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[devout snow]

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[haughty plank]

I am taking a history of mathematics class and am learning Egyptian mutliplication which is turning them into second terms, however I am stuck on this problem, do i combine to make an improper fraction or how do i even start?

[devout snow]

@haughty plank Has your question been resolved?

[haughty plank]

<@&286206848099549185>

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@haughty plank Has your question been resolved?

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[heavy current]

I haven't used the hypothesis that f is injective when proving the forwards inclusion, so something has gone wrong

[heavy current]

I can't seem to figure out what the issue is though

can somebody help me identify it?

[upper schooner]

Why not?

[wicked rover]

you started at $f(A_0)-f(A_1)$ instead of $f(A_0-A_1)$

[woven radish]

RokettoJanpu

[heavy current]

that was the intent

[heavy current]

ah, I realize the issue now lmao

hold on.

[wicked rover]

wait what inclusion are you stuck on

[heavy current]

the fwds one

I did the bwds one already

[wicked rover]

whats the forward

[heavy current]

f(A_0 \ A_1) \subseteq f(A_0) \ f(A_1)

I didn't use the fact that f was injective, so I knew I had gone wrong somewhere

let me try to fix it

[wicked rover]

that means proving if y in f(A_0 \ A_1) then y in f(A_0) \ f(A_1)

but you started by letting y in f(A_0) \ f(A_1)

[heavy current]

I did the bwds inclusion first

[wicked rover]

oh hm reading is hard

[heavy current]

[upper schooner]

(Probably a good idea to make that into a new paragraph )

[wicked rover]

mb

[heavy current]

that would make it worse actually

this problem has like

8 paragraphs already

cause there's 8 parts

[wicked rover]

a simple line break would help morons like me

[heavy current]

@upper schooner I think this works now

this is like the 15th basic set equality I've done for this class now, and I'm borderline losing it

can't even use the definitions properly anymore after frying my brain spamming them

anyways, thank you!

.solved

[devout snow]

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[cold bough]

sum of increasing function is reasonable that is increasing

[steep tide]

OK

Let's try to solve it together...

You are sure about the first one right?

why do you think the first one is right?

Ok, let me give you some insight...

When do we say some function is increasing?

Let's say there are two values x and y. And we know that y comes after x (i.e x<y)

Then we should have f(x) < f(y)

If this is true, only then the function is increasing...

But why isn't the second one true?

Yep

totally right!

Now what about the 3rd?

What does NOT increasing mean?

not really...

x < y and f(x) >= f(y)

Right!

So is the 3rd one correct?

wdym?

The 5th one?

Yep the 3rd one is correct

Oh sorry about that...

The 3rd one is wrong

I didn't read it properly

So

The upside down A means "for all x,y"

And the opposite E means "there exist some x,y"

So, the 3rd one says:
A function f is NOT increasing on an interval I if there exist x,y which belong to I such that x<y and f(x)>=f(y)

Because to show that some function is NOT increasing, we have to prove that for every x,y, if x<y then f(x)>=f(y)

We can't say it for only 2 of them

Right!

So, the 3rd is false. right?

What about the 4th?

And 5th?

Yep

What about 6th?

Just it's not totally true... It's only a partial problem of it that's true...

Not really...

Actually it's saying...

That a function f is NOT increasing if there exists some x,y such that x>=y and f(x)>=f(y)

Yeah...

But you sure that we can say that a function is NOT increasing

Only by saying that there exists some x,y such that x>=y and f(x)>=f(y)?

Oh wait...

You might be right!

Oh wait...

4th one is wrong...

Do you know why?

But I think 3rd one is correct

Because the 3rd one says:

A function is NOT increasing, if there exist some x,y x<y and f(x)>=f(y)

[steep tide]

Which means...

If even one pair doesn't satisfy the condition, then the function can't be increasing...

Right!?

So 3rd one is correct

But the 4th one isn't

Yeah...

Because it was something like this:

The function is increasing, if for all pairs of x,y if x<y then f(x) < f(y)

So, if some function is NOT increasing, even if there exists one pair that doesn't satisfy the condition, then it's not increasing...

But look at the condition of the 4th statement...

It's that, if x>=y, then f(x) >= f(y)

Which is not the condition we need...

Yeah...

But the one solution that it is telling us..

MIght not help us to make sure that the function is NOT increasing...

maybe the pair is something like this:

5 >= 3 and (f(5)=7) >= (f(3)=4)

Let's say these are the values...

Now can you say that if you are given these values, then the function is NOT increasing?

Just assuming...

Let's assume that we have some function that we need to prove that it is NOT increasing...

And according to that statement

They say that

if there is some pair x,y such that x>=y and f(x)>=f(y)

Then we can say that the function f is NOT increasing...

But I gave you the example that, let x=5, y=3 and f(x)=7 and f(y)=4

And we know that this pair satisfies the condition

x>=y and f(x)>=f(y)

BUT IT DOESN'T ENSURE THAT THE FUNCTION IS NOT INCREASING...

Please try to read the statement again and understand what it's saying...

What do you mean by it's asking?

It's not asking...

It's telling

It's a statement...

It's a statement that's stating that if there exists that one pair, then the function is NOT increasing...

Is this true?

which pair?

OMG IT'S NOT THAT...

WE CAN'T ASSUME THE PAIR...

IT SAYS THAT IF YOU ARE GIVEN SOME PAIR THAT SATISFIES THE CONDITION, THEN YOU CAN SAY THAT THE FUNCTION IS NOT INCREASING

I just said that they might give us this pair...

Oh I forgot that the caps lock was on sorry about that...

Olympiad level

Have been to IMO twice

grade?

High school...

But it's not the thing...

I do math more than the school

Yeah

IMO = International Math Olympiad... Its totally proof based

Please trust me about what I am saying...

Try to understand what the statement is telling us...

yep...

And it can be any one pair...

So, I gave you an example...

That is they might give us this:

x=5, y=3 and f(x)=7 and f(y)=4

This pair satisfies the condition...

At least one...

The might be more, but we have to assume that there is only one

[steep tide]

But it's not sufficient for us to say that the statement is true...

WAIT...

Let me give you some intuition...

Let's define two sets

The first set is some values of x!

And the other set is the values of f(x)

And we join x with f(x)

connect

Like add an edge

We are making a graph

Kind of...

Joining x with the value of f(x)

That is what a function is...

I think I am making everything complicated...

<@&286206848099549185> Can someone help this guy understand this statement!

He is confused about how the 4th one is false!

Yes

Wait let me give you some example of a function....

Let me define a function

f(x)=x

Yeah forget about it right now...

Let's say there is a function f(x)=x

OK?

Now...

Yep

So if we consider the statement for this function, it says that:

If there exists some pair x,y of this function such that x>=y then f(x)>=f(y)

then we can say that the function is NOT increasing....

BUT...

In our function f(x)=x

Let's assume x=2, y=5

We know that this pair satisfies the condition right....

So our function should have been NOT increasing...

But it's not true.

Our function f(x)=x is increasing...

You get it now?

There might be a pair that satisfies the condition... But it might not be sufficient to prove that the function is NOT increasing...

Oh sorry

Assume the opposite

x=5, y=2

You get it now right?

What about the 6th statement?

Not really...

It's saying that...

the sum of an increasing function is also increasing...

Is it?

OH yeah... Wait...

Yeah you're right!

Yeah that's false...

Surely

OK take for example:

x = 1, 2, 3, 4, 5
f(x) = 5, 6, 10, 12, 13

Let's assume there is some function like this

OK?

You see this function is increasing...

But the difference isn't...

Right!

Like a function might be increasing... But the difference might ot be increaing...

like first it's being added by 2 and then by 1...

stilll its increasing but the difference is not

You get it right?

[devout snow]

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[fathom flint]

ho

hi

[devout snow]

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[queen steeple]

how would I prove these?

[queen steeple]

I'm doing 23 and 25

[supple knot]

Just calculate the right side and factor out -

[queen steeple]

so do I just say that a = <a1, a2, a3> and b = <b1, b2, b3> and then solve for both sides?

[supple knot]

You expand the right side

Then show you get the left side

[queen steeple]

which one are you talking about?

23 or 25?

[supple knot]

Works for both

25 is factoring a vector where as 23 is factoring out a constant

[queen steeple]

for 23, I would get −b x a=(−1)(b x a)
Then what would I do?

[supple knot]

Expand the cross product

[queen steeple]

this is what I got

[supple knot]

Uh huh

[queen steeple]

ah I think I got it

I would multiply the -1 inside and then get the same vector as a x b

thank for your help

.close

[devout snow]

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[bold wigeon]

.reopen

Can someone help me with this

[bold wigeon]

\

[soft halo]

9

[bold wigeon]

how?

ohhhhhhhhhhhhhhhhh

ok thanks

[devout snow]

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.

[bold wigeon]

How do I do this????

this one has 5 points

[upper schooner]

(by the way, don't do this, or try to open threads with messages beginning with a full stop, because they won't open for you)

[bold wigeon]

oh ok

can you help me with this question

nvm

I got it

[devout snow]

@bold wigeon Has your question been resolved?

[bold wigeon]

@everyone PLS ANYONE HELP I GOT 3 QUESTIONS LEFT!

[devout snow]

@bold wigeon Has your question been resolved?

[sweet warren]

Add Rs and ts

That will be 11y+10.

Then equate

11y+10 = 76
11y=66
y=6

@bold wigeon

[devout snow]

@bold wigeon Has your question been resolved?

[devout snow]

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[umbral ivy]

I'm not sure where I went wrong here

I set the current 2005 value ($147,000) = 108000^20k

20 is the time in years between that

The 0's cancel out and we get 147/108

We can simplify the fraction to 49/36 = e^20k

e^20k = 49/36

wait.

I may see the error

I was wrong

but.. I think I dd the math right?

[devout snow]

@umbral ivy Has your question been resolved?

[umbral ivy]

I figured it out

.clse

.close

[devout snow]

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[devout snow]

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[icy glen]

I don’t know how to do 4, pls help

[mystic scarab]

,rccw

[devout snow]

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@icy glen Has your question been resolved?

[devout snow]

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[long elbow]

i tried a little but i think it’s definitely wrong

[devout snow]

@long elbow Has your question been resolved?

[buoyant rock]

is the ANS part your answer or is that just the final answer

[long elbow]

final answer to check for practicing

[quiet elbow]

bro @long elbow in which class you are?

[long elbow]

math 3 maybe can be counted as calc 3/4 but its on differential equations

[quiet elbow]

i mean in which standard you are studing i am in 9 th

i want to know these type of question comes in which class?

[long elbow]

uni?

[long elbow]

<@&286206848099549185> anyone free to help me in diff eq?

[wicked kayak]

Ai

[long elbow]

?

[devout snow]

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[crude wasp]

||notice that you can factor out x-y from the top equation and x+y from the bottom equation||

[chrome sage]

ooh

so...

[novel violet]

Both x³ - y³ and x³ + y³ are difference of squares

[chrome sage]

a^2+ab+b^2 = 7

[novel violet]

I am assuming you know how to factor them

[novel violet]

Not exactly

[chrome sage]

how do i latex this

can someone tell me how to latex

[novel violet]

Do you know how to factor difference of cubes?

[chrome sage]

so

[eager nova]

What?

[chrome sage]

a^3 + b^3 = (a + b)(a^2-ab+b^2)

[novel violet]

Sorry not squares cubes

My fault

[novel violet]

Yes

[crude wasp]

you can do it like this and then do casework but generally it is preferable to factorise

[novel violet]

Factor both difference of cubes in each equation

[chrome sage]

so...

lets see

x^2 + xy + y^2 = 7
x^2 - xy + y^2 = 5

howmany real solutions do we have

[novel violet]

Yes, but when writing it down you should most likely do every step

Anyway

[chrome sage]

x^2 + y^2 = 6?

and find solutions for that?

[novel violet]

How did you get to that

[crude wasp]

remember that you also have x=y, x^2 - xy + y^2 = 5 as a case etc.

[chrome sage]

yeah

ill put that aside for now

[crude wasp]

fair lol

[chrome sage]

add both of the equations

so we have, x^2 + y^2 = 6

and 2xy = 2, xy = 1

[novel violet]

Then you should get 2x² + 2y² = 12

[chrome sage]

divide both sides by 2

[novel violet]

Alright

Isolate and solve

[chrome sage]

hmm

can you do (x + y) ^2 = 8?

and (x - y)^2 = 4

x + y = plus/minus sqrt(8) and x-y = plus/minus 2

so we have 4 solutions of x^2 + y^2 = 6

ok so for x = plus or minus y

x = y, 2x^3 = 10x, x(x^2 - 5) = 0

x = +- sprt(5), and 0

x = -y, 2x^3 = 14x, x(x^2 - 7) = 0, x = 0, x +=sqrt7

2(5+5) + 2(7+7) + 4 x 6 = 72

so the answer should be 72

[devout snow]

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[uneven fern]

just a quick question, once i get the resultant moment the rotation is based on if the value is positive and negative which means that it is clockwise and counter-clockwise respectively, right?

[uneven fern]

also should the forces 20N and 40N be negative

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[devout snow]

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[thorny summit]

when computing integrals such the integral of x^2, I understand the answer is (x^3)/3 but i am confused on solving it from for example -2 to 1, where I keep getting a negative area. I feel like I am making a simple mistake somewhere?

I do ((-2)^3/3 - 1^3/3)) but that gives -3. the answer should be positive 3.

[restive river]

well area can't be negative

and that represents the magnitude of area

in the negative y direction

[thin tapir]

Because if you go from a to b you compute F(b)-F(a)

In this case for example 1^3/3 - (-2)^3/3

[thorny summit]

ohh

thx

:D

[twilit comet]

@thorny summit integrals don't just compute area, they compute signed area. the area below the x-axis is negative, and that above is positive. this is because integrals are generally used to compute net change, not actually area under a graph

[thorny summit]

i c

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[devout snow]

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[void parcel]

Hello guys, I can't seem to find the answer to this question.

[void parcel]

Question is: Two angles are complementary to each other. If one angle is two-third of the other, find both the angles.

[restive river]

do you know what complementary angles are

[void parcel]

Yeah, angles whose sum is 90 degrees.

[restive river]

correct

consider two angles A and B

A+ B = 90degrees

two-third of the other

means

B = 2/3A

therefore

substituing B

you get

A + 2/3A = 90

$A + \frac{2A}{3} = 90$

[woven radish]

daniel

[restive river]

this is what you have

nbvm

let one be 3x and the other 3x * 2/3 = 2x

[restive river]

solve this and you get angle A

[restive river]

that is really ambigous

[void parcel]

can't seem to solve the equatiom.

[restive river]

why?

show us your workings

[void parcel]

its in my laptop.

kind of like 2a/3 + a =90 degrees. Im doing 90*3 = 3a.

[devout snow]

@void parcel Has your question been resolved?

[void parcel]

No @devout snow .

[devout snow]

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[finite briar]

hello

[soft umbra]

yellow

[finite briar]

so wolfram spat out this

I have been trying to prove this for an hour now

and please note that here the elliptic integral is defined as this

and the second input (the -1) is actually not k but k^2

as stated here

[finite briar]

i tried IBP cus thats gunna put the sqrt(1-x^4) in the denominator

but i got stuck at separating it out into an elliptic integral of the first kind

IBP gives $-\frac{2x^3}{\sqrt{1-x^4}} + 2 \int \frac{x^4}{\sqrt{1-x^4}} dx$

[woven radish]

rak³en

[supple knot]

did you try factoring the thing under the square root

[finite briar]

(1+x^2)(1-x^2)

how does that help exactly?

thats actually how i got the idea for IBP

[finite briar]

cus if the x^4 got out smh then we actually have exactly what we need (F(sin^-1(x)|-1))

[finite briar]

another thought was even more IBP but I doubt thats what I am supposed to do

I'd like to avoid complex analysis if possible

[finite briar]

u got anything?

[supple knot]

well no not after you said this

[finite briar]

I see. whats the complex analysis route?

[supple knot]

solve for k^2 = -1

[supple knot]

then this falls out

[finite briar]

what?

falls out..?

[finite briar]

k is i yes i can see that..

[finite briar]

that becomes sqrt((1-x^2)(1-k^2x^2)) for k^2 = -1

hllo? riemann? u there?

[supple knot]

did you have a question

[supple knot]

the left side has a k^2, but you know what that equals

[finite briar]

wait what

isnt that

what i started with

[finite briar]

u need to provide me a step by step solution here, I am stupid

[supple knot]

a step by step would be to differentiate the right side

[finite briar]

right side of?

[supple knot]

the thing i replied to

[finite briar]

so i reverse engineer the solution out of that?

can i not go from LHS to RHS?

[supple knot]

you asked for a step by step solution

that would be it

[finite briar]

from LHS to RHS, not from RHS to LHS

[supple knot]

doing RHS to LHS would do the same and you'd do it yourself

[finite briar]

fine I'll try differentiating

i got $\sqrt{1-x^4}-2$ 💀

[woven radish]

rak³en

[devout snow]

@finite briar Has your question been resolved?

[finite briar]

.close

[devout snow]

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[formal quarry]

question 41

this is my working

can someone check pls

cuz textbook giving dookie answer

[finite briar]

shit man

i needed to reopen it

nvm

will use another one

[formal quarry]

oof

anyone

[deep elbow]

All your answers are correct I think

[formal quarry]

i see

then its textbook saying x =/ 0 is wrong for all domains

[deep elbow]

oh wait

I’m not sure about the domain for b and d

[formal quarry]

alrighjt

[deep elbow]

I think you cant input 0 because the function isn’t f(x)=6 it is f(x)=6+1/x-1/x

[deep elbow]

I ain’t sure tho

[formal quarry]

mhmm

could be

i guess ill ask my prof soon

anyone else? if they wanna try

[formal quarry]

even photolab is showing this as answer

so maybe im wrong

cuz im doing it from the fucntion i find at the end

whereas maybe ur suppose to do it from the function given at hand before

but can sm1 tell why? or is the question the worded weirdly

[devout snow]

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[viral sluice]

Hello

[viral sluice]

I have a question regarding notation

I need to write "If for every natural number p, the inequality m - n ≤ p holds, then m < n"

[cold bough]

just ask for asking question

[viral sluice]

Yeah sorry I was translating :D

I need to write it using ∀, ∃...

But I'm not 100% sure how to do it

[junior chasm]

if ∀ p (belongs to symbol) N

[viral sluice]

Can't use if

that is the problem, I can't use any words

[cold bough]

$m, n \in \mathbb N;:;m-n\le p;; \forall p\in \mathbb N \implies m<n $

[viral sluice]

$m, n \in \mathbb N$, $m-n\le p \forall p\in \mathbb{N} \implies m<n $

would that work?

[woven radish]

dp

[cold bough]

I would add : before m-n

: is like s.t.

[viral sluice]

s. t.¿

so that?

[cold bough]

such that

[viral sluice]

oh ok :D

thanks!

.close

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[faint gorge]

$\textbf{Exercise.}\$
Let $a,b,c > 0$ with $a+b+c=1.\$
Show that
[ \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \leq 3+2\cdot \frac{a^3+b^3+c^3}{abc} ]

$\textbf{Proof.}\$
WLOG
[1 = a+b+c > a+b > a > 0 \implies a,b,c \in (0,1) ]
Let
[x = \frac{1}{a}, \quad y = \frac{1}{b}, \quad z = \frac{1}{c} ]
Then I get
[ \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1 \implies x,y,z > 1 ]
and thus prove
[ x+y+z \leq 3 + 2\cdot \frac{\left ( \frac{1}{x} \right )^3 + \left ( \frac{1}{y} \right )^3 + \left ( \frac{1}{z} \right )^3}{\frac{1}{xyz}} ]
I will use the $\textbf{generalized mean or power mean}$ by Hölder.
[ \boxed{ \left ( \frac{1}{n} \sum_{i=1}^n x_i^{\beta} \right )^{\frac{1}{\beta}} \leq \left ( \frac{1}{n} \sum_{i=1}^n x_i^{\alpha} \right )^{\frac{1}{\alpha}} \quad \forall \alpha, \beta \in \mathbb{R}\setminus {0} \text{ where } \alpha > \beta } ]
getting the following
\begin{align*}
&\quad \frac{x+y+z}{3} \leq \left ( \frac{ \left ( \frac{1}{x} \right )^3 + \left ( \frac{1}{y} \right )^3 + \left ( \frac{1}{z} \right )^3 }{3} \right )^{\frac{1}{3}} \
&\implies \left ( \frac{x+y+z}{3} \right )^3 \leq \frac{\left ( \frac{1}{x} \right )^3 + \left ( \frac{1}{y} \right )^3 + \left ( \frac{1}{z} \right )^3 }{3} \
&\implies \textcolor{cyan}{x+y+z \leq} \left ( x+y+z \right )^3 \leq \frac{\left ( \frac{1}{x} \right )^3 + \left ( \frac{1}{y} \right )^3 + \left ( \frac{1}{z} \right )^3 }
{3^2}
\end{align*}
\begin{align*}
\implies x+y+z &\leq \frac{\left ( \frac{1}{x} \right )^3 + \left ( \frac{1}{y} \right )^3 + \left ( \frac{1}{z} \right )^3 }
{3^2} \
&\leq \left ( \left ( \frac{1}{x} \right )^3 + \left ( \frac{1}{y} \right )^3 + \left ( \frac{1}{z} \right )^3 \right ) \cdot xyz \
&= \frac{\left ( \frac{1}{x} \right )^3 + \left ( \frac{1}{y} \right )^3 + \left ( \frac{1}{z} \right )^3}{\frac{1}{xyz}} \
&\leq 2\cdot \frac{\left ( \frac{1}{x} \right )^3 + \left ( \frac{1}{y} \right )^3 + \left ( \frac{1}{z} \right )^3}{\frac{1}{xyz}} + 3\
\end{align*}

[faint gorge]

I would like to know (again) if my proof is correct or if there is some fallacy

[eager nova]

Wait why do you say that a+b+c >= a+b?

We know c>0 so that can’t be true

[faint gorge]

If I add a positive value c to a+b then a+b+c should be larger

[eager nova]

But that is >

Not >=

[faint gorge]

I see

[eager nova]

Right? Maybe i am misunderstanding

[faint gorge]

you are right

[eager nova]

Same happens with a+b>=a

[woven radish]

bacc

[devout snow]

@faint gorge Has your question been resolved?

[faint gorge]

I corrected the technicality

but I still would like someone to go over everything

<@&286206848099549185>

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@faint gorge Has your question been resolved?

[devout snow]

@faint gorge Has your question been resolved?

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@faint gorge Has your question been resolved?

[faint gorge]

.close

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[ruby rover]

How do u solve this?

[merry pivot]

newtons method?

[ruby rover]

This is exponential

Equation

[ancient sluice]

(2^x +1) = 1/3x
or 2^x = (1-3x)/3x
or x = 2^(1-3x)/3x

y = x - 2^(1-3x)/3x

[supple knot]

What techniques are you learning

[ruby rover]

Laws of exponent

[ancient sluice]

If x = 1/3 y < 0 , if x = 1 y > 0

So a root between, (1/3,1)

I don't know if there is a way to exactly find the root , without approximations

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[restive river]

pls help T^T

[still elk]

show working

[restive river]

i did

factorize x^2 - 9 and (x^2 - 5x + 6)

i did that too ;-;

[still elk]

you didn't show the working, you showed your attempted answer. How did you get to that answer?

[restive river]

okay wait

my notes T^T

hey

helppp ive been stuck on this for too long

[restive river]

first find the LCM then cross multiply it

have u tried that method

yea

for LCD i got (x+3)(x-3)(x+2) as u can see on the pic

but not sure why it's not working

what did i do wrong

<@&286206848099549185>

[solid osprey]

maybe try expanding it?

wait nvm lmao

you factored x^2-5x+6 wrong

[fervent hornet]

its (x-2) not (x+2)

[restive river]

ok

[restive river]

i dont understand pls explain ;-;

[solid osprey]

find a number that when multiplied is +6 and when added is -5

[fervent hornet]

x^2-5x+6=(x-2)(x-3)

[restive river]

,, x^2 - 5x + 6 = x^2 - 2x - 3x + 6 = x(x-2) - 3(x-2) = (x-2)(x-3)

[woven radish]

anjali

[restive river]

ab = 6
a+b = -5

a = -3 and b = -2

ty it was a sign error

this was the 10x time today

anyway ty

.close

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[potent whale]

Does anyone know how I’d progress to make this logically equivalent using the laws of propositional logic?

[steep tide]

Write the second one in this form too...

q -> (p v r) becomes not q v (p v r)

And we can just open the bracket (its a rule of v)

So

not q v p v r

[potent whale]

I see

Thank you 😊

[steep tide]

Yeah no problem

[potent whale]

.close

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[neat flax]

x2 - 9(x+2)
x3 +2x2 -9x -18
then use calculator
menu 9 then 2 then 4 then put the number into calculator

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[dark imp]

can someone explain D? i understand up to cos(pi*t/8) = -1/2

[devout snow]

@dark imp Has your question been resolved?

[rotund umbra]

$$\cos (\frac{\pi\theta}{8}) = -\frac{1}{2} \implies \frac{\pi\theta}{8} = \cos^{-1}\left( -\frac{1}{2} \right) = \frac{3 \pm 1}{3} \pi \qquad\mbox{($180 \pm 60$ degrees)}$$
$$\therefore \theta = \frac{8}{\pi} \frac{3 \pm 1}{3} \pi = \frac{16}{3} ;\mbox{or}; \frac{32}{3}$$

[woven radish]

Shuba

[rotund umbra]

does this explain it?

[dark imp]

the question asks to use a 'by hand' method

but i dont understand how my teacher did it

i understand how you did it but like i dont think thats the way im 'supposed' to do it

@rotund umbra

do you know how to explain it in the way that my teacher did it?

<@&286206848099549185>

[topaz zephyr]

Wdym

For algebra?

[dark imp]

im assuming by using the unit circle and with reference angles?

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[restive river]

Hi

[restive river]

can anyone hlep me with this

how did we know we need to subtract theta not add

[warm breach]

what's the original question

[devout snow]

@restive river Has your question been resolved?

[restive river]

i got it

nvm

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[wise arch]

can you help me?

[crude wasp]

yeah so basically this simplifies to just an integral in the y-z plane

now you can actually draw it

(also idk why the z=2 surface is there, it doesn't seem to affect anything cus z < 2 anyway)

then ur integrating

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[restive river]

need help with this question
Idek how to approach this. I tried adding the two terms and trying to find a numerator to cancel the denominator but i just get stuck and it leads to nowhere. I also tried to solve it by breaking it down to sine and cos components but it just got worse

[devout snow]

@restive river Has your question been resolved?

[crude wasp]

remember that cot = 1/tan

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[formal quarry]

How would I find the tangent line of the previous path?

[wise arch]

thanks LY 🌺

[formal quarry]

oh were you still using this channel? it was marked as open but if you still need it i can use a different one

[wise arch]

its okay my problem is solved

you can use it

[formal quarry]

I think I could plug t=2 in for the velocity, and that would be the direction vector. If I were to do r(2)+tv(2), that would be my answer right?

[formal quarry]

ok cool, just wanted to make sure

[devout snow]

@formal quarry Has your question been resolved?

[devout snow]

@formal quarry Has your question been resolved?

[restive river]

i already know that :/

[flint spire]

thats close to the correct answer

think of a reason why that is wrong

[formal quarry]

is it because i put t in it again?

should it be r(2)+v(2)?

[flint spire]

no, what 😛

[formal quarry]

I dunno, it was just a thought

[flint spire]

but youre not gonna submit an answer based solely on a thought!

lets do this the right way

"and it continues to travel with constant velocity..."

$\forall t \ge 2 : \quad \vec{v}(t) = ?$

[woven radish]

Emily

[flint spire]

i gotta go, i will be right back

write v(t) then intgrate it to get r(t)

brb

[formal quarry]

I dont think I know how to integrate that

[flint spire]

okay, i will guide you through

$\forall t \ge 2 : \quad \vec{v}(t) = ?$

[woven radish]

Emily

[formal quarry]

I don’t think I know what that means. I’ve never seen this notation before

[flint spire]

im just saying, for all time t >= 2

what will the velocity be?

[formal quarry]

Ah

It should be (0,16,2) right?

[flint spire]

no, im not substituting yet

logically, engines stopped at t=2

so it continued to travel with constant velocity

[formal quarry]

ok so (2,9t^2-2,1+4/t^2)?

[flint spire]

yeah, thats the v(t) when the engines were working

when they stopped at t=2, the spaceship continued to travel with which velocity?

dont calculate anything

just think about it logically

[formal quarry]

It’d be the same as right before they stopped?

[flint spire]

yesss

$\forall t \ge 2 : \quad \vec{v}(t) = \vec{v}(2)$

[woven radish]

Emily

[flint spire]

velocity will remain constant

[formal quarry]

Right

[flint spire]

and it is equal to the velocity at t=2

thats done

$\forall t \ge 2 : \quad \vec{r}(t)=\int \vec{v}(t) , dt$

[woven radish]

Emily

[flint spire]

just integrate that

similar to how we usually integrate

[formal quarry]

(2t,3t^3-2t,t-4/t)? But thats exactly what we started with

[flint spire]

im not asking you to substitute what were given

[flint spire]

what? no

the engines stopped, the velocity is now constant, it is equal to v(2)

its not longer (2, 9t²-2, 1+(4/t²))

$\forall t \ge 2 : \quad \vec{r}(t)=\int \vec{v}(t) , dt = \int \vec{v}(2) , dt = , ?$

[woven radish]

Emily

[formal quarry]

I suppose I was under the impression that I wasn’t supposed to enter in 2 for t yet for some reason. Let me try that again

[flint spire]

no, dont substitute

just integrate the way it is

what is the antiderivative of the constant v(2) ?

[formal quarry]

So if I’m not supposed to substitute yet, then it’s just r(2)?

[flint spire]

omg.. no

the derivative of a constant, lets say K, is ?

if you had $\int K , dt= , ?$

[formal quarry]

0

[woven radish]

Emily

[flint spire]

the integral, my bad

not derivative :p

[formal quarry]

Then it’s Kt

[flint spire]

yeah, same thing here

$\forall t \ge 2 : \quad \vec{r}(t)=\int \vec{v}(t) , dt = \int \vec{v}(2) , dt = , ?$

[woven radish]

Emily

[formal quarry]

So v(2)t?

[flint spire]

yess

but dont forget our constant of integration

[flint spire]

Kt+c

[flint spire]

$\forall t \ge 2 : \quad \vec{r}(t)=\vec{v}(2) , t + C$

all you need to determine now is the constant C

from the condition(s)

[woven radish]

Emily

[flint spire]

try to figure it out!

[formal quarry]

Well, I could rearrange the equation and solve it for C. Which gives C=r(t)-v(2)t

[flint spire]

are you kidding me? 😛

substitute an initial condition

[formal quarry]

Yes, somewhat actually

[flint spire]

from the exercise

[formal quarry]

That wasn’t intended as my actual answer lol

[flint spire]

to determine C

[formal quarry]

An initial condition?

[flint spire]

yes, whats a condition that we could use

theres only one in this exercise

and its when t=2

because at t=2, engines stop

[formal quarry]

Right

[flint spire]

so, we put t=2 in here

[formal quarry]

C=(4,-12,-4)?

[flint spire]

do not substitute

i said it multiple times

just work with letters

[formal quarry]

You said to substitute here. Did I misunderstand?

[flint spire]

t=2 only

replace t by 2

and get the constant!

[formal quarry]

Ok, so r(t)=v(2)2+C?

[flint spire]

r(2)

[formal quarry]

Ok. And since you don’t want me to use the numbers, my answer would be C=r(2)-2v(2). And i actually mean it this time since I’m not sure what other way I could solve for C

[flint spire]

no no, youre good

yes, thats C

you finally have the formula of r(t) when engines stop

[flint spire]

substitue and simplify

substitute C i mean

[formal quarry]

I figured

I appreciate you making sure I understood though

[flint spire]

final expression of r(t) is then

[formal quarry]

r(t)=v(2)t+r(2)-2v(2). Or r(t)=v(2)(t-2)+r(2). I feel like I might be able simply this further, not sure how though

[flint spire]

no no, its fine

thats the best form

do you notice something

your guess looks similar

whats the difference?

[flint spire]

r(t) = v(2) t + r(2)

and actual ans r(t) = v(2) . (t-2) + r(2)

[formal quarry]

I mean that makes sense though, since it only happens once t=2

[flint spire]

yes, its only for t>=2

using your guessed formula

if you put t=2

[formal quarry]

I’m honestly surprised I didn’t notice that was the issue

[flint spire]

you get v(2) 2 + r(2)

where you should be getting only r(2)

[formal quarry]

Right

[flint spire]

in the actual formula, t=2 gives r(2) as required

[formal quarry]

Mhm

[flint spire]

so yeah, thats about it

thats how you do it from scratch

Newton's 2nd law

good job!

[formal quarry]

I’m just glad my first answer was that close to the actual one

Shows I was on the right track

[flint spire]

yeahh

after all, it looks like a 1D uniform motion

a=0, v=constant=v0, x=v0 t + x0

[formal quarry]

Kinda. I think I see what you mean

[flint spire]

thats why you guessed v(2) t + r(2)

from x=v0 t + x0

but time axis gotta be shifted in this problem

cuz engines do not stop at t=0, but at t=2

gotta turn that t into a t-2

[formal quarry]

Mhm, makes sense

[flint spire]

last thing i wanna say

you see how better it is to just work with letters, symbols

not substitute anything

once you derive the final expression, then you can subsitute what you were given, for real

i mean v(2) and r(2)

and its done

[formal quarry]

Ehh kinda? I was taught that I should eliminate variables by filling them in when I can, so that’s why I kept trying to substitute

Oh wait you’re the person who helped me last night too

Thanks for helping me again lol

[flint spire]

yess, i am xD

i like these physics problems, they challenge my knowledge

especially if they involve vectors

[flint spire]

youre welcome!

[formal quarry]

Well I have a feeling I might be seeing you a lot then. I’m taking multivariable and vector calculus right now

[flint spire]

haha, hopefully

wish you luck, adios

[formal quarry]

.close

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[sage kernel]

What can i rewrite (ln(n))/(ln(n+1)) as?

[devout snow]

@sage kernel Has your question been resolved?

[sage kernel]

<@&286206848099549185>

[supple knot]

do you have a guess what this approaches as n goes to inf

[sage kernel]

inf/inf?

[supple knot]

the ratio specifically, not each of them separately

[sage kernel]

is it just 1?

[supple knot]

yes

if you weren't shown that in class, you could prove it using either MVT or L'Hopital

,w lim n to inf of log(n) / log(n+1)

[woven radish]

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Requested by rie.mann

[sage kernel]

Ok so is this correct?

[supple knot]

,w sum n=2 to inf 2^n / (n^2 log(n))

[woven radish]

Results provided by WolframAlpha

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Requested by rie.mann

[supple knot]

yup

[sage kernel]

Thanks @supple knot

.close

[devout snow]

Channel closed

Closed by @sage kernel

Use .reopen if this was a mistake.

[devout snow]

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[formal quarry]

I'm not certain on what the distance from P to (0,2,0) is. I believe that the distance from P to the xz plane is y though

[formal quarry]

oh wait

would it be 2? sqrt(0^2+2^2+0^2)?

[mystic scarab]

Nope

[mystic scarab]

Almost, it's |y| (because y could be negative, whereas a distance is, by definition, always non negative)

[formal quarry]

oh yeah youre right

[mystic scarab]

For this part, let's say we call (x, y, z) the coordinates of P. Then what's the distance from P to (0, 2, 0) ?

[formal quarry]

sqrt(x^2+(2-y)^2+z^2)?
i know x and z would be negative, but because theyre squared i left it out

[mystic scarab]

Yep, exactly

[mystic scarab]

Yeah, good

[formal quarry]

ok, so i have to set that equal to |y| and solve?

[mystic scarab]

So now you have $\left|y\right| = \sqrt{x^2 + (y - 2)^2 + z^2}$

[woven radish]

Alberto Z.

[mystic scarab]

Hence, ${\left|y\right|}^2 = {\left(\sqrt{x^2 + (y - 2)^2 + z^2}\right)}^2$

[woven radish]

Alberto Z.

[mystic scarab]

Therefore, $y^2 = x^2 + (y - 2)^2 + z^2 \implies y^2 = x^2 + y^2 -4y + 4 + z^2$

[woven radish]

Alberto Z.

[mystic scarab]

$\cancel{y^2} = x^2 + \cancel{y^2} -4y + 4 + z^2 \implies x^2 - 4y + z^2 + 4 = 0$

[woven radish]

Alberto Z.

[mystic scarab]

Which turns out to be the \textit{paraboloid} $y = \frac{x^2}{4} + \frac{z^2}{4} + 1$

[woven radish]

Alberto Z.

[mystic scarab]

Indeed, if you recall the definition of a parabola it was the locus of points that have the same distance from the focus (which is a point) and from the directrix (which is a line)

This exercise is the 3D version of the parabola definition, that's why you get a paraboloid 😉

[formal quarry]

that makes sense

I appreciate the help

[mystic scarab]

You're very welcome 🤗

[devout snow]

@formal quarry Has your question been resolved?

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[minor silo]

help

[minor silo]

did precalc and trig in hs. freshman yr of college i did accounting. changed majors and transferred for sophomore year and on calc 1. i just dont know what im even looking at. regarding derivatives

[dreamy mirage]

What's the context? Does Glenn start at point A?

What do the x and y axes represent?

[minor silo]

The graph below shows Glenn's distance, G(t)
, in miles, from camp as a function of time,
in minutes, as he runs from a group of zombies.

[dreamy mirage]

So she goes towards and away from camp multiple times right?

[dreamy mirage]

And about this

is it specified where Glenn starts moving?

[devout snow]

@minor silo Has your question been resolved?

[minor silo]

its his point a i believe. idk i was focusing on the gateway exam in which its pass or fail. if i failed i would have to drop the class

[flint spire]

if you read next questions, you figure out he starts where the curve starts

(0 , 0.4)

[minor silo]

thank you.

is what im doing is finding delta y/delta x?

[flint spire]

yes, the average velocity between two points P1(t1,d1) and P2(t2,d2) is given by Δd/Δt

(d2-d1)/(t2-t1) or (d1-d2)/(t1-t2) , same

[minor silo]

im sorry but may i ask u to use it in a problem? i do a lot better when understanding if its interpreted with actual numbers

[flint spire]

lets use it for the first question within b)

[minor silo]

ill show u what i have i was just plugging in numbers

the .05 is correct i believe i forgot to add the extra zero before checking answers

[flint spire]

nope, youre applying the average velocity formula wrong D:

try to do it again

[minor silo]

okay

oh wtf

its .1 y axes

axis

on 5 to 6, or C to D.
down .15
over 1.0

answer is .15

negative bc its going down.
-.15

[flint spire]

thats right

[minor silo]

i have figured it out. now onto the next part.

i believe i got the multiple choice right

second multiple choice is A

[flint spire]

its true that Glenn's average velocity between t=0 and t=5 is 1.8 mph

but was Glenn actually walking 1.8 mph all the time between t=0 and t=5 ?

[minor silo]

oh no

he wasnt

[flint spire]

"was taking it easy only walking..."

only walking 1.8 mph

[minor silo]

oh

[flint spire]

that only

[minor silo]

im sorry

im so sorry

im not stupid its just that im so preoccupied dealing with other issues atm

[pale bear]

Dudes distressed

[flint spire]

why are you apologizing, no worries 😛

[minor silo]

im about to crash out bro

with the transfer and the change of major i have had zero time to do anything

okay yes he was walking 1.8 mph the whole time

[flint spire]

he was not, thats the thing

Rick was wrong assuming that Glenn was walking at constant speed

[minor silo]

THERES 2 PEOPLE IN THIS QUESTION?

[flint spire]

lol

[minor silo]

it;s A

the answer is A

[flint spire]

right, its A

[minor silo]

and then the second multiple choice is A too

[flint spire]

having average velocity of 1.8 mph doesnt imply he was 1.8 mph the whole time

[flint spire]

but wait, before that

how about the max velocity

its not 0.03

[minor silo]

.08

something

okay actually, before that let's work on C-F?/

i am ready and focused now

At what letters is the slope of the tangent line positive?
in order to do that I have to find the slope of the points first correct

[flint spire]

no no, just imagine the tangent line at the point

if the line points up, slope is positive

[minor silo]

where do i imagine it?

i dont have time to imagine i need to wake up

[flint spire]

its tangential to the curve at the point itself

[minor silo]

okay

[flint spire]

easy to imagine/draw in your mind

[minor silo]

something like this

[flint spire]

yess

[minor silo]

okay so if i were to imagine a line

it would be

[flint spire]

if the line points up, like shown, slope is positive

if the line points down, slope is negative

if the line is horizontal, slope = 0

[minor silo]

In this case, I think the line points down

[pale bear]

Kenny are you feeling better now

[flint spire]

its pointing up

[minor silo]

im in the right mindset to learn right now

[flint spire]

its going up

[minor silo]

well that was just a graph i found on google

[flint spire]

okay, which letters do you think slope is > 0

[minor silo]

B C D E F

[flint spire]

no, how 😛

[minor silo]

HAHA

i dont know im about to freak

or