precalculus

[viscid thistle]

cuz i aint that lazy

[hoary yoke]

now Ann go to sleep is my recommendation

and bye

[viscid thistle]

im willing to work hours on hours and still wont learn

cuz abstract :3

[hoary yoke]

im 9th grader

and im lazier than koala

[marsh dew]

im not american...

whats 9th grade?

[hoary yoke]

grading system

[viscid thistle]

im not american either

but we say that

[hoary yoke]

same

[marsh dew]

but like, age?

[hoary yoke]

😐

14

[marsh dew]

not to be creepy or anything

[viscid thistle]

how are u learning that at 14

[hoary yoke]

khan academy and google

[viscid thistle]

in school?

[hoary yoke]

no

[marsh dew]

my teacher keeps telling me that ill be a big mathematician in the future

but like

cmon

[viscid thistle]

lol

u getting physics degree?

[marsh dew]

i keep getting to know 13-14 year olds that are better than me

[viscid thistle]

doing a physics major?

[hoary yoke]

i dont like physics

that much

tbh i just like math and russian as a subject

i dont rly like other

subjects

[marsh dew]

thats a weird pair

[viscid thistle]

im gat

[marsh dew]

but i agree with maths

[viscid thistle]

gay

[hoary yoke]

ok you be you

now bye

[viscid thistle]

no

dont tell me to be me.

[hoary yoke]

im getting distracted

2.2

[marsh dew]

sleep well

btw i know a 14 year old that build rockets

[viscid thistle]

@marsh dew what kinda stuff u learning

[marsh dew]

im just upper sixth

thats like

im 17

[willow bear]

that's like staring at that one person on the very right of the bell curve

[marsh dew]

does that answer your question?

kek

[viscid thistle]

so

ur learning a2 + b2 = c2

^ ^ ^

[willow bear]

so i don't think it's worthwhile to compare yourself to the best of the best?

[marsh dew]

of course not

im far from it

[viscid thistle]

i wish I was as smart AS anne

[marsh dew]

and im still the best in my class

god i hate my maths class so much

[viscid thistle]

i hate my math teacher,

she teaches us everything in her own extremely hard way

[marsh dew]

thats me

[willow bear]

math can definitely be taught poorly 😦

[viscid thistle]

take in she graphe 1 sin function using 6 tables of values when u dont even need 1 table

[willow bear]

i'm sorry you're having to go through that

6 tables? wew

[viscid thistle]

yes

[marsh dew]

...

youch

[willow bear]

i could sketch you a sinusoid without any values lmao

[viscid thistle]

xd

yeah shes kinda scary

[marsh dew]

so im guessing that the founders are genuine mathematicians?

doctorates and all that

[severe verge]

knowing younger people better than you is nice

[willow bear]

oh no far from that

lmao

i'm a 1st year uni student

😂

[marsh dew]

you had me fooled okay

[digital rock]

I have a PhD in easily detectable lying

[marsh dew]

hehe

i get it

...

[viscid thistle]

waw

i dont even wanna go into math i just need this credit 😦

[digital rock]

tbh I could probably get a PhD in shit jokes

[severe verge]

if it existed?

[digital rock]

yeah

[marsh dew]

@willow bear , how hard does math get? are you doing pure maths or applied?

[severe verge]

math gets arbitrarily difficult

the limit does not exist

[marsh dew]

i understand that

but i meant at uni

[severe verge]

hmm well i had to study for the first time of my life 😮

which ended up tragicly

[marsh dew]

oof

[severe verge]

u can't just know how to study

without ever having done it before

well i made it to second year :P

not goin so well tho haha

[patent beacon]

Haha yeah. I'm sure many students here including myself aced high school, and then did terrible at college

[severe verge]

nonono

i did terrible in highschool

my grades improved when i went to University

[patent beacon]

How get into uni? Don't they have a high grade standard?

[severe verge]

because of the heavy weighting towards exams

[digital rock]

I'm hoping my study habits improve a fuckton when I go to uni (they won't)

[willow bear]

@marsh dew technically the course is called applied but so far it's not really differentiated among the first year math courses

so i'm just basically floating in a soup of foundational higher math

[severe verge]

oh my dad was a huge pain in the butt to the administrators

[willow bear]

which i'm loving so far

[marsh dew]

yeah, forgot that thats how it works

[severe verge]

and did a bunch of roundabout stuff until they let me in

which i am very grateful for 🙏

[patent beacon]

Happy for you woog, you're very intelligent and you'll do well

[severe verge]

but yea they rejected me cause of my grades like 4 times

[marsh dew]

kudos to you for getting in

[severe verge]

i ended up missing a week of class cause of how late i was in

but ohhh

University is so great

XD

best place ever

to me

[willow bear]

can confirm, uni $$\gg$$ school

[marsh dew]

im scared because to get into uni i need atleast two B's and an A

and the only thing i might fail at is chemistry

tbh i havent been doing so hot in chem

my phys and maths are on point though, just the chemistry

[digital rock]

nex do you do further maths as well?

[marsh dew]

no, they didnt let me

[digital rock]

rip

[marsh dew]

they have these IQ tests each year a few years back

i dont remember much from them but i suspect that ive been doing very shit in them for some reason

[digital rock]

hmm

it's pretty dumb to base it on that if they did

[marsh dew]

even though im pretty good at recognising patterns

[digital rock]

my school let anyone who wanted to and got at least a 6 (B equiv) at GCSE take it

[marsh dew]

i was never given the choice outright

i guess i couldve asked if i could get into it and they would put me in it

after all i always got the highest grade in maths

but i didnt know you could do that back then

[digital rock]

o rip

[viscid thistle]

uni seems so fun

[marsh dew]

so i maybe wondered if you knew if i had a chance of getting in or not

[digital rock]

well idk your school

[tawny osprey]

is there a faster way to do binomal theorem ?

[digital rock]

but if you got A* at GCSE I reckon you would've stood a decent chance at further maths

[dense zealot]

Acelogic no .-.

[tawny osprey]

😦

[dense zealot]

I mean

[tawny osprey]

I know I could use my calc

[dense zealot]

Learn Pascal's triangle

[tawny osprey]

I know it

[marsh dew]

im gonna go make the binomial rule in paint, ill be back

as in ms paint

[dense zealot]

Then no

[willow bear]

=tex (a+b)^n = \sum_{k=0}^n \binom{n}{k} a^k b^{n-k}

[willow bear]

there, TeX'd it for ya

@marsh dew no need to waste time fighting with paint 😛

[marsh dew]

no, im just doing it for fun

[willow bear]

ah fair

[still yew]

Is there a proof for why $$(a+b)^n = \sum_{k=0}^n \binom{n}{k} a^k b^{n-k}$$

[digital rock]

yeah

I'll phrase it unrigorously because I can't be bothered to do it rigorously

if you look at n brackets, (a+b)(a+b)(a+b)...(a+b)

any term you get is created by choosing exactly one term from each bracket

if you want to create an a^k b^(n-k) term, you must choose k a's and (n-k) b's

[still yew]

Ok

[digital rock]

there are n brackets total, so the number of ways of doing this is n choose k

[still yew]

I see

[digital rock]

does that make sense?

[still yew]

Yes

[digital rock]

good

because I have a tendency to not be very clear, especially at this time of night :^)

[still yew]

Oh

[tawny osprey]

=tex (3a-4b)^5

[tawny osprey]

For the 2nd term I keep getting 405a^4

[digital rock]

should be -1620a^4 b I think

[tawny osprey]

Yeah

I don't know why I keep getting 405a^4

I do ((5C1 )*(3^4)) and I get 405

lol

[novel monolith]

5*(3a)^4*(-4b)^1

[tawny osprey]

I got (405a^4)-20b

[marsh dew]

i made it

i had fun making it

[fair valley]

just make sure you have a>b so the series converges if n is not in Z⁺

[marsh dew]

true, thank you for pointing that out

its weird finally knowing the feeling of getting out-leagued by so many people at the same time

[severe verge]

isn't it awesome?!

[marsh dew]

yes

im just getting to the final answer by my overly convoluted method

[severe verge]

😄

[marsh dew]

phew

got it

were you just supposed to find the coefficient of a^4?

because i just expanded the brackets entirely...

using my equation

[marsh dew]

I improved my statement

imma go sleep

goodnight you beautiful bastards

[dense zealot]

Oh that font

:0

[marsh dew]

i think i got a new goal in life

to get a doctorate in maths just to become a mod on this server

[willow bear]

that ain't a requirement

[marsh dew]

itd be useful though

i feel like i could help sufficiently

at that level

[willow bear]

i don't think any of the mod team are doctorates

[marsh dew]

im guessing thats because doctors in maths are actually working in offices as mathematicians

busy people

[willow bear]

probably

[hoary yoke]

@willow bear ok so like sin = 1/csc

why is that identity

1/sin is csc

[willow bear]

=tex \csc(x) := \frac{1}{\sin(x)}

[hoary yoke]

but i dont understand

is it just defined it

[willow bear]

yes that's how csc is defined

[hoary yoke]

pretty much i understand everything starting from like the right triangle

😐

which is kind of confusing to me sometimes

and i use hexagon

[willow bear]

=tex \tan(x) := \frac{\sin(x)}{\cos(x)} \ \cot(x) := \frac{\cos(x)}{\sin(x)} \ \sec(x) := \frac{1}{\cos(x)} \ \csc(x) := \frac{1}{\sin(x)}

[hoary yoke]

yes i know that graph

[willow bear]

graph?

[hoary yoke]

😐

[willow bear]

those are just the definitions of the other four functions

[hoary yoke]

uhm

i mean

uhm

the defs ye

how do i prove half angle identities

or Validate

Verrify

[willow bear]

i mean you'll probably want to reduce them to double angle identities

which ones do you want to prove

[hoary yoke]

wait

pretty much every, one of those

double angles

and half angles

wait

no

not double angles

they are just replacements

but half angles

[willow bear]

=tex \sin^2(x/2) = \frac{1 - \cos(x)}{2} \ \cos^2(x/2) = \frac{1 + \cos(x)}{2} \ \tan(x/2) = \frac{\sin(x)}{1 + \cos(x)}

[willow bear]

so these, then?

[hoary yoke]

pretty mch yes

[willow bear]

=tex \cos(2t) = \cos^2(t) - \sin^2(t) = 2\cos^2(t) - 1 = 1 - 2\sin^2(t)

[hoary yoke]

and the tg squared

[hoary yoke]

yes

by using pythagorean

[willow bear]

yes, so

now let t = x/2

=tex \cos(x) = 2\cos^2(x/2) - 1 \ \cos(x) = 1 - 2\sin^2(x/2)

[willow bear]

now you can isolate cos^2(x/2) and sin^2(x/2)

[hoary yoke]

wat

o.o

[willow bear]

=tex \cos(2t) = 2\cos^2(t) - 1 \ \cos(2t) = 1 - 2\sin^2(t)

do you understand these?

[hoary yoke]

yes

[willow bear]

okay

now replace t with x/2

[hoary yoke]

that would be cosx

and then it would continue

[willow bear]

=tex \cos(2 \cdot x/2) = 2\cos^2(x/2) - 1 \ \cos(2 \cdot x/2) = 1 - 2\sin^2(x/2)

[willow bear]

=tex \cos(x) = 2\cos^2(x/2) - 1 \ \cos(x) = 1 - 2\sin^2(x/2)

[willow bear]

is this clear??

[hoary yoke]

yes

[willow bear]

=tex \cos(x) = 2\cos^2(x/2) - 1 \ 2\cos^2(x/2) = \cos(x) + 1 \ \cos^2(x/2) = \frac{\cos(x)+1}{2}

[hoary yoke]

o.o

wut

[willow bear]

seriously?????

i added 1 to both sides, then divided both sides by 2!

=tex a = 2b - 1 \ a + 1 = 2b \ b = \frac{a+1}{2}

[hoary yoke]

oh you isolated the 2 cos

oh

o.o

ok

understood

[willow bear]

do the exact same thing with the other one

[hoary yoke]

yes

tg squared i understand

ok thank you

[viscid thistle]

Trig is so hard

[patent beacon]

Yeah, it's an extremely difficult subject. Until you learn it. Then it's extremely easy.

All it takes is practice!

[dense zealot]

I have asin(bt+c) + d

And the equation is 20sin(π/9t + π/6) + 100

It's supposed to represent the prices of shares

And if I start with $1000 how can I maximize it

To get the most money

[willow bear]

uhh

i mean no matter how much you start with, you'll want to buy when the price is lowest and sell when it is highest?

[dense zealot]

And I don't understand the book :/

Ya

So buy when it's $80 and sell at $120?

And it starts at $110

[willow bear]

yes

[patent beacon]

It doesn't start at 100 if t = 0 represents the start

[dense zealot]

Ya it starts at 110

[patent beacon]

Ya ya I guess that would be it lol

[dense zealot]

Cuz it says it takes 3 days to get to 120 in the beginning

[patent beacon]

The price is at 120 when the argument inside sin is π/2

[willow bear]

pi/2

[patent beacon]

Derp yeah

[severe verge]

if only daytrading was that easy...

[dense zealot]

Lol

[calm whale]

Yo

Ik the definition of first principle

But how do u actually use it to find the derivative of smth?

Eg (x+h)^3 from first principle

[willow bear]

as in, you're trying to find the derivative of x^3 by the limit definition?

(x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3

[calm whale]

Yes i got that as it was required to do the expansion beforehand

And then where do i take it

Do i do it for each term separately?

[willow bear]

huh?

[calm whale]

I got no idea what im doing

[willow bear]

=tex \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

[willow bear]

this is the limit you want to find

for f(x) = x^3, in your case

[calm whale]

Uhuh

[willow bear]

=tex \lim_{h \to 0} \frac{x^3 + 3x^2h + 3xh^2 + h^3 - x^3}{h}

[willow bear]

this is what that limit is going to look like for f(x) = x^3

that clear?

[calm whale]

Is that for x^3

[willow bear]

yes it is

[calm whale]

Or (x+h)^3

[willow bear]

huh?

[calm whale]

Oooh nvm

Its clear

Its asking for the deriv of x^3

[willow bear]

=tex \lim_{h \to 0} \frac{x^3 + 3x^2h + 3xh^2 + h^3 - x^3}{h} = \ = \lim_{h \to 0} \frac{3x^2h + 3xh^2 + h^3}{h} = \ = \lim_{h \to 0} 3x^2 + 3xh + h^2 = 3x^2

[calm whale]

Thanks I think I got the hang of it

[earnest nymph]

Nth term = Sn - S(n-1)

[severe verge]

A REALLY COOL CONCEPT

method of proof

[dense terrace]

yeah dominoooo

1. prove the first term to be true
2. assume that kth term is true
3. prove (k+1)th term to be true

so its like if u can topple the first domino (first term to be true) u can topple everything else

um

i guess?

its fun lol

[earnest nymph]

We can u some easy example

[quartz garnet]

I see a carrot

[earnest nymph]

For base case, it's usually n =1

But the rest is ok

Just need aome time to read through what u have written

[severe verge]

no thanks

[dense terrace]

thats an easy paypal money :/

but nah im not gonna ask for money lol

[quartz garnet]

Is this an exam question

Okay no

@severe verge kick him

[severe verge]

yea

[dense terrace]

oh

okay almost

[severe verge]

we dont allow cheatingf

sorry

[dense terrace]

are u in an exam

if ur not its okay

ohhhh

[severe verge]

im not sure if im supposed to kick or ban for cheating on tests

woops

[dense terrace]

=tex y^2 = xz \
\frac{y}{x} = \frac{z}{y}

done lol

[quartz garnet]

How do you know, @viscid thistle?

[quartz garnet]

Do you two have same teacher?

[dense terrace]

:0

ohhh

[quartz garnet]

O...kay then

Also tell them that asking for the solution of a question with money in return seems really bad and cheating

It's.... ridiculous, I'll say

Sure!

[viscid thistle]

:/

[severe verge]

ok i banned both of them

[quartz garnet]

Eh?

What happened?

[dense terrace]

lol

[viscid thistle]

He sent an inappropriate picture.

[quartz garnet]

Did the carrot post naughty pic-

[dense terrace]

pretty sure that paypal thing was a "scam"

[quartz garnet]

alright

[severe verge]

yea

oh well, theyre gone now

i was playing multi x.x

[dense terrace]

what multi

[severe verge]

taiko multi

my hands are cold tho

so wasnt goin well anyway lol

[earnest nymph]

The carrot ask another question in another group I'm in

The things is he ask arithmetic progression in the trigonometric section

[hoary yoke]

well wtf

what doi do now

Q\Q

Tan(255degrees)

[willow bear]

tan(x) = tan(x-180°)

[hoary yoke]

look

i divided into two pieces

30 and 225

used tan identity

[willow bear]

ok, that's also doable

[hoary yoke]

and it was incorrect

[willow bear]

show your work?

[hoary yoke]

i got sqrt(3) plus 3

all over 3 minus sqrt(3)

[willow bear]

=tex \frac{\sqrt{3} + 3}{3 - \sqrt{3}}

[hoary yoke]

yes which is wat i got

[willow bear]

=tex \frac{\sqrt{3} + 3}{3 - \sqrt{3}} = \frac{(\sqrt{3} + 3)^2}{9 - 3} = \frac{3 + 9 + 6\sqrt{3}}{6} = \frac{12 + 6\sqrt{3}}{6} = 2 + \sqrt{3}

[willow bear]

== tand(75) - sqrt(3)

[granite stirrup]

2

[willow bear]

so yes your answer is correct

[hoary yoke]

answer was different on khan

[willow bear]

unless there was some other requirement that you didn't mention

such as a rationalized denominator

[hoary yoke]

no

wait

ill give u answer

=tex \frac{\sqrt{3} + 1}{\sqrt{3} - 1}

[hoary yoke]

btw i converted from radians to degrees

[willow bear]

=tex \frac{\sqrt{3} + 3}{3 - \sqrt{3}} = \frac{\frac{1}{\sqrt{3}}(3+ \sqrt{3})}{\frac{1}{\sqrt{3}}(3 - \sqrt{3})}

[willow bear]

your answer is correct!!!

[hoary yoke]

o.o i dont know rationalization?

well damn

[willow bear]

i am pretty sure khan teacher rationalization??

and that wasn't even what i did

lmao

[hoary yoke]

he used 2pi over 3

and 3piover 4

[willow bear]

what do you mean

he decomposed your angle into a different sum?

[hoary yoke]

he divided 17pi/12

into 2pi/3

and 3pi/4

[willow bear]

so what???

[hoary yoke]

and he got direct answwerrr

[willow bear]

there's more than one way to solve your problem
so what???

[hoary yoke]

im dumb thats it

i dont know how to simplify

[willow bear]

well then go practice that

then come back to trig

you're learning this all in your own time, right?

[hoary yoke]

all in your own time?

[willow bear]

(you're learning this all) (in your own time), right?

as in, you're not in a rush

[hoary yoke]

no

i dont rly need it this time tbh

im just learning it cause its fun

[willow bear]

yeah well

i see you want to get better at it

but you're missing something from a lower level

[hoary yoke]

ye well

:/

[still yew]

What is $$\sin 4x$$

[viscid thistle]

Lol

[finite falcon]

$$sin(4x)$$ is the same as $$sin(2 \times 2x) $$

[viscid thistle]

Use chain rule

[viscid thistle]

Wait ur not asking its derivative sorry

[finite falcon]

This channel is precalculus lmao

[viscid thistle]

Yeah sorry

[still yew]

I think it's part of a property

[dense zealot]

It is

Use double angle

Which one u want?

it's the same as 2(sin2x)(cos2x)

@still yew

If u wanna do another double angle

2(2sinxcosx)(2sinxcosx= 8(sin^2x)(cos^2x)

And if u wanna simplify further

=tex 2 - 2\cos^2(2x)

[still yew]

Thx

[dense zealot]

Yw

[hoary yoke]

simplify :

=tex \frac {4\log_{100}(2) - (( \log_{100}(5) - \log_{100}(10)+ ( 4\log_{100}(2) ) - (2\log_{100}(2) )}{\log(100)}

[dense terrace]

\log

[hoary yoke]

wait

to state of:

=tex \frac {\log(a)}{b}

[hoary yoke]

:d

so

hopefully i wrote everything correctly

[dense terrace]

oh

=tex \log_a(b) + \log_a(c) = \log_a(bc)

[hoary yoke]

im checking and i think i wrote it incorrectly dammit

[hoary yoke]

yep

:d

[dense terrace]

yeah use the rule and simplify

=tex \log_a(b) - \log_a(c) = \log_a(b/c)

[hoary yoke]

:3

i know

[hoary yoke]

cos(acos(0.6)

what to do

[willow bear]

cos(arccos(0.6))?

that's just 0.6 lmao

[hoary yoke]

yes

so they do cancel out

thanks

[willow bear]

on the right they do

[hoary yoke]

sin(acos(0.6))

[willow bear]

it's on the left that some complications might occur

now for that

you'll want to draw a triangle

[hoary yoke]

so nothing can be canceled out

ok

[dense zealot]

There is

That's just a basic identity

e.e

[idle olive]

whata

what

[earnest nymph]

It's easier to use the triangle, at least u can see it

[ruby crater]

How in the world would I solve this? Find f(x + Δx) - f(x)/Δx for f(x) = 8x^2 + 1.

[static maple]

Solve for what? There's no equation.

[earnest sable]

you mean the limit?

[willow bear]

no

[ruby crater]

That's exactly what I'm thinking

I suppose it could be limit

[willow bear]

are you not able to plug in x+Δx though?

=tex \frac{8(x+\Delta x)^2 + 1 - (8x^2 + 1)}{\Delta x}

[static maple]

Looks like you're trying to calculate the differential quotient

[willow bear]

really it's literally as simple as playing around with and/or simplifying that

that's what you were asked to do

@ruby crater

[static maple]

Yep

[ruby crater]

Maybe I'm just dumb lmao thanks, I was over complicating it

[willow bear]

you might've

[vocal wolf]

It says to find an equation that this represents

For 55

I have an idea that it's an abs value of x

So far I got |x-3| but I dunno how it got the 2 different slopes

[willow bear]

as in, you're not allowed to define your function piecewise?

[vocal wolf]

Oh I didn't think about that

thanks

[willow bear]

what have you tried so far?

those are the local max and min respectively, yes

assuming the function doesn't extend past what's shown on the graph

because if it does and it's equal to sin(x) for all x > 0, then (π/2 + 2πk, 1) and (3π/2 + 2πk, -1) are all local maxima and minima respectively for all integer k ≥ 0

[narrow hawk]

Wait stationary points are precalculus ;(

I felt like I was doing real maths

[viscid thistle]

It's all real math 😄

Atleast that's what I tell myself 😱

[hasty wadi]

alright

im really confused on this problem

i know i have to use trig and inverse trig functions but it's been a pain for this one

[severe verge]

L = hypotenuse
h = opposite
100 ft = adjacent
sinθ = opposite / hypotenuse = h/L

then rearrange

[hasty wadi]

alright lemme write this lol

[severe verge]

oh, I guess it's a bit more complicated

[hasty wadi]

yeah lmao, i got all the questions from my assignment finished except this one

[severe verge]

dood is holding the balloon 3ft off the ground

and the height from the ground is 20+h

so opposite would actually be 20+h-3
= 17+h

[hasty wadi]

im just confused on how i should approach this

i know it's 2 triangles

but fml, this question is like "bruh, fuck u" right to my face

[dense terrace]

which question

part a is just

ur soh cah toa right

and helped u with part a in finding the height of the so called triangle

[vital tartan]

which part are you guys working on rn?

[dense terrace]

no idea lel

[hasty wadi]

a b and c, this is the only question that got me confused

[severe verge]

my time is so limited now that i have school 😒

[calm thicket]

tmw you're bored while walking in a parade and wonder how long your cable is

[grim yarrow]

what a weird problem

[hasty wadi]

tmw this is the only think you can make of this problem

[grim yarrow]

on the par of those math problems that buy 900 melons and 580 strawberries

[calm thicket]

one day the world is gonna end because they miscalculated and got 579 strawberries

but until then

we can laugh

[vital tartan]

lol

[calm thicket]

until then...

[hasty wadi]

i tried searching it up on the internet to find a solution on how it's done but my brain exploded and i've been stuck on this shitty problem for an hour

[calm thicket]

mk

so let's step back

[vital tartan]

gonna try this in a sec.

[calm thicket]

what do you know and what are you stuck on

[hasty wadi]

im stuck on how the triangles should placed and what belongs where. i don't know how to approach it

[calm thicket]

Alright

[hasty wadi]

i don't know what goes where basically

[calm thicket]

So let's start with part a. You familiar with 2d pythagorean theorem?

[hasty wadi]

a^2+b^2=c^2

[calm thicket]

Okay

Solve that for c

[hasty wadi]

but what are my adjacent and opposite (a and b) measurements? 🤔

that's what im confused about

idk what the triangle should look like

[calm thicket]

Well, what ideas do you have? Hypotheses?

*Hypothesis I was never good at vocabulary

[hasty wadi]

i thinking the triangle looks like this or something along these lines

[calm thicket]

Yea?

What do you think x is?

[hasty wadi]

20+h

?

[calm thicket]

If the rope was tied to the ground, sure, but it's not, it's three feet off the ground.

[hasty wadi]

17+h

[calm thicket]

Yep

So now you have a and b

=tex l=\sqrt{100^2+(17+h)^2}

[hasty wadi]

but solving a hypotenuse with the adjacent side having a variable?

[calm thicket]

Yea?

It says to write an equation, not a numerical solution

[hasty wadi]

that means i have to foil it

[calm thicket]

Not necessarily unless they're picky about it

But if you do

=pup expand (17+h)^2

[granite stirrup]

Query made by @calm thicket
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=(17%2Bh)^2
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/

[calm thicket]

goshdarnit wrong one 👀

h^2+34h+289

So the answer is

=tex l=\sqrt{h^2+34h+10289}

[calm thicket]

Make sense?

[hasty wadi]

yeah i get that u did pythagorean and solved for c

but the fact that L has a variable is right?

i thought it was gonna be like a number

[calm thicket]

"equation"

[hasty wadi]

ohhhhh

fml

[calm thicket]

What's the pythagorean theorem again?

c=sqrt(a^2+b^2)

[hasty wadi]

a^2+b^2=c^2

[calm thicket]

Ye

That'd be an equation, no number solution

:P

[hasty wadi]

that's technically a quadratic equation in a square root

[calm thicket]

=tex \sqrt{\frac{-b\pm\sqrt{b^2-4ac}}{2a}}?

[hasty wadi]

OH FUCK THAT'S RIGHT

THE ANSWER WAS RIGHT 😃

[calm thicket]

ye

[hasty wadi]

hell yeah

[calm thicket]

Now for part b, which is slightly fancier

Are you familiar with law of sines?

[hasty wadi]

we haven't gone over law of sines so were supposed to do this problem without applying it

[calm thicket]

w0t

Lemme think of a way to do without law of sines o.o

[hasty wadi]

square root of i? 🤔

[calm thicket]

nu

that's for a specific rotation

and one for another discussion

[hasty wadi]

no im reading ur name lmao

[calm thicket]

I know

Okay

Gonna use law of sines anyways and you can tell yer teacher yer confused later

[hasty wadi]

lmao alright

[calm thicket]

So you're not familiar with it, yes?

[hasty wadi]

i've went over it a long ass time ago but a refresher should be alright

i don't remember it

[calm thicket]

=tex \frac{\sin(A)}{a}=\frac{\sin(B)}{b}=\frac{\sin(C)}{c}

[hasty wadi]

🤔

[calm thicket]

caps are angles and lowercase are sides

[hasty wadi]

yeah lol, that's standard triangle notation if im not confused

[calm thicket]

aye

So we know the side lengths of teh triangle are l, 100, and 17+h

[hasty wadi]

yes

that's the right triangle

[calm thicket]

Aye

What's the angle opposite l

[hasty wadi]

90

[calm thicket]

And what's the side opposite theta

[hasty wadi]

100

[calm thicket]

No

[hasty wadi]

opposite theta?

[calm thicket]

Mhm

=tex \theta

[hasty wadi]

wait i need to look at the problem again

opposite of theta isn't 100?

wait, did i place theta in the wrong region 🤔

[calm thicket]

Probably not because theta is next to 100 👀

[hasty wadi]

i placed theta adjacent to 90

not opposite to 90

[calm thicket]

*100

and yea

[hasty wadi]

triangle

side opposite to theta is 100

[calm thicket]

theta is on the wrong side lel

[hasty wadi]

how do i know where theta should be placed?

[hasty wadi]

i thought it was an angle of elevation

omfg

im so fucking stupid lmao

[calm thicket]

Happens all the time

[hasty wadi]

alright

17+h

opposite to theta

[calm thicket]

in #help-3 someone pinned a mistake of mine bcuz it was so bad :c

[hasty wadi]

LOL

[calm thicket]

But anyways

Now we have our equation set up

=tex \frac{\sin(90)}{l}=\frac{\sin(\theta)}{17+h}

[hasty wadi]

to solve for l?

[calm thicket]

For theta.

[hasty wadi]

so i just cross multiply

[calm thicket]

Yep

and divide by l

[hasty wadi]

and simplify to my best ability

[calm thicket]

to get

[hasty wadi]

theta by itself

[calm thicket]

=tex \sin(\theta)=\frac{(17+h)\sin(90)}{l}

[calm thicket]

Now the fun part is gonna be getting that theta alone

[hasty wadi]

it's just using arcsin lol

[calm thicket]

yep

[hasty wadi]

arcsin of that whole thang

[calm thicket]

=tex \theta=\sin^{-1}\left(\frac{(17+h)\sin(90)}{l}\right)

[hasty wadi]

isn't sin 90 a special angle?

it equals to something already?

[calm thicket]

Good eye

Yes it does

[hasty wadi]

so it's basically "arcsin ((17+h)/l)"

[calm thicket]

Yep

You can try that but I don't know if it'll work

because it says in terms of length l

[hasty wadi]

still wrong

[calm thicket]

Yikes

[hasty wadi]

lmao

[calm thicket]

see if this were in terms of h this would be easy

but it's not :c

[hasty wadi]

i copied the "l" they used

in the question

[calm thicket]

ye

[hasty wadi]

and pasted it in the answer

it's still wrong 🤔

[calm thicket]

=tex l=\sqrt{h^2+34h+10289}

remember this thing

[calm thicket]

We need to solve it for h

[hasty wadi]

🤔

how is that gonna give us theta?

[calm thicket]

It says the equation needs to be in terms of length l

If we can solve h in terms of l, we can substitute

[hasty wadi]

alright well how are we gonna solve for "h" lmao?

[vital tartan]

...

[hasty wadi]

this is gonna be a fun time 😃

[calm thicket]

this is hurting my soul

:<

[hasty wadi]

lmao

[calm thicket]

anyways

basically

square it

quadratic formula and boom, done

[hasty wadi]

oh fml

im so dumb lmao

[calm thicket]

but i'm gonna use a shortcut

[hasty wadi]

but the l is gonna be squared also?

[calm thicket]

=pup l=sqrt(h^2+34h+10289) solve for h

[hasty wadi]

it's just gonna square root back in the end

[granite stirrup]

Query made by @calm thicket
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=+l%3Dsqrt(h^2%2B34h%2B10289)+solve+for+h
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/

[hasty wadi]

the "l" is squared so the final answer is gonna have to have a square root

because we sqaured the square root to get the quadratic equation by itself

[calm thicket]

So this means

=tex \theta=\arcsin\left(\frac{\sqrt{l^2-10000}}{l}\right)

[hasty wadi]

where is the 17?

[calm thicket]

Initially it was 17+h

h is that - 17

So it was 17+x-17

[hasty wadi]

alright my question only allows 2 more submissions

[calm thicket]

👀

[hasty wadi]

if this is wrong the next one has to be correct

[calm thicket]

this gonn' b intense

[hasty wadi]

oml

alright

[calm thicket]

Did it work? 👀

[hasty wadi]

before i submit it, im gonna show you what's in the box alright

[calm thicket]

ono

I've never felt so tense over a math problem before

[hasty wadi]

this is gonna be the answer i submit

alright?

[calm thicket]

Looks good to me

[hasty wadi]

lets fucking do this

FUCK

NO

[calm thicket]

oh shit

[hasty wadi]

IT'S FUCKING WRONG

[calm thicket]

aaaaaaa

[hasty wadi]

ONE MORE ATTEMPT

im thinking i should skip on this

and wait for my teacher to do it in class

i'll leave the last attempt for my teacher

but i think we can move on to part c

maybe

if we have enough info

[calm thicket]

hm

[hasty wadi]

this is legit the only problem that's fucking killed me

been over 2 hours tbh

lmao

[calm thicket]

rip

I'd call this a pinch 👀

<@&286206848099549185> part b has us stumped :c

[hasty wadi]

lmao

[calm thicket]

pls tell me you got that reference

[hasty wadi]

pinch?

[calm thicket]

mhm

[hasty wadi]

emp

😩

i don't tbh

lmao

[calm thicket]

rip

anyways now we wait for smert people

[hasty wadi]

u did help a lot tho lmao

thanks 😃

[calm thicket]

I didn't help a lot enough. :c

[hasty wadi]

i joined the discord cuz just for this problem lmao

[vital tartan]

theta = (sin^-1)(l / (sqrt(l^2-10000))) ?

[calm thicket]

???

[hasty wadi]

1/sqrt(-10000)

?

🤔

[vital tartan]

lowercase L

[hasty wadi]

oh "L^2 - 10000"

[vital tartan]

yeah

[hasty wadi]

are you sure that's right?

[calm thicket]

"-10000"

"1

[hasty wadi]

like sure sure

[calm thicket]

I have doubt

[hasty wadi]

this is my last submission :::::::::::::))))))))))))

[vital tartan]

oh god NO

[vital tartan]

don't do it

X

[hasty wadi]

lmao

[calm thicket]

seriously tho

[hasty wadi]

tmw pre-cal is hard af

[vital tartan]

I got it from constructing a right angle triangle where the side across from theta was h + 17

[calm thicket]

ye

[vital tartan]

and l was the hypotenuse

[calm thicket]

And adjascent was 100

[vital tartan]

so sin(theta) = (h + 17) / l

[calm thicket]

But we want theta in terms of l, which is the hard part

if it were in terms ot h, ez.

Correct

[vital tartan]

oh wait, isn't that was you guys just did?

[calm thicket]

Correct again

[vital tartan]

so you calculated h wrong or smth. ?

[calm thicket]

I only assume I calculated l wrong relative to h

[hasty wadi]

what we have so far...

[calm thicket]

In terms of h

[severe verge]

spooky

[hasty wadi]

this question is giving me vietnam flashbacks bruh

[calm thicket]

=tex \theta=\arcsin\left(\frac{h+17}{\sqrt{h^2+34h+10289}}\right)

[vital tartan]

sqrt(100*100 + (h*17)^2) = l

is that where you got that thing from?

[calm thicket]

l=sqrt(h^2+34h+10289), idk what to do with h+17

tried quadratic formula but it failed

[severe verge]

cos 35°=100/L
L=100/cos35°
(h+17)²=L²-100²

[vital tartan]

sqrt(10000 + h^2 + 34h + 289)

[calm thicket]

wait a minute

[vital tartan]

sqrt(h^2 + 34h + 10289)

[calm thicket]

woog is right

[vital tartan]

oh right back at square one ...

[calm thicket]

h+17 = sqrt((h+17)^2)

l=sqrt((h+17)^2)

[vital tartan]

l = h + 17 ??

[calm thicket]

theta = arcsin(l/l)?

[severe verge]

what are u talking about

[calm thicket]

wait nvm

[hasty wadi]

lmao

[calm thicket]

I forgot the square root was +10000

[vital tartan]

lol

[calm thicket]

l = sqrt((h+17)^2+100^2)

[vital tartan]

this seems like this would be such a simple question

yet everyone's getting it wrong

¯_(ツ)_/¯

[severe verge]

what are u trying to solve for

[calm thicket]

h

[severe verge]

ok, i just did

[calm thicket]

=pup l=sqrt((h+17)^2+100^2) solve for h

[vital tartan]

theta in terms of l ?

[granite stirrup]

Query made by @calm thicket
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=l%3Dsqrt((h%2B17)^2%2B100^2)+solve+for+h
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/

[calm thicket]

god why

[hasty wadi]

h is a number

[severe verge]

yes..

[calm thicket]

h has to be real 👀

[quartz garnet]

What is this madness

[vital tartan]

h has to be positive

[calm thicket]

gurl help D:

[quartz garnet]

What help

[vital tartan]

^^

[calm thicket]

part b

[hasty wadi]

round to 2 decimal places for h

[severe verge]

(h+17)²=(100/cos35°)²-100²

sqrt both sides, minus 17

[hasty wadi]

(h+17)^2 will give you a quadratic

there will be a variable

[severe verge]

no dude, don't solve the quadratic

it's already factored

[quartz garnet]

Err

[severe verge]

just square root it

[vital tartan]

h = sqrt((100/cos35)^2 - 100^2) - 17

[severe verge]

that should work 🤔

[calm thicket]

?????

oh for part c

but what about part b 3:

[severe verge]

lol

[quartz garnet]

=tex \theta = \arcsin\left(\frac{\sqrt{l^2-10000}}{l}\right)

[quartz garnet]

...?

[calm thicket]

== sqrt((100/cosd(35))^2+100^2)

[granite stirrup]

157.80654602

[calm thicket]

That was my first idea

But it was wrong

[hasty wadi]

can someone plug the "h" equation into the wolfram bot

[severe verge]

u added, not subtracted.

[vital tartan]

=pup h=sqrt((100/cos35)^2 - 100^2) - 17 solve for h

[granite stirrup]

Query made by @vital tartan
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=h%3Dsqrt((100%2Fcos35)^2+-+100^2)+-+17+solve+for+h
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/

[severe verge]

==sqrt((100/cosd(35))^2-100^2)-17

[granite stirrup]

53.02075382

[hasty wadi]

alright "h" is correct

b is the only one left ::::::::))))))))

[grim yarrow]

oh god webassign

[hasty wadi]

(;

it's garbage

[grim yarrow]

so glad I never had to use webassign after high school lmao

[hasty wadi]

last attempt

[calm thicket]

the pressure is real

[severe verge]

cosθ=adj/hyp = 100/L

[hasty wadi]

i think we've been going at this problem for about an hour

[severe verge]

θ= arccos(100/L)?

[hasty wadi]

cos-1(100/L)?

that's it

?

are you sure sure? this is last attempt lmao

[quartz garnet]

Er

[grim yarrow]

just accept the 0

[quartz garnet]

== 100 * tand(35)

[granite stirrup]

70.02075382

[grim yarrow]

just accept it

[hasty wadi]

LOL

@severe verge you sure? im trusting u

[grim yarrow]

own it man

put a joke answer

own the zero

[severe verge]

too many people trust me...

[vital tartan]

oh god

[severe verge]

far too many....

[vital tartan]

the pressure ...

[severe verge]

[hasty wadi]

FUCK THE ZERO

WERE GETTING THIS ANSWER RIGHT