7/x + sinx/x

yep okay this works

horizontal and vertical asymptote of 0

[wanton pecan]

Hm, is that really allowed?

You are first dividing then adding

[umbral bolt]

hmm

im adding a rational function alongside a sin function divided by x

it should be fine as long as im meeting those criteria

[wanton pecan]

I don't really understand the point of the exercise so I'll leave that to you

but the function you ended up with does indeed have both a vertical and a horizontal asymptote

[umbral bolt]

hmm alright

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[sweet obsidian]

Hello! I want to ask why the answer is infinity? Here’s how I tried to solve it, and a screenshot of a problem solver online that I used because I’ve been trying to find an explanation.

[drifting ibex]

You have to use rule of Hopital

[broken roost]

You know l’hoptial rule? You have an indeterminate form 0/0.

[sweet obsidian]

Yeah, I tried using L’Hopital’s rule in the first picture, but apparently the answer is infinity and not 4^5/5!

[sweet obsidian]

I took the derivatives of the numerators and the denominators til I ended up with only the numerator having the variable x, but it didn’t lead me to infinity

[broken roost]

If you differentiate top and bottom of sin(4x)/x^5 you get 4cos(4x)/(5x^4) now take the rhs limit to 0 we have 4cos(4x) approach 4 a constant and 5x^4 approach 0 hence +infty.

[wanton pecan]

You can't keep taking derivatives

Because after the first one you no longer have 0/0 😄

cos(0) != 0

[sweet obsidian]

That makes so much sense, I think I finally get it now :DD

Okie, so here's what I understood:

• since cos(0) = 1, L'Hopital's rule doesn't apply anymore and I can proceed with direct substitution
• and I should remember that I'm dealing with limits here so it's not actually 4/0 but approaching zero so 4 is being divided by a really small number ending up with infinity

Thanks so much, Sparky, Plegasus, and Learath2! I've been trying to wrap my head around this for hours since yesterday. Hope you all have a nice day/evening :)

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[earnest palm]

Hi

[earnest palm]

I have a doubt

I don’t know how did the 2x came in the equation

<@&286206848099549185>

[solid trail]

goes to terrance and comes back to ground floor

one trip takes x metres

two trips takes 2x metres

[restive river]

and it was already 20 meters above

therefore 2x-20 is the total distance covered

[solid trail]

basically, first trip takes x - 20 and second trip takes x

sum to get 2x - 20

[earnest palm]

Hi ok thx