help-38

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[lyric kiln]

hi

[lyric kiln]

I have a question

How do I create a functoin based on range limitations

I have to create a function with a domain of all real numbers and a range of -1,0,1

[austere cedar]

Oh, like it can only output -1, 0, and 1?

[lyric kiln]

Yes

[austere cedar]

It's hard to write some functions algebraically, but it's pretty easy to describe them. I can think of one function that can do this

[lyric kiln]

I might have learnt this but may have forgotten

[austere cedar]

There's no "trick". Just give a reasonable way to take a real number in, and output -1, 0, or 1

[lyric kiln]

fuck I dont know how to do that

I also have another question

how do I do this with an absolute value

I know that the bottom cannot be 0, and the square root cannot be negative

[austere cedar]

Right. With those two restrictions, you have:
x - 9 ≠ 0
3 - |x - 6| ≥ 0

[lyric kiln]

yes

[austere cedar]

Or, that second one rearranges to:
3 ≥ |x - 6|

[lyric kiln]

theres 2 answers righ??

x-6=3 and x-6=-3

or am I wrong

[austere cedar]

That forms a useful bound!

[lyric kiln]

what do u mean brother

[austere cedar]

But remember, we're interested in ≥, not =

[lyric kiln]

right

[austere cedar]

Those equations are when |x - 6| is exactly the same as 3

[lyric kiln]

oh

[austere cedar]

,w graph |x - 6| and 3

[lyric kiln]

so we cannot go negative

[solid kiln]

Results provided by WolframAlpha

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

Requested by kaynex

[austere cedar]

You can see the blue is smaller than the yellow over a range

[lyric kiln]

yes

what does that mean

[austere cedar]

The two equations you got are the endpoints of that range

[lyric kiln]

oh yeah

[austere cedar]

So between 3 and 9

Are all allowed points

Including 3 and 9

At least, as far as the square root is concerned

[lyric kiln]

this is a bit confusing

sorry

[austere cedar]

Np that was a lot

We're interested in
|x - 6| ≤ 3

[lyric kiln]

I kind of understand absolute value but not really at the same time

so it cant be under 0 thats the first restriction right?

[austere cedar]

|x| just takes the "positive version" of x.

So for example, |3| = 3, |-2| = 2

[lyric kiln]

then we solve the inequality?

[lyric kiln]

oh ok

theres only 1 value?

[austere cedar]

It's good to know how to graph |x - 6|

[lyric kiln]

so |-3| cant equal -3?

[lyric kiln]

I know how to do transformations

[austere cedar]

|-3| = 3 always

[lyric kiln]

right by 6

[lyric kiln]

got it!!

thanks

this one video was confusing I watched

Im gonna try one more

[lyric kiln]

brother

one more questoin

does |x-6| ≤ 3 also mean |x-6| ≤ -3

or is it only +3

[austere cedar]

Just +3

Note the left is always positive

[lyric kiln]

THANKS

so we now know that it can be 9 or larger but not below that for the absolute value

and for the other one I know that it was 9 as well

wait

do I solve the inequality after |x-6| ≤ 3 or just leave it at that

like [3,9)

@austere cedar

.close

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[rancid tendon]

Hi

[rancid tendon]

Anyone which one it is

Supposed to be a fun activity but this ain’t fun ☹️

[hard fractal]

is it multiple choice or one answer?

[rancid tendon]

Assuming mutiiple

Since it says which are correct representations

[hard fractal]

im pretty sure its everything besides b

because -4 is an open interval

[trim jolt]

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.

[hard fractal]

😭

[rancid tendon]

It’s not homework

[hard fractal]

ok so basically

[rancid tendon]

It’s a game that’s supposed to be a FUN

and this ain’t fun at all

[hard fractal]

erm ok

so

[rancid tendon]

B is wrong because it’s says it’s less than or equal to x

[hard fractal]

yeah

[rancid tendon]

Is that correct

[hard fractal]

its correct

[rancid tendon]

Cause if the number line was further than that statement would be correct right

[hard fractal]

kinda but not really because its asking what the whole line is

[rancid tendon]

But what it says that in the line there’s something less than -4 right

[hard fractal]

so if the line went to something like [3,8], you wouldnt say its 4<_x<_8, you would say that it also includes 3

[hard fractal]

no, b is saying the graph also includes -4

[rancid tendon]

Okay

Can you tell me if I’m right

( is saying it is closed for

DOT

and [ is not mention/ open dot

[hard fractal]

no, the parenthesis is for the open and bracket is for the closed dot

[rancid tendon]

So opposite

[hard fractal]

yep

[rancid tendon]

Thank you very much

[trim jolt]

@rancid tendon Has your question been resolved?

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[trim jolt]

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[vast glen]

what is the square root of 47

[marble wharf]

a little bit less than 7

you could use a calculator

[robust laurel]

a bit more than 6

[night zodiac]

Use a calculator to get the decimal approximation

[solid kiln]

Result:

2.4494897427832

[robust laurel]

,calc sqrt(47)

[solid kiln]

Result:

6.855654600401

[robust laurel]

there ^^

[vast glen]

this is the result

[night zodiac]

Yeah

That's it

[vast glen]

what is the square root of 4000?

[night zodiac]

Just use a calculator

[marble wharf]

given that you managed to enter the previous question into the calculator, you will also be able to do it with that one

[vast glen]

why do I need a calculator?

[night zodiac]

Because you're basically asking to calculate something

[vast glen]

i thought this group was supposed to help me

[marble wharf]

you are asking for the value of something

your calculator can give that to you

this server is not a calculator for you

[night zodiac]

If you can use a calculator, you don't need us

[vast glen]

k

then

what should i do when the king is the only player left in chess

[night zodiac]

This is not a chess server

[vast glen]

😭

[night zodiac]

If you don't have a legitimate math question, please close this channel using .close

[vast glen]

ok

how do i review this group?

[night zodiac]

What

What are you asking

[vast glen]

spelling error sorry-

[night zodiac]

Still what are you asking

[trim jolt]

@vast glen Has your question been resolved?

[fair bison]

If you want to evaluate this without a calculator, try breaking 4000 into factors

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[boreal vortex]

can someone help me understand how i should do this

[boreal vortex]

the integration part

[haughty oasis]

use part (a) to write it as a sum of integrals

then you can substitute

[boreal vortex]

can you explain in steps

[haughty oasis]

substitute u = 2x - 1

[boreal vortex]

ahh

[haughty oasis]

will give you $\int \left(\frac{1}{u} + \frac{4}{u^2}\right) du$

[boreal vortex]

hmm how do i know when to use the u substitution

[solid kiln]

Tushar

[haughty oasis]

intuition, whenever you see that a function and its derivative are present

you can do u = f(x) and du = f’(x) dx

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@boreal vortex Has your question been resolved?

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[wraith hinge]

find the value of tg2ϴ

It seems weird to me the problem

[chilly cloak]

here, tgϴ = BM/AB

(BM=x, AB=y)

tgϴ=x/y=y/2x

[wraith hinge]

yes

[chilly cloak]

so $2x^2=y^2?$

[solid kiln]

TimK

[wraith hinge]

let me check again

[wraith hinge]

mm yep

[chilly cloak]

there was a formula for tg(a+b)

[wraith hinge]

yes, there is a formula for that, and with that you can deduce tg2ϴ

[chilly cloak]

i know that it includes tgϴ only

in expression

like 2tgϴ/(tgϴ+1)

or -1

[wraith hinge]

mm tgϴ/1-tg^2 ?

[chilly cloak]

yeah

[wraith hinge]

I can maybe put equal tg2ϴ

using the two expressions

but idk if taht would help

[chilly cloak]

tan theta is $\sqrt{2}/2$

[solid kiln]

TimK

[wraith hinge]

omg

yes

[wraith hinge]

so it's 2√2?

?

.close

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[latent ocean]

Is this correct?

[chilly cloak]

yes

[latent ocean]

bruh image not uploading

[chilly cloak]

:)

[latent ocean]

wait no it isnt

no product rule right?

shii

.close

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[balmy nova]

Im doing lagrange multipliers and I really don’t understand them from here

Any help would be appreciated

.close nvm got it on my own

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[craggy salmon]

This is what I got. Not sure how do I determine if A should be symmetric or not

I know that if A is symmetric the equation becomes much easier and comes out as :
Ax = λx (Which means the eigenvector corresponding to the smallest eigenvalue is the optimal solution)
No clue about this determinant I got though

[trim jolt]

@craggy salmon Has your question been resolved?

[craggy salmon]

<@&286206848099549185>

[wintry stag]

Usually if A is not symmetric for a problem like this you can make it symmetric under the map $A \mapsto \frac{A + A^T}{2}$

[solid kiln]

JessicaK

[craggy salmon]

Does this help in some way? Idk what symmetric under the map mean tbh

[wintry stag]

First what you wrote isn't quite right, it should be $A^T + A -2\lambda I$. And from there you should think about it in terms of eigenvectors and eigenvalues of the matrix $(A + A^T)/2$ and maybe relook at your proof for the symmetric case to see how to proceed.

[solid kiln]

JessicaK

[wintry stag]

Namely that the minima corresponds to the smallest eigenvalue of something else other than the matrix A

[craggy salmon]

Ohk I missed I matrix there. Sure I'll look into that. Thnx a lot for the help!

.close

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[delicate ether]

help <@&286206848099549185>

[glass ridge]

Question before I tell you where to go first: did the teacher tell you the names of the new theorems?

[glass ridge]

Whoa calm down lol

[delicate ether]

all of these i JUST need answer 😦

its due in 15 min

[glass ridge]

Do u wanna work through it?

Is this a quiz?

[delicate ether]

yes i just need an answer

[glass ridge]

Sorry. Against the rules

[delicate ether]

okay then an explenation sorry

[glass ridge]

Academic honesty

[dim prism]

unlucky

[delicate ether]

plz

[jaunty scarab]

what are "the new theorems"?

[delicate ether]

thats the thing i wasnt in class and like he never got me any def

[jaunty scarab]

have you not asked any classmate for their notes?

[glass ridge]

Lets start with the last one:
A = 60
C = 45
B = 180 - 60 - 45 (because the total degrees of a triangle is 180)

[delicate ether]

ok

[glass ridge]

Did u learn sine cos tan yet? Or only Pythagorean theorem

[delicate ether]

pythagoream

[glass ridge]

Do u have access to a scientific calc

[delicate ether]

no

idk whatr that is

[glass ridge]

Do you know what sine, cos, tangent is? I’m asking this to see where to start

[jaunty scarab]

seeing the problems, the new theorems should be the sine and cosine theorems

[glass ridge]

Go online and search up a scientific calc cause in order to do these trig functions you’ll need one

[delicate ether]

did it

[glass ridge]

I’m assuming that they’re asking for the law of sines

Unless they want you to draw lines using soh ca toa

[delicate ether]

no jusst sinA/a

[glass ridge]

Draw the triangle and label your sides. Now we know that all triangle degrees add up to 180, so 180-45-60 =75 that’s angle B

[delicate ether]

oh thank

[glass ridge]

Set up the sine equation we know that the sine of 45/9 = sine 60/ BC

That’s an A on the right under the sine 60. Solve for A using algebra

[delicate ether]

oh okay ill write that down

[glass ridge]

It would be easier to make it A / sin A now that I look at it but as long as u can solve it that’s what matters

[delicate ether]

whatevrs easier im sorry but i need to leave soon and we are only on numebr 1

Theorem (Sides-angles Inequalities)

leave.

.leave

[nimble stone]

.close

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[delicate ether]

.close

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[short spear]

Why is this differentiable everywhere?

I thought if the slope was different then it’s not differentiable?

[gray rapids]

is the slope different?

2x+1 is in mx+b form so we can easily tell the slope. x^2+2 we need the derivative to tell the slope, or by finding it through other means.

derivative of x^2 + 2?

[short spear]

Oh ok I understand now thanks

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[whole dove]

How can I prove that lim(x->0) (e^ax -1)/x. = a?

[whole dove]

without using L'hopital's rule

[left oriole]

hint: do you know the definition of the derivative of a function?

[whole dove]

hmm f(x+h)-f(x) over h?

[left oriole]

yea

lim as h->0 of that

the limit in your problem is basically that in disguise

for a particular f(x)

[whole dove]

oh.. you mean in this case, f(x+h) = e^ax. and f(x) = -1 ?

[left oriole]

almost

f(x) = e^ax, and f(0) = 1

so consider f'(0), the derivative of f evaluated at x=0

write out what that would look like in terms of a limit

[whole dove]

but what's the difference from using L'hopital's rule if we use derivatives?

[left oriole]

well this doesn't require l'hopital's rule, it's just the derivative of one function, and it's just the definition

no theorem needed

[whole dove]

so if I consider derivative of e^ax -1 over x at when x=0, it is the same as the limit of it?

[left oriole]

i'm gonna write it out in a suggestive form:

$$\begin{aligned}
\lim_{x \to 0}\frac{e^{ax} - 1}{x} &= \lim_{h \to 0}\frac{e^{ah} - 1}{h} \
&= \lim_{h \to 0}\frac{e^{a(0 + h)} - e^{a0}}{h} \
\end{aligned}$$

[solid kiln]

Bungo

[left oriole]

on the first line i just renamed x to h

and then on the second line i said well h is the same as 0 + h

and 1 is the same as e^(a0)

so now suppose we let f(x) = e^(ax)

then the above is the same as:
$$\lim_{h \to 0}\frac{f(0+h) - f(0)}{h}$$

[solid kiln]

Bungo

[left oriole]

which is just $f'(0)$

[solid kiln]

Bungo

[left oriole]

does that make sense?

[whole dove]

hmm I understood untill this part

ohhh just a sec

oh I see, it's just the definition of a limit

Thank you! Now i get it!!

.close

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[placid grail]

quick lil sanity check

[placid grail]

i have to show that lim as (x,y) -> (0,0) of (x+y)/(x^2+y^2) doesn't exist

I can break it down into 3 limits (as x -> 0, as y -> 0 and as x -> y) and show that they are not equal to each other

when i try to do the first one, i get $\lim_{x\to0} \frac{x+y}{x^2+y^2} = \frac {y} {y^2} = \frac{y}{|y|} = 1 \text{or} -1$

[viscid stone]

#help-21｜아리스킨충1 plzzzzz its super easy

[wraith hinge]

i dont think youre allowed to do that

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Please do not advertise your help channel or thread in other parts of the server. There are many people who need help, and no one person can be prioritized over other people, so please patiently wait. Anyone who chooses to help you is a volunteer who is doing so out of their own kindness.

[viscid stone]

what u mean

[left oriole]

wait, how did you get y^2 in the numerator, and y in the denominator?

[placid grail]

wait i messed up

[wraith hinge]

read the rules

[solid kiln]

omgatriple

[viscid stone]

ur so smart

[placid grail]

so since this is DNE, it must then follow that the limit as (x,y) -> (0,0) is DNE as well right?

[trim jolt]

@placid grail Has your question been resolved?

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@placid grail Has your question been resolved?

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[stray junco]

hiii I need some help

[wraith hinge]

divide each side by 3/5

[stray junco]

yeah i did that

would the answer be H?

[stray junco]

?

[night zodiac]

Yes

[stray junco]

ok great thank you!

.close

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[vale spire]

ln(floor(ln(n))/n compute the limit
express 1/11 as a geometric series

[vale spire]

compute the limit to infinity

nvm i think i got it

.close

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[median solstice]

,align
\int_{1}^{\infty} \frac{\ln{x}}{x^8}

[solid kiln]

Ratatosk

[median solstice]

what's the best way to approach this?

I initially tried by parts but it doesn't seem to work

[fiery goblet]

IBP works

[trim jolt]

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

[median solstice]

hmm

so, I have this right now

,align
\lim_{a \to \infty} -\frac{\ln{a}}{7a^7} + \frac{1}{7} \left( -\frac{1}{7a^{7}} + \frac{1}{7} \right)

[solid kiln]

Ratatosk

[median solstice]

but I'm stuck with the inf / inf in the ln(a) / 9a^9

[trim jolt]

@median solstice Has your question been resolved?

[median solstice]

,align
\lim_{a \to \infty} -\frac{\ln{a}}{7a^7} + \frac{1}{7} \left( -\frac{1}{7a^{7}} + \frac{1}{7} \right)

[solid kiln]

Ratatosk

[median solstice]

1/49?

ok, yeah, cool

.close

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[heavy geyser]

Quick math question
Uhh is it

y=6x-16

Or

y=6x+-16

[full dock]

Can you provide context

They mean the same thing btw

[molten crater]

If I'm not wrong it would be the 1st one, but yes there needs to be context.

[heavy geyser]

Just for slope intercept form

but I think my homework thing accept both answer

I just wanna know which is more accurate

[night zodiac]

Both are fine

[heavy geyser]

yeah I can just correct it if I get it wrong

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[earnest breach]

https://www.youtube.com/watch?v=Dh7Pnt-HxbU
Comparison Theorem Problem
So, I tried following what he did in his video but found that I got
x^3 > x^3 + 1 which is just false. Any idea why that would be?

[earnest breach]

Assuming I did the computation correctly.

[eternal adder]

they're showing that the original integral diverges

I'm pretty sure your integral converges

[trim jolt]

@earnest breach Has your question been resolved?

[earnest breach]

ohhh okay

right

So 1 be apply here right

I actually showed that it converges

[eternal adder]

what did you compare it to

[earnest breach]

He said that you could make an easier function by just ignoring the +1 in the denom. So I constructed a g(x) that was
the def integral from 1 to infin x / x^3. Which reduced to def integral from 1 to infin 1/x^2.
Then, I took g(x) and f(x) as a inequality, cross multipled, and showed that f(x) was greater

[eternal adder]

oh, I see

sure, that should work

for the last part, I normally just say

x^3 less than x^3+1 over the domain of integration, so the fraction has a smaller denominator, so this integral is larger

[earnest breach]

gotcha

ty for that confirmation

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[jagged bough]

I dont get number 4

[drifting rose]

uhhh

[jagged bough]

Do i add 5,15,45,135,405,1215?

[drifting rose]

I think they mean something like this

[jagged bough]

Ohh

[drifting rose]

but what that means idk

¯_(ツ)_/¯

[jagged bough]

Im asking the teacher and waiting for her response

Anyways about no.5

She said it's infinite

The teacher said that the formula was ^n-k √an/a1

No wait that

That's wrongg

[jaunty silo]

you're asking with question 4 or 5?

[jagged bough]

5

Im done with 4

4

[jaunty silo]

so 4 or 5?

5?

[jagged bough]

5

[jaunty silo]

Have you learned the formula about the sum of geometric series?

[jagged bough]

Can you give an example?

[jaunty silo]

Like, formula for calculating the answer for 1+1/2+1/4+1/8+...

[jagged bough]

Sn?

[jaunty silo]

yes

[jagged bough]

Sn = a1(1-r^n) / 1 - r

[jaunty silo]

did you also learned about with infinite sequence length?

Sum when infinitly many

[jagged bough]

Haven't

Cuz of the n

nth term

Unknown

[jaunty silo]

So having problem with getting the nth term of question 5?

[jagged bough]

Yes

Teacher said it's infinite

[jaunty silo]

you learned about limits too, right?

[jagged bough]

Example?

[jaunty silo]

$\lim_{x -> \infty}{1/x}=0$

[solid kiln]

Dri111

[jaunty silo]

this kinda things

[jagged bough]

Nahh

Im at grade 10

[jaunty silo]

Sorry Im not used to foregin country's qducation courses

hmm

So, you can get the sum of the sequence from 1th to like, 100th with formula

you can, right?

[jagged bough]

Yeah

[jaunty silo]

from 1th to arbitrary kth

you can

[jagged bough]

n=?

[jaunty silo]

just, random number k

[jagged bough]

K = 1 always

[jaunty silo]

No what I mean was

you can calc the sum of 1th to nth seq

then you can think as n is getting bigger and bigger

In other words, $S_n~where~n=k$ and k gets bigger

[solid kiln]

Dri111

[jagged bough]

Not really familiar with this

[jaunty silo]

$S_n=\frac{a_1(1-r^n)}{1-r}$

In this case

$S_n=\frac{(\frac{13}{2})(1-(\frac{1}{2})^n)}{1-\frac{1}{2}}$

[solid kiln]

Dri111

Dri111

[jagged bough]

Want me to put that in a calcu rq?

Oh wait

it will give the wrong answer tho

Cuz n is missing

[jaunty silo]

Have you solved any similar questions like this?

[jagged bough]

Where'd you get 1/2?

[jaunty silo]

dealing about infinite sum

[jagged bough]

Yeah but not the n

[jaunty silo]

wdym "not the n"?

[jagged bough]

I know the finite

Not the infinitw tho

[jaunty silo]

1/2 is a common ratio of that sequence from question5

[jagged bough]

Oh

What about solving the n?

Do i put 1 there?

Or 3?

Or4?

[jaunty silo]

I'd like to say

as n gets bigger and bigger

to inifty

$(\frac{1}{2})^n$ goes close to 0

[solid kiln]

Dri111

[jaunty silo]

Idk if you'd see what I want to tell

[jagged bough]

Yeah like 9+000000000000000000.1 is still 9 and divide that 0.1 to 2 is too little that it stays as 9

But infinity

[jaunty silo]

The idea I'm saying is we only know for the finite n

So we're trying to 'expand' that finite sum formula to infinite one

[jagged bough]

So we add 1 for n till the number is too small to even add +1 to a number?

[jaunty silo]

yeah so that we think $\left(\frac{1}{2}\right)^n$ is 0

[solid kiln]

Dri111

[jaunty silo]

as it becomes smaller and smaller as n becomes bigger and bigger

[jagged bough]

Can you give a clue of what n is ?

[jaunty silo]

we can't pick a exact number for n

we just have to know, and think that n is so big

or "when n goes to infinity" for what math will way

What are you exactly learning right now?

Might be a clue for intended approach

[jagged bough]

N = 41 gives me 12.9999999999

[jaunty silo]

I think calc thinks (1/2)^42 is too small to deal with

it's not 0 afterall

[jagged bough]

I dontt gettitt

[jaunty silo]

I know it sounds wierd

Later you'll learn $S=\frac{a_1}{1-r}$ as formula

[solid kiln]

Dri111

[jaunty silo]

but you said you didn't learned that yet

Sorry if I made you more confused

[jagged bough]

It's okay

Im just not gonna solve it

Thanks tho

.close

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[terse sigil]

can some one show how this step was acheived

[snow spindle]

common denominators and simplifying the fractions

6/9 -> 2/3

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[quick tapir]

Need help understanding L Hospitals rule for this question

[nimble stone]

!status

[trim jolt]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[quick tapir]

1

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[trim jolt]

@surreal shadow Has your question been resolved?

[surreal shadow]

No

I need help with part 3

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[surreal shadow]

.reopen

[trim jolt]

✅

[surreal shadow]

<@&286206848099549185>

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@surreal shadow Has your question been resolved?

[surreal shadow]

No

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[vague pier]

Lets say a function is $$R ⊆ A × B$$
If $$R$$ is a total function, then $$|R(A)| .... |B|$$

[solid kiln]

rainy

[vague pier]

Replace ..... with <,>,=,<=,>=

I'd like to see this visually to understand it better, but I dont know how

[dull temple]

functions can be represented in many different ways

I'd say the most useful one here is lines connecting elements of A to elements of B

[vague pier]

Yeah

But how would you do it?

Currently have this but idk whats next

[dull temple]

no

elements of A on the left

elements of B on the right

R is the set of arrows

[vague pier]

ah okay

and 'total function'

basically means that

the function counts for every element

in AxB

[dull temple]

no

it means that it touches every element of A

[vague pier]

Found the definition in my book; ... is a total function if exactly one arrow departs from each element in A

[dull temple]

yep

[vague pier]

so R = |A|

And then |R(A)| ... |B| is

[dull temple]

no

|R| = |A| though

R is a set

|A| is a number

[vague pier]

Yeah i meant that

😅

I still dont know

because

:

[dull temple]

what's |R(A)| ?

actually

what's R(A)?

[vague pier]

|a|

[dull temple]

like can you explain what R(A) means?

[vague pier]

R is a total function, so R(A) means

out of every element in A

there departs exactly 1 arrow

to somewhere in B

[dull temple]

you have explained what "R is a total function" means

but R(A) is a set

what is R(A)?

[vague pier]

i dont know

🙁

[dull temple]

R(A) is the set of all outputs of R if it takes an input in A

in set builder it's $R(A) = {R(a) : a \in A}$ but that might not be meaningful to you

[solid kiln]

hayley.jpg

[vague pier]

Ah okay

Makes sense

[dull temple]

so, using the vocabulary of arrows, what is R(A)?

[vague pier]

the set of all elements in B

that are mapped

to elements

in A

If im wrong, then i dont know

[dull temple]

yeah sure

that's good enough

so then how does that compare to B itself?

[vague pier]

smaller or equal to B

[dull temple]

yep

[vague pier]

Okay im sure it would be alot easier if i drew it

Ill try

wait

In my drawing

How do i know

how many elements i need to put in A

and B

or doesnt that matter

[dull temple]

doesn't really matter, what are you trying to show?

to me it seems like

you want to show that it might be equal or it might be less

so maybe two drawings

[vague pier]

I see it

No matter what

It will always be smaller or equal to |B|

[vague pier]

yes

[dull temple]

yeah

[vague pier]

Okay wow think I made it harder than this actually is

thanks for ur help hayley

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[dull temple]

haha most of the proofs in this sort of class will be like that

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[long basin]

Find their ratio of sugar per 1 ml

So basically for example

For the orange juice

2.34/40 = x/1

What is x

[fresh snow]

x is 0.0585

[long basin]

,w 2.34/40

[solid kiln]

Results provided by WolframAlpha

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Requested by umbraleviathan

[long basin]

Yeah so you're just gonna do the same process for mango juice

9/100 = y/1

[fresh snow]

and once i got both what do i do

[long basin]

Compare them

[fresh snow]

0.09

so

0.0585 : 0.09

[long basin]

Compare 0.09 to 0.0585

You can do percent difference if you want

[fresh snow]

how would i find the % difference

would it be 3.15

[long basin]

Find the absolute value of the difference, divided by whatever you're comparing to

So |0.09 - 0.0585|/0.09

Multiply by 100

[fresh snow]

so first the subtraction then the division then multiply by 100

OKAY THANK YOU I DID IT

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[vague flare]

what did they do from teh 2nd to thrid line

[trim lichen]

factored out (csc(x)-2)

[vague flare]

ohhhhh

thank you

[sweet hare]

don't try to close someone else's channel

[vague flare]

yeh

u donut

what if i need help mr ceramic

.close

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[vague flare]

NOW I AM DONE MATE

WOAHHH

[sweet hare]

doesn't matter

[vague flare]

thats racial motivated

.ban @wraith hinge

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[neon sigil]

i don't really know what im doing

[frosty glacier]

For 3, you're just stating JK + KL = JL

And give the reason yourself

[neon sigil]

im tryna do 1 rn

kinda confused

[frosty glacier]

What reasonings have you learnt?

[neon sigil]

i dont remember all of the names but
transitive
substitution
reflexive
complementary/supplementary/linear angles(?)
vertical angles

and the a=b b=a thingy

symmetry property i think

and the addition/substraction/multiplecation/division angles postulates

for 1) is the first step SI + IT = ST / RU + UN = RN?

and then angle add. postulate?

are u still here?

nvm i gtg for now

.close

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[severe stump]

I am having issues on this one.

[severe stump]

I normally could do this kind of thing, but I just took 3 midterms today and my brain is tremendously fried. I know that the probability is equal to P(E and F)/P(E), I am just having difficulty pinning which events are E and F.

.close

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[glad falcon]

Hi I need help solving this problem, I know the answer is 2 but have no clue how to get there

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[glad falcon]

<@&286206848099549185>

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[potent smelt]

multiply the equation by xy/xy

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[solid kiln]

neither

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[shy shell]

Is there a way to get dy using Rdtheta?

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[little kestrel]

So I can do this the old substition way

but I decided to take the first system of equation and use the quadratic formula

my answer ended up being incorrect though

I want to know why the quadratic formula didn't work in this situation

[west nest]

What do you mean the first system?

[trim jolt]

@little kestrel Has your question been resolved?

[little kestrel]

As in x=2y^2 ....

[wraith hinge]

whats the answer

[little kestrel]

I got x = -0.5 or -1

which is incorrect

[wraith hinge]

you can get 1 through quadratic as well

[frosty glacier]

Show your working out

[little kestrel]

I just plugged in the quadratic equation to the quadratic formula

[wraith hinge]

u must have done some calculation error

[frosty glacier]

You could make it easier if you wrote in terms of x-1 and subbed in

[wraith hinge]

x=(y+1)(2y+1), ull get the values of y

[wraith hinge]

then just substitute in the second eq

[frosty glacier]

Wouldn't this be easier as substitution?

[wraith hinge]

ofc, but they wanted to solve it using quadratic

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[daring brook]

ok so im finding the second derivative of

[daring brook]

aka the first derivative of

and idk wtf is happening here

[vocal sail]

what part do you not understand?

[daring brook]

wait

so b is redefined as (3-4x)^2
(a/b') = (a'b - ab')/b^2
so: 2' * (3-4x)^2 - 2 * (3-4x)^2'

ok i get that part now

i didnt know we redefined the values but it makes sense now

[vocal sail]

yes that's right

[daring brook]

ok so 2' = 0

[vocal sail]

did you do the derivative yourself or is this a solution given?

[daring brook]

thats the notes

im just trying to understand the process

i understood the first derivative now im trying for this one

[vocal sail]

ok i see

[daring brook]

how does that first step turn into 2 d/dx (3-4x)^2

[vocal sail]

cuz you have a(x) = -2 and b(x) = (3-4x)² now

by applying the product formula

[daring brook]

does 2' turn into 0

[vocal sail]

you have (a'b - b'a)/b²

[vocal sail]

yes

[daring brook]

OH

that makes sense

ok and step 3 we just use product rule

[vocal sail]

this calculation is a bit lengthy for nothing imo

[vocal sail]

yep

no wait

step three is calculation of dx/dt(3-4x)^2

and for two functions f and g you have f(g(x))'=f'(g(x)) * g'(x)

[daring brook]

-12+16x + -12 + 16x
-24 + 32x = -8(3-4x)/(3-4x)^4
= 2 * -8 /(3-4x)^3
-16/(3-4x)^3

[vocal sail]

in this case f(x) = x² and g(x) = 3 - 4x

so f(g(x) = (3-4x)^2

[daring brook]

what. arent they using product rule for d/dx(3-4x)(3-4x)

[vocal sail]

oh yeah sorry

[vocal sail]

this would be a quicker way to do it tho

for later if you want

[daring brook]

where did the x^2 come from

this method makes sense now

[vocal sail]

if you apply function f(x) = x² to (3-4x) you have (3-4x)²

makes sense right

with different variables

[daring brook]

um

[vocal sail]

f you apply function f(t) = t² to (3-4x)

[daring brook]

the method they used makes more sense

[vocal sail]

i mean this also makes sense cuz it's math but that first method would be easier for you yeah i guess

[daring brook]

they probably want me to solve it that way too

[vocal sail]

yeah probably

[daring brook]

ok thx i close now

.close

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[main leaf]

would the index be 4

or something else since theres a power inside

is the index just the number on the outside no matter what

[vocal sail]

x in a nth root equals x^(1/n)

so here you have that equaling to (2^3)^(1/4) = 2^(3/4)

so the index of 2 is 3/4

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[lethal anvil]

What does the power set look like for the null set?

how many elements are in the null set?

null set is also called the "empty set"

ok yeah you have the right idea. So the power set of the null set is the singleton power set. In other words, P(∅) = {∅}

So we have 1 element right?

For the powerset? yes we do

The power set of the empty set has one element

oh right my b, you're right

yeah 2 elements

okay so then what's the power set of that?

no sorry what I mean is how many elements are in P(P{∅})? In other words, what is the power set of {∅,{∅}}?

[marble wharf]

lets not confuse P(emptyset) and P{emptyset}

[lethal anvil]

yeah wait I got mixed up lol

sorry I was right originally

[marble wharf]

the second is bad notation for P({emptyset})

[lethal anvil]

Let A = P(∅) = {∅}. So the cardinality is 1.
Let B = P(A) = P(P(∅)) = {∅, {∅}}. So the cardinality is 2.
Let C = P(B) = P(P(A)) = P(P(P(∅))) = {∅, {∅}, {{∅}}, {∅, {∅}}}. So the cardinality is 4.

there you can see a pattern

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[solid kiln]

Bourbaki

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[wise hemlock]

Idk where I even went wrong

Usually I could type infinity with just oo so like is it not towards inf?

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[wise hemlock]

<@&286206848099549185>

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[wise hemlock]

:((

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@wise hemlock Has your question been resolved?

[knotty locust]

@wise hemlock you have to consider if it tends to positive infinity, or negative infinity, depending on which side you are approaching from. Also, 00 is not infinity, I don't think your problem software recognizes that

[wise hemlock]

The software reads oo as inf idk why it dosent just give us symbols

And for this in particular I could’ve sworn it was +

[knotty locust]

well just take another look then

double check

[wise hemlock]

Yeah as f(x) moves towards + it goes to + inf

[knotty locust]

f(x) moves towards +?

the + indicates that you are approaching from the right hand side

go to your graph

follow f(x) with your finger

from the right hand side

towards pi/2

[wise hemlock]

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[wraith hinge]

Why is it that for a), they subbed in the values into an equation but for d) they didn't?

[wraith hinge]

shouldn't D) be $m^{-1}x=-1/2(x+10)?$

[fading lotus]

That's the answer only

-1/2(x+10) is same as -x/2-5

[wraith hinge]

so why didn't they leave a) to be (x-4)/3?

if technically it was the same answer

[solid kiln]

RecRio

[fading lotus]

The answer of A is infact (x-4)/3
That's what is given in the key

[wraith hinge]

they have it as 1/3(x-4)

[fading lotus]

1/3(x-4) is same as (x-4)/3

[wraith hinge]

I just wanted to know if this is some kind of rule or something because clearly they're inter-changing function form and the inverse answer

[fading lotus]

Idt there is some rule like that
Both r exactly the same

[wraith hinge]

usually textbooks would maintain the same pattern of answers unless there was a reason not to

yeah that's weird lol

hmm, yeah I can't figure it out

would any <@&286206848099549185> be able to check it out?

maybe a textbook error..?

[floral furnace]

It's the same, it doesn't matter if it looks different if the value is equal

[burnt panther]

It looks right to me

[wraith hinge]

but it's similar to simplifying, it could be right but there might be an extra step

it's just they added that for the first one but not the last one

[burnt panther]

I don’t understand. You think the textbook is wrong for 4?

Or what

[floral furnace]

No its right

It displays the answer differently compared to the first one

[wraith hinge]

I think so.. both of them are "right", it's just that for a) they put the answer into function form but for d) they just flat out gave an inverse answer

[floral furnace]

They're both in function form

[burnt panther]

They’re both right, I think it’s probably because of the way you get the answer

[wraith hinge]

not for d), they had it as -x/2 -5 vs $m^{-1}x=-1/2(x+10)$

[solid kiln]

RecRio

[burnt panther]

Because the last step was subtracting 5, not dividing by 2

It’s easier for most people to understand the way they have it

[floral furnace]

And if you use the minimum number of steps that's the answers you'd get

[burnt panther]

Yes exactly

You could definitely make it look like a) if you wanted to, you’d just need an extra step

[wraith hinge]

yeah will have to ask my teacher about this lol, just confuses me why they'd have both variations as an answer

[floral furnace]

Ok

[wraith hinge]

Thanks for clarifying it a bit more though

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[halcyon verge]

How do I find the derivative of this function/ what is the best way to go about it?

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[halcyon verge]

<@&286206848099549185>

[toxic fulcrum]

The chain rule is your friend

[halcyon verge]

I tried the chain rule on a previous attempt and got to the point where I would still have to use difference quotient

wait I might understand

Is this correct so far. Sorry handwriting kinda sucks

[toxic fulcrum]

I really can't follow the steps. You should connect things with = signs and specify when you are taking the derivative of what

Sry

[halcyon verge]

I started the chain rule first on the left then I just applied the difference quotient on the right side there and put it back into my chain rule function

[toxic fulcrum]

Mhmm. You need to review the chain rule

You multiply by the derivative of the inner function

[halcyon verge]

from 15:07 forward https://www.youtube.com/watch?v=HaHsqDjWMLU be similar to what I have to do

[toxic fulcrum]

You don't replace the inner function by its derivative

[halcyon verge]

ah I see what I did wrong. im gonna rewatch this but I would like to know if that is similar or not

[toxic fulcrum]

I don't like this guys channel. Too sloppy at times

Watch 3b1b

That's peak calc content

Also, read your textbook

You can use Wolfram alpha to check simple problems like this

[halcyon verge]

aight thanks

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[mental vault]

So, my friend taught me to do it in this way using binomial mass function. However I’ve been having some degree of doubts: as you can see in the question, it’s stated “…2 flaws”, will that affect my answer in some way? If yes, how?

[fathom folio]

I see no Binomials anywhere

Unless you mean infinite limits

[mental vault]

Maybe it’s my problem of saying it? Srry, my skills in describing things isn’t really great

[fathom folio]

Why would you approximate when given an exact

[mental vault]

Well I mean it’s the way they want us to do it anyways

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[austere echo]

-asin(x+theta) or -sin(x+theta)?

which one

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<@&286206848099549185>#

[hoary kiln]

What are you trying to do?

[austere echo]

derivative

[hoary kiln]

Then the first one so -asin(x+theta)

[austere echo]

thanks

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[orchid sonnet]

Hello is this inversion correct

[austere echo]

yes

[orchid sonnet]

Oh thank you

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[fading lotus]

Find all $x,y \in \Bbb{N}$ such that $p=\frac{(x+y)(xy+4)}{xy+13}$ is a prime

[solid kiln]

Junang / Umanied

[fading lotus]

Idk where to begin with

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[fading lotus]

<@&286206848099549185>

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