help-28

[opal pivot]

confused by red circle. Why can we write (a+1)^p = that

[ebon bison]

that's binomial theorem

[opal pivot]

Not exactly. Binomial does not contain sigma

[ebon bison]

they've created a general term

T of i and adding them all up

[opal pivot]

ok

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[reef dust]

how to take the derivative of $\frac{f(x)g(x)}{h(x)}$

[glossy valve]

taro

[reef dust]

do i express the h(x) as a negative index and apply product or is there a better way

[torn jolt]

you can do that or use the quotient rule

whichever you prefer

[reef dust]

so for quotient id express f(x)g(x) as a single function?

oh

wait

[torn jolt]

yeah sure

[reef dust]

AH ok

tysm

[torn jolt]

but take the derivative of it by applying product rule once you have to differentiate it

[reef dust]

yep

got it

tysm

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[torn jolt]

f(2x) = 2f^(-1)(x)
if f'(x)>0 and f(1) = 1
does this mean that the shape under f(x) from 2^n to 2^(n+1)
and the shape under f(x) from f^(-1)(2^(n+k)) to f^(-1)(2^(n+k+1)) similar

[south island]

$f(2x) = 2f^{-1}(x), \quad f'(x) > 0, \quad f(1) = 1$

[glossy valve]

dqvidutzul

[south island]

like this?

$\text{Does this imply that the region under } f(x) \text{ from } 2^n \text{ to } 2^{(n+1)} \text{ and the region under } f^{-1}(x) \text{ from } f^{-1}(2^{(n+k)}) \text{ to } f^{-1}(2^{(n+k+1)}) \text{ are similar?}$

[torn jolt]

yes and also

[glossy valve]

dqvidutzul

[torn jolt]

f(x) is not a polynomial

[south island]

i have no idea sorry

[torn jolt]

ill actually draw it gimme a sec

i cant explain it

well

are yellow and red similar?

[south island]

calculate the area and find out

[torn jolt]

i cant

i dont have f(x)

[south island]

oh wait you dont know f(x)

[torn jolt]

the only thing i got is F(2) - F(1) = a

and i have to express

F(16) - F(1) with a

and give a reason to the answer

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[silver blaze]

Anyone know how to do part 2?

[wild sleet]

,rccw

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[karmic eagle]

Hi, I was working on this probability problem and turned out I got it wrong. The instructor just released solutions, but I still don't understand why I can't do what I'm doing. Problem 1 solution is the typed one, mine is the handwritten one

[full forum]

@karmic eagle Has your question been resolved?

[wild sleet]

it's not what they ask

oh wait i see

what

you did everything right

you found the answer, it's in the middle somewhere

the rest is probably fine too

[karmic eagle]

ohh I guess I just started wrong initially... still, I don't understand why they went through all that when they could've just counted that...

I tried the 2nd one again as well, since all the features are conditionally independent, I thought this would work, but I'm getting a different answer again

the issue is with the P(S = Good, Cu = Italian), why is that not the same thing as P(S = Good)P(Cu = Italian)?

[wild sleet]

that's why they went thorugh that instead of counting

kinda

[karmic eagle]

I mean if I count it, it's obvious that it's 1

actually no it's not 🙃

[wild sleet]

their solution makes no sense

[karmic eagle]

is it wrong to use the Bayes Theorem as I'm using it initially compared to their P(everything)/P(given)?

maaan

[wild sleet]

i don't see any case with good service and italian cuisine

[karmic eagle]

yeah me neither

this is supposed to be using Naive Bayes Theorem if that helps

[wild sleet]

ah ok

i haven;t heard of it, but it's this, i get it

[karmic eagle]

maybe I should just go with their solutions without questioning them lol

[wild sleet]

like it's supposed to be weird

yes

[karmic eagle]

this part doesn't make sense to me at all tho

why are they adding the O = Good and O = Bad to it?

[wild sleet]

the point is the only data we're allowed to use is P(something | O = something)

and P(O = something)

and for anything else we can only derive it

and it's interesting that it actually worked for 1

no okay it should have worked, that is actually the only data we need

and then for 2 and 3 it stops telling you "the truth"

ah

but it extrapolates from the sample to other restaurants

and it makes sense that it's 1

in the sample italian guarantees good and good service guarantees good, so when we get the result that both at the same time guarantees good, that's good

except for 12th restaurant

[karmic eagle]

ohh okay I think I kinda get it now

so what would be the reason that I can't just say "there's 0 data points with these criteria, so the probability is 0"

there has to be some logical reason haha

[wild sleet]

how is it 0, if you have no data

"Alice is in 7th grade. What's the probability that she has more than 7 friends"

we have no data about amount of friends Alices usually have, and no data about what amount friends 7 graders usually have etc.

that doesn't make 0 the best guess from the data given

[karmic eagle]

okay okay I see it now

so since I don't have the data, I also can't use the regular Bayes Theorem like I tried, right?

[wild sleet]

no data can be argued to suggest 50% probability

i think you're not allowed to use all the data

[wild sleet]

idk, this seems clear, this is what you do

[karmic eagle]

I'll just follow that then 😄

thank you so much for the help!

[wild sleet]

honestly yeah, i don't get why 0 isn't a better guess either

and they don't explain why we use the last column as data

this would work if we used cuisine instead, with different answers

no wait, that's because it's what we're looking for, O = Good

it's not arbitrary

[karmic eagle]

I guess it's just the way it combines the two probabilities? maybe it somehow accounts for not having the O = Bad data in there

[wild sleet]

yeah

we want "good and (the conditions)" and "bad and (the conditions)"

and we just ignore the data we have, which says "0 cases of this" and create it instead from multiplying each condition one by one

[karmic eagle]

makes sense yeah

[wild sleet]

the first step here is redundant

they repeat it entirely in the second step

[karmic eagle]

I noticed that too lol

I was trying to do it by hand and it was the same thing

[wild sleet]

yeah it's really obvious now, "multiply probabilities to get the intersection"

maybe in 10 years they will figure out a notation that doesn't make it so impossible to learn

in plain words they have used that Italian and overall Bad never happens, to conclude that Italian and service good and overall Bad is also 0
then
good service and overall good happens
italian and overall good happens
so they even can happen at the same time
finally a/(a+0) = 1

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@karmic eagle Has your question been resolved?

[karmic eagle]

I see how it works now yepp

it's not terrible, just wasn't really intuitive at first...

thank you!

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[river wave]

For 15 I’m not sure what to compare with for the limit comparison test

1/n?

[hot wadi]

well its given in the hint

[river wave]

Oh I didn’t read it

Ok

Thanks

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[copper shell]

can someone please verify if my proof is right

[fathom saddle]

You're asserting that kv = -v, only when k = -1. This what you were asked to prove, though.

[copper shell]

what else can i do then?

[fathom saddle]

You'll need to make a statement on -1v. Algebraically, what is it?

[copper shell]

wait i will join from my phone and reply

@fathom saddle
will this work

-1 + 1)v = 0v
-1v + 1v = 0
-1v + v = 0
-1v = -v

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[torn jolt]

did a few things here

firstly, found the slope of the line

which I found to be a/4

and set that equal to the derivative of the tangent y=a/x^2

substituting in -2

but I wasn't getting it for some reason

ended up with this

$\tfrac a4 =4a$

[muted summit]

if a line approaches at a tangent, it will have a repeated solution

It may not be the case but i recon, the discriminant of a resulting quadratic has to be 0

[glossy valve]

wyldinwilliam

[torn jolt]

what can that do to help us with this question?

[muted summit]

lets try equate Ys

[torn jolt]

ok the solution is actually pretty simple

[muted summit]

oh you got it?

[torn jolt]

no i saw the solution manual

[muted summit]

ah right

[torn jolt]

you'd just have to realize that the point of tangency is (-2, a/4)

and sub in those values on the line

[muted summit]

ohh ye

[torn jolt]

yeah pretty straightforward for a question worded like this

thx

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[weary epoch]

How can i calculate sum $\sum_{n=k}^{\infty} \frac{С(n-1,k-1)}{2^n}$?

[glossy valve]

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

[weary epoch]

How can i calculate sum $\sum_{n=k}^{\infty} \frac{С(n-1,k-1)}{2^n}$?

[glossy valve]

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

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[white cape]

I need long and extensive help with trig questions. Sohcahtoa and sine/cosine law. My notes are at school and I have adhd, have to do things step by step to understand them. All word questions!

[white cape]

Oh I missed 2 d

Pause one more before word questions!

Before anything, what is an obtuse triangle?

How can I tell a triangle is obtuse by looking at it?

Nevemind this is an obtuse triangle

So I have to do 180 - my answer at the end

This is cosine law?

Ping when reply gonna load my laptop up

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[white cape]

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.reopen

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[fallow sage]

need help

[fallow sage]

uploading what i wrote so far in a sec

…

[glacial pasture]

seems mostly fine, youre just missing something

if i differentiate y^2 for example with respect to x, then i get 2y dy/dx not just 2y

[fallow sage]

so is x cos y supposed to be x cos y dy/dx

[glacial pasture]

indeed

[fallow sage]

is dat all

[glacial pasture]

pretty much, you still have to make dy/dx the subject though

[fallow sage]

do i do that to 2x^2y?

[glacial pasture]

wdym

[fallow sage]

idk when its supposed to be with respect to x like what that means im confused

[glacial pasture]

all of it is derivatives with respect to x
dy/dx means differentiate y with respect to x essentially

[fallow sage]

yeah im still confused

on where to go with what i wrote

[glacial pasture]

what do you have written currently

[glacial pasture]

okay yeah, that should be 2x^2y dy/dx

[fallow sage]

not right?

[glacial pasture]

thats fine, no need to doubt so much

actually

one sec might just be one thing

[fallow sage]

oh it said wrong

dats what i mean

like idk where i went wrong

i was applying sum from another problem i did

[glacial pasture]

yeah i just realised there was never a trig function of x

[glacial pasture]

you added a sin(x) from somewhere

[fallow sage]

wait

wth

its sin y right

[glacial pasture]

yeah

then it should be okay

[fallow sage]

ye got it\

i just miss wrote it at sum point

[glacial pasture]

all good, it happens

[fallow sage]

sum assitance on this not even sure if im going about it right

[glacial pasture]

youre missing the dy/dx 's again

if you differentiate some function g(y) then its g'(y) dy/dx

this is just the chain rule really

[glacial pasture]

-7y^3 not ^2 then its -21xy^2 dy/dx for the last term too

[fallow sage]

oh

miss wrote stuff again

it said its wrong

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[broken phoenix]

Can elimination work

For this ?

[shrewd hamlet]

try it

see what u get

[broken phoenix]

Kk

[steel grove]

you shoud change x as 2 and 3

and then solve this problem

and get f(2), g(2), f(3), g(3)

[glacial pasture]

[steel grove]

so finally can get f(2)+g(3)

[topaz valley]

blud...

[shrewd hamlet]

@steel grove why did u suggest that

[steel grove]

it is general solution

[topaz valley]

[steel grove]

and you can make it a little simpler by some method

[topaz valley]

you can make it simpler by solving for f(x) and g(x) first...

[shrewd hamlet]

bro is not seeking in the right place

[steel grove]

it can include complicated func of x

[glacial pasture]

mans still going

[steel grove]

but that is most correct method

🙂

[topaz valley]

no 😃

[steel grove]

i ve said your method is most correct

but my method is simple

[topaz valley]

ok 👍

[broken phoenix]

This is what I got

Is this correct

[steel grove]

oh its good

[topaz valley]

you didn't double g when you subbed back into 1

[broken phoenix]

Double g?

Can you show me

[steel grove]

i ve said that

f(2)+2g(2) = 62
2f(2)+3g(2) = 97

[topaz valley]

2g(x) is 12x^2 + 6

[steel grove]

g(2) = 27 f(2) = 4

[broken phoenix]

Oh shoot I’m dumb

I didn’t realize

Ok

[topaz valley]

[broken phoenix]

Wrong ?

[topaz valley]

it's 2g(x) not 4g(x)

[broken phoenix]

Omgg

WHY DID I WRITEn4

😭

I’m cooked

[topaz valley]

[broken phoenix]

DAWG

I MESSED UP AGAAAAAI

[topaz valley]

2 times 3 is 6

[broken phoenix]

3x+2

[topaz valley]

ya

[broken phoenix]

I’m cooked

[topaz valley]

keep cooking chef 🧑‍🍳

[broken phoenix]

I can’t read

But I get it

[topaz valley]

good

[broken phoenix]

Wait so

Do I plug both of these in

3x+2 into 2

[topaz valley]

yeah

but like

not like that

f(x) = 3x + 2
f(2) = 3(2) + 2

[broken phoenix]

OHH YEAH CUS ITS THE X

Dawg I genuinely lack reading comprehension today

[topaz valley]

Blud...

[broken phoenix]

No they right

F(x)

Also this right

[broken phoenix]

YESS

Also last one

How do I start this eq

I have not been able to figure it out

[topaz valley]

f is linear

so you can write down a form for f

[broken phoenix]

Yeah I don’t know how to write it like

That

[topaz valley]

f(x) = ax + b?

[topaz valley]

blud

[broken phoenix]

Is it ax+b(ax+b(3))=2

??

[topaz valley]

f(x) = ax + b
f(3) = a(3) + b

[broken phoenix]

What do I do with the 2 and 1?

Does it become negatives

[topaz valley]

no like

f(f(3)) = f(a(3) + b) = a(a(3) + b) + b = 2

[topaz valley]

do you know how functions work?

[broken phoenix]

When it’s linear I get kinda cooked

Like I know it has domain and range

Whatever f(x) is I just sub into it

Ik just ask my teacher tomorrow

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[torn jolt]

why does the exponent become x/2 and not 2(x)

[torn jolt]

i mean x/2

not 2/x

[wanton vale]

Because it's √3

Which is $(3^{1/2})^x$

[torn jolt]

okay i see

[glossy valve]

Solomaniac

[torn jolt]

thank you

it makes sense nkw

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.reopen

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[torn jolt]

.reopen

wtf

whatever

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[broken flare]

How do i simplify this...
$(A^{-1}(BA^{-1})^{-1}BA^{-1})^{-1}$

[glossy valve]

hamr

[earnest badger]

A,B are matrices?

[broken flare]

Yes sorry forgot to mention, A and B are invertible matrices

Also they are of the same order

[earnest badger]

Do you know $(XY)^{-1} = Y^{-1}X^{-1}$

[glossy valve]

94tj

[broken flare]

I didn't know that, ok I think i can do it now, is it true that (AB)C = A(BC)?

[earnest badger]

yes

[proud garnet]

yes, think of the matrices geometrically

as linear transformations

[broken flare]

How about is it true that:
A(CB) = C(AB)?

[earnest badger]

not necessarily

[proud garnet]

not always

[broken flare]

I just find it a little confusing because as I understand it, multiplication is not commutative

[earnest badger]

correct

[earnest badger]

but this is associativity

[broken flare]

$(A^{-1}(BA^{-1})^{-1}BA^{-1})^{-1}=\
(A^{-1}(B^{-1}A)BA^{-1})^{-1}=\$

[glossy valve]

hamr

[earnest badger]

no

[broken flare]

Hmm

[earnest badger]

look at this again

[broken flare]

$(BA^{-1})^{-1} \neq (B^{-1}A)$?

[glossy valve]

hamr

[broken flare]

And using this and $(A^{-1})^{-1} = A$?

[glossy valve]

hamr

[earnest badger]

well that is true but not because of what I sent

What I sent tells you how to deal with $(BA^{-1})^{-1}$

[glossy valve]

94tj

[broken flare]

This should become $\(BA^{-1})^{-1} = B^{-1}(A^{-1})^{-1} = B^{-1}A$

[glossy valve]

hamr

[broken flare]

Right?

[earnest badger]

look at this

[broken flare]

Ohhh

They switch places

Okok

gimme a second lemme fix this

[earnest badger]

To understand why they must switch places, think about the linear map represented by XY

[broken flare]

$(A^{-1}(BA^{-1})^{-1}BA^{-1})^{-1}=\
(A^{-1}(AB^{-1})BA^{-1})^{-1}=\
(((A^{-1}A)B^{-1})(BA^{-1}))^{-1}=\
((IB^{-1})(BA^{-1}))^{-1}=\
(B^{-1}(BA^{-1}))^{-1}=\
((B^{-1}B)A^{-1})^{-1}=\
(IA^{-1})^{-1}=\
(A^{-1})^{-1}=\
A\$

[glossy valve]

hamr

[broken flare]

Is there a resource i can check out for this?

[earnest badger]

im not sure

[broken flare]

i think i found some actually

[earnest badger]

this looks good

[broken flare]

math exchange someone explained it

[broken flare]

TYSM

very helpful

i appreciate it

[earnest badger]

np

[broken flare]

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[full forum]

@viral stratus Has your question been resolved?

[viral stratus]

<@&286206848099549185>

[simple totem]

what have you tried

[viral stratus]

i figured it out, ty though

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[hoary bluff]

im confused again how do I sketch these

[hoary bluff]

im trying to understand better how to find the implied domain by sketching

[pale cargo]

do you know

simple rules to follow

while graphing?

[hoary bluff]

yup

omg

i am so dumb

sorry

[pale cargo]

you're okay 😭

[hoary bluff]

brain fart

[pale cargo]

dw about it

[pale cargo]

we all get em

[hoary bluff]

i will probably come back here

for more help ahaha

.close

[pale cargo]

close it tho

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[hoary bluff]

what does the graph look like of y = 1/ sqrt x

[crisp girder]

Have you learn this before or do you not know

[hoary bluff]

i know most other ones but never this one

[crisp girder]

Try putting it into desmos

[hoary bluff]

aight

thank you

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[mighty thunder]

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[vague fulcrum]

12 i

[full forum]

@vague fulcrum Has your question been resolved?

[vague fulcrum]

<@&286206848099549185>

[sacred yarrow]

!status

[full forum]

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

[vague fulcrum]

1

[sacred yarrow]

What is vector AC?

State it with vector a and vector c

[vague fulcrum]

Oc - oa

[sacred yarrow]

No O

Only a and c

Pretty close, just remove O

[vague fulcrum]

C - a

[sacred yarrow]

yes

[sacred yarrow]

Look at the question, can you solve it now?

@vague fulcrum

[vague fulcrum]

Well I did do like this

[sacred yarrow]

Yeah, you’re doing good

[vague fulcrum]

Ok thanks

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[sacred yarrow]

Have a good one

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[fallen heart]

Am i doing the problem right? I don’t really know what the question is asking

[fallen heart]

i also know that this is not the correct answer, i want to make sure i am on the right direction

[wise gate]

$f\circ T$ is a function from $\mathbb{R}^2$ to $\mathbb{R}$. You have to write it and then find the partial derivative

[glossy valve]

d

[fallen heart]

the one under the question is the derivative of T(s,t) in respect to s

Is f o T = F(T(u,v))?

[wise gate]

Yes

[wise gate]

I think it is wrong

The second

It should be s/(1+s^2)

[fallen heart]

This is what i did

I took the partial derivatives for each the f(u,v) and the T(s,t) onto a matrix

[wise gate]

Now you multiply both matrices

And substitute (u,v)=T(s,t)

[wise gate]

No

Just multiply the vector Df with the first column of DT

Because it is the derivative with respect to s

[fallen heart]

Like this?

and then i can plug the values of (7,0) onto f(u,v) or T(s,t)?

[wise gate]

Yes

[wise gate]

You can plug s = 7 and t = 0

And (u,v) = T(7,0)

[full forum]

@fallen heart Has your question been resolved?

[wise gate]

t^2s is 0, not 7

[fallen heart]

it says that it is incorrect

[wise gate]

What was your final result?

[fallen heart]

7cos(cos(7))cos(log(sqrt50)) / 50

[wise gate]

^

It is not cos(7) but cos(0)

[fallen heart]

7 cos (log(sqrt50)) / 50

[wise gate]

No

cos(cos(0))

Is cos(1)

$\dfrac{7\cos(\cos 0)\cos\left(\log\sqrt{50}\right)}{50}$

[glossy valve]

d

[fallen heart]

ohhh, i get it now

alright thank you so much for the help

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[torn jolt]

Students Linas and Povilas decided to paint the room. At first, only Lina worked for 9 hours, then only Povilas worked for 4 hours and the work was completed. How long would it take Lina to paint the room by himself if Povilas did the job 5 hours faster by himself?

[torn jolt]

I tried this but discriminant becames terrible number

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[solemn hemlock]

what formula did they use?

[dull seal]

Formula to do what? I don't see any formula.

If you're referring on how they isolate the x, they are using log properties.

[solemn hemlock]

to how they isolate the X

[solemn hemlock]

which is?

[dull seal]

wait

$\log_a a^x = x$

[glossy valve]

Samuel

[dull seal]

This one

ln is the log in base e

so for 2^x they use log in base 2

and for e^x they use log in base e, which is ln

$x \cdot \cancel{\log_a a}$

[solemn hemlock]

woah what happened

[dull seal]

wait a typo

[glossy valve]

Samuel

[dull seal]

now

log_a(a) = 1, you know this right?

for example, ln(e)=1, or log_2(2) = 1

[solemn hemlock]

oh yeh i know this why

[dull seal]

and u know log(a^x) = xlog(a)?

[solemn hemlock]

because of this ?

[dull seal]

no, that is consecuence of the other 2

[solemn hemlock]

is this another formula?

[dull seal]

that is the property of logarithm i explained before

but with different bases

and exponents

but in your exercises b and M are equal

that's why u can simplify them

because they are equal to 1

in your exercises you choose carefully the base of the log so you can simplify it

here u choose log_2 so u can make xlog_2(2) = log_2(5)

and log_2(2) = 1

so x * 1

[solemn hemlock]

ahhhhh ok now I get it

so if it was 3^x

we would chose log3?

[dull seal]

yes

log_3

cause log_3(3^x) = xlog_3(3) = x * 1 = x

[solemn hemlock]

I am solving these rn

[solemn hemlock]

so i would use this right?

[south island]

you use exponentiation for c) and d)

[solemn hemlock]

what is that

and why?

[south island]

to get rid of the natural logarithm

use the base of e

for example at c)

$e^{\ln(x)} = e^{-1}$

[glossy valve]

dqvidutzul

[south island]

and remember that

$x^{log_x{y}} = y$

[glossy valve]

dqvidutzul

[solemn hemlock]

so it will be x=e^-1?

[south island]

yup or as a fraction $\frac{1}{e}$

[solemn hemlock]

bro so many formuals

idk which is which 😭

[glossy valve]

dqvidutzul

[south island]

yes logs have a couple of formulas

[solemn hemlock]

so when should I use each?

[south island]

here are the most common formulas

“formulas” in quotation marks because these are not really formulas just rules

[south island]

that is the logic part of math

[solemn hemlock]

let me look a tit

[south island]

you should always think “damn, there must be a rule for this, what can i change so a rule can apply?”

[solemn hemlock]

I got an exam tmrw bro

[south island]

damn good luck you got this

[solemn hemlock]

i thought this topic would be simple

[solemn hemlock]

thank you bro 💪🏽

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[south island]

one more thing

the underlined ones i would say are the most common @solemn hemlock

[solemn hemlock]

perfect, thank you so much bro!

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[sinful totem]

can anyone help with this bio

[sinful totem]

ik this is math but there are no actie bio servers

I believe it would be A

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[granite hollow]

I need some help with this one. I need to solve for y and then once I have y =, I plug in 0 and sqrt3 and solve for c. Not sure how to solve this for y though. I might’ve done something wrong in my math, or I just don’t know the next step. I believe the answer is y=arctan(x) + sqrt/3

[full forum]

@granite hollow Has your question been resolved?

[devout valley]

You could substitute x = 0, y = sqrt{3}/3 in the point you have, arctan(y) = arctan(x) + c, but regardless you can apply tan to both sides and use tan addition formula

[granite hollow]

What’s the tan addition formula? Yeah I tried doing tan to both sides, so I’d have y = x + tan(c). Then I’d have sqrt(3) = tan(c) and not sure what to do from there.

[devout valley]

this thing

[granite hollow]

I never seen that

[devout valley]

And it doesn't become y = x + tan(c) because of the above formula, you don't have tan(arctan(x) + c) = tan(arctan(x)) + tan(c)

[granite hollow]

What’s my A and B here

Is A = x and C = B

[devout valley]

"A" is arctan(x) and "B" is C

[granite hollow]

Ok so (arctanx + tan(c))/ (1 + arctanx • tan(c))

[devout valley]

Well tan(arctan(x)) and those cancel out

[granite hollow]

Let me write this

[granite hollow]

What’s this thing

Is it the same as the others

Why is it flipped

[devout valley]

Basically it's saying if you have tan(A + B) you want that as -, you want the sign to be opposite to what was there before

[granite hollow]

Ahh ok

[devout valley]

So in this case $\tan(A + B) = \frac{\tan(A) + \tan(B)}{1 - \tan(A)\tan(B)}$ with $A = \arctan(x)$ and $B = c$

[glossy valve]

@devout valley

[granite hollow]

Gotcha that makes sense

And the regular +/- just means to keep the sign the same

So don’t my tan(c) cancel too

y = 1?

[devout valley]

The tan(c) doesn't exactly cancel (unless you choose to relabel it as a new constant, which would be fine)

It would be easier to find the constant $C$ first though, from $\arctan(y) = \arctan(x) + C$ use that $x = 0, y = \sqrt{3}$ to get what $C$ is, then use the formula as per before

[granite hollow]

Ok let me do that

Why sqrt(3)/3?

Where the /3 come from

[glossy valve]

@devout valley

[devout valley]

Misread

[granite hollow]

Ah ok np

So not really sure what to do now

I still don’t know how to solve for c

[devout valley]

What's arctan(0)?

[granite hollow]

No idea

[devout valley]

(argubaly you should also know what arctan(sqrt{3}) is too, but that isn't as urgent!)

[granite hollow]

I don’t know anything

I did arctan(sqrt3) in my calculator and got a decimal

[devout valley]

do you know the trig ratios of special angles?

[granite hollow]

No

[granite hollow]

0

I don’t even know the unit circle

[gritty rose]

,tex .unit circle

[glossy valve]

riemann

[granite hollow]

How do I read inverses on unit circle

Nvm idk what I’m doing

[devout valley]

Something worth reading up on eventually

Anyways, we have that C = arctan(sqrt{3}) from the fact that arctan(0) = 0

[granite hollow]

Yea

[devout valley]

(there is an explicit value that arctan(sqrt{3}) takes, but we don't really need it, finding it is "an exercise for the reader" )

[granite hollow]

🧐

[devout valley]

Putting that all into the formula from before, with A = arctan(x), B = arctan(sqrt{3}), what do you end up getting?

(remember that tan(arctan(d)) = d for any d you choose)

[granite hollow]

Ok let me try

What’s my A and B now

A is 0?

[devout valley]

"with A = arctan(x), B = arctan(sqrt{3})"

[granite hollow]

Oh

But we got rid of arctanx

Cuz we plugged in 0

[devout valley]

That was only to find what the constant c was

so we have the constant c and put that back in

[granite hollow]

Oh ok

[devout valley]

it'd be y, not arctan(y), but basically yea

$y = \frac{x + \sqrt{3}}{1 - x\sqrt{3}}$

[glossy valve]

@devout valley

[granite hollow]

Why just y

[devout valley]

Because you applied tan to $\arctan(y) = \arctan(x) + \arctan(\sqrt{3})$, which gets you $\tan(\arctan(y)) = \tan(\arctan(y) = \arctan(x))$

[glossy valve]

@devout valley

[granite hollow]

Oh ok

[devout valley]

But as per before...

[granite hollow]

Got it

Thank you

I think I’m all good now thx!

[devout valley]

Have a good one

[granite hollow]

U too

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[wicked zodiac]

im not sure what im suppose to do

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@wicked zodiac Has your question been resolved?

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[oblique skiff]

how did they get 114.4

[torn jolt]

180 - angle

[oblique skiff]

why not 360 - angle

[glacial pasture]

,w graph y=sin(x) and y=0.911 and x=pi/2 from x=0 to x=2pi

imagine a line at y=0.911, and it should be clear

[oblique skiff]

its not

help me understand

[glossy valve]

Results provided by WolframAlpha

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

Requested by aldrnari_

[glacial pasture]

okay, the green line is the symmetry between 0 and 180

if the yellow line (your 0.911) crosses it at that point there, then you can see it crosses it again at that otherside of said line

they are equal distance from 0 and 180 respectively

so if the first value is 0+a then the second is 180-a

[oblique skiff]

what about when the yellow line is negative

[glacial pasture]

if its negative, then say you have one solution at pi+a, then there is another at 2pi-a

[oblique skiff]

so am i always going to do 180 - a if positive

and 180 - (-a), and 360 - (-a)

if negative ^

[glacial pasture]

i would just try to mentally visualise it like this, or if you need to then just quickly sketch the graph

itll help see why

[oblique skiff]

honestly this graph isnt helping because what is the point of it

[oblique skiff]

i get that its equal distance

but im not understanding why we're adding 180

[glacial pasture]

we arent adding 180

were saying okay theres a symmetry between 0 and pi
if theres a point at (0+a) then there must then be another at (pi-a) due to this symmetry

[glacial pasture]

this graph just shows those two intersections clearly

[oblique skiff]

but 180 + (-14.12) is around 166

so that means theyre doing 180 - a

and there is no solution at 166 and 374

[oblique skiff]

so im confused does this apply or not

because im trying to understand when to apply 180 + a or 180 -a or 360 -a or 360+a

[glacial pasture]

where is the -14.12

[oblique skiff]

do the sin of -0.244

well, inverse to find the angle

[glacial pasture]

alright, so you have a solution outside the 0 to 360 range

you can get one in that range my adding 360

this will be a (360-a) one where a is 14.12

the other will be 180+14.12

[glacial pasture]

this is because the previous one was 0-14.12

sine repeats every 360 so 360-14.12 has the same sin value

just pretend im writing degrees

[oblique skiff]

sure

[oblique skiff]

but the angle isnt 14.12 its -14.12

thats why im not understanding

[glacial pasture]

i never said it was see

im saying you can express it that way

im saying that sin(-14.12)=sin(360-14.12)

[oblique skiff]

okay so

[oblique skiff]

if my goal is to find all the solutions within the interval

so the reason we cant do 360 - a when its positive is because its outside the parameters

for example 360 - 65.6 doesnt work here because the next interval happens at 425.6

so is that why 360 - a only works for negatives?

because the next interval happens between 0,360

[glacial pasture]

i seriously think it would help for you to look at a graph you draw yourself so you can see the logic of it

and put these values youre thinking of on that graph

[oblique skiff]

i mean im watching this guy on youtube do 360 - 68.13 for example in a quadratic equation

which is similar to the problem im doing

[glacial pasture]

indeed it is

[oblique skiff]

but im still trying to figure out why this doesnt work

[oblique skiff]

yea idk its not making sense

is there another way to explain it

[glacial pasture]

if you had a solution at -65.6 there is another at 360-65.6 thats not a problem

[oblique skiff]

im thinking of it in terms of quadrants?

and like the unit circle

[glacial pasture]

i think graphically is the most straight forward, otherwise its hard to see
you can use the unit circle if you want to, look at the points where sin is equal at different angles

[oblique skiff]

so waves are just applications of the unit circle right

if the domain of sin is -90 to 90

[glacial pasture]

what do you mean by 'waves'

[oblique skiff]

sin wave cos wave

is that not what theyre called?

[glacial pasture]

they are the graphs

the graphs represent to functions

[oblique skiff]

^

mb

[glacial pasture]

sure

they come from the unit circle

[oblique skiff]

hold on im onto nothing

im trying to grasp this using the unit circle 😦

[red blade]

@oblique skiff

[oblique skiff]

yep this is what i meant

[full forum]

@oblique skiff Has your question been resolved?

[oblique skiff]

so why is it done differently here

[oblique skiff]

why is it not okay to do -14.12 + 180 here

[oblique skiff]

but you can do it here

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@oblique skiff Has your question been resolved?

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[karmic vapor]

please someone tell me whats wrong with the black

[karmic vapor]

the orange is correct , i just expanded it first before using derivative rules

no, i didnt multiply 4 to both the denominator and numerator it got cubed

im losing my mind

am i just doing math wrong

does it have to be expanded first

nvm gotta expand unless im using chain rule apparnetly

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[tribal flax]

when you're sketching a function based on its derivative

[tribal flax]

how do you know if the function is on the positive or negative y axis

ik when the derivative function has turning points its a point of inflection, a f'(x) = 0 is a stationary point and when f'(x) = + the function is increasing and f'(x) = - the function is decreasing

but how would you know when the function is above the x axis

I drew one based off of the f'(x) and I had it below y = 0 the whole time but the answers showed otherwise

[deft briar]

iirc you're not supposed to tell that based on the derivative, here we're taught to always see if we can calculate the collisions with the x axis (by putting y = 0) and the collisions with the y axis (by putting x = 0) but im also like just a high schooler myself so i suppose you should wait for a more learned person to answer too 😅

[tribal flax]

i dont have the equation for f(x)

so i couldn't really make x = 0 or anything but i didnt think you could tell either, i was just hoping it was something i didnt know

thank you though

i'll just hope that this still gets marked right in the test tomorrow

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[deft briar]

ah well
if there was indeed something you were meant to know please do tell me too im curious now!

i dont know much math but id at least like to know functions right 😦

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[torn jolt]

What theorem did they use?

Det of the following is 0

9 9 9 9 9
9 6 9 9 9
6 9 6 9 9
9 9 9 6 9
9 9 9 9 6

[languid junco]

if two column vectors in a matrix are the same, the determinant is 0

(since this means they are linearly dependent)

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@torn jolt Has your question been resolved?

[deft briar]

what does that mean exactly?
we were taught that if two rows or columns are the same the det is 0 but no one ever told us why 😅 just had us calculate a few times to see its true

[torn jolt]

Oof I tried this and checked like 3 tims to find if any of the rows are the same
And guess what col are the same

Thank you so muchh bothh

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[wet wave]

I'm pretty confused on this question and the given solution. Why is the function f given by that integral expression?

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@wet wave Has your question been resolved?

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@wet wave Has your question been resolved?

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[hushed barn]

Hi, I'm so sorry if this is not math related but I was wondering about an experiment. Say people have to pick 7 integers between 1-100 (repetitions are allowed), and the winning number is decided by the largest value of n/(number of attempts of that number) for each n. What would be the best number to pick from and why, given that the people in the experiment are fairly smart? Sorry for the inconvenience

[hushed barn]

I would like to know the probability and statistics behind this

[wild sleet]

unknown number of people who each pick seven numbers?

[hushed barn]

say 100 people

[wild sleet]

what's 7 then

[hushed barn]

each person can pick 7 integers between 1-100 (they can be the same)

[wild sleet]

there's no point repeating youself right?

[hushed barn]

yeah theres no point

[wild sleet]

it would likely involve introducing randomness, like flipping coins

so there's a chance that you avoid everyone else's pick

can't say anything more, too hard

[hushed barn]

alright thanks for helping

ill see what i can do

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[dark cedar]

Stuck with the last question

I honestly don’t know what it’s trying to say

:(

It’s for the last question

Can anyone point me in the right direction?

[dusky locust]

r is the area of the paralelogram

[dark cedar]

Alrighty

[dusky locust]

You are maximizing the area

to find the upper bound

I actually don't know how to do that, but shouldn't be hard

[dark cedar]

Umm

[dusky locust]

it's a determinant

[dark cedar]

Any ideas cause I don’t either

[dusky locust]

First of all

[dark cedar]

Could possibly use the formula from above

And using the magnitude of both sides

[dusky locust]

Yeah

then it's easy

[dark cedar]

I don’t know what to do at this point tho haha

[dusky locust]

cause | a x b | = |a| |b| |sin theta|

|a| = |b| = ?

[dark cedar]

Ye ye

1

?

[dusky locust]

why?

[dark cedar]

Just going off the points they gave

They all seem to have length 1

Cause it’s the radius perhaps?

[dusky locust]

What is a unit circle

[dark cedar]

Oh man

[dusky locust]

all vectors in it have unit size

[dark cedar]

Radius 1

Yep

It says unit circle omg I skimmed over that

[dusky locust]

so | a x b | = | sin theta|

[dark cedar]

Sorry

[dusky locust]

What are the possible values for |sin theta|

[dark cedar]

0 to 1?

[dusky locust]

yes

So what can we say about r

[dark cedar]

It’s between 0 and 1?

[dusky locust]

hmm

look at z

[dark cedar]

Or am I missing something

Mhm mhm

Which z we talking

[dusky locust]

the one in the question

OA x OC

[dark cedar]

Yep yep

Alrighty

What about z are we looking at?

[dusky locust]

What is r for z

[dark cedar]

By r do you mean radius

[dusky locust]

yes

[dark cedar]

1

[dusky locust]

By r, i mean the r in the last question

[dark cedar]

The r <0,0,1>

[dusky locust]

yes

[dark cedar]

-1 for z

[dusky locust]

yes

[dark cedar]

Omg

I just..

I’m sorry

That was

Brain rot from me

I just understand what it was asking

😭

Got it right

Thank you so much

[dusky locust]

nice

[dark cedar]

Way more difficult then that should of been

Have a good night

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[raw ruin]

I need help with B.

Is it (1,2,3,4,5,6), (1,2,3) or (1,2)?

[velvet sedge]

what part of B

[raw ruin]

First number

[velvet sedge]

well you're tossing a fair coin 3 times

you can use binomial distribution to figure this out, or just write out all possible events with 2 heads and 2 tails out of 3 trials

[raw ruin]

How do I write that?
Like?

1. HHT
2. THT
3. HTH
   ...

[velvet sedge]

yea

there's 8 possible ways in total for a trial of 3 coin flips to go anyways so its not much work, so even less if you're considering the cases with 2 heads and 2 tails

[raw ruin]

Repitition isn't allowed right?
So no

1. HHT
2. THH

[velvet sedge]

it is allowed

[raw ruin]

Ohh

I don't get the idea

How many
HTH
THT

Are there in total?