help-38

[inland orbit]

I think it's clearer with this example

[lone mauve]

OK

Alright

[inland orbit]

Does it make sense how we just plugged G(x) into F?

[lone mauve]

I’m going to ignore f(x) to start with if that’s possible
So..
g(x) = x+1
f(g(x)) = f(x+1)

All I’ve really done is add f function to the beginning of the g function now right?

[inland orbit]

Basically yeah

So far so good

But this is when you need to stop ignoring f

[lone mauve]

Yeah

I’m trying to figure out how f gets from the front to the end now

[inland orbit]

if f(x) = x^2

Then what is f(x+1)?

[lone mauve]

(x+1)^2

[inland orbit]

Almost

but you don't need the f there

[lone mauve]

Hmmm

[inland orbit]

if f(x) = x^2
then f(x+1) = (x+1)^2

we just replaced x with x+1

[lone mauve]

Right

Kept the parentheses

That’s very important right?

[inland orbit]

well, that's just because x+1 is more than one symbol

f(x) = x^2
f(4) = 4^2
f(a) = a^2
f(b+c) = (b+c)^2
f(x+1) = (x+1)^2
f(4t) = (4t)^2

a variety of examples

Some require parentheses because they're more than one symbol

[lone mauve]

Yeah.. I’m thinking of this like declaring variables as code and I think it’s kinda confusing me. I’m used to seeing the variable name when I call it but in math it’s replaced by a real value and I need to try and remember where that variable was

[inland orbit]

in math it can be replaced by anything lol

yeah the notation is different than programming

[lone mauve]

If that makes sense.. math doesn’t seem to have the liberty of colour coded IDE like VSCode.. or if it does I need to download it 🙂

[inland orbit]

You can download different colored pens lol

[lone mauve]

Haha

I do wonder though if there is a program that will do this? It would be like VSCode but for mathematicians

Typing is more challenging on keyboards.. I’m using GoodNotes on iPad and seems to be pretty decent for math

Anyways I do hear what you are saying..

Let me think this over

With these

[night zodiac]

f(x) = x^2
f(4) = 4^2
f(a) = a^2
f(b+c) = (b+c)^2
f(x+1) = (x+1)^2
f(4t) = (4t)^2 

Technically this

[lone mauve]

With these examples are you overwriting the old function f(x) at the very top?

Should the following ones be g,h,I,j,k, etc?

[night zodiac]

No

It's not getting overridden

[lone mauve]

Or you are showing exactly what f(x) at the very top can do to literally anything you plug in

I see I see

I get it now

[night zodiac]

It's the same function with different things plugged in

[lone mauve]

So FoG

[night zodiac]

Like f(2) means plug 2 into the function f(x)

Same idea as f(3)

But if you change the letter to g(x), that means plug 2 into the function of g(x), which is a new function

[lone mauve]

f(x) = 3
g(x) = (x + 5)

f(g(x)) = …

[inland orbit]

If f(x) = 3

That's a constant function

[lone mauve]

({3} + 5)({3})?

I’m calling 3 twice

[inland orbit]

I'm not sure what this means?

[lone mauve]

{3} to show it’s from f(x) = 3

[night zodiac]

As mentioned, f(x) = 3 is a constant function, mean it does not change with x, it's constant

So there's not a solution to f(g(x))

[lone mauve]

Oh

[night zodiac]

If it was f(x) = 3x, that's different

[inland orbit]

If f(x) = 3, then f(anything) = 3

It's a constant function

f(g(x)) = 3 for any x

if f(x) = 3

[split chasm]

(where g(x) is defined)

[inland orbit]

yes

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[inland orbit]

I think maybe you got mixed up between f(g(x)) and g(f(x))

[night zodiac]

@lone mauve What even is the question you have still?

[inland orbit]

If we take f(x) = 3 and g(x) = x+5, then

f(g(x)) = f(x+5) = 3
and
g(f(x)) = g(3) = 3+5 = 8

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[lone mauve]

.reopen

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✅

[lone mauve]

If we take f(x) = 3 and g(x) = x+5, then

f(g(x)) = f(x+5) = 3
and
g(f(x)) = g(3) = 3+5 = 8

This explains it perfectly! Thank you @inland orbit

I finally understand now

It’s like using variables in programming

Calling a function with a variable

function(var){}

they are literally just what they equal

I was over complicating things

[night zodiac]

You shouldn't compare things to programming like that. Yes, CS and math are similar but they are also different so you won't be able to find a perfect analogy that makes sense

[lone mauve]

https://tenor.com/view/why-gif-23768806

[lone mauve]

Actually in this case I think it’s a perfect analogy. But instead of calling a var inside of a function it’s calling a function inside of a function.

[inland orbit]

Yes, it is like calling a function in a function, that's a good way to think of it

But also @night zodiac is right. You won't always be able to find a clean analogy between mathematical ideas and programming

But in this case, yes, it is quite a nice analogy

[lone mauve]

Well, I would agree: that’s a better way of putting it. “Won’t always be able to”, rather than “shouldn’t compare” and “won’t be about to find”

[night zodiac]

It was suppose to say "won't be able"

[inland orbit]

Well

Analogies are always helpful in understanding

But programming and mathematics are not the same

[night zodiac]

This was my point

[inland orbit]

The analogies will break down at some point. You cannot count on them. You may use the analogies to guide you but you should try to understand the mathematical logic as it is

[inland orbit]

Yes, I agree with you

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[hazy light]

$\cos(\pi-x)=-\cos{x}$

[solid kiln]

QuantumBee

[hazy light]

For any acute angle x, pi-x lies in the second quadrant

[cloud canopy]

Oh ok, ASTC?

[hazy light]

And, Cos(x) is negative in second quadrant

yeah

[cloud canopy]

I understand it now. Thanks mate🥰

[hazy light]

you can also check with $$\cos(a-b)=\cos{a}\cos{b}+\sin{a}\sin{b}$$

[solid kiln]

QuantumBee

[cloud canopy]

That's cool

Thank you!

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[hazy light]

nice username :)

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[frail current]

You flip a weighted coin that comes up Heads 40% of the time and Tails 60%. If you flip this coin 5 times, what is the probability that you see at least 3 Tails, rounded to the nearest percent?

[frail current]

anyone wanna help me?

would it be this (5c3) + (5c4) + (5c5) / 2^5?

if this was a fair coin it would be .5 right?

[amber python]

why /2^5?

[frail current]

thats the total number of flips

[amber python]

but getting heads has a different probability to getting tails

so its not just simple counting thing anymore

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[frail current]

I said assuming we’re talking about a fair coin in this case

[amber python]

well in that case its fine

[amber python]

but you already have the answer dont you

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[onyx dock]

"Given directed line segment RT below, find the coordinates of such that the ratio of RS to ST is 5:1. Plot point S"

[onyx dock]

R=(7,-1)

T=(-5,-5)

8th Grade geometry class, my grade is already an 80, begging for help to learn 😭

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[wraith hinge]

any hints?

from rank nullity, i see that i need a list of n^2-1 elements

and eij where i/=j comprises of n^2-n elements which can be in the basis

so gotta find independent list of length n-1

with only diagonal entries ofc

i can find floor(n/2) more, if we take (e11-e22),(e33-e44)....

floor(n/3) if we do something similar like (e11+e22-2*e33)...
seems like i can do it so on, floor(n/4), floor(n/5)...

<@&286206848099549185>

ping me if anyone answers

[wraith hinge]

nvm, if this was true rank nullity would get violated

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[latent pelican]

needa help

how do i read this

what does it mean

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[latent pelican]

bruh

<@&286206848099549185>

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[primal musk]

hey all, having a bit of trouble with #9 here, I'm fairly certain I have to use substitution or elimination but I'm not really sure how to get that started

[primal musk]

oh and the question is, use the definition of continuity to find the values of k and/or m that will make the function continuous everywhere

actually i believe i figured it out ill try and come back to see if im right

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[rough hinge]

From this equation I need vo in terms of g and D but I can't do it

[rough hinge]

I'm getting here but I can't solve further

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<@&286206848099549185>

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[tidal igloo]

hi. The numbers a and b satisfy the equation 56a = 65b. Prove thata+ b is composite. Is this reasoning incorrect: a=65b/56 composite numbers is a concept defined only for natural numbers so a+b is a natural number, therefore b+65b/56=121b/56 ∈ N therefore 56 l a+b

[tidal igloo]

<@&286206848099549185> plz help lol thx im an idiot

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[quiet remnant]

r^4 - 6r^3 -89 r^2 -294r+2401 = 0

[quiet remnant]

I already tried with rational root theorem

and neither +- 1, 7, 49, 343, 2401(divisors of 2401) give 0 in the final one

I want to clarify I already know the result I just have to do it by hand.

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<@&286206848099549185>

[burnt mulch]

,w r^4 - 6r^3 - 89r^2 - 294r+2401=0

[solid kiln]

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Requested by enclave wangedrocht#1465

[quiet remnant]

Ok

But how do you get to that

by hand

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[keen plover]

You probably don't

There's a formula, but good luck remembering it

It's big

[quiet remnant]

Lol

[viscid flower]

root finding method maybe but idk how youre gonna end up at these numbers

[quiet remnant]

I already did it

[viscid flower]

where are the centers of these roots

no i mean like

eulers or something

[quiet remnant]

Wdym

[viscid flower]

its not gonna get you an exact answer anyways

since the roots are so bad

hrm

that symmetry of the 17 one

[quiet remnant]

I did it with synth division

But I didnt get the result

Which is 3.67 btw

Well there are two results 3.67 and 13.32

But yeah

[viscid flower]

well i guess its symmetry around 8.5

maybe you could guess

the solution takes the form (x-8.5-a)(x-8.5+a)(Bx^2 + Cx + D)

where the final quadratic is nonfactorable

youd have to inform this guess with something

you could do like

bisection or something idk

somehow or another youd end up that it looks like you have those two roots about the point

guess its 8.5

[quiet remnant]

This si confusing tbh

Heheh

Is*

[viscid flower]

youll have to inform a guess some how

[quiet remnant]

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[wraith hinge]

Isn't the inequality wrong here?
I'm getting x - 2 - (2/x+1) > 0
Am I wrong?

[acoustic trout]

how did you get x-2-2/(x+1)?

[wraith hinge]

I took the minus sign out from the LHS, reversing the inequality
I then brought the RHS Fraction term to the left.

The original question:

This is how I got my answer:

[solid kiln]

~~

~~

~~

~~

[wraith hinge]

<@&286206848099549185>

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[smoky sundial]

not sure how to do this. any help would be appreciated

[edgy cedar]

Express 9 interma of 3

[smoky sundial]

i understand that but im not too sure how to do that

[mystic grotto]

It might be the wording tripping you up

I'll just say that 9 =3^2

It just wants 3^(something to do with x)

[smoky sundial]

oh i see

ok that makes a lot of sense now

nvm i still dont know what to do from there

[smoky sundial]

i understand this but dont know what to do next

[mystic grotto]

Ok so

Imma write it a bit weird (9)^4x+6

9=3^2

(3^2)^4x+6

@smoky sundial

What happens when you raise a power to another power

[smoky sundial]

do you just multiply them together?

[mystic grotto]

Yeh

So 3^...

[smoky sundial]

so would it be 3^8x+12?

[mystic grotto]

Yep

[smoky sundial]

Perfect, thanks!

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[humble yew]

Guys is making this type of statement valid?

[stark bison]

Not that the statement is true but sure that is still a statement nonetheless

[humble yew]

would this be considered true then?

[random citrus]

No

[stark bison]

No, that's false

Could think of millions of counterexamples

[humble yew]

yeah, i realized that after i posted it

[sharp arrow]

Wait hold up how’s that false

Also the first statement, i feel its true

@stark bison enlighten me

Umm wait

[random citrus]

Pick x and y at random then plug them in. See where that gets you

[sharp arrow]

Da

I’m dumb

I was reading it wrong

Yes both wrong

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[humble yew]

what about

[sharp arrow]

No

[humble yew]

why not?

[sharp arrow]

Think about x=4680 y=1

[stark bison]

x,y = 0 is a counterexample

[humble yew]

4680 can be written as 4680/1

[sharp arrow]

,w 4680/(4680^2 +1^1)

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[humble yew]

0 is also part of Q right?

0/1. 0/2

[stark bison]

4680/0 is not

[sharp arrow]

Yes

[stoic iron]

this looks very rational

[sharp arrow]

Yees my b

[humble yew]

but any value in Q is of the form p/q where p and q are integers

0 is also considered an integer

[stoic iron]

no, q must be non zero

thats in the definition

[humble yew]

so x and y would have to be in the set of natural numbers for the statement to be true

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[vast flint]

So I have a function $f:(X,d)\to(Y,\rho)$ between two metric spaces with $a\in X$ a limit point of $X$. I define the set $A_a(f)\subseteq Y$ by $b\in A_a(f)$ iff there exists a sequence ${x_n}\subseteq X\setminus{a}$ such that $x_n$ converges to $a$ and $f(x_n)$ converges to $b$. Suppose $A_a(f)={f(a)}$. I am trying to show that $f$ is continuous at $a$. I have shown that for any ${x_n}\subseteq X\setminus{a}$ such that $x_n$ converges to $a$, if $f(x_n)$ converges in $Y$ then it converges to $f(a)$. I have not been able to show that $f(x_n)$ converges in general though.

[solid kiln]

Beous

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[neat knoll]

Hello, I am studying discrete mathemathics and I am preparing myself for an exam, I am stuck on the set notations (Union, Intersection etc.) I have a question what does the bar above the variable means?

[stoic iron]

can you show it in context

[neat knoll]

Yes, is ok If I post a pic of the HW?

[stoic iron]

sure

[stoic iron]

so its most likely the complement of C

[neat knoll]

Can you give me an example please?

[stoic iron]

its U-C

everything in the universal set that isnt in C

[neat knoll]

Oh ok

So I forgot to remove 5 on the answer

[stoic iron]

indeed

[neat knoll]

Good

Thanks

How do I put the channel in available?

[stoic iron]

``.close`

[neat knoll]

.close

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[neat knoll]

Thanks

.close

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Please don't occupy multiple help channels.

[bright spire]

Wdym

[vagrant marsh]

#help-29

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[wraith hinge]

i need help with b ive done work for 9. a)

[wraith hinge]

a) e because it is after the letter d in the english alphabet

confused on b tho

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[wraith hinge]

its fine

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[molten comet]

This is embarrassing, but someone please help me

[keen plover]

Try factoring by grouping

Aka, for the cubic, factor the first two terms and second two terms separately and see if it can be factored further from there

[molten comet]

How?

x(x^3 - 3x^2) + x(x - 3) = 0

Like this?

[keen plover]

Dont distribute the x

The objective is to factor, not unfactor

[molten comet]

Should I use synthetic division?

[keen plover]

That's unnecessary

Look at x³ - 3x² + x - 3

[molten comet]

Sorry, I didn't understand this

[keen plover]

Try factoring x³ - 3x² and factoring x - 3 (which can't be factored anyways)

[molten comet]

x^2(x - 3)

(x^2 + 1)(x - 3)^2 = 0?

[keen plover]

Why (x - 3)²?

[molten comet]

Because there's two of them

?

[keen plover]

Then why didn't you get x⁴ when factoring the first time?

[molten comet]

(x^2 + 1)(x - 3) = 0

So like this?

[keen plover]

Yes

And don't forget the original x that was there

And now it's factored as much as possible, at least in the reals

[fluid gulch]

Hey can someone help me with question b

[molten comet]

No

[keen plover]

#❓how-to-get-help

[molten comet]

Oh okay

x(x^2 + 1)(x - 3) = 0

[keen plover]

Yep

[molten comet]

So x equals to 0, 1i, -1i, 3?

[keen plover]

Remember the problem asks for real solutions

[sharp arrow]

They asked for real roots right

[molten comet]

Ooo they did

So 0 and 3

Ayyy

Tysm guys 🙂

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[proven bobcat]

did i do something wrong?

or is this my answer

[viral shore]

u can simplify this more

theres still a common factor of 2

[proven bobcat]

right, so

2

2(3x^2 - x +1)

?

[viral shore]

id apply rational roots test at the start after setting it = 0 to get (2x-1)(3x^2-x+1)=0 first

[proven bobcat]

ok

where did you get the 2x-1 @viral shore

oh, 2(x-1/2

right

[viral shore]

its the results of factoring this with rational roots test

[proven bobcat]

so what does that mean then?

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[short helm]

Not sure if I’m doing these right, #4 and 7 are especially confusing for me

[lament jewel]

what is the question for 4

[short helm]

Decide whether or not there is enough information ti determine if p is parallel to q

[short helm]

If so state the theorem you would use

[lament jewel]

yea um

u can use any but the most straight forwards is the

u shape thing

[short helm]

what

[lament jewel]

the 3rd 1

its u shaped

[short helm]

wtf

am I blind

I don’t see a u

looks like a crippled H

[cerulean hemlock]

look at it sideways

[short helm]

oh it’s consecutive interior angles

I see now

[lament jewel]

lol

[short helm]

Hmm what about 7

[short helm]

if that image didn’t load it’s in this one

there we go

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[sly spoke]

f(x) is 1/x btw

[sly spoke]

Is this correct?

Or should it be 1/(x+h)

[stark bison]

f(x + h) = 1/(x + h)

[sly spoke]

Instead of (1/x)+h

Oh, aight

thanks

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[clear zodiac]

I'm trying to write up some tensor equations where the indicies, and the number of indicies are both arbitrary

[clear zodiac]

Example a $\sum$ that can go over any set of indicies

[solid kiln]

FargothUrIsOnline

[clear zodiac]

Is it proper to write something like $\sum_{i_1, \text{ ... }, i_n}$ to represent how generic I'm being

[solid kiln]

FargothUrIsOnline

[clear zodiac]

so I'm writing i_1, ..., i_n here but is there some general symbol I can use to represent a set of arbitrary indicies?

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[acoustic bay]

i have drawn z < 5 but forgot how to do z< z-6i

surely like the z<z-6i line would be above the z<5 circle

[edgy willow]

|z-point| is the distance to a point in the complex plane

so |z| is distance from (0,0) and |z-6i| is distance from (0,6)

and you have to figure out what it means to be closer to (0,0) than (0,6)

[acoustic bay]

so would the line be above the circle?

[edgy willow]

yup

[acoustic bay]

but where would i shade it then

all inside the circle?

[edgy willow]

yea inside the circle and below the line

[acoustic bay]

ok, thank you for your help :D

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[supple furnace]

.close

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[lofty charm]

The OP arm rotation is
controlled by slotted guide A
that slides horizontally
as the picture shows. If the
speed with which A moves to the
right is 1.2 [m/s] and with a
acceleration of 9 [m/s2], determine
the angular velocity of the OP arm
when x = 50 [mm]

[lofty charm]

i really have no idea how do to this problem, if someone can help me out

!close

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[lofty charm]

!close

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[shut edge]

Need help with this please

[cerulean hemlock]

What is exactly the problem?

[thick kayak]

what are you trying to do ?

[cerulean hemlock]

Bruh don't give away answers

[shut edge]

That's it? I thought it'd be more complicated

[cerulean hemlock]

@shut edge What made you think it was more complicated? Have you done any attempts?

[thick kayak]

actually I think you drew wrong

Is the drawing in the problem itself or a product of your attempt to solve it?

[shut edge]

I have and came to the conclusion the entire C circle could be the answer

Before I posted the problem here

[shut edge]

I didn't alter the arrangement of the circles

They're separate like that

[cerulean hemlock]

Ok, start by drawing the intersection of A and C, then the intersection of B and C, then see what their union looks like

[thick kayak]

okey then it is the answer

https://www.cuemath.com/algebra/sets/

[shut edge]

Haha, now I understand

[thick kayak]

read this article it will be more easy for you good luck dude

[shut edge]

Then I had another confusion. Same figure, but the question this time is (A ∩ B) ∩ C

[cerulean hemlock]

Same thing, start by drawing A ∩ B

[shut edge]

A ∩ B can't be drawn

[cerulean hemlock]

Are you sure?

You are supposed to have all circles overlapping

[thick kayak]

∅

answer is this

null set

[cerulean hemlock]

Bro, don't give answers

[shut edge]

As in?

I don't get what you meant by overlapping

[cerulean hemlock]

Unless you are told otherwise, you can't assume that A and B don't share elements

[shut edge]

Ah, no. The figure isn't like that

[shut edge]

It's the same figure as this

A and B don't overlap

[cerulean hemlock]

Ok, if you are supposed to work with the one you have, then that's good, A and B don't share elements, so their intersection is empty

[shut edge]

But C overlap A and B

[shut edge]

Whew

Then I have one last question. It's A ∩ (B ∪ C). This time I'm assuming the answer is A ∩ C?

Oh, and the figure is the same as the last two times

[cerulean hemlock]

Yes that's good

[shut edge]

😔
🙏

I shall close this channel

.close

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[viscid flower]

Not sure this is the place to ask, but I have a small confusion about leapfrog method for numerical DE solving

[viscid flower]

this is the image im working from, with leapfrog on the bottom

I have code, which works, too

% [tvec,Uvec] = LeapfrogSys(f,tspan,U0,k)
% Leapfrog method
% Inputs
% f: name or function handle of the right-hand side function f:(t,u)->f(t,u)
% tspan(1),U0: initial condition (U0 can be s-vector)
% tspan(2): end time, so that number of steps N = (tspan(2)-tspan(1))/k
% k: stepsize
% Outputs
% tvec: vector of t values
% Uvec: vector (or matrix) of corresponding U values

U0 = U0(:);          % make sure U0 is a column vector
s = length(U0);      % number of equations in system
tvec = tspan(1) : k : tspan(2);   % a row vector
N = length(tvec);
Uvec = zeros(s,N); % REMEMBER OFF BY 1 ISSUE
                   % FOR STARTING INDEX ON LOOP
                   % concerned about transpose here
                   % check :-swap anu eulersysec :)
Uvec(:,1) = U0;

% start up with one Euler step
fval = f(tvec(1), U0);
Uvec(:,2) = U0 + k*fval; % 0.5 from half-step needed
for idx = 2:(N-1) % this should be n-1 i think ...
    curr = Uvec(:,idx);
    fval = f(tvec(idx), curr);
    back = Uvec(:,idx-1);
    Uvec(:,idx+1) = back + 2*k*fval;
end
tvec = tvec';        % to match MATLAB output
Uvec = Uvec';```

what im confused about is the end of this, following "start up with one Euler step"

shouldnt some component of this method involve a half-step of time?

why are all the time inputs just pulled directly from an evenly spaced time array? ( tspan(1) : k : tspan(2) )

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[supple furnace]

what is a homogeneous equation?

[supple furnace]

separable equation is of the form dy/dx=M(x)N(y) but idk about homogeneous

[light crystal]

In the context of differential equations.
If the differential equation
P(x, y)dx + Q(x, y)dy = 0
has homogeneous coefficients, i.e., P(x, y) and Q(x, y) are homogeneous functions of the same order then it can be shown that y = xv will transform the differential equation into a separable one*.
If a function f(x, y) is homogeneous of order n then
f(λx, λy) = λ^n f(x, y).
Depending on the book λ can be any number or just positive. For example, for x > 0, λ = 1/x >0 gives us
f(1, y/x) = f(y/x) = x^(-n) f(x, y) or x^n f(y/x) = f(x,y).
You may see a definition which uses this as a alternative definition of homogeneous.
If P(x,y) = x^n f(y/x) then it's homogeneous of order n.
*See page 57 of Ordinary Differential Equations by Morris Tenenbaum and Harry Pollard.

[stark bison]

You could use google btw

[supple furnace]

Ooo okay i'll read that, thank you!

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[hollow steeple]

How to solve this question without guessing functions?

[wise sinew]

Try to substitute 2 in f(x) in the limit and see whether you can factor something in the numerator

[wise sinew]

By that I mean to substitute f(x) with f(2)

@hollow steeple

Ping me if you need more help

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[sacred fern]

How do I proof, using induction, that $\sum_{i=1}^{k} \frac{1}{2^i}=1-\frac{1}{2^k}$

[left oriole]

it's not true, take k=0, then the LHS is 1 but the RHS is 0

probably the sum should start at i=1

[sacred fern]

yeah

my bad

[solid kiln]

𝙹o

[left oriole]

ok, i assume it's clear that it is true for k=1

[sacred fern]

Yes

[left oriole]

now assume it holds for some general k, and the goal is to show that it holds for k+1

[sacred fern]

If you want write it on LaTex

[left oriole]

to get started, try writing $\sum_{i=1}^{k+1}\frac{1}{2^i} = \frac{1}{2^{k+1}} + \sum_{i=1}^{k}\frac{1}{2^i}$

and then apply the induction hypothesis to the sum on the RHS

[solid kiln]

Bungo

[sacred fern]

So $1-\frac{1}{2^{k}}+\frac{1}{2^{k+1}}=1-\frac{1}{2^{k+1}}$

[solid kiln]

𝙹o

[left oriole]

yes

[sacred fern]

And now we substract the $\frac{1}{2^{k+1}}$ term on both sides and get that indeed they're the same

[solid kiln]

𝙹o

[sacred fern]

Thanks man

[left oriole]

sure

[sacred fern]

I'll close

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[opaque bay]

I got -75.2

ight so

I need to know what order to work through this problem in

[sick jasper]

You should first start with the deepest parentheses if you know what I mean

In that case with (3²*2)

[split chasm]

I got -75.2
show work

[sick jasper]

Yes please

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[terse geyser]

any ideas how to solve this?

i mean, i know d is 4

other than that, i have no clue

[edgy cedar]

What are the roots of this cubics? And how can you use them to form an equation.

[terse geyser]

-1 and 2

idk what to do with them, i mean i know it's also the points where the function changes its growth

[edgy cedar]

Yeah and why are there only two roots when it’s a cubic?

[terse geyser]

i mean -1

[terse geyser]

idk exactly

[edgy cedar]

You can clearly see two turning points

[terse geyser]

yes

[edgy cedar]

Because one of them is a repeated root

Can you see which one?

[terse geyser]

uhh

2?

[edgy cedar]

Nope it’s -1

Because the curve touches the x-axis

And then just go up

So there are two roots there

-1 and -1

[terse geyser]

ohhh

so i divide the function to x+1

or that's unneeded

[edgy cedar]

Well you technically have 3 roots

-1,-1 and 2

Can you form a factorised equation using these?

When you plug these values in your equation you must get 0

Like when you factorise a quadratic

$x^2 + 3x + 2 = (x+2)(x+1)$

[solid kiln]

Pure

[edgy cedar]

This means -2 and -1 are the two roots

You can do the same thing for this cubic

Because you know the roots are -1,-1 and 2

[terse geyser]

ohh

well if i divide it by (x+1) since -1 is a root, i can multiply the quotient by (x+1)

to get the factorized equation

right?

[edgy cedar]

You can factorise the cubic into three brackets of linear.

And you know the factors are (x+1)(x+1) and (x-2)

Because you know the roots of the cubic

[terse geyser]

ohhhhh, yess

i see

what do we do next

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[digital path]

Three cities A,B and C are located as shown in figure 3.33

City B is located 10 km east of city A, and city C is located 3 km north of city B. A new road is to be planned between A and C, as shown dotted on Figure 3.33. there is an existing road between A and B as shown with a solid line. If the road is built on the existing road between A and B, the cost will be 4 million *((USD)/(km))otherwise the cost will be 5 million *((USD)/(km)).

[edgy cedar]

Find total cost?

[long basin]

What are you trying to solve

You just gave us a diagram and the costs

[digital path]

oh

i need to decide where point P should be located, so that the cost of building the road is as cheap as possible.

[stark bison]

Try parameterizing the cost by introducing some variable

[digital path]

pure :)))

[long basin]

Yeah I would try parameterizing this

[stark bison]

For example you can find an expression for the cost in terms of AP

[digital path]

so x=4 mil/km and y=5mil/km

[digital path]

AP=AB-(PB)

[stark bison]

Yeah, solve for PB though

[digital path]

AB=10

so

AP=10-(PB)

PB=10-AP

[stark bison]

Yes, now can you solve for PC?

Using the Pythagorean theorem

[digital path]

a^2+b^2=c^2

[stark bison]

In this case PC^2 = PB^2 + 3^2

[digital path]

PB^2+3^2=c^2

yeag

exacly

[stark bison]

Now plug in PB = 10 - AP

[digital path]

(10 - AP)^2+9=c^2

[stark bison]

So PC = sqrt((10 - AP)^2 + 9)

[digital path]

yes

[stark bison]

Now you can determine the price with AP and PC

[digital path]

yes

with a min

min is when f'(x)=0

$f'(x)=01

[solid kiln]

Mxrcus

[digital path]

and f'(x)=PC

so

sqrt((10 - AP)^2 + 9)=0

[stark bison]

Yeah, so the cost is 4AP + 5PC (in millions)

Let's define the function to be f(x) = 4x + 5sqrt((10 - x)^2 + 9)

Now just find the min of this f

[digital path]

how do i know that?

[stark bison]

Each kilometer from A to P costs 4 millions, so the total cost of the road from A to P is 4AP

Same thing with the road from P to C

[digital path]

why is it 4AP

[stark bison]

Okay look if AP was 1km long, how much would the road cost?

[digital path]

i am sorry for being dumb❌

[digital path]

4 mil+PC

[stark bison]

No, just the road AP

So if it was 2km, then it'd cost 4*2

[digital path]

yes

[stark bison]

If it was 3km, then it'd cost 4*3

In millions

[digital path]

yes

[stark bison]

So whatever is the length of Ap

[digital path]

agree

[stark bison]

Its cost will be 4AP

[digital path]

yes

agree

[stark bison]

Can you now see why PC road costs 5PC?

[digital path]

idk what i was thinking

[digital path]

yes

[stark bison]

Okay, so the total cost is 4AP + 5PC

Which is the same as 4AP + 5sqrt((10 - AP)^2 + 9)

[digital path]

✅

[stark bison]

Now all we need to do is to find value of AP for which this is minimum

So basically find the minimum of the function f(x) = 4x + 5sqrt((10 - x)^2 + 9)

[digital path]

0 = 4x + 5sqrt((10 - x)^2 + 9)

= 4x + 5\sqrt((10 - x)^2 + 9)

[stark bison]

No, you should set the derivative equal to 0

Not the function itself

So first differentiate f(x)

[digital path]

4x + 5\sqrt((10 - x)^2 + 9)

4x= 4

5\sqrt((10 - x)^2 + 9)= umm

[digital path]

how do i solve his

[stark bison]

Differentiate 4x + 5sqrt((10 - x)^2 + 9)

[digital path]

what about the 5

[stark bison]

mb

[digital path]

mb?

[stark bison]

My bad

[digital path]

oh

np

idk how to Differentiate that

[stark bison]

Are you having trouble with differentiating the sqrt?

[digital path]

yes

[stark bison]

Use the fact that sqrt(x) = x^(1/2)

[digital path]

omg yes

sry

4 + 1/2(10 - x)^2 + 9)^-1/2

[stark bison]

Looks like you forgot to differentiate the 4x

And to also apply chain rule

[digital path]

4 + 1/2(10 - x)^2 + 9)^-1/2

[digital path]

upsi

[stark bison]

Don't forget about the 5 too

[digital path]

isnt 5 a constant?

[stark bison]

It's a coefficient of the sqrt

Just like the fact that 4 was coefficient of x and thus didn't disappear

[digital path]

idk

i am lost

I GIVE UP

too dumb

thx for trying @stark bison

.colse

close

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[valid glacier]

got it wrong when i put those Vavg in for the answer

[valid glacier]

<@&286206848099549185>

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[valid glacier]

<@&286206848099549185>

[wraith hinge]

dawg

you need to specify what specific ones you're having trouble with

let me see what you did

unless you're saying, that you typed in the exact values and still got marked wrong?

[valid glacier]

well the whole table, it doesnt tell me which ones are wrong

[valid glacier]

pretty much and used Vavg formula, plugged in values and still wrong

here is a better picture of a similar problem with the work

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[brittle kite]

logic and proofs

[brittle kite]

can this rule also be p and q -> q?

or just p and q -> p

I'm looking at my professor's notes and he pulled q(a) from the simplification

instead of p(a)

upon inspecting. i believe that is a mistake and p(a) should be pulled for simplification

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[grave zenith]

I just need a clarification if this is right because im not sure

[dense breach]

How many vertical stripes of fence are there?

[grave zenith]

6?

[dense breach]

Count the drawing

[grave zenith]

7?

[dense breach]

Ye

[grave zenith]

What about it

[dense breach]

So

7x+2y is your perimeter

Or total fence length or whatever

[grave zenith]

But the word problem says 6 strips i dont get it

[dense breach]

But look at the drawing

You drew 7

7 vertical lines of fence

6 strips of fenced in field

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[vague walrus]

any help on 76?

[molten comet]

,rccw

[vague walrus]

how is trigo ratio applicable in a non right angled triangle?

[long basin]

You can make right triangles from non right angled

[vague walrus]

can u elaborate?

[long basin]

I mean you can legit just draw right triangles inside the non right angled

[vague walrus]

but it asked about that trianfle

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[silent kiln]

Can someone tell me which angle would be the one between the sum and difference of the vectors

[sturdy leaf]

the sum of the vectors is the blue vector

the difference is the red vector

[silent kiln]

what is the angle between them though

[silent kiln]

wait so lemme write something

How do i get to this

This was the answer but i dont understand how they said that was the angle

[random forum]

listen close

you will soon know

every door that has opened close one close soon know youll know what in the othert side

who i am fighting for

[sturdy leaf]

?

[silent kiln]

he got kicked

anyway can u continue

[sturdy leaf]

my guess would be to use the scalar product

since that defines angles

but we wont get arctan from that

so im confused

im gonna think a bit

[silent kiln]

here imma send a pick of the problem adn thread

[silent kiln]

he attached an image of this

[sturdy leaf]

this is so weird

it is given that a and b have the same magnitude

[silent kiln]

yea

which meants that the intesection would have to be 90 percent

[sturdy leaf]

oh wait

[silent kiln]

he got the right answer btw i just dont understand how he decided that angle

probably something with trianges but i cant figure it out

[sturdy leaf]

from this angle

the opposite would be |a-b|/2

the adjacent would be |a+b|/2

so we would get

tan(theta)=|a-b|/|a+b|

[silent kiln]

but why that angle?

OH SHITTTT

[sturdy leaf]

🙂

[silent kiln]

YOU are my savior

Thank you so much

[sturdy leaf]

your welcome

[silent kiln]

have a good night :)

[sturdy leaf]

u2

[silent kiln]

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[karmic acorn]

can anyone help me with this, this is not for a graade or anything learn what my sister is learning

[ionic kindle]

X=4 y=2

Pretty much the asymptote is where the curve doesn’t hit

So x approaches 4 as y approaches infinity

Y approaches 2 as x approaches -infinity

[karmic acorn]

?

[uncut walrus]

Intake is wrong, this function has 1 asymptote, not 2

[karmic acorn]

thats what got me confused

[uncut walrus]

An asymptote is when a line gets closer and closer to some value but never quite gets there

[karmic acorn]

and what represents the value?

just a coordinate?

[uncut walrus]

a line

[karmic acorn]

the line is the value and then the asymptote would be a number on the x or y axis?

[uncut walrus]

So in your specific case the asymptote is y = 2

which is graphed as a horizontal line 2 units above the x-axis

[long basin]

The asymptote is just an equation

[ionic kindle]

on my phone screen it looked like 2 myb

[long basin]

Not a value on the axes

[karmic acorn]

is that becuase the line gets close to 2 but not quite at two?

[long basin]

Yes

[karmic acorn]

no worries

[uncut walrus]

Specifically an equation describing a line

[karmic acorn]

ahh i see

and would that lead to part b?

it says describe the end behavior of the graphed function

[long basin]

Not all asymptotes need to be lines

You can have curved asymptotes

[karmic acorn]

?

[uncut walrus]

They're not understanding

[long basin]

Don't worry about curved ones

[uncut walrus]

your asymptote is represented by the blue dotted line

[karmic acorn]

yes

the line is closest to 2 at the curve

y = 2

correct

i understand that part now, but part b of this question is what im looking at now

[uncut walrus]

okay

[ionic kindle]

Well what does x approach

As y approaches 2

[uncut walrus]

intake, you need to explain the process of why you ask that specific question.

[ionic kindle]

I thought that would just simplify it

[uncut walrus]

Since we're talking about the asymptote we're interested in the side of the line having to do with said asymptote.

don't skip steps in an explanation, it only causes confusion in the future.

[karmic acorn]

yea sorry intake i actually havent learned any of this, im just trying to learn ahead

[uncut walrus]

Question B is asking what the asymptote tells us about the function

[karmic acorn]

is it right to assume since we know the asymptote is 2 then the answer would start with f(x) ->2?

[uncut walrus]

for the first part of B ofc

yes, because the asymptote is y=2 and f(x) just means y

then we need to describe the second part

[karmic acorn]

nice and how would i go about that

[uncut walrus]

well

do we have a reason to believe the graph will stop going further to the right?

[karmic acorn]

no

i dont think so

[uncut walrus]

correct

as far as we know x can in theory go all the way to infinity

so the answer is going to have something to do with x -> infinty

[karmic acorn]

so that would be positive infinity

[uncut walrus]

in this case we know that f(x) approaches infinity as x gets larger because all of the answers provided say so

[karmic acorn]

yes

[uncut walrus]

the only question we needed to answer was how far would x go, and we came to the conclusion of infinity

[karmic acorn]

so f(x)->2 as x->inf?

[uncut walrus]

no

remember, each answer is composed of 2 parts

[karmic acorn]

thats right mb

also that wouldnt be an answer anyways

[uncut walrus]

yeah

so we know on the left side y is just going to get infinitely closer to 2

[karmic acorn]

yea

[uncut walrus]

and we have no reason to believe it will stop going to the left

so we start with f(x) -> 2 as x -> -infinity

[karmic acorn]

true

[uncut walrus]

and we have no reason to believe y will stop growing or that we would stop going to the right on the opposite side

[karmic acorn]

oh since infinity is going left it would be a negative

yea

[uncut walrus]

So..?

the answer would be..?

[karmic acorn]

answer would be a

[uncut walrus]

yes

[uncut walrus]

If you want I can come up with another question like it so we can verify your understanding

[karmic acorn]

well i have a bunch of random questions from my sisters calculus reviews

[uncut walrus]

Do you want another like this one to verify your understanding? (or maybe you happen to have one?)

[karmic acorn]

i dont know if this is similar since i havent worked it

[uncut walrus]

different type of question

[karmic acorn]

what type would this be?

[uncut walrus]

this isn't an asymptote question, it's asking where the line breaks.

[karmic acorn]

what does that mean

[uncut walrus]

you see at x = -1 where it jumps up 1 unit without a connecting line?

[karmic acorn]

no....

[karmic acorn]

oh it jumps one unit from (-1, 2.5) to (-1, 3)

[uncut walrus]

yes

it's actually half a unit, I didn't realize that

[karmic acorn]

yea

[uncut walrus]

but the function is discontinuous at x = -1 because the line isn't connected

[karmic acorn]

so why does x = -1

[uncut walrus]

because the disconnection is at the coordinates (-1, 2.5) and (-1, 3)

[karmic acorn]

ohh