precalculus

[neon trench]

okay thats what i thought

now what lmao

[sonic jay]

x = (log(216)(1-x))/log(36)

I would do this

which can also be written as

x = (log(216)/log(36)) * (1-x)

tell me if ur confused

[neon trench]

okay so you just divided the log36 over

[sonic jay]

yeo

and then i said

this

is the same thing as

[neon trench]

okay

[sonic jay]

Why?

because its easier to use the calculator to solve for that messy log side on the left

[neon trench]

yeah that s what i thought

[sonic jay]

and then multiple it into (1-x)

[coarse fjord]

you can also rewrite 216 as 6^3

[uncut mulch]

^. don't really need logs for this

[coarse fjord]

and rewrite 36 as a 6^2

[neon trench]

UHHH

i got 1

tf

[sonic jay]

for X?

[neon trench]

yeah lmao

i put the log part in my calculator and got 1,5

1.5

and then distributed to 1-x

[sonic jay]

X value is wrong

[neon trench]

got 1.5-1.5x

[sonic jay]

what u solved for log is write

*right

jesus i have dyslexia

also what ramonov and august suggested maybe the method preferred by ur professor

[uncut mulch]

36^x = 216^(1-x)
6^(2x) = 6^(3(1-x))
and then equate exponents

(or apply log_6 results in the same thing)

[sonic jay]

btw if u are looking at the way we were doing

upto this part is correct

u are messing up some algebra

towards the end

[neon trench]

nani

[sonic jay]

x=−1.5x+1.5

[neon trench]

so thats right?

[sonic jay]

u gotta solve it a little further

[neon trench]

subtract 1.5 to the otherside?

[uncut mulch]

not quite

[sonic jay]

'

how about u get variables on 1 side and constants on the other

not exponents

[neon trench]

so just add the 1.5x over

[sonic jay]

*constants

sorry

right

[neon trench]

1.5x+x = 1.5

[sonic jay]

yes what do u get once u take X in common?

[neon trench]

huh

[sonic jay]

we are focusing the left hand side

[neon trench]

yeeee so 2.5?

[uncut mulch]

2.5x

[sonic jay]

yes now u can solve for the rest easily

what ramonov said

^

[neon trench]

1.5/2.5

OHHHHHHHH

.6

which is the answer

wowie

THANK YOU

[sonic jay]

yup

[neon trench]

I UNDERSTAND NOW

[sonic jay]

But

If u are not allowed calculators on the test

[neon trench]

nah i can 🙂

[sonic jay]

Okay

[neon trench]

❤

[sonic jay]

dam this math server got some weeaboo emojis

Im happy

[neon trench]

@sonic jay hehe one more?

[sonic jay]

yeah sure

gonna head to bed after this one tho

[neon trench]

ln (x+6) - ln (x+4) = 4

yeee

lmao

thankls

since its subtracting can you do ln(x-6/x+4) = 4

OHMY GOD

wrong question lmao

there we go

[sonic jay]

sec

was this the original question or did u simplify it a bit?

[neon trench]

ln (x+6) - ln (x+4) = 4

was the original

[willow bear]

you forgot parentheses

[neon trench]

where

[willow bear]

ln(x-6/x+4)

[sonic jay]

yeah Im a bit confused on this as I gotta remind myself of all the log rules

[willow bear]

no

it's not about that

\verb&x-6/x+4& reads as $x - \frac6x + 4$, not $\frac{x-6}{x+4}$

[obsidian monolith]

Ann:

[neon trench]

waitWTF

[willow bear]

order of operations is still a thing yknow

[neon trench]

so um

how do i do it?

[willow bear]

do what

type $\frac{x-6}{x+4}$ in plain text?

[obsidian monolith]

Ann:

[willow bear]

(x-6)/(x+4).

[neon trench]

no

like solve for x

in

ln (x+6) - ln (x+4) = 4

[willow bear]

the step you attempted to do - rewriting the two logs as one by means of a log rule - is fine.

i was specifically commenting on your fuckup in how you typed it out.

[sonic jay]

yeah im a bit stumped on this

one

[neon trench]

oh

gotcha

[sonic jay]

might msg u later if i solve it i think i know what to do just seems like a little long process

[neon trench]

so from my fuckup, what do i do next?

e^4 = (x-6)/(x+4)

[willow bear]

yes good keep going

[neon trench]

@sonic jay all good thanks though!

okay so now algebra?

[willow bear]

...

as if you weren't doing algebra already?

[neon trench]

well yeah

so multiply x+4 to the otherside

[willow bear]

you were doing algebra already
the equation just doesn't involve logs anymore

and no, you multiply by (x+4) on both sides.

[neon trench]

yeah i know

so

(e^4)(x+4) = x-6

distribute?

[willow bear]

you've now made this into a linear equation

solve it as such

[hexed bolt]

How would I do this

[willow bear]

what have you tried so far

how did you get those answers that you wrote up there

[hexed bolt]

Just -1 and 0

I thought it was a reflection but i guess not

[willow bear]

what do you mean by "a reflection"

a reflection of what

i repeat, how did you get a=-1 and d=0

[hexed bolt]

x axis

reflection

But it's wrong

It's not a reflection

I don't know how to find d

[willow bear]

d is the point halfway between the maximum and the minimum values of your function.

both of which you are meant to read off the graph.

[hexed bolt]

ok

[viscid thistle]

is ln(| (x^4) + 7 |) = ln( (x^4) + 7 )?

[willow bear]

yes

the parentheses around x^4 are redundant but yes

[green zenith]

I have a question about inverse sines

Why is y=sin^(-1)(x) equals to sin(y)=x?

Could you actually divide sin^-1 on both sides?

[viral meteor]

Sin^-1 is the arcssin

[serene heath]

Ur not dividing by anything

[green zenith]

So its just how it is?

[willow bear]

do the words "inverse function" ring any bells

[green zenith]

Yeah

Oh

Its a function

Because it looked like its divided xD

[green zenith]

Hello again.

In this picture, i want to know why g'(x)=1/(cos(sin^-1x)) has cos in it but f(x)=sin(x)

And what does the ' means?

[viscid thistle]

' means prime in geometry but idk if it relates to Precal

[spring thunder]

' means derivative wrt x here

[green zenith]

Wrt?

[spring thunder]

with respect to

[green zenith]

Oh

[proud wraith]

could someone please walk me through this?

[patent beacon]

Can you do that division, first of all? Check out synthetic division if you haven't yet @proud wraith

[viscid thistle]

can i write, y = 1 / (-e^x + C[1]) as y = -e^-x + C[2], where C[1] and C[2] are different constants?

for context, this is the general solution to a differential equation

[patent beacon]

No, those are very different things

You need to go from one to the other, with specific substitutions

[viscid thistle]

ok. thanks!

[odd helm]

Why can't I just use distance formula to find distance between (3,7) and (-1,8)

[gentle vigil]

You can

[odd helm]

kalfefew it didnt work but ok I guess I typed it wrong

Thank you

Wait what no

It didn't work

3--1=4 and 4^2=16

and7-8= -1 and -1^2=1 and 16+1 =17

So shouldn't the distance be sqrt 17?

[proud wraith]

could somone please walk me through this?

[odd helm]

<@&286206848099549185>

[latent tinsel]

Uhm you're supposed to find distance from end to start

Between 3, 7 and - 1, 8 would mean you found end and middle

I reckon you find his x and y displacement, so he goes back down 15 points and goes left 2 points. Now you have your x and y, your hypotenuse is your distance @odd helm

@proud wraith hmm 🤔 I see that if you put those to one side, you can form (a^3 + b^3) and a perfect square, but I dunno if that'd be of help

[hoary valley]

What's the inverse of a number that has an exponent?

For Example this number...

[karmic wave]

what's the inverse of a number that doesn't have an exponent? 🤔

[hoary valley]

@karmic wave 1 in the numerator, and the number in the denominator

[sleek pawn]

or the negative version of it?

[karmic wave]

there's additive and multiplicative inverses. I'm guessing the question is about multiplicitave ones @sleek pawn

[sleek pawn]

oh that's interesting

what other inverses are there?

if any

[karmic wave]

quite a few
https://en.wikipedia.org/wiki/Inverse

@hoary valley how does that rule effect exponents?

[hoary valley]

@karmic wave I'm asking you.. but nvm you are just wasting my time, unfortunately.

[sleek pawn]

woah

well I want to take the given opportunity to ask

[hoary valley]

Bye

[sleek pawn]

because I'm not sure how you were able to figure out that they were talking about a multiplicative inverse number

@karmic wave

isn't the inverse of an exponent a logarithm?

[karmic wave]

depends on the effect you're trying to achieve. If you want to add two numbers together to get zero, that's the addititve inverse. If you want to multiply two numbers together to get one, that's the multiplicative inverse

Josh described the multiplicative inverse, so I continued with that train of thought

[sleek pawn]

hmmm

I'm thinking in terms of an equation, and forgive me if this all sounds confusing but I'm myself a bit confused, that's the purpose of the question, but for example, if we had a square number then the inverse operation would be the square root

then we have logarithms, which I've also been told are the inverse operation of exponents

so yeah, what's the deal with that?

[karmic wave]

those are inverse functions

[sleek pawn]

hmmm

so... I get that, but I'm still quite confused

[karmic wave]

because addition and multiplication are commutative, there's only one inverse function for each; subtraction and division respectively

but because order matters when taking exponents, order matters in the inverse function(s) of it, too

generally, $a^b\neq b^a$

[obsidian monolith]

Googol30:

[sleek pawn]

mhm?

[willow bear]

the functions $x \mapsto x^c$ and $x \mapsto c^x$ are very, very different

[obsidian monolith]

Ann:

[sleek pawn]

yeah, that's not what confuses me really. It's more about what's the relationship of roots and logarithms to exponents

[unborn pond]

anyone here able to like, dm me and help me out? I have a test tomorrow and don't understand any of the content

i'm doing the trig portion of college precal

[willow bear]

if you have a test tomorrow and don't understand any of the content then you are what's known as "absolutely fucked"

[plain flicker]

can someone help me with 10

[willow bear]

what's giving you trouble here

[plain flicker]

the 8%

80%

[willow bear]

can you show what you've got so far?

i am afraid i don't understand why the 80% bit specifically is giving you trouble

[plain flicker]

11 liters of 15% antifreeze means you have...
15%*11=1.65 L of antifreeze and 11-1.65=9.35 L of water

im sorry im not to good with ratios

[willow bear]

yeah but that's only at the beginning, isn't it?

you're them removing some amount from your system

and you're asked to find it

so let's call it x.

[plain flicker]

okay

[willow bear]

we're removing x liters of mixture and replacing it with another x liters of 80% antifreeze solution.

before we refill the system, but after we've removed x liters from it, how many liters of mixture remain in it?

[plain flicker]

11-x?

[willow bear]

are you unsure of your own answer?

[plain flicker]

im confident

if i have 11L and removing x L from it

leaves me with the 11L minus the portion i took out

[willow bear]

yeah but then why the question mark

[plain flicker]

oh sorry i meant i as answer

an

[willow bear]

anyway you're right

so there are (11-x) liters of old mixture (15%) and x liters of new mixture (80%)

can you find the total amount of antifreeze in the system, in terms of x?

[plain flicker]

is it (11-x)(x)=95%

wait

i think its (11-x) times 0.15

• x times 0.8

= 25

[willow bear]

are you sure it equals 25 specifically?

[plain flicker]

no

i think it will produce 20%

Instead of 25

[willow bear]

where did you even get 20 from

[plain flicker]

0.8 times 0.25

[willow bear]

why are you multiplying those

[ionic timber]

decreasing

[willow bear]

@ionic timber this channel is occupied, please move to #❓how-to-get-help

@plain flicker

[plain flicker]

yes sir

i should have multiplied those

i thought maybe i was looking for 80% of he 25% of the result

shouldnt*

[willow bear]

why are you calling me sir.

[plain flicker]

my aplogies dr ann

[willow bear]

i'm not a doctor either

just.

don't.

[plain flicker]

okay im sorry

[willow bear]

anyway...
what does the expression (11-x)*0.15 + x*0.8 represent?

[plain flicker]

the amount of anti freeze?

[willow bear]

the total amount of antifreeze in the system

so if we want a concentration of 25%

and there's 11 liters of mixture

how much antifreeze SHOULD there be

[plain flicker]

0.25*11

2.75

[willow bear]

well there you have it

now you have an equation in x

0.15(11-x) + 0.8x = 2.75

[plain flicker]

ohhh okay

thanks so much ann

[spice surge]

I have been trying to understand this, but I can’t work out how to do the problem. Could anyone help me? Thank you

[patent beacon]

If x = 1/3, then cos(θ) = 1/3

You can use that to find sin(θ) pretty easy. The other trig functions follow from there

[spice surge]

So the p(x,y) means (1/3 , 0)?

Since only x is given?

So what I’m getting is let’s say it was in quadrant 2

That means instead of x = 1/3, it would be x = -1/3?

Since it’s in the negative?

Ahh I see

I found it out hopefully

[warm crescent]

For sets $A$ and $B$ prove that $A-(A \cap B)=A-B$

[obsidian monolith]

Raftaar:

[limber bone]

take an arbitary element in A-(A intersects B)

unfold definitions

and show its in A-B

and do the otherway around

to show double inclusion

[warm crescent]

Okay

Thanks for the hint.

[limber bone]

np

if u get stuck ping

[warm crescent]

Sure.

[willow bear]

i see you've adopted "unfold definitions" as a catchphrase

[warm crescent]

$ A-(A \cap B)={x|x\in A \wedge x \notin (A-B)}$

Ann,Whats the conjunction and disjunction symbol

$\hat$

[obsidian monolith]

Raftaar:

[warm crescent]

$ A-B={x|x\in A \wedge x \notin B}$

[obsidian monolith]

Raftaar:

[warm crescent]

$A-B=A \cap B'$

[obsidian monolith]

Raftaar:

[limber bone]

'i see you've adopted "unfold definitions" as a catchphrase'

its totally from you

😄

let x be in (A-(A intersects B) )

unfold ur definitions

what does this mean for x

@warm crescent

[warm crescent]

I proved it to be equal to $A \cap B'$ then proved that $A \cap B'$ equal to$ A -B$

@limber bone

[limber bone]

i cant understand this

[obsidian monolith]

Raftaar:

[limber bone]

yea

thats nice

gj

[harsh cipher]

Hi guys I have a question

not calculus, but it's to do with calculations on a spreadsheet

If I want to divide in a cell.

=A2/(A2-A1)+(C3-C2).

this gives me wrong answer, eventhough I get the correct answer doing the division on paper.

[willow bear]

uh

what exactly are you trying to accomplish here

[harsh cipher]

Im trying to make a poker calculation.

Ill show you what I'm trying to do

In cell H2, I need 28.5%

My opponent is risking 3.67/(3.67+9.16) = 28.6

[gentle vigil]

Put your brackets right

[harsh cipher]

I did?

or is it wrong?

[gentle vigil]

How did you get 9.16?

[harsh cipher]

Pot before villain raise = 2 assuming no antes

I mean =3

Pot before= 3, villain raises 2, pot = 4

I mean 5

[gentle vigil]

(G2-B2)/((G2-B2)+(G2+C2))

I think you want to write it like this

[harsh cipher]

k let me try that

doesn't work

[gentle vigil]

It's not giving 0.28?

[harsh cipher]

nope

[gentle vigil]

What is it giving then?

[harsh cipher]

10.17

[gentle vigil]

Can you share the screenshot?

Of expression you wrote

[harsh cipher]

nm It worked

[gentle vigil]

Great!

[harsh cipher]

Thank you.

[gentle vigil]

You're welcome!

[harsh cipher]

What do I search in google to put in proper brackets in excel?

[gentle vigil]

You just need to use bodmas

[harsh cipher]

ok

[gentle vigil]

If you write 1/(1+1), and if you write 1/1+1, they are different

First one is 1/2 and second one is 2.

[viscid thistle]

How is 13 wrong

ummm rotate it

[fading token]

It's a rather sly question

The idea is that x = -4 is always a solution to that equation, regardless of what y is

In particular, y = 0, which is an x-intercept

So that's why A is the correct answer

[viscid thistle]

help

[willow bear]

what is this a picture for ants??

$\lim_{n \to \infty} n \left( \sin\left( \frac{\pi}{n} \right) + \sin\left( \frac{3\pi}{n} \right)\right)$

[obsidian monolith]

Ann:

[willow bear]

typed up more legibly for everyone else's convenience

[primal karma]

OHHH there is an n in the beginning

[willow bear]

@viscid thistle what is giving you trouble here

[viscid thistle]

everything

what to do

i'm dumb

i have to do 200 ex for tmrow

and like i can't think straight

for tihs one

pls help

or i die

[willow bear]

i have to do 200 ex for tmrow
what

[stuck lark]

whoever said you "have" to do so many?

[serene heath]

$\pi \cdot \frac{\sin\left(\frac{\pi}{n}\right)}{\frac{\pi}{n}}$

[obsidian monolith]

the one n only:

[serene heath]

Do something similar for the other part

[viscid thistle]

so i

do

the

formula of sin x/x = 1

but if i do sin(pi/n)/pi/n i have to multiply with pi/n

$\frac{\pi}{4}\right)} \cdot \frac{\sin\left(\frac{\pi}{n}\right)}{\frac{\pi}{n}}$

[obsidian monolith]

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

[viscid thistle]

wait

no

$\frac{\pi}{n}\right)} \cdot \frac{\sin\left(\frac{\pi}{n}\right)}{\frac{\pi}{n}}$

[obsidian monolith]

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

[viscid thistle]

and that would get me after all to something like

n x (4pi/n)

[serene heath]

What

[viscid thistle]

you take

lim(sin(x)/x)=1

now i have lim of sin(pi/n)

if i divide and multiply with sin(pi/n)

would get to sin(x)/x

x being pi/n

anyway

i think i figured it

out

tnx

[blissful wadi]

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

[obsidian monolith]

dvaix:

[blissful wadi]

someone help me find y in term of x please

[viscid thistle]

@blissful wadi do you still need help

If so just simplify the left equation

(you can do so by factoring)

[indigo fiber]

yep

[spice surge]

I’m stuck in this

Could anyone help me?

[uncut mulch]

where are you stuck?
are you familiar with the unit circle, ASTC?

[spice surge]

Well when I saw the work done, they used sin^-1(0.7)
I was wondering why you can’t just do sin(0.7)

[uncut mulch]

because 0.7 is the ratio
you apply arcsin to get the principal angle

[undone drift]

@spice surgeTake a look at inverse trigonometry. Why and when we take an inverse. Bijective functions and all that...

You'll understand it better

[viscid thistle]

????????

[undone drift]

What the hell?

There is a way simpler way to understand this

,rotate 90

[undone drift]

Check it☝

[viscid thistle]

oh thankyou

[blissful wadi]

@viscid thistle yes i do need help , i tried factoring left side and everything possible ( at my level ofc ) but no result

[languid crane]

@viscid thistle post that kind of questions in #calculus in the future

[blissful wadi]

@languid crane sure

[vague zephyr]

im not sure how its wrong

i have tried a few different methods and this is what i get each time

[odd helm]

Why does the function go up at x=0

Ik the range is infinity

But why there?

[uncut mulch]

the asymptote doesn't necessarily need to be at x=0

@vague zephyr for x>0, (2x-6) > -6

[odd helm]

Ah ok

For the neither column

Can it be any value besides the one in the even and odd column

Like could I put 1000 or -1 or whatever

[willow bear]

yes

[odd helm]

Ok thank you

[warm crescent]

@split loom Find the tangent and then use slope formula

[ripe dust]

https://gyazo.com/1f02321266afdc1ce842e4eb11dd13ed

for finding the domain i have to find denominator restriction

however i haev something like x^2 = -1

so when i have an equation like that I just don't take in account the denominator for finding the domain?

[limber bone]

there is no real number that satisfies x^2=-1

so the domain here is all real numbers

what do u not understand

[ripe dust]

oh yeah right,

[limber bone]

if the function is rational

then domain problems can happen at the denomaintor

you dont want to divide by 0

so you take in account the denominator ofc

for finding domain

[ripe dust]

ok so it has to be a real number alright thx ig

[limber bone]

well no actually depends on like

the book or the class or whatever

but yea you are on real numbers ig

[acoustic geyser]

Why do we differentiate the equation while finding range?

Plz @ me if you answer

[rigid sun]

Range?

It’s so then we can find the critical points to where the max and min can occur

@acoustic geyser

[harsh cipher]

Hi guys where does "a" come from?

[stuck lark]

just a tacked-on arbitrary constant

[harsh cipher]

in the video he says, "because constant term is 80 "a" has to be.

I don't get that?

[harsh cipher]

Hi

Consecutive odd integers... 2,4,6 is even numbers?

nm 1 1+3 , 1+4 , 1+6

[proud sparrow]

it's probably better if you parametrize in terms of the median @harsh cipher

[odd helm]

Guys for this I distributed the 3 to x and -9 individually

So instead of (x-9)^3

I did x^3-9^3

Even though I think I know what will be the answer

Is that ok?

[viscid thistle]

No, you cannot distribute like that.

Consider a simple example: (a-b)^2

This means (a-b)(a-b)

Which results in a^2 - 2ab + b^2

On the contrary, if we considered (ab)^2, this becomes (ab)(ab), or a^2*b^2

[rigid sun]

Stop violating math laws

Distribute exponents to multiplication and division, not addition and subtraction

[tame wedge]

Hey I have a question: p(z)=z^3+az^2+bz-32

I'm told that z=4i is a zero, does this mean that the conjugate of 4i is also a zero, i.e. -4i?

The question asks to solve for a and b given 4i as a zero, I figured that -4i would be a zero and multiplied the conjugate factors to get (z^2-16)(px-q)=z^3+az^2+bz-32 and tried to solve for a and b to no avail

[uncut mulch]

it should be z^2 + 16
also z not x

[tame wedge]

Ahhh wtf yeah it is z^2+16

The stupidest shit sometimes

Cheers

[willow bear]

@tame wedge are a and b known to be real

if yes then yes -4i is a zero too

[tame wedge]

Yes a and b are real, I worked them out to be -2 and 16 I think

[candid onyx]

Did I do this right ?

[gentle vigil]

Looks right

[candid onyx]

Having trouble with this one, specifically how to factor it

[serene heath]

It's a quadratic in terms of e^x

[candid onyx]

Yeah I’m not understanding how to break it down )

[eternal cosmos]

Well, that's write e^(2x) as (e^x)^2, and let y = e^x. Does that give you a clue as to how to solve it?

[candid onyx]

Thank you. We are able to rewrite because of exponential rules ?

[eternal cosmos]

Yes.

[candid onyx]

🙏🏽 thanks ! The answers I was looking for !

[eternal cosmos]

No worries.

[viscid thistle]

Why does
cos^2(x)-sin^2(x)/cos^2(x)+sin^2(x)
equal cos^2(x)-sin^2(x)

And not just 1-1?

[tardy crescent]

The denominator, cos^2(x) + sin^2(x) equals 1, so it can be rewritten as cos^2(x) - sin^2(x) / 1, which is just the same as the numerator

[viscid thistle]

Oh

Pythagorean identity?

[tardy crescent]

Yes, because sin = Opp/Hyp and cos = Adj/Hyp, so Opp^2/Hyp^2 + Adj^2/Hyp^2 = Hyp^2 / Hyp^2

[viscid thistle]

Yeah

Okay thank you

[tardy crescent]

np

[leaden stratus]

Can this be done without using an auxiliary term?

[serene heath]

A what?

[leaden stratus]

@serene heath auxiliary term. For example you call "t" 4^x

[prime prawn]

why do you think it cant be done without using an auxiliary term

[winter isle]

does anyone have a proof that tan theta = cos theta/sin theta

[stuck lark]

flip that fraction, then you have a valid question

[tardy crescent]

You can think of tangent as the opp / adj sides of a right triangle. Since sine is the opp / hyp, and cosine is the adj / hyp, then sine / cosine = (opp / hyp) / (adj / hyp). Dividing by a fraction is like multiplying by its reciprocal, so sine / cosine = (opp / hyp) * (hyp / adj) = Opp / Adj, or tangent

[hoary valley]

And how am I supposed to know what is ln 57/ln 3 ?

[eternal cosmos]

Well, you can use a calculator. Most calculators don't have a log_3 option.

But they do have a ln option.

[hoary valley]

calculators are not allowed.

[eternal cosmos]

Oh.

[hoary valley]

how this happened? ( i know log_b (x) = ln (x)/ ln(b) )

But how it became, ln (x) ?

[uncut mulch]

what's the whole question?

[hoary valley]

Graph the given function log_1.5 (x) and describe how it is related to y=ln x

[uncut mulch]

the 2nd line is just stating the equation
y=ln(x)

[hoary valley]

Dude you're good.

Thanks

[heady jewel]

lol

[prisma prairie]

lol

[native flower]

i need help with trigonometry

plz

if cos (Θ) = 1/6, what is csc (Θ) and cot (Θ)?

[uncut mulch]

what have you tried?

[native flower]

what do u mean

[uncut mulch]

where are you stuck?

[native flower]

just where to start

[uncut mulch]

draw a triangle

also are you told the domain of theta?

[native flower]

yeah

between 0 and pi / 2 inclusive

[uncut mulch]

ok, that makes things simpler.

draw a triangle where cos (Θ) = 1/6,

[native flower]

ok

done

[uncut mulch]

use pythag to find the 3rd side, and you should be able to figure out the rest

[native flower]

ok

thanks man

appreciate it

[bold hearth]

Hey. I can't for the life of me figure out how to simplify 2(1-4x)/(2x-4)^2. Probably missing something really obvious. Could anyone point me in the right direction?

[uncut mulch]

$\frac{2(1-4x)}{(2x-4)^2}$?

[obsidian monolith]

ramonov:

[bold hearth]

Yeah. That's it.

[uncut mulch]

uh no

you'd have to factor out 2^2 from the denominator

[willow bear]

$(2x-4)^2 \neq 2(x-2)^2$

[obsidian monolith]

Ann:

[willow bear]

$(ab)^2 \neq ab^2$

[obsidian monolith]

Ann:

[coral valve]

damn my calc gave me that

[willow bear]

gave you what

[bold hearth]

So basically I have to write it as (2x-4)(2x-4) and then take out 2 from each of those?

[eternal cosmos]

Technically, you are doing (2x - 4)² = (2(x - 2))² = 2²(x - 2)².

[bold hearth]

Excellent. I understand it now. Thanks a lot!

[distant nova]

Given , 2^x + 3^x + 4^x - 5^x = 0
Find the number of solutions

Can I habe some help plis

[heady jewel]

any conditions on x?

like real,integer,positive

@distant nova

wait

number of solutions

ok

so

we can rewrite

it as a decreasing function

of x

thus 1 real solution

[rigid sun]

Bro

[heady jewel]

x<=2

[rigid sun]

Just plug in 0

And then find some limit

[heady jewel]

uhh

for x<=2 its positive

x>=3 its negative

so yeh one solution

I GUESS tho

[rigid sun]

or you could solve

Easy

It usually helps to find the more extreme values and analyze them

When you have some exponential function, the lower extreme is usually at 0 and some min/ max is at infinity

[rare zephyr]

in what way derivatives are applicable in modern careers?

like what can I find in real life with f'(x)?

[languid crane]

they are used in physics for example

generally problems with rates of change involved, eg. motion

[rare zephyr]

motion?

oh

i get it

what about integration

[languid crane]

well when u want to find areas n things like that

[distant nova]

Finding areas of irregular surfaces?

Irregular ‘objects’?
I don’t really know the word tbh

[rare zephyr]

irregular polygon areas?

[languid crane]

"To calculate the velocity and trajectory of an object, predict the position of planets, and understand electromagnetism."

irregular shapes?

also took this from somewhere: "Biologists use differential calculus to determine the exact rate of growth in a bacterial culture when different variables such as temperature and food source are changed."

[distant nova]

Yep,shapes was what I was looking for

Oof I’m drained today

[blissful wadi]

guys for a function to be continue by extension , the lim on (x0)- should be equal to the lim on (x0)+

right ?

x0 ∉ Df

[heady jewel]

@rare zephyr you can find the area of anything

with an integr

al

[blissful wadi]

right ?
x0 ∉ Df ```

<@&286206848099549185>

[serene heath]

U mean continous?

[blissful wadi]

yes 😄

[serene heath]

Well yea the limit has to exist

And for that to happen both sides of the limit must be equal

[blissful wadi]

exactly what i was thinking but i got the right limit equal to 1

and the left one equal to sin(a)/3

a \in IR

[serene heath]

Uhh

What's ur function

[serene heath]

Bruh

Ok

Let's see

Is that a 2 at the bottom

[blissful wadi]

yes

[uncut mulch]

are those wedge shapes for x>1, 1s?

[serene heath]

Ok well that limit should be a/3 I believe

How did u get sina/3

[blissful wadi]

factorising the bottom

then simplifying

[serene heath]

Yea but sina tho?

Can u show working

[uncut mulch]

what's the original question?
is it meant to be sin( a(x-1))?

[blissful wadi]

https://media.discordapp.net/attachments/363224154469826562/640935128721391617/20191104_162625-1.jpg?width=1082&height=301

1 ) calculate lim 1+ and lim 1-

2. find a value for ( a ) such as f is continious by extension in 1

[uncut mulch]

sin a(x-1) is ambiguous

[blissful wadi]

its sina * (x-1)

[serene heath]

Oh

Really?

Weird

[uncut mulch]

because the way you interpreted it gets you sin(a)/3 at x→1^- which is bounded

[serene heath]

Ok then ur limit is fine

[uncut mulch]

and won't ever be equal to 1

also the function isn't defined at f(1) so it can't be continuous unless your missing some info

[blissful wadi]

yes so the original expression is wrong

that what i was thinking too just wanted to verify

[uncut mulch]

do you have a pic of where you got it from?

[serene heath]

Then a must be in the argument

[blissful wadi]

@uncut mulch my maths teacher 😂

[uncut mulch]

overall bad question

[blissful wadi]

yeah

just like the question before it where i have to workout a cubic equation ( that we didnt study yet ) 🤦

[uncut mulch]

right and left limits aren't enough to determine continuity

[blissful wadi]

its continuity by extention in a single point \notin Df

its enough i guess ?

[uncut mulch]

df?

[blissful wadi]

domaine

of definition

[uncut mulch]

didn't read the end part of the question. never seen it asked like that before

but yeh do it with a(x-1) in the argument

$\frac{\sin( a(x-1))}{(x-1)(x+2)}$

[obsidian monolith]

ramonov:

[blissful wadi]

one sec

i got it equal to plus or minus infinity

@uncut mulch

cant simplify bottom

[serene heath]

What is

Limit as x goes to 1?

[blissful wadi]

@serene heath x goes to 1- with a(x-1) in the argument

[uncut mulch]

do you know the:

$\lim_{x\to 0} \frac{\sin(x)}{x}$

[obsidian monolith]

ramonov:

[blissful wadi]

yeah

[uncut mulch]

good. do you think you can apply that here?

[blissful wadi]

how tho

[uncut mulch]

as x→1, (x-1)→0
which is very similar to your limit identity

but you have a in the argument, so what could you do to introduce a into the denominator?

[blissful wadi]

i could put a = x-1

but what about

1 / (x+2 )

[uncut mulch]

wdym

put a = x-1

[blissful wadi]

change in variables ?

or nvm that wont work

[uncut mulch]

the argument is a(x-1)
so you want to try and get a(x-1) in the denominator

which you can do by multiplying by a/a

$\lim_{x\to 1^-} \frac{\sin( a(x-1))}{(x-1)(x+2)} = \lim_{x\to 1^-} \frac{\sin( a(x-1))}{a(x-1)} \cdot \frac{a}{(x+2)}$

[obsidian monolith]

ramonov:

[blissful wadi]

oh

i see now

thank youu so muhc

much*

[harsh cipher]

Hello

In each factor, depending on their multiplicity it determines the behaviour of the graph.

question: How do I know if it will go through the y-intercept or goes back up.

[harsh cipher]

It's only different when y intercept is (0,0). 😛

[unkempt bone]

hey everyone

[hexed ermine]

@vague zephyr do you have work done

[vague zephyr]

I finished the problem

thank you though

[finite nimbus]

can anyone help me with this problem?

I assume you do 250=30t

since 30 is the max

but looks like im doing something wrong

[eternal cosmos]

Your time needs to be in seconds, but your speed is in miles per hour.

[finite nimbus]

So i first have to convert the mph to second per hour?

[eternal cosmos]

Well, your distance is in feet and your time is in seconds, so which unit should your speed have?

[finite nimbus]

X ft/sec?

[eternal cosmos]

Right.

So you'd need to do some conversion.

[finite nimbus]

alright, Im gonna try once more real quick

Nope, it still didnt work out for me

[eternal cosmos]

Did you round to two decimal places as specified? Your original answer was only to one.

[finite nimbus]

yes I did

i just got a huge number

that doesnt seem right

[eternal cosmos]

30 miles per hour = 30/3600 miles per second = (30/3600) × 5280 feet per second, right?

[finite nimbus]

yes

[gusty igloo]

So basically

My mind is broken I guess

My answer is there but it’s not correct

This is what mine looks like

The bottom has to stay how it is because both sides are heading in the same direction of both of the asymptotes

The only thing that needs to change is the horizontal asymptote needs to be 1 not 0, but when I try to even out the degree of the top and bottom everything else gets messed up

[hexed ermine]

what is your original question

[gusty igloo]

Use the graph to create a function

My friends are also stumped

@hexed ermine

[native sequoia]

try the form $\frac{p(x)}{(x+1)^2(x-2)^2}+1$

[obsidian monolith]

EpicGuy4227:

[native sequoia]

@gusty igloo

changing the squares into abs' is a bit better

[hexed ermine]

why +1

I have one that looks very similar

but its (x+1)^4(x-2)^2 in the denominatore

[native sequoia]

so that if deg(p)<4 then the limit as x to +-∞ is 1

[hexed ermine]

but its 0

oh wait im looking at the wrong pic

okay youre right

seems as if there is no symmetry around x=-1

so (x+1)^2 might not work

[native sequoia]

it does if you only care about the asymptotes and intercepts

as I said, using abs is a bit better

if you care about the shape

[hexed ermine]

wait nvm

no

no abs

just use a third degree in the numerator

[native sequoia]

abs is a bit better in terms of shape. if you use a third degree won't your derivative at x=-1 or 2 be bad?

[hexed ermine]

hmm

maybe making it (x+1)^4

yeah I got one that looks just like it

[native sequoia]

this is what I have, what about you?

[hexed ermine]

oh actually the one you have is really good

||I had (-x^3/((x+1)^4(x-2)^2)+1) but if I wanted to make the points line up im sure I can do a system solving for the coefficients a b and c of ((-x^3+ax^2+bx+c)/(x+1)^4(x-2)^2)+1)||

[gusty igloo]

It shouldn’t cross the horizontal asymptote on the right tho

Thank you guys for the help btw

[native sequoia]

@hexed ermine That actually gave me an even better result :o

[hexed ermine]

nice

I am close to doing it

[native sequoia]

@gusty igloo oh right my mistake, the first one should be good then

[hexed ermine]

Pretty sure I got it

idk

how does this look

yours looks better tho

because the hump on the left curve is positioned more accurately in yours

so the denominator is probably (x+1)^2(x-2)^2

so yeah, (-ax^3+bx)/((x+1)^2(x-2)^2)+1

use a system to solve for a and b

[willow bear]

what's h

[elder edge]

u can call e^x as T

and solve for T

[willow bear]

you can take the log of both sides but it'll be largely useless as you'll end up with $$\ln(2e^x + 5) = -x + \ln(3)$$ with no real way to move forward from there

[obsidian monolith]

Ann:

[willow bear]

no you don't have to do anything ever

[elder edge]

you dont have to

it just makes it simpler to look at

[willow bear]

a substitution is just the most straightforward method

[proven dove]

say I was looking for the domain of 6x/x^2+1

i can see the domain is (-∞,∞)

but if i needed to show work, would what -i and i mean?

would getting -i and i be the same as -∞ and ∞?

[willow bear]

no

first off

parentheses

\verb|6x/x^2+1| reads as $\frac{6x}{x^2} + 1$, not as $\frac{6x}{x^2 + 1}$.

[obsidian monolith]

Ann:

[willow bear]

second, no, -i and i are not the same as -∞ and +∞

if you solved the equation x^2 + 1 = 0 and got ±i, all it means is that x^2 + 1 does not equal 0 for any real x.

and that means that the only thing that could exclude points from your domain does not exclude points from your domain.

[proven dove]

ok

thanks

[spark lance]

Hello! How do I simplify tan(x)sin(x) + sec(x)cos^2(x)?

[patent beacon]

@spark lance
Convert everything to sin and cos first

[flint river]

sin x image is [-1; 1]... that is
-1 <= sin x <= 1

what about sin^2 x???
I can't just do: (-1)^2 <= sin^2 x <= (1)^2

[patent beacon]

It's awkward to think in terms of inequalities here

If something is in the range of sin(x), then its square is in the range of sin²(x)

@flint river

[spark lance]

here's where I'm at now

wait I think I figured it out

[flint river]

ill be at #help-0

[vocal pivot]

hi

quick question

[rigid sun]

69

That’s funny

[vocal pivot]

A circle has a radius of 7in. Find the radian measure of the central angle θ that intercepts an arc of length 15in.
Do not round any intermediate computations, and round your answer to the nearest tenth.

lol

its been my name ever since i saw avengers

[rigid sun]

Wdym

just do it

How many radians is 15in?

15/7 radians

Done

Remember, 1 radian corresponds to the angle formed by 1 radius on the circumference of the circle

no

Decimal form

15/7

[vocal pivot]

2.14

well 2.1

[rigid sun]

sure

[vocal pivot]

thank you man ❤️❤️❤️❤️❤️❤️❤️❤️❤️

[opal swallow]

https://gyazo.com/23ddefa1cada53da24f3272315d33a52

ayone help me please

[calm kraken]

Have you tried plotting/graphing the equation?

[frail zephyr]

https://gyazo.com/290510acdba2dbe5e960deda3bffb6e6

anyone know how to solv ethis?

[calm kraken]

Try drawing it out

Essentially you have a diagonal line which you can use to create a triangle in the 3rd quadrant

[novel vortex]

yeah what he said

[final mauve]

Geez

I suck at reading

If only I remembered the term terminal

Otherwise I would have gotten the answer

[swift wagon]

@opal swallow Use implicit differentiation. Factor out dy/dx. Set it to 0. Then solve.

[opal swallow]

@swift wagon thank you bro

been waiting

for osmeone to help me out

[swift wagon]

Yea that should work

[opal swallow]

wanna be my tutor?

ill compensate you for your time

[swift wagon]

Lol, I’m only a year 11 student

[opal swallow]

ima year 10 student

wait

11 years old?

[swift wagon]

No

17

Haha

[opal swallow]

youre a junior?

hs?

[swift wagon]

High school yea

[opal swallow]

dude

[swift wagon]

In Australia

[opal swallow]

youre set man....

you can implicit at junior

damn

ap calc im guessing?

[swift wagon]

Umm, We have different things in Australia

But I’m doing calc rn

Extension 2 math we call it here

[hexed bolt]

Just a simple question

[willow bear]

what is giving you trouble

[hexed bolt]

I just don't know how to do it

Like i dont know how I should write it in terms of x

[willow bear]

does the "SOH CAH TOA" mnemonic ring any bells

[hexed bolt]

Yes

[willow bear]

okay so then surely you understand that cos(θ) = 4/x

[hexed bolt]

Yep

[willow bear]

so can you rewrite this so that it uses an inverse trig function

[hexed bolt]

How would I do that

[willow bear]

do you know what the inverse trig functions are

[hexed bolt]

Ye

[willow bear]

so

[hexed bolt]

Arcsin, arccos and arctan

[willow bear]

well

what do you know about arccos

what is the definition of arccos

[hexed bolt]

Just the inverse function of cos

[willow bear]

there are some technicalities behind that but yes

so

you know what an inverse function is

right

[hexed bolt]

Yes

I just need help applying it

[willow bear]

you either know it or you don't. if you know it then you should be able to apply it.

$f(x) = y \iff x = f^{-1}(y)$

[obsidian monolith]

Ann:

[hexed bolt]

Oh ok

[willow bear]

was this news to you

if yes then you did not in fact know what an inverse function is

[hexed bolt]

So if cos(θ) = 4/x then θ = cos^-1 (4/x)?

[willow bear]

well there you have it

[hexed bolt]

Ok thanks

[swift wagon]

Umm yea. But after doing a lot of trig questions, it becomes easy.

[lilac pier]

or

you can learn the graph 😛

i do it that way

nvm

the graph mainly focuses on values of 0, pi/2, pi, etc

in b/w them, like angles of 30 or 60 you still have o learn

RA?

make it easier

pi is 180 degrees

so they're asking for 300 degrees

yeah

and we know that sine is negative in the fourth quadrant

and we know that the whole cycle is 360 degrees

cancel that from 300, you get 60 degrees, that small angle you left out

we know sin(60) = sqrt(3)/2

but since it's negative in the fourth quadrant, it'll be -sqrt(3)/2

sqaure root

ill show yuo

what do you mean?

you dont even need the calculator

pi is 180, so you divide that by 3, gives you 60

then 60 x 5 = 300

and we know the whole cycle is 360 degrees

[willow bear]

"pi is 180"

[lilac pier]

cancel that out, we get 60 degrees

180 degrees* 😛

[willow bear]

pi radians = 180 degrees yes

[lilac pier]

^

[willow bear]

the relationship between radians and degrees is almost the same as between feet and inches

they're two different units for the same quantity

the conversion rate between them is 1 degree = π/180 rad

but ppl overthink this a lot

[lilac pier]

and then sin(60) = sqrt(3)/2, but since it's in the fourth quadrant, you get -sqrt(3)/2.