help-10

[devout solar]

The first one also has to be true with infinite too

oh wait I am just dumb

[long sinew]

Wut

[gray creek]

So how would we write it on the solution set indicator?

[devout solar]

$well I tried some numbers and got that \sqrt{-1}=1$

[warm shale]

.,..

[devout solar]

but im dumb

[long sinew]

Sus

[devout solar]

yes

[long sinew]

No clue how and what that is

[gray creek]

The line thing

[devout solar]

but wait

$\sqrt{x^2-4x+4}=x-2$

[warm shale]

.,..

[devout solar]

if x=1 it is sqrt1=-1

[long sinew]

Oh

Then there must be a range of answers

[gray creek]

So it wouldn’t be infinite?

[devout solar]

correct

[gray creek]

But how would you graph it?

[devout solar]

no clue

we dont do that type of stuff

But first we need to figure out the range

[gray creek]

X^2 - 4x + 4 = X^2 - 4x + 4

How would we find the range?

[devout solar]

I know the range but I'm trying to figure out how to do it by hand

We went through these like 6 months ago but I don't think I saw one like this

[gray creek]

Oh

Weird question huh

[devout solar]

Or then I dont remember

did pretty bad on these

[gray creek]

How about number 7

[devout solar]

What do you think it is

[gray creek]

Can’t be c or d

[devout solar]

And why is that

[gray creek]

Is it b?

[devout solar]

Nope

do you see where it goes wrong

[gray creek]

Oh wait it’s the opposite

Yeah it’s c

[devout solar]

No its not

@gray creek

[gray creek]

Damn it’s d

[gray creek]

Could we do one more ?

[devout solar]

sure

[devout solar]

the first one?

[gray creek]

Yeah it’s with question 7

[devout solar]

What you thinking

[gray creek]

I think b is solution

[devout solar]

thats one of them

[gray creek]

E is another

[devout solar]

what about D

[gray creek]

It wasn’t a solution for D?

[devout solar]

how did ypu get E

[gray creek]

I simplified X^2 - 4^2

[gray creek]

Could we do the last one?

Wait wouldn’t C be solution?

[devout solar]

hmm

[gray creek]

?

[devout solar]

sorry i gtg

[gray creek]

Okay thx for the help

.close

[obtuse pebble]

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[lilac loom]

i cant get this, i mult

[teal prawn]

divide by √(ab) on both sides

its a positive value anyways

$\frac{ab}{\sqrt{ab}} = \frac{(ab)^{1}}{(ab)^{\frac{1}{2}}}$

[warm shale]

kinglacto

[teal prawn]

use exponent laws

[lilac loom]

i will try thx

[obtuse pebble]

@lilac loom Has your question been resolved?

[lilac loom]

@teal prawn sorry but i cant find the same result 😩

[timid silo]

kasparov

i assume a,b are positive reals

[teal prawn]

yeah and now simplify further

[timid silo]

(rt(a)-rt(b))^2>=0 right?
now expand it

[lilac loom]

yes a, b > 0

[teal prawn]

seems like youre trying to prove the A.M G.M inequality?

[timid silo]

ye

[lilac loom]

this is the exercice;

[timid silo]

^

[obtuse pebble]

@lilac loom Has your question been resolved?

[obtuse pebble]

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[sly folio]

Working on 102, I’m not sure how to continue. Do I plug in t between 0 and 2pi?

[sly folio]

<@&286206848099549185>

[obtuse pebble]

@sly folio Has your question been resolved?

[sly folio]

<@&286206848099549185>

[obtuse pebble]

@sly folio Has your question been resolved?

[warm shale]

dackid

[jade hornet]

The arclength formula of an ellipse is famously one that can not be solved directly, as there is no known antiderivative

[warm shale]

dackid

[obtuse pebble]

@sly folio Has your question been resolved?

[jade hornet]

.close

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[alpine trout]

if i have to integrate 1/(x^2 + 9)

[alpine trout]

can i not just get the reciprocal and integrate that?

or do i have to use substitution

[slim lake]

Hear me out, you can substitute x=3u

[brave bramble]

No you can't just integrate the denominator haha

[slim lake]

Take out 1/9 and then you have the derivative of inverse tan

So $\int {\frac{1}{x^2+9}dx = \int {\frac{3}{9u^2+9}du = 3/9 \int {\frac{1}{u^2+1}du = 1/3arctan(u) +c = 1/3arctan(x/3) +c$

[warm shale]

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

[alpine trout]

yeah well you see our professor told us nothing bout that so it would be pretty weird for me to use such method on a assignment

like ive got no clue what arctan is

[slim lake]

Well it only integrates to one function

arctan = inverse tan

tan^-1

[alpine trout]

the actual question is a definite integral if that makes any difference

[slim lake]

Are the bounds in terms of pi?

[alpine trout]

sorry definite

no

the limits are 1 and 0

[slim lake]

You're missing a negative somewhere in the integral then

1/(x^2-9) ?

[alpine trout]

oh whoops

yeah that not +

[slim lake]

That can be done via partial fractions

[sly folio]

@jade hornet So there is no perimeter formula of an ellipse?

[slim lake]

$\int \frac{1}{ x^2-9}dx$ we can use the fact $\frac{1}{ x^2-9}= \frac{1}{ (x-3)(x+3)}= \frac{A}{ x+3}+\frac{B}{x-3}$

[warm shale]

Max..

[slim lake]

Then you can find A and B by reputting the fraction together and equalling the numerator

[sly folio]

Also @jade hornet, it asks for the area??

[alpine trout]

right yeah

i always understood that part but never understood what id do with A and B after finding them

[slim lake]

Well you can split the integral

[jade hornet]

No, there certainly is not

[slim lake]

$\int \frac{1}{ x^2-9}dx= \int \frac{A}{ x-3}dx+\int \frac{B}{ x+3}dx= A\int \frac{1}{ x-3}dx+B\int \frac{1}{ x+3}dx$

[warm shale]

Max..

[sly folio]

Okay. But the formula I should be using is $\int ydx$, right?

[slim lake]

And then the integral of 1/(x+a) = ln|x+a|

[warm shale]

FivePixels

[sly folio]

Between 0 and 2pi

[jade hornet]

Based on what it says, yes

[jade hornet]

No, polar form is good. But use the one for the area, not the perimeter.

[alpine trout]

ahhhh i seee

[sly folio]

I'm confused, polar form? What formula would that be?

[alpine trout]

okay that makes much more sense appreciate it 🙂

[slim lake]

Np

[alpine trout]

do i close the channel lol or let the other guys keep going

[jade hornet]

Nvm, that's not helpful. I think $2\int y dx$ is the way to go. I have $2$ there becaue that integral only evaluates half the area

[warm shale]

dackid

[jade hornet]

Let us finish. He was here before you (I closed it earlier)

[sly folio]

So the form I suggested is just half of the ellipse, doubling gives the full ellipse? On an example I found online, it suggests that $\int ydx$ only gives 1/4 of the circle

[warm shale]

FivePixels

[jade hornet]

Depends on the bounds

The bounds I was going for is -a to a, which gives half

[sly folio]

So from 0 to 2pi would be a full revolution right?

Meaning my most recent screenshot

[jade hornet]

You are NOT in polar coordinates here

[sly folio]

Right, these are cartesian

I guess I don't understand how that's relevant to the problem though.

[jade hornet]

The bounds 0 to 2pi don't mean jack since you aren't in polar coordinates

[sly folio]

Should we be in polar coordinates?

[jade hornet]

Not necessary

[sly folio]

Then if 0 to 2pi doesn't matter, what should our bounds be?

Just take the indefinite integral?

[jade hornet]

Feel free to start there

[sly folio]

Here's what I've got so far

Does this look on the right track? @jade hornet

[obtuse pebble]

@alpine trout Has your question been resolved?

[sly folio]

<@&286206848099549185>

[tardy epoch]

That's not how integrals work

An integral of a product is not the product of integrals

You just want to find the antiderivative of sin^2(theta)

[severe dune]

The antiderivative of that can be found in 2 ways right

Just cos 2theta substitution

Or reducing it as 1/2- sinxcosx/2

[sly folio]

So I have that sin^2(theta) = 1-cos(2theta) / 2, and I've solved using u sub

[severe dune]

Yes that works

[sly folio]

Simplified = -theta + 1/2 divided by sin(2theta)

[jade hornet]

Small mistakes

[sly folio]

I think I see it

[jade hornet]

You forget to multiply the -2 into -cos(2theta)

[sly folio]

-2 should be multiplied across both integrals

[jade hornet]

Correct

Sin(2theta) is also not in the denominator

[sly folio]

Right

It should be -2theta + sin(2theta), right?

[jade hornet]

Yes. Make sure you can understand your own writing.

[sly folio]

Right sin(2theta) should have been on top, since we're multiplying the whole thing by it.

[jade hornet]

No we are not

[sly folio]

Well

I meant on my original solution

Where I had the mistake

sin(2theta) should have been in the numerator is my point

In my picutre

Right?

When I say the whole thing, I just mean multiplying (sin(2theta)/1) times everything. So it multiplies the top by that.

@jade hornet

[jade hornet]

That negative concerns me a bit. I understand the logic, but your area should end up being positive

Oh, actually it's right. So because you multiplied by 4, your bounds are from 0 to pi/2.

However, if we integrate with respect to x, then that is going from right to left. So the true area starts at pi/2 and ends at 0

Then you'll get the area

[sly folio]

I don’t understand this. What makes this the bounds?

[jade hornet]

Think about it. The original integral evaluates $\int_0^a y dx$. But once you change your parameters, $a\cos(\frac{\pi}{2})=0$ and $a\cos(0)=a$. So your new bounds go from $\frac{\pi}{2}$ to $0$.

[warm shale]

dackid

[sly folio]

Do you mean from 0 to pi/2?

With 0 on the bottom and pi/2 on top?

[obtuse pebble]

@alpine trout Has your question been resolved?

[sly folio]

@alpine trout you can respond yes so you stop being pinged

[alpine trout]

but it would close the channel no?

[sly folio]

Just until I respond next

@alpine trout

[jade hornet]

No, the other way around

[sly folio]

Right, since the definite integral should return positive

[jade hornet]

Yes, but also setting $x=a\cos(\theta)$ gives us our bounds for $\theta$ with u-substitution

[warm shale]

dackid

[sly folio]

Could you explain further?

[jade hornet]

This explains it further

[sly folio]

What do you mean by parameters?

Ah wait. So cos(pi/2) = 0, and cos(0) = 1, okay. So the arithmetic is checking out now.

[jade hornet]

Yes, more importantly, acos(pi/2)=0 and acos(0)=a

[sly folio]

Right

[jade hornet]

So the bounds reverse

[sly folio]

We just plug in what I typed.

[jade hornet]

We evaluate the integral from pi/2 to 0 instead of from 0 to pi/2

[sly folio]

Right, because when we evaluate the x at the original 0 to pi/2, we get the inverse as the result of the bounds.

[jade hornet]

Yep, right idea

[sly folio]

I kind of understand that but it doesn't seem the most grounded. Is there a different way of thinking about it?

[jade hornet]

You're misunderstanding. If we solve acos(theta)=0 for theta, we get theta=pi/2. Similarly, acos(theta)=a gives theta=0

Well, there is a periodicity thing, but we only care about the range [0,pi) for inverse cosine

[sly folio]

So what happens to a in the case of the first one? We just divide 0/a and are left with cos(theta) = 0 right? And take the inverse cosine of 0, to get the theta of pi/2?

[sly folio]

Right, so we wouldn't need to worry about -pi/2 since it's not in the inverse cosine range

[obtuse pebble]

@alpine trout Has your question been resolved?

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[gray creek]

I need help on this problem

[fierce lagoon]

what have you tried

@gray creek

[gray creek]

I solved it I just don’t understand the graph part

[fierce lagoon]

just plot then as a dot

if its equal to

[gray creek]

How would I plot X^2 - 4x + 4 = X^2 - 4x + 4?

[fierce lagoon]

Oh wait

if theyre the same function, its just all numbers

basically

<---------------------------------------------------------------------------->

well no

actually

because you have a square root

[gray creek]

Oh

[fierce lagoon]

what are the x intercepts of $x^{2}-4x+4$

[warm shale]

Umbraleviathan

[gray creek]

(x - 4) (x-2)

[fierce lagoon]

check your work

thats not correct

[gray creek]

Oh no it’s (x-2)^2

[fierce lagoon]

yeah

knowing behavior function, that also means that (2, 0) is also ....?

[gray creek]

Yeah it’s been a while since I’ve done this 😞

[gray creek]

Is it the intercept

[fierce lagoon]

Well yes

But its also the vertex

and because x^2 has a positive coefficient, 0 is also the abolsute minimum

Meaning that for all x values, $x^{2}-4x+4\ge0$

so you verified that

which is good

hold on

[gray creek]

God it’s been a while since I’ve done this

[warm shale]

Umbraleviathan

[fierce lagoon]

yeah you need to check for extraneous solutions

so, that means youve verified that x can be all numbers

so it is <----------------------------------------------------------------------------------------------------------------------------------------------------------->

You want to make sure that some numbers will not make the sqrt function have complex solutions

[gray creek]

What’s the difference between closed and open again?

[fierce lagoon]

closed is inclusive of the endpoint

open is not inclusive of the endpoint

but because x is all real numbers

its just gonna be a line

two rays, per se

[gray creek]

So how would the rays be on 2 and go to the right?

[fierce lagoon]

well

<---------->

like that

[gray creek]

.close

[obtuse pebble]

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[obtuse pebble]

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[vital brook]

how do i prove that 0 < a*b < 1 when 0 < a < 1 and 0 < b < 1?

[strong vale]

it should be obvious

if two numbers

lie between 0 and 1

then obviously there product will be between 0 and 1

[vital brook]

But for what reason?

is it because you're essentially doing a fraction multiplied by a fraction

I guess I'm a bit confused putting it into words or a statement that proves the reason why

[strong vale]

choose any two numbers between 0 and 1

let's say

0.2 and 0.4

,calc 0.2 * 0.4

[warm shale]

Result:

0.08

[strong vale]

when u multiply two numbers between 0 and 1

the numbers just get smaller

[vital brook]

Then would my justification be for the proof that my values are two fractions less than one being multiplied reducing their values?

[strong vale]

If $0<a<1$ and $0<b<1$ then $a=\frac{m}{n}$ where $m<n$ and $b=\frac{p}{q}$ where $p<q$ $ab = \frac{mp}{nq}$. Since $m<n$ and $p<q$ then $mp < nq$ and hence $0<ab<1$

[warm shale]

azeem321

[vital brook]

ohh I see this makes a lot of sense thank you!

[obtuse pebble]

@vital brook Has your question been resolved?

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[strong vale]

(ii)

So for this, I got $(z_4-z_2)e^{i\frac{\pi}{2}} = z_6-z_3 \implies (z_4-z_2)i=z_6-z_3$

[warm shale]

azeem321

[strong vale]

Is this what they are after?

[obtuse pebble]

@strong vale Has your question been resolved?

[hushed moat]

if this is true then that directly implies they are perpendicular

[strong vale]

cool, thanks

.close

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[timid silo]

Someone help pleasee

[obtuse pebble]

Please don't occupy multiple help channels.

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[plush valley]

guys i cant simplify this question

[plush valley]

i need to solve it, but idk how to remove the sqrt inside a sqrt

[scarlet locust]

you need to express whatever is inside the outer sqrt in the form of (a+b)^2

try to think how you can do it

hint: || 6root2 will be the 2ab term ||

[plush valley]

so smth like this?

now it makes sense

thanks

.close

[obtuse pebble]

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[versed zinc]

I dont get any of that

why 90

[brazen tusk]

Hi again. When two lines intersect, the opposite angles are equal, agreed?

[versed zinc]

yea

hii

[brazen tusk]

So, y (the yellow angle) is made up of two angles. We know one is 90 degrees because it's opposite from a given 90 degree angle.

The other angle making up y is opposite from x, so it is also x.

[versed zinc]

oh

[brazen tusk]

So 90 + x = y

[versed zinc]

all makes sense now

[brazen tusk]

🙂

[versed zinc]

:smileystare:

no nitro lol

thank youagain

.close

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[unkempt swift]

hi

[obtuse pebble]

Please don't occupy multiple help channels.

[unkempt swift]

.close

[obtuse pebble]

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[brazen tusk]

The division you're doing is finding out how many cups you need for 5l of oj.

Try writing out an equation and introduce a variable for the number of cups.

[timid silo]

can 5/0.45 = 11 1/9

oh

.close

[obtuse pebble]

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[crisp dune]

for this question

to make it equal to f1

what do we state type of transformation is required??

[lethal sand]

Well in order for 1/(4x^2) to become 1/x^2, what do we have to do to 1/(4x^2)?

[obtuse pebble]

@crisp dune Has your question been resolved?

[crisp dune]

multiply it by 1/4

[lethal sand]

No, multilying it by 1/4 gives 1/(8x^2)

You are close

[obtuse pebble]

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[rugged void]

Can anyone help me prove this theorem?

[obtuse pebble]

@rugged void Has your question been resolved?

[gilded needle]

did you try using fubini's theorem?

[rugged void]

Yes but Fubinis theorem only works if intergral is finite

Tonellis theorem works for non-negative functions

[gilded needle]

fubini's (or tonelli's) applied to |f| and |g|?

[rugged void]

For the second part we can use Tonelli but how do I change \int|\int f(x-y) g(y) dy| dx to \int\int |f(x-y)g(y)| dx dy

to

[gilded needle]

$\int \int |f(x-y) g(y)| dx dy = \int |g(y)| \int |f(x-y)| dx dy = \int |g(y)| |f| dy = |f| \int |g(y)| dy = |f | |g|$ and the RHS is finite since f and g are integrable. Then observe that $\int |fg| = \int |\int f(x-y)g(y)dy| dx$ is less than or equal to the LHS of the first equation I wrote. Therefore $|fg|$ is integrable so in particular it is finite a.e.

[warm shale]

OurBelovedBungo

[rugged void]

Thank you that solves the second part of the theorem. Do I prove the first part by using fg < |fg| and hence is finite then we can use Fubini's theorem?

[gilded needle]

yeah, integrability of |f*g| implies integrability of f*g (assuming you've established that f*g is measurable), so Fubini applies with f and g (not just with |f| and |g|)

[rugged void]

Also, in the theorem it states f, g measurable and integrable implies f*g is measurable. Is it some famous result that I don't know of?

[gilded needle]

ah, I couldn't recall how to prove that f*g is measurable, it's in several of my textbooks but that's not easy to copy/paste here. but you can check this PDF on page 16: https://people.math.gatech.edu/~heil/6338/summer08/section4c_convolve.pdf

[rugged void]

Thank you so much!

I was so confused with all of this

.close

[obtuse pebble]

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[lyric rune]

how do i solve a quadratic equations

[short spruce]

there's a number of ways

if it's factorable, that can help, you could factor and use the zero product property

if applicable

[lyric rune]

ok can you help

a sec

[short spruce]

you could use the quadratic formula, which never fails

you could also complete the square

[lyric rune]

i dont have that formula

[short spruce]

might want to get it

check notes/online

[lyric rune]

i have a book for that stuff

but i dont undarstand it

i think i have it in my book

[woven citrus]

1. Rearrange the equation to the form of ax^2 + bx + c = 0
2. Use the formula

[lyric rune]

hmmm

[lyric rune]

ok?

bc i have a qestion that is 3x2+3x-6=0

the 2 is like tiny and beside the x and its over it to idk

[short spruce]

that's an exponent...

that function is also factorable

[lyric rune]

?

[short spruce]

why are you working with quadratics if you don't know what an exponent is

[lyric rune]

bc i want

i just need to understand it

[short spruce]

well if you don't understand the basics, don't expect to understand things that rely on knowing the basics

you can't learn something without having a foundation

[lyric rune]

ok cant you learn me the basics

ik

[short spruce]

if you want to learn concepts, go to youtube

[lyric rune]

ok

[short spruce]

i'd suggest a review of arithmetic

[lyric rune]

what is arthmaetic

[woven citrus]

Emm

Basically

Calculation

[lyric rune]

ok

[ashen dove]

...

[nocturne minnow]

You should probably learn the other material that leads up to quadratics

[obtuse pebble]

@lyric rune Has your question been resolved?

[obtuse pebble]

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[bright geyser]

idk how to do dis

[obtuse pebble]

@bright geyser Has your question been resolved?

[bright geyser]

@heloers

<@&286206848099549185>

[plain owl]

Consider the function f(x)-x

[bright geyser]

Why?

[plain owl]

We can show via intermediate value theorem that this function definitely achieves value of 0 for some x in [0,1]

[bright geyser]

I don't really understand or see how

[bright geyser]

Is this correct?/

like the definition of invt

[plain owl]

How did you conclude f(0)≠f(1) ?

Suppose there is no c in [0,1] such that f(c)=c. That means f(0)-0>0 and f(1)-1<0. Precisely our function f(x)-x evaluated at 0 and 1

[bright geyser]

isnt that the definitiion

[bright geyser]

Why does that mean that?

I dont really understand

[plain owl]

Not necessarily. Consider the function f(x)=0 for all x in [0,1] it satisfies all the conditions in the question

[bright geyser]

huh

that doesnt make sense then does it

if f(x) = 0

[plain owl]

Nowhere in the question does it say f(0)≠f(1)

[bright geyser]

then f(a) and f(b) bound nothing

[bright geyser]

Is it not implied by the actual definition of IVT that im using?

[plain owl]

But it is a valid continuous function from [0,1] to [0,1]

[bright geyser]

That doesnt mean that f(c) = k

i don't think we're on the same page im quite confused now

[plain owl]

IVT does require f(a)≠f(b) sure. But not necessarily the question

[bright geyser]

reposting q for referencing

from the intermediate value theorem

we have to show that f(a) < k < f(b) or f(b) < k < f(a) right

[plain owl]

We are applying IVT on the function f(x)-x as i said earlier

[bright geyser]

Why?

[plain owl]

Ok then tell me how does that show f(c)=c for some c in [0,1]

[bright geyser]

thats my point idk lol

[plain owl]

it's not some specific constant value analogous to "k" in the IVT

But if we do show there is some "c" in [0,1] such that f(x)-x achieves 0 there it is equivalent to saying f(c)=c

[bright geyser]

hmm maybe im starting to get it

so youre proiving that f(c) - c = 0 essentially

which is equivilant to f(c) = c?

[plain owl]

In our question, we need to show f(c)=c for some c in [0,1]. IVT directly on the function f doesn't work

[bright geyser]

ywyw

yeye**

[plain owl]

Yep

[bright geyser]

hmm okay i am kinda understanding the goal ate last now

[plain owl]

And now, we start with this

[bright geyser]

proof by contradiction?

[plain owl]

Well, you could also do direct proof

[bright geyser]

maybe id like to try direct proof

im not completely sure where the inequalities are from im sorry

[plain owl]

They are easy to see actually. Codomain of our function f is [0,1]. Which means f(0)≥0 implying f(0)-0≥0 but if f(0)=0 we already have a contradiction. Thus we assume f(0)-0>0. And similarly for f(1)-1<0

[bright geyser]

Why does f(0) >= 0 true?

oh wait

sorry

[plain owl]

1≥f(x)≥0 for all x in [0,1]

[bright geyser]

ye i get that

hmm

I don;t really understand

[plain owl]

Try this

[bright geyser]

is this the direct proof>

?

[plain owl]

No. It was my attempt to explain the proof by contradiction earlier. Direct proof is also similar to this actually.

Ignore all that lets start the proof from scratch

[bright geyser]

Thank you for your patience btw 🙂

[plain owl]

First of all, f(0)≥0

[bright geyser]

Yepp

Is it also neccesary to say that 1 >= f(0) <= 0

[plain owl]

Well ofcourse that's where f(0)≥0 comes from

[bright geyser]

okiee

[plain owl]

Well, 1>=f(0)>=0

Anyways, if we do have f(0)=0, we already have our required c

[bright geyser]

Ye sure

[plain owl]

So we assume f(0)>0 which is same as saying f(0)-0>0

[bright geyser]

sure

[plain owl]

And we do something similar for f(1)≤1

If we have f(1)=1 we have c=1

So, f(1)<1

[plain owl]

And ofcourse this also comes from 0≤f(1)≤1

[bright geyser]

okie dokie i get that

[plain owl]

And that's equivalent to saying f(1)-1<0

[bright geyser]

yee

[plain owl]

Now, we consider the function f(x)-x on domain [0,1]

[bright geyser]

ye]

[plain owl]

Now we actually have f(1)-1<0<f(0)-0 combining the two inequalities from earlier

Which means we also have f(1)-1≠f(0)-0

All the required conditions for IVT are satisfied for our function f(x)-x on [0,1]

f(x)-x is a continuous function on [0,1] where f(0)-0>0 and f(1)-1<0. It means there exists some c in (0,1) such that f(c)-c=0

@bright geyser

[bright geyser]

sorry just trying to read it over

i think i kinda get it

[plain owl]

If you are uncomfortable, you can call our new function g(x). g(x)=f(x)-x

g(0)>0>g(1)

[plain owl]

It looks just like the IVT here

[bright geyser]

alright this is what i dont understand

why are you assuming this

[plain owl]

Well, we definitely have f(0)≥0

[bright geyser]

Yep

[plain owl]

It was just broken into two cases f(0)=0 or f(0)>0

[bright geyser]

ahh

the goal here is

f(x) - x < 0 < f(x) - x right?

sorry

f(0) - 0 < 0 < f(1) - 1

[plain owl]

Well, yes just the other way around

f(0)-0>0>f(1)-1

[bright geyser]

so youre saying ignoring the cases where f(0) = 0 and f(1) = 1

we have that^^

Now i dont fully understand why we ignore those cases

[plain owl]

Well, we did address those cases so it's not exactly ignoring

[bright geyser]

by adressing do you mean

that there is automatically f(c) = c from those cases?

[plain owl]

Yes!

[bright geyser]

hmm

so like f(1) <= 1 and f(0) >= 0

ahh i understand that there is a f(c) = c from the cases where f(1) = 1 and f(0) = 0

but i dont understand the function of the exclusion

or how you can just freely do it

[plain owl]

By definition, a ≤ b means a < b or a=b

[bright geyser]

yep

[plain owl]

So it naturally breaks into cases

[bright geyser]

omg wait okay okay okay

one sec lemme try and write this out

Case 1: f(1) = 1 or f(0) = 0

[plain owl]

In one case we address f(0)=0 and in other we can take f(0)>0 with worrying about f(0)=0

[bright geyser]

we automatically have f(c) = c if this is the case

the next case

f(1) < 1 and f(0) > 0

implies

f(1) - 1 < 0 and f(0) - 0 > 0

which omplies

f(0) - 0 > 0 < f(1) - 1

which now for the function f(x) - x

[plain owl]

f(0) - 0 > 0 > f(1) - 1

[bright geyser]

implies that there is some c such that f(c) - c = 0 by IVT

and therefore in this second case f(c) = c?

is that it

[plain owl]

Yep you got it!

[bright geyser]

THANK YOU SO MUCH

you are the best

im really sorry i took so long

[plain owl]

Glad I could help

[plain owl]

Also, saying c in [0,1] is important since that's where f(x)-x operates

[bright geyser]

ttytytytytytytytyytytytytytytytytytytytytyytyyt\

[bright geyser]

.close

[obtuse pebble]

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[bright geyser]

!close

[obtuse pebble]

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[lament osprey]

can someone check if my answer is correct?

[lament osprey]

im not confident with answer on the piecewise function

[desert sparrow]

What is 'they'

[lament osprey]

where

[desert sparrow]

Your description of the discontinuity is almost there

But in general I think you need to be more specific in what you mean

[lament osprey]

because the functions are not equal

is the piecewise correct though?

[desert sparrow]

The function is not correct by the way

[lament osprey]

ohh

how can i fix it

[desert sparrow]

What is the charge for someone who parks 4.4 hours?

Your function does not correctly specify that

Also, the charge does not linearly increase (for any part of its domain) like how your function is currently being defined

So first, I'd think about how to define the function

[lament osprey]

Im lost on how to add 6 dollars additionally after half an hour

[desert sparrow]

Are you aware of $\mathbb{N}$

[warm shale]

pepper

[lament osprey]

no i wasnt taught of that :(

[desert sparrow]

You also need a floor/ceiling function

But first let's check your intuition

[lament osprey]

ik the floor and ceiling onee

[desert sparrow]

I park for
25 minutes - what is the charge?
45 minutes - ?
70 minutes - ?
100 minutes - ?
500 minutes - ?

[desert sparrow]

Ah good, you'll need it

[desert sparrow]

Try these values

[lament osprey]

ok hold on

idk how to subtract 5 mins from the 6 dollars

[desert sparrow]

It's not 'subtract 5 minutes'

25 minutes is 6 dollars

45 minutes is 12 dollars

70 minutes is 18 dollars

[lament osprey]

isnt it 30 mins 6 dollars

[desert sparrow]

You need to pay 6 dollars for every multiple of 30 broken

no

Read the question

[lament osprey]

or part

okok i did not see that

let me try to make a function

[royal basin]

may i suggest a slightly different way of looking at this pricing scheme

[lament osprey]

yes plss

[royal basin]

you pay $6 for each half hour of parking time, rounded up to the nearest half-hour multiple.

[lament osprey]

6 ; 0 < x < 1
6 + 6x ; x > 1
im lost

ohh

hmm

i dont get it :(

[royal basin]

is that supposed to mean that you don't get what i am saying at all? or do you not get why my phrasing refers to the same thing as the problem? or do you not get how to translate my phrasing into a formula?

[lament osprey]

phrasing into formula

[royal basin]

have you heard of the floor and ceiling functions?

[lament osprey]

yess

[royal basin]

okay

so now just to be clear, what unit do you want to measure time in? hours or minutes?

neither choice is less valid than the other but it'll affect what the formula will look like

[lament osprey]

im not sure but the problem says half an hr im assuming hr?

it didnt tell me what specific unit

[royal basin]

kat, do you think that was a question with a right and a wrong answer?

because it wasn't.

i specifically said there was no wrong answer here.

[lament osprey]

i guess hour

[royal basin]

if you want to go with measuring time in hours then just say hours lol

okay, hours it is

[lament osprey]

ok sorry i wasnt sure huhu

[royal basin]

now the following is a question with a right and a wrong answer:

since we are rounding up, what rounding function do we use? floor or ceil?

[lament osprey]

ceil

[royal basin]

correct

[lament osprey]

oh wait my friends used mins

can i change?

[royal basin]

sure

[lament osprey]

ok ty!

[royal basin]

we didn't really get to the part where the unit matters anyway

well, we were about to get there but you stopped just moments prior

but ok, minutes it is

now we need to find how many half-hours there are in x minutes

as i am sure you know there are 30 minutes in a half-hour, so...

[lament osprey]

yes

[royal basin]

when you convert x minutes into half-hours, what do you get?

[lament osprey]

30
60
90

[royal basin]

....................

what

[lament osprey]

am i right?

oops

HAHAHA

[royal basin]

no, that's nowhere near it

[lament osprey]

well

[royal basin]

are you familiar with the concept of unit conversion at all

[lament osprey]

no

sorry im so dumb rn

[royal basin]

so you do not know how to do things such as convert lengths between feet and inches?

[lament osprey]

oh yea i meant yes

yes ik unit conversion my bad

[royal basin]

so you are familiar with this concept. ok

[lament osprey]

yes

[royal basin]

so let's try to jog your memory a little

there are 12 inches in a foot.
if you have a measurement listed in inches and you want to convert it to feet, what do you do?

[lament osprey]

do i divide it by 12

so 1 foot

[royal basin]

"divide the number by 12" and "always say it's 1 foot no matter what" are very different things to do, don't you agree?

[lament osprey]

yes

[royal basin]

"a measurement listed in inches" does NOT mean "a measurement that says 12 inches" despite the presence of the number 12 in the previous sentence...

[lament osprey]

okok

[royal basin]

would recommend not latching onto the first number you see out of desperation to use something as the input to some arithmetic.

but other than that, yes, to convert a measurement from inches to feet you divide the number by 12.

now in the same token

[lament osprey]

okk

[royal basin]

there are 30 minutes in a half-hour.
if you have a length of time listed in minutes and you want to convert it to half-hours, what do you do?

[lament osprey]

30 = 1 half hr
60 = 2 half hrs
so basically i divide it by 30

[royal basin]

those two first sentences were entirely unnecessary

i was intending for you to follow purely by analogy with the foot/inch thing.

[lament osprey]

i meant divide by 30

[royal basin]

two units, one of them a multiple of the other.

but yes, so to convert from minutes to half-hours you divide by 30.

and now, to bring in the "rounded up" thing

a parking time of x minutes is counted as ceil(x/30) half-hours.

[lament osprey]

okk

how do i say that in a piecewise function?

in a formula

[royal basin]

is your teacher requiring you to write it as a piecewise function?

[lament osprey]

yes

[royal basin]

that's strange.

i mean, you could shoehorn a piecewise definition in, but there's no need to

the price for $x$ minutes of parking is simply $f(x) = 6 \ceil{x/30}$

[warm shale]

Ann

[royal basin]

this works regardless of whether x is less than 30 or greater than 30

[lament osprey]

she prefers me to put a piecewise function and conclude what the discontinuities mean

[royal basin]

so just to be clear, despite this, your teacher is FORCING you to write down a PIECEWISE FORMULA for the function, and will REJECT OUTRIGHT any answer that doesn't write it like that

is that correct?

am i understanding you correctly?

[lament osprey]

yes

[royal basin]

okay, then i guess we will have to do something stupid like this:

[lament osprey]

my friend put like
6 ; 0<x__<__30
so on

[royal basin]

$f(x) = \begin{cases} 6 & 0 < x \leq 30 \12 & 30 < x \leq 60 \ 18 & 60 < x \leq 90 \ 24 & 90 < x \leq 120 \ \vdots & \vdots \end{cases}$

[lament osprey]

yeaa

[warm shale]

Ann

[royal basin]

like this

and this despite your class knowing about the existence of the ceiling function

presumably

[lament osprey]

ooo ok tysm i get it now

[royal basin]

what is there to get? i didn't write anything new.

[lament osprey]

i just didnt know how to form the piecewise function

[royal basin]

all i did was transcribe the definition that you just said your friend wrote down, no?

i guess this one has the benefit of explicitly highlighting the first few points of discontinuity in the pricing function.

namely the ones at x=30, x=60, x=90, etc.

[lament osprey]

yess

[royal basin]

every half hour the price jumps up by $6

[lament osprey]

thank you so much for the help! sorry if i wasted any of your time

[lament osprey]

ty!

.close

[obtuse pebble]

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[tired gull]

Hi, Im struggling a bit with this problem

problem description/context:
Silvia has n dresses and decides which one to wear by flipping a symmetric coin. If the coin shows tails, the dress is ready for the next turn, however if the coin shows heads then it goes back into the wardrobe. This is repeated for all n dresses until the first turn is over and repeat for the dresses left in the selection.

question:
Determine a random variable X (in terms of X_i, i € N) that counts how many times Silvia flips her coin totally (NOTE: it is already n times the first turn).

my attempt:

[tired gull]

but it feels like i have done something terribly wrong

feels like it should be way simpler

[desert sparrow]

I'm not sure if I'm right, but consider only 1 dress. What is this distribution?

It generalises to n dresses by independent sums.

[tired gull]

binomial?

[desert sparrow]

1 dress is binomial?

[obtuse pebble]

@tired gull Has your question been resolved?

[tired gull]

oh, yeah sorry.

[desert sparrow]

So is that a yes or a no?

[tired gull]

well my reasoning was that it is always binomial, no matter the amount of dresses

[desert sparrow]

By the binomial random variable

I suppose you mean the number of success and failures for a turn

Instead of that

focus on the total number of coin flips

Consider one dress

What is the random variable

That represents how many coin flips there are?

[tired gull]

X = n?

I suppose

[desert sparrow]

Why are there n

What do you mean by n?

There's only one dress

[tired gull]

n coin flips representing n dresses

[obtuse pebble]

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[wispy brook]

hei, how do I go on to solve these problems

[wispy brook]

any tips ?

[obtuse pebble]

@wispy brook Has your question been resolved?

[obtuse pebble]

@wispy brook Has your question been resolved?

[obtuse pebble]

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[wanton dagger]

I need help with this question

[obtuse pebble]

@wanton dagger Has your question been resolved?

[royal basin]

this misprint makes the question impossible to solve

one of the four inequality symbols (<, ≤, >, ≥) was mistakenly replaced by °

but there's no telling which one it was

[wanton dagger]

Oh

[royal basin]

so the only correct course of action is to let your teacher know about this so that they can tell you what the inequality symbol was.

[wanton dagger]

Sure

Ok then thanks

.close

[obtuse pebble]

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[swift heart]

mainly looking for someone who can look over my answers, pretty sure this is right since if your under 30 it doesnt matter, andsince charlie has feature A it doesnt matter if they are below 30

[royal basin]

yes, that is correct

[obtuse pebble]

@swift heart Has your question been resolved?

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[tulip aspen]

Why is it that when you square root both sides the 2 on the denominator is under the square root and not the whole fraction?

[robust sleet]

because

sqrt(1)=1

[tulip aspen]

ohhh ok yea that makes sense thx

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[obtuse pebble]

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[timid silo]

What I've tried so far:

I moved the triangle to the origin (see attached image)

Where I'm stuck:

I don't know what to do next

[gloomy valve]

can't you literally just count the boxes?

[timid silo]

I wrote this equation ||$z-3-2i=e^{i \theta}(3+4i).$|| trying to relate $z-3-2i$ and $3+4i$

[warm shale]

Prem

[timid silo]

no because its not exactly on a point

[gloomy valve]

oh i see now

that would have been funny

[timid silo]

lol

<@&286206848099549185>

[brazen tusk]

Since no one is helping, I'll throw out a suggestion, but I don't know if it will work. Write the two equations for distance from z to each of the other points.

[obtuse pebble]

@timid silo Has your question been resolved?

[jade hornet]

Yea, that's all it is this time around. @timid silo you can find the point just like you would by finding an (x,y) pair: determine the real part,which corresponds to the x-axis, and the imaginary part, corresponds to the y-value. Since the grid is there, it is pretty easy to determine what z is

[devout solar]

what do you even have to do

[jade hornet]

They need to determine what z is

[devout solar]

oh

[jade hornet]

The problem is generous, since all you need to focus on is the grid.

No need for any fancy equations, just count the boxes

[devout solar]

But z is slightly off of it being 2+5/2i or whatever it iw

so just \approx 2+5/2 lol

[obtuse pebble]

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[vague scroll]

for 5a is the only way to find the eigenvector of A^2 - 3I to calculate A^2 - 3I and then use the eigenvalues we know to figure out the eigenvector?

[vague scroll]

<@&286206848099549185>

[obtuse pebble]

@vague scroll Has your question been resolved?

[obtuse pebble]

@vague scroll Has your question been resolved?

[obtuse pebble]

@vague scroll Has your question been resolved?

[obtuse pebble]

@vague scroll Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[short spruce]

can you show work?

[inner raft]

I’m having trouble with the domain intervals

[short spruce]

sin(x) is not equal to 1 at x=0

[inner raft]

where

no. i added 1 to the other side

thats where i got sin(x) = 1

[gilded needle]

your answers for when sin(x) = -1/2 are fine, but you don't need the +(2 pi k) because only for k=0 will those values be in the interval [0, 2pi)

[inner raft]

ik i didn't include it. i just wrote it as a habit

[gilded needle]

i don't understand what is happening starting at the line with the 2*sqrt(3), where did that come from?

[inner raft]

look more to the right under the graph. those are my solutions

that's another problem dont mind it

[gilded needle]

i see

well as @short spruce said, x=0 is not a solution

the other three are fine

[inner raft]

is that my only mistake?

[short spruce]

the x=0 solution needs to be replaced with a correct solution

[inner raft]

i don't really get the [0, 2pi). I thought it should include 0

[gilded needle]

there is only one correct solution for sin(x) = 1 in that interval, no?

[inner raft]

pi/2?

[gilded needle]

right

which is already listed

but sin(0) is not 1, so x=0 is not a solution

(it's not a solution to 2sin(x) + 1 = 0 either)

[inner raft]

so just get rid of 0 from my list of solutions?

[gilded needle]

correct

[inner raft]

alr thanks for the help

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[royal valley]

is the 8 positive in this equation: -3 - 8

[warm pine]

?

you mean -3-8

[idle swallow]

i think so

[short spruce]

why would it be positive

[idle swallow]

uh

[royal valley]

no theres space between the minus sign and 8

[short spruce]

it's the same thing

[warm pine]

exactly

[short spruce]

whitespace doesn't matter

[warm pine]

you can read it as -3+(-8)

[royal valley]

ohh

[idle swallow]

-3+(-8)

[warm pine]

which makes 8 a negative

[royal valley]

thanks

[idle swallow]

yeah

[short spruce]

don't just tell people answers :/

[royal valley]

.close

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[neon loom]

how can i make an implicit equation for 3 dimensions out of 2 points?

[neon loom]

i know the vectorial form for a plane in 3 dimensions is something like t(a,b,c) + s(d,e,f) + P where P is a point

so, if i have these two points P=(-2,3,4) and Q=(-1,3,1) i could maybe start from one of these two points, say, P a,d build something like t(a,b,c) + s(d,e,f) + (-2,3,4)

but how could i find the scalars t and s and the directors (a,b,c) and (d,e,f)?

if it was in 2-d i could just calculate the slope and it would be easy but i don't think there's such thing as slope for 3d

[obtuse pebble]

@neon loom Has your question been resolved?

[neon loom]

.close

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[brisk arrow]

given that cosB=5/9, how'd i get sinB

[brisk arrow]

is it possible to get the value of sinB with all the information on this picture

the final goal is the area of triangle ABC

and i know heron's formula could be very useful in this case

but i want to do it another way

the area will also be (1/2) * 6 * 3 * sinB

in order to get the answer, i have to get the sinB

do you guys have any idea about sinB

[surreal zephyr]

could you do reverse sinB to find the angle of B

angle B would be 56.4427 degrees

[twilit loom]

can you recall an identity linking sin and cos?

[brisk arrow]

sin^2+cos^2=1

[twilit loom]

nice, that would work

[brisk arrow]

how'd you get angle B would be 56.4427 degrees?

[brisk arrow]

that sounds fantastic, let me try

[surreal zephyr]

I did sin^-1(5/6) but I realized thats not a right triangle so im wrong

[brisk arrow]

it does looks like a right triangle tho, even its not

5/9=(5^2/9^2) right

[twilit loom]

5/9 = 25/81?

[jade hornet]

They like to trick ya like that

[brisk arrow]

(5 * 5)/(9 * 9)

you cancel out those 9 and 5 on both denominator and numerator

then you get 5/9

[jade hornet]

No

That equality is false

[brisk arrow]

cuz i have to multiply those first right?

[twilit loom]

yup, I was verifying if that was what @brisk arrow meant

[jade hornet]

And no, that logic is wrong Grammar

[twilit loom]

the 9 and the 5 can't be cancelled out like that

[jade hornet]

Sounds like dino's got this. I'll leave y'all be

[twilit loom]

if that were the case, does 1/2 = (1^2)/(2^2)?

[brisk arrow]

no

[twilit loom]

you can only cancel out factors that are present in both the numerator and denominator

[brisk arrow]

i think i got it, i have to follow the calculation rule

i think the question is solved

i have found either the sinB's value or the area of the triangle

thank you guys

.close

[obtuse pebble]

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[lapis mural]

Is $f: \mathbb{R} \to \mathbb{R}$ a bijection if $f(x) = x$? I mean, $f(a) = f(b) \implies a = b$, and for every $b \in \mathbb{R}$, there is an $a \in \mathbb{R}$ (which is $a = b$) so that $f(a) = b$

[warm shale]

Taiga

[timid silo]

yep

[lapis mural]

:o

so there's always a bijection $f: X \to X$ whether $X$ is an infinite set or a finite set?

[warm shale]

Taiga

[timid silo]

yea, the identity function works

[lapis mural]

that's so coolll

thanks a lot yumeee

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[obtuse pebble]

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[timid silo]

np 🙂

[obtuse pebble]

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[next trellis]

so if I want to prove that 2 sets have the same cardinality, I set up a function, I show that it is one-to-one , and onto. But here is the problem. How do I come up with this function (f(x) = ??) ? Do I have to use my imagination, or is there something I am missing?

[timid silo]

sometimes coming up with a function for that can be tough 😵‍💫

[next trellis]

yes thats why im confused

[timid silo]

so yea, I'd say you have to use your imagination sometimes

there isn't one trick that will always work

[next trellis]

how do I do that on a 1 hour exam lol

[timid silo]

maybe the problems that could come up on your exam will be somewhat related to other problems you've seen already?

there aren't any universal tricks but some of the same ideas could work for different problems

what kind of class is this for?

[next trellis]

discrete math

[timid silo]

any examples in particular you are struggling on?

[next trellis]

https://imgur.com/gallery/gX2TKYh

[timid silo]

ok, cool

an idea you could have for this is to make a linear function that sends a to 0, and b to 1

[next trellis]

what?

[timid silo]

one thing that might sounds a little off about what I said is that 0 and 1 are not in S, and a and b are not in W

is that confusing or is it something else?

but the idea is that W and S are both like lines

if you think of them on the real number line

and you can transform one into the other by sending the endpoints of one to the endpoints of the other

and transforming all the other stuff in each of the sets

[timid silo]

but that part kinda happens naturally when you do this

[next trellis]

so f: W -> S and f: S -> W are the same?

[timid silo]

hmm I'm not sure exactly what you are asking?

[next trellis]

like the definition of the function

that I will use in this problem

[timid silo]

uhhh

maybe I'll say... no, but I'm still not sure what you mean 😕

[next trellis]

how would you solve this?