help-13

[shut pier]

how can i solve the integral:

x sqrt (1 + (-1/x^2)^2) dx

[neon void]

$\int x \sqrt{1 + (\frac{-1}{x^2})^2} dx$

[wraith dagger]

mochaccino

[shut pier]

yes

i need to find the area. of revolution and i end up with this integral

[dusk finch]

hmm

i think you could at least simplify it a bit

with some algebra

[shut pier]

yea and i end up with sqrt x^2 + 1/x^2. dx

[dusk finch]

um

maybe try adding the fractions instaed

this

it will open door to a nice simplification of the square root

and later to a very tasty substitution

[shut pier]

ok i will see now

[dusk finch]

(ill take a shower rq, brb, if any helper comes, you may take over if you wish to)

[shut pier]

ok thanks man

i will try to solve now

[dusk finch]

back

[dusk finch]

How did you progress?

[shut pier]

i am stuck at eleminatin the square root

iam thinking of what to add and substrac

[dusk finch]

it cant entirely be eliminated

sqrt(x^4+1)/x

[shut pier]

aa i see

[dusk finch]

Have you gotten there?

if so, try substitution

it should simplify quite nicely

[shut pier]

now i am here

[dusk finch]

okay cool

do you know hyperbolic functions btw?

tanh

you will need it i think

[shut pier]

no i havent learned this in calc2 yet

[dusk finch]

damn

well

you can get it to something like sqrt(u-1) / u

but then it will be hard for you to proceed

or impossible

because you need tanh^-1

[shut pier]

look friend the question is to find the area od revolution of xy= 1

around x axis

[dusk finch]

xy=1?

around x axis...

[shut pier]

yea

[dusk finch]

you got your integral incorrect then

im pretty sure

xy = 1

y = 1/x

[shut pier]

i dont think so

[dusk finch]

ye, you dont need it

because you got the integral incorrec

t

[shut pier]

omgg

[dusk finch]

What are the bounds

of your integral

or the solid

defined by the revolution

[shut pier]

x from 1 to 2

[dusk finch]

alright

so since xy = 1

y = 1/x

so it will be this thing

revoluted around x axis

can you set up the integral now?

[shut pier]

i see

[dusk finch]

do you know how its done?

[shut pier]

yea ofc

[dusk finch]

cool

[shut pier]

and i end up with the same thing

[dusk finch]

nope

you definitely wont

[shut pier]

finding f’x

[dusk finch]

f'x?

why f'x?

thats unneeded

[shut pier]

its the formula

to find the area

[dusk finch]

what method are you using

to set up the integral

[shut pier]

S = 2 pi R dl

[dusk finch]

2piR dl?

[shut pier]

where dl is the arc length

[dusk finch]

that sounds incorrect

shouldnt it be pi R^2 dl

wait are you finding

volume

or surface

[shut pier]

no area

surface

[dusk finch]

surface area

[shut pier]

right

thats my notes

[dusk finch]

well then the integral seems correct

but you will need tanh

[shut pier]

and i am using the second formula in terms of Y around X axis

[shut pier]

thank you

and what is tanh equal to?

[dusk finch]

tanh is hyperbolic tan

and ull specificlaly need its inverse

that is tanh^-1

[shut pier]

what is its derivative

[shut pier]

oh even worst hahah

[dusk finch]

1/(1-x^2)

is derivative of tanh^-1

[shut pier]

and how can i solve this with tanh

what substitution will i need to do

[dusk finch]

well from here

you can try substitution u = x^4

that itself should simplify it significantly

you'll need one more substitution then

[shut pier]

ok thank you

i think i will stop here

[dusk finch]

it's doable, but uneasy tbh

it requires some thought

and experience with hyperbolic stuff

[shut pier]

anyway i am leaving this haha

thank you for your time

how can i exit now

.end

[dusk finch]

.close

[cedar kiln]

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[shut pier]

.close

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[wind sentinel]

Where did they -3/4pi from as a point of intersection without using the graph

[hollow totem]

Just due to intersection: sin(x) = cos(x)

sin(x)/cos(x) =1

tan(x) = 1

In range -pi to pi, only 2 values of x satified: x = pi/4 and x=-3pi/4

[cedar kiln]

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[gray wedge]

I really dont understand how to solve the 3 questions w highlighted answers in red. I ultimately just summed up the numbers in the given range of ages. Im not sure if i just calculated things wrong, put in the answers wrong, or if i just went about the questions completely wrong.

[gray wedge]

this was my work

[vagrant topaz]

for question b, you have calculated a proportion, you must then multiply this by 100 to get the percentage

[gray wedge]

so just 10?

[vagrant topaz]

you’ve written 4/30 = 0.1 to get to that result, is this correct?

[gray wedge]

mhm

[vagrant topaz]

4/30 = 0.133… not just 0.1

[gray wedge]

ah i see, let me recalculate then

[vagrant topaz]

for part c, consider looking at the cumulative relative frequency column

[gray wedge]

13.3 percent..?

[vagrant topaz]

that sounds correct to me, assuming it has the correct number of significant figures

[vagrant topaz]

also note that the third row of data contains those of age 40-49, these should not be included when calculating those under 40

[gray wedge]

im supposed to round to the 1st place after the decimal point, im not good w rounded decimals unfortunately... not when the numbers are bigger.

would it still be 13.3?

[vagrant topaz]

yes

[gray wedge]

for c. i got 0.4001 when i summed up the numbers in the range it asked for. Would that be the answer?

Can i also ask why i would look at the cumulative frequency for this question?

[vagrant topaz]

the cummulative relative frequency tells us the sum of relative frequencies for all lower age groups

so when looking at the relative frequency for those aged under 40, the 2nd row of the cumulative frequency is the sum of relative frequencies for ages under 40 and will therefore be our answer

[gray wedge]

so it would be 0.0667 + 0.2667 (which are the age range) = 0.3334?

relative frequency of ages under 40 = 0.3334?

[vagrant topaz]

yes

[gray wedge]

so then for question d. the answer would be 0.4334?

I got that from adding the all the relative freq under 50, or just by lookin at the cumulative freq of ages 40-49

[cedar kiln]

@gray wedge Has your question been resolved?

[gray wedge]

<@&286206848099549185>

[wind sentinel]

How does -3pi/4 satisfy it

[gray wedge]

.close

[cedar kiln]

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[marsh pond]

I'm in need of help with this question

I have tried <1,1,1> and <0,0,0>

here is my work involved:

notice how i answered basis for the 2 vector eigenspace. the basis is uncovered by the free variables, s and t

but there are no free variables for the 1 vector eigenspace, so I'm not sure how to answer that?

[royal loom]

how would a basis for it's associated eigenspace by (0,0,0)?

(0,0,0) isn't a basis for anything

[marsh pond]

i tried <1,1,1> first

[royal loom]

Well, when you input (1,1,1) into this matrix you get out (0, 5, -5)

which is not = 5*(1,1,1)

Have you found the eigenvector associated with the eigenvalue 5?

[marsh pond]

yes

[royal loom]

What is it?

[marsh pond]

oh the eigenvector, not the eigenvalue

ummm

[royal loom]

There's a few ways that you can solve this, which I can tell you the other ways after, but the simplest (and the way you should begin with) is just finding the eigenvector(s) associated with the eigenvalue

[marsh pond]

are these considered the eigenvectors associated with the eigenvalue 0?

[royal loom]

Yes, if you were to input them into the matrix, it would output (0,0,0) which is equal to 0*(1,0,0) or 0*(0,0,1)

[marsh pond]

so that's what i'm trying to find

[royal loom]

Yes, but for the eigenvalue 5 instead

[marsh pond]

for eigenvalue 0 i found the eigenvectors by using RREF

but for eigenvalue 5 i cannot seem to find them by using RREF

[royal loom]

What are you putting into RREF?

[marsh pond]

here:

[royal loom]

You made a silly mistake

[marsh pond]

where?

[royal loom]

you were supposed to subtract 5 off the diagonal right?

[marsh pond]

oh

[royal loom]

but the middle entry was 5 to begin with, 5-5=0

not 10

[marsh pond]

dang, ty

i just noticed that now too

[royal loom]

No problem

Here's an easier way you can do this too

If you have 3 distinct eigenvalues, then you know your eigenspace is R3

in this case, you have a repeated eigenvalue, and one distinct

so things could go wrong, if your repeated eigenvalue did not have two LI eigenvectors associated with it

but you already computed that it does have those

(1,0,0) and (0,0,1)

so, since 5 is distinct from 0

[royal loom]

Actually, please ignore my blabbering above, your work is correct there good job

[marsh pond]

does this look correct?

[royal loom]

Yes looks good, but you can always verify by substituting into the matrix

A*(0,-1,1) = (0,-5, 5) = 5* (0,-1,1)

[marsh pond]

hmmm, OK

tyvm!

.close

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[gray wedge]

could I get a check on my relative frequencies? I want to know if they’re correct so i can go ahead and calculate my cumulative frequencies. I’m horrible at rounding :/ I’m not sure what i’m supposed to be rounding to either

[cedar kiln]

@gray wedge Has your question been resolved?

[cedar kiln]

@gray wedge Has your question been resolved?

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[gray wedge]

If i remember correctly the mean means average. I know im supposed to be finding the mean of the ages, but i dont know how to start doing so when theyre intervals, or how im even supposed to in the first place.

[rigid lava]

they probably want you to use the middle of each interval

so like use 26 for the interval 24-28

[gray wedge]

But im supposed to give one answer. For the next interval 29-33 i assume the middle would b 31, but how does that cooralate to 26?

[rigid lava]

mean is to add up all the values and divide by total values right?

well there are 20 people who are 26

there are 17 people who are 31

etc

so add up all of those values

20 * 26 + 17 * 31 + ...

then divide by 144 at the end

[gray wedge]

i got 348/144 = 2.416666667 ( answer to 2 or more decimals) = 2.47?

[rigid lava]

you added wrong

did you just do 20 + 17 + 18 + ... ?

[gray wedge]

nope

[rigid lava]

how did you get 348 then

[gray wedge]

i thought i was supposed to find the middle of each interval, then add all those intervals. thats from what i got from ur explanation at least D:

[rigid lava]

you need to multiply by the frequencies

consider this example

what is the average of the 1,1,1,1,2,2?

[gray wedge]

1....?

[rigid lava]

no

it's (1+1+1+1+2+2)/6

add all the values and divide by the number of values

[gray wedge]

411/144...?

thats also wrong

uhm

.close

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[dire gazelle]

can someone please check my answer

[dire gazelle]

I also need hlep with this I've been stuck on it for a while

I found the POI but that's not what this question asks for

I found the derivitive of f(x)

[dull oxide]

You got a few of these wrong

[dire gazelle]

Is 3 false and I wasn't sure if 2 was true or false since the second derivitive leads to acceleration

[dire gazelle]

I then subbed in f'(x) into y after moving everything to one side

[dull oxide]

I can see your confusion on 2 with how you're describing it. But 2 would be false, I'd argue

[dire gazelle]

do I have to inverse the reciprocal and then do y = mx+b?

[dull oxide]

You can prove 3 is true by differentiating $f(x)=\sum_{k=0}^n a_k x^k$

[wraith dagger]

SWR

[dire gazelle]

thenw hich ones would be wrong other than 2?

[dull oxide]

Let's start with 6.

[dire gazelle]

ok for 6 how is it false?

[dull oxide]

Explain to me how it is true

[dire gazelle]

oh wait we have to solve for the derivitive it'll be like x^2 to 2x so it wouldn't go in oposite direction

oh wait I just have to think it out properly to understand it

is there anythign else wrong there?

<@&286206848099549185>

.close

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[dire gazelle]

.reopen

[cedar kiln]

✅

[dire gazelle]

can i please get some hlep with this

[coral jewel]

!1q

[cedar kiln]

It is suggested that you limit yourself to one question per help channel, opening a new one once your original question is answered and your original channel has been closed. This is to make your channel easier to follow for potential helpers and can bring attention to the fact that your question has changed.

[dire gazelle]

.close

[cedar kiln]

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[cedar kiln]

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[quartz crypt]

quick question

how does this work

(x^5+5)^5

how do the exponenets cancel?

[digital cliff]

they dont

did you read carefully?

f is the outer function

g is the inner function

h is f(g(x))

[quartz crypt]

oh then why is it giving me h

[digital cliff]

because it wants you to find g??

[quartz crypt]

some I am plugging g into h?

[digital cliff]

no

h is g plugged into f

f(x+5)=(x+5)^5=h

[quartz crypt]

I don't get it

I don't get it

[digital cliff]

why

[quartz crypt]

theres 3 things

it wants me to plug 2?

f= x^5 and g(x)

that should be be x^5

[digital cliff]

it states very clearly that h(x)=(x+5)^5=(f o g)(x)

and that f(x)=x^5

[quartz crypt]

right so I plug g into that

f(x)=(x)^5

[digital cliff]

sure

[quartz crypt]

okay

where did my exponent go

[digital cliff]

???

(g(x))^5=(x+5)^5

[quartz crypt]

the answer is then x^5

[digital cliff]

that is f

[quartz crypt]

right

[digital cliff]

it wants g

[digital cliff]

this is true

so g is x+5

[quartz crypt]

but it give mes me g already

[digital cliff]

no it doesnt

[quartz crypt]

g(x)

[digital cliff]

what are you talking about

it tells you to find g

[digital cliff]

just because the answer is already there now doesnt mean it was given

[quartz crypt]

okay I don't get it

I've been solving these problems by plugging in

f of g

okay f equation

plug g

[digital cliff]

you plug g into f

yes

[quartz crypt]

yes

okay same page

[digital cliff]

then?

[quartz crypt]

that should be the answer

[digital cliff]

why would that be the answer

[quartz crypt]

because I plugged to solve the equation

[digital cliff]

you havent found g

youve just said (fog)(x)=(g(x))^5

thats not finding g

[quartz crypt]

okay how do I find g

[digital cliff]

what is h

[quartz crypt]

(x+5)^5

[digital cliff]

which is equal to what

[quartz crypt]

I plug into that

[digital cliff]

no

[quartz crypt]

(x^5)+5)^5

[digital cliff]

stop getting plug happy

no

[quartz crypt]

that's what I have been doing lol

[digital cliff]

back it up

[digital cliff]

this is h, what is h equal to also

[quartz crypt]

h(x)= (x+5)^5

[digital cliff]

what else is it equal to

read the question

[quartz crypt]

f o g (x)

but

that would be x^5

[digital cliff]

no thats just f

f o g is (g(x))^5

(g(x))^5=(x+5)^5

if a^c=b^c then a=b

so g(x)=x+5

[quartz crypt]

you lost me on the third line

okay does h equal this

[digital cliff]

equal what

[quartz crypt]

okay

f o g is (g(x))^5

[digital cliff]

yeah

[quartz crypt]

where do we get this

(g(x))^5=(x+5)^5

[digital cliff]

because (f o g)(x)=h(x)

its stated in the question

[quartz crypt]

okay but said x^5 is f

not g

[digital cliff]

you even reading?

what are you talking about

[quartz crypt]

I am reading you are not explaing

[digital cliff]

i very much am

[quartz crypt]

then tell me how f suddenly becomes g

[digital cliff]

where did i ever say that?

show me one instance

[digital cliff]

thats me saying thats not the answer, that is just f

[quartz crypt]

and now it's also G?

[digital cliff]

i never said it was g

[quartz crypt]

right there

[digital cliff]

where did i say g=x^5

i said (g(x))^5=(x+5)^5

[quartz crypt]

OKAY

lets walk this through

I plug g into f

x^5 is f

[digital cliff]

yes

[quartz crypt]

WHERE DO I GET G

[digital cliff]

you are finding g, my word

just say f(g)=g^5

that is it for now

as ive been saying since the start

[quartz crypt]

okay so f and g are the same thing?

[digital cliff]

no

[quartz crypt]

I don't understand

[digital cliff]

why do you think thats what im saying, i dont get it

[quartz crypt]

WHAT IS THE NEXT STEP

how

[digital cliff]

f(g)=h

g^5=h

[quartz crypt]

okay stop

where did you get g from

[digital cliff]

i dont even know what youre talking about, g is just g for now until we find what g actually is as a function

[quartz crypt]

well how is it g^5

where before g=(x)

[digital cliff]

f( g )= g^5

[quartz crypt]

g=(x)

[digital cliff]

?????????

[quartz crypt]

okay we start with g=(x) correct

[digital cliff]

no, where did that come from

[digital cliff]

doesnt say that anywhere

[quartz crypt]

find the inner function g(x)

[digital cliff]

yeah, that doesnt say g=(x)

[quartz crypt]

plug g into f(x)=x^5

[digital cliff]

means there is a function g with input x

[digital cliff]

again, yes

[quartz crypt]

I think we're not understanding on the variable

[digital cliff]

g isnt a variable

you seem to think whenever i say f(g)=g^5 that im saying g=f

and i dont see how

[quartz crypt]

slow down

[digital cliff]

if i said f(2)=2^5 would that mean 2=f?

[quartz crypt]

go back

[digital cliff]

to where

[quartz crypt]

f(g)=g^5

[digital cliff]

okay

[quartz crypt]

we are not understanding the variable

[digital cliff]

there are no variables there

[quartz crypt]

the letter

[digital cliff]

f and g are functions

[quartz crypt]

okay so why aren't we using x

im saying if you use the variable then f is g

[digital cliff]

how

[quartz crypt]

but you use g instead

slow down

[digital cliff]

this makes no logical sense to me at all

[quartz crypt]

f(g)=g^5 or f(g)=x^5

[digital cliff]

the only way f(g)=x^5 is if g(x)=x
and we know that f(g) is not x^5

we just know f(g)=g^5

=h

[quartz crypt]

okay you're suppose to use g instead of x

that's why you kept saying it was f not g

[digital cliff]

yes because x^5 is f, not g

if i input something into a function, x doesnt stay x

[quartz crypt]

even though g=x which is also technically correct

but anyways it equals h

where does the exponent go on H

[digital cliff]

thats just a dodgy and misleading way to write that, but i digress

[digital cliff]

nowhere

g^5=(x+5)^5

[quartz crypt]

well I am trying to get on the same page but it is interchangeable

g^5=(x+5)^5

where does ^5 go

[digital cliff]

its pretty obvious the bases must be equal if the exponents are the same, if a^5=b^5 then obviously a=b

if you dont like that then just take the 5th root of both sides

[quartz crypt]

I've never heard of either but basically this would explain it a^5=b^5 then obviously a=b

Okay thank you lol

the variable thing though lmao

so unfortunate

[digital cliff]

my poor eyes

you were saying g(x)=x

tut tut

[quartz crypt]

lmaaaooo

[cedar kiln]

@quartz crypt Has your question been resolved?

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[upbeat pond]

how do i solve for x?

@knief

@dusk goblet

[dusk goblet]

😹😹😹

what is ln

[upbeat pond]

PLEAAAAAAAAAASE

[dusk goblet]

log base what

[upbeat pond]

Log base e

[dusk goblet]

so

[dusk goblet]

yes

[upbeat pond]

IM the 1 FOR YOE

[dusk goblet]

just do e^ on both sudes

it’s the anti log

if it was log base 10

you’d do

[upbeat pond]

wdym like

[dusk goblet]

10^

e^(lnx)=e^1

[upbeat pond]

what does lnx mean

[dusk goblet]

x=e

[upbeat pond]

as a log

[dusk goblet]

log base e

natural log

[upbeat pond]

how tf do you konw this

[dusk goblet]

how do you not

[upbeat pond]

i guess math is not my favorite

[dusk goblet]

this is something you should’ve learned like 2 years ago

[upbeat pond]

yeah and i did i just got bad memory

[dusk goblet]

you didn’t do exponentials in algebra?

or precalc

[upbeat pond]

i did do precalc

wow theres a buildup to calc

im listening to funky town by alvin and the chipmunks

[dusk goblet]

well logs are just functions that are the inverse of exponentials

they tell you the exponent needed to raise a base to to get a certain result

[dusk goblet]

get funky

[upbeat pond]

oh ok

[upbeat pond]

my calves jsut grew 5 inches

[dusk goblet]

https://tenor.com/view/the-squeakquel-chipettes-gif-17382312219462998101

[upbeat pond]

which one am i

[dusk goblet]

need that

[dusk goblet]

you tell me

[upbeat pond]

realistically Im left realistically im the right

[dusk goblet]

🤔

i was out here playing games

top of the morning

i’ll keep ya best interest

🫱🏻‍🫲🏽

[upbeat pond]

why are the chipettes built to be the chipmunks types this is actual patriarchy

[dusk goblet]

not sure that’s how it works

[upbeat pond]

Oh

[dusk goblet]

you’re the only power

keep it math related

[upbeat pond]

sorry

[dusk goblet]

the middle one is alvin’s counterpart

[upbeat pond]

what is lne

[dusk goblet]

1

[upbeat pond]

why

[upbeat pond]

why

[dusk goblet]

look at them

[upbeat pond]

I hate the short one

https://tenor.com/view/eleanor-chipmunks-chippettes-theodore-smile-gif-5142748

bro i rmr watching this 3d ass movie they look so goofy

[dusk goblet]

the fat one mirrors theo

[upbeat pond]

https://tenor.com/view/travis-scott-apology-gif-25547371

[dusk goblet]

and tall nerdy one is just counterpart to simon

[upbeat pond]

simons literally u but whatever

[dusk goblet]

🙏🏻👍🏻

[upbeat pond]

I HATE THIS SO MUCH THIS MAKES ME MAD

[dusk goblet]

probably understands logs

[upbeat pond]

https://tenor.com/view/the-squeakquel-chipettes-gif-10275117322110343028

HOLY WHERE ARE THEIR

[dusk goblet]

they’re chipmunks

[upbeat pond]

actual sexism

[dusk goblet]

oh ig the males do

have clothes

yea interesting

[upbeat pond]

https://tenor.com/view/the-squeakquel-chipettes-gif-12909464334382822719

omfg i need to rewatch this movie its my actual childhood

.close

[dusk goblet]

do it then

[cedar kiln]

Channel closed

Closed by @upbeat pond

Use .reopen if this was a mistake.

[upbeat pond]

bruh u aint gonna ask to come watch

[dusk goblet]

🇨🇦

[upbeat pond]

🎑

[dusk goblet]

🗽

[cedar kiln]

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[limber geyser]

why does nothing come up with this eqn

lets juist say we input x=1

shouldnt y=-2

etc

[weak otter]

Well yes but what if you put in x=2/2

Then maybe it’s positive 2?

Negative bases are already a bit ambiguous for rational numbers

But then they’re completely undefined for some of them

And undefined for every irrational number

[limber geyser]

but should they still graph every possibility for every x input

[weak otter]

🤷‍♂️

[pliant atlas]

because

x can have an infinite number of values

and for many of those values, the equation just cannot be graphed

[limber geyser]

such as what value for example?

[pliant atlas]

Like

lets say

1/2?

When we're computing like

(-2)^x, we're essentially finding the square root of a negative number

which is not defined in the real numbers

You can't really square root a negative number now can you?

Unless you go onto the complex plane

[cedar kiln]

@limber geyser Has your question been resolved?

[pliant atlas]

@limber geyser

what has not been resolved 😭

Did you not get it or something

[limber geyser]

oh lol

i misslciked mb

.close

[cedar kiln]

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[limber geyser]

wait wouldnt that just be a restriction tho

.reopen

[cedar kiln]

✅

[limber geyser]

if its x=4

you can grraph that

[cedar kiln]

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[arctic torrent]

The answer I got isn’t in the question. Can anyone tell me what did i do wrong?

[arctic torrent]

<@&286206848099549185>

[royal finch]

You haven't even calculated any eigenvectors in your work

[arctic torrent]

Oh how do i calculate them?

[royal finch]

You can check your notes or youtube for that.

[arctic torrent]

I checked my notes and thats what i got from them. I’ll check youtube

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[primal dust]

i know i have to use tangent but im not sure how to solve it

would i just take 23/404 then take the inverse of tan to get the angle?

[cedar kiln]

@primal dust Has your question been resolved?

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@primal dust Has your question been resolved?

[primal dust]

<@&286206848099549185>

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[vernal violet]

I have a deck of 30 cards. I am trying to pick a magic card. I can choose any of the cards in my hand. I am given two options:

1. I have 5 cards in my hand and there are 8 magic cards in my deck

2. I have 7 cards in my hand but there are only 5 magic cards in my deck

Which option has the higher chance to have a magic card in your hand?

[pastel vault]

For number 1, how many ways of choosing the cards are there in total?

[vernal violet]

5 random cards drawn from the 30 with 8 of those 30 being the card I want

[pastel vault]

Yeah so can you express that as n choose r?

[vernal violet]

I'm not sure. My area of knowledge is low when it comes to probability

All magic cards are equal and the other 22 cards don't matter

More magic cards would be better, but I'm trying to keep it simple, at least at first

[pastel vault]

Okay so that's 30 choose 5

Or $\frac{30!}{5! (30 - 5)!}$

Have you not learned about combinations (and permutations)

[wraith dagger]

south

[vernal violet]

It has come up vaguely but I haven't done much in the way of probability

[pastel vault]

Yeah then you probably don't have the knowledge to answer this question

[vernal violet]

That's why I hope someone can help me

[pastel vault]

https://www.youtube.com/watch?v=4tyzC-MkzOY

Watch this video then

[vernal violet]

Thanks. Checking it out. I should specify a few things. This isn't for a class. This is my own curiosity as applied to a video game trying to decide between two options the game gives me. If the idea is to lead me to the answer rather than tell me, I'm happy to do so

[pastel vault]

Ah okay cause it sounded like a textbook problem

So the way you do it is that you find the denominator first, the total number of ways

And then you find how many ways you can choose the magic cards, so 8 magic cards and you choose 5

[vernal violet]

Ha, I was going for that rather than trying to explain the video game. It's Battle Network 2, but I'll spare you the details unless you want to know and instead focus on the math

[pastel vault]

Wait the 30 doesn't matter for 1

This is confusing hold on

[vernal violet]

So that's the total number of hands of 5 possible? 30 choose 5?

[pastel vault]

Yes

[vernal violet]

Nice

Good start

[vernal violet]

All magic cards are equal here, so it's not order based

So let's see

[pastel vault]

Ah wait it's much easier to find the probability that you don't have any magic cards

[vernal violet]

So it's a combination not permutation

[pastel vault]

And then do 1 - that

[vernal violet]

Ah, yes! That's often how I go about these

Ah, so let's see. First draw is 25/30 then 24/29

So all that multipled together and subtract that from 1?

[pastel vault]

So it's (22 choose 5)/(30 choose 5)
22 of the cards are not magical, and 22 choose 5 is the number of ways you could choose 5 cards that are not magical

[vernal violet]

(meaning the first fractions like that)

[pastel vault]

I think this also works, so yeah 25/30 * 24/29 * 23/28 * 22/27 * 21/26

,w (22 choose 5)/(30 choose 5)

[vernal violet]

Gotcha!

[wraith dagger]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by sooouth.

[pastel vault]

,w 25/30 * 24/29 * 23/28 * 22/27 * 21/26

Shit

Oh wait

[vernal violet]

This is the way I generally go about doing probability questions, breaking it down step by step rather than trying to memorize these formulas. It's why I like other areas if math way more

[pastel vault]

,w 22/30 * 21/29 * 20/28 * 19/27 * 18/26

[wraith dagger]

Results provided by WolframAlpha

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

Requested by sooouth.

[pastel vault]

Yes they give the same answer, just started with 25 by accident

So you would need 1 minus that

1 - 209/1131

[vernal violet]

So it's like 81% of having a magic card with option 1?

Better than I expected

[pastel vault]

And then for 2 you would have:
,w 25/30 * 24/29 * 23/28 * 22/27 * 21/26 * 20/25 * 19/24

,w 25/30 * 24/29 * 23/28 * 22/27 * 21/26 * 20/25 * 19/24

[wraith dagger]

Results provided by WolframAlpha

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

Requested by sooouth.

[pastel vault]

1 minus that

[vernal violet]

Ah, a bit worse

[pastel vault]

So 1 - 0.236 or around 0.764

[vernal violet]

Aha

[pastel vault]

Yeah so 1 is better

But the probabilities are still fairly high

[vernal violet]

I see. But they're quite close. Now, what if we expand it to say "the more magic cards, the better"?

I'm not sure if we can really rigidly define that

Ah, wait. Expected value!

Which option gives the larger expected value of magic cards in the hand?

[pastel vault]

Well the expected value just depends on the probability and the number of trials

So if you keep the number of trials the same

The larger expected value is just whatever probability is larger

Since the probability is fixed each time

[vernal violet]

Right. So trials isn't relevant. But now how many magic cards are in the hand is relevant rather than just having at least one

[pastel vault]

This is when you would need to calculate it

[pastel vault]

Yes

[vernal violet]

Okay, cool. That makes sense

[pastel vault]

It depends on the number of cards in your hand, the number of magic cards in the deck
As well as the number of cards in the deck

[vernal violet]

Gotcha

[buoyant latch]

(expectation isn't the only thing you can ask of it! you can ask about variance, skewness etc)

[pastel vault]

For more information look up the hypergeometric distribution

[vernal violet]

Hyper what now? Gotcha, I'll see what that is

[buoyant latch]

it's a name for the set up you have

[vernal violet]

Ah, wow

[pastel vault]

When you're choosing objects without replacement

There's a general formula actually

[buoyant latch]

there are many very common set ups so we give it names

so it's easier to talk about

[vernal violet]

Ah, yes!

[vernal violet]

For sure. Very mathy

That sounds sarcastic but I mean it

I know about other non probability areas of math so I get the idea of having set ups

[buoyant latch]

someone asked me recently about how to find the probabiliy/expected number of trials needed to get a full collection or near full collection from a random population

it's very hard to try do this but if you've seen a problem like this it's just a variant of what's called the coupon collector's problem

and a lot is already known about this problem

so the name is very good for being able to refer to the exact/similar set up and so you dont have to go through the maths if someone else already has, and you just want an answer

[vernal violet]

Ah, right. I still want to know the basic idea of how it's found, but I find myself getting lost for probability in particular unless I can see every small step like how I was working on the initial problem

[vernal violet]

Also, I love that name

I think I get the basic idea of hypergeometric distributions. It seems that it's for finding an exact amount, like 2 magic cards for my example. How would I convert this to expected value?

I see that adding up x=1 , x=2, ..., x=5 would give me the probability for option 1 in the initial question

[buoyant latch]

there's just a formula for the expected value of a hyper geometric distribution

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

look on the right side

there's one that says "mean"

[vernal violet]

Ah, wow. It somehow ends up being n * K/N? Seems a lot simpler than I would expect it would simplify to

So ours would be 5x8/30 for our option 1, or about 1.33 magic cards per hand?

Option 2 would be 7x5/30,which is 7/6 or about 1.16 and thus worse for this as well?

[cedar kiln]

@vernal violet Has your question been resolved?

[vernal violet]

Almost

[cedar kiln]

@vernal violet Has your question been resolved?

[cedar kiln]

@vernal violet Has your question been resolved?

[cedar kiln]

@vernal violet Has your question been resolved?

[covert cipher]

is it solved?

[cedar kiln]

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[bold pond]

im pretty sure $3x^2 dx = x^3 + C$ but idk what to do after that

[livid hound]

missing the integral operator,
but the antiderviative is x^3

evaluat your bounds

[bold pond]

i thought i had to substitute 1 - 6 into x and add all of them

but thats the wrong answer

[wraith dagger]

wakamole

[livid hound]

$$\int_1^6 3x^2 \dd{x} = x^3\eval_1^6$$
$$[F(x)]\eval_a^b = F(b) - F(a)$$

[bold pond]

oh fuq

ok and how do you know when to put the C

[wraith dagger]

ℝαμΩℕωⅤ

[livid hound]

you'd put it for indefininite integrals

[bold pond]

whats that

[livid hound]

integrals without bounds or infinite bounds

technically there's a C when doing indefinite integrals
there's just no issue whether you include it since it ends up getting cancelled when subtracting

[bold pond]

ok

makes sense

what if it's like this?

[livid hound]

what concerns do you have with this

[bold pond]

would you make it $x^{9/5}$?

[wraith dagger]

wakamole

[livid hound]

not quite

[slate lintel]

(yes, but you need a coefficient)

[bold pond]

so $5/9 x^{9/5} + C$

[wraith dagger]

wakamole

[slate lintel]

yeah (btw use $\frac{5}{9}$

[wraith dagger]

هايلي

[bold pond]

ahh ok

i will next time

[slate lintel]

now do the bounds

[bold pond]

the bounds were just 1 and 0

so its 5/9

i dont get this one

whaats the difference between antiderivative and an integral?

.close

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[wintry roost]

It marked my answer wrong, what am i doing wrong?

[wintry roost]

i also tried y/2 for the y lower bound

[slate lintel]

well your bounds right now describe a rectangle

and that region is very much not rectangular

[wintry roost]

mhm

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[fringe totem]

The help

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[lean cave]

can someone help me with this question?

[lean cave]

not sure where to start

[cedar kiln]

@lean cave Has your question been resolved?

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@lean cave Has your question been resolved?

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@lean cave Has your question been resolved?

[ebon juniper]

well since cos(nx) is well defined on all R, u can tell that it is integrable on R right?

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[cold sundial]

why is this not the answer

[humble karma]

You're looking for c

They give you the area between -c and c

[cold sundial]

i thought the 0.9 was the c

and im supposed to find the z

because this was the other problem on this one

[upper abyss]

The differences between these questions are subtle, but important

The "other" question is asking you to highlight the area (probability) where z < -1.7

The current question is saying "there's an area (probability) of size 0.9. It's between which two values?"

[cold sundial]

ohhhhh ok

hmm

yeah im still a bit confused on how exactly to solve

im not great at stats

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[cold fulcrum]

can u help me with this applications of integration

[cold fulcrum]

idk what to do help

[void glen]

what have you tried

[void glen]

looks right

[cedar kiln]

@cold fulcrum Has your question been resolved?

[cold fulcrum]

how abt this, the second one

[void glen]

so we don’t really need the absolute value here

we just need to do the right function minus the left function

the bounds are correct btw

[cold fulcrum]

alr, thanks a lot

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[balmy yoke]

is sinxcosx/1 + cos^2x = cotx ?

[balmy yoke]

this is the original question

I don't know if my simplification so far is correct

[muted bear]

redo your somplification for the denom

[balmy yoke]

= 2sinxcosx / 1 - 2cos^2x - 1
= 2sinxcosx / 2 + 2cos^2x
= 2sinxcosx / 2(1 + cos^2x)
= sinxcosx / 1 + cos^2x

wait that should be negative

2 - 2cos^2x

[muted bear]

ok

do you know any identities of
1-cos^2

[balmy yoke]

yup sinx^2, and then the equation will simplify to cosx/sinx which is cotx

the sign was my mistake

[muted bear]

nice

yeah, trig identities are funky

[balmy yoke]

true

thanks 👍

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[dire tundra]

Are there any inf series that this rule wouldn't apply for?

I'd say it wouldn't work for an alternating series, no?

[main needle]

it's just a definition, anything works here

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[cosmic quest]

.close

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[hollow dragon]

Hey does anyone know if wolframalpha has an expansion system for quaternions like it does for complex numbers?

Like for complex, I can just run complexExpand() to convert a complex formula into algebra

[native dirge]

I am confused on this question, its not for school obviously but it seems so simple:

red is a cosine wave of the blue circle, the purple is tangent to y=-1, the red cosine wave, and the blue circle, what is the radius and center of the purple circle?

[cedar kiln]

@hollow dragon Has your question been resolved?

[hollow dragon]

This channel is already occupied >_>

use an available one

[cedar kiln]

@hollow dragon Has your question been resolved?

[hollow dragon]

.close cause no ones bothering to give me anything

[cedar kiln]

Channel closed

Closed by @hollow dragon

Use .reopen if this was a mistake.

[cedar kiln]

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[lucid locust]

Finding the derivative of 3(cscx)^(3/2). Im confident with my chain rule differentiations and Im confused where I went wrong comparing what I got to my solutions manual

[lucid locust]

Like I do not understand the transition from the second to last step to the last step in the answer key

[amber oriole]

do u get chain rule?

[lucid locust]

yeah pretty good i think

i dont get what’s going on here tho

like

y’ = n(u)^n-1 right

so 3/2 - 1 = 1/2

which makes sense

but why does it go back to ^3/2???

[amber oriole]

cuz csc is to the power of 1

(csc(x))^1/2*(csc(x))^1

so (csc(x))^(1/2+1)

so its 3/2

[lucid locust]

oh wait lollll

the 2 csc’s got put together

mb bruh i’m just slow

.close

[cedar kiln]

Channel closed

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[cedar kiln]

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[wraith trellis]

i need help on this problem

i need help setting the equation up

[weak otter]

Draw radii to the tangent points

See if you can get it from there

[wraith trellis]

i would be 90 degrees right

ya but how do u set it up from there

[normal shore]

whats x referring to

[wraith trellis]

the smaller arc

the equation for the angle outside of the circle is

1/2 (big arc - small)

right

[normal shore]

idgi

x is an angle

30 degrees* not 25

[wraith trellis]

how do u solve for x from that

[normal shore]

is x referring to the yellow angle.?

[wraith trellis]

no

its refering to the arc degree

on the answer key it says x = 120

no work tho

[normal shore]

doesnt make sense

[normal shore]

this would give x =120

[wraith trellis]

o ok

how do u solve for that

[wraith trellis]

tho

[normal shore]

this is wrong

[wraith trellis]

ok

[normal shore]

triangle total is 180 right

u have 30 and 90

[wraith trellis]

yup

[normal shore]

and x/2

[wraith trellis]

so 60 x 2

ok

[normal shore]

sum all and equate 180

[wraith trellis]

ohhh

ty bro

have a good night

[normal shore]

morning here but gn anyways

[wraith trellis]

lol

good morning

then

.close

[cedar kiln]

Channel closed

Closed by @wraith trellis

Use .reopen if this was a mistake.

[weak otter]

Can jusr use the fact that quadrilateral adds up to 360

Same thing saves a step

[wraith trellis]

oh

thanks

.close

[somber wren]

sorry

[wraith trellis]

ur good idk why its not closing

.close

[cedar kiln]

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[vocal oak]

how to prove that x²+4x+7 has no zero?

[vocal oak]

Like the process to start without the quadratic equation

without using discriminant part

[dire thorn]

Can you find the minimum?

[cedar kiln]

@vocal oak Has your question been resolved?

[vocal oak]

?

minimum of?

[dire thorn]

The quadratic

If the minimum of the quadratic is positive then the entire quadratic must be positive

[vocal oak]

i couldnt understand

[dire thorn]

What does it mean for a quadratic to have a zero graphically

[vocal oak]

it should cut the x axis

[dire thorn]

Ok so what happens if the lowest point on the graph is above the x axis?

Will it ever cross the x axis?

[vocal oak]

no

but the real problem is

i need to prove algebraically

i know the graphical method

[dire thorn]

Yes you can algebraically find the minimum and show its positive

And by definition if the minimum is positive every other point is positive so no point is zero, aka no zeros

[cedar kiln]

@vocal oak Has your question been resolved?

Channel closed

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Use .reopen if this was a mistake.

[cedar kiln]

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Remember:
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[rancid wasp]

Need help finding X

[rancid wasp]

this is 10th grade geometry

The unit is

Circle- interior and exterior angles

[livid hound]

alt seg theorem

alternate segment theorem

[rancid wasp]

Idk what that is

[livid hound]

look it up

its a circle theorem