help-0

[stiff badger]

u so patient as well

[north hemlock]

idk

I'm sleep deprived as hell

but I have to go to class in like 20 min

[stiff badger]

ow, we have our exams the day after tomorrow so this helps a lot

[north hemlock]

nice nice

good luck

[stiff badger]

byebyee, tenkss!

gudluck as well

[north hemlock]

just spam more problems to get super good at it

real

bye

[stiff badger]

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[sinful dragon]

apparently this is something I should take as a "fact"

but how do I prove it

[lone heart]

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[past current]

is this identity true?

[past current]

I know if inside the bracket is multiplied, it can be pulled out into loga+logx. but does it work in the other way ?

[vale wigeon]

no

[past current]

ok thanks.

is there anything I can do about log(a+x)?

[vale wigeon]

no

[past current]

great, thanks very much!!

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[sonic hull]

trig ish? i think… i cant do word problems help

[bright tundra]

Your diagram is drawn wrong

[sonic hull]

oh

ion know what im doing

[bright tundra]

Idk how to guide you into drawing the diagram correctly ngl

Wanna see how the diagram should look?

[sonic hull]

yeah

[bright tundra]

Sorry for the bad drawing, but something like this

[sonic hull]

oh wth ive never seen that before

[bright tundra]

The ship travels from point A to B right, it has to be closer to the lighthouse for the angle with the water to be higher

[sonic hull]

oh

ty

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[echo depot]

Determine the number of integer values of such that the fraction has an integer value.

[echo depot]

Determine the number of integer values of k such that the fraction has an integer value.***

[frozen laurel]

where can I get help with my exam question for tomorrow?

[echo depot]

its supposed to be easy and im here sweating i feel dumb

[livid sage]

divide 2k + 1 into 6k + 9

[echo depot]

just go to any available help chat

#❓how-to-get-help

[frozen laurel]

ok thanks man 🙂

[echo depot]

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[earnest fern]

Pls help me

[earnest fern]

I need to find surface area and volume

[vale wigeon]

do you know how to find the SA and volume of a cylinder in general

[earnest fern]

i forgot

[vale wigeon]

ok how about

the area of a circle

[earnest fern]

pi multiple radius power 2

[vale wigeon]

pi * r^2 yes

now consider

what is the surface of a cylinder made of

[earnest fern]

rectangle?

[vale wigeon]

it is made of the two circular bases on either side,
and a lateral surface between them which can be unrolled into a rectangle.

as in "What does the word rectangle mean?" or "How does a rectangle happen here"?

[earnest fern]

is this right?

wait i have to take the SA of 1 circle then multiply 2

[vale wigeon]

yes, but the width of the rectangle is not 12m here

[earnest fern]

oh

[vale wigeon]

also you might want to not use a decimal for pi

[earnest fern]

alright

so to find the width i take 2xpix3

[vale wigeon]

don't use the letter x as a multiplication symbol

but yes

[earnest fern]

okok my mistake but ty, i get it now

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[gritty harbor]

Ignore the corrections with blue ink (its cos3x and sin3x)
How does the 4th step equal the 5th step?

[remote escarp]

,rccw

[gritty harbor]

https://byjus.com/question-answer/prove-the-following-frac-text-cos-4x-cos-3x-cos-2x-text-sin-4x-sin/

Found the same one on web

[lone heart]

@gritty harbor Has your question been resolved?

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@gritty harbor Has your question been resolved?

[ancient saddle]

Hello. cos(θ)/sin(θ) = tan(θ)

[gritty harbor]

What about other terms?

[ancient saddle]

you mean 2cosx + 1 ?

[gritty harbor]

Yeah

Cos3x/sin3x will become cot3x right?

[ancient saddle]

yes

[ancient saddle]

this is cancelled out because it's both in numerator and denominator

$\frac{\cos 3x (2\cos x + 1)}{\sin 3x (2\cos x + 1)} = \frac{\cos 3x}{\sin 3x}\cdot \frac{2\cos x + 1}{2\cos x + 1}=\frac{\cos 3x}{\sin 3x}$

[ocean seal]

ELeonardo

[gritty harbor]

Cosx/cosx gets cancelled?

[ancient saddle]

no

[gritty harbor]

Where did the tanx go?

[gritty harbor]

Cos3x/sin3x

[ancient saddle]

$\frac{\cos 3x}{\sin 3x}\cdot \frac{\cancel{2\cos x + 1}}{\cancel{2\cos x + 1}}=\frac{\cos 3x}{\sin 3x}\cdot 1 = \cot 3x$

[ocean seal]

ELeonardo

[gritty harbor]

There's two cos3x/sin3x

[ancient saddle]

no, there's an equal sign
= cos3x/sin3x

[gritty harbor]

Wait how did it change to 2cosx+1

[ancient saddle]

oh ok

they factored out common term cos3x in numerator

and common term sin3x in denominator

$2\cos 3x \cos x + \cos 3x\
= \cos 3x (2\cos x) + \cos 3x (1)\
=\cos 3x (2\cos x + 1)$

[ocean seal]

ELeonardo

[gritty harbor]

Why cos3x*1?

[ancient saddle]

I just wrote it to make it more obvious

it's just a 1, it doesn't change anything

[ancient saddle]

this is all just about reordering terms

[gritty harbor]

2cos3xcosx gets rearranged into cos3x*2cosx right?

[ancient saddle]

yes

[gritty harbor]

How to add it?

3rd step

[ancient saddle]

that uses the distributive property of multiplication

a(b+c) = ab + ac , right?

so ab + ac = a(b+c)

[gritty harbor]

Oh so when we expand the brackets we get cos3x2cosx+cos3x

Which is same

[ancient saddle]

yes

[gritty harbor]

Right i got it

Thanks

[ancient saddle]

ok great 👍🏽

[gritty harbor]

Same applies for sin3x right?

[ancient saddle]

yes

you can do the same whenever you have a common term

like

ab + acd + am = a(b + cd + m)

a is the common term

[gritty harbor]

I thought it was a result or something

Thank you so much

[ancient saddle]

No problem 👍🏽

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[violet spindle]

knowing the coordinates of M and M` and the radius of both circles, i want to derive a formula for the coordinates of point L (without finding U and U` first, just a formula for L)

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@violet spindle Has your question been resolved?

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@violet spindle Has your question been resolved?

[lavish sedge]

L is just simply the midpoint of the straight line M'M, you can use geometry to prove this

so M'x+Mx/2 = Lx and the same thing for the y coordinate

wait, do the two circles have the same radius?

[lavish sedge]

because this wouldn't be true if that wasnt the case

[lone heart]

@violet spindle Has your question been resolved?

[violet spindle]

Lets make one circle bigger

Boom, L moves

L being in the middle of the points only works if the circles are the same size

I am looking for a general solution

Not just the easiest case

If you imagine 1 circle with a radius slightly larger than the distance between the 2 M points and the other being relatively small, point L will be the same as point M

I'm looking for a formula for L

Because of course there is a formula for finding the points and then we can find the point in the middle but I want to find one specifically for L (sorry that I'm repeating myself, but this is a very important constraint

[lone heart]

@violet spindle Has your question been resolved?

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[long umbra]

How do I fill this out?

[lone heart]

@long umbra Has your question been resolved?

[long umbra]

<@&286206848099549185>

[fervent kestrel]

Trial and error is your best best tbh

look for larger numbers in the top row and smaller numbers in the bottom row

[exotic belfry]

4 unknowns, 4 equations, should be solvable

or you choose any value for no taco tongue/no evil eyebrow (<=120) and calc the rest.

[fervent kestrel]

That might take away from the independent clause

[exotic belfry]

why?

[long umbra]

.close

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[golden ferry]

f(x) = |x| has a limit at x = 0 right?

[golden ferry]

but isnt diffrentiable

[exotic belfry]

yes

[golden ferry]

since the difference quotient isnt the same for both sides

ok so the limit exists though correct? just wanna confirm

[exotic belfry]

youre talking about two different lim

[golden ferry]

ye

ok

thanks

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[zenith fog]

HOW is the answer not 21m?????

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Show your work, and if possible, explain where you are stuck.

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[fluid crow]

How would I attempt to solve this problem?

[tacit arch]

,tex .double angle

[ocean seal]

riemann

[whole lion]

alr, part of my statement stands

you're not supposed to find the value of theta

@fluid crow try to use the pythagorean theorem to draw a full right triangle

[cinder sundial]

Ayy

Yea

[cinder sundial]

Ayy

[lethal belfry]

use $sin(2x) =2sin(x)cos(x)$

[ocean seal]

Why am. I here

[lone heart]

@fluid crow Has your question been resolved?

[tacit arch]

@fluid crow what are you stuck on

[fluid crow]

I'm assuming I should plug the theta value where theta is in the formula right? This is a calculus class so we're not supposed to be using calculators and I've been trying to look through my notes to find what the next step is after

(Sorry if I have a lot of questions, it's been 2 years since I did any real math)

[tacit arch]

no you don't need to find theta

read the double angle formulas

[remote fulcrum]

@tacit arch

can u pls help me w further integration in help 12

[fluid crow]

cos = 1/13
2sinθcosθ = 2(12/13)(1/13)
24/149
Does that look right?

[tacit arch]

where does cos(theta) = 1/13 come from?

[fluid crow]

Ah wait I've completely messed up pythagrorean

cos = 5/13
2sinθcosθ = 2(12/13)(5/13)
= 120/169
Is that on the right track?

[tacit arch]

,calc sqrt(1-(12/13)^2)

[ocean seal]

Result:

0.38461538461538

[tacit arch]

,calc 5/13

[ocean seal]

Result:

0.38461538461538

[tacit arch]

looks right

[fluid crow]

Thank you

I’ve resolved this problem

[tacit arch]

.close

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[forest bay]

How do I know if two triangles are congruent ?

<@&286206848099549185>

Does the rotation of the triangle matter

[fervent kestrel]

!15mins

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Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[fervent kestrel]

and no

[forest bay]

How do I get it to help me ok thanks

Does the bot work for math images?

[fervent kestrel]

I don't think so

[forest bay]

Ok thanks

.close

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[forest bay]

.reopen

[lone heart]

✅

[fervent kestrel]

Congruent just means the triangles are the same in terms of ratiometrics

[forest bay]

Yea but if it orders it like ABC and BCA does that matter?

[fervent kestrel]

not really

those are placeholders to know about what sides or angles you are talking about with that specific triangle

[forest bay]

My teacher goofy he doesn’t teach thanks

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[final topaz]

Uh

[final topaz]

Idk if I put in the answer wrong or what

[sleek girder]

🤔
what are the marked sides anyway

[buoyant saddle]

opposite sides are congruent for a parallelogram

[final topaz]

Probably AD or BC

Ik

I got the answer

[sleek girder]

fair

[final topaz]

AD=27?

But it doesn’t work when I put in

[sleek girder]

not 27

[buoyant saddle]

4x+3

[final topaz]

O

Ok

[buoyant saddle]

it is 27

no?

[final topaz]

I got 27??

[buoyant saddle]

x=6

[sleek girder]

wait

[final topaz]

Yea

[buoyant saddle]

he’s right

[sleek girder]

lag 💀
sorry, 27 is right

[final topaz]

Idk if I put it in wrong or what

AD=27 doesn’t work

BC =27 doesn’t work either

[sleek girder]

maybe just type "27"?

[final topaz]

Ok

😐

That was it

lol

Thanks

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[buoyant saddle]

💀

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[wispy sable]

could someone help with the calculation of the expected value of a probability game i have to make?

The game instructions:
pick 1 card and roll 2 dice

Rewards:
if card value = sum of dice ; win $6

if card value < sum of dice ; win $3

if card value > sum of dice ; win$0

Costs 2 dollars to play

[wispy sable]

currentl my calculations show an expected value of around -0.55 but it is very wrong after running some code

i calculated the probability of getting the same card value as sum of dice as 1/13

and card value < sum of dice as 1/6

however 1/6 seems wrong not sure where i went wrong though

[brittle ember]

card trump card?

[wispy sable]

wut?

[brittle ember]

what card you using in game?

[wispy sable]

a deck of 52

Ace as a 1 so 1-13

[runic iris]

you want to find the expected value the player gets per game right?

[wispy sable]

yeah

[foggy pecan]

what about kings, queens and jacks?

[wispy sable]

11,12,13

so sum of dice can be 2-12 and cards can be 1-13

[foggy pecan]

but when you lose?

[wispy sable]

whoops typo

changed it up top

[foggy pecan]

ok

[runic iris]

the probability of getting the same card value as sum of dice is not uniform since there are different ways u can get a sum. For example, the sum of 7: 6+1, 3+4, 2+5, etc.. so u need to get the probability per sum

[wispy sable]

is it still not 1/13

i know it peaks at sum of 7 at 6/36

[runic iris]

nope its not the same for all

like if u get 2 as the sum the probability would only be 1/36

[wispy sable]

yeah but the value of the card that matches the sum is always 1/13

so since probability has to equal 1

then 1/13 (1) = 1/13

<@&286206848099549185>

[runic iris]

probability of drawing any card (1/13) * probability of the sum = the probability of drawing a card with the same sum

the probability of the sum shouldnt simply equate to 1 since each sum has different probabilities

[wispy sable]

is it not (1/13)*(each probbaility of sum added which should equal 1)

[runic iris]

The probability of the sum is always 1 because when u roll each time u get a sum. However, the probability of u getting the same sum as the card number is different. It would only equate to one if the probability of getting each sum were equal.

[lone heart]

@wispy sable Has your question been resolved?

[runic iris]

its just

(1/13)*(1/36)

[lone heart]

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[cerulean thistle]

Need help on rational exponents and graphing polynomials

[cerulean thistle]

Just like basic understanding

Have midterms tmr

[quasi vector]

[cerulean thistle]

hello

I need help like understand the basic structure of it

like how to solve

[quasi vector]

do you have a specific question you would like help with?

[cerulean thistle]

Let me find my old test I did

Alr look

Need to review this test paper

[vague mirage]

Step 1 : ignore y⁰

[cerulean thistle]

huh

[vague mirage]

Step 2 solve the parenthesis

[cerulean thistle]

I don’t know how to do any of that

[vague mirage]

Wait what

[cerulean thistle]

Ye

[vague mirage]

Are you familiar with exponent

[cerulean thistle]

Like wdym

[vague mirage]

Like we call it "powers of numbers"

[cerulean thistle]

Okay

[vague mirage]

Wait you aren't familiar with exponents then you should learn that first then approach to solve this problem

If you*

[cerulean thistle]

Okay let me watch a YouTube video

[vague mirage]

Aghh what happened to my keyboard

[cerulean thistle]

Brb

[vague mirage]

Yeah that would be a great idea

[cerulean thistle]

.close

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[chrome tapir]

I just want to confirm this is right, what I did was I factored it and then it become 2a+1 = 0 or a-1 = 0

after thant I got x = pi/6 or x = pi/2

[chrome tapir]

did i do it right?

[quasi vector]

not quite

[vestal plume]

Yeah there's a small mistake

[quasi vector]

$2\sin x+1=0$ does NOT imply $x=\frac{\pi}{6}$

[ocean seal]

kheerii

[chrome tapir]

yeah, thats what i thought

what I did was 2sinx=1

then I divided each side by 2

[quasi vector]

no.

$$2\sin x +1 &= 0\newline
2\sin x &= -1\newline
\sinx &= -\frac1{2}$$

[ocean seal]

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

[chrome tapir]

oh right

its neg

[quasi vector]

yes

sorry for my bad texit

[chrome tapir]

so it would be sinx= -1/2

meaning its in the lower 2 quadrants

aka at 270 or 3pi/2

alright thx

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[raw jetty]

if a,b,c are diffrent positive integerd where abc=16, then whats the maximum value of ab-bc+ac

[raw jetty]

the answer to this is somehow 263 and i dont know how thsts even possible

[vapid shuttle]

yeah what

there's not many options for what they can be

[raw jetty]

A. 253
B. 63
C. 249
D. 263
these are the options

[vapid shuttle]

you can say WLOG a<=b<=c

[raw jetty]

what

[vapid shuttle]

without loss of generality

by different do you mean distinct

then you could even say a<b<c

[raw jetty]

i think so?

[vapid shuttle]

Okay so yeah then you can say without losing any generality a<b<c

then the maximum on c is 16, you'd think, except then you'd need a=b=1 which isn't possible

so you have to factor 16

other possible options 8, and 2

c=8, b=2, a=1

[raw jetty]

why a<b<c

[vapid shuttle]

that's fine

[vapid shuttle]

well because in the problem it clearly doesn't matter if you relabel "a" to be "c" or to be "b"

multiplication is associative

abc=cba=bac...

and if they're all positive distinct integers

one is the biggest

and one is the smallest

and one is in between

so for the sake of solving the problem you can just say

call the biggest one a

[raw jetty]

uhh ok then

[vapid shuttle]

Is that confusing?

[raw jetty]

so the best would be like a=8 b=2 c=1 right

[raw jetty]

nah

[vapid shuttle]

I would think so

[raw jetty]

that would lead to 30

[vapid shuttle]

well, not the best for ab-bc+ac specifically

because you have to think about

you want to maximize ab+ac

and minimize bc

maximize a(b+c) and minimize bc

clearly then you just want a to be the largest

then think about how you can make (b+c) big while bc small

[raw jetty]

to do so you maximize a

[vapid shuttle]

Actually

!original

[lone heart]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[vapid shuttle]

I think that the only distinct positive integers a,b,c such that abc=16

[raw jetty]

its not in englisj

[vapid shuttle]

are 8, 2, 1

I think those are the only options

[vapid shuttle]

post it anyways

because I think you must've mistranslated it

[raw jetty]

8. Jika a, b dan c adalah tiga bilangan bulat positif berbeda yang memenuhi abc = 16, berapakah nilai terbesar yang mungkin dari ab – bc + ca?

running it to a translator i got
8. If a, b and c are three different positive integers that satisfy abc = 16, what is the largest possible value of ab - bc + ca?

[vapid shuttle]

Yeah this is odd

the only 3 distinct positive integers whos product is 16

are 8,2 and 1

that's it

you can run through the options of like, what to make a,b, and c from 8,2 and 1 and see what is maximal, but I don't think any of those option choices are right

@raw jetty is this from a class of yours?

[raw jetty]

uhh

im prepping for an olympiad and searched some questions online

[vapid shuttle]

Okay this is a bad one then

[raw jetty]

k

[vapid shuttle]

this question is incorrect or like someone is trying to troll with it

[raw jetty]

thanks for coming tho

[vapid shuttle]

you can solve it, but the correct answer is not one of those options

no problem

[raw jetty]

dont think this is a troll, there are 40 questions and the questions look legit (like i tried and there were answers

.close

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[astral holly]

How do i correctly note the derivation process? f.e.:

f(x)= 1/x

My process is x^-1 -> -x^-2 > -1/x^2

but if i write it like this:

f'(x) = x^-1
= -x^-2
=-1/x^2 that would be wrong technically

[foggy pecan]

look at it:

[mortal trellis]

well just dont write the first equality

[foggy pecan]

$f\left( x \right)=\frac{1}{x^{n}}\\f'\left( x \right)=\left( \frac{1}{x^{n}} \right)'=\left( x^{-n} \right)'=-nx^{-n-1}=-\frac{n}{x^{n+1}}$

[ocean seal]

Joanna Angel

[astral holly]

Perfect thanks!

[foggy pecan]

yw

[astral holly]

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[wanton nova]

???

[wanton nova]

wdym its wrong

its correct

oh you wanted the general case?

or what

[astral holly]

f'(x) does not equal x^-1, i just wanted to know how to properly note it down

[wanton nova]

ok

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[gaunt dirge]

Short question, how is all of this calculated? (Taken from https://dropchance.app/)

[gaunt dirge]

I need to know for a spreadsheet I'm working on

[mortal trellis]

"at least once" is the opposite of "never"

[gaunt dirge]

ok

but like how are those chances and expected runs calculated

like how do i take a percentage chance p (in the case above =0.03%) and n runs (in the case above, n=10, 50, and 100 respectively) to get the percentage chance of success

[prime badge]

you're looking for the opposite of failing n times

[gaunt dirge]

i don't understand

i figured it out

or at least i figured out the chance for the item to drop at least once in N runs

its 1-(1-p)^n %

i haven't figured out the bottom part yet

[prime badge]

the second part is the same thing, they just solve the equation to find N that corresponds to 50%

[gaunt dirge]

ok so solve for N

ok so that appears to be something like N = log(1-x)/log(1-p) where x = that 50/90/99% figure in the image

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[edgy relic]

2^2022 mod 2022

[vale wigeon]

!status

[lone heart]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[edgy relic]

1

[vale wigeon]

do you know modular arithmetic in general?

[edgy relic]

Yes

Since the divisor is big , I cant get a clue

I thought of doing (-2020)^2022 mod 2022 but it didn't work out

[vale wigeon]

that won't do much, no.

do you know Chinese remainder theorem?

[edgy relic]

Yeah ig

Ok no I dont know that

[vale wigeon]

ok in that case

repeated squaring

work out 2^1, 2^2, 2^4, 2^8, 2^16, etc. all mod 2022

[edgy relic]

Ok?

[vale wigeon]

then multiply some of them together to get to 2^2022

or wait. hold on

[edgy relic]

Mmmhm

[vale wigeon]

🤦‍♀️

ah wait nvm that would need crt still

yeah do the repeated squaring thing

[edgy relic]

Ok lemme try

Didnt help

[lone heart]

@edgy relic Has your question been resolved?

[edgy relic]

Nope

[lone heart]

@edgy relic Has your question been resolved?

[lone heart]

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[mild fractal]

my initial thought was to think of them as vectors on a 3d plane

since i,j,k cannot be equal simultaneously;
I thought that the series should go on with
i,j,k being:
(0,1,2), (1,2,3), (2,3,4),(3,4,5),...
and we multiply this by 3(symmetry) so that i, j and k get all values from 0 to infinity without all of them being equal simultaneously.
we can write the summation as 1/3^(i+j+k)
so I get 3x infinite summation of a GP with 1st term=1/3 and common ratio=1/3
the answer I got is 3/2

however my solution is wrong...why?

[vale wigeon]

for staters, this would give you 1/27 as the common ratio.

and it feels like you are missing a lot of terms

in a lot of places

[mild fractal]

wait how would it be 27

oh ur right

nvm

[vale wigeon]

i think it would be much easier to first calculate the sum without the consideration that i, j and k are all different

then subtract away the terms in which two or more of them are

[mild fractal]

I did think of that too but then why am i missing terms in my solution?

[mild fractal]

in that approach I'll also have to do something about the terms in which
i=j=/=k for all three variables, right?

while subtracting

when I correct the common ratio to 1/27, my answer is 3/26

[vale wigeon]

the example you gave had j=i+1 and k=i+2, which does not need to be the case

[mild fractal]

ohh you're right

right it could be anything from +1 to any natural no

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[karmic ravine]

Hey i just need some help with question 2

[vale wigeon]

which letter?

[karmic ravine]

Can we start with A

[vale wigeon]

ok, what's troubling you with a)?

[karmic ravine]

im just not sure how to translate the open and closed circles

[vale wigeon]

open circle means the point is excluded, filled circle means it's included

[karmic ravine]

alright

im gonna try have a go

at A

{x|2<x<4}

im not quite sure though

[lethal belfry]

{x|2<x} not {x|2<x<4}

where did you get the 4 from?

[karmic ravine]

oh right

well i just thought bc 2 was the halfway point

[lethal belfry]

question 2, right?

[karmic ravine]

yeah

i see it now ty

im gonna try b

{x|1<x(less or equal then)5}

i dont know how to type that part

do i have to include the "x E Z" thing too?

[lethal belfry]

if you're working on the set of integers yes

[lethal belfry]

looks right

[karmic ravine]

how do i know if im wokring on the set of intergers or not?

alr nice

[lethal belfry]

have you learnt real numbers?

[karmic ravine]

is that just 1 2 3 4 5 6...

[lethal belfry]

no, those would be natural numbers

[karmic ravine]

oh whoops

[lethal belfry]

usually the set you're working on should be given in the question

so I'm not too sure

if you work with stuff like $\sqrt2$ you're probbly working on reals

[ocean seal]

Why am. I here

[lethal belfry]

any number that can be plotted on the number line is real

so you're most probably working on reals

[karmic ravine]

ohh ok thanks

so i should incluse the x E Z stuff

[lethal belfry]

x E R

not z

if you're working on reals

what set are you working on?

[karmic ravine]

so would A be
{x|2<x<4, x E R}

[lethal belfry]

not less than 4

[karmic ravine]

oh yeah mb

[lethal belfry]

just {x|2<x , xER}

try c now

[karmic ravine]

adn would B be
{x|1<x(less or equal then)5, xER}

sure

[lethal belfry]

wrong

for c?

[lethal belfry]

is this for c?

[lethal belfry]

for b this is right

[karmic ravine]

{x|-2(>or equal)x, 3<x, x E Z}

im not sure about this one

[lethal belfry]

the best way to write it would be ${x| x \epsilon {R- [-2,3]}}$ IMO

[ocean seal]

Why am. I here

[quasi vector]

$\in$

[ocean seal]

kheerii

[lethal belfry]

oh, thanks!

[karmic ravine]

would that be valid in an exam?

[quasi vector]

[lethal belfry]

should be , I'm not sure if your teacher would accept it though

[karmic ravine]

sorry what does the xeR mean again?

[lethal belfry]

x belongs to the set of Reals (x E R) . I meant to type R-[-2,3]

that means the set of reals minus the elemenst between (inclusive) -2 and 3

[karmic ravine]

so this is the answer?

its throwing me off how that part is first

[lethal belfry]

I can't help for now, I have some work to do , sorry

[karmic ravine]

oh alr thanks for everything

[lethal belfry]

OK, I'll jsut clarify what I just said

write it as x| xE R-[-2,3]

[karmic ravine]

thank you

Is anyone else able to help me with 2 D. and on please

[vale wigeon]

bad

[karmic ravine]

is that incorrect?

[vale wigeon]

no just ugly and redundant

wait no

it's also wrong

R - (-2,3) would be the correct "subtractive" description

[karmic ravine]

Are you able to correct it please

[vale wigeon]

{x | x <= -2 or x >= 3} is the set builder notation

i just did in two ways

[karmic ravine]

this looks alot more right to me based on the other formatting of the other questions

are you able to help me with D?

[vale wigeon]

for D you need to assume that these dots are spaced exactly one unit apart, i think.

[karmic ravine]

alright

[vale wigeon]

and you will need to just list them and enclose the list in {}

comma-separated obviously

[karmic ravine]

yep

theres no arrows which is throwing me off

and can i ask, do you always read it from left to right?

[vale wigeon]

wym?

as in, should you always arrange the numbers from low to high? yes. for consistency.

[karmic ravine]

{x|-1(< or equal)x(< or equal) 5, x E Z}

im not that confident

[karmic ravine]

Alright thank you

[vale wigeon]

<=

but also i had something simpler in mind

[karmic ravine]

thank you lol i wasnt sure how to write it

[vale wigeon]

{1, 2, 3, 4, 5}

[karmic ravine]

would that be the answer?

[vale wigeon]

for e

[karmic ravine]

is that acceptable for an exam?

[karmic ravine]

or is this more suited?

Ohhh for e sorry i thought we were still on d

would that be correct for D?

[vale wigeon]

should stop at 3, not 5.

[karmic ravine]

oh yea mb

but apart from that all good?

[vale wigeon]

seems so.

[karmic ravine]

Yay

im just noting some down 1 sec

[karmic ravine]

and E is deffinitely this?

its so simple i dont wanna believe it lol

[vale wigeon]

why would it be anything else

it is the set consisting of the numbers 1, 2, 3, 4 and 5, and nothing else

[karmic ravine]

is F
{x|0>x, x E R}

[karmic ravine]

oh yeah thats true ty

[vale wigeon]

technically correct but you might want to turn that "0>x" around to x < 0

[karmic ravine]

oh lol its always the little things'

Alright awesome thats it

thank you so much for helping me and being patient i rlly appreciate it

.close

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[nova estuary]

how am i supposed to solve this?

[livid sage]

you can solve for b1 by plugging in the means

then, you can solve for r by using the relation b1 = r * sy / sx

[nova estuary]

oh waiiit

is it because regression passes through the means?

hence why i can plug it in?

[livid sage]

yep!

[nova estuary]

shiiiiii i feel stupid

[livid sage]

😅 i was clueless at this (type of) problem when i first saw it too

[nova estuary]

what about this?

how do i go to -0.7 from a single percentage

i get why its negative

[livid sage]

so the coefficient of determination is 49%

the coefficient of determination is how much of the variation in y can be explained by the variation in x

it's also equal to r^2

knowing that, the correlation coefficient r must be either 0.7 or -0.7

[nova estuary]

ahhhhhh

thanks

.close

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[hidden zenith]

i didnt get it why it didnt give me same answer

[chilly crag]

I think the chain rule is not used like that.
Example:
e^(e^x))=f(g(x))
With f=g=e^x
But here we subbed it into an a

Im not sure thats valid

[ancient temple]

What are your f and g when you use the chain rule?

[hidden zenith]

man wait

inner function is g e^x

and outer is f

[quasi vector]

it doesnt work when you have $f(x)^{g(x)}$

[hidden zenith]

x^x

[ocean seal]

kheerii

[quasi vector]

you're complicating it too much

[hidden zenith]

can i ask why?

[quasi vector]

because differentiating $(e^x)^x$ as you have written is not easy

[ocean seal]

kheerii

[quasi vector]

you cannot use both the rules you state

since both the base and the exponent are functions of x

[hidden zenith]

oww making sense

man how i can learn the basics like this

[hidden zenith]

hey man can you give me some recourses to learn calc like you

ok if not . then

i should close it

.close

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[hidden zenith]

.reopen

[lone heart]

✅

[hidden zenith]

@quasi vector man

igone all the orange text

here the chainn rule has used for f and g both function of x

[quasi vector]

the chain rule applies when one function is the argument of another function

that's not what's happening wiht $f(x)^{g(x)}$

[ocean seal]

kheerii

[quasi vector]

here, one function is exponentiated to another function

[hidden zenith]

owww'

thanks man

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[scenic patio]

im trying to understand how i can solve for X but im having a bit of troubble?

[gray isle]

in this context, it seems those marked angles are congruent
in which case you can consider similar triangles (or view it as trig function of the same angle gives the same ratio)

[scenic patio]

yes, they are similar, i understand that the right triangle has a side of 14 and the left one has a side of 8

other than that im quite clueless as what to do

[gray isle]

set up your ratios

[scenic patio]

so 2x-2=3x?

[gray isle]

no

just because you see certain expressions doesn't mean you should set them equal to each other without proper justification

do you know properties of similar triangles?

[scenic patio]

not really, other than the fact they're the same pretty much, like same angles and stuff just different sizes

[gray isle]

i'd recommend first looking up an article on similar triangles

[scenic patio]

okay, thank you for helping 😄

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[mystic swallow]

no, actually; here it's f(g(x)) where g(x) = x^2 and f(x) = e^x

[mystic swallow]

.clsoe

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[vast sonnet]

.close

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[vale wigeon]

!status

[lone heart]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[verbal galleon]

1

[vale wigeon]

do you know how to solve already-factorized equations

[verbal galleon]

I missed most of it since I couldn't go to school for 3 weeks so not really

[vale wigeon]

so if i gave you the equation

(z-5)(z-8) = 0

could you solve it

[verbal galleon]

z=5 z=8? Not sure

[vale wigeon]

let me remind you of the zero-product law

if the product of two (or more!) numbers equals 0, then this could ONLY have happened by means of one of the original factors being 0.

so in your case
(x = 0) or (6x-5 = 0) or (6x+1 = 0)

[verbal galleon]

Alr

[vale wigeon]

do you understand this? Y/N

[verbal galleon]

Y

What's the next step?

[vale wigeon]

.

this should let you find the three roots of your equation...

[verbal galleon]

I'm not exactly sure what the question is asking can u tell me what they mean by asking for the sum of all the solutions

.close

[lone heart]

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[mystic nebula]

tan is opposite over adjacent

5.77 is adjacent to the angle (the camera)

12.39 is opposite to the camera

so it should be tan^-1(12.39/5.77)

[civic void]

They have asked for angle of depression though

[mystic nebula]

which is the hypotenuse right

[civic void]

thats the angle between the hyp and the parallel line above it/perependicular to the base

[vale wigeon]

the sum of all the solutions means the sum of all the solutions. the sum of a list of numbers is what you get when you add them all together...

(sorry for interruption op)

[mystic nebula]

u good

[civic void]

angle of depression is how much the object (camera in this case) is tilted from looking straight

[mystic nebula]

wouldnt it be parallel to the base since the ceiling is parallel to the floor

[civic void]

yeah mb parallel to the base

not perpendicular

[mystic nebula]

the angle of depression is the hyp tho since it clearly functions as a hyp

[civic void]

do you know what the angle of elevation is?

[mystic nebula]

positive slope?

[civic void]

think about it like this

say the camera is originally looking straight or along the x axis, angle of depression/elevation is the angle by which the camera adjusts to look to its new target

so the angle of depression or elevation is always with respect to x axis or any parallel line in this case

if they had asked for the angle of elevation, what would you have done?

[mystic nebula]

the same exact thing i guess

[civic void]

talk me through it

[mystic nebula]

hold on

so the angle of depression here is 90- the hypotenuse

the ceiling is technically the x axis right

is that what you mean?

[civic void]

by hyp i assume you mean the angle the ceiling makes with the hyp?

[mystic nebula]

and im measuring the distance it moved from parallel to the ceiling to the angle its at right now?

[mystic nebula]

yea mb

[civic void]

yeah youre right

although if you look at the angle at the 'head teller', you can see that its the same as the angle of depression

[mystic nebula]

oh then that makes sense i guess. so i could still do tan^-1(12.39/5.77) id just have to subtract it from 90

[civic void]

vertical angle theorem

if you use that, you would just get tan inverse (5.77/12.39)

[mystic nebula]

havent learned this yet

[civic void]

you should have

one sec

do you recognise this?

[mystic nebula]

well did in like sophomore year

but i forgot

and it wasnt mentioned after

and this is pre calc trig rn

yea it looks familiar

[civic void]

an application of that is

[civic void]

if you slide angles C and D up, you would go back to this sort of figure

in this case the difference is that the angles are across parallel lines

[mystic nebula]

i see

[civic void]

Do you see where you can apply that over here?

do you see how the angles in green are equal?

if you do, then this question would just be simplified into tan^-1 (5.77/12.39)

[mystic nebula]

that makes more sense now thanks

[civic void]

if they ask for angle of depression, you should just be able to apply this everywhere

@mystic nebula could you do .close

[lone heart]

@mystic nebula Has your question been resolved?

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[signal halo]

anyone know how to solve the 1st order differential equation (1+x^2)dy/dx +y^2 = 4

[quasi vector]

isn't this separable?

[signal halo]

i didnt think so

[quasi vector]

it is

[signal halo]

thought it was integrating factor

[quasi vector]

no, integrating factor only works for linear DE's

of the form $y'+P(x)y=Q(x)$

[ocean seal]

kheerii

[quasi vector]

which this is not, because of the y^2 term

[signal halo]

what would i do then?

[quasi vector]

separate the variables

[solid yacht]

Sorry, I have seen this question and i'm now interested but how would ypu seperate the variables

[quasi vector]

just isolate dy/dx in the equation

it seems pretty straightforward, no x's and y's are connected anywhere

[solid yacht]

when you divide by 1 +x^2 then y^2 is divided by x no?

[quasi vector]

first take the y^2 to the other side

[solid yacht]

what about secondly?

[quasi vector]

what?

[solid yacht]

whats the next step

[quasi vector]

$(1+x^2)y'=4-y^2$

[ocean seal]

kheerii

[quasi vector]

divide both sides by $(1+x^2)(4-y^2)$ to get $$\frac{y'}{4-y^2}=\frac1{1+x^2}$$

[ocean seal]

kheerii

[quasi vector]

then integrate both sides with respect to x and isolate y in the resulting equation

[solid yacht]

then? sorry about this i just don't understand

[quasi vector]

well you have now isolated the variables..

should be pretty straightforward from here

[solid yacht]

bro i am sorry but i really sont know what to do

please type out next step

[quasi vector]

abusing notation, we can write this as $$\frac{dy}{4-y^2}=\frac{dx}{x^2+1}$$ or $$\int \frac{dy}{4-y^2}=\int\frac{dx}{x^2+1}$$

[ocean seal]

kheerii

[quasi vector]

both of these integrals should be fairly simple to solve

[slim idol]

sorry mate mind explaining how you would solve these integrals?

[quasi vector]

well, the left one is a simple logarithmic integral, which you can compute by partial fraction decomposition

the right one is just arctan(x)

[solid yacht]

show me how please

sorry I just don't get it

how does the one on the left not integrate like this

is it cos its a negative?

[quasi vector]

yes

,w integral of 1/(a^2-x^2) dx

[ocean seal]

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[lone heart]

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[upper brook]

How would you factor the polynomial a³ + 5a² + 3a - 9 using common factor by groups? The answer is: (a - 1)(a + 3)².

[lone heart]

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@upper brook Has your question been resolved?

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[glass tide]

What does the vertical line mean?

[tacit arch]

Probably conditional expectation

[glass tide]

@tacit arch can you dumb it down more

[livid sage]

Probability (M = m), given that (C = c) ?

[glass tide]

Okay, thank you @livid sage

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[alpine sable]

Can you convert pound-mass to newtons without using the decimal conversion factor

[alpine sable]

i think its like 4.42 or something

[lone heart]

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[alpine sable]

.close

[lone heart]

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[night bane]

is someone here belgium ?

[night bane]

pls ;answere

i need quick help for a competitive assesment

[zealous valley]

I can speak Dutch if that is the question?

[night bane]

today there was the qualifications of math olympiads of belgium and i was supposed to do it today but they closed school bc of snow and people in belgium could do the assesment so i am looking for someone who did it and can tell me more about it bc i will do the test tomorrow

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@night bane Has your question been resolved?

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@night bane Has your question been resolved?

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@night bane Has your question been resolved?

[solid smelt]

if its the same r you trying to cheat

[lone heart]

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[lone heart]

@unkempt anchor Has your question been resolved?

[unkempt anchor]

<@&286206848099549185>

[fickle sentinel]

do you see that LJ lies along the line JD

@unkempt anchor

[unkempt anchor]

yea but idk how to solve it after that

i forgot 💔

[fickle sentinel]

think of LJ as one part of the line

and LD as another part

those two lengths form JD

if you can visualise it

[unkempt anchor]

ok so how do i add x+4 and 6x-10??

[fickle sentinel]

x + 4 + 6x - 10

[unkempt anchor]

sooooo

7x-6??

[fickle sentinel]

mhm

thats it

[unkempt anchor]

ohhh ok

you were wrong

💔

oh well i got the other questions right anyways

[fickle sentinel]

uhm it is supposed to be 7x - 6

[unkempt anchor]

ty for the help

.close

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[high tartan]

dont know how to answer g(x)=log(f(x))

[high tartan]

0 clue dont remember doing it in class

[unkempt mica]

what is the domain of log(x) ?

[high tartan]

(0,∞)

[unkempt mica]

yessir

[high tartan]

lets gooo

thanks

[unkempt mica]

for what values of x is f(x) in that range ?

[high tartan]

x>0

[unkempt mica]

no

look at the graph

when is f positive

[high tartan]

[1,5]

[unkempt mica]

no

[high tartan]

[-5,-3] U[3,5]

[unkempt mica]

yes but -3 and 3 shouldnt be included

because 0 is not in the domain of log

[high tartan]

[-5,-3)U(3,5]

better

[unkempt mica]

yes and that is your answer

[high tartan]

thanks

alot

.close

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[vapid steppe]

$\int\left( b'(x)\cdot g(b(x))-a'(x)\cdot g(a(x))\right)\dd{x}$

[ocean seal]

Jash

[safe tartan]

should equal g[b(x)^n+1] - g[a(x)^n+1]

[vapid steppe]

wait isn’t it $\int_a^b g(t)dt$

[ocean seal]

Jash

[safe tartan]

wait is g a function

or a constnat