help-23

[delicate sphinx]

Like with $\sqrt{x + 3}$ you can clearly see that $\sqrt{x}$ is the outer and x + 3 is the inner

[flat frigate]

CaptainNova22

[delicate sphinx]

With (5x+3)^4, it's a bit harder but it's got an exponent, like x^4 is a function

So something like (3x^2 + 7x + 23)^34, the inner would be the stuff in the parentheses so 3x^2 + 7x + 23 and the outer is x^34

[viscid crane]

Last question

So the only way we use the chain rule

Is if it is composite ?

[delicate sphinx]

Yeah chain rule is applied to composite functions

[viscid crane]

Thank you : )

I appreciate ur help a lot

.close

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[grand shoal]

can enyone tell me what has happened to the negative

[grand shoal]

<@&286206848099549185>

[severe pond]

they switched the order of subtraction

[grand shoal]

is it not lower integral minues uper integral tho?

[severe pond]

$-\frac{1}{2}(e^{-b^2} - e^{-1}) = \frac{1}{2}(e^{-1} - e^{-b^2})$

[flat frigate]

knief

[severe pond]

no it’s upper limit - lower limit

always

and it’s not upper/lower "integral"

[grand shoal]

oh my days, you are right my brain rot is somewhere else, and im in seconed year uni doing maths

thanks

[severe pond]

you’re welcome

[grand shoal]

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[desert iron]

how to solve this

I don't know how to do it using counting principle

there's too many branching

[plucky elk]

#old-network for bio server

[desert iron]

bro that's a course in statistics 😭

it's on mit ocw

got nothing to do with bio

I'm a physics major

lol

[rain gale]

How would you think about the problem if it was just 1 of each nucleotide?

1 A, 1 C, 1 G, 1 T

[desert iron]

well you can't do that because it's a sequence of 8

[rain gale]

I'm giving you a different problem

A sequence of 4 using 1 of each

[desert iron]

permutations

432*1

[rain gale]

Ok cool

How about sequence of 5, with 2 A and 1 of the rest?

You don't have to give me an answer, just how you would think about solving it

[desert iron]

donno

it depends where to put the A's

[rain gale]

Ok

But if the As were different somehow it would be easy right?

It would just be 5 x 4 x 3 x 2 x 1?

[desert iron]

aha

oh

it doesn't matter

?

[rain gale]

No it does matter

It matters because the As are not different

Im just trying to relate this to a problem that we already know how to solve

So let's first pretend that the As are actually different

Call them 1 and 2

I am ordering a sequence of CGT12

You agree that there is 5x4x3x2x1 ways to order this right?

[desert iron]

yup

[rain gale]

Ok

I will now claim that for every ordering where the 1 comes before the 2

There exists exactly 1 other ordering where the 2 comes before the 1, but the 2 and the 1 occupy the same positions

For example, 1CG2T and 2CG1T

If we change this back to As, we get the same sequence; ACGAT for both

There fore we double counted

[desert iron]

yeh right

so it's gonna be 2 * 5p5 ?

right

[rain gale]

So therefore the amount of orderings of AACGT is the same as orderings of 12CGT / 2. This is 5! / 2

How would you order something like AAAGG?

[desert iron]

oh it's divided because they're the same

oh I see

[desert iron]

I donno 😭

oh wait I think I'm onto something

[desert iron]

here you said those two are the same

[desert iron]

but here for A it'll not be two but 6 right ?

so in terms of A's we should divide by 6 somewhere

alright

I geuss it's $\frac{5!}{3!*2!}$

right ?

[rain gale]

very close!

acutally exactly close but

i would be more precise here and put 2!

instead of 2, just becauase it shows more logic

but yes

[desert iron]

yes

[flat frigate]

Hamdy Hisham

[rain gale]

ok so you should be able to solve AAAACCGT now, right?

🙂

[desert iron]

so in case of the first one it'll be $\frac{8!}{4!*2!}$

[flat frigate]

Hamdy Hisham

[desert iron]

am I right ?

[rain gale]

yep!!

good job you caught on fast 🙂

[desert iron]

thnx bro

that helped alot

[rain gale]

glad to help

[desert iron]

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[hasty wagon]

I literally do not get this

Please help

[severe pond]

$A = A_0(0.82)^t$

[flat frigate]

knief

[severe pond]

if t = 4

$1.52 = A_0(0.82)^4$

[flat frigate]

knief

[hasty wagon]

I already did that

[severe pond]

then what did you get for A_0

because you wrote 4/(0.82)^0

[hasty wagon]

I was trying a different method

[hasty wagon]

I got 1.07

[severe pond]

,calc (1.52)/(0.82)^4

[flat frigate]

Result:

3.3619262209366

[hasty wagon]

Oh

I actually might be brain dead

Thank you

[severe pond]

all good

you’re welcome

dr phil

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[livid bronze]

How to find centroid or orthocenter or invented or circumcenter whag are the differences

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[prisma wren]

https://www.mathsisfun.com/geometry/triangle-centers.html

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[exotic charm]

to show an isomorphism exists between groups, if i can show a bijection between the two groups exists then can i just show that the order of each element has a corresponding element of the same order ?

[mortal sandal]

For infinite groups definitely not: take an infinite product of Z2s with one Z4 and an infinite product of Z4s

Very unlikely to be true for finite groups either but thinking of a counterexample might at least be amusing

Can't find any with |G|<12

[exotic charm]

aa

[mortal sandal]

In any case the bijection itself certainly isn't guaranteed to be an isomorphism

[exotic charm]

ye

so if i had a bijection and i could show that each element has a corresponding element of the same order

then that is not enough to show they are isomorphic i assume

[mortal sandal]

Presumably not

[exotic charm]

does that mean im going to have to work through every element to show it aaaaaa

as in show a homorphism

[mortal sandal]

Depends

On what the groups are and what description of the function you have

[exotic charm]

eg im working with a group of 6 elements and another group of 6 elements and i know they are isomorphic and ive defined a bijection already but to prove the homorphism property will i need to verify that theta(xy) = theta(x)(theta(y)

for every element

[mortal sandal]

are they D_3 and S_3?

[exotic charm]

ye

[mortal sandal]

There's a visual argument about why the symmetries of a triangle are S_3 though it's not a formal proof

Basically you're permuting the vertices around

Do you have a set of properties which you know is sufficient to "define" D_3

A presentation

So like <s^2 = 1, r^3 =1, rs = sr^-1>

[exotic charm]

uhh not sure

[mortal sandal]

ah

maybe if you choose one element of S_3 to be f(r) and one to be f(s)

you can show that they satisfy the defining properties of D_3

So like multiplication in D3 is defined a certain way

Use that definition to show that $f(r^as^b)f(r^cs^d)=f(r^as^br^cs^d)$ directly

[flat frigate]

Dreyuk

[mortal sandal]

you'll probably need to manually verify $f(r)^3 = 1$, $f(s)^2 = 1$, and $f(sr)=f(r^{-1}s)$ manually to do this proof, which it sounds like you've done the first 2.

[flat frigate]

Dreyuk

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[fresh birch]

Given these equations, how come the one w/ 2x can be written both ways, but not the one w/, say, 7e?

[fresh birch]

If that makes sense

[quiet plume]

It's the same thing happening.
The first one is really $1 \cdot \frac{-2x - 3}{x-1} = 1 \cdot \frac{-1(2x + 3)}{x-1} = -1 \frac{2x+3}{x-2}$

The second one is $\frac{3e^5 - 1}{4} = \frac{1}{4} (3e^5 - 1) = \frac{1}{4}\cdot -1 (-3e^5 +1) = \frac{-1}{4} (-3e^5 + 1)$

[flat frigate]

Azyrashacorki

[fresh birch]

What abt (x-7)/-2 and -(x-7)/2?

@quiet plume

[quiet plume]

This is just $\frac{-a}{b} = \frac{a}{-b} = - \frac{a}{b}$

[flat frigate]

Azyrashacorki

[fresh birch]

U can do this?

Is that the case for both (x-7)/-2 and (-2x-3)/(x-1)?

[quiet plume]

$\frac{-a}{b} = \frac{a}{-b} = - \frac{a}{b}$

[flat frigate]

Azyrashacorki

[quiet plume]

Essentially, yes.

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[magic patio]

can yall guide me step by step cuz I dont know

[swift loom]

recall each quarter is 3 months

that should be your growth period

[magic patio]

is this future value of simple annuity?

[swift loom]

yes

[magic patio]

my teacher said use this formula F = A ((1+j)^n-1/j) if future in simple annuity

[swift loom]

yes

just plug in the values

convert 6 years + 3 months all into months

and devide

to get the number of deposits

[magic patio]

let me plug the values

whats gonna be my j

[swift loom]

its compounded interest quarterly

there are 4 quarters per year

[magic patio]

0.07 divided by 4 then?

[swift loom]

yes

[magic patio]

139,623.56618234253 pesos

i got this

oh wait

can u check

i think my calcu broken

[swift loom]

139623.566182

yes you are correct

[magic patio]

should I round it off

to 57

[swift loom]

i believe so

[magic patio]

okay thank!

s

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[strange trout]

How should I approach this problem?

[quiet plume]

Let's say you were given some function $F(x)$ such that $\frac{d}{dx} F(x) = f(x)$.

[flat frigate]

Azyrashacorki

[quiet plume]

What would the integrals yield?

[strange trout]

it would yield the area under the curve?

[quiet plume]

Have you learned something called the fundamental theorem of calculus?

It kinda looks like $\int_a^b f(x) = F(b) - F(a)$

[flat frigate]

Azyrashacorki

[strange trout]

yeah

OH wait i think i have the answer but idk if it’s right

actually wait nvm im still lost

[quiet plume]

Well let's say you use that theorem

With the F as I defined it

[strange trout]

both of them?

[quiet plume]

What do you get for $\int_a^x f(t) dt?$

[flat frigate]

Azyrashacorki

[strange trout]

F(x)-F(a)

[quiet plume]

Ok. Now you take the derivative of that. What's left?

[strange trout]

so i do (F(x)-F(a))’ ?

[quiet plume]

Yes

[strange trout]

f(x)???

[quiet plume]

Yep

[strange trout]

wait so the question is asking us to evaluate 2 equations?

[quiet plume]

They just want the derivatives of $\int_a^x f(t) dt$ and $\int_a^b f(t) dt$

[flat frigate]

Azyrashacorki

[quiet plume]

The idea is that it gets you to see that derivatives "cancel" integrals out

At least for the first one

[strange trout]

ohh

i see

[quiet plume]

What do you think you get for the second one?

[strange trout]

F(b)-F(a)

but they’re both constants

[quiet plume]

Hmmm

So then what's the derivative

[strange trout]

OHHHHH

derivative of constant: 0

[quiet plume]

So the exercise is showing you that (1) derivatives and integrals are "inverses" and (2) definite integrals with constant bounds are constants

[strange trout]

i think i need to learn how they write the question bc that’s why i got confused

ty

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[lean otter]

Can I get help with this?
\sum_(k=0)^(\infty ) ((-1)^(k)\pi ^(2k))/(3^(2k)(2k+1)!)

[lean otter]

I want to make sure I did it correctly

I rewrote it as this...
(3)/(\pi )\sum_(k=0)^(\infty ) ((-1)^(k)((\pi )/(3))^(2k+1))/((2k+1)!)

And got this as my sum
(3)/(\pi )*sin((\pi )/(3))

[hasty marsh]

$\sum_{k=0}^{\infty}((-1)^{\frac{k}{pi ^2k})/(3^(2k)(2k+1)!)$

[flat frigate]

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

[lean otter]

Ill post the ss

Original problem

I approached it like this

Did I find the sum correctly?

(3^(3/2))/(2pi)

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@lean otter Has your question been resolved?

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[verbal trout]

Is the answer 14?

h'(x) = f'(f(x)) * f'(x)
h'(1) = f'(3) * f'(1)
h'(1) = 7 * 2
h'(1) = 14

Idk if I did smth wrong in step 2?

[severe pond]

good

[verbal trout]

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[bitter steppe]

how'd he compute this

.close

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[bitter steppe]

aight

.close

[eternal carbon]

it is already closed…… lol

[noble kiln]

Is closed alr

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[lean otter]

Cn someone help me do 2)a)

[noble kiln]

Think of rolling out the cone

What shape would you get

[lean otter]

Uhm

A circle and a triangle😭🥲

[sleek sentinel]

were you given formulas?

[lean otter]

Yes

[sleek sentinel]

which formula do you think might be relevant here

[lean otter]

πrl+πr²

[sleek sentinel]

that looks right yeah

[lean otter]

But im getting the wrong answer

It should be 204cm³

When I use that equation im not getting it

[sleek sentinel]

that's the next bit - the question is asking for the curved surface area

ie. without the base

[lean otter]

Ohhhh so just πrl

[sleek sentinel]

yeah

[lean otter]

Ooo

Thank u very much

Have a nice day

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[rocky sedge]

help with probability question?

"A thief was seen jumping into a taxi and disappearing. An eyewitness told the police that the taxi was yellow.
To verify if the testimony is useful, the police investigator analyzed the situation and produced the following data:
in that city, 80% of the taxis are black and 20% are yellow,
from past experience, eyewitnesses are reliable in 80% of the cases
assuming the taxi was reported as yellow, what is the probability that the testimony is correct?"

i tried calling a reliable/true testimony T and a yellow car Y, where we are looking for P(T|Y) and we know that P(T)=0.8, P(Y)=0.2 but i got nowhere
i don't know how to solve it

[compact rampart]

id think the factthat eyewitness are 80%reliable has anything to dowith this

cuz the eyewitness predicting the correct taxi

should be an independent event ?

[rocky sedge]

idk, this question really stomped me

[compact rampart]

well whats the answer

i mean if u got an answer key or smthn

[rocky sedge]

supposedly 0.5 but take it with a grain of salt

[rigid inlet]

I think you'll have better luck if instead of considering the result of the testimony, your events be "the testimony is yellow/black" and "the taxi was yellow/black"

a true testimony envelops two events: the taxi is black and the eyewitness says it's black, and ditto for yellow

[rigid inlet]

so this hopefully segregates the information a little better

actually having thought about it for literally 90 seconds it's perhaps not so simple

maybe this question isn't great

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[rocky sedge]

.reopen

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✅

[rocky sedge]

wait i think it worked for me

i called a testimony of yellow car T and actually a yellow car Y, we want to find P(Y|T), we know P(Y)=0.2, P(T|Y)=0.8 using the complete probability law we get for P(T)=P(Y)*P(T|Y)+P(Y^C)*P(T|Y^C) which is equal to 0.2*0.8+0.8*0.2=0.32 and then using Bayes theorem we get P(Y|T)=P(T|Y)*P(Y)/P(T) -> = 0.8*0.2/0.32 = 0.5
and this is suposedly the correct answer

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@rocky sedge Has your question been resolved?

[rigid inlet]

Fair enough. Admittedly it was 3:30am when I typed that (and 5:15 atm) so I can't necessarily verify, but if it matches then I guess we're good

[rocky sedge]

bro get some sleep, what are you doing awake at these hours?

thanks a lot btw

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[lean otter]

Im solving for systems of linear equations in 2 variables but idk how do to this? I don't know where to start supposedly I could use a substitution method or the elimination method but I'm not sure how to implicate it into this

[frozen marlin]

express one variable in terms of the other, and substitute that expression into the second equation

although you can also see a very obvious discrepancy here

[desert juniper]

you're given the graph, so i'm assuming they ask you to draw the equations of both lines

[frozen marlin]

"solving for systems of LE in 2 variables"

hm

actually imma let felix handle this

[desert juniper]

Do you know how to draw each of the equations, butter?

[frozen marlin]

(the graphs for each of the equations)

[cedar owl]

This is a nice one you probably need to put the things into y=mx+b form

[lean otter]

Umm no

This is a new unit we just started and the teachers not here

[desert juniper]

okay, so you know that you have a linear equation, aka, its graph is a line

to draw a line, you need two points

two easy-to-obtain points on a line are the x-intercept (when y=0) and the y-intercept (when x=0)

let's focus on the first equation,
3x-4y=12

if x=0, how much does y need to be for the equation to hold?

[lean otter]

Uhh

It equals 12 right?

So 12?

Should I just solve using to find x and y intercept and then plug it into slope intercept form

[desert juniper]

if x=0, you have -4y=12, which does not hold for y=12

[lean otter]

O

.

[lean otter]

-3?

[cedar owl]

Yay

[desert juniper]

good. So you have one point, (0, -3)

now, can you draw that point?

[lean otter]

This?

[lean otter]

^

[desert juniper]

yeah

and i see that you have also the other intercept

[lean otter]

NEVERMIND I GOT IT

[desert juniper]

now do the same for the 2nd equation, and draw both lines

[lean otter]

Is this correct

Bc parallel lines are no solution

[desert juniper]

i mean, your parallel lines are crappily drawn, but yes

[lean otter]

Yeah erm I'll fix that 😭😭

.close

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[lean otter]

How do I look for x? Do I just plug it in? 😭 idk

[lean otter]

I just need affirmation that this is correct 💀 thanks

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[lean otter]

u must create a function ?

@modern bloom

[honest perch]

x^2

[safe radish]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[frozen marlin]

provide an example of

[modern bloom]

Aah shoot I'm stupid, tysm

[safe radish]

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[honest perch]

💀

[modern bloom]

Uhh what's wrong

[safe radish]

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[lyric belfry]

I'm having trouble with question B here

[lyric belfry]

I've determined the kernel. its a set containing 2 matrices, so the dimensionality of the kernel is 2

the rank nullity theorem says that the dimension of the kernel and the dimension of the image should add to the dimension count of the domain vector space

I'm not entirely sure what the question wants from me?

the domain vector space is of shape R(2x3)

the dimensionality of that is just 6

2 + 4 = 6

rank nullity theorem is not so difficult, surely thats just not it?

[safe radish]

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[lyric belfry]

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[full gust]

can anyone help solve this?

[full gust]

@vocal rampart

guys anyone?

i would really appreciate it

<@&286206848099549185>

<@&286206848099549185>

[safe radish]

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

[full gust]

i dont know where to begin

and i have to submit the answer in 8 minutes

so i just need a solution, and i can understand how to do it later

?

what did u ask?

[full gust]

.

nvm, the times over, it doesnt matter now

.close

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[earnest rain]

Usually i change the x to y by drawing it

But I cant draw it like this

<@&286206848099549185>

[safe radish]

@earnest rain Has your question been resolved?

[earnest rain]

Is like this fine <@&286206848099549185>

Given the domain

Make it based on y

Idk how to say

[earnest rain]

<@&286206848099549185>

[earnest rain]

.close

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[livid vector]

Hi. Im stuck on this problem. I'm supposed to sketch the graph of f' below it, but I don't know how. Where do I start?

[median vigil]

what sort of features from f might we expect to show up on the graph of f' ?

[livid vector]

An increase and a decrease?

[median vigil]

yes, how would the fact that f is increasing show up on the graph of f' ?

[livid vector]

im not sure

[magic tendon]

If a function is decreasing in an interval, what would be the sign of f'(x)?

[livid vector]

negative on the interval?

[magic tendon]

Yup
And what would the sign of f'(x) if f(x) is increasing in an interval?

[livid vector]

positive?

[magic tendon]

Yup
Do you know what are these points called?

[livid vector]

would those be the maximum and the minimum?

[magic tendon]

Yeah they are local minimal and local maxima
They are called critical points in general
Do you know what would f'(x) be equal to at these points?

Tip: the slope of the tangent of curve at these points is 0

[livid vector]

would it be 0 then?

how would i apply this to the graph of f'?

[obsidian jetty]

Like find how the slope of the tangent line moves

[livid vector]

Is there an equation for that?

Or is it just by seeing like the curves and the increase/decrease?

[obsidian jetty]

Here your function is like a 4th power polynomial so the f prime would be a 3rd

[livid vector]

ohhh

[obsidian jetty]

Wait

I ll do it

The blue line is the f prime

See like when you start to see the red line from the left you can see that it starts to decrease really quick but it slowly stops at like -1
There the slope of the tangent becomes 0, the red line then increases a bit but it stops at 0, where it starts decreasing again. After 0 it does the same but opposite way

f'(x) represents the slope of the tangent line of f at every x, as a function of x

You get that?

You must be able to visually draw the tanget lines at every point

[livid vector]

Ohhh okayy

I'm gonna try to apply that to another problem

Tysmm

[obsidian jetty]

Alr bro

https://youtube.com/shorts/8Q5HsVrcu6g?si=5QFBRr0__9imRvaJ

Check out some yt videos

Basically it shows that if the tangent is pointing down from the right then the slope is negative, if it is horizontal the slope is 0
If it points up it is positive etc

[safe radish]

@livid vector Has your question been resolved?

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[urban eagle]

Am I just doing
(sinh(x+h) - sinh(x))/h = sinh(h)/h
?

[plucky elk]

no it says "calculate the derivative"

[urban eagle]

So the identify the point part means nothing I’m so confused

[plucky elk]

It does

you're supposed to find what the point is

did you learn the formula for the difference quotient ?

[safe radish]

@urban eagle Has your question been resolved?

[urban eagle]

Is it just
(f(x+h) - f(x))/h

[plucky elk]

yes

find x

that's "the point"

but really the point is (x, sinh(x))

[urban eagle]

Find x so that the difference quotient is just sinh(h)/h ?
So x is just 0?

[plucky elk]

correct

[urban eagle]

Ok and then am I just differentiating sinhx plugging in 0 and then saying the lim h -> 0 … = 1

[plucky elk]

sounds right.

[urban eagle]

Ok ty

[plucky elk]

,w diff sinh(x)

[flat frigate]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by rie.mann

[plucky elk]

,w cosh(0)

[flat frigate]

Results provided by WolframAlpha

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Requested by rie.mann

[urban eagle]

.close

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[lyric haven]

where dufuq do i begin

[lyric haven]

this is all i got

im going to assume i need to use the pytag equation at some point and know the area of a triangle

yeah im kinda lost after that

[near shoal]

Let me see

I will just try typing formula with my phone don’t want to get up

[lyric haven]

💀

[near shoal]

\begin{equation} 4^2- (\frac{x-4}{2})^2=h^2 \end {equation}

\begin{equation} 4^2- (\frac{x-4}{2})^2=h^2 \end{equation}

[flat frigate]

meow

[near shoal]

x is the base basically and then you find the derivative I think

Because you have the area formula too

$\frac{x+4}{2}$

[flat frigate]

meow

[near shoal]

Do you get it already

[lyric haven]

im probably going to need a fat min, gimme some time and i’ll try to work it out now.

[near shoal]

Yes it’s shaped I am sorry I made it more complicated then needed

[random cobalt]

Are those triangles 45 45 90?

[lyric haven]

bro i got no clue

[near shoal]

The height can be fixed though

Other probably arbitrary

[safe radish]

@lyric haven Has your question been resolved?

[lyric haven]

trust, im working still

all i got so far

[lyric haven]

hmmm that cant be right

im going to change B-b to b-a cuz that confused me

i think im cooked ngl

[lean otter]

whats the original quesiton

[lyric haven]

#help-23 message

[lean otter]

is this maximazation/minimization in calc?

[lyric haven]

si

[lean otter]

what is h supposed to denote

a?

height of trapezoid?

[lyric haven]

this

[lean otter]

do you know what the area of trapezoid is?

its

A=h/2 (L + l)

also your equaiton is wrong for the height

use pyth. theorem

[lean otter]

try again

[lyric haven]

ok

yeah i give up ngl

.close

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[mortal ore]

what does the := mean here?

[mortal ore]

if I want to calculate cosine similarity and I have A and B do I need to mess with the cos(theta) part?

[austere cypress]

:= means "is defined as"

[mortal ore]

okay

seems close enough to = to treat it as such

[austere cypress]

it acts the same way as =, just a notation to indicate that it is a definition and not a derived formula or something

[mortal ore]

thank you

.close

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[halcyon fulcrum]

The quadratic equation x^2 - 2x - 1 = 0 has solutions x = r and x = s.
Determine integers b and c for which the quadratic equation ax^2 + bx + c = 0 has solutions x = 2r + s and x = r + 2s

what I have so far:

x^2 - 2x - 1 = 0, roots are 1 + sqrt(2) and 1 - sqrt(2)
r = 1 + sqrt(2), s = 1 - sqrt(2)

r1 + r2 = -b/a, r1r2 = c/a

so: r + s = (1 + sqrt(2)) + (1 - sqrt(2)) = 2, r * s = (1 + sqrt(2))(1 - sqrt(2)) = -1

(x1) + (x2) = 2r + s + r + 2s
= 3(r + s)

(x1)(x2) = 2r^2 + 4rs + rs + 2s^2 = 2r^2 + 5rs + 2s^2

then i put in -1 into the rs

[halcyon fulcrum]

this is where im stuck

[lavish lichen]

Hope this helps

[halcyon fulcrum]

can I ask a question

[lavish lichen]

Sure

[halcyon fulcrum]

did you go 2r + s = 2(1 + sqrt(2)) + 1 - sqrt(2)

for the first new root

@lavish lichen

[halcyon fulcrum]

shouldnt it be 2sqrt(2) instead of 4sqrt(2)

[lavish lichen]

Where exactly?

[halcyon fulcrum]

2r + s
r is 1 + sqrt(2) s is 1 - sqrt(2)

[lavish lichen]

Oh yeah sorry man , you're right

My answer is completely wrong then

[halcyon fulcrum]

<@&286206848099549185>

[halcyon fulcrum]

a

[lavish lichen]

Oh wait , i did it again , only the constant c changes to 7 instead of 1.

[halcyon fulcrum]

i reviewed

-6 and 7 should be the solutions lol

for b and c

why 11

what did you do the last steps

for the 7?

I got the -6 value but idk how to get the 7

yeah

I have that

and then i subbed in -1 for rs

can I ask a question

since we know the r and s values. why cant you just slap them into the new expressions you have. it works for the 3(r + s) but you get 6 if you do it for the 2r^2 + 2s^2 - 6

2.718

[halcyon fulcrum]

@languid pawn mb can you help me here

why not

its not 0 lol

its any real number

wait i thought he put 10x

[halcyon fulcrum]

@languid pawn do u understand my questoin

before the other person asked his problem

im sorry if i didnt clarify it

lol

i didnt do anything to offend you

do you not understand my question? or are you offended by the fact that I didnt answer his question seriously which he didnt bother to put into his own channel

the issue here is that i was cut off

i was just joking lol

there's no reason to be mad; im sorry

69 is not a insult

[hard crest]

yeah sorry we can't have 12yo here by discord policies

i feel bad when that happens

yeah

i'd say i see about one every couple of weeks

yeah

think the youngest i've seen was 8 years old

[halcyon fulcrum]

<@&286206848099549185>

@languid pawn lol clearly ur not willing to help me

For making a 69 joke

Did that strike too close?

.close

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[lapis pebble]

.

Help

[lapis pebble]

idk why my mind goes blank while doing math MCQs I could do the non MCQ ones but the same one when is asked in MCQ some other way i can't get it also on my exam there's to be 45 marks of MCQ is there any faster way to do MCQs.
Also I'm having some doubts on this one

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@lapis pebble Has your question been resolved?

[lapis pebble]

<@&286206848099549185>

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@lapis pebble Has your question been resolved?

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[molten pawn]

Hi! I was redoing some hw and forgot how to get Total Fixed Costs.

[safe radish]

@molten pawn Has your question been resolved?

[molten pawn]

<@&286206848099549185>

[young nexus]

fixed overhead times production per year calculated per month

[molten pawn]

Thank you!

[molten pawn]

Can you put it in the formula for me? I’m a visual learner

[young nexus]

its a formula: what is the fixed overhead?

[molten pawn]

3

[young nexus]

whats the production per year?

[molten pawn]

572,000

[young nexus]

so fixed overhead times production per year is?

[molten pawn]

1,716,000

[young nexus]

how many months do you have in one year?

[molten pawn]

12

OHHHH

Thank you sm!!

[safe radish]

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@modern bloom Has your question been resolved?

[safe radish]

@modern bloom Has your question been resolved?

[eternal carbon]

think about what happens when both of these values are positive, or if one is negative

[safe radish]

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[fervent pecan]

I ain’t got a clue on how to solve question 16

[lean otter]

use distance formila

or discriminant

[fervent pecan]

How would I go about discriminating I haven’t been taught distance formula yet

Discriminant

[lean otter]

change the y in the circle eq

[fervent pecan]

Change the y to 12?

[lean otter]

no mx + 12

[fervent pecan]

Oh okay

If I square mx do I get mx^2 or m^2x^2

[lean otter]

the second one

[fervent pecan]

Idk if I’m making a mistake is this right

Oh wait nvm I think I know what I done

.close

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[obsidian violet]

does this proof work?

[safe radish]

@obsidian violet Has your question been resolved?

[nimble panther]

yes its correct

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[sleek comet]

Draw the domain of the function: $f(x,y) = \sqrt{\frac{1}{x^2+(y-1)^2}-1}$

First have to solve the denominator being different than 0

$x^2+(y-1)^2=0$

$x^2=(y-1)^2$

$|x|=|y-1|$

$x=y-1 \lor x=-y+1$

$y=x+1 \lor y=-x+1$

Second, the stuff under the square being >= 0

$\frac{1}{x^2+(y-1)^2} - 1 >= 0$

$\frac{1}{x^2+(y-1)^2} >= 1$

$1 >= x^2+(y-1)^2$

Here I can use the equation for a circle $(x − a)^2 + (y − b)^2 = r^2$

Which means it's a crircle with a center at (0,1) and a radious of 1.

Now I can combaine the two answers and get this (the straight lines are out of the domain):

[sleek comet]

The problem is the first part where according to the answer it's wrong

[flat frigate]

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

[tall bluff]

It's not x²=(y-1)²

U made a mistake there

It's just the point (0,1) which is out of the domain

@sleek comet

[sleek comet]

Oh, you're right. I forgot the minus. So it's $ $x^2= -(y-1)^2$.

Now I noticed I can also use the circle equation here to get that single point (0,1) but I'm wondering about how could I solve:

$x^2=-(y-1)^2$

[flat frigate]

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

[tall bluff]

Well if x²+(y-1)² is zero then the only solution is that the individual terms are zeros

As both of the terms cannot be negative

[sleek comet]

Makes sense, thanks!

.close

[safe radish]

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[acoustic nymph]

please help me solve this question

[patent flame]

it has no elementary solution

[nimble panther]

$u = x^n$

[flat frigate]

DW1

[near shoal]

You have to do partial decompression

First

Since the difference is one this is a bit easier

[near shoal]

It has

[nimble panther]

you mean partial fractions?

[broken forum]

yep

[acoustic nymph]

yep

[near shoal]

So it became $\frac{1}{x^n}-\frac{1}{x^n+1}=I$

For first one it is trivial

But second term

[patent flame]

the integral of 1/(x^n-1) will output a hypergeometric function

there is no elementary solution

[near shoal]

Yes it’s right

No elementary solution

Second term I felt like impossible

Either way it won’t have elementary solution

[patent flame]

if there are bounds on the integral, then techniques in complex analysis can be used, i.e. finding the poles - residues of those poles and then Cauchy's integral formula

[near shoal]

Yea

Residues to fine poles

[flat frigate]

meow

[near shoal]

I thought it might have one

🫠🫠🫠should have done more integral

Yes

Sure

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[bronze yoke]

how do I even start solving this

[young nexus]

solve them.

[fast bane]

maybe solve for a variable in terms of a for system M, then plug that into the other

[broken forum]

that sounds like the way to go

[bronze yoke]

I tried with the first equation because it looked easier

can I get anywhere with that

[young nexus]

as M has two equations with 3 unknowns its underdetermined. if N should be consistent it has also to be underdetermined, which means ine equation has to be a linear combination of the two others.

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@modern bloom Has your question been resolved?

[peak estuary]

what are all the possible jordan forms that A could have

if A^2=0

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@modern bloom Has your question been resolved?

[modern bloom]

Yea that's exactly my doubt😭

[peak estuary]

what are all the possible jordan forms of a 3x3 matrix

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[jaunty root]

I’m a bit unsure of what my answer to this question should be. Give me a minute and I’ll share my work

[jaunty root]

P=sin(x)+y
Q=(3x+y)
Py=1
Qx=3
Py-Qx=(3-1)=2
area of circle=pir^2
area of half circle= pir^2/2
area= pi(2)^2/2=2pi
2pi(2)=4pi

along the part of the curve that i added to close the curve:
r(t)=<0,6-4t> 0<t<1
r'(t)=<0,-4>
dx=0
dy=-4
(sin(0)+6-4t)dx+(3(0)+6-4t)dy
(6-4t)(-4)
integrate this over 0<t<1
integtaring gives -16.
4pi+16

I was told that it could also be 4pi for my final answer

I’m just wondering which one is correct, and if it’s 4pi, I’m wondering why

If it’s of any help, here’s where I asked about it before: #help-25 message

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@jaunty root Has your question been resolved?

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<@&286206848099549185>

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@jaunty root Has your question been resolved?

[jaunty root]

<@&286206848099549185>

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@jaunty root Has your question been resolved?

[jaunty root]

Welp, I tried

.close

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[lusty breach]

I need a general formula to find D in terms of the angle phi_2 as well as the radii of both circles

So far I have D = h cosec(phi_2) but I don’t know how to solve for h in terms of the radii and phi_2

<@&286206848099549185>

Where knief at bro help me out 😭🙏

<@&286206848099549185>

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[winter crag]

i dont know where to go

[obsidian oracle]

since you already went the way to find F(x), you can try this: $\$
Since P is a monic, F is a monic too, so $F(x) = x^2 + bx + c$. Thus use $F(-2) = F(1) = 1$ to find $b$ and $c$

[flat frigate]

rafilou is not not born in 2003

[winter crag]

ok ill try

[obsidian oracle]

(you could also go two times faster if you tried to find F(x) - 1)

[winter crag]

i got it

thank you so much!!!!

.close

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[bitter tangle]

just question 3, not really sure where to even start

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@bitter tangle Has your question been resolved?

[bitter tangle]

i got 3, need help with a diff problem now

[patent moon]

@bitter tangle Oh ok, I was just about to help you with Q3, but you figured it out

[bitter tangle]

,rccw

[patent moon]

Well we know for sure that (a) goes only from 1 to 2 for x

so (a) must be III

Also, in (c), y is always positive, so can't be anything other than IV

So far we have (a) = III and (c) = IV

[bitter tangle]

it looks like x is going from 0 to 1 on x though

oh wait

nvm

[patent moon]

Nope, it goes from 1 to 2

Ye

[bitter tangle]

the graph is throwin me off

[patent moon]

Yeah it's a bit confusing

So now we need to match b and d

[bitter tangle]

i think d is 1

since it looks like x goes from -2 to 2

same with the y

[patent moon]

Well that can't be the correct justification because that also applies for b

[bitter tangle]

ah ur right

i mean

d looks like 2

but

thats not a good enuff reason lol

[patent moon]

Yes, it does.

Yeah exactly

I think II is going in a counterclockwise, meaning that the parametric arrow is pointing in a counterclockwise direction

Oop no it's not.

it's going in clockwise

For graph (d), if you look at the section where x and y is both getting greater as t increase, it's clear that the slope in the middle for II is that section

[bitter tangle]

both x and y start negative, then climb which is in line with graph 2

is that

[patent moon]

Yes exactly.

But, x start positive actually, so I think the graph starts at the right bottom corner.

and goes clockwise direction to form the triangle-like shape.

[bitter tangle]

oh right

[patent moon]

.close

Oh I can't close it nvm you gotta

[bitter tangle]

wait for uhhhhh

b = 1 what can i say

nvmi think i got that one

thanks for ur help!!

.close

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[vague rain]

how

[vague rain]

i solve that if cos2x is cos^2 (x) - sen^2(x)

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[proper moth]

I'd take 9 common
And then put 1+cos2x=2cos^2x

@vague rain

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[sharp idol]

Without giving answers my best advice for each problem would be to write down what information you do know each pair has in common

[shut sky]

yes

i need that information

i already got 3 done

11. SAS
    Additional information needed:

The length of one of the other sides or the measure of the included angle between the marked sides.
12) SSS
Additional information needed:

The length of the remaining side to ensure all three sides are equal between the two triangles.
13)
The measure of the included angle between the two marked sides.

[sharp idol]

I believe on 13 you don’t need the angle S

[lean otter]

!STATUS

[safe radish]

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

[sharp idol]

#14 is essentially the same as #13 but mirrored

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[shut sky]

i got htis

good guys?

[sharp idol]

I think 13 and 14 should be very very similar answers, with the same reasoning basically

[sharp idol]

This applies to both of them

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[vale hearth]

Hi, I was reading about the MAPE (Mean Absolute Percentage Error) defined as : $MAPE = \frac{100}{n} \sum_{i=1}^n \frac{|y_i - \hat{y_i}|}{|y_i|}$ on this website https://medium.com/@vinitkothari.24/time-series-evaluation-metrics-mape-vs-wmape-vs-smape-which-one-to-use-why-and-when-part1-32d3852b4779. They say that MAPE favors models that under predict and give an example. In the example, they say that if the actual value is 600 and the model gives 500, the MAPE will be less than if the actual value is 500 and the model gives 600. I agree with this but to me, it has nothing to do with overestimating or underestimating. It's just that an absolute error of 100 represents a bigger part of the expected value if the expected value is 500, that's exactly how you would expect a ratio to behave. If you expect to get 400 and that you have two models ; one giving you 500 and the othet 300, the MAPE will be the same. I ran a simulation with many points of a known function and a prediction systematicly giving f + 0.1 and the other giving f - 0.1 and I found equals MAPEs. Could you tell me if I'm wrong or not in my reasonning or what the author meant when he said that.

[flat frigate]

weareinthematrix

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.reopen

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✅

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[cobalt pebble]

Any tips?

[cobalt pebble]

Diagram to accompany

I know the angle <JPR=tan-1(4/3), but cant prove <JPQ=<JPR, or maybe there is a more efficient synthetic solution

Through the software AP=PQ, but i cant arrive at that on my own too

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<@&286206848099549185>

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[cobalt pebble]

<@&286206848099549185>

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[faint seal]

could someone just confirm my answers for these

[faint seal]

(a) and (d) are subspaces while (b) and (c) are not

(b) does not contain the element 0 and (c) is not closed under addition

since (0,1,1)+(1,1,0) does not lie in the set

[obtuse jackal]

yeah, (b) is an affine hyperplane
i.e. a n-1 dimensional affine space

theorem: any affine hyperplane can be defined by an equality of this kind

and will be be a hyperplane iff it's = 0

[faint seal]

is that a theorem or a definition of what a hyperplane is?

[obtuse jackal]

a hyperplane is a n-1 dimensional subspace of a n-dim space

[faint seal]

oh, okay

hmm I think I would need to learn dimensions properly before trying to prove that

[obtuse jackal]

yeah that's why I stated it but didn't offer you to prove it

it's definitely linalg 1 material, but more advanced

[faint seal]

I think I'll encounter dimensions in chapter 2

[obtuse jackal]

in general, it's having a 1-dimensional supplement

which lets you define them in infinite dimension

[faint seal]

for this, the zero function (f(x)=0) satisfies these conditions, and if f and g are two functions which are differentiable and have f'(-1)=3f(2) and g'(-1)=3g(2) then any linear combination of them also satisfies the conditions

say the combination is h(x)=af(x)+bg(x)

[obtuse jackal]

yes

[faint seal]

then h is differentiable (which I can 100% prove I just don't feel like writing it all out) and h'(x)=af'(x)+bg'(x), so h'(-1)=af'(-1)+bf'(-1)=af(2)+bg(2)=h(2)

hence this is a subspace

[obtuse jackal]

in some time you'll be able to just say "f -> f'(-1) - 3f(2) is linear"

[faint seal]

is linear?

[obtuse jackal]

is a linear function

[faint seal]

why?

[obtuse jackal]

you've got ways to go little man

just look at the definition and it should be clear

[faint seal]

I don't get what function you defined

[obtuse jackal]

once you get there

[faint seal]

what's the domain and codomain?

[obtuse jackal]

phi: {differentiable functions on (-4, 4)} -> R

[faint seal]

oh, okay

so phi(f)=f'(-1)-3f(2)

ahh okay I see what you mean now

[obtuse jackal]

and you'll see any kernel of a linear function is a vector space

well you can prove that already, it's easy

[faint seal]

i don't know what a kernel is yet :/

[obtuse jackal]

phi^-1(0)

[faint seal]

so all functions with phi(f)=0?

so it's like a null space of phi

[obtuse jackal]

kernel is another name for null space

[faint seal]

ooh okay

once again f(x)=0 gives us b=0

and then that set is a subspace because all linear combinations of functions also satisfy integral from 0 to 1 = 0

we can take h(x)=af(x)+bg(x) again

[obtuse jackal]

(or f -> int f is linear)

[faint seal]

for this one, I assume C^2 is a vector field over C, in which case R^2 is not closed under scalar multiplication

so it's not a subspace

[obtuse jackal]

where equality of this kind is "H = Ker(phi) for some linear phi: V -> F"
This defines all hyperplanes

and all phis define a hyperplane this way

[faint seal]

oo

[obtuse jackal]

(by the rank-nullity formula)

[faint seal]

I see

[faint seal]

but there's nothing preventing C^2 to be a vector space over R is there?

[obtuse jackal]

R² is a subspace of the R-vector space C²

[faint seal]

yeah

oh wait they said complex vector space

okay

[obtuse jackal]

the third dimension is so relevant here damn

[faint seal]

(a) is of the form {(a,a,c)\in R^3}

so it should be a subspace

but B is not

[obtuse jackal]

now same but with squares instead of cubes

[faint seal]

it won't be a subspace then

[obtuse jackal]

correct

[faint seal]

because we could have (1,1,0) and (1,-1,0) as elements but (2,0,0) is not an element

so it's a subspace for odd integers only

but the complex one is never a subspace

except for if the power were 1 i guess