precalculus

[vital tartan]

then?

[vital tartan]

yeah

lol

[lost pawn]

That's what I thought two

[viscid thistle]

the first one isn't infinity, the second is

[hollow obsidian]

^

[viscid thistle]

you can graph it if you don't believe me

or calculate for like 0.9, 1.1 and you'll see why

[vital tartan]

nah it's fine but is it ^^^^ that you're testing it with?

I'm pretty sure that limit doesn't exist

=tex \lim_{x \to 1^{-}} \frac{x}{x - 1} = -\infty

[lost pawn]

Does this have something to do with the even exponent?

In which case is the following limit true for any n?

=tex \lim_{x\rightarrow 1} \left ( \frac{x}{x-1} \right )^{|2n|}=\infty

[vital tartan]

=tex \lim_{x \to 1^{+}} \frac{x}{x - 1} = \infty

[vital tartan]

oh wait

[vital tartan]

what's the question then ?!

I saw multiple equations

I'm confused at which one it is

[lost pawn]

f_1 and f_2

@viscid thistle

[viscid thistle]

f_2 goes to infinity

f_1 diverges

[lost pawn]

Look at the limit i posted tho

[vital tartan]

okay, the first limit doesn't exist, then the second goes to infinity

^^^

[lost pawn]

Would the same function raised to the power of 6 go to infinity?

[viscid thistle]

yes

[lost pawn]

So this is true+

=tex \lim_{x\rightarrow 1} \left ( \frac{x}{x-1} \right )^{|2n|}=\infty

[viscid thistle]

depends what n is

[lost pawn]

As a general term?

Integer

[viscid thistle]

hmm

is it true for n = 0

[lost pawn]

No

Okay

So

Integers > 0

Or less than

[viscid thistle]

$$\lim_{x\rightarrow 1} \left ( \frac{x}{x-1} \right )^{|2n|}=\infty \ \forall \ n \in \mathbb{N}$$

[lost pawn]

Yes

[viscid thistle]

I think so

[lost pawn]

Alright

That's quite nice

[viscid thistle]

as long as it's even so everything is positive

[lost pawn]

Thanks for the help 😛

[gritty blaze]

n = 0

[proper flicker]

how do like graph log equations?

for ex: y=log _3x

[fringe stream]

can you graph 3^x?

3^x is the inverse of log_3(x), so that's the way i'd do it.

graph it, then use y = x as the axis of symmetry.

[proper flicker]

but how do I graph log_3(x) after?

[fringe stream]

since 3^x and log_3(x) are inverses of each other, their graphs will be symmetric about y = x.

so, graph 3^x, then use y = x as the line of symmetry to graph log_3(x).

=pup graph 3^x and log_3(x)

[granite stirrup]

Query made by @fringe stream
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=graph+3^x+and+log_3(x)
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/

[proper flicker]

what do you mean by using the line of symmetry?

sorry I don't get how to graph the log part

[unborn mist]

So imagine a mirror was on the y=x line

And all you gotta do is create the image shown by the mirror

[proper flicker]

So the point for 3^x would be (-2,1/9) , (-1,-1/3) , (0,1) , (1,3) , (2,9)

and I'd just flip x and y?

[unborn mist]

Uhuh

Since you know log1 is 0

When we do inverses we essentially swap the range and domain

[proper flicker]

Ahh I see

thanks for explaining

[real fog]

cos^4+2cos^2sin^2+sin^4 = (cos^2+sin^2)(cos^2+sin^2) i cant seem to factorise it what method

x^2+2xy+y^2

[true vigil]

what

you just did

x^2 + 2xy + y^2 = (x+y)^2

[real fog]

kyizo did it

[fringe stream]

what do you not understand about that?

[real fog]

how do i factorise this cos^4+2cos^2sin^2+sin^4 after changing to this x^2+2xy+y^2

[true vigil]

you complete the square

but it's already complete

[real fog]

ohhh now i get it thanks you guys

[true vigil]

👌

[slate hill]

how do i "Write the statement as an absolute value equation. Let x represent the unknown number.The distance between 5 and a number is 2."

[dense zealot]

Hi

|5-x| = 2

[young fractal]

For negative tangent graphs

[viscid thistle]

If a function is neither odd nor even, what would it be?

@viscid thistle go algebra

pls

Uhhhh... ok.

[young fractal]

Do I set (x-c) to +-pi/2

Or 0 and pi?

[viscid thistle]

what

[young fractal]

For negative tangent graphs

I’m trying to find out where I’m putting my asymptotes

[viscid thistle]

I don't know honestly

@true vigil

Idk and i’m not a helper :/

loses honorable

[young fractal]

@viscid thistle you at least know what I’m asking?

[true vigil]

What

[young fractal]

Oh

[true vigil]

Whats a negative tangent graph

[dense zealot]

@viscid thistle neither

There's odd, even, and neither

@true vigil -tan(x)

???

Idk

[true vigil]

=pup -tan(x)

[granite stirrup]

Query made by @true vigil
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=-tan(x)
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/

[young fractal]

@true vigil it’s meant to be like cotangent

But

I’m new to this so

I don’t know if the cotangent rule where you set (x-c) to 0 and pi to find the asymptotes apply

[true vigil]

What

[hasty wadi]

hi, im confused on how im wrong for this question. i'll post my work so you can look at it cuz im confused. i need help lol

never mind, i got it

[viscid thistle]

Someone can help?

[deft granite]

$$0.5 = r/(1/2)^\theta$$

[deft granite]

!!!

1.2 not 1/2

What is a "best description"?????

[golden oriole]

Did I get these correct?

[deft granite]

Try again for the fill in the blank

[stark trench]

Help a retarded fellow out

Thanks

<@&286206848099549185>

[viscid thistle]

that's pretty cool

didn't know quadratic solar troughs were a thing

also that's really smart

[stark trench]

Finally

So can you help me out?

[viscid thistle]

so how I would solve it is I'd put the bottom of the thing at the point (0, 0)

then the end points are (+- 90, 56)

right?

[stark trench]

Yep I got that but the equation was so diff from mine

[viscid thistle]

wait actually the equation would differ based on where we put this solar panel on the graph

[dense zealot]

Parabolic solar panels are smart

Since beams bounce around

and stuff

And all hit where u want them tk

[stark trench]

So...?

[viscid thistle]

yeah I know

the design is amazing

[hasty wadi]

can i please get some help with this question

[granite stirrup]

289.85507246

[granite stirrup]

Query made by @pine kindle
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=solve+-14000+-+10.35n+%2B+58.65n++%3D+105000
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/

2463.76811594

[granite stirrup]

91000

1845.84178499

1884.05797101

2463.76811594

[stark trench]

Someone please tell me the steps to solve this

I just need this solved because Idk how quadratic function word problems work

[void patio]

it's asking for 2 roots at -90, and 90 with a vertex at (0, -56).

@stark trench

just write the general formula that represents it

[rotund mantle]

^ with these three points you can simply use the vertex form to solve it, or if you’re fancy express it as a Laplace polynomial

[stark trench]

The a comes out to 0.0069

I just don't know how

Mine is like 0.002

[viscid thistle]

Can someone help?

[unborn mist]

Volume=500=pir^2h

Solve for r

[brave cedar]

imagine there are n cans

n times the volume of 1 can = total volume of liquid

[viscid thistle]

I solved it already thanks.

V = pir^2h

make r^2 subject of the formula

and solve

after getting 15.9, I square root, since it is r^2

hence I get 3.9, which is 4.

The Answer is A, thanks for the assist though.

[unborn mist]

Np

[viscid thistle]

I NEED HELP FOR MY PRACTICE TEST TO SKIP PRE CALC OMG I NEVER STUDIED

its tomorrow

[final sigil]

Why skip Pre-Calc?

[viscid thistle]

lol cuz it was an available option

[final sigil]

I don't think that's a good idea.

[viscid thistle]

i kinda understand how to do the question but not fully

[rocky bison]

Do you know your rules of indices

[viscid thistle]

i still want to try the test tho

[rocky bison]

Or exponents you americans all them

[final sigil]

There's a lot of precalc that then transfers to calculus.

[viscid thistle]

lol yeah

i was like whats indices

[rocky bison]

So just apply those

So rewrite the squareroots

as powers

[viscid thistle]

oooohhhhhh

[rocky bison]

Then the solutions are kinda obvious

[viscid thistle]

gah ok

[rocky bison]

for the first two

[viscid thistle]

but the third one?????

[rocky bison]

First of all

I'd move that square inside the brackets

on the left bracket

After that I'd move the bottom terms up to the top

and make them negative powers

Then just add the powers as they multiply together

make sense?

[viscid thistle]

yuuuhh

ill try it

and see if i get it right lolol

thanks thanks

[rocky bison]

np

[viscid thistle]

Sin theta = opp/ hyp

Hyp * sin theta = opp

Hyp * sintheta./ opp = 1

Amirite?

[hexed ermine]

yes

You can prove that by doing Hypotenuse/Opposite = csc(theta) and a trigonometric function multiplied by it's inverse will equal one

csc(sin)=1; sec(cos)=1; cot(tan)=1

[past jay]

Yeah what you've really done here is make a reciprocal identity. since sin(theta) = y/r then it follows r/y * sin(theta) = 1 since r/y * y/r = 1

Of course, what pjs said is just as true. Though I'd be careful calling csc and sin "inverses" but.

[hexed ermine]

Haha true, you can confuse inverses. for example sin^-1 you would think is the 1/sin or the reciprocal but it's the arcsin instead :v a bit confusing at first

[past jay]

personally I never use the ridiculous -1 notation for trig

I try to avoid it for functions entirely

I just write arcsin and arccos, etc

[hexed ermine]

same lol

[grim yarrow]

-1 forever

[past jay]

HERETIC

[clever inlet]

i like the arc notation

but like school is picky

and they require the -1

[onyx basalt]

Hey everyone! This ought to be super simple for you, but I can't seem to figure out what to do.
How would you find out the result of the following expression on paper?

[patent beacon]

= log1/4(3*8 / 6)

[onyx basalt]

I see, I am an idiot. thanks 😄

[patent beacon]

Not at all! Just new to these rules

[onyx basalt]

actually, not new to this at all. I have just forgotten about this rule 😊

[onyx basalt]

alrighty, so I have another one here:

[onyx basalt]

My attempt:
looking at the denominator, it's domain is

[onyx basalt]

since argument of square root can't be negative and the denominator can't yield 0. The next thing is that the argument of log must be > 0. Since the square root in denominator returns both positive and negative results, I am not sure how to proceed.

[onyx basalt]

I am an idiot again. the domain of f in whole numbers is -6, -5, -4, -3, -2, -1, 0, 1, 2

sorry for spamming here

[onyx basalt]

Since I havem't gotten the answer in #help-2 yesterday, I am trying my luck today here.
How do I find such q which will make the line a tangent of the parabola?
https://cdn.discordapp.com/attachments/359052604149465088/419142317853638656/latex.png

[clever inlet]

with only precalc i assume

well substitute them into each other

resulting in a new quadratic

you need to force this new quadratic to have a discriminant of 0

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

[onyx basalt]

🤔 that was my train of thought as well, but then I have gotten that q^2 is a negative number. But it's nice to see my reasoning was right and I just screwed up with numbers.
Thank you 😄

also - if you get only one result for intersection of parabola and line, it does not have to be the tangent IMO.

lemme draw an image

oh no, it does have to be a tangent line, I see

$$(x+q)^2-x-1=0$$

$$x^2+2xq+q^2-x-1=0$$

$$x^2+x(2q-1)+(q^2-1)=0$$

$$D=(2q-1)^2-4(q^2-1)$$

$$D=4q^2-4q+1-4q^2+4=0$$

$$-4q=-5$$

$$q=\frac{5}{4}$$

[clever inlet]

yeah

[onyx basalt]

😊
nice, thanks 😄

[gritty pewter]

anyone mind an assist? Got one making my head hurt a tad, lol

System of three numbers problem

can't seem to eliminate anything

[viscid thistle]

[gritty pewter]

lol

labelling these as t, s and f

t = 3f

f = s - 10

and t +f+s = 125

[viscid thistle]

Yep.

So we can now substitute/add equations in our system together to get values for t, s and f

vich von first zo

[unborn mist]

try using eqns 1 and 2 to sub into 3 and make sure you only get either t,f or s on the LHS

[gritty pewter]

write this out as t + s - 10 +s = 125?

[unborn mist]

I think the easiest is f

[viscid thistle]

^

Isolate f

[gritty pewter]

effort is being made not to set this pc on fire

[viscid thistle]

Which PC 0.o

[gritty pewter]

my own

[viscid thistle]

Jeez.

[gritty pewter]

lol

[viscid thistle]

That ain't right

Having any luck with your problem though?

[gritty pewter]

Not really yet no

[viscid thistle]

Try rewriting eq. 3 in terms of f.

[gritty pewter]

just keep writing and rewriting this out , hoping it clicks

so then f = s - 10 and 3f = t

[viscid thistle]

Yes, yes.

So we want to replace the s and t in the third equation.

[gritty pewter]

f = t/3

[viscid thistle]

So we need to find forms for s and t in terms of f.

Right.

[gritty pewter]

f = 3f/3

errr

[viscid thistle]

Wait wut

[unborn mist]

hold on

you have t=3f and s=10+f

and also that f+s+t=125

so from here

try and find a way to use those first two equations

to make the third equation simpler

so that you would have 1 unknown for 1 equation

[viscid thistle]

Have you seen substitution before?

[gritty pewter]

I have, just been at this too long this mornig

[viscid thistle]

I see.

[gritty pewter]

may be its break time

[viscid thistle]

Maybe.

This is the most important piece of information.

you have t=3f and s=10+f
and also that f+s+t=125

[gritty pewter]

rofl

f = 25

excuse my looking like a buffoon

[viscid thistle]

Are you sure about that?

Could you show your working?

[gritty pewter]

so this becomes

f + s-10 + 3f

= 125

f + s - 10 + 3f =125

f + s +3f = 115

[viscid thistle]

Where did you get the s-10 from?

f+s+t = 125
f+s+(3f) = 125

I understand this, because t = 3f

[gritty pewter]

so 4f+s =125

[viscid thistle]

Right.

And f=s-10

[gritty pewter]

4(s-10)+s=125

[viscid thistle]

Sure.

So now we rearrange to get s.

And since we know f = s-10, and t=3f, it's pretty easy to get the other two values from here.

[gritty pewter]

4s=155/4

errr

s - 155/4

s = 155/4

[viscid thistle]

4(s-10)+s=4s-40+s=5s-40

So, 5s-40=125

[gritty pewter]

add 40 to both sides

5s=165

[viscid thistle]

Okay.

[gritty pewter]

s = 33

makes jokes about writing problem incorrectly always leading to the wrong solution

[viscid thistle]

Sure.

So now we know f=s-10

And t=3f

[gritty pewter]

23

t = 69 (3f

[viscid thistle]

== 69+23+33

[granite stirrup]

125

[viscid thistle]

And it works.

Hurray.

[gritty pewter]

Jichael, thank you mate

[viscid thistle]

You're welcome.

[hasty wadi]

I'm confused on this problem

I get two lengths but idk how to differentiate between which one is the right length

In order to find the width

[dense zealot]

Hi

This is so easy

2(x+y) = 30

xy=56

(x+y)^2 = 225

xy = 56

2xy = 112

[hasty wadi]

?

[dense zealot]

Oh u just want dimensions

[hasty wadi]

yes

[dense zealot]

x+y = 15

xy=56

make a quadratic

x^2 -15x + 56=0

[hasty wadi]

i did but im confused on how you can get 2 x's

two lengths?

[dense zealot]

?

wdym

[hasty wadi]

if ur "x" is length and ur width is "y"

how can you get x^2

[dense zealot]

Wdym

How did u get x^2

[hasty wadi]

quadratic

i end up with x^2-15x+56=0

[dense zealot]

Cuz

Vietas formulas

[hasty wadi]

i don't know that lol

[dense zealot]

sum of roots of quadratic= -b/a

product = c/a

[hasty wadi]

im only supposed to be getting the dimentions by means of using system of equations

im confused on how i can get two lengths

[dense zealot]

Then use substitution

x+y=15

xy=56

x= 56/y

y+ 56/y = 15

(y^2+56)/y = 15

y^2 + 56 = 15y

y^2 - 15y +56=0

[hasty wadi]

yes

but

you end up with y^2

[dense zealot]

yes

[hasty wadi]

which is only the width

[dense zealot]

Doesn't matter

Find the solution

U'll see why

[hasty wadi]

7 and 8

[dense zealot]

(to the quadratic)

Ya so

Any way u put it

Its 7 and 8

[hasty wadi]

so how to you know which one is the right answer?

[dense zealot]

I said try them!

7+x=15

x=8

[hasty wadi]

ohhhhhhhhhhh

[dense zealot]

8+x=15

Lol

[hasty wadi]

so it doesn't matter which one i put as width or length?

[dense zealot]

Ya doesn't matter

[hasty wadi]

🤔

[dense zealot]

Just test ur answer at the end

2(8+7)=30

8*7=56

[hasty wadi]

ahhhhhhh

ok

thanks dude 😃

[dense zealot]

Yw

[gilded echo]

hey, I'm wondering how I could find the h (missing in the parentheses) in the following problem:

it should be pretty easy, I just can't figure it out

[viscid thistle]

[viscid thistle]

can someone help with that

[rugged marsh]

What did you try?

Y intercept means x=0

So g(0)=4

[pliant tide]

yo wanna do my homework

[rugged marsh]

Depends

[pliant tide]

wait fr

im just too tired rn to do the last two problems

[rugged marsh]

I guess

[pliant tide]

YO

UR A GOD

[rugged marsh]

The problems?

[pliant tide]

d and e

[rugged marsh]

Hmm

[pliant tide]

plz?

[rugged marsh]

How much time do you have?

[pliant tide]

i have the class

second period

tomorrow

so not a lot

[rugged marsh]

Ok

I will send a pic just a sec

[pliant tide]

THANK U

[rugged marsh]

Im only doing this because you told me you were tired

[pliant tide]

yea

thanks man

I havent been getting much sleep recently

[rugged marsh]

Oh

[pliant tide]

my grades are

kinda bad

i've been

staying up late trying to work on assignments

[hexed ermine]

Moving to other channel I see lmao

[rugged marsh]

Then you should definitely get rest

[pliant tide]

Agreed

Im done after this math assignment

[rugged marsh]

Done

[pliant tide]

alright

pic?

[pliant tide]

is the bottom part e?

[rugged marsh]

Ya

[pliant tide]

thanks bro

[rugged marsh]

Hey

[pliant tide]

?

[rugged marsh]

What's going on in the general voice channel?

[pliant tide]

no clue

im not in it

[rugged marsh]

Anyway

Wish you a good sleep!

[pliant tide]

thanks! you too!

[rugged marsh]

It's morning here...

Your homework was the first thing I did in the morning

[past jay]

@frosty hound

$$\frac{\cot(x) + \cot(y)}{\tan(x) + \tan(y)} = \cot(x)\cot(y)$$

[frosty hound]

lol hey spider

[past jay]

Aaaanyway,

[frosty hound]

lol!

[past jay]

Tbh, trying to figure out best way to do this,

[frosty hound]

i mean that dude already found the answer

i just need to know how to do it

[past jay]

Yes but he just bulldozed through

[frosty hound]

xD

[past jay]

I'm sure there's probably a much neater way of it

[frosty hound]

ok

[past jay]

So here's what I'm gonna try,

$$\frac{\cot(x) + \cot(y)}{\frac{1}{\cot(x)} + \frac{1}{\cot(y)}} = \cot(x)\cot(y)$$

[frosty hound]

ok

[past jay]

Just reciprocal identity

Now I'm going to get a common denominator for the two bottom fractions

$$\frac{\cot(x)+\cot(y)}{\frac{\cot(y)}{\cot(x)\cot(y)} + \frac{\cot(x)}{\cot(x)\cot(y)}} = \cot(x)\cot(y)$$

[hexed ermine]

LOL

[past jay]

$$\frac{\cot(x) + \cot(y)}{\frac{\cot(x)+\cot(y)}{\cot(x)\cot(y)}} = \cot(x)\cot(y)$$

[livid birch]

Ahhhhh

[past jay]

Hush boi

[analog hawk]

lmao

[frosty hound]

hi pjs long time no see

[hexed ermine]

Heylo

[frosty hound]

ok so i get that...

wow got croweded here

[past jay]

So now, multiply the left by cot(x)cot(y)/cot(x)cot(y)

[livid birch]

What is this proof

[past jay]

$$\frac{\cot(x)+\cot(y)}{\frac{\cot(x)+\cot(y)}{\cot(x)\cot(y)}} \times\frac{\cot(x)\cot(y)}{\cot(x)\cot(y)} = \cot(x)\cot(y)$$

[livid birch]

You could just shorthand it right?

[hexed ermine]

The ended doesn't have { i think

[livid birch]

Just imply how many operations you wanna do

[past jay]

asdf

My internet went poof

[past jay]

$$\frac{\cot(x)\cot(y)(\cot(x) + \cot(y))}{\cot(x)+\cot(y)} = \cot(x)\cot(y)$$

[past jay]

And, as you can see, we now have cot(x) + cot(y) in the numerator and denominator, simplify and you get,

$$\cot(x)\cot(y) = \cot(x)\cot(y)$$

[past jay]

1 = 1

The most boring equation in the world

[frosty hound]

but the best answer

lol

i mean best equation

alright i wrote that all in my notebook

as reference

[past jay]

True.

I thought of doing that but figured it would take more steps... clearly I was wrong.

[viscid thistle]

Does anyone know a trig eqns pdf resource or something? Not in my textbook

[dull hinge]

Tbh you can work all of them out from sin^2+cos^2=1

I still don't know most off the top of my head and I can function fine

[viscid thistle]

Really? That's cool, I'm gonna try it now

[patent beacon]

@viscid thistle
You can work them all out from
sin(a + b) = sin(a)cos(b) + cos(a)sin(b)
cos(a + b) = cos(a)cos(b) - sin(a)sin(b)
sin²(a) + cos²(a) = 1

That's enough to get everything.

[viscid thistle]

Thanks, I will write them down.

[patent beacon]

Try an app for trig equations, if you want a quick resource

[viscid thistle]

Thanks, my trig unit didn't cover trig equations. Would like the see how they are derived.

[dull hinge]

do you know complex numbers?

[viscid thistle]

Yes a bit, for creating a Mandelbrot set once

[dull hinge]

so you know the e^ia=cosa+isina?

this thing

or maybe this representation

[viscid thistle]

No, I know e^i pi tho lol

I'm just going to find a textbook which has it :P, mine doesn't

[dull hinge]

hang on

[viscid thistle]

Oh my gosh, it makes a circle in the complex plane?? That's cool

[dull hinge]

yes exactly

now from this you can derive all the tirg fomulae

[viscid thistle]

Thanks for the help! My goal is to learn trig equations by next week.

Keeps popping up everywhere, kind of annoying that I can't solve them.

[dull hinge]

so to find cos(a+b) you note that cos(a+b)+isin(a+b)=e^(i(a+b))=(e^ia)(e^ib)=(cos(a)+isin(a))(cos(b)+isin(b))

=cos(a)cos(b)-sin(a)sin(b)+i(sin(a)cos(b)+sin(b)cos(a))

and you just want the real part

so you end up with cos(a+b)=cos(a)cos(b)-sin(a)sin(b)

it's quicker in your head tbh

it also means you can get the formulas for multiple angles too pretty easily

[viscid thistle]

Oh yea I see it now

[dull hinge]

so you can do things like that for all the formulas

[viscid thistle]

That's pretty neat, thanks again for showing it.

[dull hinge]

np

it's how I learned the formulas

[hasty wadi]

i need help with this problem, this is my work

i did it but im confused on how my answer is wrong

@earnest nymph i see ur a helper, can you please help me on this problem if you can

[void patio]

csc(cos^-1x)= ????

need help

[viscid thistle]

@void patio create a triangle

[void patio]

ok how will that help me

i'll just label the sin from csc and the inverse of cosine?

i'll try it but i have no idea where you're pointing me to

[dense zealot]

Hi

use some algebra

[hasty wadi]

damn, this question is killin me

[dense zealot]

1/sinx * 1/cosx = 1/sinxcosx = 2/sin(2x)

@hasty wadi double the second

Subtract from first

Double the first

Subtract from the second

[hasty wadi]

im redoing the problem right now

[dense zealot]

triple the middle

[hasty wadi]

do you mind if you check my work real quick

[dense zealot]

Double the bottom

K

[void patio]

@dense zealot was that for me?

"use some algebra"

😂

[dense zealot]

Looks ok

[hasty wadi]

my answer problem on my phone says it's infinite solutions

for some reason

but that's what i got

i checked my answer

it's wrong again @dense zealot

[dense zealot]

Put it in wolfram alpha then

uwu

Don't they have step by step solutions?

[hasty wadi]

not if ur a member

im no member so i cant

[viscid thistle]

@void patio what would be the csc of this?

[void patio]

idk, the opposite is missing.

if i can't figure out sin then i can't say what csc is.

[viscid thistle]

how do you find the opposite given the adjacent and the hypotenuse?

[void patio]

well i got sqrt of x-1

not sure what this is.

[void patio]

after doing the a^2+b^2=c^2 stuff

yeah pythagorean theorem

i remember the name

oh ok i see alright thanks.

[viscid thistle]

i like the quadratic formula

);

np

[viscid thistle]

@hasty wadi i think i got it, let a be any integer, and the solutions are x = 3(7-a), y =5(a-3), and z = a

[hasty wadi]

i tried that, didn't work

[viscid thistle]

fuck

[hasty wadi]

it doesn't matter anyways. i had 5 attempts and used them all lol

(the assignment isn't graded, don't worry)

[viscid thistle]

haha thats pretty lucky, my ap physics teacher gives us only 2 attempts

[hasty wadi]

i tried the factored out version, and the simplified version

fml, i gotta talk to my teacher

shietttttttttt 2 attempts?

that dude need to chill

[viscid thistle]

^^

[hasty wadi]

well, im heading out

thanks for attempting it tho

i appreciate it 😃

[viscid thistle]

learned more linear algebra doing this problem than i ever have from my school :p

[hasty wadi]

lmao

this question was a pain]

i'll learn the solution tomorrow tho 😎

[sacred eagle]

help

https://gyazo.com/53d4571b12b868a13800fa7f06563622

how can i solve this

[unborn mist]

you been taught CAST?

[sacred eagle]

probably yes since i have this homework

[unborn mist]

https://www.onlinemathlearning.com/image-files/xtrigonometric-ratios5.jpg.pagespeed.ic.pmDOsdR1Ej.jpg

[sacred eagle]

yes

[unborn mist]

that should help you decide

where theta must be

[sacred eagle]

oh i get it CAST

thats easy

[unborn mist]

yeah

so sintheta is negative in the tan and cos quadrants

and tan is negative in the cos and sin quadrants

they both intersect negative in the cos quadrant

[sacred eagle]

thx

[unborn mist]

np irene

[sacred eagle]

nooo

its not irene

[unborn mist]

oh

my bad

[sacred eagle]

its seulgi

[unborn mist]

seulgi

yeah thats the one

[sacred eagle]

the best

[unborn mist]

mb

aha ^_^

[dense zealot]

Hi

K

[viscid thistle]

Hello 😄

[calm thicket]

So yesterday our teacher wanted us to add the values of some sinusoidal function from f(0)..f(100) using our calculator

and got no credit because I used summation

so that's my little mini-rant

manually type it out apparently

(slightly exaggerated, only went to 20, not 100)

Me or her?

[dense zealot]

Her

[delicate fractal]

Me

[past jay]

Well.... did the argument within the sine have a pi?

Because if that's the case she may have wanted you to look at the repeating values and make a conclusion based on that

i.e. if it was similar to

$$f(x) = 5\sin(\pi x) + 7$$

[past jay]

Then you could calculate f(0) + f(1) + f(2) +... + f(20) by noticing that ... well this was a poorly chosen example but every value in this case will be 7

so you'd have 7*21 = 147

Again, poorly chosen example but you get the idea.

[viscid thistle]

so, im struggling with matrices and identities in trig. anyone willing to help out for a few?

[tall granite]

just post your question

[viscid thistle]

i've got 1 and 2 done

there

[viscid thistle]

this was the resulting matrix from number 2:
[cos(beta) * cos(alpha) + -sin(beta) * sin(alpha)]
[sin(beta) * cos(alpha) + cos(beta) * sin(alpha)]

[tall granite]

ok so what are you having trouble with in 3

@viscid thistle '

[viscid thistle]

im not sure how to use the matrix to modify the equations

like, im so lost with all of this

[tall granite]

you don't need to use the matrix anymore

you're done with it

the point is that you've derived new identities

for sin(x+y) and cos(x+y)

[viscid thistle]

ahaaaaa

ok

[tall granite]

and you can use them

[viscid thistle]

hmmm

give me a moment to process it all

[tall granite]

"this was the resulting matrix from number 2:
[cos(beta) * cos(alpha) + -sin(beta) * sin(alpha)]
[sin(beta) * cos(alpha) + cos(beta) * sin(alpha)]"

this is a vector

matrix would have 4 entries

[viscid thistle]

so for problem 3 a i would basically be choosing a negative rotation of theta from the unit circle to make the sin(pi/2 - theta) into sin(pi/2 + theta) which I would then turn into cos(theta)

right?

or am I completely off track

the vector did have 4 entries

[viscid thistle]

It's linear programming

[subtle narwhal]

Hey

Does anybody know what are these kind of problems called?

[delicate fractal]

precalculus or something

manipulating logarithms

[subtle narwhal]

@delicate fractal i tried looking that up but nothing related came up

[delicate fractal]

google "logarithm problems"

[rocky bison]

@subtle narwhal Logarithmic Simplification

[viscid thistle]

Hey Iam relearning logarithmic inequations, can anyone point me in the right direction? here is a screenshot

the result is suposed to be ]1,2/3] U ]2,3]

[tall granite]

is this really the right answer?

seems suspicious as 1> 2/3

[thick raptor]

what's wrong with 1 > 2/3?

oic

ugh

why can't you just attach images like normal people 👀

[tall granite]

just rename .jpg to [anything].jpg ;D

[viscid thistle]

sry Iam new here 😦

[tall granite]

:D

[thick raptor]

:P

[tall granite]

i got that the answer is ]3/2, 3[

[viscid thistle]

😮

[tall granite]

you basically do the same thing as with polynomials

as ln is an increasing function

do the difference of squares

[thick raptor]

There is an assumption made

[tall granite]

ah yes

[thick raptor]

when you multiplied both sides by ln(x-1), you assumed it was positive

If its negative, the inequality flips into ≥

[viscid thistle]

oh

wow

u guys are great

thx 😄

[thick raptor]

np

[tall granite]

i guess 2 case investigation would work

[thick raptor]

Also

=tex x^2\le y^2\\implies|x|\le|y|

[thick raptor]

So you really want to check |ln(x-1)| ≤ ln(2)

which comes out to
-ln(2) ≤ ln(x-1) ≤ ln(2)

with the restriction that ln(x-1) is positive

then do the same with ln(x-1) being negative

[tall granite]

imo just taking the difference of squares works better here

like with polynomials

ln((x-1)/2) ln(2x - 2) ≤ 0

you check where each factor becomes 0

which is not hard

[thick raptor]

I personally don't like factoring approaches for inequalities but whatever suits you

[tall granite]

your method seems more complicated

[viscid thistle]

well its cool to knw both ways 😄

[thick raptor]

¯_(ツ)_/¯

[tall granite]

but idk

whatever works i guess

[patent beacon]

You can avoid some case work by setting one side of line 2 equal to zero via subtraction.

Bridge the denominator, and find when positive.

[dense zealot]

Someone halp!

Y is da unit vector of vector w = w/||w||

[unborn mist]

divide by the norm

so you get unit length

say you had the vector (1,1)^t

then the norm is sqrt(2)

[dense zealot]

Oh

[unborn mist]

dividing by that the unit vector is (1/sqrt(2),1(sqrt(2))^t

[dense zealot]

k thx

[unborn mist]

np

[viscid thistle]

help

[dull hinge]

it's hard to tell from the sideways but it looks fairly linear

[viscid thistle]

there

[dull hinge]

find the y intercept and the gradient

[viscid thistle]

what does gradient mean

[dull hinge]

the slope

[viscid thistle]

ah

[dull hinge]

change in x/change in y

[viscid thistle]

i got y = 2x + 3

[dull hinge]

what do you get when x=0?

[viscid thistle]

3

[dull hinge]

what do you get when y=0?

[viscid thistle]

-1.5

[dull hinge]

seems right then

[viscid thistle]

but there are 2 points of discontinuity

[dull hinge]

oh is that what those are?

[viscid thistle]

yea

[dull hinge]

then you can just find the x values and the y values and say this is the equation except for these points i guess

[viscid thistle]

iight

[subtle narwhal]

hey

can someone please help me with domains of functions?

[loud swan]

What is the problem?

[subtle narwhal]

Sry for bad quality lol

I dont know when it is negative infinity

or positive

[loud swan]

it's positive infinity as x tends to infinity

since the x+2 term increases at a faster tate than the sqrt(x-10) term

your domain is correct

[subtle narwhal]

can u give an example of a similar problem where the solution would be - infinity please?

[loud swan]

Simple

$$ \frac{-x-2}{\sqrt{x-10}} $$

[subtle narwhal]

i see, thanks alot 😃

[fringe stream]

your domain is actually not right.

just $$x > 10$$ instead of $$x \geq 10$$.

[quartz garnet]

Yea bc denominators are not supposed to equal 0

[loud swan]

Oh shit, you're right @fringe stream

Can't believe I didn't notice that

[viscid thistle]

$$z=\frac{15cos(\sqrt{(x-10)^{2}+(y-10)^{2}})}{\frac{1}{10}((x-10)^{2}+(y-10)^{2})+6}+\frac{15cos(\sqrt{(x+10)^{2}+(y-10)^{2}})}{\frac{1}{10}((x+10)^{2}+(y-10)^{2})+6}+\frac{15cos(\sqrt{x^{2}+(y-10)^{2}})}{\frac{1}{10}(x^{2}+(y-10)^{2})+6}$$

[dense zealot]

Someone!

Help me!

I don't get this!

For the use substitution part!

[dense zealot]

I already got it it's k

[viscid thistle]

@dense zealot Is that Art of Problem Solving?

[dense zealot]

Ya

Matrices r so confusing ;-;

Y is AB ≠ BA ugh

:/

[dull hinge]

Tbh in most things AB=/=BA tho

At least in 3d

[deft granite]

@dense zealot have you seen how matrices connect with linear maps?

[true vigil]

things don't have to commute @k, flipping and then rotating will be different from rotating and then flipping

[viscid thistle]

https://www.youtube.com/watch?v=kYB8IZa5AuE&list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab&index=4&t=0s

https://www.youtube.com/watch?v=XkY2DOUCWMU&list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab&index=5

[sacred eagle]

hey

[loud swan]

Hey

[sacred eagle]

im gonna need some help lol

always get confused when something should be positive or negative on the sin or tangent stuff

[hexed ermine]

Okay

Do you have examples?

[dense zealot]

Is this actually precalc.. ;-;

[hexed ermine]

Yes

[sacred eagle]

yeah

[hexed ermine]

Trig is classified as precal

[sacred eagle]

its in my precalc class so i think

[dense zealot]

No, linear algebra stuff

[hexed ermine]

???

Anyways, if you give me an example, I can walk you through why it would be positive or negative

[sacred eagle]

ok

im getting started

[hexed ermine]

Alright

[sacred eagle]

https://gyazo.com/f1f0cae27d171aa9189395247b24f6e4

i use the CAST but idk where cot is negative

[loud swan]

Cot is negative where tan is negative

[sacred eagle]

i would dsay 4th quadrant

https://www.onlinemathlearning.com/image-files/xtrigonometric-ratios5.jpg.pagespeed.ic.pmDOsdR1Ej.jpg

i use that

[hexed ermine]

So cot is negative

and cosine is positive

So for cot, its negative in 2 and 3

4*

2 and 4

[sacred eagle]

and cosine 4 so 4

[hexed ermine]

And cosine is positive in 4 and 1

yes

[sacred eagle]

what was the formula for these ?

https://gyazo.com/4904fa14d351df01cb98700f42462a43

25pi /180

then that times 45 ?

i get D like that just not sure if its the right formula

[hexed ermine]

yes

s=r(theta)

where theta is in radians\

so 25 in radians is just 25pi/180

So you get as your final 45(25pi/180)

Which you are correct

[viscid thistle]

decimal places

future engineer

[sacred eagle]

ah

what about this

https://gyazo.com/9bac3d62c6d3d08ecb87e2945954704c

i get 9 sqrt(97) /97

[hexed ermine]

No

[sacred eagle]

sqrt 97

[hexed ermine]

tangent is defined as sin/cos

or when dealing with the unit circle

y/x

so we do 9/4

[sacred eagle]

ooh just y/x

[hexed ermine]

yes

[sacred eagle]

thx man

[hexed ermine]

you are doing sine of theta

sin(theta) = 9sqrt(97)/97

[sacred eagle]

yeah i was so confused

https://gyazo.com/b279abc5b8bb22b9a592a44b8b7e28f9

its to the left since its + ?

[true bone]

factor -2

would make it neg

[sacred eagle]

i thought so

so its to the right

[true bone]

yep since it is neg

[sacred eagle]

but i thought it just depended on the begining being +

not the end result

loll

[true bone]

-2(x-pi/4_

[sacred eagle]

the professor just said if its + in the equation its going to the left

[true bone]

wait

you might be right

[sacred eagle]

idk

[sacred eagle]

whats that

[hexed ermine]

my bad lol

wrong channel

Okay for Ayzen, A is correct

If it's (x+something)

[dense zealot]

@hexed ermine ur evil

[hexed ermine]

it's shifted to the left

LOL

I like to use an inequality to show if it's being shifted

[sacred eagle]

ok

is this positive or negative

https://gyazo.com/a6e35e465e0c50e6fe205ad640464952

only tangent is negative right ?

[hexed ermine]

405-360

[sacred eagle]

45

yeah

[hexed ermine]

Okay so it's just 45 degrees

So you are correct

[sacred eagle]

https://cdn.discordapp.com/attachments/373558752655310848/420986181476679680/ea119f500fe64a0a9ba2e33f3ad3b833.png

i make this

[hexed ermine]

That works

[sacred eagle]

if u dont use that how do u do it

[sacred eagle]

https://gyazo.com/6fda6e1f98da81c5cdbdb0a419b73912

i got 9/17 for sec

[hexed ermine]

Okay so tangent is positive

and sine is negative

so it's in quadrant iii

[sacred eagle]

but i got x = 17 y = -8 and r=9

[hexed ermine]

That means that 8 is negative

no no

quadrant iii, cosine is also negative

So hypotenuse is 9 and leg is 8

so we can say the other is 81-64

[sacred eagle]

yes

17

[hexed ermine]

sqrt(17)

[sacred eagle]

oh sqrt ?

[hexed ermine]

Remember it must be squart root

Yeah the pythagorean identity

a^2+b^2=c^2

a^2+-8^2=9^2

a^2=81-64

a=sqrt(17)

So we have the horizontal component now, so sec is inverse of cosine

cosine would be -sqrt(17)/9

[sacred eagle]

ok so then https://gyazo.com/da34c6535d61cd6678ea2fc4ce7dfb92

[hexed ermine]

That means sec is -9/sqrt(17)

yes

[sacred eagle]

i onkly got 17 for x

didnt get squart

[hexed ermine]

Using pythagorean's identity?

$$ a^2+b^2=c^2 \Rightarrow a^2+(-8^2)=9^2 \newline \Rightarrow a^2+64=81 \Rightarrow a^2=17 \newline \Rightarrow a=\sqrt{17}$$

[sacred eagle]

yeah

i just got rid of the sqrt and the ^2

but is this negative cuz its in the 2nd quadrant

https://gyazo.com/58adfe7bdabdbb3e9e7f9767c03e4517

also wondering if this is negative as well always get mixed up on it

https://gyazo.com/ab0a389772295a4f3343d8fe2d322a44

[clever inlet]

Thats fine