precalculus

[patent beacon]

Yes that's right

[main ledge]

tyvm

[thick raptor]

yw

[patent beacon]

You'd have less steps like so:
ln(2) - 0.5ln(2⁴)
ln(2) - 2ln(2)
-ln(2)

[distant flume]

I got 0 as a result, fair enough

[thick raptor]

Definitely not 0

[distant flume]

If I have (1/t+1/sqrt(t))(sqrt(t+1)-1) With Lim of 0, I rationalize the last part so that I can then divide 1/t+1/sqrt(t) by (sqrt(t+1)+1)/t. I then multiply t and 1/t+1/sqrt(t), which if I put the limit, it gives me 0; the denominator is sqrt(t+1)+1 aka 2

If you can, tell me what I didn wrong please

[thick raptor]

what

What are you multiplying by to rationalize?

[distant flume]

(sqrt(t+1)+1)*(sqrt(t+1)-1)

I am using (sqrt(t+1)+1)/(sqrt(t+1)+1) to rationalize (sqrt(t+1)-1)

I don't get where I got the mistake

[thick raptor]

Okay, so you have this:

$$\left(\frac1t+\frac1{\sqrt t}\right)\frac{(\sqrt{t+1}-1)(\sqrt{t+1}+1)}{\sqrt{t+1}+1}$$

[obsidian monolith]

Simple_Art:

[distant flume]

Yup

[thick raptor]

So what does the numerator become?

[distant flume]

Everything except for sqrt(t+1)+1

Oh you mean that I have to multiply all the denominators together

[thick raptor]

What no

What did the numerator become

Not denominator

[distant flume]

The numerator of the last part is t

Because we get t+1-1

[thick raptor]

Okay

$$\left(\frac1t+\frac1{\sqrt t}\right)\frac t{\sqrt{t+1}+1}$$

[obsidian monolith]

Simple_Art:

[distant flume]

Yes

I wish I could do that whole string of input

[thick raptor]

Okay so what's sqrt(t+1)+1 approach?

[distant flume]

2?

If t gets close to 0 then sqrt(1)=1 which means 1+1=2

And that's the denominator part

[thick raptor]

Okay

So what does the rest become?

[distant flume]

You have 0/2 on the last part though, and given that the first part has infinite, I don't know

I think that the logical answer would be infinite since the first part approaches infinite faster than the second part, given how we can simplify the first part to (t+sqrt(t))/(tsqrt(t)) and we multiply that by t/2, getting (t^2+t)/(2t(sqrt(t))

This problem isn't supposed to be hard man

@thick raptor I give up

[cursive cave]

@distant flume there?

[distant flume]

I'll read

[thick raptor]

:l

$$\left(\frac1t+\frac1{\sqrt t}\right)\times t$$

[obsidian monolith]

Simple_Art:

[thick raptor]

$\frac1t\times t=~?$\$\frac1{\sqrt t}\times t=~?$

[obsidian monolith]

Simple_Art:

[thick raptor]

@distant flume

[distant flume]

But why was the denominator of the second part 1 instead of 2, as I said before?

[thick raptor]

??

the limit of √(t+1)+1 is 2, not 1

[waxen ruin]

It's been almost 4 years since I took calculus and longer since I had trig/algebra, is this a good place to ask for help? I'm starting a calc class this semester since I want to be better at it for my cs degree and this is just review stuff that I have trouble with, should it go in a different channel?|

https://i.imgur.com/0KbwLtP.png
https://i.imgur.com/aQkcsec.png
https://i.imgur.com/dWHcon2.png

[patent beacon]

@waxen ruin. Nah, can do here. You just want to solve these for x?

For 3, you want to clear the denominators. Just multiply everything by x(x - 5)

[waxen ruin]

Ooooooooohhhhhh yeahhhhh, I remember that now. Thanks a bunch! @patent beacon I figured out the other ones too

[potent viper]

how the FUCK do i do this mf

[patent beacon]

@potent viper
Please keep it to one channel.

[potent viper]

o ty

i still dont get it

for the first one right

i' d get like log2M^6+log2N^3

@patent beacon

[patent beacon]

That's a good first step. Then, use the power rule

6log(M) + 3log(N)

[hexed ermine]

Oh Kaynex is helping you lel

[potent viper]

wait

THATS IT

damn the properties thing rlly helps

oh theres another thing

i cannot like understand no matter what

this

like i'd get to the part where you do the 1/2+ 1/4 over and over again

i think this is the answer but i dont understand

how it gets to that

waikt nvm

but um

when it gets to this part, is there a way to get them faster, or the only way is by looking at the pattern

[spiral vale]

Hey guys I have been having trouble solving this problem, could anyone help me? Thanks

Part b

[sharp pagoda]

Anyone care to help sketch this word problem?

I missed a few classes while I was gone for vacation and I regret it haha

[shrewd flame]

I hated log

[distant flume]

@sharp pagoda positive Cosine function with distance of 12 seconds between periods

You can work out the function on your own I suppose

Negative Cosine function, sorry

[formal fable]

Yo quick question because I'm dead tired and am not figuring this out

If I have a function of time and the unit of my time is, say, months
How do I convert that to weeks

Ie my function is $2\cdot3^{n-1}$ for n amount of months

[obsidian monolith]

LikeALol:

[earnest nymph]

We assume that there is 4 weeks in a month

So 1 month = 4 weeks

[formal fable]

Of course

[earnest nymph]

No, it's actually n/4 - 1

[formal fable]

4 years in uni and I don't realise this shit
Thanks

[earnest nymph]

👍

[sweet wind]

Heeey guys, does the straight line equation go here? I don't know if that's what it's called in english

[patent beacon]

y = mx + b? Linear equations

[sweet wind]

Yeaaah but i have learned its like y = kx + m

[patent beacon]

Just labels. Same thing

[sweet wind]

I feel like a complete fool but im a first year sicence student and im desperate

Yeah okay

I need help with a problem

[patent beacon]

y = qx + λ

[sweet wind]

a:”the rectangle can be puzzled together by the parts to the right. Solve how many pieces of each you need” and b:”Prove that it’s impossible to make a chessboard out of these pieces” there are 77 squares. 39 Black and 38 white.

[gritty blaze]

,rotate left pi/2

[native sequoia]

,rotate 180

[craggy burrow]

^what?

[spice wadi]

please stop spamming channels

[vernal lantern]

Does anyone know what is included in the precalc cirriculum?

[rocky bison]

precalc mostly

[limber bone]

search up precalc exams

and see

[prime prawn]

review from a2

2nd half is precalc

[viscid thistle]

=tex \Re

[granite stirrup]

Command disabled

The sever owner has disabled that command in this location.

[viscid thistle]

It cost a bus company $225 to run a minibus on a ski trip, plus 30$ per passenger. The bus has seating for 22 passengers, and the company charges $60 per fare if the bus is full. For each empty seat, the company has to increase the ticket price by $5. How many seats should the bus run with maximize profit from this trip?

how do i solve this

[obsidian monolith]

cix:

[compact galleon]

Cos(-x)/1+Sin(-x) = Sec(x)+Tan(x) Proof Got left side to Cos(x)/(1-Sin(x)) and right to (1+Sin(x))/Cos(x)

Don’t know where to go

[swift glacier]

Have you tried seperating the fraction on the left?

[compact galleon]

In what way?

[slow wharf]

How would you say

the answer is 2 when n is an even number

$x = 2, 2n \in N$

[obsidian monolith]

Autistic Hoodie:

[slow wharf]

I would assume?

[flint lichen]

Would someone be willing to help me out with this? I’ve tried solving it my self but haven’t gotten anything close to what the answer is. I’ve also tried mixing up the numbers to make sure I didn’t flub something but nothing matches any of the answer choices.

[flint lichen]

(Just specifying what I tried doing) For #1 I tried using 58cos/sin(14) and
196cos/sin (270). I then added the vector points together and used inverse tan to find the angle. However this gave me something like -63.

[static vapor]

separate into horizontal and vertical components

then add them

[flint lichen]

I’m sorry I don’t quite understand.

If you mean like vector point 1 is <a,b> and vector point 2 is <c,d> and add c and a along with d and b that’s what I did

[flint lichen]

<@&286206848099549185>

[static vapor]

@flint lichen yes

but for the diagonal vectors you need to use trig to separate the vectors

[flint lichen]

@static vapor sorry phone ran out of charge so I didn’t get the notification so like 196cos(270) and 196sin(270) ? Or do you mean something else and I just don’t understand

[static vapor]

yes like that

but where did the 270 come from?

[flint lichen]

I did that sorry I did cos/sin because I got lazy totally backfired and wasted even more of yours and my time. It says it comes due south which is 270 on the unit circle

I’m fairly certain I also tried it with 90 degrees and just got a negative value instead of a positive one on the vertical component

I’ll try again though

Yeah just a positive value

I tried doing the problem again and ended up with 273 (well -87 but I added 360) the same I got before. I have no clue what I’m doing wrong

[clever inlet]

some say it's necessary, other's say not

[viscid thistle]

Depends on what you already know

[clever inlet]

^

[viscid thistle]

If you know how to work with functions, took a class on trigonometry and did some analytic geometry, you're basically set already.

[static vapor]

@flint lichen sorry for the VERY LATE response i went to sleep

notice how angles D and E are given?

you will have to form a vector "triangle" using this

and use trig

ive done a triangle for vector a for you

all you need to do is fill in the numbers whip out ur calculator and use some trig

remember that EVERY VECTOR is made up of a horizontal and vertical component

and thus a right angle triangle can ALWAYS be formed

because horizontal is PERPENDICULAR to vertical

hence a RIGHT ANGLE is formed

and theres 3 vectors so its a triangle (horizontal, vertical and resultant technically theres only 1 actual vector but i hope you get my point)

I got C as my answer

You need to:
Separate Vectors
Add Vectors
Use Pythagoras (for vector size)
Use arctan (for the angle)

[buoyant ore]

for this problem, would i need to use this formula?

[trim coyote]

https://gyazo.com/9d82ad1d3e1d1830b48e9494309f10a4

can someone explain how you'd write a function for this?

I'm seriously too lost

answer: https://gyazo.com/184c0bd6b5aaf50701bb04589dac464e

[strange prism]

Bruh

[trim coyote]

what? 😦

where in the world does that 12/x come from??

[slow wharf]

Any idea how to do this shit?

$y+cos^2(\frac{x}{y}) = \frac{9}{2}$

[obsidian monolith]

Autistic Hoodie:

[patent beacon]

What do you mean "do"?

[slow wharf]

How do you solve for y...

[patent beacon]

No chance. Won't happen for this function

[slow wharf]

Really?

How am I supposed to graph it

Actually

I am supposed to get the function for the tangent and normal of the function...

At a certain point

Hm

Hmm

[patent beacon]

That's a good question, what are you supposed to do with it?

[slow wharf]

It basically says find the function of the tangent at the point T(pi, 4)

[patent beacon]

That's much easier, but it needs implicit differentiaton

[slow wharf]

What about this

I solve for x

So it is an inverse function

I find the tangent function for the inverse function

and I inverse the tangent function

[viscid thistle]

Someone answer this question

https://gyazo.com/3a134d0adc9da8d6c353955cab6f9232

cause it makes me want to punch my fucking computer screen

[patent beacon]

No, you're going to want to apply implicit differentiaton

It's easy if you know how

[viscid thistle]

?

These questions are so boring I can't concentrate

My mind repels against it

so can you answer it for me

ah I'll take it some where else

[slow wharf]

lol

$y' + 2cos(\frac{x}{y})(-sin(\frac{x}{y})) = 0$

[obsidian monolith]

Autistic Hoodie:

[slow wharf]

That ain't working out

Oh

sec

Incorrect derivative

[swift glacier]

if two vectors are parrellel, then the dot product would be one, right?

[swift glacier]

someone pls?

wait nvm

[strange prism]

i wonder if its possible to get an A in precalaus

[limber compass]

Guys i think this is precalclulas

but https://www.youtube.com/watch?v=yhwbO_wDqm0

Can someone frking explain thsi too me

i dont understand it

and veererererrey new to calc

--
the guy says :

find the steepest point sloping upwards. What does he mean? by sloping upwards

and what is a gradient function ?

[static vapor]

the most "straight" like close to vertical point

where the graph is moving up

so from left to right

the graph is increasing

[serene heath]

points where the gradient is maximum

so the second derivative is 0

[static vapor]

its precalc i doubt he knows what the derivative is

i think this is more of the start of calc so i think it would be better to describe it graphically

and doesnt second derivative =0 just mean possible point of inflexion?

it should be second derivative >0

@limber compass

[thick kayak]

Quality paint.net curves

[limber compass]

@static vapor thnx dude 🙂

[prime prawn]

"You intercepted the following cryptogram and have intel suffesting it was encoded with a 2x2 matrix."

8 21 -15 -10 -13 -13 5 10 5 25 5 19 -1 6 20 40 -18 -18 1 16

The last word of the message is _RON. What is the message?

How do i do this problem, all I have done is transformed the series of numbers into a matrix w/ 2 columns, and converted _RON --> 0 18 15 14

the letters follow the alphabet, and _ is 0.

it's part of the matrix intro in our precalc book.

<@&286206848099549185>

[prime prawn]

I'd be really grateful if anyone could help me, ping me or DM me please. thanks

[rare zephyr]

No. 8

Topic is Quadratic Equations

I need help on "translating" it into a quadratic equation

<@&286206848099549185>

[fallen cloud]

!15m

[spring thunder]

rekt

[rare zephyr]

pls help ;(

pls

<@&286206848099549185>

[viscid thistle]

his speed increases by 5 km/h

then the time he needs decreases by 15 minutes

which is also 1/4 of an hour

so if his speed increases by 5km/h then he needs 25% less time to get there

[spring thunder]

1/4 h less doesn't mean -25% tho

[viscid thistle]

but it should in this case

because if his new speed is 20km/h then he needs 0.75h (=45 min) to get to the post office

so it fits the -15 min part

[spring thunder]

how do you know it takes him 1h originally?

[viscid thistle]

its based on the assumption that if he increases his speed by 5km/h then the time decreases by 15min

and given that an hour has 60 minutes

the time decreases by 1/4

when the speed increases by 5

[spring thunder]

yeah that works

[viscid thistle]

what did you have in mind @spring thunder

[spring thunder]

actually doing quadratic equations lel

[viscid thistle]

uhhh XD

i missed that point, nice

[rocky bison]

[viscid thistle]

but i don’t know how i would make it quadratic

[spring thunder]

we know distance = speed*time

so if we apply this to the two situations we get

$$\begin{cases}15=vt \ 15=(v+5)(t-\frac14)\end{cases}$$

[obsidian monolith]

emeric75:

[spring thunder]

(v and t being the original speed and time taken)

[viscid thistle]

yea that’s what i had as well

[spring thunder]

$$\text{ie }\begin{cases}\frac{15}{v}=t \ 15=(v+5)(t-\frac14)\end{cases}$$

[obsidian monolith]

emeric75:

[spring thunder]

substitute and expand

[viscid thistle]

i see

[viscid thistle]

Regarding the ambiguous case, my text book says if a<b there are zero triangles, h < a < b there are two, and if a> b there are one

But for a>b

Can’t there be two possibilities? If a is long enough it could stretch over b and form a narrow triangle

Not a right triangle tho

So really only a = b has one because it’s either line or triangle

[odd yarrow]

15 = tv = (v + 5)(t - 0.25)

tv = vt + 5t - 0.25v - 1.25

tv = t(v + 5) - 0.25(v - 5)

t(v+5) - 0.25(v - 5) = (v+5)(t-0.25)

= 15

idk I'm in high school Geometry I shouldn't even be posting in this channel :/

[static vapor]

@viscid thistle for sine rule?

its not really related to length

its more like angle sum

so say u had a triangle

35 degrees is the given angle

youre given 2 sides

you get a result of say 20 degrees

but because of the ambiguos case it could be 160 or 20

but it cant be 160

because 160+30 is =190>180

larger than the max angle sum of triangle

35*

but say if the result was something like 36 degrees

then you get either 36 OR 144

and 144+35 = 179

which is less than 180

so if you get a result

where a 3rd angle exists

i.e its "degrees/angle" is more than zero

[viscid thistle]

Can someone post a problem you were given in your Pre-calc. class. I just want to know what it looks like

[open star]

How do?

[viscid thistle]

,rotate 270

[viscid thistle]

How do you solve these nested rectangle problems?

well you know the area inside is 35cm²

,$ 35 = a*b \ a = 7-2x \ b = 9-2x

[obsidian monolith]

yami:

[viscid thistle]

thanks weeb

u welcome

[viscid thistle]

What if it’s? A cube

so you know that V = 128cm³

they wrote cm² but thats a typo

the height of the thing is 2cm, so you can divide the volume by 2 to get the area of the square

do you know how to get the rest

also its not a cube because the height would have to be like 3 or 4 times of what it is

w*l = 128/2

And to find w * l

That’s the part I’m asking abt

well you have a square so W = L

ohhh

but what if a rectangle

then you have to do the thing you asked above

Okay

https://cdn.discordapp.com/attachments/363224154469826562/535938086262145035/133376600275156992.png

Thanks

so what do you get for the length of the cardboard thats needed

or size, whatever

[viscid thistle]

Why is log base-10^-1000 undefined? Shouldnt the answer be 3 since -10^3 = -1000.

<@&286206848099549185>

[sullen shoal]

logarithms do not allow the base to be less than or equal to 0

or 1

[viscid thistle]

Why not?

Couldnt we define them for negative values in some instances like this?

[polar lodge]

You can't because of how logarithm function is defined.

[viscid thistle]

Fair enough

just was curious what the motivation was for defining the log function that way

Also, does the notation "a sub n" denote a series or a sequence? Or is it ambigous?

@polar lodge

[polar lodge]

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

But, most of the maths they discuss will probably be out of scope for you depending on your mathematical background.

@viscid thistle, it depends on where it's used so, it's contextual.

[viscid thistle]

So in the case of "the limit as n approaches infinity, of a sub n, equals 8"

Does the context of the limit clarify it or is it still ambigous?

[polar lodge]

Well, you ask yourself then where does limit work on?

[viscid thistle]

In my mind I could concieve of it two ways. The limit is the value the sum of all the terms is approaching (series), or the limit could be the value of the nth term (sequence)

[polar lodge]

So, here, a_n can be either sequence or function whichever terms the writer uses.

If it's series it will likely have summation symbol but again, it depends on the pre-assumption before u see that.

[prisma scroll]

t!remindme breakfast in 30 minutes

[brisk mica]

⏰ | Got it! I'll remind you in 30 minutes!

[viscid thistle]

That makes a lot of sense. Thanks! @polar lodge

[knotty garden]

Hi there. Looking for some help with a logarithmic equation question, can i ask here?

[spring thunder]

no?

[knotty garden]

I tried to use the latex formatting but I dont think it fully worked

$\frac{1}{log_y(4)}=log_\frac{1}{4}\frac{1}{8x}$

[obsidian monolith]

peterdorvay:

[knotty garden]

oh it worked! awesome. So I'm trying to express y as a polynomial

what would be a good way to approach this? change of base? I'm lost.

[knotty garden]

ooh im getting closer

[royal gull]

1/log_y(4) = log_4(y)

and then try to have same base rhs and lhs?

yeah

y=8x

@knotty garden you got it?

[knotty garden]

hmmm i got y=2x

let me put everything i did down and maybe someone could show me what I did wrong

$\log(y) = log(4) log_\frac{1}{4}(\frac{1}{8x})$

$\log(y) = log(4) \frac{log(\frac{1}{8x})}{log(\frac{1}{4})}$

$\y=4\frac{\frac{1}{8x}}{\frac{1}{4}}$

[obsidian monolith]

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

[knotty garden]

ok so this is what i did

the bottom line should have y= but im not sure how this bot works

and after evaluating it I get 2x. But I think 8x is the correct answer @royal gull

[royal gull]

@knotty garden the way I did it: 1/log_y(4) is just log_4(y) (y is the base)

then you change RHS to be a log in base 4

RHS after changing the base is - log_4(1/8x) = log_4(8x)

so you have the same logs on both sides with constant base, so you just need to compare the things inside

getting y=8x

at least that's what I got

and you cant just multiply the things inside the logs like you did

log(a) * log(b) =/= log(ab)

[knotty garden]

yeah i just tried getting them all log base 10

[royal gull]

ooh

but still, get them to base 4

as I said

[knotty garden]

yes ok ill try that now

when you change the base you do log_1/4(1/8x) = log_4(1/8x) / log_4(1/4) right?

like the change of base formula.. log_a(b) = log_c(b) / log_c(a)

because doing this i got y=0.5x 😦

[royal gull]

umm yeah, you get that fraction

but log_4(1/4) is just -1

so its -1 * log_4(1/8x) = log_4(8x) (because you take -1 into the log and you get (1/8x)^-1 )

[knotty garden]

oh ive just been trying to cancel the logs

by getting them all to the same base

[shrewd flame]

https://prnt.sc/m9ikpc

help 😢

[shrewd flame]

https://prnt.sc/m9ioif

Did i do it right?

[eternal lotus]

$cos(2x+x) \ne cos(2x)+cos(x)$

[obsidian monolith]

stephen:

[eternal lotus]

Instead use the additional formula: $cos(a+b)=cos(a)cos(b)-sin(a)sin(b)$

[obsidian monolith]

stephen:

[shrewd flame]

oh okay

what do i do after?

https://prnt.sc/m9iwxi

[eternal lotus]

What exactly are we trying to prove here?

$cos(3x)=cos(x)(1-2sin(x))(1+2sin(x))$ \
or $cos(3x)=cos(x)(1-2sin(x))(1+sin(x))$

[obsidian monolith]

stephen:

[eternal lotus]

Also the picture you just sent, you forgot to multiply the cos(2x) and cos(x)

[steady cove]

In this question I've been asked to find the rate of change in this function

How do I find the rate of change?

[patent beacon]

Usually the derivative

What's the function? Taking calculus?

[steady cove]

Yea

One sec

[patent beacon]

The rate of change is the slope of the tangent line. The derivative gives those

[steady cove]

So I just do the first differentiation

[patent beacon]

dy/dt means the derivative of y in terms of t

[steady cove]

Ok

[patent beacon]

So they want you to find dy/dt with the derivative

[steady cove]

So I just do first differentiation then plug in the t value?

[patent beacon]

No t value necessary, dy/dt is just the function from differentiaton

[steady cove]

So that's the rate of change of that function?

[patent beacon]

You'll want to curve your t's, mistaking that for a + is inevitable!

That derivative looks right

[steady cove]

Yea

Just realised

Thanks!

[steady cove]

Same question but now I'm trying to find the max temparture

Is 15 the max temparture?

[patent beacon]

The max temp happens when t = 15. Remember that t represents time.

@steady cove

[steady cove]

Ok

Do I sub t=15 into the first function

[shrewd flame]

@eternal lotus it's cos3x=cosx (1-2sinx)(1+2sinx)

[eternal lotus]

$cos(3x)=cos(x)[1-2sin(x)][1+2sin(x)]$ \
LHS: $cos(3x)=cos(2x)cos(x)-sin(2x)sin(x)$ \
$cos(3x)=[1-2sin^2(x)] \cdot cos(x) - 2sin(x)cos(x) \cdot sin(x)$ \

[obsidian monolith]

stephen:

[eternal lotus]

@shrewd flame Can you continue from here?

[vast star]

Alright, third time posting this, but it seems like I can't get an answer for this.

I just can't find how to perform it.

[viscid thistle]

I mean. You said it - the thing is already symmetric

[vast star]

Ugh, I meant elementary symmetric polynomials.

[viscid thistle]

Ye

Expand my guy

[vast star]

I did.

[viscid thistle]

It's that easy

[vast star]

It isn't.

[viscid thistle]

The expansion is elementary lmao

[vast star]

You sure?

[viscid thistle]

Check if you don't believe me

[vast star]

I did.

Still sure this is elementary?

I'm pretty sure it isn't.

Unless your definition is different.

[viscid thistle]

Nah I goofed

[vast star]

No problems.

I just want an answer, this was supposed to be so simple. .-.

Yet neither I can do it nor I can get an answer.

[viscid thistle]

Woah woah woah

Hold up a Dusseldorfing minute

(a-b)(b-c)(c-a) = (a-b)(bc-c^2-ab+ca)

= abc-ac^2-a^2b+ca^2-b^2c+bc^2+ab^2-abc

Check the terms

[warm nebula]

Easier way is just to expand all 3 at once

[vast star]

I did.

[viscid thistle]

And what do you get?

[vast star]

They aren't elementary and cannot be factored into elementary factors.

Thing is, you don't have to factorise.

[warm nebula]

Factor what question

[vast star]

You just have to rewrite in terms of them.

You can see it above.

[viscid thistle]

We get the premise

ab(a-b) should be ringing alarm bells though

(b-a) sry

[vast star]

Like, I suppose you know the identity,

a^3 + b^3 + c^3 = (a + b + c)^3 - 3(a + b + c)(ab + bc + ac) + 3abc

It'll be something like this, won't be factored but rewritten.

[viscid thistle]

Ye it's made with the ring of symmetric polynomials of n-variables

[vast star]

And since its square is well known, although a lot more messy, I'd thought this would be easy to rewrite.

[viscid thistle]

Might be easier to attack from the factors

For instance how would you write "a-b" in terms of elementary symmetric polynomials?

[vast star]

Well, good idea, it can definetly help to start from simpler things.

[flat turret]

Guys tf are you even doing

[viscid thistle]

Idk Gonzo

Enlighten us

[flat turret]

You can't write a non-symmetric polynomial in terms of symmetric polynomials

a-b change variables b-a

it's not equal

[viscid thistle]

ab(b-a)

[flat turret]

(a-b)(b-c)(c-a) is not symmetric either

change a<->b

[vast star]

I'm a complete idiot.
Well, thanks for daring to lighten me!

It took a while.

[flat turret]

Kek it happens

[vast star]

Happens a lot more often when you are me.

:)

Since I managed to get some of you here,

https://cdn.discordapp.com/attachments/527899096883789860/534800806021431316/image0.jpg

Would you like to take a look at this one?

The first line tries to say a, b and c are in the interval [0, 2pi).

[flat turret]

Everything is symmetric and the product says cos(x+y)=-1 for some x, y among A, B, C

So we can WLOG assume cos(A+B)=-1

which means A+B is an odd multiple of pi

which means $3(A+B)=n\pi$ for some odd n

[obsidian monolith]

Gonzo17:

[flat turret]

so $\cos 3A=\cos (n\pi -3B)=\cos (\pi-3B)=\cos (3B-\pi)=-\cos (3B)$

[obsidian monolith]

Gonzo17:

[flat turret]

from parity of cos and the fact that $\cos(x-\pi)=-\cos x$

[obsidian monolith]

Gonzo17:

[flat turret]

Wait but if $(A, B, C)=(0, \pi, \frac{\pi}{2})$ it doesn't work I think

[obsidian monolith]

Gonzo17:

[flat turret]

$\cos 3A+\cos 3B+\cos 3C=1+(-1)+0=0=12\cdot 1\cdot (-1)\cdot 0=12\cos A\cos B\cos C$

[obsidian monolith]

Gonzo17:

[flat turret]

and $\cos (A+B)+1=\cos (\pi)+1=-1+1=0$

[obsidian monolith]

Gonzo17:

[vast star]

Don't the equations still hold?

[flat turret]

Oh wait, sin(pi/2)=1 so it's okay

So we have $\cos 3C=12\cos^2 A\cos C$

[obsidian monolith]

Gonzo17:

[vast star]

Hm-hm.

[shrewd flame]

how does cos3x become cos2xcosx? @eternal lotus

[elfin night]

cos(3x)$\neq$cos(2x)cos(x)

[obsidian monolith]

Colen:

[rocky bison]

but what if x = 0 @elfin night

[fallen cloud]

ye forgot the -sin(2x)sin(x)....

with the cos(3x)

[elfin night]

do u wanna die

@rocky bison

caus this is how it ends

triggering a coLEN

[serene heath]

trigering?

trig?

[slender river]

smgh

simplify cos3x to cos(2x+x) then do a sum simplification and then some double angle stuff

idk

ok lol idk the context so ignore me maybe?

[elfin night]

Idk the Q but still gonna give my opinion that SA should get the biggest available

[fallen cloud]

colen

nvm

o wait, you can play chess with me

[elfin night]

no

why

[gritty blaze]

Baduk ?

[elfin night]

just say go you pleb

[gritty blaze]

Weiqi ?

[elfin night]

[eternal lotus]

@shrewd flame

[obsidian monolith]

stephen:

[viscid thistle]

hello i need some guidance pls

[shell salmon]

For first one, use trig ratios to get horizontal distance between b and c

For second one, use triangle area formula to find the perpendicular line length, then trig ratios to find x

For the third one, you can create a triangle with a leg through c, perpendicular to ad. Use lengths and trig ratios to find d

[viscid thistle]

thanks @shell salmon

much appreciated

[viscid thistle]

Review for exam not worth marks,,, how do I do this question

[tawny nacelle]

=pup maximize x(150 - 3x/2) fro x > 0

[granite stirrup]

Do more with Wolfram|Alpha pro

Query made by

@tawny nacelle

[tawny nacelle]

so the maximum area is 3750

[serene heath]

set up variables and 2 equations

[tawny nacelle]

oh he asked HOW to do the question

[serene heath]

yea

thats how u do it

[odd yarrow]

what kind of trigonometry would go here as opposed to #geometry-and-trigonometry?

[rocky bison]

What u confuse about with integral

[odd yarrow]

the concept of it as opposed to derivates/differentiation

[rocky bison]

You know how derivatives work and such?

[odd yarrow]

kind of

its the rate of change of x

[rocky bison]

Well integration is the opposite to that

So

,$ \int\left(\frac{d}{dx}f(x)\right)dx=f(x)

[obsidian monolith]

Pseudo:

[odd yarrow]

thats the notation for integrals?

[rocky bison]

The integral is the curvy boi

,$ \int

[obsidian monolith]

Pseudo:

[odd yarrow]

o

[rocky bison]

,$ \int f(x) dx

[obsidian monolith]

Pseudo:

[rocky bison]

Is the integral of f(x) with respect to x

Integration of polynomials is pretty ez

like derivs

[odd yarrow]

so lets say f(x) = x^3

d(f(x))/dx = 3(x^2)

[rocky bison]

,$ \int\left(x^n\right)dx=\frac{x^{n+1}}{n+1}+C\n\neq-1

[obsidian monolith]

Pseudo:

[rocky bison]

Ok

So

,$ f(x)=x^3\iff f'(x)=3x^2

[obsidian monolith]

Pseudo:

[rocky bison]

Now let's do

,$ \int\left(3x^2\right)dx

[obsidian monolith]

Pseudo:

[frozen needle]

ouch

[rocky bison]

wot

[frozen needle]

not an equivalence relation

[rocky bison]

huh?

[frozen needle]

,$ f(x)=x^3\implies f'(x)=3x^2

[obsidian monolith]

Tuong:

[rocky bison]

[frozen needle]

but no iff

[rocky bison]

PEDANTIC

[odd yarrow]

cool

so d/dx(f(x)) = f'(x)

[rocky bison]

d/dx(f(x))

[odd yarrow]

yeah

[elfin night]

Pedantry, my favourite

[stuck palm]

Anyone have any idea? Lol

Wouldn't this be 4?

[patent beacon]

Yes

The limit can come into the square if that doesn't force a limit to not exist

[stuck palm]

So all I do is literally plug the f(x) function which is 2 and just square it in that bracket for this example? No other steps in this instance?

[patent beacon]

Formally,
lim [f(x)]²
= [lim f(x)]²
= [2]² = 4

[stuck palm]

Sweet, that's exactly how I did it. Lol the 1 and 0 functions are just in there to confuse people... smh

[patent beacon]

Indeed, they don't seem related. Are they part of another question?

[stuck palm]

Nope thats all it is LOL

I appreciate your help as always 😃 @patent beacon

[patent beacon]

Np! Feel free to ask if you have anything else

[stuck palm]

Thanks will do! 😃

[rotund surge]

tanx-secx=-1, solve by squaring both sides

[patent beacon]

Okay. Have you squared both sides?

[rotund surge]

yes, and i have tan^2x+sec^2x-2tanxsecx=-1

[patent beacon]

Should be +1 on the right

[rotund surge]

oh

so it should

[patent beacon]

Hmm. Let me suggest doing this instead
tan(x) = sec(x) - 1

[rotund surge]

yeah

[patent beacon]

Then square that
tan²(x) = sec²(x) + 1 - 2sec(x)

[rotund surge]

and tan^2(x)+sec^2(x)=1?

[patent beacon]

No, tan²(x) + 1 = sec²(x)

[rotund surge]

how do you do that

how do you type exponents

[patent beacon]

So you could write
tan²(x) = sec²(x) + 1 - 2sec(x)

tan²(x) + 1 = sec²(x) + 2 - 2sec(x)

sec²(x) = sec²(x) + 2 - 2sec(x)

0 = 2 - 2sec(x)

My phone has some exponents on it

[rotund surge]

ah

[patent beacon]

And thus
sec(x) = 1
cos(x) = 1
x = 0 + 2kπ

[stuck palm]

So on this graph for problem g) what would the answer be since there is a point on both sides of that? Would it be 1,-1 or undefined? So confused about that

[patent beacon]

Undefined. The limit does not exist there.

Remember, the limit only exists if the left and right limits agree

[stuck palm]

They would both have to be either open or closed circles? Not one or the other?

[rotund surge]

how do i set it so that wolfram alpha gives me 0-2pi

[stuck palm]

Wait so how would I know on that problem if the left and right limits agree?

[patent beacon]

The left limit at x = 2 is -1
The right limit at x = 2 is 1

[stuck palm]

So just to be clear there can't be 2 limits on the same spot of the x axis?

[patent beacon]

Since the left limit and the right limit are not the same, the total limit does not exist

[stuck palm]

I see i see

And if a limit is infinity do I just put undefined or infinity?

[patent beacon]

A limit to infinity is technically undefined, but your teacher probably expects you to put infinity

[stuck palm]

Gotcha I will do infinity/undefined just to be safe. Thanks

[stuck palm]

I got one more for you, I have no idea where to even start on this problem. The entire thing is confusing me

[stuck palm]

@patent beacon Just whenever you have time, no rush

[patent beacon]

Oh boy ε-δ definitions

[rocky girder]

Pls help with 42

[stuck palm]

Like I am so screwed

[patent beacon]

@stuck palm
The ε-δ definition is the rigourous definition of the limit.

Imagine we're playing a game. We start with where the limit is, and we want to build a rectangle around the limit point such that the function is contained inside the rectangle.

I start by giving you a height. Call it ε.

I want you to give me back a length, δ.

Use this length and height to create a rectangle with a center on the limit point. If the function leaves the top or bottom of the rectangle, you "lose". If you can contain the function with appropriate choice of δ, you "win".

If you can always win, no matter what ε I give you, then you've proven the limit. If there's a case where you can't win, then the limit must be false.

[viscid thistle]

help help help

[patent beacon]

There's definitely a lot there, this isn't the easiest concept

[stuck palm]

@patent beacon See I understand the concept, it's just the functions and finding out what point they want me to write down for epsilon, delta, b & C

[patent beacon]

From that picture
ε = 0. 05
f(b) = 2 - 0.05
f(c) = 2 + 0.05

[stuck palm]

Epsilon makes sense to me, but how did you come up with B & C?

[patent beacon]

You can see it on the picture. b, when drawn up to the function, has a height 2 - 0.05

f(b) = 2 - 0.05

[stuck palm]

So that's what the answers would be for B & C? Since anything between those 2 points is epsilon?

[patent beacon]

I don't know about that, but from the picture, f(b) = 2 - 0.05

This is enough to find both b and c

Then, δ is the shortest distance necessary to bound the function within ε. In this case, that's a - b

[stuck palm]

Then so how would you go about finding B and C based on the 2+- 0.05?

[patent beacon]

If you start at x = b, then draw up to the function, then you go to a height y = 2 - 0.05

Ergo, f(b) = 2 - 0.05

[stuck palm]

Oh I see!

[shrewd flame]

Question: Modelling Exponential Growh and Decay Problems.

• In 1985, there were 285 cell phone subscribers in the small town of Centerville. The number of subscribers increased by 75% per year after 1985. How many cell phone subscribers were in Centerville in 1994?

My question is, how does 75% become 1.75??

[patent beacon]

Increase by 75% means take the old amount, multiply by 1.75

@shrewd flame

[shrewd flame]

but why 1.75?

[patent beacon]

Or, you can think about adding 75% on.

That is, increasing x by 75 is the same as doing x + 0.75x

[stuck palm]

@patent beacon Btw if you scroll up to my message where I sent the last picture. That definition sucks 😂

[patent beacon]

ε-δ

?

[stuck palm]

The definition they gave me on my problem lol yeah the delta epsilon definition. Could be so much easier to explain like you did

[viscid thistle]

How do i do e?

[patent beacon]

You're doing fine with what you've written

That method should work

[viscid thistle]

I dont know what to do next though

[thick raptor]

Don't advertise your questions in other channels. #❓how-to-get-help for rulez.

[viscid thistle]

Sorry

[patent beacon]

You'll need tan(π) = 0

[viscid thistle]

Wont do again

Tan pi equals 0?

Oh

[patent beacon]

Know your unit circle, of course

[stuck palm]

So I basically squared my epsilon’s since y = square root of X. Do these figures I put on the right look accurate? @patent beacon

[patent beacon]

ε = 0.05, not 2 ± 0.05

,calc (2 - 0.05)^2

[obsidian monolith]

Result:

3.8025

[stuck palm]

Sweet! So everything else seems fine apart from that? (I changed it)

[patent beacon]

,calc 3.8025 - 4

[obsidian monolith]

Result:

-0.1975

[patent beacon]

Everything looks good!

[stuck palm]

Yesss I am so glad I am getting the hang of this. Would mean absolutely everything if I can get an A in Calculus.

[patent beacon]

Definitely worth going for. Anything else bugging you?

[stuck palm]

That's everything 😃 I'm sure my professor will assign more difficult problems this week. I got him tomorrow morning and Thursday morning. I'll let you know later this week if I need help! Thanks! 😃

[viscid thistle]

x^0 (the constant) controls the point at which the polynomial starts (y-intercept), x controls the slope at y-intercept, what about x^2 and higher degree terms, what do they control in a polynomial?

[limber bone]

they do not act like lines with like degree 1

ax^2 + bx + c

c is y intercept

[deft flume]

I need help with finding the exact solution of this equation 8sinxcosx=1 I set the equation to 0 and got stuck Any help would be appreciated.

[patent beacon]

8 sin(x) cos(x) = 1

4 sin(2x) = 1

That help? Need further steps?

[deft flume]

oh okay

thanks, i will try to evaluate it further

wait @patent beacon , wouldn't it make sense to not convert the sinxcosx into sin2x?

[patent beacon]

4 sin(2x) = 1

sin(2x) = 1/4

2x = sin¯¹(1/4)

x = 1/2 sin¯¹(1/4)

If you can make your equation have only one x in it, you can always solve for that x

[deft flume]

for the inverse sine, how exactly do you approach that

[patent beacon]

Take the inverse sin of both sides

You can use a calculator to get it

If you want. The above is an answer as well

[deft flume]

so i would do inverse sin times the sin on the left side?

which cancels it out?

[patent beacon]

Not "times" there's no multiplication there

[deft flume]

that's what im confused about

[patent beacon]

sin is a function. It takes a number as an input and gives a different number back as an output

As well, there's sin¯¹ which is another function, that does sin in reverse

If
y = sin(x)

Then
sin¯¹(y) = x

[viscid thistle]

What is the difference between cos^2(pi) and cos(pi)^2

Same thing

@viscid thistle

Just different way to write

Hmm

ok so cos(pi)=-1

cos^2(pi)= -1^2= 1?

@viscid thistle

cos(pi)=-1 indeed

cos^2(pi)=cos(pi) * cos(pi)

-1 * -1 = 1

confirm

@viscid thistle

=pup cos^2(pi)

[granite stirrup]

Do more with Wolfram|Alpha pro

Query made by

@languid wind

[viscid thistle]

OK got it! Thanks man 😃

Just a confusion of the notation

so

x^2 = x * x

cos^2 = cos * cos

@viscid thistle

just remember that

Got it, that is intutive

Thank you for the hlep

[swift glacier]

How would you find a vector that is parallel to a given vector?

[viscid thistle]

@swift glacier cross

oh wait

no thats the orthogonal

[rocky bison]

[viscid thistle]

whats the problem

[rocky bison]

Parallel

[viscid thistle]

if <-1,-1,-1> is a vector

[rocky bison]

[viscid thistle]

<1,1,1> is parallel

[swift glacier]

Like, is there a particular formula

[rocky bison]

if u got two vectors

a and b

[deft flume]

im having trouble with factoring this: 2cos^2x-1+cosx=0 any help?

[rocky bison]

and a=cb

for some value c

Then they parallel

[deft flume]

just 2(cos^2x)-1 + cosx = 0 ?

[rocky bison]

Two vectors are parallel if they're scalar multiples of each other

wtf

@viscid thistle overcomplcating it

@deft flume do you know what quadratics are?

[swift glacier]

wouldn't the angle be zero?

[rocky bison]

could also be 180

could be opposite directions

[viscid thistle]

Theres 2 reasons they would be paralell,
one is a scalar multiple of the other

[swift glacier]

tru

[viscid thistle]

orrr

(looking through notes 1 sec)

[rocky bison]

There is no or tbh the first case always hold

unless one is 0

or both

cos then it make no sense

[viscid thistle]

cos(theta)= u * v / (||u||)(||v||)

that one @swift glacier

?

[rocky bison]

[viscid thistle]

denominator = norm of u * norm of v

the | wont write out

it either boils down to -1 or 1

[deft flume]

oh wait

i can do abc ight

[rocky bison]

wot

[swift glacier]

So the angle between the two vectors is 1 or -1?

[deft flume]

well for the quadratic formula

[viscid thistle]

no

the angle is either 0 or 180

paralell

[rocky bison]

so not -1 or 1

[deft flume]

oh nvm, just factor using the values

[viscid thistle]

this

is it

[deft flume]

nvm that doesn't even work either

[viscid thistle]

@swift glacier

if you get the last line

to theta = cos^-1 (-1)

or to

cos^-1(0)

then they are parallel

[swift glacier]

Isn't that just the formula for the angle between two vectors?

[viscid thistle]

because cos^-1 (-1) is pi (180 degrees)

[rocky bison]

how is this becoming so convoluted

he literally agreed like 10 mins ago plz

[swift glacier]

So, would I just use that formula and set it for 1 or -1

[viscid thistle]

If needed yeah you could

[swift glacier]

I understand the pi180 part

I see

Thanks

[deft flume]

2cos^2x-1+cosx=0 -> cosx(2cosx+1) -1 =0

is that correct?

[rocky bison]

That's not quite factored tho

Are you familiar with quadratics?

[deft flume]

i can't seem to get -1 in tho

yea

[rocky bison]

So you know how to solve

,$ ax^2+bx+c=0

[obsidian monolith]

Pseudo:

[rocky bison]

,$ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

[obsidian monolith]

Pseudo:

[rocky bison]

So let's imagine

,$ X=\cos(x)

[obsidian monolith]

Pseudo:

[rocky bison]

Therefore you have

,$ 2\cos^2(x)+\cos(x)-1=0\implies2X^2+X-1=0

[obsidian monolith]

Pseudo:

[rocky bison]

Solve for X

[deft flume]

i got stuck at x^2+x = 1/2

[rocky bison]

You said you knew how to solve quadratics

[deft flume]

lmao, im kind of trash at it

[rocky bison]

Use quadratic equation tbh if you can't see an obvious factor

[deft flume]

but sometimes i know what im doing

i guess that's where my flaw is

[viscid thistle]

factor out an x

x(x+1)=1/2

[rocky bison]

doesn't help

[viscid thistle]

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

[rocky bison]

not sure how you got from that factor to that tbh but ok

[deft flume]

lol

[rocky bison]

also somehow got three roots for a quadratic

,$ \left(2x-1\right)\left(x+1\right)=0

[obsidian monolith]

Pseudo:

[rocky bison]

That's our facotered version

[viscid thistle]

which

[rocky bison]

But we need to know how to get there from quadratic

[viscid thistle]

oi i see

[rocky bison]

So are you familiar with quadratic equation

[viscid thistle]

u rite pseudo

[rocky bison]

i always rite

[deft flume]

lmaoo

i didn't think to realize to factor the form right off the back

[rocky bison]

,$ ax^2+bx+c=0\implies x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

[obsidian monolith]

Pseudo:

[rocky bison]

This our quadratic form

And we have

,$ 2X^2+X-1=0

[obsidian monolith]

Pseudo:

[rocky bison]

So

a=2
b=1
c=-1

[deft flume]

ye

[rocky bison]

We just plug those bois in and use calculator

or work it out in brain

just takes bit longer

And you get the solutions

x=1/2

and x=-1

Ye?

[deft flume]

didn't even touch a calc yet

[rocky bison]

make sure you get same answers

[deft flume]

just gonna do it on paper

[rocky bison]

good idea

[deft flume]

i got .911 and -.41

[rocky bison]

That doesn't sound like it

oke we go thru the process of reducing fraction

,$ \frac{-1\pm\sqrt{1^2-4\cdot2\cdot\left(-1\right)}}{2\cdot2}

[obsidian monolith]

Pseudo:

[rocky bison]

So we got this

We do sqrt first

,$ \frac{-1\pm\sqrt{9}}{4}

[obsidian monolith]

Pseudo:

[deft flume]

oh no wonder

i wrote c as 1

lmao

btw, there's no point in using the quadratic formula

[rocky bison]

why not

[deft flume]

since i just needed to factor the expression

[rocky bison]

So?

This is how to factor it

If you can't see how to factor it straight off

[deft flume]

oh nvm

its the same thing

thank you

[swift glacier]

is it possible for a dot product to be greater than the product of it's magnitudes?

[eternal lotus]

Hmm maybe not

$a \cdot b > |a||b|$

[obsidian monolith]

stephen:

[eternal lotus]

The angle between two vectors wouldn't hold I suppose as it would lead to cos(theta)>1?

[swift glacier]

oh true

I see

Thanks!

[eternal lotus]

A second opinion would be helpful as im not 100% sure

[nocturne wraith]

I think this is precalc, just trying to make sure my understanding is correct. So if I have a function f(x) with two given points in the form of (a, b) , if it's one to one, if i were to plot it's inverse, I just have to plot the two points (b,a) right?

[viscid thistle]

Yes

Domain becomes range

Range becomes domain

[nocturne wraith]

Gotcha, just wanted to double check before I inputed lol

[slender river]

yes

there's a lot of cool stuff on inverses hehe

[nocturne wraith]

Taking calc 2 online, little rough remembering some of the previous stuff

[hexed ermine]

Nice, Lagrange Inversion Theorem is pretty cool to learn when you get to a point of learning power series for inverse functions

[nocturne wraith]

It's name sounds like i'll be back in the future for a more in depth explanation xD

[hexed ermine]

If you are given f(x) you can find the power series of f^(-1)(x)

[thick raptor]

Basically Taylor expansion for inverse function

[viscid thistle]

trying to find where these functions intersect, y^2=x , x=2y. setting them equal to each other and solving for 0. i keep getting stuck, would anyone be able to help?

nvm, i cant believe i didnt see it before, i got it

[hexed ermine]

:+1:

[viscid thistle]

how do I solve p

@Everyone

[serene heath]

really

@in everyone in a server of 10k smh

which one

u need help with

[viscid thistle]

the one in the middle of the third pic

[serene heath]

3x+6 one?

[viscid thistle]

the one above 3x+6

I can't factorize the numerator

[clever inlet]

this one is a nice enough case

[viscid thistle]

?

Chu can help?

[clever inlet]

actually nvm

it's really not a nice case

[viscid thistle]

what am I supposed to do?

[clever inlet]

do you know how to long divide?

[viscid thistle]

I do

[clever inlet]

well

you know x - 1 is a factor

[viscid thistle]

yes

[clever inlet]

divide by that

[viscid thistle]

then
?

[clever inlet]

h

[viscid thistle]

The limit tends to 1

[clever inlet]

it's a factor twice

yes

[viscid thistle]

sth over 0 is undefined

[clever inlet]

you have a 0/0 indeterminate form

just long divide