precalculus

[desert socket]

The equation is |u • (v X w)| and they decided to add the terms instead of getting the root of terms squared

For final answer

[rare pagoda]

Well the dot product is a scalar so you can't find the magnitude of that anyway

[desert socket]

Ooh

That explains it

I just found it odd they decided to put it within a magnitude thing

[rare pagoda]

It's an absolute value

[desert socket]

Okay

[rare pagoda]

Lol

[desert socket]

Yep that clears it up

[rare pagoda]

Gotta hate those tricky notations

[desert socket]

I was going nuts thinking somehow my process was wrong

Cross referencing examples etc.

[foggy lake]

wat does the symbol sigma mean?

[past meadow]

Capital or little sigma? And in what context

[foggy lake]

$\sum_{k=0}^{n}$

[obsidian monolith]

Goosy<3:

[foggy lake]

like this one

[past meadow]

Summation

Like $\sum_{k=1}^{3}k^2$ is $1^2+2^2+3^2$

[obsidian monolith]

Sneaky:

[foggy lake]

oooh

i get it now

[azure sinew]

yo

[hard mulch]

how do i do this

Im not sure where I am messing up

[past meadow]

F isnt a number. Its a function

[mental cove]

If f(x) = [expression], then f(4) = [that expression, but with al of the x's replaced with 4].

[hard mulch]

F isnt a number. Its a function
@past meadow ahhhhhhhhhh

[past meadow]

But what red herring said is how youre suppoed to solve it

[hard mulch]

Okay, Thanks a lot

[sullen canyon]

if i know cos2x

and sin x

what can i do to find cos4x

[patent beacon]

Namely, you care about
cos(4x) = 2cos²(2x) - 1

[sullen canyon]

cos 4x = 2 cos^2 (2x) - 1

right?

oh cool

so then i got 2(4cos^4 - 4cos^2+1)-1

8cos^4 x-8cos^2 x +2-1

[sullen canyon]

Also

if i have f(x) = (sin x)(tan x) + cos x

what can i do to get it to 1/cos x

i've gotten to ((1 - cos^2 x)/cos x) + cos x

[patent beacon]

Can you put everything over that cos(x) denominator?

[sullen canyon]

yep just realised

i didn't need to make 1-cos^2x

cuz cos^2 x / cos x = cos x

so i can just put everything over denominator

thanks @patent beacon

[tame wedge]

Hey guys can anyone help with this sum? I'm trying to get it into a form where I can use the formula for the sum of i^2 on it but I'm getting stuck.

The i on the bottom should be a j

[north crescent]

Expand (i-1)^2

And then split it up into 3 sigmas

Each one would have a formula to solve for it

so i^2-2i+1

1 would be n

2i would be 2* i(i+1)/2 I think

[tame wedge]

Yeah that would work hey, I didn't notice that thanks !

[north crescent]

And i^2 would be i(i+1)(2i+1)/6

Yeah no probs

[viscid thistle]

hm?

can't you just cancel i^2 and 1

[willow bear]

i is not the imaginary unit here

[viscid thistle]

oh obviously lmao

[viscid thistle]

hi

How do i solve this?

i have a 30 question bonus and i can't seem to solve some questions

@viscid thistle seems like a test

it ain't a test

it's a mathspace homework task

@viscid thistle

@viscid thistle have you heard of cos(A+B) or cos(A-B)

yes

For example for the first one apply it

With 3x being A and 2y being B

And same with cos(x-3y)

@viscid thistle

[coral garnet]

Just mathematica that sit

Nobody got time for dis

[exotic spindle]

i was absent on the day this was explained and i dont get how it works?

[viscid thistle]

@exotic spindle Still need help?

[exotic spindle]

no i got it, thank you though

[smoky needle]

Can someone help me with this problem?

[viscid thistle]

what is $\tan(\theta)$ equal to?

[obsidian monolith]

Botnuke:

[viscid thistle]

in terms of sines and cosines

[smoky needle]

is it: tan = sin/cos

?

[viscid thistle]

yea

[smoky needle]

and from there, i can just solve for sin correct

?

[viscid thistle]

yes

[smoky needle]

i don't know why i was overthinking this problem

thank you so much

[viscid thistle]

lol

[smoky needle]

can i show you my work after i attempt to do it?

[viscid thistle]

sure

[smoky needle]

@viscid thistle I got this. lol

[fluid shore]

That is incorrect

[viscid thistle]

yea

[smoky needle]

yikes

[fluid shore]

The thing in your denominator is 61

Not 6

[smoky needle]

oh shoot sorry

i forgot to put 1

i meant 61

[fluid shore]

If it was just 6, then sin(theta) = sqrt(61) and that wouldn't make sense

[smoky needle]

thanks for catching that!!

[fluid shore]

You're welcome. Be very careful about such things.

[smoky needle]

yeah, it can screw me up a lot!😅

[fluid shore]

🙂

[smoky needle]

Hello again, i'm stuck on the same kind of problem. I tried to use reciprocal identities or even tried to manipulate some. But I couldn't get it

[fluid shore]

Well

$\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \sin(\theta) \cdot \sec(\theta)$

[obsidian monolith]

Abhijeet Vats:

[fluid shore]

Does that help?

[smoky needle]

yes, i tried to use that, but i don't have cos theta

[fluid shore]

But you have sec(theta)

Remember:

$\sec(\theta) = \frac{1}{\cos(\theta)}$

[obsidian monolith]

Abhijeet Vats:

[smoky needle]

yes, will i just have to manipulate that somehow to get cos theta?

[fluid shore]

You don't need cos(theta), my dear

$\frac{\sin(\theta)}{\cos(\theta)} = \sin(\theta) \cdot \frac{1}{\cos(\theta)} = \sin(\theta) \cdot \sec(\theta)$

[obsidian monolith]

Abhijeet Vats:

[smoky needle]

Ohh i see. i was on the right path! lol but i tried to solve for cos(theta). this helped so much, i will try and redo it again! thank you!!!

[fluid shore]

You're very welcome.

[smoky needle]

hello, again. I'm still on the same problem and i can't seem to figure it out.. 😦

[fluid shore]

What is the issue?

[smoky needle]

i think i am getting confused on the identities

like how does it result in sin (theta) multiplying by sec (theta)?

sorry i'm kinda slow

[fluid shore]

Well, $\sec(\theta) := \frac{1}{\cos(\theta)}$

[obsidian monolith]

Abhijeet Vats:

[smoky needle]

Yes, i have that written down. I guess where i'm having trouble is using the identities in general.

[fluid shore]

Well, so:

$\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \sin(\theta) \cdot \frac{1}{\cos(\theta)} = \sin(\theta)\sec(\theta)$

$\sin(\theta) = \frac{\tan(\theta)}{\sec(\theta)}$

[obsidian monolith]

Abhijeet Vats:

[fluid shore]

This is just algebra that I'm doing here

@smoky needle

[smoky needle]

ohhh!!! it was probably my algebra. thank you! that helps a lot. i can see it in a different way now! thank you!

[fluid shore]

You're welcome.

[compact tide]

Are any of you good with solving trig functions

[fluid shore]

Nope

[rare pagoda]

what do you need help with?

[compact tide]

I need help with about 5 problems if thats okay

finding a set of solutions

[rare pagoda]

lol i'll see

[fluid shore]

Magnus has a PhD in trigonometry

He'll be able to help ya

[rare pagoda]

xP

[compact tide]

sinx + cosx = -1 in the interval [0, 2pi)

[thorn mountain]

What have you tried?

[compact tide]

im not exactly sure how to do it is the thing

[thorn mountain]

But what have you tried?

[fleet yew]

Just try plugging in a value

See what happens

[thorn mountain]

I mean, that’s not wrong, but not really useful

[rare pagoda]

yeah technically you can combine sinx + cosx into the form acos(x-t) but plugging in should work just as well

[thorn mountain]

Square both sides gives you something to work with tho

[rare pagoda]

in this case sinx + cosx = sqrt(2)cos(x-pi/4)

[fluid shore]

Not if you want to find all solutions

[compact tide]

that would help

doing that now thanks

i got 2 of them like that

i have 2 more

I keep getting domain error when i attempt -3csc(theta) = 3root2

any ideas?

@rare pagoda

[fluid shore]

You're trying to solve $-3\csc(\theta) = 3\sqrt{2}$? What's the interval you're interested in?

[obsidian monolith]

Abhijeet Vats:

[viscid thistle]

Why not solve for all intervals.

[compact tide]

same interval

[0, 2pi)

[fluid shore]

Well, yea you can solve it for all intervals if you want to

Okay:

$\sin(\theta) = -\frac{1}{\sqrt{2}}$

[obsidian monolith]

Abhijeet Vats:

[fluid shore]

So, in the interval $[0,2\pi)$, does this have any solutions?

[obsidian monolith]

Abhijeet Vats:

[compact tide]

the thing is i dont know the formula

its multiple choice tho

[fluid shore]

?? Can you take a picture of the problem?

[thorn mountain]

Then either learn the right way to do it or just try the solutions offered

[compact tide]

right trying the solutions was my main way

but it keeps saying domain error

[fluid shore]

Can you post a picture of the problem?

[compact tide]

yes 1 sec

[thorn mountain]

Note: $ \csc x \neq \sin^{-1} x $

This might be the source of your domain errors

[obsidian monolith]

Malix:

[fluid shore]

Okay, so literally

It's just solving $\sin(\theta) = -\frac{1}{\sqrt{2}}$

[obsidian monolith]

Abhijeet Vats:

[thorn mountain]

Yep

[compact tide]

oh i was putting in sin^-1

so whats the what from there

i did terribly on this unit and i have a retake soon

[thorn mountain]

Are you familiar with the unit circle?

[fluid shore]

Okay, so you need to look at the quadrants of the unit circle where $\sin(\theta) = -\frac{1}{\sqrt{2}}$. You already know that $\sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$. So, you just need to make sure that the sine itself is negative

[obsidian monolith]

Abhijeet Vats:

[compact tide]

ok

so thats 7pi/6 and 11pi/6

wait thats not right

this is my last problem of the day

im genuinely confused on what im being asked to complete, since 1/root2 isnt on my unit circle anywhere

@fluid shore @thorn mountain

[thorn mountain]

You probably know it as $ \frac {\sqrt{2}}{2} $

[obsidian monolith]

Malix:

[compact tide]

uhh yes yes

[fluid shore]

Eh malix, can you help him out

[compact tide]

that makes sense

[fluid shore]

Sorry lol, i'm busy

[thorn mountain]

It’s fine, I got it

[compact tide]

so wherever sin value is that, is my answer

[thorn mountain]

Just rocking a kid to sleep

[compact tide]

7pi/4, 5pi/4

so its C?

[thorn mountain]

Yes. Those are where sin(x) is negative pi/4

[compact tide]

ok tysm

[thorn mountain]

👍

[gilded mirage]

nvm i see

[orchid hearth]

hi

precalc help?

[acoustic laurel]

@orchid hearth sure

[orchid hearth]

@acoustic laurel

[wraith kernel]

Im doing implicit derivatives I want to know if this is right

[thorn mountain]

are those two minus signs in the numerator?

[wraith kernel]

for the f'(x) is an =

[wraith kernel]

this is correct?

[gritty sandal]

No, if f(x) was xy then you have to apply product rule

But usually implicit difs are equal to something

So it’s a hard case

Like x^2 + y^2 = 1

But the derivative of xy is y + (dy/dx)x

derivative of $xy = y + \frac{dy}{dx} x $

Getting used to Texit

I still suck

[obsidian monolith]

Parking Ticket:

[viscid thistle]

Excuse me can someone help me with verifying and proving trig identities

I have a quiz tomorrow and a test Friday and I’m definitely not ready for them

So if someone could help me understand it better I’d be very grateful

[past meadow]

Hey just ask your questions. #geometry-and-trigonometry might be better though

[pale bison]

To be fair precal and trig are very intertwined

But

[fluid shore]

I was pretty intertwined with your mum

Prank

Relax

Memes

Chill

[past meadow]

im not from america so idk what the different classes are.

[fluid shore]

Same tbh

[gritty sandal]

same

[mental cove]

B-b-but... if you're not from America, then where else could you be from? :^)

[fleet yew]

if america's so great why isn't there an america 2

[mental cove]

https://en.wikipedia.org/wiki/America_II

czechm8

[fluid shore]

Yea but

There's no america 3

[past meadow]

https://en.wikipedia.org/wiki/America³_(1992_yacht)

[slate scroll]

can you guys help me with my homework

my professor mixes our topics into homework all the time

its so fkn annoying

[craggy dune]

well 16 + 3 = 19

I think

[slate scroll]

ok

thankyou

[viscid thistle]

🗿

[mental cove]

Who would've guessed?

Thought 16 + 3 = 163...

[craggy dune]

well I feel 2 would have been legit as well

[north crescent]

How is this pre uni pre calc

[mental cove]

You know what? I don't think 19 or 163 make sense. I don't think there's any number past 2.

I remember learning this in pre uni pre calc.

[misty heath]

Why is problem 1 even there

[viscid thistle]

Your teacher is a savage.

[slate scroll]

yes we started the semester with addition

we had an exam on that

and now we have started complex analysis

but I don't know why he is still putting the addition stuff in the homework

[viscid thistle]

Addition

R u kidding?

[slate scroll]

no

we didn't do any numbers over 100

so it wasn't too bad

[viscid thistle]

@slate scroll ur a funny guy

[viral trail]

I tried it for n^3 and n^4 and got 1/3 and 1/4 so i think answer should be 1/2020 but how to prove

[willow bear]

you can write the sum in the limit as the riemann sum with $n$ subintervals for $\int_0^1 x^{2019} \dd{x}$

[obsidian monolith]

Ann:

[willow bear]

i.e. $\frac{1}{n} \sum_{k=1}^n (k/n)^{2019}$

[obsidian monolith]

Ann:

[viral trail]

Right ty

[maiden pebble]

https://i.imgur.com/Z0kSY4A.png

-8sin^2x + cos^2x

[uncut mulch]

what's that supposed to be?

[past meadow]

is that supposed to be the derivative of that function

[viral trail]

Every option looks correct to me

[willow bear]

no

take f(x) = x^3

C will not work

[viral trail]

Ty

I have no idea except that f(n) = 0 is a solution.

f(n) = n also works

So are there infinte sol or finitely many?

[maiden pebble]

@uncut mulch https://i.imgur.com/bRanF45.png

Sorry, is that the derivative of that function

[uncut mulch]

no

[thorn mountain]

Hint: $ 8 \sin x \cos x = 4 ( 2 \sin x \cos x ) $

[obsidian monolith]

Malix:

[uncut mulch]

if you tried to differentiate by product rule, you didn't apply it properly

[thorn mountain]

Lost parentheses

[maiden pebble]

I did (Deriv 8sinx)(cosx)+(8sinx)(deriv cosx)

il redo

@thorn mountain are you saying that sin x cos x =/= (sinx)(cosx)

[thorn mountain]

No

The stuff inside my parenthesis should look familiar

[maiden pebble]

I dont see how the hint helps me

It doesn't

[thorn mountain]

Then check your trig identities

Or do it the way you were going to before but don’t lose pieces

[maiden pebble]

I cant find any trig identities that look like what you gave me so il do that

[thorn mountain]

$ \sin (2 x) = 2 \sin x \cos x $

[obsidian monolith]

Malix:

[maiden pebble]

(Deriv 8sinx)(cosx)-8sinx^2

good so far?

[thorn mountain]

Yes

[maiden pebble]

i believe the derivative of 8sinx is cosx

😭

ok yea i get it

[fluid shore]

$\frac{d}{dx} cf(x) = c \frac{df}{dx}$, where $c \in \bR$.

[obsidian monolith]

Abhijeet Vats:

[fleet yew]

what if $c\in\C$

[obsidian monolith]

AMD:

[stuck lark]

that's fine too

[maiden pebble]

hope im not meant to know waht yall talking about

[fleet yew]

yeah that's what i was thinking

[thorn mountain]

Other than AMD, yeah you should

[stuck lark]

no bridge, you SHOULD know what it means

[maiden pebble]

no clue what this c trident R is

[fluid shore]

It means that the derivative of a constant times a function is the constant times the derivative of the function

It's not a trident wtf

[thorn mountain]

Never seen set notation apparently

[maiden pebble]

https://i.imgur.com/6lxgcsE.png

[fluid shore]

How dafuq does that look like poseidon's trident?

[maiden pebble]

Trident!!!

[fluid shore]

Lmao HAH

[fleet yew]

he ain't wrong lmao

[thorn mountain]

Ignore that one little piece, the rest you should know

[fluid shore]

Okay, $\in$, very loosely speaking, just means "belong to"

[obsidian monolith]

Abhijeet Vats:

[fluid shore]

I hate to attach a meaning to it but you can think of it like that

[maiden pebble]

Well then whats the wierd font for R and C

[fluid shore]

So $\bR$ is the set of real numbers

[maiden pebble]

I assume that means something too

[stuck lark]

teach bridge notation naive set theory later, abhi

[obsidian monolith]

Abhijeet Vats:

[fluid shore]

Okay yea fair

[maiden pebble]

yea i have no clue what u guys are talking about

[thorn mountain]

The derivative of (a constant times a function) is the constant times the derivative of the function

[fluid shore]

How tf are you doing derivatives without knowing what the real numbers are?

[maiden pebble]

ok, so your trying to tell me that i keep the 8 in

[fluid shore]

Surely you must've seen this notation somewhere?

[thorn mountain]

Yes

[maiden pebble]

Yea, in #calculus

[thorn mountain]

I’m still not clear how your precalculus class is doing so many derivatives but whatever

[maiden pebble]

8cosx^2-8sinx^2

BUT

[thorn mountain]

No, for notation reasons

[maiden pebble]

I can factor out that 8

[fluid shore]

Okay first of all

[maiden pebble]

AND

[thorn mountain]

And as I said earlier, lost parenthesis

[fluid shore]

(cos(x))^2

[maiden pebble]

that simplifies

[fluid shore]

Parentheses!!

[maiden pebble]

8(((cosx)^2)-((sinx)^2))

[thorn mountain]

Over correction, but yea

Wait, now no

[maiden pebble]

tf

Yea i see what ur saying

[thorn mountain]

$ 8 ( \cos^2 x - \sin^2 x ) $

[obsidian monolith]

Malix:

[maiden pebble]

yes

I have that now

So that simplifies further to 8

[thorn mountain]

No

[fluid shore]

$\cos^2(x)-\sin^2(x) \neq 1$

[obsidian monolith]

Abhijeet Vats:

[thorn mountain]

The sum is not the difference

But $\cos^2(x)-\sin^2(x) = \cos (2x) $

[obsidian monolith]

Malix:

[maiden pebble]

rly?

[thorn mountain]

Trig identities

[uncut mulch]

^ is something you should be well versed in by the time you're doing differential calculus

[thorn mountain]

And had you used the one I told you earlier, we would have gotten here much quicker

[maiden pebble]

I mean, we wouldve got here much quicker in many different ways

[thorn mountain]

And trig identities are actually an appropriate topic for this channel

[maiden pebble]

8cos2x

assuming that cant be simplified

[willow bear]

[sly osprey]

can someone help me w surface area

[acoustic laurel]

lol

I saw that deletion

@sly osprey I can

[silk sail]

Can someone help with this problem? I dont think I understand what to do with a negative exponent for the (4+h)

[thorn mountain]

What do negative exponents mean?

[silk sail]

means put it all over 1 i thought

so i get 1/(4+h)-1/4

[thorn mountain]

$ \frac{1}{4+h} - \frac{1}{4}$

[obsidian monolith]

Malix:

[thorn mountain]

So what can you do when you need to add/subtract fractions with different denominators?

[silk sail]

need a common denominator, so add h to both sides of 1/4?

[thorn mountain]

That’s not how that works

[silk sail]

ya this is where im getting stuck i guess

[thorn mountain]

https://youtu.be/bcCLKACsYJ0

[silk sail]

ok looks like i need to review

thanks

[thorn mountain]

Yep

[silk sail]

so i realized that i need to multiple the left side by 4/4 and the right by (4+h) but then i still get an equation that comes to 0 when plugging in 0 for h.

[craggy dune]

you prolly forgot the h

$ \frac{1}{4+h} - \frac{1}{4} \neq \frac{\frac{1}{4+h} - \frac{1}{4}}{h} $

[obsidian monolith]

deekaan:

[silk sail]

$ \frac{4}-{4+h}{4}{4+h}{h}$

[obsidian monolith]

admrhds:
Compile Error! Click the reaction for details. (You may edit your message)

[silk sail]

ugh idk how to type it

i got 4-(4+h)/4(4+h)/h

[craggy dune]

yes and now simplify

[silk sail]

ya and then i take the h out of the denominator by making it 1/h. so now i have 4-(4+h)/4(4+h)h

[craggy dune]

okay now divide nom and denom by h

[silk sail]

ya that lost me... i dont understand how or why... 😦

[craggy dune]

well you need to send h to 0

right now that would cause a lot of trouble because the denom has an h in it killing the whole thing

so you need to get rid of that

you divide everything by h and then everything will work out fine

[silk sail]

that makes sense. So it becomes 4-(4+h)/h/4(4+h)?

i get 0/0 up top?

[craggy dune]

well in the nominator you have 4 - 4 - h

that gets you -h

now divide -h by h

and the denominator in the nominator is h

so if you divide that by h you get something as well

[slow roost]

sorry if im interrupting something. couldnt tell if it concluded or not, but can i get some help with this. Im not sure if theres a formula i should be using

[silk sail]

you're good thank you @craggy dune i can figure it out from here i think

[fluid shore]

@slow roost Draw a picture

Don't think in terms of formulas, draw a picture first. We'll worry about the formulas later.

[slow roost]

uh i cant rlly visualize this. not sure if this image is right

[fluid shore]

That isn't a proper diagram

The point of drawing a diagram is so that you can have access to something that can help you visualize the situation.

Right now, you know that the plane is flying in a 2D plane and you know that it is flying with direction angle 245. Do you know how to convert that into a diagram?

[slow roost]

no

these are the notes i got so im guessing the |v| represents 530

and theta represents 245

[fluid shore]

Why are you guessing? Are you not actually sure?

Based on that, can you draw a diagram that reflects the information in the question?

[slow roost]

Nvm I got it

[viscid thistle]

God I feel like such a failure for not understanding trignometic functions

No matter how much I practice it I just can’t grasp it

[past meadow]

you aren't a failiure for that

[viscid thistle]

I always get stuck midway through a problem and it sucks

[craggy dune]

thats a stupid reason for feeling like a failure

just ask for help if you're stuck

[viscid thistle]

Well here’s the thing

[craggy dune]

sooner or later you'll find a way

[viscid thistle]

I get help

But no matter how much I get I still don’t get it

Like I understand it

But I just can’t physically do the problem if that makes sense

[craggy dune]

write out the steps

like make a guide for yourself

before you do something ask yourself what do you have to do and why

and when you ask for help think about every step

if you can't do it by yourself that means there is a step you haven't fully grasped

and that means you'll have to observe yourself when solving a problem

so you can find that point where your understanding fails you

[lethal oracle]

I’m kinda stuck on where to get started here

[past meadow]

well we can start by setting up some equations with those points, right? for instance, you can get c very quickly by just substituting -1=a* 0^2+b*0+c

[lethal oracle]

So c=-1?

[past meadow]

yeah. now set up equations with the other points using the same method.

it might not seem as obvious how to figure out a and b from that, so come back if you aren't sure after you've set those up

[lethal oracle]

So wouldn’t one of the equations be

4=a+b+c

[past meadow]

yes, but remember c=-1

so 4=a+b-1

a+b=5

[lethal oracle]

And the other be 13=4a+2b-1

Alright I see

[past meadow]

yeah so 14=4a+2b

[lethal oracle]

Ya

[past meadow]

do you know how to solve these?

[lethal oracle]

Do I just use substitution?

[past meadow]

or elimination

up to you

[lethal oracle]

Or any method to solve a system of equations really

I got a=2 and b=3

[past meadow]

sounds right

[lethal oracle]

Now I just plug those into my parabola

[past meadow]

yeah

[lethal oracle]

Thank you so much

[past meadow]

no worries

[lethal oracle]

If I was solving this system

Can I do it by turning it into an augmented matrix and then multiplying the inverse

By the matrix of 5,5, -1

[mental cove]

The augmented matrix representing this system is

     2    -3     1     5
     4     0    -1     5
    -2     6     2    -1

but if you wanted to solve it by taking inverses, you'd want the inverse of the coefficient matrix

     2    -3     1
     4     0    -1
    -2     6     2

[lethal oracle]

Yes

I think I know why it wasn’t working

[mental cove]

But yeah, you'd get the inverse of the coefficient matrix and multiply on the left of

     5
     5
    -1

[lethal oracle]

I wrote the rows in the wrong order

[lethal oracle]

I know that both sides of the equations are equal

[willow bear]

no they are not

not in general

[lethal oracle]

You mean since they are matrices?

[willow bear]

for matrices you do not have $AB = BA$ generally

[obsidian monolith]

Ann:

[lethal oracle]

Oh

[willow bear]

yes it's because they're matrices

[lethal oracle]

So I need to explain that AB=BA doesn’t work for matrices

[willow bear]

if A and B commute, i.e. if they happen to be such that AB does equal BA, then the equation is true

but otherwise, no.

and it was probably explained at least once in class. i don't know if you're expected to endlessly repeat yourself.

[lethal oracle]

Maybe it slipped

With all classes being remote

Cause I don’t remember this

[willow bear]

have you learned matrix multiplication at all

because any teacher worth their salt would mention that matrix multiplication is not commutative

[lethal oracle]

Yes I have

I mean he did explain that

But maybe I just didn’t connect it to this example

This is the correct start, right?

[willow bear]

no, because your two denoms are identical

there's something you're missing

[lethal oracle]

does one need to be (x)

[willow bear]

no

there's something you're missing

[lethal oracle]

Oh does one need to be (x-1)^2

[willow bear]

yes there we go

[lethal oracle]

Oh alright I think I got the rest

Thank you

Alright apparently I don’t know how to do it

I tried some stuff but it wasn’t working

[viscid thistle]

can somoene tell me what a left continious graph looks like?

[gritty sandal]

use common denominator

[lethal oracle]

How do you mean

[gritty sandal]

well just get rid of the denominator

wait hmmmmm there isn't a function next to B

[lethal oracle]

I’m doing partial fraction decomposition

[gritty sandal]

Looks to me like normal partial fractions

If u multiply A by x-1 then you get

$2x-3=A(x-1) + B$

[obsidian monolith]

Parking Ticket:

[gritty sandal]

And then you can just expand and equate coefficients

So like $2x-3=Ax+(B-A)$

[obsidian monolith]

Parking Ticket:

[gritty sandal]

So A = 2

And then we get $-3=B-A$ for the constant

[obsidian monolith]

Parking Ticket:

[gritty sandal]

But we know A=2

So B=-1

Hope that helps

[lethal oracle]

Ya it does

So my answer would be

2/(x-1)

[gritty sandal]

Well you also need to sub B back in as well

[lethal oracle]

Ya and

b would be -1/(x-1)^2

?

[gritty sandal]

Ye

So you would write

$\frac{2x-3}{(x-1)^2} = \frac{2}{x-1} -\frac{1}{(x-1)^2}$

[obsidian monolith]

Parking Ticket:

[trim karma]

How would I go about graphing this?

[gritty sandal]

$\frac{1}{2x-2}$?

[obsidian monolith]

Parking Ticket:

[rich light]

Could someone help me with a + e

[willow bear]

i said #calculus, not #precalculus.

[rich light]

I'm in precalc tho 🙈

[willow bear]

ok whatever

i already told you what mistake you made in (e)

[rich light]

and it heads towards infinity I'm assuming?

[willow bear]

"it"

yes, x^2 -> +∞ as x approaches infinity.

so why'd you write zero?

do you think $\frac{+\infty}{99} = 0$?

also, in (e), you seem to have thought for a moment that $+\infty = 0$ too

[obsidian monolith]

Ann:

[willow bear]

do you think $\frac{+\infty}{99} = 0$?

[obsidian monolith]

Ann:

[willow bear]

aside from your insistence that x/x = 0, that is.

@rich light

[rich light]

nah, I see what you're saying

I just did it as sqrt x^2/99

so + x/11

which makes no sense either way

🙂

[willow bear]

i mean

x/11 would still approach +∞

but 11 isn't even the square root of 99

so yeah it's nonsense

[rich light]

🙄

so it would be infinity, okay

for e, I did this

which I think is also incorrect

but not sure why

like idrk how I'm supposed to be solving this exactly

[willow bear]

ok now

take a close look at the limit

and without jumping ahead

answer me this

what is x approaching?

[rich light]

-inf

[willow bear]

no

[rich light]

Positive

[willow bear]

yes, positive infinity.

so why are you plugging in zero?

do you still think $+\infty = 0$?

[obsidian monolith]

Ann:

[rich light]

No

[willow bear]

then why are you plugging in 0 for x when x -> +∞?

[gritty sandal]

u could l'hopital rule

[willow bear]

no

get out

[rich light]

I just plugged it in like that cuz that’s how they did it in an example I saw 💁🏻‍♀️

[willow bear]

WHY ARE YOU PLUGGING IN ZERO

X ISN'T APPROACHING ZERO

[gritty sandal]

kind of aggressive man

[willow bear]

WHY ARE YOU PLUGGING IN ZERO WHEN X ISN'T APPROACHING ZERO

WHY

[rich light]

why are you surprised?

[gritty sandal]

ye u plug in infinity

not zero

[rich light]

kk

give me 1 sec

[willow bear]

l'hop is EXCESSIVE here.

also, plugging in infinity as is will lead to ∞/∞, which is indeterminate

[gritty sandal]

so you l'hop

[willow bear]

NO!!!!!!!!!!

TO HELL WITH THAT!

[gritty sandal]

that's kinda the point of l'hop

[willow bear]

that's like smashing a fly with a hammer

you can do it without l'hop much easier

[gritty sandal]

I mean deriving linear functions is rly ez

[willow bear]

NOT EVERYTHING NEEDS TO BE DONE WITH FUCKING L'HÔPITAL'S RULE.

anyway, you're better off first dividing the num and denom by x, and THEN plugging in infinity.

[rich light]

exactly what you just said

[willow bear]

yes

exactly what i just said

[rich light]

I did that the first time tho didn't I?

when I posted in prealg

but still got a shit value

[willow bear]

you incorrectly simplified x/x as 0

and also thought +∞=0 for a moment

[rich light]

give me a sec

[gritty sandal]

plug in infinity

[willow bear]

$\frac{2}{x} - \frac{x}{x} \neq \frac{2}{x}$

[obsidian monolith]

Ann:

[willow bear]

and also

WHY

ARE

YOU

STILL

PLUGGING

IN

FUCKING

ZERO

FOR

X

WHEN

X

IS

APPROACHING

[rich light]

I didn't update it brah

[willow bear]

INFINITY!!!

AND NOT ZERO!!!

[rich light]

I was just posting it so we could see where we went wrong lmaooo

[willow bear]

i already told you several times

[rich light]

I**

okay hold on

so I don't plug in zero

okay lit lit

I plug in infinity instead?

[gritty sandal]

yes

[willow bear]

yes AND ALSO REALIZE that $\frac{x}{x} = 1$ and not 0 so $\frac{\frac2x - \frac{x}{x}}{\frac{x}{x} + \frac{4}{x}}$ does \textbf{\textit{\underline{NOT}}} simplify to $\frac{2/x}{4/x}$

[obsidian monolith]

Ann:

[rich light]

look at this

[gritty sandal]

this whole conversation is kinda funny

[willow bear]

...

[rich light]

Is this right or nah

[gritty sandal]

forgot the minus

[willow bear]

WHY

ARE

YOU

STILL

[gritty sandal]

it should be $\frac{\frac{2}{x}}{-1}$

[willow bear]

PLUGGING

IN

ZERO

FOR

X

(╯°□°）╯︵ ┻━┻

[rich light]

woman, that's literally what they did in the example I linked

that's how I learned it

[gritty sandal]

yeah plug in infinity, might help you understand the problem

[past meadow]

is that supposed to be 2/infinity? the reason they have 0 in your example you posted is because its n/infinity

[gritty sandal]

that did not work

[past meadow]

which approaches 0

[willow bear]

woman, that's literally what they did in the example I linked
no they didn't

and don't you fucking take that tone with me

they replaced 1/x with 0

NOT X ITSELF WITH 0

AT NO POINT DID THEY WRITE 1/0 OR 7/0 OR WHATEVER OTHER NUMBER/0

are you DELIBERATELY ignoring or CHOOSING to misunderstand this or what

[gritty sandal]

also remember its $\frac{2}{x} - 1$ not $\frac{2}{x} +1$

[obsidian monolith]

Parking Ticket:

[willow bear]

and also that.

[rich light]

yeah I changed the symbol^

[gritty sandal]

and also plug in infinity please

[willow bear]

you're refusing to listen to me

[rich light]

so you guys want me to plug x with infinity?

[gritty sandal]

yes

[willow bear]

finally.

[rich light]

so inf/inf right?

[gritty sandal]

oh god

[willow bear]

no!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

(╯°□°）╯︵ ┻━┻

[gritty sandal]

x/x simplifies to 1

[willow bear]

yknow what

i'm out

[gritty sandal]

so no need to evaluate x/x

[willow bear]

since you are clearly just gonna continue not listening to me at all

[rich light]

how about you just do it on paper so I know wtf you're talking about

instead of mentioning hospital rule

[gritty sandal]

ok ok wait a bit

[rich light]

tf is that?

no covid patients allowed?

[gritty sandal]

l'hopital is very useful

[past meadow]

l'hopital rule is a rule for computing limits

but its very unnessecary here

[rich light]

I'd love to use it if I had an idea as to what it actually was

also yes, sorry I didn't listen to you @willow bear

[gritty sandal]

do u want me to do it and send u a photo

[rich light]

sure

[willow bear]

apology rejected

[rich light]

🤷‍♀️

[gritty sandal]

Do you get it now?

[rich light]

not to trigger you again

but in this example

with the exception of the +

did I do that incorrect?

[gritty sandal]

Plug infinity not 0

Anything divided by 0 is undefined

[rich light]

so what I did wrong was I didn't plug in the infinity

which would have resulted in 0 values anyway, right?

for that particular fraction?

[gritty sandal]

Subbing 0 doesn’t give u 0

Cause of undefined

[past meadow]

don't just mindlessly plug in 0

for no reason

[gritty sandal]

The limit does not approach 0

[rich light]

I got my inspiration from this

[gritty sandal]

It approaches infinity

[past meadow]

yes i know, and they didnt do what you did

[gritty sandal]

They subbed infinity which got them to 0

[past meadow]

they didnt plug in 0 for x

the got the limit as x of 3/x and all the other fractions as x approaches infinity.

That is the limit that is equal to 0

[gritty sandal]

This probably belongs in #calculus tbh

[rich light]

nah it's chill

I solved the question haha

thank you boys

and ty Ann 😁

Denom is x^4, do I apply the same to the numerator? Or do I do x^2 for the numer

[gritty sandal]

Everything the same

Whatever you do to the denominator, you do the same for the numerator

Fraction laws

It’s like if you have $\frac{2}{4}$

[obsidian monolith]

Parking Ticket:

[gritty sandal]

And you divide the numerator by 2

You still have to divide the denominator by two

You can’t just divide the denominator by a diffferent number

Different*

[patent beacon]

In this case, the denominator gets larger than the numerator, and the function goes to 0

[rich light]

Yeah?

[gritty sandal]

why negative infinity

nvm

yeah its right

probably another line of working out evaluating all those fractions to something neater would be nice but not needed

[patent beacon]

Oh the limit is negative infinity haha, I didn't see that either

[gritty sandal]

but u have "square" it

so technically its positive

[rich light]

this correct too?

[rare pagoda]

should be -infty

[patent beacon]

Yes, make sure that when you plug -inf into x⁴, you get positive infinity

[rich light]

uhh

gotcha, thanks for the tip

and cute dog btw haha

[patent beacon]

Thx Suki is the best

4x³, as x approaches negative infinity

Doesn't come out to 0

[rare pagoda]

^

[gritty sandal]

so divide by x^6

instead

to avoid that

wait then we get 0/0

or something/0

[rare pagoda]

nah his/her work is right, they just didn't plug in for x correctly

[patent beacon]

There's no great solution here because the limit does approach -inf. The numerator is larger, and so the function acts like x³

[rare pagoda]

^

[gritty sandal]

ye

[rare pagoda]

its like a pseudo squeeze theorem kinda thing

[rich light]

right so, what do I change?

from x^3/1-8/x^3 blah blah

[patent beacon]

You have it right that the function is
4x³ / (1 - 8/x³)

And near the limit, it acts like
4x³

[gritty sandal]

this definitely should be in calculus

[rich light]

I can move it for the next question

so do I change anything?

[patent beacon]

So, let's think about what a cubic looks like

,w graph 4x^3

oh

Uh

[gritty sandal]

texit is running slow today

[obsidian monolith]

Display results online and refine query

Requested by Kaynex#7520

[gritty sandal]

idk why

but texit is very slow

[patent beacon]

For x that is very negative, the limit gets very negative.

[rich light]

okay yes

[patent beacon]

And that's the answer lol. The limit is -inf

[rich light]

algebraically, can you pinpoint where I went wrong

like in my algebra where I fucked up

[gritty sandal]

u subbed in -infinity to x^3

and said it was 0

[patent beacon]

Just saying "= 0" where that's not true haha

[rich light]

okay so what should x^3 have been

-inf?

[gritty sandal]

ye

so overall answer is -inf

[rich light]

hold on

so why would 8/x^3 be 0

is it because /-inf = 0?

[gritty sandal]

if u divide something by infinity, it's pretty small, so we say its 0

including -infinity but -0 doesn't rly change 0

[rich light]

alright

so just to clarify

all the other examples in which I ended up with zero values in similar equationm

s*

was cause of /inf/-inf right?

[thorn mountain]

what?

[gritty sandal]

what

[patent beacon]

what

[rich light]

for example

my 0s on the numerator/denominator were all due to dividing by inf/-inf

[gritty sandal]

yes

[patent beacon]

1/xⁿ approaches 0, for large (or largely negative) x.

[thorn mountain]

You should never plug an infinity into an equation

Infinity isn’t a number

[gritty sandal]

'approaching'

[thorn mountain]

Not correcting you parking ticket.

Looking at the work written down

Has stuff like $ \frac{6}{\infty} $

[obsidian monolith]

Malix:

[rich light]

that's it yeah?

[viscid thistle]

Yeah.

[rich light]

thank you all for the help

that was my last question so your eyes will now be spared 😊

[viscid thistle]

:0

[polar oyster]

Does anyone know how to write the linear factorization of this function?

[mental cove]

Well, you're given that x = 5 is a root. What does that tell you about the linear factorization of f?

[fleet yew]

Bro

you can factpr out an x

[gritty sandal]

so you have two roots

[sullen canyon]

i need to explain the transformations used to get f(x) = x^2 to g(x) = x^2 +4x+11

i get that +11 is a vertical shift

but what is the 4x? is it a horizontal stretch?

[willow bear]

actually no

there will not be a vertical shift by 11 at all

complete the square to get (x+2)^2 + 7

[sullen canyon]

ah

vertex form, totally forgot

so its 2 to the left, and 7 up?

[willow bear]

yes

[sullen canyon]

i don't rly understand how to do C

[willow bear]

write out the areas of OBC and sector OAB in terms of r and θ

[sullen canyon]

i wrote out OBC = 1/2 (r cos (x)) (r sin (x))

but then how would i write out the sector

if i don't know the radius

[willow bear]

do you know the formula for the area of a sector in terms of its radius and angle

...

[sullen canyon]

oh

[willow bear]

if i don't know the radius
somehow this didn't stop you from writing out the area of the triangle lol

[sullen canyon]

allow it

i got it

sometimes i don't know what i say

3/5 (1/2 r^2 x) = 1/2 (r cos (x)) (r sin (x)) right?

[willow bear]

assuming you're using x for θ, yes

[sullen canyon]

wait, but then how would i eliminate r

from the equation

[willow bear]

i mean you could just divide out by r^2...

[sullen canyon]

but then wouldn't it leave r^2 as the denominator

in 3/5

[willow bear]

no it wouldn't

[sullen canyon]

so i would have 0.3θ = 1/2 cos θ sin θ ?

[willow bear]

yes

[obsidian monolith]

Disabled_Skooter:

[viscid thistle]

$ABC$ is an equilateral triangle with side length 4. $M$ is the midpoint of $\overline{BC}$, and $\overline{AM}$ is a diagonal of square $ALMN$. Find the area of the region common to both $ABC$ and $ALMN$.

[obsidian monolith]

Disabled_Skooter:

[viscid thistle]

may i get help with this?

You may draw it @viscid thistle

And what have you tried

nvm lol I got it just a simple arithmetic error

[full garden]

hey guys can someone please guide me through this

I don't know how to do it with no double angle formula

[viscid thistle]

divide each side by 2 and use unit circle values

[full garden]

so I have $ sin 2 theta = 1 /2 $

[obsidian monolith]

iRazur:

[full garden]

and then i dont know what to do after this

[past meadow]

$\sin(2\theta)$

[obsidian monolith]

Sneaky:

[full garden]

yes how do i solve this bro

[past meadow]

anyway, what if it was sin(theta)=1/2

could you solve that?

without the 2?

[full garden]

yes i think 30

yes sneaky

[past meadow]

so what if you just did

[full garden]

sorry what?

[past meadow]

if the solution to sin(theta)=1/2 is theta=30

[full garden]

yes

[past meadow]

the solution to sin(2theta)=1/2 is 2theta=30

do you follow the logic?

we just have 2theta instead of theta

[full garden]

yeah....

[past meadow]

so divide both sides by 2

[full garden]

yes theta = 15

wait

idk how to do that sorry

[past meadow]

👍

[full garden]

oh

[past meadow]

you just did it no?

[full garden]

so the angle is 15?

[past meadow]

yes, so long as you're only looking for an acute angle

[full garden]

but if the angle is 15 that means 120-15 is also a solution right bro?

[past meadow]

sorry yeah

and there are more solutions, right?

[full garden]

yes bro im sorry I for some reason despite having my status online

didn't get any notifications

[past meadow]

don't worry about it

[full garden]

thank u

I thought 120-15 would give the right answer

180-15 sorry

thank u so much bro ❤️

if it wasn't for you I would've not been able to solve it

thanks a lot

[past meadow]

you sure about 180-15? maybe thats meant to be 180+15 😉

[full garden]

oh

why bro

when I take the sin of 165 I get 0.258

and when I take the sin of 15 I get 0.285

maybe it's different because I don't know how to do it for double angle formulas?

@past meadow u there bro?

[past meadow]

What you should be doing is more along the lines of finding solutions for 2sinx=1 between 0 and 720 and then dividing them all by 2

Just using the unit circle from here will confuse you

[full garden]

does that mean 75 is another solution..?

[past meadow]

It most certainly does

[full garden]

thank u very much

thank u bro

[wide lynx]

How would I factor 6x^3+17x^2+11x+2

[radiant ibex]

could someone plz explain how they get to that final answer

[craggy dune]

well a and d arent dependent on r so so you just keep summing them up and get na and nd

in the middle they just pulled the d out

[radiant ibex]

im good up to there

[craggy dune]

the bottom doesn't have anything to do with the calculation up top except that you need it to get the solution

see its the only sum term left