help-39

[safe prairie]

just power rule

[exotic plume]

ok

also

now I want to now how to add this up

this is based on the previous problem

.close

[pearl pond]

Channel closed

Closed by @exotic plume

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[outer hare]

Hi, im trying to learn the chain rule

[outer hare]

how would you solve this

[hexed crypt]

You know how to do chain rule?

[outer hare]

Yeah, its just mostly splitting it up into steps right

like

differentiating f(g(x))
and it would become f’(g(x)) g’(x)

[hexed crypt]

Yes

So in this case, f(x)=sqrt(x) and g(x)=5-4x²

[outer hare]

hm ok

so split them up and do them separately

ok i think i understand

give me a second to work it out

so f(x) = x^1/2

f'(x) = 1/2x^-1/2

idk if im doing this right

[hexed crypt]

That is correct

[hexed crypt]

And so you plug in g(x) into x here

[outer hare]

and then gx = 0-8x

[hexed crypt]

Yes

[outer hare]

or just -8x

[hexed crypt]

So just -8x

[outer hare]

1/2(-8)^1/2

[hexed crypt]

And now just multiply

[outer hare]

so 2

[hexed crypt]

What

No

[outer hare]

i thought to plug in the x

[hexed crypt]

No, it's f'(g(x))*g'(x)

So the derivative of f, with g as x, times the derivative of g

So [1/2f(x)^(1/2)]*-8x

Or rather:
[1/2(5-4x²)^(1/2)]*-8x

[outer hare]

[1/2(x-4x^3)^-1/2]*-8x?

yeh

damn ok

[hexed crypt]

?

[outer hare]

2

not 3

[1/2(x-4x^2)^-1/2]*-8x

[hexed crypt]

Where did the other x come from

[outer hare]

mistake

[1/2(5-4x^3)^-1/2]*-8x

sorry sorry i get it im just typing it in my head

[1/2(5-4x^2)^-1/2]*-8x

[hexed crypt]

Yeah

And so the -8 and 2 cancel out

So you have

[outer hare]

wait how would they

[hexed crypt]

$\frac{-4x}{\sqrt{5-4x^2}}$

[jolly parrot]

Estebson

[outer hare]

ahh

it didnt require me to simplify

but yeah i got the process

f’(g(x)) g’(x)

[hexed crypt]

Ah ok

[outer hare]

alr rememvering that

[hexed crypt]

Good to know

[outer hare]

im gonna do one more just to make sure

thank you so much for your help

[hexed crypt]

That is going to be a very useful tool

You're gonna see it a bunch in derivatives

[hexed crypt]

No problem!

[outer hare]

yeah i immagine lmao

this next one makes less sense tbh

do i do the same thing but just use the graphs outputs

[hexed crypt]

Oh dear what is that

[outer hare]

i think its the same thing just idk

graphing lol

im trying it work it out

f’(g(x)) g’(x)

g(1) would be

2

so

f'(2)*g'(1)

hopefully

-3 * 2

-6

does that seam right?

bam

lmao

[hexed crypt]

Well good job, because I was completely stumped lol

[outer hare]

yeah its just a huge mindfuck

basiclly its the same thing just

using the graphs outputs

[hexed crypt]

But yeah, it seems to be right

Just take the slopes

[outer hare]

and using the actual slope

yup

i hate these already lol

f(x) = cos^4x

g(x) = 4x^2+1

[hexed crypt]

Oh so this is a double

[outer hare]

f'(g(x))*f(x)

bruh

your scating me

scaring

whats a double

[hexed crypt]

f(x)=x^4
g(x)=cosx
h(x)=4x²+1
f(g(h(x)))

[outer hare]

jesus christ ok

[hexed crypt]

Ok so do derivative of x^4

[outer hare]

how did you figure out f(g(h(x)))

[hexed crypt]

Right

[outer hare]

yeah 4x^-3

ops

4x^3

^^

then sinx

then 8x

[hexed crypt]

Ahh

Clipboaaard

[outer hare]

im so lost rn

[hexed crypt]

So replace x with cos(4x²+1) here

[outer hare]

can you hop in a vs by chance lol

[hexed crypt]

Sure ig

[outer hare]

vc

kk 8kbs?

f'(g'(h'(x)))

[hexed crypt]

Oh dear hold on

My mom's in the room

Can't talk

[outer hare]

kk no worries

ill try to figure it out and send you a dm if i need help :D

[hexed crypt]

Ok

See if you can follow
f(x)=x^4
g(x)=cosx
h(x)=4x²+1
f(g(h(x)))
[f(g(h(x)))]'=f'(g(h(x)))*g'(h(x))*h'(x)

So
f'(x)=4x³
g'(x)=-sinx
h'(x)=8x

When plugging in everything:
{[cos(4x²+1)]^4}'=4[cos(4x²+1)]³*-sin(4x²+1)*8x

=-32x[cos(4x²+1)]³sin(4x²+1)

[pearl pond]

@outer hare Has your question been resolved?

[pearl pond]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[languid vigil]

Hello! I'm working on a YouTube video discussing the hardest Geometry Dash level and the hardest one requires a FPS count of this number. I'd really like to be able to at least attempt to be able to understand what this means and I'm wondering if any of you understand this lol

[midnight haven]

those are meaningless letters without a definition of what they are

[languid vigil]

oh damn

ok lmao i wasnt rly expecting to understand it even if it did mean smth anyways 😭

thx for informing me its not actually anything tho

.close

[pearl pond]

Channel closed

Closed by @languid vigil

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[lucid brook]

the ratio of LCM (2,4,6,8 . . . 26) the smallest first 6 prime numbers (2,3,5,7,11,13)

[lucid brook]

vroom vroom

If u do this question

u are legend

[young ermine]

$\frac{8}{2}}$

[jolly parrot]

Ken_

$\frac{8}{2}}$
```Compilation error:```! Extra }, or forgotten $.
l.57 $\frac{8}{2}}
                  $
I've deleted a group-closing symbol because it seems to be
spurious, as in `$x}$'. But perhaps the } is legitimate and
you forgot something else, as in `\hbox{$x}'. In such cases
the way to recover is to insert both the forgotten and the
deleted material, e.g., by typing `I$}'.

Preview: Tightpage -1310720 -1310720 1310720 1310720
[1{/usr/local/texlive/2020/texmf-var/fonts/map/pdftex/updmap/pdftex.map}]```

[young ermine]

$\frac{8}{2}$

[jolly parrot]

Ken_

[lucid brook]

Options 11 13 24 25 26

[jolly parrot]

Ken_

[lucid brook]

so basically the least common multiple (2,4,6,8,10,12,14,16,18,20,22,24,26) divided by (2,3,5,7,11,13)

[pearl pond]

@lucid brook Has your question been resolved?

[pearl pond]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[verbal snow]

How do you solve (5 - 3i) divided by (2+5i) where i is the square root of negative one, i know how to add substract and multiply, but i have no idea how to approach division

[cinder flower]

omgggg

[verbal snow]

Omgg

[cinder flower]

does solve (5 - 3i) divided by (2+5i) mean write it like a + bi?

[verbal snow]

Yeah

That is what form the answer is supposed to be in

[cinder flower]

I think the most straightforward way is to start with writing $\frac{5-3i}{2+5i}=a+bi$

[jolly parrot]

layla💜

[cinder flower]

then $5-3i = (2+5i)(a+bi)$

[jolly parrot]

layla💜

[verbal snow]

Then 5-3i = 2a + 2bi + 5ai + 5bi^2?

[cinder flower]

yep, perfect

[verbal snow]

i squared is -1 right

[cinder flower]

yep

[verbal snow]

So you move that over

But what you do with 2bi and 5ai

[cinder flower]

so 5-3i = 2a -5b + 5ai + 2bi = 2a - 5b + (5a+2b)i

[verbal snow]

You lost me😭

[cinder flower]

the first equality is just this with 5bi^2 = -5b

[verbal snow]

Oh

[cinder flower]

then the second one is from factoring out the i from 5ai + 2bi

also oops had a typo

following so far?

[verbal snow]

No

I dont know how to use 3 way equations

At least we dont use them in class

[cinder flower]

oh ok well we just want 5-3i = 2a - 5b + (5a+2b)i anyway

[verbal snow]

Oh ok

[cinder flower]

do you follow up to there?

[verbal snow]

I think so

[cinder flower]

so now if two complex numbers are equal then their real and imaginary parts are both equal

in other words, 5 = 2a - 5b and -3 = 5a+2b

[verbal snow]

Then you do system of equations to find a and b?

[cinder flower]

sure

[verbal snow]

Then once you have the value of a and bi, you just plug it in to a+bi

and that is solution?

[cinder flower]

yep

[verbal snow]

I think i still dont get the 2nd step

5- 3i = 2a + 2bi + 5ai - 5bi

5 - 3i = 2a + 5ai - 3bi

Then im lost

[cinder flower]

asdufhaesfuh sorry I made a typo earlier

[verbal snow]

Oh wait no

5 - 3i = 2a + 2bi + 5ai - 5b*

[cinder flower]

I meant 5bi^2 = -5b, not 5bi^2 = -5bi

[verbal snow]

There

Yeah

[verbal snow]

So what is next from here

[cinder flower]

5 - 3i = 2a - 5b + (5a+2b)i

because 2bi + 5ai = (5a+2b)i

[verbal snow]

Oh

I see it now

Then how do you get the 2 equations for the system

[cinder flower]

the real part of 5 - 3i is 5 and the real of part of 2a - 5b + (5a+2b)i is 2a - 5b

and the real parts must be equal

the real part of a complex number like x + yi is x

and the imaginary part is y

[verbal snow]

Then im parts equal?

[cinder flower]

yep

[verbal snow]

-3i = (5a+2b)i

Divide by i?

[cinder flower]

the imaginary part of 5 - 3i is actually just -3, not -3i

[verbal snow]

Oh

[cinder flower]

so we can go right to -3 = 5a+2b

[verbal snow]

It wouldnt be bad if you did this though right? Cause you get the same thing if you divide by i

[cinder flower]

yea that's fine too 🙂

[verbal snow]

I think i got it now

Thx sm!

[cinder flower]

np!

[pearl pond]

@verbal snow Has your question been resolved?

[pearl pond]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[tough void]

Linear Algebra Question: If an indexed set of vectors S = {v1, v2 .... vn} contains a vector vj where j > 1 such that vj is a linear combination of the other vectors, why can it not be said that the set of vectors are linearly independent? Furthermore, let's say that the set of vectors S form the columns of an nxn matrix A. Why could it not be said that the columns of A span R^n, given that the above conditions are true?

[tough void]

So I kind of understand one part of this question, but I am wondering more on how these concepts tie together. The set of vectors would be linearly dependent because if we set Ax=0 and solved for x, we could have some non-zero solution x. But why does this not allow us to say that the columns of A span R^n?

This is the essential part that I'm missing

[rancid hearth]

Fix S as the canonical basis of R^(n-1) plus a linear combination, it's clear that there's a vector in R^n that's not a linear combination of vectors in S

[tough void]

What do you mean by canonical basis? Are you saying that if that vj vector exists within the set S, then it (the set) would span R^(n-1)? Also just to clarify, in my textbook, it says that R^n would just be a column vector with n x 1 entries. So when you say that there's a vector in R^n that's not a linear combination of the vectors in S, do you mean that c1v1 + c2v2 + c3v3 + cnvn != some(b)

From my understanding, if was to say that 2 column vectors in a 2x2 matrix span R^2, this would mean that there is some linear combination of these columns that would yield a result for every value in the set R^2

[mental hinge]

im gonna suggest the 3b1b eola video on linear combination span and basis

[rancid hearth]

Doesn't really matter which basis, i just set an example where S would span R^(n-1) but not R^n

It might be easier to see concretely, let n=3 and say S={v1=(1,0,0), v2=(0,1,0), v3=(1,1,0)}

It's clear that v3=v1+v2, but there's no combination that would be equal to (0,0,1) which is in R^3

[tough void]

do you mean v3 = x1+x2?

[rancid hearth]

No, I wrote v1=(1,0,0) and v2=(0,1,0)

[tough void]

oooh

thats what you mean

got it

Ok ok i see what you're saying

So is this generalizable to all cases?

This is very interesting

[rancid hearth]

The fact that you have a vector that is a linear combination means that a set of cardinality n that contains it cannot span a space of dimension n

[tough void]

Let me think for a sec

So... just to be clear, when you say "set of cardinality n that contains it cannot span a space of dimension n" do you mean that if I can find a single value in R^n that is unobtainable as the linear combination of the vector set S, then it necessarily cannot span a space of dimension n?

Is there a generalized proof for this?

Like, now I understand what's going on, I just don't know that it's true in every case where that vj exists in the set

Like I haven't seen it

[rancid hearth]

It's the definition of span; the smallest vector subspace that contains the set

[rancid hearth]

Yes, being in the span of a set is the same as being a linear combination of elements in the set

https://en.wikipedia.org/wiki/Linear_span

[tough void]

This was the definition in my textbook

https://i.imgur.com/f8wwPKg.png

[rancid hearth]

It's equivalent

[tough void]

It's missing the "smallest vector subspace" part

[rancid hearth]

"Smallest" there means that nothing but the linear combinations are counted in

[tough void]

ooo

IC

Okay

One last thing

I think I get it now, but, I want to see the proof that if vj exists within the set then the set cannot span R^n (assuming nxn matrix A)

Is there a proof for this? What should I look up to find it?

[rancid hearth]

It is a theorem that the span of the set including linear combinations is the same without them, then you can construct a vector in R^n that is not obtainable by taking the orthogonal of the set S, which necessarily exists

Sorry if it's still too hand-wavy that's as detailed I can phrase it without checking a book lol

[tough void]

Hahaha no it's okay you have helped out a lot

I'm going to be doing more research on it, maybe I'll go to office hours and ask my professor for a proof

Really appreciate the help

[mental hinge]

just gonna mention it again because i rly do believe it to be a valuable resource

3b1b's essence of linear algebra is a super intuitive way of looking at the subject

and its extremely helpful

[tough void]

I will watch it rn

The whole series?

[mental hinge]

each video is about 10-15 mins ish

i would recommend the whole series to supplement your coursework

[tough void]

Will do

I'm about to start watching

Thanks for the recommendation

[mental hinge]

np and gl with ur work!

[tough void]

.close

[pearl pond]

Channel closed

Closed by @tough void

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[rose hazel]

Suppose the inflation rate of a country in 2009 was 20%. If a dress costs $150 at the beginning of the year, how much would it cost at the end of the year?

[inner cairn]

Convert the percentage to a decimal by dividing it by 100

Then multiply 150 by your answer

[rose hazel]

k

[inner cairn]

And then add 150 onto that

Or just add 1 to the decimal before multiplying

[rose hazel]

so 150.3

but the answer options are these

https://gyazo.com/78bc9450acf1e8e209cc40d063c13549

[inner cairn]

0.2 * 150 = 30

[rose hazel]

o

i am a dumbass

[inner cairn]

30 + 150 = 180

[rose hazel]

I did it incorrectly

ty

[inner cairn]

All good

[rose hazel]

I got another one :

the bruh

https://gyazo.com/ada8d23651b3afedf229b506ce18099f

[inner cairn]

It’s the same working out

[rose hazel]

so

[inner cairn]

Almost

[rose hazel]

-2%*150

I mean

-2/100

then *150

[inner cairn]

Do 100% - 2%

Then multiply that by 150

[rose hazel]

is that minus?

[inner cairn]

Yeah

[rose hazel]

98%*150?

[inner cairn]

yeah

[rose hazel]

147

[inner cairn]

Yeah

[rose hazel]

and the only one close is 1470

[inner cairn]

1500*

I was still thinking of the last question my bad

[rose hazel]

o

[inner cairn]

98% * 1500

[rose hazel]

oh ok

1470

[inner cairn]

Yeah

[rose hazel]

trol

ok

tysm

.close

[pearl pond]

Channel closed

Closed by @rose hazel

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[naive musk]

How do I group this 3z^2 - 9z - 2z + 6?

[naive musk]

I'm workin on quadratic formula

[fiery vortex]

wdym group

[fleet sky]

well that would be 3z^2 - 11z + 6 to combine like terms

[naive musk]

welllll like how 3x^2 + 9x + 1x + 3 can be grouped and simplified into (3x + 1) (x + 3)

[fleet sky]

ah you mean factored

[naive musk]

factorered mb

factored*

[fiery vortex]

eh this

doesn’t seem

right

[fleet sky]

well you need to find 2 values that multiply together to 18 (6*3) and add up to -11.

[naive musk]

i did tho, thats how i made the firs part of 3z^2 - 9z - 2z + 6

the problem is 3z^2 - 11z + 6

[fleet sky]

oh I see. Well then I would look for common factors of your equation

for example 3x^2 and -9z share a factor of 3z

[naive musk]

so 3z would be put in front of somethin like 3z( )( )

[fleet sky]

so 3z^2 - 9z - 2z + 6 = 3z(z - 3) - 2z + 6.

then 2z + 6 has a common factor of 2

so then we get 3z(z - 3) -2(z-3). Then you can factor out a (z - 3) to get (z-3)(3z-2)

and there's your factor

and that's really just playing around with different things you can factor out until you get one that fits

[naive musk]

this is makin sense just have another question, am i lookin for common factors in ANY part, or did ya purposefully look for common factors between 3z^2 and 9z because theyre neighbors?

[fleet sky]

nah I just normally start with "what's the most I can factor out". I could factor out a z, and a 3, so why not factor out 3z and see what happens?

it's a bit of trial and error at first then you get good at spotting the stuff

[naive musk]

can a common factor be negative?

does that make sense or nah?

is that just relative to the equation im finding common factors in?

nvm all good ty

.close

[pearl pond]

Channel closed

Closed by @naive musk

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[arctic horizon]

Hello!

-Assume that 18% of the population is left-handed. Determine the probability of having terms within a group of 50 individuals at least 2 lefties

[arctic horizon]

How can i do? I did like this: 9/50*9/50=3.24%, its correct?

[cinder flower]

that sounds wrong

I'd calculate the probability of 0 or 1 lefties

[hybrid dome]

hmmm

[hybrid dome]

it says atleast 2

so that means 2-50

[arctic horizon]

how?

[cinder flower]

yea but I meant for complementary probability purposes

sounds a lot easier that way

[hybrid dome]

oh

but

wouldnt it be

wait howd i explain it

@cinder flower so how would you assume to do it?

[hybrid dome]

??

can you explain your thought process

im trying to see how you did it

[cinder flower]

the number of lefties follows a binomial distribution

[arctic horizon]

9/50*9/50 =81/2500 = 3.24%

9 = the number of lefties

[hybrid dome]

wait so its 18% right?

[arctic horizon]

50 = the group

yeeh

[cinder flower]

for what?

[arctic horizon]

50*0.18 = 9

[hybrid dome]

ohhhhhh

i see

wait so @arctic horizon

its like

48/50?

wait no

according to layla's logic

[cinder flower]

I can confirm the answer is not 3.24%

nono that's not my logic 😭

[hybrid dome]

mybe 96.76

dude idk

[cinder flower]

not that either

[hybrid dome]

im more confused

than i was

-Assume that 18% of the population is left-handed. Determine the probability of having terms within a group of 50 individuals at least 2 lefties

so

like

ohhhhhh

what would it be

1 <2> atlest 2 >50

what is the question asking HELP

WHATS THE probability of getting atleast 2 lefties

[arctic horizon]

yesssss

[hybrid dome]

i need some music

ill be back

ok im back

[cinder flower]

I'm gonna guess it's translated which is why it sounds a little off

[arctic horizon]

at least two peoples need be lefties, what is the probably?

[hybrid dome]

,w solve9/50*8/50

but is that how probs work?

[jolly parrot]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by idk man#0434

[hybrid dome]

myb?

[cinder flower]

no

[hybrid dome]

but is it a sample spaced roup or

[arctic horizon]

2.88%?

[hybrid dome]

nono

probability is 0-1

layla said it

,w 1-0.0288

oh wait it doesnt work lol

[jolly parrot]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by idk man#0434

[hybrid dome]

thats like not the answer

ok so

[cinder flower]

what's 9/50*8/50 supposed to be a calculation of?

[hybrid dome]

-Assume that 18% of the population is left-handed. Determine the probability of having terms within a group of 50 individuals at least 2 lefties

0.0288

oh

like

from 50 the probs is 18

9

is the probable no of lefties

from 50

thats what i though anywats

imma look at some google to learn how youd do this type of question

sorry dudes

[arctic horizon]

im looking here

[hybrid dome]

Part A:

Given that 18% of the population is left handed, in a sample of 6 customers, we expect to see 0.18 x 6 = 1.08 lefties.

Part B:

Given that the proportion of lefties is 18% or 0.18 and the proportion of non-lefties is 100 - 18 = 82% or 0.82.

Since the sample size is 6, the sample standard deviation is given by:

s=\sqrt{npq} \ \ =\sqrt{6\times0.18\times0.82} \ \ =\sqrt{0.8856}=0.94

[arctic horizon]

https://www.quora.com/Suppose-that-15-percent-of-the-population-is-left-handed-What-is-the-probability-that-in-a-group-of-50-individuals-that-there-will-be-at-most-10-left-handers

[hybrid dome]

holy i was almost right

wait i was right

[arctic horizon]

i didnt understand

[hybrid dome]

this is for 10%

[arctic horizon]

it

[hybrid dome]

yeye

[arctic horizon]

10%?

[hybrid dome]

i got the answer for 18

[hybrid dome]

this is for 15

but i doubt the answer is 0.95%

[cinder flower]

none of the answers proposed so far have been right

[hybrid dome]

yeah but howd you know

ok so

[cinder flower]

because I did it correctly and nobody got the same answer as me 🥲

[hybrid dome]

ahuhhhh

🤷‍♂️

[arctic horizon]

how you did?

[hybrid dome]

second that

[cinder flower]

does binomial distribution ring any bells?

or binomial anything

[hybrid dome]

so i am on that calculator

https://stattrek.com/online-calculator/binomial

i cant figure that shi out

so i didnt bother

[cinder flower]

lolll

on there it would be...

[cinder flower]

.18 probability success, 50 trials, and number of successes = 2

and P(X>=2) is the probability

[hybrid dome]

ohhhhh

[arctic horizon]

@cinder flower can you do the count on a paper and send a picture?

please

[hybrid dome]

0.99942

BOONGA

[cinder flower]

$1 - (1-0.18)^{50} - 50(.18)(1-0.18)^{49}$

[hybrid dome]

@cinder flower SO WHICH IS THE ANSWER

[jolly parrot]

layla💜

[hybrid dome]

bruh i need someone to explain this shi to me

[cinder flower]

,calc 1 - (1-0.18)^(50) - 50(.18)(1-0.18)^49

[jolly parrot]

Result:

0.99941252383037

[arctic horizon]

hmmmmmmmmmmmmmmmmmmmmm

[hybrid dome]

jesus im getting lost

[cinder flower]

that's what I was suggesting at the start

1 - (probability of 0 lefties) - (probability of exactly 1 lefty)

[arctic horizon]

now the final anwser, the prob is 0.99941252383037 or 0.00058747616?

[hybrid dome]

0.99

one

this 0.99941252383037

i think

or so

for the right handed peeps

[cinder flower]

that precision 💀

[hybrid dome]

ofc

there is no probabiliy of it being 0

or all the 50 peeps are lefties

🥲

[arctic horizon]

thank you so much guys ♥

[hybrid dome]

awww its closing

[arctic horizon]

i go to sleep now

its 3am

[hybrid dome]

see ya chief

good nighr

[arctic horizon]

bye bye!

= D

[cinder flower]

gn

[arctic horizon]

.close

[pearl pond]

Channel closed

Closed by @arctic horizon

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[hot niche]

Can someone help me verify that what i have on this question is right? And also find the missing angle

[hot niche]

The picture wont upload?

omg

i cant upload pictures wtffff

It just says queued

[toxic lichen]

try cancelling and reuploading

[hot niche]

I did

[toxic lichen]

it's probably just a glitch on your end

[hot niche]

I agree

Ok we good now

alright so the problem gives us angle 2 as 92 degrees, and angle 12 as 74°.

First thing I did was look at my vertical angles, which are congruent regardless of parallel lines

That gives me number 23 on the left, which asks for angle 8

Next thing i did was find corresponding angles

Angle 2 and 10 are corresponsing, so angle 10 is also 92°

Next i had to find angle 9

We can see that its a linear pair, so i took 180-92 to get 88

Giving me the measure of angle 9

but the last angle im asked to find is angle 5, how do i find that?

It cant be corresponding with angle 2 because the lines arent parallel

I have another question over a problem using a quadratic

A large majority of my issues come from barely being able to do the algebra

I think this one will be a linear system and a quadratic combined

[pearl pond]

@hot niche Has your question been resolved?

[pearl pond]

@hot niche Has your question been resolved?

[hot niche]

Algebra help please

[midnight haven]

Do you need z?

[hot niche]

Yes

@midnight haven sorry for slow reply

[cursive fulcrum]

What issue are you facing after this?

[hot niche]

Z is squared btw its a quadratic

I didnt write it properly

I dont know how to solve 3z^2+33=0

[midnight haven]

alright

Let's subtract 33 on both side

Then divide by 3 both side

what do you have?

[hot niche]

-33/3

-11@midnight haven

[midnight haven]

z²=-11

[hot niche]

Ok wait

[midnight haven]

but z² is always positive for z real

[hot niche]

We can divide it just by 3?

[midnight haven]

so you probably got something wrong

[hot niche]

ahhh man

This is what i got here

[midnight haven]

alright

z is obviously a real

[hot niche]

i got x=18 and y=7.5

which i suppose y has to be wrong

[midnight haven]

So your calculations are wrong somewhere for z

[hot niche]

8x-7+3x-11=180

Gets you x=18

And i took 2y+23=4y+8

[midnight haven]

ok

I'm a bit confused about where that's coming from but alright

[hot niche]

Why?

2y+23=4y+8 because of alternate interior angles theorem

[midnight haven]

ok

so for z²

[hot niche]

Yeah?

[midnight haven]

what did u use

[hot niche]

3z^2-5

[midnight haven]

how did you find your 3z²+33=0

[hot niche]

I got 7.5 as y

By putting alternate interior angles equal to each othee

So

4(7.5)+8

Is 38

and 38 plus 3z^2-5 will equal 180

[midnight haven]

Sounds good

3z^2-5+38=180 not 0

[hot niche]

Thats my issue

thank you now i have to solve

[midnight haven]

You probably can find z² and apply square root to find z

[hot niche]

Can you give me the answer and explain how you get there? I dont know how to use imperfect squares

[midnight haven]

alright

You do it

3z^2-5+38=180

find z²

[hot niche]

Ok ill start with 180

And bring it across

3z^2-142

[midnight haven]

3z^2-142=0

Alright let's take it easy

3z^2-5+38=180

You bring across you have:

(you can say you substract 180 on both side)

[hot niche]

Yeah

so 3z^2-5+38-180=0

[midnight haven]

Alright, keep going

[hot niche]

3z^2+33-180

3z^2-147=0

3z^2=147?

[midnight haven]

yeah

z²=

[hot niche]

49

Right?

[midnight haven]

yes

Now you have

z²=49

[hot niche]

So z^2=7

Z*

Not z squared

[midnight haven]

yes

You applied square root on both side

z=7 OR -7 (because z=-7 also means z²=49)

here z is clearly not -7 so z=7

[hot niche]

Im whacked out bro this class is murdering me

Thank you so so so so much

your a lifesaver man

[midnight haven]

We've all been here

Wlc

[hot niche]

.closw

.close

[pearl pond]

Channel closed

Closed by @hot niche

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[indigo sapphire]

I'm stuck can someone please help me what I did wrong here

[indigo sapphire]

<@&286206848099549185>

[light helm]

your y at x=1 is wrong

[indigo sapphire]

Should I just substitute the x=1 into the original equation?

[midnight haven]

Yeah it's not ln(3)

[indigo sapphire]

So it's just gonna be In(3)-e^-2 ?

[light helm]

yeh

[indigo sapphire]

Oh okay

Is it suppose to be like this now?

[light helm]

no

check your signs in the part with
y - y_1

[indigo sapphire]

-In(3)+e^-2?

[light helm]

yeh

[pearl pond]

@indigo sapphire Has your question been resolved?

[indigo sapphire]

This is what I got but I just wanna make sure if I still did any mistakes on it?

Like in the part where y=(2 x 1/e^2 +1)

[light helm]

looks fine

[indigo sapphire]

okie

.close

[pearl pond]

Channel closed

Closed by @indigo sapphire

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[grim patrol]

help

sqrt((b-x)(x-a)) = t in [0,-ing)

(b-x)(x-a) = t^2

-x^2+(a+b)x-ab=t^2

x^2-(a+b)x+ab+t^2=0

x = (a+b+-sqrt((a+b)^2-4(ab+t^2)))/2

$x = (a+b)/2 \pm \sqrt{(a-b)^2-t^2}/2$

[grim patrol]

so when t=0, x = a,b

so like this

where 1 is branch point, 2 is branch cut

and they connect at inf

im tired ill b back

lol.

[pearl pond]

@grim patrol Has your question been resolved?

[pearl pond]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[grim patrol]

.reopen

[pearl pond]

✅

[rustic gate]

[grim patrol]

hii

was that right so far

ok idk what to do now, i tried to let sqrt(z-a)sqrt(z-b) =t and square it again

but i realise its the same as whats shown in the option? whats the diff

[rustic gate]

,w (z - a)(z - b) = t

[jolly parrot]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by 夢雪#2250

[grim patrol]

idk i skipped this

now im on wiki reading about keyholes

so i can become a future locksmith

[rustic gate]

like

this is meant to be

a dogbone contour

[grim patrol]

LOL

u refering to this or keyhole

[grim patrol]

but the other?

wait actually

[rustic gate]

?

[grim patrol]

how did u even get dogbone

dog bone as in

like a dumbell right

no weird bit goes to inf

[rustic gate]

this

[grim patrol]

this is a

dual keyhole contour

[rustic gate]

so a dogbone lol

[grim patrol]

so C goes to inf

[rustic gate]

no you dont even need C goes to inf

you can just remove the C

[grim patrol]

??

[rustic gate]

like

its a contour around infinity

it encircles infinity

[grim patrol]

thats

the same

anw

no how did u get this

[grim patrol]

i got this

:c

[rustic gate]

no like

you gotta solve

,w (z - a)(z - b) = t

[grim patrol]

how

[jolly parrot]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by 夢雪#2250

[grim patrol]

hm

[rustic gate]

so like

[grim patrol]

whyd u get t^2

whyd i

i squared right

[grim patrol]

oh urs doesnt hv sqrt tho

[grim patrol]

ur original

[rustic gate]

so like

[grim patrol]

change t to t^2?

[rustic gate]

no

you need t to be (-inf, 0]

so you get a line on the real axis somewhere when -(a - b)^2/4 < t <= 0

and then

you get some imaginary line

i think its like

a vertical line

when t < -(a - b)^2/4

[grim patrol]

wow

so weird

just say

(a-b)^2+4t>=0

[rustic gate]

bruh

anyway

you need to have

[rustic gate]

to calculate the branch cut

[grim patrol]

ok but

this is for

z-a z-b

werent we doin sqrt

also

[rustic gate]

no thats the branch cut of

[grim patrol]

hb the non real bits

aka when sqrt is neg n gives im

[grim patrol]

wldnt it differ

[rustic gate]

??

[grim patrol]

or na

like

[rustic gate]

differ what

[grim patrol]

brach cuts for

z-c

[rustic gate]

what

[grim patrol]

is same as

(z-c)^q

q in Q

[grim patrol]

?

wait

[rustic gate]

its entire

[grim patrol]

TnT

what r we doin then

we =that to t for wot

[grim patrol]

(z-c)^q

has branch cuts

[grim patrol]

when z-c in t

yay?

[rustic gate]

yes

[grim patrol]

bit confusin

ill look thru again ltr

[rustic gate]

just

https://tenor.com/bpZ4K.gif

[grim patrol]

BREAKTHROUGH!!!

💕

[grim patrol]

yes

naruto

.close

[pearl pond]

Channel closed

Closed by @grim patrol

Use .reopen if this was a mistake.

[grim patrol]

layer 7

mhmmm

[rustic gate]

time to

layer 8

[grim patrol]

nearly don w this realm

no later

ive to

[rustic gate]

[grim patrol]

stabilise my foundation

lol

hey

i learntab

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[glass cliff]

Can someone pleas help me re-arrange this equation i will be eternally grateful :WOWOWOWOW:

-log(x - T) + c1 = kt + c2
Rearrange to be in terms of T=f(t)

[grim patrol]

as in how to

or just rearange

[glass cliff]

So i have to sub some values in once ive rearranged but i end up with a negative log and no real solutions so i've gone wrong somewhere but im not sure where

[grim patrol]

,w solve for T where -log(x - T) + c1 = kt + c2

whatd u get

[jolly parrot]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Circle#0025

[glass cliff]

i can show the working

[grim patrol]

lol

lol

[pearl pond]

@glass cliff Has your question been resolved?

[pearl pond]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[indigo sapphire]

Idk what to do next tryna differentiate using quotient rule

[shrewd basin]

check dv/dx

[indigo sapphire]

Yes?

I don't get it

[shrewd basin]

Your dv/dx isn’t right

[indigo sapphire]

Ooh okay I fixed it

So 4(2x-1)

And idk what comes after that which I'm confused about

[shrewd basin]

Cancel one of the (2x-1)

Then just expand brackets and simplify

[indigo sapphire]

Idk

[shrewd basin]

,w derivative of (3x^2)/(2x-1)^2

[jolly parrot]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Pure#9389

[shrewd basin]

You’ve already cancelled the (2x-1) in the second term

[indigo sapphire]

aaaah

Okay I get it now

.close

[pearl pond]

Channel closed

Closed by @indigo sapphire

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[tough token]

the first two terms are your first 3 terms, the last 2 terms look similar but not entirely sure if they are exactly the same. cant see anything wrong in the algebra so looks right

symbolab

oh wait no its just derivative calculator

https://www.derivative-calculator.net/

[arctic nexus]

I think you're correct @delicate dock

[tough token]

its $\sinh^{-1}{x}$

[jolly parrot]

Duh Hello

[tough token]

(x^3(1+x^3)^(e^x)(sinh^-1(x))^2022/(log (pi)+cscx)^(1/5)

my absolute value symbol doesnt work on my keyboard so u gotta fill those in youself lol
(3/x+(3e^x x^3)/(1+x^3)+e^x ln(1+x^3)+2022/(sinh^-1(x)sqrt(x^2+1))+(cscx cotx)/(5 log(pi)+5cscx))((x^3(1+x^3)^(e^x)(sinh^-1(x))^2022)/(ln(pi)+cscx)^(1/5)

mine times out as well

think ur just gonna have to go with ur instinct

either go through and make really sure the algebra is right, or try and manipulate yours into being the same as the one this calculator gives

wouldnt take long to try and get them to be the same, the first 3 terms in yours are trivial to see if its the same, can probably use wolfram or something for the last two to check if they are the same

no idea, i havent used anything other than natural log for years

but this calculator i used treated the log as ln

same rules apply anyways

so shouldnt matter, just keep it as ln on both sides

you need to find a function which checks if they are the same. pretty sure its just using 2 equal signs, so "your expression"=="derivative calculator expression"

then you should get true or false

ye

"your expression"=="derivative calculator expression"

[delicate dock]

this?

[tough token]

you can remove the first 3 terms in your expression and set that equal to the last two in the derivative calculator expression

because we already know those are equal, will maybe make the computation time a bit lower, either way u can also just put the full expressions in and it will probably be fine

[delicate dock]

aight wish me luck haha

[tough token]

i gotta go now tho. good luck

[delicate dock]

lol it didn't return true or false

I wish I had mathematica

[delicate dock]

yoo I will trust you okay?

😂

[pearl pond]

@delicate dock Has your question been resolved?

Channel closed

Closed by @delicate dock

Use .reopen if this was a mistake.

[delicate dock]

.reopen

do I have to delete all my messages before I close

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[junior pawn]

an idea to integrate $\int \frac{1}{1 + sin^2x}$?

[jolly parrot]

Max.cc

[grim patrol]

residue theorem

[junior pawn]

never hear about it

[grim patrol]

:c

substituition

[junior pawn]

ok

i was thinking about u = tan(x)

but i'm block

[grim patrol]

sry im bad at this

but

integral calc is good

look at this

[toxic lichen]

@grim patrol you're suffering from hammer-nail syndrome

[grim patrol]

wot

oh using op stuff to do easy stuff?

[toxic lichen]

when you wield a hammer everything looks like a nail

[junior pawn]

$\int \frac{dx}{1 + sin^2x} = \int \frac{dx}{2 - cos^2(x)}$

[grim patrol]

sry i just slayed the demon king

[jolly parrot]

Max.cc

[grim patrol]

idk it gave me weierstrass sub vibes

or res lol

[junior pawn]

hey what the website

[grim patrol]

https://www.integral-calculator.com/

[junior pawn]

it can be pretty useful

[grim patrol]

u can google integral calc w steps

[grim patrol]

it nearly always is

except when they give u smt which antiderivative doesnt exist

[junior pawn]

idk why but worlfram alpha bug with me and you must pay to see the step sx)

[grim patrol]

probably why they r more profitable

LOL

in a capitalist society

companies do what they can to maximise profit

even if it harms the consumer or the workers

[fierce oxide]

mega stonk

[grim patrol]

hey u look diff

[fierce oxide]

just my pfp
whchi depicts my current mood

[junior pawn]

calm calm x)

[rustic gate]

LOL

[grim patrol]

yes

[severe quartz]

for integrations where theres trig functions squared, dividing the num and denom by (cosx)^2 then taking tanx=t works most of the times

[grim patrol]

most is the keyword here

[severe quartz]

this one is solvable by that

write 1 as sin^2 + cos^2

[grim patrol]

sometimes ud have to resort to more powerful weapons

[severe quartz]

what is that

[junior pawn]

it looks simple say like that x)

[rustic gate]

you cant do indefinite integrals with residue theorem lol

[grim patrol]

an integration technique

[grim patrol]

D:

oh its from -inf to inf right

who even does indef integrals smh

[junior pawn]

so i found arctan(u), is it right?

[grim patrol]

well id say just check it w integral calc

im bad at int

mayb if u wait some1 else will come along

if u dk a specific step u can ss it n i can try to help

if im still here

[pearl pond]

@junior pawn Has your question been resolved?

[pearl pond]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[naive musk]

Can I look at 5x(3x-2)-3(3x+2) as 5x(3x-2)+3(3x-2)

[toxic lichen]

no

[naive musk]

ok, im tryna factor the first equation, and im wondering what step im missing to end with (3x-2)(5x-3)

the 3x+2 is throwing me off

cause i know i can throw 3x-2 up front if it had another 3x-2

[peak mist]

suggestion - just multiply it out then factor?

[naive musk]

oh like multiply out 5x(3x-2) and 3(3x+2) and then factor the result of those?

[peak mist]

y

multiply them out

add like terms

then factor if possible

[naive musk]

oh wait but i had it multiplied out to begin with, like (15x^2-10x)-(9x+6)

or is that not the multiplied out version?

[peak mist]

?

then add the x terms

what are you trying to do, factorize?

[naive musk]

yes im tryna factor

[peak mist]

divide the full thing by 3

[naive musk]

the original expression was 15x^2 - 19x + 6

[peak mist]

all of that should equal to 0 right?

[naive musk]

mmm there is no = 0

if thats what ya mean

[peak mist]

what the question say?

[naive musk]

"factor the given expression"

[peak mist]

well

i guess we can factorize this then

[naive musk]

im goin off the ac method