precalculus

[brave frigate]

so you multiply the cos x on the bottom with the one on the right?

[winter tide]

well you multiply and divide the cos(x) term by cos(x) so that there’s a cos(x) in the denominator

[brave frigate]

you only multiply one side?

[winter tide]

Sorry I don’t know how to typeset equations but I hope this picture makes it clearer

It’s technically a multiplication by one

[viscid thistle]

is there anyway to use a calculator to find a derivitive

and keep the x

[true talon]

Can someone help me I have no clue how to factor trinomials

Its 2x^2+3x-2

Can someone explain it

[brave frigate]

Ty

@winter tide ty

[winter tide]

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

I dont have a set way of doing it, but in general:
I. The coefficients of x should multiply to the lead coefficient (so since 2 = 2*1, I immediately checked (2x+a)(x+b) for some placeholder numbers a and b)
II. The constants should multiply to the constant of the trinomial (so since -2 = -1 * 2, I tested (2x-1)(x+2) and replaced those placeholder constants from part I)

Then it comes down to properly FOILing to make sure that your arrangement gives the correct trinomial.

I dont know if my way is the best way @true talon but it works for me

no prob @brave frigate

[true talon]

@winter tide thanks really helped!

[viscid thistle]

derivative of: 2sec(e^2x)

[rocky bison]

Is that a question @viscid thistle

[viscid thistle]

yes

[rocky bison]

Ok

Are you familiar with chain rule?

[viscid thistle]

yes

[rocky bison]

Do you know how to apply it in this situation?

[viscid thistle]

no because the e is in the inside of the parenthisis im not sure what to do

[rocky bison]

Just apply chain rule

So let's say

,$ f(x)=e^{2x}

[obsidian monolith]

Pseudo:

[rocky bison]

We want the differential of 2sec(f(x))

in terms of f(x)

Which is..?

[viscid thistle]

e^2x

[rocky bison]

,w differential of sec(x)

[obsidian monolith]

Display results online and refine query

Requested by Pseudo#2703

[rocky bison]

It's actually sec(x)tan(x)

[viscid thistle]

😮

[rocky bison]

But in this case

[viscid thistle]

what is that

[rocky bison]

x is our f(x)

[viscid thistle]

is that a identity

[rocky bison]

I guess?

It's just the differential of sec(x)

So we have

,$ f(x)=e^{2x}\\frac{d}{df(x)}\left(2\sec\left(e^{2x}\right)\right)=2\sec\left(e^{2x}\right)\tan\left(e^{2x}\right)\cdot f'(x)

[obsidian monolith]

Pseudo:

[rocky bison]

Now you just need to work our f'(x)

[viscid thistle]

the derivitive of sec is sectan?

[rocky bison]

sec(x)*tan(x)

Yes

[viscid thistle]

is there a sheet of these i should look over?

[rocky bison]

Probably

If you're english

You'll get given them

[viscid thistle]

also why do you multiply by f'x

we did this months ago haha

however it is 15 percent of the test

[rocky bison]

We multiply by f'(x)

Because that's how chain rule works

[viscid thistle]

i thought chain rule is bringing down the exponent

and multiplying by it

[rocky bison]

,$ \frac{d}{dx}\left(f\circ g\right)(x)=\frac{d\left(f\circ g\right)\left(x\right)}{dg(x)}\cdot\frac{d\left(g(x))}{dx}

Tbh realising now that this is a useless way to write it most likely

Urm

t!yt chain rule

[brisk mica]

📽 | ** https://www.youtube.com/watch?v=XIQ-KnsAsbg **

[rocky bison]

That's probably better for you to follow

[viscid thistle]

thank you for your time and help

[rocky bison]

np

[keen root]

This question has me stuck pls help

[grave tartan]

So okay

What are the solutions to f(x)

When does f(x)=0?

@keen root

[keen root]

2

no

4

f(4)=0

[grave tartan]

Uhh yes

So what can we kinda infer?

[keen root]

x-4 in the numerator

just figured out the whole thing

thx

[grave tartan]

:>

[viscid thistle]

How do you do 15?

[grave tartan]

Well you wanna write Z in the form re^itheta?

[viscid thistle]

What's re^itheta?

[grave tartan]

Re^ix

If you prefer

Theta is just like the angle formed

r is the modulus

[viscid thistle]

Wait whats re?

Okay

[grave tartan]

e^itheta= cos (theta)+isin(theta)

Right?

[viscid thistle]

Like r cis theta?

[grave tartan]

No..?

We’re in the complex plane right?

[viscid thistle]

Yeah

[grave tartan]

Oh wait

Okay it’s just notation

Yeah cis

[viscid thistle]

Okay

[grave tartan]

Cuz I don’t use cis notation

[viscid thistle]

Okay

[grave tartan]

So yeah express it in that form

[viscid thistle]

Okay

So rcostheta + irsintheta?

[grave tartan]

Yeah

alternatively you could write it as

$ee^i/theta$

[obsidian monolith]

Itadakimasu!:

[grave tartan]

please work

nNO

[viscid thistle]

Uhh

[grave tartan]

$re^{i/theta}$

[obsidian monolith]

Itadakimasu!:

[grave tartan]

ok whatever

the /theta

is just your angle

your argument

[viscid thistle]

Okay

How do i solve for z from there?

[grave tartan]

Ok so now you have the equation

Ok wait first what is your current equation?

[viscid thistle]

The one i have written?

[grave tartan]

Yup

[viscid thistle]

Wait jus a sec

[grave tartan]

Ok okay holll up

[viscid thistle]

Ok

[grave tartan]

Don't you want to express in the form e^itheta

instead of tediously writing cos theta +i sin theta

[viscid thistle]

Okay

Like this?

[grave tartan]

What's arg z

[viscid thistle]

Not given

[grave tartan]

Yeah but what form do you express it in

[viscid thistle]

Just theta isnt it?

[grave tartan]

Yup

but you wrote it as

2cis(theta)

[viscid thistle]

Wait im confused

I wrote 2rcis(-theta) because thats the conjugate pair

[grave tartan]

for argument z?

But agr z

[viscid thistle]

So what should i write instead, im a bit slow sorry

[grave tartan]

Is just an angle

right?

[viscid thistle]

Yeah

[grave tartan]

So we can express agr z

as simply theta

Because it's just an angle

[viscid thistle]

Yeah

[grave tartan]

arg

Yeah but you've expressed it in the form cis (theta)

[viscid thistle]

Wait im confused

[grave tartan]

?

[viscid thistle]

The z* is just z conjugate

So what would thst have to do with the arg?

[grave tartan]

Maybe our notation is diff eff

Because I always use z* as arg

Okay nvm

[viscid thistle]

Ohh

[grave tartan]

If it's just the conjugate

Then yes okay

[viscid thistle]

Our z* is conjugate

[grave tartan]

In that case right

probably good to express it in the form

[viscid thistle]

X + yi?

[grave tartan]

rcis(theta) ya

or that^

both work because one of them is just

cos (theta) -isin (theta)

[viscid thistle]

Yeah

Im confused abour how to solve for z

If we plug the value x + yi

[grave tartan]

Okay so we know the modulus of z

[viscid thistle]

Is root 3

Yeah

[grave tartan]

and then we form a equation

so essentially what we wanna do is form an equation and simplify

after that manipulate the equation

So like in this case

[viscid thistle]

Okay

[grave tartan]

yeah okay that's okay but

You're not fully utulising the information at hand

[viscid thistle]

Okay

[grave tartan]

utilising

Because we know the modulus is sqrt 3

So we should write it into the form

sqrt 3 e^itheta

[viscid thistle]

So the sqrt 3 cistheta?

[grave tartan]

$\sqrt{3}e^{i\theta}$

[obsidian monolith]

Itadakimasu!:

[grave tartan]

Yeah

oh shit

I did it

wooo

[viscid thistle]

Wait what is that

[grave tartan]

your cis theta

[viscid thistle]

Okay

[grave tartan]

is essentially the same as e^itheta

Like have you heard of eulers equation

[viscid thistle]

Yeah

[grave tartan]

e^ipi=-1

Yeah that's where it comes from

[viscid thistle]

Oh okay

Thats cool

Is cistheta also y + ix?

[grave tartan]

Yes but it's pointless to express it in that form

[viscid thistle]

Okay

[grave tartan]

Because we alr know the modulus

[viscid thistle]

So just express it in cis theta form?

[grave tartan]

yup

[viscid thistle]

Should i expand out the cis theta

[grave tartan]

kjf sdjkhsjkhjk

You should express in the e^itheta formm

[viscid thistle]

So it becomes costheta + sintheta?

[grave tartan]

ok look

cis theta right

to a large extent

[viscid thistle]

Yeah

[grave tartan]

is redundant notation

because it'll just confuse you in the long run

[viscid thistle]

Okay

[grave tartan]

$re^{i\theta}=cos(\theta)+isin(\theta)$

[obsidian monolith]

Itadakimasu!:

[grave tartan]

Remember this

[viscid thistle]

Okay

[grave tartan]

oh wait shit

with the r in front

$re^{i\theta}=r(cos(\theta)+isin(\theta))$

there

[obsidian monolith]

Itadakimasu!:

[grave tartan]

Because when you use this notation like

things get a lot easier trust me

[viscid thistle]

OHHHH

So ik the modulus

[grave tartan]

yesss

[grave tartan]

yup ok that's great

[viscid thistle]

Something like this?

[grave tartan]

improvement from before at least haha

[viscid thistle]

Lol

[grave tartan]

so like remb

$re^{i\theta}=r(cos(\theta)+isin(\theta))$

[obsidian monolith]

Itadakimasu!:

[grave tartan]

$re^{i-theta}=r(cos(-theta)isin(\t-heta))$

[obsidian monolith]

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

[grave tartan]

Then we have this

oh shitu

$re^{-i\theta}=r(cos(-\theta)+isin(-\theta))$

[obsidian monolith]

Itadakimasu!:

[grave tartan]

yeah so we have this right

[viscid thistle]

The conjugate pair

[grave tartan]

then remb cos - thea is just cos theta

so we have

$re^{-i\theta}=r(cos(\theta)+isin(-\theta))$

[obsidian monolith]

Itadakimasu!:

[grave tartan]

And then sin - theta is just -sin theta

so

$re^{-i\theta}=r(cos(\theta)-isin(\theta))$

[obsidian monolith]

Itadakimasu!:

[grave tartan]

which is our conjugate

[viscid thistle]

Yeah

[grave tartan]

so essentially we're gonna write the conjugate in the form above^

and then the z

in the e^itheta

instead of

$cos(\theta)+isin(\theta)$

[obsidian monolith]

Itadakimasu!:

[grave tartan]

Yeah

Then we will be able to manipulate the equation easily

[viscid thistle]

Ok

[grave tartan]

So you should have something like this

[grave tartan]

yup exactly

so we can simplify the second term

[viscid thistle]

Wait whered your i on the bottom go?

[grave tartan]

ah shit

yeah

it's there mb

[viscid thistle]

Ok

[ruby otter]

There are 2 functions F,G
Root(3) f + g =4
Find
(1-f)^3 + (g-3)^3
Given f ,g are continuous and range of both of them is rational

[willow bear]

...do you have a picture of the problem exactly as it is stated

[hybrid pewter]

hello

that’s the first part of the question.

[limber bone]

,rotate 90

[hybrid pewter]

here is where i’m stuck

[limber bone]

,rotate 180

[limber bone]

integrate 5sin^2(x)cos^3(x) from 0 to pi/2

u need help with the integral?

[hybrid pewter]

yes

using the substitution

[limber bone]

ok

[hybrid pewter]

i can’t get rid of the cos x

marking scheme shows only the first bit, involving sin

[limber bone]

5sin^2(x)cos^3(x) = 5sin^2(x)cos^2(x)cos(x)

5sin^2(x)(1-sin^2(x))cos(x)

(5sin^2(x)-5sin^4(x))*cos(x)

u = sin(x)

[hybrid pewter]

yeah

[limber bone]

got it?

[hybrid pewter]

yes

that’s where i stopped

[limber bone]

oh rly?

[hybrid pewter]

exactly where u stopped is where i stopped

[limber bone]

u = sin(x)

du = cos(x) dx

dx = du/cos(x)

integral (5sin^2(x)-5sin^4(x))*cos(x) dx

= integral (5u^2-5u^4)*cos(x) * du/cos(x)

= integral (5u^2-5u^4) du

can u integrate this?

[hybrid pewter]

sorry one min i’ll try

[limber bone]

tyt

[hybrid pewter]

du would be - cos (x) dx

right?

[viscid thistle]

tfw can't integrate polynomials

[limber bone]

wdym du would be -cos(x)

dx

if u = sin(x)

du = cos(x) dx*

[hybrid pewter]

oh fuck

differentiate

[limber bone]

ye ye

[hybrid pewter]

sorry sorry

[limber bone]

not integrate

its ok

keep going

[hybrid pewter]

wow

ure smart

or i’m dumb

thank u

[limber bone]

got it ?

[hybrid pewter]

yes

[limber bone]

good job

anything else?

[hybrid pewter]

thank u

[limber bone]

np

[hybrid pewter]

for now, no

but thank u

[limber bone]

np

[viscid thistle]

What did you get as the answer though

[hybrid pewter]

i didnt know how to get rid of the cos x

[limber bone]

did u do the definite integral?

[hybrid pewter]

oh not yET but im glad i got through the first bit lmaooo

[limber bone]

ye gj

[hybrid pewter]

ahh yup got it

thank you!!!

[limber bone]

gj

np

[valid vector]

Why are the answers to SOHCAHTOA these?

[limber bone]

lets choose the first question

sin(theta) in the smaller trinagle

sin(theta) = opp/hyp

[spring thunder]

,calc sqrt(5)

[limber bone]

opp to theta here is

[obsidian monolith]

Result:

2.2360679774998

[spring thunder]

they just wanted exact answers ig

[limber bone]

oh y

[valid vector]

But 1 doesn't equal sqrt 5

[spring thunder]

1/sqrt(5) = sqrt(5)/5

[valid vector]

Wait, wut?

You mean converting it from sqrt to exponents

[spring thunder]

so your answer is p much right

[valid vector]

I'm confuzzled, sorry.

[royal gull]

@limber bone do you still eat pen ink?

[limber bone]

i just did a few hours ago

funny coincidence

[spring thunder]

bloody hell

[valid vector]

If I'm looking for c to find the hypothenus I get a different answer

[royal gull]

Yeah, I saw you, I was checking if you admitted

[limber bone]

lol

[spring thunder]

wdym?

[valid vector]

Why are these answers wrong? When I use Pythagorean theorem for the adjacent side I get 11.618

Yet they are wrong

[spring thunder]

they want the exact answers

ie you shouldn't have a 11.618 in here

[valid vector]

So how do I get the exact #s

[spring thunder]

well how did you get 11.618?

[valid vector]

By doing b^2 = sqrt(12^2 - 3^2)

[spring thunder]

well b=sqrt(12²-3²) yeah

just keep sqrt(135) instead of going in your calc

cause that's not what they want

[valid vector]

Oh, so I just leave it at the square root?

[spring thunder]

ye

[valid vector]

Thanks so much!!

[viscid thistle]

Is Pie/25 the right answer for #1?

[patent beacon]

,w 7° 12' in radians

[obsidian monolith]

Display results online and refine query

Requested by Kaynex#7520

[patent beacon]

@viscid thistle

[serene heath]

pie

[viscid thistle]

Thx

-_- *sigh bye.

lol

[viscid thistle]

Is this correct?

[willow bear]

almost but not quite

[viscid thistle]

All of them or one in particular?

Oh because I didnt account for the negatives?

[willow bear]

yes indeed

[viscid thistle]

so the answers for cos and csc become negative right?

[willow bear]

oh and another thing

csc isn't 1/cos

[short sorrel]

is that supposed to be a spotify link or

did you just think it was topical

[viscid thistle]

Is this correct? The answer I'm least sure about is the phase shift

[grave tartan]

Isn’t the amplitude -2

[hexed ermine]

Nah

amplitude is magnitude

the - just means its been flipped

and I would say +pi/2 is the phase shift

[stiff rune]

https://cdn.discordapp.com/attachments/557873274302758923/558207092998275094/unknown.png

help

[hexed ermine]

Split the log by property of multiplication and then do change of base formula

[viscid thistle]

how am i supposed to find it if i can't divide it... ?? any help is appreciated.

[willow bear]

use the factor theorem

your polynomial equals 0 when evaluated at x = 4/3

[hybrid pewter]

dy/dx is k(y^2/x)

where’d the k come from

[willow bear]

it's the coefficient of proportionality

[hybrid pewter]

sorry?

what's that

[willow bear]

do you know what "A is proportional to B" means

[hybrid pewter]

np

no*

i dont

[willow bear]

how come you're doing calculus and don't know what that means o.O

[hybrid pewter]

cause im dumb

obv

i’m confused as to how i didn’t fail math

[tawny nacelle]

$A \propto B \implies A = kB$

[obsidian monolith]

soap:

[tawny nacelle]

k is the coefficient of proportionality

[hybrid pewter]

thank you

[tawny nacelle]

[viscid thistle]

Is this correct?

[willow bear]

might wanna double-check your signs

[viscid thistle]

if adjacent/ hyp = - 3/5 either ajd or hyp must be negative

how do I determine which one is negative though?

[willow bear]

this is quadrant 3. in quadrant 3, cos(θ) and sin(θ) are both negative

alternatively, you can take the hypotenuse as always being positive no matter what

[viscid thistle]

Ahhh

I see

How come we always treat the hypotenuse as positive? Just convention?

[willow bear]

i mean, this whole signed-side triangle thing is basically full of "by convention" things

[viscid thistle]

what are some other examples of that

[swift wagon]

(-3)^2 + (-4)^2 > 0

Hypotenuse is always positive cause of Pythagoras

[viscid thistle]

would those be double zeros at points -4 and 2

[hexed ermine]

yes

[viscid thistle]

okay thnx

would this be wrong then? :/

[hexed ermine]

nope

thats good

[swift wagon]

In line 2 you have (-3+4)^2 twice

@viscid thistle

[viscid thistle]

accident @swift wagon i meant -2

but i think the math's good right?

ty

[viscid thistle]

What does the order of a factor in an algebraic expression mean?

[willow bear]

context?

[distant flume]

Find the dimensions of the rectangle of largest area having fixed perimeter 100?

[patent beacon]

P = 2x + 2y = 100
A = xy

So you want to maximize xy, given x + y = 50

[distant flume]

Indeed

[patent beacon]

y = (50 - x)

A = x(50 - x)
A = 50x - x²

[distant flume]

Is that 25=x and 25=y?

I see

[patent beacon]

Yes it is, lel. But you can prove it by finding the vertex of the parabola

[distant flume]

You've been very helpful, thanks

[patent beacon]

Np. Feel free to ask if you have anything else

[viscid thistle]

anybody have recomendations?

[viscid thistle]

<@&286206848099549185>

[serene heath]

for the first one consult the double angle formula for tan

,w expand tan(2x)

second one factor out an 8 from both terms

then use the cos double angle formula

[viscid thistle]

awesome got the first one.

[serene heath]

epic

[viscid thistle]

Ok second one is still kicking my but. My answer was 1.6/65, or -8sqrt41/65 and its saying its wrong. What mistake did i mae

[serene heath]

how are u getting a number

it should be interms of x

factor out an 8

then use cos double angle formula

[viscid thistle]

oohhhh youre looking at 3

anyways im done for today.

[viscid thistle]

hi

can anyone help on this pre calculus question

find all the soultions on the interval of [0, 360) for the equation cos²(θ) – 3sin(θ) – sin²(θ)= –2

so i am at the part of 3-sin^2(theta)-3sin(theta)-sin^2(theta)=0

i think i add -sin^2(theta) with -sin^2(theta)

then we get the equation 3-2sin^2(theta)-3sin(theta)=0

but then im stuck at this point

[barren hedge]

That's pretty good.

[viscid thistle]

thx but im stuck now

[barren hedge]

Now do the quadratic formula for sin(θ)

[viscid thistle]

no its still not in qudratic form

or do i rearange it as -2sin(theta)-3sin(theta)+3

[barren hedge]

Say x=sinθ,
-2x²-3x+3

[viscid thistle]

oh so we do rearange it?

[barren hedge]

Yep.

[viscid thistle]

just like a rule then?

[barren hedge]

Well so you know the A, B, and C's

[viscid thistle]

x^2+x+1=0

okay im put it in quadratic formula and tyr to solve it now

thx

[viscid thistle]

i think i made an error

-2x^2-3x+3=0 so ax^2+bx+c=0 would formulate x=-(-3)+-Sqrt-3^2-4(-2)(3)/2(-2) simplified to x=3+-Sqrt-9+24/-4 then we have x=3+-Sqrt15/-4 so now we have 3+Sqrt15/-4=-1.71 or 0.21=3-Sqrt15/-4

[barren hedge]

Hm

[viscid thistle]

i made an error

do you think it was with the way i reranged the equation?

[barren hedge]

I cannot understand that since I'm tired but...

x=sin(θ), right?

[viscid thistle]

sin=y right?

cos=x

but do you mean as a quadratic

[barren hedge]

No no, from what I defined up earlier

[viscid thistle]

yea

[barren hedge]

So the quadratic principle root gives you when the quadratic=0

[viscid thistle]

yea sin(0) is pretty much x in this equation

[barren hedge]

But we want θ so we

Take a guess lol

[viscid thistle]

i rearanged 3-2sin^2(theta)-3sin(theta)=0 it as -2sin(theta)-3sin(theta)+3

i think the mistake his here right?

everything was correct up till this point

[barren hedge]

-2sin²(θ)-3sin(θ)+3, right?

[viscid thistle]

yea

[barren hedge]

That should be correct.

sin(θ)≈0.686140...

Should be what you get from your quadratic

[viscid thistle]

i did not get that

hmm i got -1.71 or 0.21

yea i made an error

im look through my work to see errors

dosnt -(-3) turn to 3

[barren hedge]

Yes

[viscid thistle]

oh i replaced c with 2

should be 3

oh nvm

thought it was an error lol

shouldnt i then get x=3-Sqrt15/-4

[barren hedge]

Not 15

[viscid thistle]

what

what would it be

how come its not 15

[barren hedge]

25

[viscid thistle]

-9+24

[barren hedge]

Can't be -9

Since the b is squared

[viscid thistle]

o

so waht do we do about -9 so it does not be a negative

[barren hedge]

I mean

(-3)²=9 not -9

[viscid thistle]

oh

i forgot the paranthesisi

sneaky paranthesis

[barren hedge]

Heh.

[viscid thistle]

so 9+24=33 so 3+-Sqrt33/-4

[barren hedge]

Yep.

[viscid thistle]

oh yea don't we do sin^-1(x) for our 2 soultions to get the degree

[barren hedge]

Yep.

[viscid thistle]

ah ok

[barren hedge]

After that, you check if they "make sense" and remove the one that is extraneous (non-sense answer), and keep the good one as your final answer.

[viscid thistle]

ah ok

well -2.18 gave me an error since its less than -1 on the interval (0, 360) guess its removed

[barren hedge]

That's good.

[viscid thistle]

and sin^-1(0.68)=42.84

[barren hedge]

So that should be the angle.

Nice

[viscid thistle]

should i subtract it to 360 or somethin

to find another angle

like with the quadrants

like 42.84 is in quadrant 1 so thats an angle

and sin is y

so would that mean quadrants 2 is also an angle?

so like 180-42.84=137.16

right

[barren hedge]

Should be.

[viscid thistle]

since sin is y and its positive its only in quadrants 1 and 2

[barren hedge]

I had to think about it, lol.

[viscid thistle]

yay we did it

"plays dora music"

[barren hedge]

@viscid thistle Hold on, the value you got for arcsin seems fishy.

[viscid thistle]

should look like this ? as final answer {42.84 degrees, 137.16degrees}

okay dokey

[barren hedge]

wait maybe I got my calc in the wrong mode

Odd, my arcsin gives me 43.325

[viscid thistle]

im using googles calculator lol

do you think it will matter?

[barren hedge]

Kinda

You did arcsin(0.686140661635)?

[viscid thistle]

no

i did 0.68

is that wrong?

[barren hedge]

It could get significant figures off by a bit. Let me check.

Okay yeah, that's what happened.

[viscid thistle]

cause thats how it was in the book they just used the first two decimals

[barren hedge]

Oh well, then I guess that's okay.

Nice work.

[viscid thistle]

aite thanks

[viscid thistle]

i am stuck on this question

A plane takes off from an airport at an angle of 14.5°. If the plane is traveling at a velocity of 220 feet per second, how high off of the ground is the plane at 10 seconds?

[willow bear]

what have you tried so far

[viscid thistle]

honestly i think it has to do with like how ships and bearing right?

thats the closest thing i can come up with

but im so confused on where to start

like i never seen a question like this before

[willow bear]

well the angle refers to angle with the ground, surely

[viscid thistle]

oh i see

its because the wording and concept confused

wait would ?=220 then?

[willow bear]

would it?

[viscid thistle]

no

it would have to be the hyptonuse

[willow bear]

the hypotenuse is 2200

[viscid thistle]

wait how did you get 2200?

[willow bear]

says it right there in the picture 😛

[viscid thistle]

oh XD

okay i so we have to find the oppsite angle then?

[willow bear]

"opposite angle"?

[viscid thistle]

but in order to do that we must find the adjacent?

[willow bear]

you know the hypotenuse and you want the opposite side

and you know the angle

there's a mnemonic some people use to remember what ratio is called what

it might've come up in class

[viscid thistle]

hmmm

[willow bear]

SOH CAH TOA

[viscid thistle]

oh

yes

i remmber

oppsite over hyptonuse

adjacent ove rhyoptunse

yea i remmber

but then how do we fine oppsite?

[willow bear]

well

let's just call it h, for convenience, since it stands for height

[viscid thistle]

okay h=oppsite

[willow bear]

$\sin(14.5^\circ) = \frac{h}{2200}$

[obsidian monolith]

Ann:

[viscid thistle]

hmm im still not sure how to find h

wait

maybe 14.5-180=165.5

[willow bear]

don't overthink this!

sin(14.5°) is just a number!

[viscid thistle]

oh ok

[willow bear]

if i asked you to solve the equation $17 = \frac{x}{5}$ for $x$, would you be able to do it?

[obsidian monolith]

Ann:

[viscid thistle]

no

wait if 14.5 would that mean the other angle across it would be 75.5?

[willow bear]

yes, but that is irrelevant

[viscid thistle]

oh

[willow bear]

we have the equation $\frac{h}{2200} = \sin(14.5^\circ)$ and we need to solve it for $h$

[obsidian monolith]

Ann:

[viscid thistle]

wait is it always going to be sin if its a triangle?

[willow bear]

no

[viscid thistle]

opp/hop then?

[willow bear]

which trig ratio it is depends on what two sides you care about

[viscid thistle]

so oppsoite/hyoptonuse=sin

[willow bear]

yes that's the definition of sin

anyway

[viscid thistle]

ah o

k

[willow bear]

you seem to be having trouble with solving basic equations

[viscid thistle]

yea

i usuually throw away everything i learn so i can focus on what im leanring atm but that came back to bite me

[willow bear]

if i gave you the equation $7z = 43$ and asked you to solve for $z$, would you be able to do it?

[obsidian monolith]

Ann:

[viscid thistle]

yea

[willow bear]

how would you solve it then?

[viscid thistle]

z=6.14

[willow bear]

no

[viscid thistle]

i would divide 7 by toh sides

[willow bear]

you would divide both sides by 7

and get z = 43/7

[viscid thistle]

yea to get x alone

[willow bear]

z

anyway

ok

[viscid thistle]

i mean z

[willow bear]

now

if i gave you the equation $\frac{x}{4} = 29$ and asked you to solve for $x$, would you be able to do it?

[obsidian monolith]

Ann:

[viscid thistle]

the fraction is throwing me off

would i just divide it both by 4 anyway?

[willow bear]

in the previous multiplication, you successfully undid multiplication using division

[viscid thistle]

or multiply it by 4?

[willow bear]

here, you have a division, so how do you undo division?

[viscid thistle]

multiplication

so i multiply both sides by 4

[willow bear]

yes indeed

[viscid thistle]

so x=12.5

12.25

[willow bear]

you said, just a message ago, that you are multiplying both sides by 4

[viscid thistle]

yes

[barren hedge]

29·4 has to be bigger than 29, lol.

[willow bear]

how did you get 12.5 from multiplying 29 by 4?

[viscid thistle]

no i did 29*4

the i did 4*4

and divided both by that

[willow bear]

what

[viscid thistle]

29*4=116

[barren hedge]

That's... An interesting error.

[viscid thistle]

4*4=16

[willow bear]

can you show your work

[viscid thistle]

so 16/116

right?

x=12.25

116/16

[barren hedge]

Oh I think I see..

[willow bear]

$\frac{x}{4} = 29 \ \frac{x}{4} \cdot 4 = 29 \cdot 4 \ x = 29 \cdot 4$

[obsidian monolith]

Ann:

[willow bear]

this is what you should have done

instead, you did something downright weird

[viscid thistle]

oh so when x/4*4 it becomes x alone

[barren hedge]

,, 4*\frac{x}{4}

[obsidian monolith]

𝓗𝑒𝓃𝓇𝓎 𝓒𝒶𝓈𝓉𝓁𝑒:

[barren hedge]

Exactlyy

[viscid thistle]

ah cool

[willow bear]

you are in DIRE NEED of a basic algebra review!

[viscid thistle]

XD

yea

i was going to review for algebra and geyomtry

but then i got caught up in pre-calculus

[barren hedge]

That's what this course is meant to do, before one goes into actual Calculus.

[viscid thistle]

i can't review my algebra or anything till i finish this at the moment

[willow bear]

i mean ok

in $\frac{h}{2200} = \sin(14.5^\circ)$, multiply both sides by $2200$ to get $h = 2200 \sin(14.5^\circ)$

[obsidian monolith]

Ann:

[viscid thistle]

anyway so x=h/sin14.5 right?

oh ok

lol

do i simplify sin(14.5) before i multiply it by 2200?

[willow bear]

sin(14.5°) is not a nice value

[viscid thistle]

yea it gave me error

[willow bear]

what

what do you mean "it gave you error"

[viscid thistle]

oh i did inverse sin

[willow bear]

...

[viscid thistle]

lol sorry

so 2200*14.5=31,900

[willow bear]

no

[viscid thistle]

so now we have h=sin(31900)

what

[willow bear]

NO!

[barren hedge]

sine of 14.5 pour flavour.

[willow bear]

sin(14.5°) is not 14.5!

[viscid thistle]

oh

we gotta put it in radians

[willow bear]

no

[viscid thistle]

what

[willow bear]

you don't need to

like

just put the thing into a calculator

make sure it's in degree mode

,w 2200 sin(14.5°)

[obsidian monolith]

Display results online and refine query

Requested by Ann#8192

[viscid thistle]

oh ok

[willow bear]

there

[viscid thistle]

im try it

[barren hedge]

Oh that's a pretty nice number.

[viscid thistle]

yea i got 550.836009

btw can you take the SAT after highschool?

[willow bear]

idk i'm not american

[viscid thistle]

oh ok

[barren hedge]

SAT 2 I believe

[viscid thistle]

the act?

[barren hedge]

I don't think so.

[viscid thistle]

i never heard of a sat 2

[barren hedge]

Neither did I until after I graduated high.

[viscid thistle]

anyways now we got a nnice number

what do we do with it

do we put it in the formula now?

oppsite over hyptonuse

sin(theta)=opp/hyp

[barren hedge]

It lets you select topics to test in, mainly so you can transfer to colleges after you got an A.A. degree.

[viscid thistle]

whats aa mean?

[willow bear]

no, @viscid thistle

[barren hedge]

Basically the first degree you get in college, Associate of Arts (doesn't have to be an art major, it could be a degree is Math, etc.)

[viscid thistle]

what we don't

[willow bear]

you got h = 550.8

that is the answer

you're done

that's it

[viscid thistle]

i thougth that was the oppsite

OH

[willow bear]

the opposite is what we were finding

[viscid thistle]

LOL

yea i jsut remmber

oppsite was the height

how far it was from the ground

ah thanks for everything

yea i have pretty trash memory and im bad at focusing and learning/studying

[barren hedge]

Not to scare you or anything, but it takes about 5,000 hours before you can master something.

[willow bear]

5000? last i heard it was 10000

[barren hedge]

Oh, for me personally I've calculated it takes about 10⁹⁹⁹ hours to master things, but studies don't like my personal outliers.

And according to Google, you're right.

10,000 hours. Wow, if you got a life, lose it and study.

[viscid thistle]

so to master studying i need to study for about 10k hours to be really good at it

atleast you can master it

[barren hedge]

Okay maybe you don't need THAT long to master Pre-Calculus

[viscid thistle]

its not just pre-calculus i think im bad at studying in general

like my memory is super bad

like its riduclously bad

[barren hedge]

What's my name?

[willow bear]

@viscid thistle have you seen a professional about it

[viscid thistle]

no

[willow bear]

i think it wouldn't hurt

[viscid thistle]

but like i forgot some stuff like im like where di i put it just now or what did i eat yester day or did i shower yesterday?/

[willow bear]

if you at least found out what's going on and how much of it is in your control

[barren hedge]

"where do I put it"

[viscid thistle]

XD

[barren hedge]

Youtube videos are sometimes interesting. It tends to have many subjects that can be nice to watch. Same with Khan Academy.

[viscid thistle]

maybe i have cognitive impairments

[willow bear]

yeah again

[viscid thistle]

like my grandma has Alzheimer's from my dads side and my dad also has really bad memory as well like he forgets some of the stuff i tell him

[willow bear]

please see a professional about this

see how much of it is in your control

[viscid thistle]

yea i will

like i legit have to take notes just so i don't forget to do something

and there like small things

[barren hedge]

Sounds a little like ADD.

[viscid thistle]

yea probs XD

also my younger brother also has trouble focusing/concentrating

[viscid thistle]

@barren hedge @willow bear Thanks for all your help and concern I have to go sleep now.

[barren hedge]

Good luck with everything.

[viscid thistle]

thanks

[viscid thistle]

i kinda need help again im not sure if its to do with my calculator or not since it gives me error but i feel i did everything right

–5sec(θ) = 10 have to find all the soultions

so i divided both sides by -5

then i i have sec(x)=-2

so now x=cos^-1(-2)

[slender river]

right

then wat did u do

[spring thunder]

then the calc said ERROR

[slender river]

haha yes

[viscid thistle]

but i can't solve for x since it gave me error

[slender river]

lol

ok

u just

uh

use ur identities i g

make it cosine

[viscid thistle]

i turned sec to cosine

[slender river]

okey

wat u got then

[viscid thistle]

what u mean

it gave us error

[slender river]

i i mean

like

howd u turn it into cosine wat

[viscid thistle]

cosine not inverse cosine?

[slender river]

ya cosine

show ur work kek im gettin confused on wat u did

[viscid thistle]

the rule sec(x)=1/cos(x)

[slender river]

sure

did u do that to both sides

[rocky bison]

,$ \sec\left(x\right)=-2

[obsidian monolith]

Pseudo:

[rocky bison]

Then you went to

,$ \cos^{-1}\left(-2\right)=x

[viscid thistle]

then i wanted to solve for x

[obsidian monolith]

Pseudo:

[rocky bison]

Which is not correcto

You just pretended cos=sec

[viscid thistle]

oh

[rocky bison]

which is not true

[viscid thistle]

they don't?

[slender river]

yeah um

ur supposed to do 1/-2

[viscid thistle]

then what is cos reciprocal

[slender river]

cos u gotta do the 1/ bit for both sides

no no reciprocal correct

ur c alculator's notation is just dumb

[viscid thistle]

oh

ah

[slender river]

when it says cos^-1

[viscid thistle]

ye

[slender river]

it does not mean 1/cos

[viscid thistle]

oh

[spring thunder]

ho

[slender river]

it means secant

this notation is heccin weird but we gotta live with it

[viscid thistle]

yea lol

[slender river]

anyway u dont really need calculator for this problem lol

especialy if doin general solution

[viscid thistle]

oh ok

[slender river]

so let's start from where u were correct

sec(x)=-2

[viscid thistle]

yea

[slender river]

so do 1/stuff for both sides

[viscid thistle]

so 1/sec(x)=1/-2

[slender river]

ye

so what's 1\sec

[viscid thistle]

oh

cosine

[slender river]

yup

[viscid thistle]

so now we have x=cos(1/-2)

[slender river]

only then u do the inverse cosine thingy

😡

shoo

[viscid thistle]

huh

[slender river]

lol

dwai

just listen to my voice

lol

[viscid thistle]

so now i got 2.09

[slender river]

make sure it's inverse cosine and not cos like u wrote

[viscid thistle]

so puting it in radians

[slender river]

for that last bit

yeah rad is fine

oi \cos

kek

[viscid thistle]

so x=120?

[slender river]

yeah x is 120 degrees

but general solution

[viscid thistle]

so thats in the second quad

and its a positive

[slender river]

find all the places on the unit circle where the x coordinate is -1/2

or actually

u have to consider the domain

of sec(x)

[rocky bison]

👁‍🗨

[slender river]

what was ur original problem?

[viscid thistle]

quads 1 and 4

but it can't fit in quads 1

[slender river]

no hold on fren

wait a min

wat was the original problem

cos that's where the domain matters

[viscid thistle]

–5sec(θ) = 10.

[slender river]

ok

[viscid thistle]

cosine is domain right

[slender river]

so it's just a matter of getting the domain of secant

yeah and cosine cant be 0 for this else bad

[viscid thistle]

yea which is positive so quads 1 and 4

and quads 2 because thats were 120 fits

[slender river]

120 degress fits

oi shoo wait like 2 secnds

[viscid thistle]

so only quads 4 and 2 then?

[slender river]

wat

where u gettin 4 from

[viscid thistle]

since the domain is positive there

[slender river]

i mean

there's also

quadrant 3 tho

[viscid thistle]

but its negative

it has a negative domain

[slender river]

wym

[viscid thistle]

and our cosine is has a positive domain

[slender river]

cos(120) gives -1/2

[viscid thistle]

no

[slender river]

wat

[viscid thistle]

it gives 1/-2

[slender river]

same thing fren

[viscid thistle]

but then i put it in cosine inverse

and gave me positive

[slender river]

ech dwai

[viscid thistle]

oh yea

[slender river]

we only lookin at the original bit

[viscid thistle]

i frogot it have to be before cosine inverse

yea mb

so cosine is negative

[slender river]

yeah

[viscid thistle]

so quads 2 and 3

[slender river]

120 is a solution but there's also anotheron e in q3

[viscid thistle]

okay that makes sense

tea since quads has domain of a negative

so 180-120?

[slender river]

yeah quadreants

uh

leme th0nk

[viscid thistle]

so quads 3 gives us 60?

[slender river]

wait w0t

[viscid thistle]

oops

i meant to add