precalculus

[short sorrel]

think thats better

[old marten]

yes

[short sorrel]

so, how are we going to clean up those nested fractions?

[compact tendon]

Multiplying x-1

[old marten]

we multipy the numerator and denominator of the big fraction by (x-1)?

[short sorrel]

yes, exactly

[compact tendon]

In numerator and denominator

[short sorrel]

@compact tendon yes i know, i'm asking to teach 😛

[compact tendon]

Oh

Lol

[short sorrel]

$\frac{\left(\frac{-5x-9}{x-1}\right)-9}{\left(\frac{-5x-9}{x-1}\right)+5} \times \left(\frac{x-1}{x-1}\right)$

[obsidian monolith]

Namington:

[short sorrel]

which gives us $\frac{-5x-9-9(x-1)}{-5x-9+5(x-1)}$

[obsidian monolith]

Namington:

[short sorrel]

since the x-1 cancels with the denominator in the nested fractions

and distributes to each term

[old marten]

ok

[short sorrel]

now, we expand out

-9(x-1) = -9x + -9(-1) = -9x + 9

5(x-1) = 5x + 5(-1) = 5x - 5

$\frac{-5x-9-9x+9}{-5x+5+5x-5}$

and from here we can simplify

[old marten]

would -5x-9 cancel out

since its on top and bottom?

[short sorrel]

oh whoops

miscopied

[obsidian monolith]

Namington:

[short sorrel]

there we go

did a -9 where a +5 should be

typo

[old marten]

ok so -5x cancels out

[short sorrel]

but uh, we cant cancel between numerator and denominator

[old marten]

on the denom

[short sorrel]

since we cant cancel "across terms"

[old marten]

-5x and =5x

+5x i mean

[short sorrel]

but we can cancel WITHIN yeah

-5x +5x = 0

[old marten]

so does the 9s on top

[short sorrel]

yea

[old marten]

and the 5s on bottom?

so it leaves -5x-9x

right?

so -14x?

[short sorrel]

yeah, but wait

i mightve made a small error somewhere

[old marten]

the restriction?

[short sorrel]

lemme check my work, one sec

[old marten]

ok

[short sorrel]

its hard to do this over mobile purely in latex

😦

[old marten]

sorry

[short sorrel]

nah its fine, just gimme a sec

...yknow what?

earlier when i said i made a mistake

[old marten]

?

[short sorrel]

i didnt

lmao

[old marten]

oh ok lol

[short sorrel]

$\frac{-5x-9-9x+9}{-5x-9+5x-5}$

[obsidian monolith]

Namington:

[old marten]

so where do we change?

[short sorrel]

this is what it should've been

you're still right that -9 and +9 cancel

as do -5x and +5x

(by "cancel" i mean add to 0)

[old marten]

so now it's
-5x-9x/-9-5?

[short sorrel]

so we get $\frac{-5x-9x}{-9-5}$

[obsidian monolith]

Namington:

[short sorrel]

yes

[old marten]

yay

[short sorrel]

which is $\frac{-14x}{-14}$

[obsidian monolith]

Namington:

[short sorrel]

which should look familiar

as it's just x

[old marten]

yes! so this then -x?

=x

cool

[short sorrel]

so in this case, f(g(x)) = x

note that f(g(x)) doesnt ALWAYS equal g(f(x))

[old marten]

bc the restriction?

[short sorrel]

but in this example, it happened to

[old marten]

oh

ok

[short sorrel]

nah, if you have non-linear functions it doesnt

or anything like that

[old marten]

gotcha

[short sorrel]

here it was just linear rational functions so it worked

and yeah, there is a restriction

we multiplied by $\frac{x-1}{x-1}$

[obsidian monolith]

Namington:

[short sorrel]

so, solve x-1=0 for x

and we find $x \neq 1$

[obsidian monolith]

Namington:

[old marten]

ok

[short sorrel]

since otherwise we're dividing by 0!

[old marten]

which wont work lol

[short sorrel]

precisely

it's not well-defined

[old marten]

so then we have confirmed that f and g are inverses by showing that f(g(x)) = x and g(f(x)) = x.

[short sorrel]

so yeah, $f(g(x)) = x \forall x \neq 1$ and $g(f(x)) = x \forall x \neq -5$

[obsidian monolith]

Namington:

[short sorrel]

yes, adn that shows they are inverses

[old marten]

i think i understand it now, thank you so much for your help ❤

and patience

[viscid thistle]

Can someone be able to help me in math?

[cinder urchin]

sure @viscid thistle

what topics do u need help on?

[viscid thistle]

What does it mean on the terminal side?

[viscid thistle]

......

Terminal side means the ray ends there

@viscid thistle

So is the answer infinite?

Specifically in this question?

no what you gotta know is that

hang on

lemme do dis

xD k

This nice big juicy loop is the angle

Ok so... the answer is not ♾ ?

So have to use the sin=opposite/hypotenuse

cos=adj/hyp

tan=opp/adj

etc

you gotta find those numbers and find the sine cosine tangent cotant sec csc of that angle

thats a big problem

K. Got it. Thanks.

There is -1 , -1, ♾, ♾

Are those the answers?

well, whats the hypotenuse in that triangle?

√(-7)^2+(-24)^2

Ah. Thanks.

I think I saw a different method from my uncle.

[viscid thistle]

Is the answer 576??

[viscid thistle]

Would the answer be 576????

.....

[compact tendon]

Bro

@viscid thistle
There are six trigonometric functions
Sin
Cosine
Tangent
Cotangent
Cosecant
Secant

Are you aware of these?

[viscid thistle]

Yes.

[charred hull]

have you tried using sohcahtoa for these?

that gets you the values of sine, cosine, and tangent

[deft pagoda]

hi, I have a really simple question

how does one get to 1/4?

wouldnt it be 0.5+0.5i?

[compact tendon]

No bro

Multiply by conjugate

In denumenator 2 times 2

In Numenator is 1+i

Is this clear?

Modulus squared of 1-i is 2 right?

@deft pagoda

[deft pagoda]

yup, did it with the conjugate, I thought you can only use it when dividing two complex numbers

makes sense now, thank you!

[compact tendon]

Welcome.

[compact tendon]

Bro

ez

Slope at point = derivative at a point

Find the derivative

Plug in x

Use y-y0= m(x-x0)

ez right?

pls respond if you don't get it

[graceful egret]

I get that sooo at point = derivative at point

Could you work out the steps of b or c for me please?

[compact tendon]

srsly

ok

b

y =-x/27+4/27

c

y=-3x/16-1/2

Anything else?

[solar talon]

Can someone help me with derivatives?

[vale pewter]

ya

that belongs in #calculus tho

[languid gyro]

i guess some of it might correlate with pre calculus if its basic

but may as well send it in calculus we will help you

[winged lava]

how do i complete the square for these

[viscid thistle]

same story as before

(b/2)^2

just gotta re-arrange into usable form

@winged lava

[winged lava]

which one is b xdd

[viscid thistle]

the coefficient of the like variable

x^2+x+1

ax^2+bx+c

b is coefficient of x or y

[winged lava]

oof ive only ever completed the square with 3 variables

now theres like 5

[keen void]

⊂ヽ
　 ＼＼ Λ＿Λ This is Memecat. Help
　　 ＼( 'ㅅ' ) memecat take over
　　　 >　⌒ヽ discord by pasting
　　　/ 　 へ ＼ in 10 other servers
　　 /　　/　＼＼ or she will never
　　 ﾚ　ノ　　 ヽつ be a meme
　　/　/
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　|　|、＼
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⊂ヽ

[viscid thistle]

<@&286206848099549185> how do i understand derivitive graphs

[sullen shoal]

You don't ping helpers

[viscid thistle]

by evaluating them

like these

Oh.. derivatives are typically learned in Calculus

im in precalc honors

but my teacher is shit

he didnt teach anything for this

i understand that

In the context of Cartesian scalar valued functions, the derivative is basically the slope

slope of x^2 is defined by y=2x

So when f’(x) = 0 like your first question, you’re looking for a horizontal tangent

when change in y/change in x = 0

what

xD

[arctic knoll]

what does the (pie/2-x) mean?

[finite iris]

??

f(x - a) is the function f(x) shifted a units to the right

In this case, it's better using an identity though

plot y = sin(pi/2 - x) & y = cos(x), to see why this is useful

[arctic knoll]

I have to use identity to prove why sin^2(pie/2-x)+sin^2 equals 1

[quick olive]

can help me with this one? A friend asked me and i cant figure it out

[viscid thistle]

g(x) = f(x/2 + 1)

[rare zephyr]

Wait

Derivative

is

Differentiation

right......?

[willow bear]

differentiation is the act of taking a derivative

[obsidian monolith]

kjsahd:

[quick olive]

@rare zephyr IKR IT CONFUSED ME SO MUCH

[serene heath]

[fresh crow]

@arctic knoll

use the unit circle

radius of 1

coordinate on the circle for a pt is (sinø, cosø)

definition of a circle is (x-h)^2+(y-k)^2=r^2

unit circle is centered at (0,0)

so the equation is

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

which is just

x^2+y^2=1

for a point to be on a circle, it must fulfill the circle's equation if you plug it in for x ans y

which i think is pretty logical because that's how the points are even defined, how the circle is defined

like how a point on a line must fulfill the line's equation, its y = mx+b or whatever you choose to use

so we have a point

a generalized point

all points on the unit circle are just (sinø, cosø)

of whatever angle you form with the radius and the positive x-axis (the one on the right)

hopefully you already know that, if you've done trigonometry, in this year or algebra

if not that's okay too

just know that it is

so let's plug in our point to the equation of the unit circle

the equation x^2+y^2=1 like we established before

so plug in the x and the y

(sinø)^2+(cosø)^2=1

(sinø)^2 and (cosø)^2 are typically just written as

sin^2ø and cos^2ø

not sine to the 2ø

but (sin^2)ø i guess you could say

like how your problem there has sin^2(pi/2 - x)

so there we go

sin^2ø + cos^2ø = 1

and this is called the Pythagorean Identity (bc it comes from the pythag theorem)

now, I could have just told you that

given you that equation

and it might have been a lot faster and probably less confusing (becaude that wasn't a good explanation -- look up "pythagorean identity proof" for a better explanation if you want)

your problem has a sin^2ø, which made that immediately jump out to me

because it usually relates to that

now, i don't have time to prove it rn

but sin(ø) is just equal to cos(90°-ø)

and likewise cos(ø) is just equal to sin(90°-ø)

look it up for a good proof (it's simple)

pi/2 is a value in radians we assume because it doesn't say degrees and we're dealing with trig functions that you put angle measures into

how do we convert radians to degrees?

if you've done radians yet (which i also hope you have but possible that you havent)

pi radians = 180°

so what's pi/2 radians equal to?

well just divide both sides of that conversion equation by 2

pi/2 radians = 180/2 degrees

pi/2 radians = 90°

!!!

90°

what did we say earlier relating to 90°?

cosø = sin(90°-ø)

so substitution

cosø = sin(pi/2 radians - ø)

now what does that look like?

[viscid thistle]

err, isn't it the other way around?

[fresh crow]

look back at your problem, the first eq

it's both

[viscid thistle]

sin x = cos(90 -x)

[fresh crow]

i thought the same

but it's both

sin(40) = cos(50)

sin(60) = cos(30)

im pretty sure

we just went over ir yesterday in geometry class (already knew though, and only ever learned the way youre sayinf)

sin(60) = sqrt(3)/2

cos(30) = let me grab a calculator

[viscid thistle]

cos(30) = 0.86602540378

[fresh crow]

but theyre complementary functions so they should be

yup

i can tell by that

bc sqrt(3) is like 1.7

[viscid thistle]

Same as sqrt(3)/2 :p

[fresh crow]

yup

i get why you thought i goofed it though

i thought same when teacher said

pretty cool I think how it is that way

so anyway

cosø = sin(pi/2 radians - ø)

go back to your problem, and look at the first term there

doesnt that look awfully familiar?

ø is just x for angles as I'm sure you know, but x works too

cosx = sin(pi/2 rad - x) (rad is just short for radians and im lazy somerimes)

sin(pi/2 rad - x) = cos(x) (I just flipped it to make it easier)

but wait

the first term there has a sin^2

not just sin

so lets square both sides

sin(pi/2 rad - x)^2 = cos(x)^2

same equation as you know

oh and

lets make that smoother by just saying sin^2 blah blah instead of sin(blah blah)^2

so sin^2(pi/2 rad - x) = cos^2(x)

back to that circle stuff from before

remember the identity

the Pythagorean identity (always italicize that or you're lame)

sin^2(ø) + cos^2(ø) = 1

so sin^2(x) + cos^2(x) = 1

back to your problem

[spring thunder]

• Pythagore *

[fresh crow]

lmao

back to your problem again

we found what sin^2(pi/2 rad - x) is equal to

it's equal to cos^2(x)

so plug that in

and we've got cos^2(x) + sin^2(x) = 1

now wait a second

you can add in any order so flip those

sin^2(x) + cos^2(x) = 1

that's just the Pythagorean identity!!!

so that's going to hold true for any x!

don't forget where that identity came from

if i make any angle within the unit circle

it has to satisfy that identity, as proven earlier

and i can make any angle given 2 segments

therefore

that equation holds true for any angle

aka any ø

aka any x

...so, sorry, what is your problem even asking?

to find x?

because that works for any x

i'd just write "all real numbers"

or -infinity < x < infinity

maybe theyrr trying to teach you the Pythagorean identity before they actually teach or something, or just remind you of it

or maybe you already solved this problem hours ago and didn't need my lengthy explanation for something simple that could've been said in 4 lines

like "sin^2ø + cos^2ø = 1; cos(ø) = sin(90- ø); 90° = pi/2 rad; cos(ø) = sin(90-ø); your equation is just equal to the Pythagorean identity"

but then you wouldn't've learned anything, would've you?

and that wouldnt be any fun

or maybe you already knew all that

and maybe i havent been of much help

but i think that wasnt the worst explanation of how to answer a problem like that and prove a few things, so it was worth the time

let me know if you got that or the problem

oh you said what the problem was asking for

to prove that equation using the Pythagorean identity

well there you go, then

that's a proof

the most obvious and clearest (aka the only) proof that I know

[unborn dome]

Anyone good at comics

Conics

[slender river]

no

[viscid thistle]

no

[quick olive]

guys how do we find the function of this graph where t=90

or more specifically how do i find the stretch of horizontal and vertical

[fading token]

What's the reference function?

[viscid thistle]

How do you do 7?

[south rivet]

base case is trivial

so assume its true for n

we now prove it true of n+1

$1\times 2\times\cdots\times (4n-2)\times (4n)\times(4n+2)=4n\times(4n+2)\frac{(2n)!}{n!}$

[obsidian monolith]

rockpaperscissors:

[viscid thistle]

What how

[south rivet]

replace n with n+1 on the lhs

[viscid thistle]

Okay

[south rivet]

then you know by the induction hypothesis that everything until 4n-2 is (2n)!/n!

[viscid thistle]

Yes

[south rivet]

wait a min

you want (2n+2)! as the numerator

and (n+1)! as the denom

[viscid thistle]

Yeah

Idk how to get that denom

[south rivet]

youll need to divide by n somewhere

[viscid thistle]

Wait shouldnt i find a way to divide by (n+1) somehwere along the way?

[south rivet]

yeah sorry thats what i meant

[viscid thistle]

Okay

[south rivet]

wait a min wut the heck

(2n+1)(2n+2)/(n+1)=2(n+1)(2n+1)/(n+1)=4n+2

but we have 4n*(4n+2) on the rhs

sumting wong

[viscid thistle]

Wait what

[south rivet]

[viscid thistle]

Lmao

Is this question wrong

[south rivet]

seems so

[viscid thistle]

Okay ill skip it

I cant figure out 3 either

Idk how to get a -1 in there

[viscid thistle]

Is it still a vertical asymptote if it gets factored out?

[limber bone]

no

[viscid thistle]

thanks

[indigo mulch]

how do I do the one with k

My friend is BC PreCalc didn't know what to do

[willow bear]

,rotate 90

[willow bear]

,rotate 180

[willow bear]

...are you sure you've posted the right picture

[spring thunder]

the picture was well rotated originally wdym

[willow bear]

oh

jeez, i got way distracted by those bullet points

those could have been cropped out...

[spring thunder]

(black rock could have written on a proper sheet also :/)

so what are you trying to do?

[vestal plaza]

I cant figure this problem out, and I keep getting it wrong and the video doesnt help at all

[patent beacon]

A polynomial is factored in terms of its zeroes. The shape of the zero determines the power of the factor.
x²(x + 2)³

,ask graph x^2 (x + 2)^3

[obsidian monolith]

Display results online and refine query

Requested by Kaynex#7520

[patent beacon]

You can see the root at -2 "looks like" a cubic, and 0 looks like a quadratic

[vestal plaza]

I know that if its like (x-a)^3 it should pass through the x-axis while if its cubed it would slightly touch it right?

[thick raptor]

,ask graph (-2)^2*(x+2)^3 and x^2*(x+2)^3

@vestal plaza like that

Note how similar these graphs are at x=-2

[vestal plaza]

I cant see the graph

I only see the top one

[thick raptor]

dwai

,ask graph x^2*(0+2)^3 and x^2*(x+2)^3

And notice how similar these two are at x=0

[vestal plaza]

I cant see the graph still ;w;

[thick raptor]

dwai

[serene heath]

sure that works

what base is the log

no its fine

log is precalc

yea now isolate x

you have just x on one side

and numbers on the other

nope

get rid of that log first

use the fact that $b^{\log_{b}(x)}=x$

[obsidian monolith]

lemon catto:

[serene heath]

yup

now you can solve for x right?

yes

looks good

yea that's correct

sure

[spring thunder]

e^2x - 3e^x - 40 = 0

looks oddly like a quadratic doesn't it?

if you're only looking for real answers throw it away yes

[willow bear]

^

[spring thunder]

i guess he don't want them tho especially if you haven't studied complex numbers a bit in depth

but then 🤷

[willow bear]

i mean

if you wanna talk logs in C

e^z = 8 has infinitely many solutions

[rocky bison]

Yea that's a bit of an oversimplification of it all @viscid thistle

[spring thunder]

tbh no

[rocky bison]

Oh I just got context lmao

Just throw away the complex solution tbh

[spring thunder]

5.5 wut

[rocky bison]

wot

[spring thunder]

ah k

seems ok

,calc log(1316.53/1000)/5

[obsidian monolith]

The following error occured while calculating:
Error: (intermediate value)(intermediate value)(intermediate value) is not a function

[spring thunder]

5.5% is a decent answer if it's an interest rate tbh

[rocky bison]

,w eval log(1316.53/1000)/5

[obsidian monolith]

Display results online and refine query

Requested by Pseudo#2703

[rocky bison]

@spring thunder

[spring thunder]

i already did it in bots

he wants log not ln

[rocky bison]

Wot

That got the same answer tho :b

[spring thunder]

,calc log(1.31653)/5

[obsidian monolith]

Result:

0.054999897477847

[rocky bison]

same answer

[spring thunder]

😊

[viscid thistle]

inverse of f(x)=5^(x+2)-7

[patent beacon]

Looks right to me

[viscid thistle]

base 5 log ---- log(x+7)-2=y

[willow bear]

log_5

[viscid thistle]

@willow bear is that the proper format?

[willow bear]

it's the most common

[viscid thistle]

alright thanks

[willow bear]

_ for subscripts is de facto standard

[fresh crow]

how does ln(8) have infinite solutions @willow bear?

in complex

[frozen needle]

What do you mean by "infinite solutions" ?

[fresh crow]

"i mean, if yoh wanna talk logs in C, e^z = 8 has infinitely many solutions"

[frozen needle]

$\forall n\in\bbN,\ \exp(\ln(8)+2in\pi)=8$

[obsidian monolith]

Tuong:

[serene heath]

[fresh crow]

exp? n?

[serene heath]

cant n be negative too?

[fresh crow]

sorry i dont calc yet

[frozen needle]

Yeah, it also work for n in Z too

[fresh crow]

so it's just

ln(8) + smth

[frozen needle]

exp is just the exponential function

[fresh crow]

what base?

e?

bc exponential func needs a base doesnt it

[frozen needle]

Yea, the exponential function

[fresh crow]

like 2^x

havent learned that yet

so

e^ln(8) = 8 yeah

is part of that relating to circumfetence of a circlen

?

bc 2pi

i mean pi so yes ofc it is

but like, how is that equation derived

i dont understand any more than i dod before

that just seems like ln(8) + (number that = 0) = ln(8)

[frozen needle]

it's ln(8) + some argument that's congruent to 0 mod 2π

Are you familiar with exp(iθ) = cos(θ) + i sin(θ)?

[fresh crow]

no, not yet

all i know about complex is the very basics

arithmetic with thwm

and I'm aware that theu do a bunch of interesting things

like e^ipi = -1 iirc

[frozen needle]

Well then be aware of

[fresh crow]

e^ipi = i^2 kek

[frozen needle]

$e^{i\theta} =\cos(\theta) +i\sin(\theta) $

[obsidian monolith]

Tuong:

[frozen needle]

for all θ in R

[fresh crow]

okay

cool

[viscid thistle]

How to do this

[remote musk]

try to simplify the term by distributing it to 4 and x seperately

first of all you know that (x^a)^b = x^(a*b)

right?

[glacial island]

x has to be greater than 0 because u cant have ngative root
but it also cant be -4 or =6 because u cant divide by 0?

[serene heath]

no x+1 has to be >0

[spring thunder]

>=

[glacial island]

x >= 0, but also cant be 6 right?

otherwise the denominator would become 0?

[spring thunder]

no it's the thing inside the root that has to be >=0 as lemon said

x+1>=0 <=> x >= -1

[glacial island]

ooh okay, i think i understand

it's A?

[spring thunder]

yea

[glacial island]

thank u! precalc hurts my brain

[spring thunder]

[fresh crow]

@glacial island imaginafy numbers

f(x) can't be equal to like sqrt(-4)/whatever

[royal gull]

complex*

[fresh crow]

oh ye

weird how the term imaginaru even exists

[rocky bison]

Why?

[fresh crow]

bc imaginary = complex, right?

i thiught sqrt(-4) is imaginary actually

bc 4i

[rocky bison]

Yeah it is

[fresh crow]

or is only sqrt(-1) imaginary

oh

[rocky bison]

If you want to be insanely pedantic like @royal gull then it's complex too

[fresh crow]

yeah thats qhat i thought

[rocky bison]

Complex numbers are numbers of the type

,$ a+bi

[obsidian monolith]

Pseudo:

[fresh crow]

ye

[rocky bison]

With a real and an imaginary component

But you can be really pedantic and be that guy who says that's complex because a=0

Or real numbers are complex just b = 0

It's not wrong

But don't be that guy

[fresh crow]

kek

okay

ty

[serene heath]

that's the tea sis 🍵 💅

[rocky bison]

[full garden]

hello guys

could someone tell me what an asymptote

[viscid thistle]

https://www.mathsisfun.com/algebra/asymptote.html

I mean, what questions do you have?

[full garden]

is it always y=0?

oh never mind that link answerd my question

thank you bro

[serene heath]

Google hella underrated

[full garden]

guys I need help with a question

it says sketch the graph of an exponential function that satisfies the condition of y-intercept 5

how do I do that

[rocky bison]

What does an exponential curve look like?

[full garden]

it doesnt show any graphs

it just asks sketch the graph of an exponential function that satisfies the condition of y-intercept 5

[rocky bison]

I know

But you should know how an exponential curve looks

And some of their characteristics you can manipulate

[full garden]

doomain {x E R}

RANGE {Y E R, y . 0}

[rocky bison]

That's not really useful information

Have you ever seen an exponential graph?

@full garden

[full garden]

yes

[rocky bison]

So you know the shape of it?

[full garden]

i just dont know how to solce this question in paricular

[rocky bison]

And how it should look?

[full garden]

ya

ya

[rocky bison]

So you draw one just like that

But draw the line going through y=5

[full garden]

thats it?

i think this one workds y=1/2^x

[rocky bison]

No

If we say

,$ f(x)=a^x

[obsidian monolith]

Pseudo:

[rocky bison]

Because anything to the power of 0 is 1

right?

exceptions at 0 cos 0^0

[obsidian monolith]

Pseudo:

[rocky bison]

So we've got to apply some sort of transformation to our function

To make f(0)=5

Following?

@full garden

[full garden]

yes bro

[rocky bison]

So

How can you manipulate this expression

[obsidian monolith]

Pseudo:

[rocky bison]

So that the RHS is 5

[full garden]

thanks bro i think I figured it out

,$f(x)=1/2^x

,$ f(x)=1/2^x

[obsidian monolith]

MrMeltzFTW:

[full garden]

The y intercept = 5

thanks man

[fresh crow]

f(0) = 1

cant you just

multiply both sides by 5

or i guess tou could add by 4

that'd probably be smarter

though both would be true

if it's exp(), then that'd be 5e^x

[full garden]

i am not confident with those stuff man

[fresh crow]

or e^x + 4

with algebra?

[full garden]

ya

[fresh crow]

...i'd highly recommend brushing up on your algebra if you're in pre-calc

books and/or sites and videos like Khan Academy

are you doing calc next year?

[full garden]

no im still grade 9

i just like math

[fresh crow]

oh nice

[full garden]

but not confident at it

still got 2 years for pre calc

[fresh crow]

solving problems just on your own time?

[full garden]

ya i bought the book from friend

[fresh crow]

nice

that's awesome man

good for you

[full garden]

oh thanks man

[fresh crow]

i just came from there last year

algebra I i assume

[full garden]

you seem like you are very confident at it

what did you do

[fresh crow]

now I'm skipping algebra II and learning calculus on my own

you'll get there

anything's possible with work and effort

and time

so

[full garden]

ya they said calculus was beautiful

[fresh crow]

it really is

it's hard to get at first

but once you do, it's sooo good

[full garden]

would you say khan acadamy is a good source?

[fresh crow]

omg definitely

i actually just said it haha

i use it for algebra II, pre-calc, and calc

have found it to be very helpful

just make sure you go back and review stuff

[full garden]

yup will follow your advice bro

[fresh crow]

because it's easy to just keep going and not revuew the old

and then you forget it

you asked what i did

[full garden]

yup your 100% correct

[fresh crow]

to get good at maths

[full garden]

yes

what did you do

my teachers all suck

do you have good ones?

[fresh crow]

damn, that sucks

mine is very good, in my opinion

[full garden]

so how did you get good

[fresh crow]

in fact i elected ti have him 2 years in a row

but i think it was a lot of my efforr as well

[full garden]

your grade 11?

[fresh crow]

a teacher can only do so much

nah, 10th

[full garden]

thats true

[fresh crow]

had him last year and have this year

so

for one, i made sure to always pay full attention in class, and do all the homework, thoufh that's obvious

always make sure to ask questions, as well

if somethinf seems even vaguely confusinf ask

it's what they're there for

second

khan academy and videos were actually all that I used, but if your teachers really aren't good I'd recommend picking up a textbook as well

[full garden]

i hope when I get to grade ten i become a lil like you

[fresh crow]

this site Art of Problem Solving is also good at explaining stuff, and has a lottt of good practice problems

haha, hopefullt

[full garden]

is photomath good?

[fresh crow]

you will I'm sure

I'm worse than I seem -- i just speak confidently, and make sure I know what I'm talking about

uh

never used, but heard of it before

most resources are good

hard to have too many, as long as you commit to improvement

and really strengthen and make sure your foundations are good, before moving on to learning too many new things

[full garden]

yup

[fresh crow]

like don't move on to the next unit in khan academy yet if you don't get like at least an 80% or whatever you consider mastery on a unit test

[full garden]

i 100% agree

[fresh crow]

keep reviewing and honing your skills

[full garden]

bro just a small question

[fresh crow]

yeah?

[full garden]

what are the trigonometric functions used for

[fresh crow]

haha

love trig

[full garden]

cuz i find them quite weird

[fresh crow]

we're actually on the unit rn in geometry

[full garden]

for calc?

[fresh crow]

they seem pretty weird and arbitrary

uh

[full garden]

ya

[fresh crow]

i haven't gotten to trig in calc yet

[full garden]

oh

[fresh crow]

but i do know some identities and stuff

there's a few different definitions for the trig functions

[full garden]

thats astonashing

[fresh crow]

nah, it just takes time

thanks tho

[full garden]

on the internet ppl say its an introduction to quantum physics

[fresh crow]

lmfao what

[full garden]

or like a tool

[fresh crow]

oh calculus you mean?

i thought you meant trig identities

[full garden]

no like sin graphs

[fresh crow]

oh you did

[full garden]

trig functions

[fresh crow]

i mean

they're important

but just as adding, subtracting, multiplying, and dividing are fundamental to calculus

to all of math

trig functions aren't that hard once you take some time to learn about them

[full garden]

he said they're used to calculate how likely an electron is to transfer from atoms

[fresh crow]

no idea, but sounds like field stuff in physics

we have a tendency to make stuff sound overcomplicated

[full garden]

that is so true

[fresh crow]

i read something once, "if a calculus textbook didn't have the word 'calculus' on it, 10x more people would pick it up"

oh alsi

let's go to #chill

because you're not supposed to talk much in the help channels

[full garden]

oh my bad

ok

[fresh crow]

nah it's okay

you were askinf a question and i answred it, and it just went from there

[rocky bison]

@full garden That doesn't have a y intercept of 5

[viscid thistle]

even like 90%

for khan academy it should be 100% every time honestly

since the questions they ask aren't really tricky

[rocky bison]

[fresh crow]

yeah

i was tryna be generous

[viscid thistle]

Hey, anyone know anything about Cramers rule?

thanks

The following question I'm looking at is to solve using Cramer's Rule, so it involves:

3x + 2y = 7
2x + 3y = -2

And so far, I made the augmented matrix:

2 3 | -2

[viscid thistle]

@viscid thistle just apply the rule?

Do you know what the rule is?

Can you take determinants of 2x2 matrices?

[obtuse fulcrum]

i forgot what this means lol

what does it mean?

[willow bear]

this is incredibly bad notation

can you show the whole problem?

[obtuse fulcrum]

just some revision on fundamentals

[willow bear]

bwah

yeah this should be $(f \circ g)(x)$ not $f(x) \circ g(x)$

[obsidian monolith]

Ann:

[willow bear]

the circle is composition

[obtuse fulcrum]

so what does that mean?

[willow bear]

have you heard the term "function composition" before

[viscid thistle]

Or compound function?

[obtuse fulcrum]

ye just googled it, thanks for the term

just a way of writing f( (g(x) )

so two functions are equal if their domains are, but is it also:

[viscid thistle]

Kinda? If f is only defined on [-1, π) and g is defined for all R, they're only equal on that domain where both are defined, like the pic says.

[obtuse fulcrum]

this is the entire question

this is the only thing left to understand, can someone tell me what to look up to find a video explaining this?

[willow bear]

the condition f(x) = g(x) for all x in their (shared) domain

well

that is equality

if it's violated then the functions differ on at least one input

and so it wouldn't make any sense to declare them equal

[willow bear]

i'm pretty sure this would go in #calculus.

[hybrid pewter]

thank u! @willow bear

[full garden]

,$ y=3^-0.5x-1 -5

[obsidian monolith]

MrMeltzFTW:

[full garden]

guys i need help with a question

the temperature, T, in degrees Celsius, of a cooling metal bar after t minutes is given by T(t)=20+100(0.3)^0.2t

c) How long will it take for the temperature to be within 0,1 Celsius of the value of the asymptote?

how do I do that

ok I got this equation

20+100(0.3)^0.2t = 20.1

100(0.3)^0.2t = 0.1

is there a way i can solve this without log?

,$ 100(0.3)^0.2t = 0.1

[obsidian monolith]

MrMeltzFTW:

[full garden]

#helpers

[rocky bison]

@full garden Sorry what are you trying to solve for?

t?

[full garden]

he temperature, T, in degrees Celsius, of a cooling metal bar after t minutes is given by T(t)=20+100(0.3)^0.2t
c) How long will it take for the temperature to be within 0,1 Celsius of the value of the asymptote?
how do I do that
ok I got this equation
20+100(0.3)^0.2t = 20.1

100(0.3)^0.2t = 0.1
is there a way i can solve this without log?
,$ 100(0.3)^0.2t = 0.1

[rocky bison]

In the equation you just gav?

Answer is no, not really

[full garden]

so I am trying to solve for C

c) How long will it take for the temperature to be within 0,1 Celsius of the value of the asymptote?

and I got this question when I was trying to solve it

20+100(0.3)^0.2t = 20.1

100(0.3)^0.2t = 0.1

and I tried it on photomath

but it can't solve it

[rocky bison]

Use logs

,w plot 20+100\left(0.3\right)^{0.2t}

[obsidian monolith]

Display results online and refine query

Requested by Pseudo#2703

[rocky bison]

So this is your curve

You want to solve it using logs

,$ 100\left(0.3\right)^{0.2t}+20=1

[obsidian monolith]

Pseudo:

[full garden]

i dont know how ot use logs bro

thats the issue

[rocky bison]

Oof

Well

,$ a^b=c\iff \log_a\left(c\right)=b

[obsidian monolith]

Pseudo:

[rocky bison]

That's the principle

Might wanna learn about logs before doing questions reliant on logs

[full garden]

i mean like

I dont know why the book would give me a log question if i didn't get logs yet

is logs gr 12?

[potent badger]

No I believe it is grade 11.

[full garden]

i was loking for it in the book and I didn't find it

[potent badger]

I'm gonna be honest with y'all I'm in second year and I don't remember the Log rules, only ln lmfao

I would just search them up and learn them tbh @full garden

[full garden]

khan Acadamy?

[potent badger]

Like this.

You have to manipulate some of them

[full garden]

this looks complicated bro

its like exponential laws @potent badger

?

[potent badger]

Yessir

[full garden]

thank you sir

[full garden]

hello guys please i need someone help

I got the equation of y=250(1/2)^x

how do I answer this question

C) how long will it take for the sample to decay to 20% of its initial mass?

[small grail]

well

its initial mass would be at x=0

which is 250

[full garden]

yes i am here

[small grail]

you need to find at what value x is 250(1/2)^x = (20% of 250)

find 20% of 250

which I'm assuming you know how to do

[full garden]

yes

[small grail]

then solve for x

[full garden]

50?

[small grail]

yep

so 50 = 250(1/2)^x

probably a good idea to simplify and get it down to 0.2 = (1/2)^x

now use logs and solve

[full garden]

yes bro

the things

is

I dont know how to use logs or what they are composed of

I still didn't see a lesson ab them

[small grail]

oh

that makes things complicated

because unless you're allowed to use a calculator

logs are how you'll have to solve this

if you are allowed to use a calculator

[full garden]

oui je suis

soryr

yes am i

[small grail]

graph each side of the equation

[full garden]

sorry I mean i can

[small grail]

and find the point where they intersect

[full garden]

is that a good idea

that sounsd rational but too hard

is there a way to guess the approx value @small grail

[spring thunder]

well it's between 2 and 3

[small grail]

yes

because

[spring thunder]

(1/2)^2 > 0.2 > (1/2)^3

[full garden]

how did you know @spring thunder

[small grail]

you know that (1/2)^2 = 0.25

and (1/2)^3 = 0.125

[spring thunder]

ah eyah lel

[small grail]

yeah anyway @full garden since 0.2 is in between 0.25 and 0.125, x has to be between 2 and 3

but your teacher or whomever you're doing the problem for

is probably expecting a more precise answer

in which case you'd have to use logs or a graphing calculator

[spring thunder]

alors comme ça tu es @full garden

[full garden]

oh la la

[small grail]

oh la la le francais

qu'est-ce qu'on va faire

hon hon hon baguettes baguettes baguettes

[spring thunder]

too many french fags in here

[full garden]

je vais bien et vous

[spring thunder]

ça va lel

[small grail]

un drapeau blanc

lmao

[full garden]

Je déteste l'anglais

[small grail]

je deteste ta mere

gotem

[full garden]

je viens de Montréal

elle est morte..

[small grail]

oh

uhhhh

rip

desole

[full garden]

ya rip

[spring thunder]

rip the honor of 8bit

[small grail]

yeah that was kind of an oof

[full garden]

Crois tu en la vie après la mort?

[small grail]

euhhh un peu

also I barely speak french

[frozen needle]

[full garden]

anyway

is log gr 11?

[small grail]

je pense qu'on renait apres la mort

log gr?

[spring thunder]

grade

[full garden]

grade

[small grail]

oh

[spring thunder]

in france it's grade 11-12 lel

[small grail]

idk how schools work wherever you are but I learned logs in Alg II

I took that class in 9th grade

[full garden]

idek man

bc I am not following canada's stupid ass education system

this is why you should vote for bloc Québécois

[small grail]

but the thing is, at least where I'm from, precalculus expects logs to be understood

[full garden]

is it needed for trigonometric functions?

[small grail]

no

not at all

[full garden]

bc the next unit is trig

and I haven't learned ab logs yet

[small grail]

yeah trig and logs are separate dw

[full garden]

are logs hard?

[small grail]

no

[full garden]

to comprehend

[small grail]

they're just a bit weird at first but if you understand exponents and stuff logs should be easy

[full garden]

ok what does this mean

log of 49

does this eman what is x times x =49

?

[small grail]

no

basically

first off

when you use logs

it's always with a base

so like

idk how to format this in latex

but logbase b (49) = y

means

[spring thunder]

$\log_7(49)$

[obsidian monolith]

emeric75:

[small grail]

ok perfect

thank you

basically

it's saying

[full garden]

how did you do that

[spring thunder]

i called gods to teach me

[small grail]

$\log_7(49) = y$

[full garden]

oh

[obsidian monolith]

8BitRobot: