help-23

[clever storm]

Going back to this, how did you incorperate x back into the equation?
Did you just multiply it in soo x^ becomes x^n+1?

That makes sense

[drowsy karma]

yes, we always do that, we move "x" to inside of the series

if you read problem books with series

you always notice answers

are writtin in such way

bu tyou also may transform it to yoru needs

[clever storm]

Okay, and 9/15?

[drowsy karma]

yes

becasue

constant alaso belongs

to the general term

of the series

but remeber

if you have x^n, then all next to it, we call c_n

but if you have

x^(n+1)

then all you call as: c_(n+1)

so the name

depends on the epxoent

of the x

but there is another trick , also sued in such sum notations:

look

[clever storm]

ok

[drowsy karma]

$\sum_{n=0}^{\infty }\frac{x^{n+1}}{n+1}=\sum_{n=1}^{\infty }\frac{x^{n}}{n}$

[flat frigate]

Joanna Angel

[drowsy karma]

pzl consider it

and thikn if you agreed

i want you to see, that both sides shwos the same

onyl in two diffeernt ways

[clever storm]

Yes, that makes sense

[drowsy karma]

this way , allowws you for manipualtions

with indexes

and n_terms inside the series

=

when you subtract one from n nside the terms

you add one to index

and opppsite

memorize it

[clever storm]

Yes, that makes sense, let me write this down

[drowsy karma]

k

[clever storm]

got it

[drowsy karma]

look:

$\sum_{k=1}^{n}c_{n}=\sum_{k=1+m}^{n+m}c_{n-m}$

[flat frigate]

Joanna Angel

[drowsy karma]

that is fo rfinite sums

but infite sums alslo work wiith it

in discrete mathemtics, you do oftensuch manipulations

[clever storm]

Okay, I'll keep this in minid for my final exam for calc 2, I think this could really help out there

Since there are many series like this

[drowsy karma]

yes )

lotsa

[clever storm]

Alright, thanks a bunch, I'll get back to studying but I appreciate it

[drowsy karma]

kk Good Luck! 🙂

[clever storm]

Thanks again! 😄

.close

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[worthy frost]

NO idea what to make of it

[wind shoal]

Still Need help

?

Have u seen Hôpital rule

[worthy frost]

sure do

[wind shoal]

Ok so basically when u plug in 2 u get 0/0

[worthy frost]

again forgot to specify

[wind shoal]

Now derive the top and the bottom

[worthy frost]

i can't use it in this exercise, my proff asked me

[wind shoal]

Ohh

[worthy frost]

my bad

[wind shoal]

Hmm ok

Gimme a sec

[safe radish]

@worthy frost Has your question been resolved?

[worthy frost]

<@&286206848099549185>

[safe radish]

@worthy frost Has your question been resolved?

[worthy frost]

.close

[safe radish]

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[worthy frost]

.reopen

[safe radish]

✅

[upper rivet]

$\sin{\pi x} = \sin{\pi x - 2\pi} = \sin{\pi (x-2)}$

[flat frigate]

nebula40

[fleet condor]

does this help though

[sage vault]

it might

[fleet condor]

the term is still stuck in the sin

[upper rivet]

you use a^3-b^3 in the numerator and sub u = x-2

[worthy frost]

Still dont understand

The answer is 12/pi

If that helps

[sage vault]

do you know how to factorise x^3 - 8?

[upper rivet]

the basic concept is that we want to bring it of the form sin(x)/x because we know that approaches 1 as x -> 0

[worthy frost]

You mean (x-2)(x^2 + 2x + 4?

[upper rivet]

or x/sin(x) in this case

[upper rivet]

yes

[worthy frost]

but x—> 2, no?

[upper rivet]

thats why you do what I said

[upper rivet]

.

[sage vault]

I think the brackets in that chain of equalities is confusing

there's no brackets in the sines

[upper rivet]

yeah idk why tex doesnt do it automatically lol

but everything is the argument of the sin

[sage vault]

then that means you should do it manually I guess

[worthy frost]

So ill have x->2 (x^2 + 2x + 4)/sinPi

[upper rivet]

$\lim{x \to 2} \frac{(x-2)(x^2 + 2x + 4)}{\sin{(\pi (x-2))}}$

[sage vault]

nebula I think you should write those chain of equations again

with the brackets inserted manually

also it's \lim_{x\to 2} (you're missing the underscore)

[upper rivet]

$\sin{(\pi x)} = \sin{(\pi x - 2\pi)} = \sin{(\pi (x-2))}$

[flat frigate]

nebula40

[upper rivet]

thanks

[sage vault]

nw

[meager plover]

Why sub if 2 isnt a root of x^2 + 2x + 4

[upper rivet]

that part doesnt matter

$\lim_{x \to 2} \frac{(x-2)(x^2 + 2x + 4)}{\sin{(\pi (x-2))}}$

[flat frigate]

nebula40

[upper rivet]

the left part is where we bring to x/sin(x) form

[meager plover]

Oh you want to get rid of - 2

Whatvever

[upper rivet]

just substitute 12 for x^2 + 2x + 4

so now we have $\lim_{x \to 2} \frac{12(x-2)}{\sin{(\pi (x-2))}}$

$\lim_{x \to 2} \frac{12(x-2)}{\sin{(\pi (x-2))}}$

[flat frigate]

nebula40

[upper rivet]

this should be clear enough

[worthy frost]

Could you do one more step? Maybe im tired but i dont see what i should do next

[upper rivet]

alright so we substitute u = x - 2

like bishop said it's not really necessary but it helps makes things clearer

$\lim_{u \to 0} \frac{12(u)}{\sin{(\pi (u)}} = \lim_{u \to 0} \frac{12(\pi u)}{\pi \sin{(\pi u)}}$

[flat frigate]

nebula40

[upper rivet]

forgot the multiply denominator

[sage vault]

yeah I see where this is going now. This is pretty good

[sage vault]

John, do you remember that infamous sin(x)/x limit that equals 1?

[worthy frost]

sure

[sage vault]

do you think you can use it here?

meanwhile, I think I learned something new here thanks to @upper rivet

[worthy frost]

How did we get to x-> 0 from x-> 2?

[sage vault]

basically we let u = x - 2

as x -> 2

[meager plover]

Sub

[sage vault]

you can see that u -> 0

[worthy frost]

Oh i see

I dont know what do with sin being in the denominator, if not that i would know how to solve it

I think i got it

[meager plover]

L hopital

[worthy frost]

like that?

[meager plover]

What are you doing

[worthy frost]

oh

[meager plover]

limit of x/sin(x) is 1 because of l hopital

1/cos(0)

= 1/1

we assume it exists

Because if it didnt then your original limit doesnt either

[worthy frost]

So that would be easier

but its still correct, no?

[meager plover]

I dont even know what are you doing

It makes no sense

[worthy frost]

I didnt know x/sinx also -> 1, so instead I created it in the denominator dividing and then multiplying it by (piu) and then reduced piu in the nominator

[sage vault]

@meager plover He's not allowed to use l'hopitals. Scroll up to see the bit where he says it

[sage vault]

very well done mate

[worthy frost]

thank you!

[meager plover]

Whats the proof of sin(x)/x one

[sage vault]

long story

[meager plover]

well technically you can probably prove x/sin(x) without l hopital either but if you insist you have a different proof then...

[sage vault]

https://www.khanacademy.org/math/ap-calculus-ab/ab-limits-new/ab-1-8/v/sinx-over-x-as-x-approaches-0

[worthy frost]

alright, i will close it now, again thank you very much for your help, i learned something new

.close

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[sour karma]

can someone explain why these are equal?

[meager plover]

Let $a^x$ be some exponential function , so all the conditions like $a > 0$ etc , then its obvious that $e^ln(x) = x$ and that $ln(x^k) = kln(x)$ thus $e^{ln(a^{x})} = e^{ln(a) \cdot x}$

[flat frigate]

Bishop

[meager plover]

in general

$\dv{}{x} e^{f(x)} = \dv{f}{x} e^{f(x)}$

[flat frigate]

Bishop

[meager plover]

the derivative of e^f(x) is f'(x) e^f(x)

[meager plover]

The derivative of ln(a)x is ln(a) , because it's a simple linear function

[sour karma]

i think im kind of getting it, so the derivative of an exponent is the whole thing multiplied by the derivative of its exponent?

[meager plover]

Yes

of an e^f(x) exponent

[sour karma]

i see thanks, could you explain this a little more? i'm not too sure on why the x is being removed

wait actually i think i kind of get it now

[meager plover]

f(x) = ln(a)x

Derivative of f(x) is ln(a)

With respect to x

i dont think i need to prove that

[sour karma]

oh okay, thanks that makes sense now

thanks for the help and sorry if it was a little hard to get me to understand

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[safe hawk]

would anyone by chance know how to solve this?

[normal moss]

This requires knowledge of the properties of logarithms

[normal moss]

Do you know them?

[safe hawk]

I actually do not, no

[normal moss]

I would suggest learning them, as they are crucial when solving these equations

Like how to change bases, what's the logarithm of a product, etc.

Here

[safe hawk]

I missed the crucial days due to me being sick, would you be able to walk me through that one problem?

[normal moss]

I unfortunately can't because I don't have time right now as I have to leave in a few minutes. I'm sure someone else will hop in soon. Sorry about that 😦

[safe hawk]

no worries! I'll just wait hopefully someone else can help me

[normal moss]

Okay. Best of luck. Have a good day

[safe hawk]

you as well, thank you again

would anyone by chance know how to solve this problem?

[craggy falcon]

i’m trying to figure it out rn i’ll walk you through it if i figure it out

[safe hawk]

thank you!

[craggy falcon]

i’ve looked through the properties of logarithms image enterprise provided and i don’t know if the answer lies there

but idk

wait the product rule for the right side of the equation

log_3(x) = log_9(6) + log_9(x)

idk if that helps i’m just throwing ideas around

with my understanding of the logarithm properties that’s the only thing i can think of

idk where to go from there

hope i helped

[safe hawk]

thank you so much for trying to help me out, i might have some sort of idea

[craggy falcon]

alright

[safe hawk]

I’ll ask my teacher tomorrow as I’m returning to school thank you again

[craggy falcon]

yeah no problem

[tranquil drum]

raise 9 to both sides of the equality so 9^(log_3(x)) = 9^(log_9(6x)) which can be rewritten as (3^(log_3(x))^2 = 9^(log_9(6x)) which from there you can cancel the logs

and check for extraneous solutions

[willow gust]

the 9 in the RHS, i.e the base of the log

convertt it to a form of 3

like 9 = 3^2
log_(3^2) (6x) = 1/2 log_3 (6x) = log_3 (6x)^(1/2)

[safe radish]

@safe hawk Has your question been resolved?

[normal moss]

Change bases on the right, it changes to base 3 very nicely; use the product rule; move all terms with x to one side; collect like terms. Then you've got your answer on a silver platter

[safe radish]

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[sour karma]

Is there a better way of finding the second derivative when x = 8?

[sour karma]

I messed up somewhere but this seems really long

[hasty wagon]

yes there is

consider cubic root of x as
x^(1/3)

and also get the constant out

$\frac12\left(x^{\frac53}+x^{-\frac13}\right)$

[flat frigate]

Biscuity

[hasty wagon]

now you can differentiate 2 times

[sour karma]

How did you get 5/3?

[hasty wagon]

2-1/3

[sour karma]

I see, thanks I’ll try that

Did I do something wrong here

<@&286206848099549185> am I doing this right

[dapper venture]

you don't need the product rule but ok

looks fine

[sour karma]

How do I do it without the product rule?

My answer is off from the answer key though

[dapper venture]

treat 1/2 as a constant instead of a function

[hexed panther]

as you noticed, the derivative of a constant is 0

so you can just skip writing the 0x part

[sour karma]

Oh yeah I just like writing it out sometimes so I know what I did

[sour karma]

How do I do that?

[hexed panther]

if c is a constant, then the derivative of cf is c times the derivative of f

basically what you did, again, just shortcutting

[sour karma]

Oh okay, I’m just a little lost now because I’m not getting the right answer

[hexed panther]

bc 5/2-1 is 3/2

there is at least this mistake

[dapper venture]

what's the answer in answer key?

[hexed panther]

and 1/3-1 is -2/3

you made mistakes when applying the power rule

[dapper venture]

???

[sour karma]

161/576

[dapper venture]

how did you not get the right answer

you do get 161/576 when plugging in x=8

[sour karma]

Oh okay thanks, I might’ve plugged it in wrong because I’m using my calculator on my phone rn

.close

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[quartz wasp]

Need help with this

[supple plaza]

X=-5, x=-3

Oh wait it says that below

[quartz wasp]

ignore the below part

how do i do the equatio

[supple plaza]

No I just found the solutions cause its kinda obvious

[quartz wasp]

yea but i need to do it

[supple plaza]

Thinking on it

Well there is a hard way to do it

You can rewrite (x+5)^4 as (x+5)^2 * (x+5)^2 then use the formula

Doing the same for x+3

Not sure if it will get you far tho

[quartz wasp]

i did that alr but the what

[supple plaza]

Another idea could be taking (x+5)^4 to the other side of =

[quartz wasp]

its still a ^4

[supple plaza]

So you get 16 - (x+5)^4

Which is the same if we write 4^2-((x+5)^2)^2

And then we use a^2-b^2 formula

This should get you somewhere

Try it, you can also bring (x+3)^4 after some simplifications

[quartz wasp]

can i just substitute it?

[supple plaza]

Whats that

[quartz wasp]

like let something be a

[supple plaza]

Thats what im suggesting

Without the use of a tho, you can do it if it helps you

Let (x+5)^2 be a then (x+5)^4 is a^2

Take a to the other side and you get 16-a^2

16 is 4^2 so we get 4^2-a^2 which is (4-a)(4+a)

Insert value of a

That is (x+5)^2

[quartz wasp]

Can i just rewrite as this?

[supple plaza]

No

You cant take square root of sum

[quartz wasp]

Ill lose answers right

[supple plaza]

Root of a+b is not root of a + root of b

[supple plaza]

You wont even get right answers

[supple plaza]

Do this its getting us somewhere

[quartz wasp]

Ok wait

[supple plaza]

Then let b=(x+5) then (x+5)^2 is b^2

We have 2^2-b^2 which is (2-b)(2+b)

And so on

[quartz wasp]

Like this?

[supple plaza]

No

Check again what I did

You take (x+5)^4 to the other side of the equation

[quartz wasp]

can you write it on paper?

[jaunty owl]

do u know binomial theorem

@quartz wasp

[quartz wasp]

no

[jaunty owl]

It's ok

what u can do is try to let u = x+4
then u get

$(u+1)^4 + (u-1)^4 = 16$

[flat frigate]

WOWWA

[jaunty owl]

u would still have to expand but it's easier

and alot of stuff will cancel out

like alot

[quartz wasp]

ok ill try and let you know

do i just square it twice?

[jaunty owl]

yeah so like
$(u+1)(u+1)(u+1)(u+1)$

[flat frigate]

WOWWA

[jaunty owl]

$(u+1)^2(u+1)^2$

$(u+2u+1)(u+2u+1)$

[flat frigate]

WOWWA

[jaunty owl]

and then do the distribution

from that

or if you know $(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab +bc + ac)$ then u can apply that to
$(u+2u+1)^2$

[flat frigate]

WOWWA

[quartz wasp]

ok now i do the tecond one right?

[jaunty owl]

not sure what u mean but u just multiply it out

or if you mean (u-1)^4

then yeah same thing

but u need to fully expand it

once u fully expand both terms and then do the canceling let me know what you get in terms of u

[quartz wasp]

Like write it like that

(U+2u+1)(u+2u+1)

Do i just multyple to ech other or what next

[jaunty owl]

yes

[quartz wasp]

What now?

[jaunty owl]

you forgot the plus sign

[jaunty owl]

should be a plus sign there

[quartz wasp]

yea

forgot it

[jaunty owl]

also

need to multiply it out

both sides of.the plus sign

so far it's good

but need to multiply the trinomials together

[quartz wasp]

ok wait

This good right

[jaunty owl]

hmm

don't think so

after you fully expand it and simplify you should get
$2u^4 + 12u^2 + 2 = 16$

[flat frigate]

WOWWA

[quartz wasp]

what how

[jaunty owl]

you probably messed up some arithmetic

,w (u+1)^4 + (u-1)^4

not what I wanted lol

,w expand (u+1)^4 + (u-1)^4

[flat frigate]

Results provided by WolframAlpha

Click here to refine your query online
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Requested by wowwa

[quartz wasp]

oh yea wait

did it

[jaunty owl]

great

[quartz wasp]

now what

[jaunty owl]

so you should have

$2u^4 + 12u^2 + 2 = 16$

[flat frigate]

WOWWA

[quartz wasp]

yes

[jaunty owl]

this is a quadratic in terms of u^2

so factor the quadratic

you can divide out a two

[quartz wasp]

i can substitute again right

[jaunty owl]

should get
$u^4 + 6u^2 - 7 = 0$ I think

yeah u can

[quartz wasp]

and solve this

i will wait

[flat frigate]

WOWWA

[quartz wasp]

i substituted it

let u ^2 be a

and got

a=1

a=-7

[jaunty owl]

great

so now sub back in

and then sub back in again

[quartz wasp]

so u would be root 1

[jaunty owl]

so you can get it in terms of x

[quartz wasp]

?

i need u first

[jaunty owl]

yeh

u is root 1

[quartz wasp]

and what about -7

[jaunty owl]

complex solutions

[quartz wasp]

isint root 1 just one

[jaunty owl]

yes

root 1 = 1

[quartz wasp]

ok so u is 1

[jaunty owl]

or u = root(-7)

what is sqrt -7

[quartz wasp]

idk

nothing

[jaunty owl]

do u know sqrt(-1)

[quartz wasp]

1?

[jaunty owl]

1*1 = 1 so nah

it's called $i$

[flat frigate]

WOWWA

[jaunty owl]

have u heard of imaginary numbers or nah

[quartz wasp]

i dont need tht we havent lernd it yet

[jaunty owl]

oh ok

[quartz wasp]

i just need the normal 2 answers

[jaunty owl]

so if ur only looking for real solutions

then yes

u = 1

u only need to consider that case

[quartz wasp]

so

x+4=1

[jaunty owl]

oh wait no

u^2 = 1

so what happens if u solve for u

[quartz wasp]

yes u eqauels root1

wich is 1

[jaunty owl]

it's $\pm 1$

[flat frigate]

WOWWA

[jaunty owl]

if u take the square root of both sides u need the plus or minus

so u = 1, or u = -1

[quartz wasp]

oh yea

i get it

answers are -3 and -5

[jaunty owl]

yeh

[quartz wasp]

Can you also help me do this?

[jaunty owl]

factor the quadratic

s

well

what is the question

pol

lol

[quartz wasp]

Like how do k do it

[jaunty owl]

do what

find the zeroes?

[quartz wasp]

answers like the graph

This are the answers

I need to find them

[jaunty owl]

oh the domain

the main thing for the domain is that none of the denominators can = 0, and the inside of square roots cannot be negative

[quartz wasp]

so do i write it like this?

[jaunty owl]

nah the inside of the root

has to be greater than or equal to 0

[quartz wasp]

This good?

[jaunty owl]

yeah but the fourth root part isn't needed

[quartz wasp]

what dose that even do?

[jaunty owl]

wdym

it takes the fourth root of a number

[quartz wasp]

ok what do i do next

[jaunty owl]

solve the inequalities

[quartz wasp]

like the roots?

[jaunty owl]

have you ever solved a quadratic inequality

[quartz wasp]

nope

what do i do

[jaunty owl]

https://youtube.com/watch?v=OP91XWBRI1w

I would reccomend this video

watch the video you will understand much better than if I explain through text

he is good teacher ong

[quartz wasp]

ik this

i can solve it

[jaunty owl]

ok then solve the inequality

[quartz wasp]

2 of them

like the ones that are in the roots

?

[jaunty owl]

yes

[quartz wasp]

4root throws me off tho

[jaunty owl]

it doesn't matter

if u raise both sides to the 4

what happens

0^4 = 0 and the 4th root vanishes on.the left hand side

also u need to understand that we specifically don't want to inside of the root to be negative

[quartz wasp]

so it dose nothing

[jaunty owl]

so u shouldn't have written the inequality with the fourth root in the first place

bc the point is we don't want the inside of the root to be negative

bc taking the nth root of a number which is negative when n is even is undefined

over the reals

[quartz wasp]

Do i slove this first?

[jaunty owl]

it's the fourth root and also again there's no point writing the inequality with the root in the first place

bc it's the inside of the root that should be greater than or equal to 0

[quartz wasp]

Ok withouth te root

Do i solve that?

[jaunty owl]

yes

[quartz wasp]

Did it

-2 and 8

Next what do i solve

[jaunty owl]

well u need to find the interval

what values of X satisfy the inequality

it's not only -2 and 8

[quartz wasp]

Next wich one

[jaunty owl]

well u still didn't answer the question

[jaunty owl]

which values of X satisfy this inequality

[jaunty owl]

yes

[jaunty owl]

now u need to solve the denominator not equal to 0

[quartz wasp]

do i solve that normally

[jaunty owl]

not sure what normally means to you

but maybe

sure

[quartz wasp]

like get 2 answers

[jaunty owl]

ye

[quartz wasp]

-1 and 1

[jaunty owl]

ok

so now u know x cannot equal those values

oh but actually it's more intricate than that

[jaunty owl]

if both the numerator and denominatouplr are negative, then the inside becomes positive

so u need to figure out the interval where they are both negative

and the interval where they are both positive

[quartz wasp]

how do i do that?

[jaunty owl]

find the interval where the numerator is positive, find the interval where the denominator is positive and take the intersection

find the interval where the numerator is negative and where the denominator is negative, and take the intersection

making sure not to include the values where the denominator is zero

[quartz wasp]

i gtg now will you be on in like 3 hovers so we can continiu?

[jaunty owl]

nah I'll be in school lol sorry

there's lots of ppl here

someone can help u fs

[quartz wasp]

ok wait

i got one more thing

so in a club there are 320 seats they make lines in each line there are same amount of seats the club owner added 4 seats to each line and he also added 1 more line after that there where 420 seats in total

i need to find how many lines there are now

i think i need a equation

[safe radish]

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[ashen parcel]

is thsi right?

[safe radish]

Please don't occupy multiple help channels.

[willow storm]

yep

[dapper venture]

yes

[willow storm]

the timing xd

same second

[ashen parcel]

lmfooo

thx gabf

gavg

gang

[willow storm]

nop gang

[ashen parcel]

wait r these also right

[safe radish]

@ashen parcel Has your question been resolved?

[safe radish]

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[thin bridge]

(sqrt(x))^2 = x
by definition

[wild copper]

Is this supposed to a complex number problem?

[thin bridge]

i suppose you'd need to be working with complex numbers here though

$\sqrt{x^2}$ is different from $(\sqrt{x})^2$

[flat frigate]

ℝαμΩℕωⅤ

[safe radish]

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[umbral swan]

can the cauchy-product converge if the two series don't converge absolutely?

[obsidian oracle]

Of course it can

but it won't always

[umbral swan]

kk

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[neon citrus]

@safe radish I'm in serious need of help, i'm in middle school and I still don't understand fractions, and it's embarassing,

[frozen veldt]

I recommend looking up fractions on Khan Academy

[neon citrus]

No,

I'm not a fan of using websites or online tutors that my parents paid for

[frozen veldt]

Khan Academy is free

[neon citrus]

Nah,

I appreciate you trying to help me but I don't want that kind of help.

[peak wraith]

or oganic chemistry its also free ...

[peak wraith]

online helped me a lot with math

[frozen veldt]

well is there something specific about fractions that you're struggling with?

[neon citrus]

the whole thing

[frozen veldt]

how deeply do you wish to understand fractions

[neon citrus]

like deep into my heart

[peak wraith]

can u provide a example of a qn

[frozen veldt]

The way I like to think about fractions is that the defining property of a/b is that when you multiply it by b you get back a

So for example 5/3 is such a number that when you multiply it by 3 you get 5, in other words three of that number equals 5

[neon citrus]

ok

[frozen veldt]

From this point of view one can understand the multiplication of fractions as not a definition but as a consequence of the properties that we want fractions to satisfy

So if we consider
3/4 * 5/6 then we can think about what properties that number has

If you multiply it by four and then six then you get

(3/4) * (5/6) * 4 * 6
= ((3/4) * 4) * ((5/6) * 6)
and now by definition, the first part becomes 3 and the second part becomes 5

Therefore, even though we don't know what (3/4) * (5/6) is, we know that it has the property that if you multiply it by (4*6) then you get 3*5. But hmm, this is by definition (3*5)/(4*6)

By the same token we have that multiplying any two fractions together amounts to multiplying the numerator and the denominator together because the result satisfies the same defining property as the initial expression by definition

[neon citrus]

This isn't what I was studying last year...

it's 5th grade fractions not whatever that is,

[frozen veldt]

You asked to understand fractions deep in your bones, this is what you got

[neon citrus]

Ok fair enough.

[frozen veldt]

if you're not satisfied then feel free to consult Khan Academy

[neon citrus]

Just a bit of it made sense

but i'll think about it

Anyways thanks.

[frozen veldt]

no problem

As a fun sidenote, the rabbit hole for what fractions actually are actually goes a lot deeper than that. In fact I've basically just assumed that a certain number with certain properties exists with no justification to my claim. But this turns out to be a reasonable assumption to make, and the technical details don't add any real additional understanding

The real usefulness of fractions is not in what they are, in fact in middle school they're often taught as representations of sliced pizzas or pies, but rather what makes fractions useful is how they play nice with multiplication

This turns out to be a common pattern in math. Often times the true nature of a mathematical object is hidden away behind a million layers of abstraction, kind of like how a computer is made up of a million highly technical smaller parts, but then ultimately what mathematicians, or computer users, care about is what you can do with your objects of study and how they behave

[snow haven]

hey, just think it as a part of something

or

2 person shared 3 pizzas

[safe radish]

@neon citrus Has your question been resolved?

[safe radish]

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[sour karma]

Does doing this make sense if I’m trying to find the derivative?

[sour karma]

And if it is what do I do next

Would I just do the product rule

[drowsy moss]

What is the variable here? x or k?

[sour karma]

We’re trying to find k

[drowsy moss]

in what context? you should post the whole original question

[sour karma]

Oh there isn’t really context, my friend just made me a random question

Maybe it isn’t really a great question

Well thanks I think I’ll just try a diff problem

.close

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[silent bloom]

hiya, so im writing a paper about some statistical stuff and im attempting to use the empirical rule to justify why it's a normal distribution. as you can see from my findings, it does not meet the first part of the rule. im wondering how ""lenient"" the empirical rule is, whether im still able to justify it being a normal distribution, and if not what my next-best alternative is... dont assume i know a lot of stats, im having to teach myself as i go for this project as i havent actually learned any stats past linear regression. thanks in advance

[stoic dune]

Not usually too ""lenient""

What's the data from?

[silent bloom]

deviation from the mean climotogical climate betweeen 1961 and 1990 (in kelvin)

here's all the graphs ive made using the data provided, if any are of use

i'm using exclusively the natural forcings

don't know if i should use the simulated forcings, i should look more into that most likely

did i just rubberduck myself

should i just try applying the empirical rule to the simulated forcings

[safe radish]

@silent bloom Has your question been resolved?

[safe radish]

@silent bloom Has your question been resolved?

[silent bloom]

okay back to my question... i replotted and redetermined the data, it's very close to the boundaries... what do i do?

[silent bloom]

.close

[safe radish]

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[wooden umbra]

would the derivative of f(x)=-3000,97^(x)+300 if x = 8 be
f'(8)=-300*0,97^(8)*****1ln(0,97^(8))?

[wooden umbra]

do you also remove the -300?

[candid oar]

yes

the derivative of a constant is always zero

[wooden umbra]

but do you multiply by 0 in that case?

or just +- the 0

[thin bridge]

do you have a pic of the original

$f(x) = -300 \cdot 0.97^x + 300$

[flat frigate]

ℝαμΩℕωⅤ

[thin bridge]

is that an accurate representation of your function?

[wooden umbra]

yes it is

[wooden umbra]

nope

[thin bridge]

if so, that -300 doesn't get removed/erased

[wooden umbra]

because it's inn front of the thing we want to find the derivative of

right?

cuz f'(x)=-300******[0,97^(x)******x'****ln(0,97^(x))]

[thin bridge]

yeh, also you didn't differentiate 0.97^x correctly

[wooden umbra]

what do you mean?

[thin bridge]

* ln(0.97)
not *ln(0.97^x)

[wooden umbra]

ooooooooooooh right

cuz it's [a^(u)] = a^(u)*u'**ln(a)

thank you

.close

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[paper cove]

The assignment is to "find a polar equation for the curve represented by y=x". I'm getting 1=tan(theta)? Could someone just confirm if that's right, it seems too simple 😂

[paper cove]

Because y=r•sin(theta) and x=r•cos(theta) so if I see them equal then I get 1=tan

[median vigil]

you could also solve for a particular value of θ that satisfies that

but in general a line passing through the origin can be described in polar by a constant angle and nothing else

[paper cove]

Makes sense

So it's not right then? How do I get to the single angle?

Oh I see what you mean

Solve tan=1

π/4?

[median vigil]

yes

[paper cove]

Cool, thanks 🙏

[median vigil]

since tan(θ) is periodic on period π you can find infinitely many other angles that also work by adding multiples of π, but θ=π/4 is the simplest

[paper cove]

Oh yeah, makes sense

Thank you

.close

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[patent vault]

how in the world am i supposed to do this lol

[lusty spear]

is "L2" of importance here ?

dont know what that means

[patent vault]

no its just a name of the group

[lusty spear]

the only reasonable answer for me is monish took a look at the card, if this has a mathmatical explanation I wanna hear it as well

but i think I once heard about a familiar problem and it was quite crazy

[fleet condor]

there is almost certainly a math way to do it - the ones like this are usualyl soeme crazy tricks

[patent vault]

im just confused how he could possibly know

it sounds pretty random

[fleet condor]

the important part is "picks"

each of the places it mentions "picks" has to be deliberate

[buoyant shadow]

it's kinda easy

[lusty spear]

enlighten us please

I wanna know

[patent vault]

but dont give the full answer

[buoyant shadow]

it's easier than i expected from first glance to be precise

[fleet condor]

oh can you just do every other, sandwich it, then third card gives the last hint for the 38

[patent vault]

but shuffling?

[buoyant shadow]

idk how to give just a hint

[patent vault]

ok then

you can just give the solutoin ig

[trim swan]

the pigeonhole principle is a hint. though I'm not sure if frownyfrog and I are thinking of the same solution. there might be more than one way to do it

[lusty spear]

but how in the world would the pidgenhole principle be applied when you could have any card out of 38 ?

[trim swan]

It's more about Alex's two cards

[buoyant shadow]

if you pair up all cards, then Alex can pick a complete pair

[lusty spear]

how does that info help?

[buoyant shadow]

then the third card will stand out

[lusty spear]

wdym

alex gets to look at the cards ?

[buoyant shadow]

yes

[patent vault]

oh

[trim swan]

Alex gets to choose two of the 40 cards the student gives him

[lusty spear]

bruh

[buoyant shadow]

picks is about looking in this question

[lusty spear]

...

ok but still

even if so

monish still has to figure out what card out of the 36 were chosen

how ?

[patent vault]

does monish look at the 3 cards

[buoyant shadow]

yes

[trim swan]

after they're all chosen, yes

[lusty spear]

boooh

[trim swan]

otherwise he is just picking randomly

[lusty spear]

I thought only from the 2 chosen he can predict the third

[patent vault]

so what is he predicting

[trim swan]

He looks at the three cards, and he can tell which one was added by the student

[patent vault]

oh

[lusty spear]

since the other 2 were paired right ?

[patent vault]

couldnt monish and alex have talked before hand and talked about which 2 cards he would choose

[lusty spear]

probably what they did

cheap trick

[patent vault]

wait no

[trim swan]

No, because Alex only has access to 40 of the cards, and he doesn't know which 40 they will be

[lusty spear]

but they could impose some pairing

or something like that

[patent vault]

but what pairing lol

[trim swan]

"what pairing" is the whole question

what pairing is always possible given any 40 of the cards?

it's always possible for Alex to choose two cards that have some property together

[patent vault]

maybe something with mods?

or primes

[lusty spear]

damn it thats the exact problem I talked about in the beginning, now I am remembering it lmao

I think mod sounds right

also any third card must not have that property

[patent vault]

yeah

i see where pigeonhole principle comes in

[lusty spear]

ye

[buoyant shadow]

the exact pairing doesn't matter, it only changes how much they have to think

in this case even odd makes it easy

[lusty spear]

wait what

[patent vault]

that would cause too much ambiguity though

[buoyant shadow]

i mean (56)(78)...

[patent vault]

?

are those card numbers

[buoyant shadow]

yes, Alex can always pick an odd card X and an even card X+1

[patent vault]

oh

ooh

and the student cant choose anything else to mess it up

and there will always be at least 2 consecutive cards by pigeonhole principle

[trim swan]

ahh yeah, I was thinking of two cards whose sum is 79, but yeah any predetermined pairing would work

[patent vault]

are there always garunteed 2 of those?

how would you prove that there will always be 2 that sum to 79

[buoyant shadow]

because it's also a pairing

[patent vault]

wdym

[buoyant shadow]

1 goes with 78
6 goes with 73

you want to split them in to 39 groups of 2, then the proof is the same whatever you did

[patent vault]

oh

thanks

thanks so much for the help

.close

[safe radish]

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[foggy cloud]

For this example I did was using -b/2a

Which is wrong

So how to know when using complete the square

[thin bridge]

show work

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@foggy cloud Has your question been resolved?

[foggy cloud]

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[manic heart]

what did i do wrong

[subtle robin]

you shouldve factored out the two at the start

[manic heart]

so i should have divided everything by 2?

[subtle robin]

yeah

[manic heart]

oh

ok thanks

.close

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[near ginkgo]

Any idea what [z(x)]8 mean? The square bracket?

[dreamy hornet]

square bracket is the same as () usually

[near ginkgo]

oh so its basically power 8?

[dreamy hornet]

yes

[near ginkgo]

ohkk thanks

[dreamy hornet]

no prob

[near ginkgo]

.close

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[native robin]

Let $r, s, t$ be the solutions to the cubic equation $x^3+7x^2=1$. What is the value of $\frac{1}{r^3}+\frac{1}{s^3}+\frac{1}{t^3}?$

[flat frigate]

Jash

[native robin]

no calculater

[quasi sentinel]

1+7/x = 1/x³

[native robin]

then what

[quasi sentinel]

think about it

[native robin]

you get 1/1+7/x-1/x^3 = 0

[quasi sentinel]

no not like that

1/r³ = 1 + 7/r

1/s³ = 1 + 7/s

1/t³ = 1 + 7/t

[native robin]

ohhhhhhhhhhhhhhhhhhhhhhh

but then still u dont know r,s,t

[quasi sentinel]

that doesnt matter

[native robin]

the sum is 3+7/r+7/s+7/t

[quasi sentinel]

you can still get the values of r+s+t, rs+st+rt, rst

[native robin]

so combine the 3 fractions

[quasi sentinel]

yes

[native robin]

3+(7s+7r+7t)/(rst)

[quasi sentinel]

no

[native robin]

wait i know

[quasi sentinel]

ok

[native robin]

3+(7sr+7st+7rt)/(str)

[quasi sentinel]

yes

[native robin]

then what

[quasi sentinel]

gimme a sec

when ax³+bx²+cx+d = 0

i gtg

[native robin]

x^3+7x^2-1=0

isnt it just 3+0

[safe radish]

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[lean otter]

if i have a table, like this
Tide Height(ft) Time
first high tide 12.95 4:25 AM
first low tide 2.02 10:55 AM

a. Determine the amplitude, period, phase shift, and vertical shift of a sine function that models the height of the tide. Let x represent the number of hours that the high or low tide occured after midnight

b. Write the a sine function that models the data

c. according to said model, what is the height of the tide at 8:45 PM at night?
(mostly just A and B here)

[safe radish]

@lean otter Has your question been resolved?

[lean otter]

<@&286206848099549185>

[drifting mountain]

!show

[safe radish]

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

[drifting mountain]

whats a sine functions

[lean otter]

i dont have work dude
thats the thing
idk how to do this problem at all

[drifting mountain]

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

google is your friend

<@&268886789983436800>

<@&268886789983436800>

[lean thorn]

So youtube is a great rsource for questions like this

[white umbra]

cyanide, yeah only ping helpers once after 15 minutes, ils stop posting here if you're not gonna help

[lean otter]

yeah ik-im sorry abt that

[lean otter]

do tell

[lean thorn]

this is where google comes into rescue

https://www.youtube.com/watch?v=UdwRwdLvfMI

[lean otter]

no like ehres the thing

i know what those are

i just dont know how to turn a table into that

[lean thorn]

ah I see

ok so in part (a), it tells you what we should represent for x. What should our x line be?

[lean otter]

time-hrs past midnight

which means the function will have to have a pi in it somewhere since its whole numbers and not fractions with pi in them

[lean thorn]

well not necessarily. The way I see you breaking down your function is by doing the hour and then the mintutes / 60 for the fraction portion

So someting like 4:25AM would be 4.4167, right? It's 4 hours and 25/60 = 4.16666667 (simplified here to 5 sig figs)

so there's your first x value

[lean otter]

i see ok ok

[lean thorn]

are those the only 2 values they provided in the table?

[lean otter]

yes

[lean thorn]

ok bet

since one represents high tide and one represents low tide, it should be enough info

[lean otter]

ye...

[lean thorn]

Ok so what would our x value be for the low tide?

given the logic above?

[lean otter]

10.91666

i think

[lean thorn]

,calc 55/60

[flat frigate]

Result:

0.91666666666667

[lean otter]

ye

[lean thorn]

yep!

jsut round that last number to a 7

[lean otter]

yeah ik binary and decimal do a funny when you get small enough

[lean thorn]

like 10.917 if we want to keep to 5 sig figs

[lean otter]

alr so 4.4167 and 10.9117

[lean thorn]

yep. Now just plot those somewhere like desmos with the corresponding y value and then figure out the requested items 🙂

[lean otter]

heres the thing

there are multiple unknowns

a,b, h, and k

[lean thorn]

well we can infer them with the info we have

[lean otter]

mm do tell

[lean thorn]

so let's start off with the amplitude

[lean otter]

ok

[lean thorn]

the amplitude represents the distance to our given crest (aka the max) and trough (aka the min) to the symmetrical line where they're equidistant from one another

[lean otter]

yes

that i know

[lean thorn]

in other words, amplitude is the distance between the crest and the trough divided by 2

[lean otter]

ye

[lean thorn]

so what would that be? 🙂

[lean otter]

wait you're fucking joking

[lean thorn]

I hardly do lol

[lean otter]

im actually about to

oh my god

5.465

i feel like such a moron

oh my g o d

[lean thorn]

nah not a moron

you're learning 🙂

[lean otter]

no like
THAT SEEMS SO OBVIOUS NOW

[lean thorn]

we all go through this too don't worry

math is obvious once you figure it out!

but until then you struggle lol

[lean otter]

i just assumed for whatever reason that the values were not top and bottom

and now i realize

THEY WERE

ueiryngeukriyfngvh8uwriejs

anyways

[lean thorn]

oooh oh yeah

yeah that's what they mean by high tide and low tide. They represent the min and max height of a shoreline

anywho

[lean otter]

so now B

[lean thorn]

ok so that's the amplitude

[lean otter]

yeah

[lean thorn]

ok so the period represents what?

[lean otter]

the period is 2pi/b

[lean otter]

1 full cycle

[lean thorn]

bingo

[lean otter]

period is 6.945

no wait

[lean thorn]

so since sine waves are cyclic, then we can assume that the distance between the trough and the crest will be equidistant themselves

so in other words, the distance from one crest to another is twice the distance from a crest to a trough

[lean otter]

yes

one up->down is 6.945

so 2 of them

is 13.89

[lean thorn]

very good

[lean otter]

or just 13 (according to my book)

oh my god this seems so simple now

RAHHH

[lean thorn]

huh that's sus haha

[lean otter]

it has the amp as 5.465

but period is 13

¯_(ツ)_/¯

weird

anyways

[lean thorn]

yeah that's strange lol

[lean otter]

yeah that is hella weird

but for vertical shift thats simple

2.02+5.645/2

[lean thorn]

,calc 2* ( 10.91666 - 4.4167)

[flat frigate]

Result:

12.99992

[lean thorn]

oh that's why

I think your arithmetic has an error

[lean otter]

oh weird i got 13.89

no i used uh

10.9117

not wait

wait

thats so f*cked up LMFAO

[lean thorn]

Oh that's why lol. Sorry my bad I didn't catch that. Your value should be rounded to 10.9167, not 10.9117

[lean otter]

yeah

wait

if period is 2pi/b

were is the pi

[lean thorn]

I dunno hidden somewhere in that number probably

lol

it's not important for these purposes

[lean otter]

alr

so all that aside

vertical shift shoudnt be that bad

i THINK its amp/2 + 2.02

no wait

LMAO

just amp + 2.02

[lean thorn]

so a normal sin function has a crest at 1 right?

[lean otter]

yeah

i got the vertical shift

7.485

book concurs

[lean thorn]

ok excellent

[lean otter]

and wait

the phase shift

is just 4.417

lmfao this all seems so simple now

i h8 myself

[lean thorn]

I haven't done it yet but I'll take your word for it 🙂

[lean otter]

phase is just 4.417 because the smallest x value is 4.417 and thats where trough is

like in the table

also this is completely unrelated

but your PFP is hilarious

[lean thorn]

lmaooo thanks

[lean otter]

but wait part B now

[lean thorn]

it's from the movie Sister Act

[lean otter]

so we have
A = 5.645
period = 13
H = 4.417
K = 7.485

afaik

i think this is where the 2pi/B comes in