help-38

[vague walrus]

and take the limit in the power

right

i got it the notation was confusing thats all

i think

[vestal summit]

They wanted you to check whether f is continuous at 0

[vague walrus]

nah

they told f is continous on R

[vestal summit]

oh ok

W

[vague walrus]

i guess ill remember it

wait

they wanted it at 0

i think i did right

that log sub no?

[vestal summit]

Perhaps

I didn't check the log sub

[trim jolt]

@vague walrus Has your question been resolved?

[sour tree]

y=x+1?

i'm joking

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[hybrid egret]

Hey Im just looking for help with algebra review

[hybrid egret]

specifically fractions

[sour tree]

ok

[hybrid egret]

I have (x+3)/x...im just confused what i need to do with this

does it just turn in x/x + 3/x?

[sour tree]

nothing you cant do anything

you cant take out x

[hybrid egret]

ok ok so next one is (bare with me idk how to type it out in math format)

[sour tree]

yeah but why...

[hybrid egret]

3/x-1 + x/x+2 =

[sour tree]

cross multiply

[hybrid egret]

idk its the only formula i used so i figured thats what he wanted me to do

let me try and do it first

[sour tree]

its simplified to the max

no need

[hybrid egret]

ok

so I would get 3x/(x-1)(x+2) then simplify the bottom

3x/x^2+x-2?

wait nvm

i got the cross multiply

i just forgot i was adding not subtracting

[sour tree]

u good?

It's just x^2+4x+6/(x+2)(x+1)

[hybrid egret]

i need help w next question lol

im just confused

so for this photo i fucked up clearly

i looked it up for help and im wondering why I randomly just transfer the negative to the other side

like i get the -2 on the right side can be put on the whole fraction, but why would you want to (and what law allows it to happen) that you bring the negative to the left side

how does 1/4-x/2 become 2/4-x?

like i just dont get how that works lol

[zinc ginkgo]

$\frac{1}{a/b} = \frac{b}{a}$

[solid kiln]

pizzanator

[zinc ginkgo]

a = 4-x and b=2

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[winter summit]

Exercice 9 :

Conjecture an expression for Un as a function of n, then demonstrate the thus demonstrated

[winter summit]

how am i supposed to do it please i haven't seen it in class yet ?

[marble wharf]

can you conjecture an expression for un-2?

[winter summit]

why Un-2 ?

[marble wharf]

well 50000002 looks quite a bit nicer if you subtract 2

[winter summit]

i dont know how to do it im new to sequences could you show me ?

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[winter summit]

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[thorn heart]

So, this was a question someone actually posted before, but asking about a different part of it

[thorn heart]

The part I'm struggling with is figuring out the analytic expression of the smaller circle

All the info we've got is a single point, fairly sure a circle can't be defined by one point so, yeah, been stuck here for a while

(oh, and as a heads up, the photo's not mine, took it from the guy who originally posted it to try the problem myself)

[final coyote]

the circle is defined by two tangents and one point (where the small circle touches the big circle).

[thorn heart]

We do not have the coordinates of the second tangent though

All we know is that it's an x intercept, so it's of the form (x,0)

[final coyote]

i do not understand, according to your picture, the two tangents are x = 0, and y = 3/4 x

[void tide]

Hi

[thorn heart]

I'm talking about the smaller circle, that one has no y intercept

[void tide]

You know where I can get my homework helphelp

[thorn heart]

So x≠0 for any point in that one

[final coyote]

i am talking about the small circle, too.

[void tide]

ICan somebody help me out I'm stuckI do not know what to do

[whole coral]

Go to #❓how-to-get-help and then look underneath it for the unoccupied channels

[thorn heart]

Then I might be falling behind, Don't see how the small circle could contain any point such that x=0

[void tide]

Ok

[final coyote]

ST is perpendicular to y = 3/4 x

[thorn heart]

Yeah, just read that the tangent at a point is perpendicular to the vector ST

That's a property I didn't know

I'll try it again with that, brb

[trim jolt]

@thorn heart Has your question been resolved?

[thorn heart]

Hm, I've gotten to something strange

Since ST is perpendicular to the line(and as such to its direction vector) and SR perpendicular to the direction vector of the x axis(1,0), the dot products of these with their perpendicular vectors should be 0

However, the latter one doesn't give me any information, as I already know the point is an x intercept and any y makes the vector perpendicular

While the first one gave me a relation between the coordinates of the center of the circle, but it's still not enough to determine it

[final coyote]

why should the dot product of ST with x-axis be 0?

[thorn heart]

No, the dot product of ST and the line's direction vector should be 0

I may have worded it weirdly though

[final coyote]

and whats the problem now?

[thorn heart]

The dot product of SR and (1,0) doesn't give me any info (as any y satisfies the condition)

And the other one gave me a relation between x and y

[final coyote]

again why do you calc the dot product from ST with (1,0)?

[thorn heart]

I don't, I calculate the dot product of SR(not ST) with (1,0)

ST is not perpendicular to (1,0), I know

R is the x intercept of the small circle

[final coyote]

an why dont you use the information about ST?

[thorn heart]

Tried to, I assumed that C, T and S are aligned(that is, they belong to the same straight line) since it seems like that in the picture

And substituted into the line's equation, but doing so, I got that the center of the circle was the point T, which is obviously false

[final coyote]

ST is perpendicular to y = 3/4 x, so it has a slope -4/3.

[terse pivot]

i mean OT and ST are perpendicular, and OT and CT are perpendicular, due to the properties of tangents, so ST and TC must be parallel

[thorn heart]

The slope wouldn't help me though, right?

I don't know which of the infinite points lying on the line is the center

Nor do I know the radius, so I can't determine the center with the slope

[final coyote]

with the slope and a point (T) you can get the equation of the line for example. or you can use that OTS and ORS are congruent triangles, ....

if you know OT you know OR.

[thorn heart]

The congruent triangle part is something I hadn't thought about, actually

Still don't see how the slope could be helpful though

[final coyote]

you started with dot-product calculation, not i.

[thorn heart]

Well, Thought it might be useful, turned out not to be so much, didn't want the slope though

Imma try again with the triangle thing

Got another wrong answer

Well, pretty sure the x coordinate is right, 10

But still gotta figure out half of the problem

[final coyote]

what coordinates for T do you have?

[thorn heart]

T's (8,6)

[final coyote]

what length is OT then?

[thorn heart]

10

Which is equal to the length of OR, and thus x is 10, that's what I've done so far

[final coyote]

and whats the remaining problem?

[thorn heart]

Finding y

Which will also be the radius

[final coyote]

maybe now the slope will help you. you have a point T and a slope -4/3, so you can write a formula for the line ST. if you have ths, you use x = 10 and calculate the y-coordiante from S.

[thorn heart]

Oh, true, since real functions can't be multivalued, that should give me the y I want

I got exactly the same line that contains C and T

[final coyote]

yes of course. it woudl be wrong if not.

[thorn heart]

Didn't this guy say that was impossible?

Two parallel segments cannot be contained within a single straight line

[final coyote]

no he didnt. CT and ST are both perpendicular to y = 3/4 x, they have a common point (T) and therefore they are on the same line

[thorn heart]

Alright, seems like I just got the right circle, geez, I really found that harder than I should've

One last question though

How did you know the triangles OTS and OSR were congruent?

We're lacking some of its sides and only have one angle

[final coyote]

2 sides with the same length, and on angle with the same size.

[thorn heart]

So that's enough to affirm congruency, huh? Alright, thank you

And thanks for the help, really, this problem really did cost me more than It should've

[final coyote]

OS in both, ST = SR and the right angle, what else do you need?

[thorn heart]

Alright

Thanks again, and sorry for taking so much of your time

.close

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[snow sierra]

Guys

when finding angles for cos2x=0 in part bi) we have to double the range given in the question right

[urban copper]

<@&286206848099549185>

[snow sierra]

so, cos2x(tanx-root3)=0 is the simplified form in part bi) btw

And when finding tanx-root3=0 do we use the normal limits given in the question?

Range**

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[snow sierra]

Ok

[urban copper]

how do I compute the orthogonal complement of the subspace H

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[urban copper]

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[languid token]

Why is the left picture not correct at being DNE while the right Is? (In my mind I feel it is still DNE for the left when its -2) Any reason why?

[viscid flower]

The short answer is that the value of a limit doesnt have anything to do with the functions value at that point

actually, the function could not even be defined at the point towards which youre taking the limit

What the exercise is telling you is that the point doesnt really matter, but what does matter is that we're approach the same thing from each side

thats why they ask you those 4 questions:

1. whats the limit from the left
2. whats the limit from the right
3. whats the limit (are 1 and 2 the same?)
4. whats the value of f at that point?

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@languid token Has your question been resolved?

[languid token]

So for the left picture, the limits from left and right are the same. But noted as x approaches 2, why is it -2 for the answer if there's a positive value and negative Y value for the 2 (opened and closed circle) Would we not choose the closed circle in this case?

[viscid flower]

you have to react ✅ or the channel will close

[viscid flower]

because the limit as an operator doesn't care about the value of the function at some point

it only cares about the region nearby, usually called the neighborhood of the point

[languid token]

Ok Thanks!

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[lucid cypress]

So because this is a continuous function, I should be able to plug in 2 and the limit is 4. But my classmate got 8?

[lucid cypress]

Am I missing something here?

If I'm remembering it right, 2 is on the domain.

[zinc ginkgo]

,calc 2 * sqrt(20-2^2)

[solid kiln]

Result:

8

[zinc ginkgo]

Show your work so someone can find your error

[lucid cypress]

Hang on

OH MY GOD

Nevermind I think I just saw it

I forgot all about the x in front

Sorry about this.

.close

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[olive egret]

Isn’t it decreasing from -2,1

@bright quarry

[acoustic flint]

Yes, but that’s not the only range it’s decreasing on

[olive egret]

But why isn’t that an option?

Looks like none of em correct

[acoustic flint]

Because it’s asking for all x values for which the function is decreasing

[2,-1] is only a subset of all the x values for which f is decreasing

[olive egret]

I don’t see where else it’s decreasing

Beside there

Everywhere else is increase

[acoustic flint]

Well what makes you think its decreasing between -2 and 1

[olive egret]

Well it’s decreasing from the X points

-2

And 1

[fervent bloom]

The graph is going by intervals of 2

[olive egret]

Oh

So it’s

-4

And 2

[fervent bloom]

👍

[olive egret]

And open circle

So parenthesis

[fervent bloom]

I’m pretty sure

[olive egret]

Cause those points aren’t included

For this do I graph it out

Or solve

[fervent bloom]

I would solve

But you could do both

[olive egret]

How do I solve it

Do I plug in the nunbers

Intoned

Into X

[fervent bloom]

Do you have a calculator/are you allowed to use one

[olive egret]

Yes

[fervent bloom]

Graphing calculator?

Put the equation in the graph and graph it

Press (sec) then press the trace button (above it says calc)

[olive egret]

I have a TI-30Xs multi view

[fervent bloom]

Oh i don’t think you can do it

[olive egret]

How do I solve it

[fervent bloom]

Is the course a calculus course or does it require a graphing calculator and if you’re able to get one/rent it

You’ll need to graph it out

[olive egret]

Pre calculus

[fervent bloom]

Does your school loan graphing calculators?

Cause you’ll need it for this type of question iirc

[olive egret]

We never use calculators in class

[olive egret]

I can buy one

What kind

Casio?

-1, infinity

I think

[fervent bloom]

I would recommend a ti84 or 83(if you’re on a budget) and if you can’t access a laptop

They’re the most common

Or see if your school loans one

[olive egret]

Is it this

With parenthesis

[fervent bloom]

Yes that looks to be right

[olive egret]

In class we don’t use calcualtors

And still solve it

[fervent bloom]

for relative max/min without given a graph?

[olive egret]

Yes

[fervent bloom]

is it only for quadratic functions

for those you can

for a quadratic function the relative max/min will always be the vertex

if you have a cubic or higher i think you'll need to calculator

[olive egret]

Hard to fucking tell

Without seeing whole graph

Is it 12.5,17.5

[fervent bloom]

i think its decreasing there

[olive egret]

Man this bullshit

Can’t see whole graph

[fervent bloom]

you can try to deduce it with patterns since youre given the period

since you can kinda see 12.5 to 17.5 is decreasing, then you can deduce 17.5 to 22.5 is increasing

[olive egret]

This one tricky

Cause of the straight lines

Nvm that’s easy

.close

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[past cargo]

CAD is congruent to ACB right?

By SSS

so why is angle <C not bisected?

[left oriole]

why should it be?

try drawing the same situation with a very narrow rectangle (say w much smaller than l)

[past cargo]

Yep, i’m aware that the diagonals are not angle bisectors in a parallelogram unless all sides are congruent

I’m curious why my explanation doesn’t work

[fervent bloom]

in a rectangle, yes they are congruent

[past cargo]

What, no

[left oriole]

well if C were bisected, you would expect angle ACB to equal angle ACD, right?

each would equal 45

how does your argument imply that?

[fervent bloom]

the two triangles have equal sides and angles i thought

[fervent bloom]

because all sides of a rectangle are not necessarily equal

if im understanding what youre asking

[past cargo]

Most of the time, I use this argument but i guess it only works if the triangle isosceles right?

Here the angles are not the same and so it doesn’t have to be a bisector?

[left oriole]

yep

[hearty plank]

Note that CAD is congruent to ACD but NOT to CDA

[past cargo]

Thanks

[past cargo]

What

[hearty plank]

Yep

[bright quarry]

do i know you

[past cargo]

you just wrote that the same triangle is congruent to itself

[hearty plank]

The order in which you write the vertices matter for congruence

[past cargo]

Yes but i wrote it correctly

[hearty plank]

What I wrote is correct but slightly off, let me correct that

[gaunt nest]

@hearty plank if you were to do gcse maths would you get a grade 9?

[hearty plank]

Note that CAD is congruent to ACB but NOT to CBA

@past cargo

[hearty plank]

I'm unsure what that is

[past cargo]

Now look at what i wrote

[past cargo]

lol

[gaunt nest]

gcse uk exams? in high school

[hearty plank]

Yes, but they you applied congruence parts equivalence theorem on CBA

[gaunt nest]

u from usa?

[hearty plank]

No

[heady osprey]

Hi

[hearty plank]

You assumed Both the angles C are equal. You implied that CAD and CBA are the corresponding order

But you'd observe angle ACD is equal to CAB from congruence of CAD and ACB

[past cargo]

Okay sure, thanks

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[iron valley]

is this not 0

I put A and got it wrong

please help guys

[simple haven]

@iron valley 0 happens when 4^x gets very large

But what happens if x gets very negative?

[iron valley]

OHHH

then it approahces -3

[simple haven]

Careful

Show your work, why -3?

[iron valley]

oh wait 3

[simple haven]

Still, show your work, why 3?

[iron valley]

wait so

4

will get smaller

and smaller

when the exponent decreases

so basically once its super duper small

it should be 3 right?

[simple haven]

3/(1-0) = 3 yeah

Gj

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[tardy hamlet]

uh how is this not 7

plug in any number you want....

the answer checker said i was wrong

[zinc ginkgo]

Show your work

[split chasm]

was there more info?

[flat steppe]

does p = 84 and q = 96?

[split chasm]

not necessarily, those values satisfy the division property, but there are infinitely many other values.

[flat steppe]

wdym, what other number when divided by 11 will give a remainder of 7 other than 84?

[split chasm]

just keep adding 11 and you'll get another

[flat steppe]

oh

my

[sacred sapphire]

7, 18, ...

[flat steppe]

your righht

[carmine mauve]

Hint: "P leaves a remainter of 7 when divided by 11" means P = 11p+7.

[flat steppe]

i need to get better at word problems

[tardy hamlet]

do i have to

i think the key is wrong

[zinc ginkgo]

If you don't want to explain then don't

Just gonna make it harder to help you

[tardy hamlet]

but is the key wrong or no

im willing to explain

but i dont want to waste time thinking im wrong

.close

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[bronze sigil]

x^2 + y = 16; y is a function of x right? I saw a source for the solution and it said that it wasn't a function. Just wanna make sure I'm not trippin.

[split chasm]

yes, y would be a function of x

do you have the original question and what the source is saying exactly

[bronze sigil]

this is from vaia

idk if it's ai generated or what

but it says that the parabola doesn't pass the vertical line test?

[split chasm]

they're wrong
could be human error

[bronze sigil]

alr thanks

.close

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[iron valley]

can someone hep with c

[pliant wagon]

what is your current thinking?

[iron valley]

well the deminator is greater

right?

so 0

[pliant wagon]

well that's one of the horizontal asymptotes

No fraction can equal zero, hence there is going to be a horizontal asymptote at 0. But there's another one.

[iron valley]

yeah i dont get it

[pliant wagon]

try finding the limit from both sides

[iron valley]

okay so

when x approaches +infinity e will get smaller

right?

[pliant wagon]

Yes it will

[iron valley]

so wouldnt it be infinity

wait im even more confused now

[pliant wagon]

if the term gets smaller and smaller as you approach infinity, what happens to it eventually?

[iron valley]

it will go to 19

10

ohhh

[pliant wagon]

there you go, that's the other asymptote

[iron valley]

wait so when i get one of these problems

what should be my first approach

to find limit from both sides?

can you give me approach list in order that would help

[pliant wagon]

yes I'd find the limit from both sides for horizontal asymptotes. For hyperbolas, it is usually easier, like m/x + n always has an symptoms at x = n. For vertical asymptotes, that's about which values of x have division by zero,or root of negative unless dealing with imaginary numbers

In certain cases, you can find the inverse and determine its implied domain, but that's a bit more work and I think limit usually works. Asymptotes are approached values.

[iron valley]

ohhh okay

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[lucid owl]

js a quick question why do you not multiply the 13 by 7 too??

[stray drum]

Its not multiplying

[lucid owl]

oh okay thank you

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[green perch]

help with this

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[zinc ginkgo]

Did you show it for small n?

Induction eventually probably

[green perch]

yeah induction but I want to kinda understand it in a combinatorial way

[zinc ginkgo]

Understand it for small n first then

n = 1 and 2 should suffice

[green perch]

I cant see it

can u just bring it up in a clear way?

for the small cases

[dry copper]

check with n=0

its the quickest

[green perch]

checking with 0 doesnt explain anything

the equality holds, okay, but It doesn't really show anything important

[dry copper]

oh nvm, i thought you were doing the 1st step of induction proof

[green perch]

txh anyway

[worthy tree]

rewrite the identity as $$\sum_{k=0}^{n} {x+k \choose x} = {x+n+1 \choose x+1}$$ and consider a way to count the number of ways to choose x+1 distinguishable objects from x+n+1 objects

[solid kiln]

Bob the Builder

[worthy tree]

this identity is called the ||hockey stick identity||

[green perch]

Yeah its the bijection way to see the problem

The rest that gets chosen by nor choosing it

Not*

Okay, I totally understand it now

[green perch]

Thank u, it really helped

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[solid kiln]

Architect

Architect

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[gusty agate]

u can get sin60sinxsin10=sin20sin20siny
x is abn and y is nbc
siny is sin(40-x)

by appyling trignometriv version of cevas thm

but i cant simplify

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[gusty agate]

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[turbid violet]

Am I right in saying that if they are perpendicular

tangent(1) = -1/tangent(2)

at the point of intersection

[median torrent]

You mean like slope of the tangent line right?

[turbid violet]

yeah

[median torrent]

Yea

Correct

[turbid violet]

sorry that was poor wording

how can it be proved tho

[median torrent]

i.e. their derivatives

[worldly wing]

first find x

hmm that might be a bit complicated actually

[turbid violet]

do I just find x and sub it in

and show lhs=rhs

[worldly wing]

if you can find x in a nice form then yes

[turbid violet]

if not...?

[worldly wing]

say cosx = tanx = t at the relevant x value, x=alpha

then find the slopes of the two tangents at this x value

[median torrent]

You're in calculus?

[turbid violet]

me?

[median torrent]

Yea

[turbid violet]

i mean part of my course is calculus

i do hsc

and the extension 1 math course

[median torrent]

Ohh ok so this is like a calc unit then

[turbid violet]

yeah

well this is j a revision Q

[median torrent]

Just making sure

[median torrent]

This is where you gotta start

[turbid violet]

would t be an actual value

[worldly wing]

it will be

[median torrent]

Find the solution

cosx = tanx

[hasty peak]

you dont actually have to find the solution

you dont need to figure out x

[median torrent]

Suggestion?

[hasty peak]

Try taking the derivative of cosx and tanx, and giving an equation for when their slopes would be perpendicular dependent on x

[worldly wing]

you cannot figure out x, that's the problem

[median torrent]

Ooohhh

It's so obvious now xD

[turbid violet]

i did that and got sin(alpha) = cos^2(alpha) ??

[worldly wing]

correct

[hasty peak]

try to rearrange this equation

[worldly wing]

first find the derivatives of the two functions and write down explicitly what you want to prove

[worldly wing]

in terms of t like we defined it here

[turbid violet]

do you put it in terms of a quardatic in sin(alpha)

[hasty peak]

no

[median torrent]

tan x = cos x

Vs

sin x = cos^2(x)

[turbid violet]

oh

its just the same thing

[median torrent]

If you do 1 thing...

You can show they're equal

[turbid violet]

divide by cosx?

[median torrent]

Yea

So you have

"If their derivatives are negative reciprocals at alpha, then cos x = tan x"

[turbid violet]

im a bit confused

[worldly wing]

hold on

[median torrent]

Or cos(alpha) = tan(alpha)

[turbid violet]

how does that prove that the intersection is perpendicular

[worldly wing]

this is what you need to prove..

[turbid violet]

like i get we want to prove thatt he derivatives are negative recipricals

[median torrent]

Ohhh I'm sorry yea my bad

[turbid violet]

but how does showing its the same as the orignial functions do that

[median torrent]

I worked in reverse

Start from

If cos(alpha) = tan(alpha)

[turbid violet]

oh wait

i get it

ok thank you

[worldly wing]

Let $\cos\alpha=\tan\alpha := t$, then the statement we need to prove is\ $f'(\alpha)=-\frac1{g'(\alpha)}$, where $f(x)=\cos x$ and $g(x)=\tan x$

[solid kiln]

kheerii

[worldly wing]

the point is that you can prove f'(alpha)= -1/g'(alpha) without explicitly finding alpha by restating it in terms of t

[turbid violet]

what does ':=' mean

[worldly wing]

it means we set that value equal to t

[turbid violet]

ah ok

i understand now

ty guys

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[turbid gazelle]

8b

i get the general gist of the q

its the squares of the vectors which cancel out and give 0

but idk how to write it out in algebra

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[turbid gazelle]

<@&286206848099549185>

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[coarse vortex]

i think its non-separable but idk how to set it up such that i can do integrating factor

[limpid dawn]

I dont think you can obtain an explicit solution

[coarse vortex]

yea

but idk how to make that in the y' + p(x)y = q(x) form

[limpid dawn]

I dont think that's possible either without trying to find a good sub

[coarse vortex]

what do you suggest

[sleek canopy]

y=-x ?it seems to be a solution, can't you see it ?

[coarse vortex]

ermm

idk what thats supposed to mean

like y=-x as a substition? orrr

[sleek canopy]

no

[coarse vortex]

like something from my working

[sleek canopy]

like y=f(x)=-x

in your problem, y is supposed to be a function in the x variable

[coarse vortex]

yes

[sleek canopy]

so y=-x shouldn't create any issue with you

[coarse vortex]

where are you getting -x?

[sleek canopy]

i guessed that y=-x could be a solution from this writing

[coarse vortex]

thats the answer

[limpid dawn]

tbh, i dont think the integration factor method is here applicable in the first place, because your equation has a very different form

[coarse vortex]

youre talkign about the black one right

[limpid dawn]

?

[coarse vortex]

sorry for the confusion but the blue text is the final answer and the black text is the original question

[limpid dawn]

yea i mean the "black text"

[coarse vortex]

yea

so i did the other method

but im stuck at the end

[limpid dawn]

So that you can write it in the form $$y' + a(x)y = b(x)$$

[solid kiln]

bacc

[limpid dawn]

You just need to bring x²/2 to the left side and multiply both sides by 2 and then you can define a new final constant C

[coarse vortex]

ok let me try

[limpid dawn]

C = 2(C_2 - C_1)

SO your whole issue was how to get from your step to the final solution? 😭

[coarse vortex]

yes

[limpid dawn]

i thought the whole time you wanted to see if it's possible with integrating factor method

[coarse vortex]

oh bruh

i just got it

im so stupid

thanks for enlightening me guys

[limpid dawn]

Also you can keep in mind when integrating both sides, one constant is enough cause you can always do C = C_2 -C_1

[coarse vortex]

yea i just realised

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[tawny phoenix]

For number 2 I put 6 is greater than or equal to x and x is less than or equal to 2?? But the text in green is what it says in the markscheme

[tawny phoenix]

Why is it only one or the other

[dull crystal]

which

.rotate

[dull crystal]

$x \leq -2 \quad \text{or} \quad x \geq 6$

[solid kiln]

Aestusy

[dull crystal]

I mean how can you be smaller than two but greater than six?

[dull crystal]

How can it on a number line

[tawny phoenix]

Ohhh

That makes sense

Thank you!!

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[vague walrus]

howww

[vague walrus]

i learnt it yseterday

but we have sqquared term here

dont wanna use by parts

[bright quarry]

is this chatgpt

[vague walrus]

real i just didnt want to format it

correctly

hence i used that to ask here

[fleet bear]

Why not?

[vague walrus]

too long?

besides i wanna see if its possible like that

[fleet bear]

Hmm okay

I doubt that form would be too useful tbh

Plus you only need two levels of by-parts to solve this

[vague walrus]

yea

ik

[fleet bear]

I see a way but it's a little smelly

$$ e^x \cos^2 x = e^x {\left(\frac{e^{ix} + e^{-ix}}{2}\right)}^2 $$

[vague walrus]

nahhhh

too messsy ur right

its best for simple linear x

[solid kiln]

StrangeQuarkAL

[fleet bear]

Not too bad from here actually

[vague walrus]

i mean what they did in the book was

they had P= e^x cos ax

and then considered P+ i Q

then considered

the addition and took the real part

[fleet bear]

Ohhh

[vague walrus]

Q was the conjuagte sin function

[fleet bear]

Q = e^x sin(ax) ?

[vague walrus]

yea

very clever lol

[fleet bear]

Mhm very

[fleet bear]

Although, it doesn't end up being too different from ^ (albeit there's less work for you to do)

$$ e^x \cos^2 (x) = \frac{e^x cos(2x)}{2} +\frac{e^x}{2} $$

[vague walrus]

hmmm

[solid kiln]

StrangeQuarkAL

[fleet bear]

Focusing on just the first part of rhs

$$ e^x \cos (2x) = e^x (e^{2ix} + e^{-2ix}) = $$
$$ e^{(1+2i)x} + e^{(1-2i)x} $$

[solid kiln]

StrangeQuarkAL

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[main sigil]

How would I prove that sum of remainders of 1, 2, ..., k-1 mod q <= sum of remainders of k, k+1 ..., 2k-1 mod q. (q is a prime power)

[vagrant prism]

huh

[main sigil]

I wrote it badly, sorry

[vagrant prism]

ohh

[main sigil]

now it's corrected

(one more correction, the first one ends at k-1)

[vagrant prism]

also is it $\sum (i \text{ mod } q)$ or $(\sum i) \text{ mod } q$

[solid kiln]

artemetra

[main sigil]

first one

[vagrant prism]

okay okay

[main sigil]

$\sum_{i=0}^{k-1} (i \text{ mod } q) \le\sum_{i=k}^{2k-1} (i \text{ mod } q)$

[solid kiln]

MæthIsAlwaysRight

[main sigil]

this

oh wait

i got it

im dumb

the first one has k terms, starting at 0 and increasing by 1

the second one has k terms, starting at anywhere between 0 and q and increasing by 1

[main sigil]

possibly going back to 0 during that

but I think I see it

[main sigil]

thanks for this, I just needed it to be written clearly

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[iron valley]

Can some help with 16 d

[zinc ginkgo]

Use average rate of change around 8

[iron valley]

how-

like rate of change is slope right

[zinc ginkgo]

Yea do this

[iron valley]

but how do i find slope at single point

[zinc ginkgo]

Around 8 means 6 and 9

[iron valley]

ohhhh

then -2

aso can you help with 15. d

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[icy hedge]

Can someone check these please

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[molten comet]

Hello

[molten comet]

How do I find the ratio

[raw magnet]

why not find the inverse function first

as the question proposes

[molten comet]

I did that

[raw magnet]

did you get the correct answer/

[molten comet]

Yes

I'm the person who asked about curl and divergence

[raw magnet]

now $\frac{E(M+2)}{E(M)}= \frac{10^{1.5(M+12.7)}}{10^{1.5(M+10.7)}}$

[molten comet]

rip

[solid kiln]

nosqldb

[raw magnet]

can you simplify this

[molten comet]

Nop

[raw magnet]

uh

$\frac{10^{a}}{10^{b}}$ how would I simplify this in general

[solid kiln]

nosqldb

[molten comet]

So distribute?

[raw magnet]

no

subtract

use the negative exponent rule

to say $10^{a} * 10^{-b}$

[solid kiln]

nosqldb

[raw magnet]

and proceed from there

[molten comet]

10^1.5(aouwehfa) * 10^-1.5(aw;eoifja;)

Like this?

[raw magnet]

😭

what's aouwehfa

[molten comet]

Whyyy

Com'on

M + 12.7 and M + 10.7

[raw magnet]

ohhhh

yes

sorry

and use the addition rule for exponents

to proceed from here

[molten comet]

10^(1.5(M+12.7)-1.5(M+10.7))

[raw magnet]

you should get 3

in the exponent

[molten comet]

???

Oh wait

Hold on

Yes

3

[raw magnet]

yes

aouwehfa

🙏

[molten comet]

Therefore 1000

[raw magnet]

great job

[molten comet]

Thank youuuuuuu

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[molten comet]

Hello

[molten comet]

Can anyone explain this solution for me

[raw magnet]

hmm

what are you not understanding

[molten comet]

The squeeze theorem looking part

[raw magnet]

well basically

this part??

[molten comet]

Also, "take log of this to get"

Yeah

[raw magnet]

well to get 0 as e^x

we note that lim x-> -inf e^x = 0

[molten comet]

Ye

[raw magnet]

and e^0 = 1

and e^x is increasing

[molten comet]

[0, ∞)

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[raw magnet]

uh

(-inf ,0]

[molten comet]

?

Wait

Isn't the domain for x all R

[raw magnet]

nope we restricted it to -2 to 2

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[molten comet]

What is this then?

[raw magnet]

that's the domain under our 2nd restriction

but you have to combine both the first and second restriction

to get ur final domain

😄

[molten comet]

How do I do that

Finding domain isn't my forté

[raw magnet]

well what is the intersection

of (-inf, 0] and [-2,2]

like where do they uh match up

[molten comet]

[-2, 0]

[raw magnet]

EXACTLY

wow amazing

[molten comet]

I don't understand

Where -inf is coming from

[raw magnet]

well e^x is like greater than 0

[molten comet]

Yes

[raw magnet]

like

but it decreases

[molten comet]

?

[raw magnet]

when we go negative

[molten comet]

Yes

[raw magnet]

like e^{-3} < e^{-2}

etc etc

[molten comet]

Yes

[raw magnet]

but it doesn't reach zero

[molten comet]

Yes

[raw magnet]

until you consider the limit

as it approaches -inf

I'm prob butchering this

[molten comet]

Nah you're not

[raw magnet]

ye I'm a little ill

so my brain is quite bad rn

but basically

[molten comet]

You taught me calc 3, you got this

[raw magnet]

yeah

so (-inf, 0]

[raw magnet]

idk how I'm goign to TA tmr

ahh kms

[molten comet]

$\lim _{x\to -\infty }\left(e^x\right)=0$

[solid kiln]

Chaewon

[raw magnet]

yes

[molten comet]

That is range

[raw magnet]

mhm but that is what is being inputted into g

[molten comet]

$\lim _{x\to \infty }\left(e^x\right)=infty$

[solid kiln]

Chaewon

[molten comet]

Why how

[raw magnet]

it's g o f

g(f(x))

[molten comet]

How does domain work in composite functions

CE

[raw magnet]

but basically you need to worry about f's domain

[molten comet]

Are you Canadian

[raw magnet]

and if ur in canada it's amazing tuition

no but I was considering waterloo CS

[molten comet]

80k CAD per year 😭

Why not come here

[raw magnet]

IN CANADA?

[molten comet]

Yes

[raw magnet]

have you seen the meme

[molten comet]

Elon Musk?

Ohh yea lol

This is ECE first year schedule

Like help

[raw magnet]

WHAT THE SHIT

😭

okay my life is good

nvm

[molten comet]

We do discrete math, calc 3, set theory, electricity and magnetism in first year

[raw magnet]

ehh it's very varied for my uni

I brought in calculus 3 credz from hs

but hmm

back to the problem at hand

[molten comet]

Damn

[molten comet]

Okay

LOL

Do you do make up

I mean like

This make up

^

[raw magnet]

oh who doesnt

[molten comet]

Waterloo CS people

[raw magnet]

I put on the moisturizer

[raw magnet]

eh but I never enrolled

[molten comet]

But ye

Back to topic

[raw magnet]

back to topic

how am I going to do it

[molten comet]

Um

[raw magnet]

haha sorry I was just joking

okay

basically g o f

we have to consider the intersection between the domain of f and the intersection of (the domain of g and the range of f)

[molten comet]

Why range of f?

What does that mean 😭

[raw magnet]

bc that's the input to g

[molten comet]

Ohhh

I think I get it now

Thank you so much!

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[molten comet]

Hello

[molten comet]

Can anyone explain the latter part

<@&286206848099549185>

[raw magnet]

hmmm

[molten comet]

Ahhhhh

Why isn't it [0, 3]

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[restive elbow]

i know i have to use law of sines but idk how to do it when i don't know the magnitude and the direction of F

[sacred sapphire]

well you have the components of F dont you

[trim jolt]

@restive elbow Has your question been resolved?

[restive elbow]

yeah but idk how to use them to find F

[sacred sapphire]

well whats the formula for magnitude

[restive elbow]

they're not at a right angle though

<@&286206848099549185> por favor

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@restive elbow Has your question been resolved?

[sacred sapphire]

hmm i found an answer using cosine law

but idk if its right

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[vale abyss]

For question c it says estimate but I was wondering if the formula I had would give the exact probability?

[vale abyss]

if we let n be 6 and r be 4

actually

I could probably check

by running some experiments

no nvm

I get 2

oh wait

I didnt take into account the different probabilities of getting a color into account

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[wraith hinge]

calculate the area of a triangle whose medians are:
11.23610253, 12.16552506 and 12.97112177.

[dry copper]

use the formula of the median, for the 3 medians

youll get a non-linear system of 3 equations in 3 unknowns, which are the sides of the triangle

[wraith hinge]

k.

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[copper acorn]

can someone give me an example of a non-open set? I can't wrap my head around it

[nova spire]

well for example take [1,2)

on R

open set is the same thing as "neighborhood of each of its points"

meaning if I take any point in the set

I'm allowed to wiggle a tiny bit in any direction I want

take the point x = 1

Well no matter how tiny is my wiggle to the left

[copper acorn]

ohhhhhhhhh

[nova spire]

I'm always gonna get out of [1,2)

[copper acorn]

Thank you, I get it

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[quaint ruin]

Can someone help me understand this, I felt like I understood the first about Markov's, but this threw me off

[nova spire]

well

markov tells you that $\mathbb P(X\geq \varepsilon) \leq \frac{\mathbb E[X]}{\varepsilon}$

[solid kiln]

rafilou2003

[nova spire]

this you understand?

[quaint ruin]

yes, I think I do

[nova spire]

well

if X >= epsilon

then f(X) >= f(epsilon)

[nova spire]

so this event is more likely than X >= epsilon

(f is monotonically increasing)

[quaint ruin]

But is that in terms of the extended version on the lecture slide?

[nova spire]

?

[nova spire]

we're only going from the usual Markov's inequality

and we took f some monotonically increasing function

(oh and they forgot the assumption that f is a positive function)

[quaint ruin]

Doesn’t that just tell you that when X is at least excellent then the probability will be upper bounded by the expected value divided by epsilon and does this mean that in most cases when X is greater than epsilon, the probability will be less than the expected value or something or is that wrong to think like that

[nova spire]

if you want to start from Markov's inequality

the correct way to think about it

is that if you know some random variable X