precalculus

[sleek ruin]

In one of my homeworks, I had to use the Law of Sin to find the sides and angles of a triangle. And whenever an inverse of sin happens, I would have to do 180 - the answer of arcsine.

or it might be from a video I watched

[willow bear]

that's horribly one-dimensional

[sleek ruin]

Hm weird

[viscid thistle]

can someone give me the steps to solving for x? Would greatly appreciate it.

[willow bear]

what have you tried so far

[astral mantle]

you can reduce both sides by 7

e^x = 2

then use logs to solve

[viscid thistle]

this is what I did so far

[astral mantle]

take the log of both sides

[peak badge]

Are you familiar with logarithms in the first place?

[viscid thistle]

kinda but there is definitely more I can polish up on

[willow bear]

uh

why did the x jump outside the parentheses?

[viscid thistle]

ln(e) = 1, isn't it ?

[sleek ruin]

Something like this. Sorry I had to ask again. Still a little confused and all. Appreciate it

[willow bear]

that isn't the issue @viscid thistle

it's $\ln(e^x)$ not $\ln(e)^x$

[peak badge]

...definitely review on logarithm properties then......

[viscid thistle]

oh for ln(2)?

[obsidian monolith]

Ann:

[willow bear]

when you take the log of e^x you don't suddenly get to just plop the ^x on the outside of the log

[viscid thistle]

oh

[peak badge]

Any ideas on how do you proceed from here, 403? It's just one step away from the answer IMO

[viscid thistle]

let me think'

[willow bear]

no no 403

[sleek ruin]

my bad

[willow bear]

the top line in your latest image

it's just wrong

[viscid thistle]

ok

[willow bear]

it should never have been there

it is the result of a mistake

as i have UNSUCCESSFULLY been trying to point out

but you never understood, it seems

[viscid thistle]

lol

wait

where did I make the mistake again?

[sleek ruin]

That's tuff

[willow bear]

ok let's back up

let's back up

you had the equation e^x = 2

this is what you came to, everything up until now was ok

the next most sensible step is to take the natural log of both sides

thus, ln(e^x) = ln(2)

[willow bear]

yes! there we go.

[viscid thistle]

ok

[willow bear]

and it is now that you can simplify ln(e^x) to just x.

why aren't my messages sending

[peak badge]

Did you get the intuition/reasoning behind the procedure though?

[viscid thistle]

yeah

because e^x is togehter

so there is a parenthesis?

at least

tahts what I'm thinking

[peak badge]

Yes. e^x is an entire number that you would have to take it together

[viscid thistle]

yeah thats what I was thinking after

after @willow bear told me my mistake

[peak badge]

why aren't my messages sending

Internet issues?

[viscid thistle]

ln(e^x) = x?

[sleek ruin]

There a video that explains this as well.

Text can get you so far.

[peak badge]

Okay. @viscid thistle I would advise revising logarithm properties here on out. Because to get the next step for the answer - you need to apply 1 of them.

Here's a starting material if you don't have any to refer to
http://www.montereyinstitute.org/courses/DevelopmentalMath/TEXTGROUP-1-19_RESOURCE/U18_L2_T2_text_final.html

https://www.mathsisfun.com/algebra/exponents-logarithms.html

[sleek ruin]

oof

[peak badge]

For completeness sake - I will finish up with the problem here -

$$\ln (e^x) = \ln(2)\newline\rightarrow x \ln(e) = \ln(2)\newline\rightarrow x = (\ln(2))/(\ln(e))\newline\rightarrow x = \ln(2)$$

[willow bear]

do you really need more than 1 step to go from ln(e^x) to x?

i thought familiarity with the definition of log could be assumed

also it's \\ for new lines

[peak badge]

Err..sorry with that. My LaTeX skills are wack

[obsidian monolith]

HarshlyDOOM:

[viscid thistle]

I forgot that I can move the exponent to the front lmfao

now it makes perfect sense

[harsh smelt]

but why you need this tho

i mean it is correct

[viscid thistle]

yeah but

I want better understanding

oh you mean to further simplify?

[harsh smelt]

but if e^x = e^x which is definitely true, by the definition of ln, x = ln(e^x)

[viscid thistle]

just for better understanidng I want to furhter simplify

mainly

[harsh smelt]

sure no problems

[viscid thistle]

hello what is the derivative of sin(x^3)

?

using substitution we can put x cube = t

then sin t derivative is cos t

then substitute t again

that gives answer cos (x^3)

AM I RIGHT ?

CHAIN RULE IS DUMB ?

[willow bear]

no, you're not actually applying chain rule here

and to be clear, chain rule is the tool to use to differentiate sin(x^3) you're just not using it properly

in your notation, you forgot to multiply by dt/dx

which would give you d/dx sin(x) = 3x^2 cos(x^3)

[sleek ruin]

Expanding use the Laws Of Logarithms. Not sure what I got wrong on this one. Would appreciate any feedback, thanks.

[viscid thistle]

no, you're not actually applying chain rule here
@willow bear ik

i am USING SUBSTITUTION

[willow bear]

parentheses around ( ln(sqrt(5)) + ln(y) ) in the second line @sleek ruin

[viscid thistle]

just tell me why am i wrong

why is my method wrong ?

[willow bear]

using your method you can prove the derivative of any function is 1

[viscid thistle]

yess replace that function by t then diffrentiate that gives 1

[sleek ruin]

Thank you.

[willow bear]

@viscid thistle do you not see the problem here?

[viscid thistle]

yes i see

[willow bear]

clearly the derivative isn't just an operation that takes in a function and always returns 1.

[viscid thistle]

but why am i wrong

You forgot what's chain rule

i can replace anything by t and derive that its answer is one

WHY DO I ALWAYS HAVE TO USE CHAIN RULE

isnt substitution valid

x^3 = t
d(sint)/dt = cos(t) * dt/dx

ok

[willow bear]

isnt substitution valid
chain rule is the calculus version of substitution

[viscid thistle]

diffrentiation is something that is not a function

wot

yes

What does that even mean lol

[viscid thistle]

so i am supposed to do it in terms of dt not dx

[pastel pecan]

Guys this problem is seriously stumping me and my friend

[willow bear]

this is a matter of incomplete simplification

4 - 4sin^2(θ) = 4(1 - sin^2(θ)) = 4 cos^2(θ)

[pastel pecan]

owww

ok

Thank you very much it makes complete sense now

[sleek ruin]

thats what i am going to relearn that in the next chapter. i remember the struggle.

[pastel pecan]

yeah lol

[sleek ruin]

Question:

My Work:

To be honest, this one was tough for me. A lot going on and the fractions + the exponents threw me off. I guess my question is what would the first step be?

[vale widget]

just take the root

[sleek ruin]

Oh. I forgot you can do that

It was a mess and had to be done quickly

[pastel pecan]

here we go again boys lol

spent another 20 mins on this one

still lost

as hell

[echo wagon]

Well, show your work how you got to 2

[pastel pecan]

ok

I factored the top and bottom

which I got

(sin+4)(sin+4)/sin(sin+4)

to which I crossed out both sins'

to which I was left with 4+4/4

8/4

=2

OMG

I just realized I should of crossedd out (sin+4) and (sin+4)

shoot I got it

thanks

[pastel pecan]

Anyone understand how they got a positive 1 in the denomiator

[harsh smelt]

@pastel pecan write down denumenator explicity

[pastel pecan]

so what does that mean?

in that context

[harsh smelt]

i mean what have you got when you expand denumenator

in particular

[obsidian monolith]

Commander Vimes:

[pastel pecan]

I get csc^2x-cot^2x=-1

so there should be a negative 1 in the denominator

But in the anwser show in the second part it is positive

You see where Im lost?

[harsh smelt]

how you get -1?

,w csc^2x-cot^2x=-1

[obsidian monolith]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Commander Vimes#7532

[pastel pecan]

because its a pythageorum identity

so it equals to -1 on the bottom

[harsh smelt]

no

can you please expand expression

and write in down using all definitions

[pastel pecan]

1=csc^2x-cot^2x

[harsh smelt]

so?

[pastel pecan]

Wow

I see it now

Thank you

[harsh smelt]

yw

[pastel pecan]

It was backwards so I didnt know why

thank you very much

[harsh smelt]

np

[toxic coral]

Is the vertical asymptote of csc(pi)x just n or (pi)n?

[willow bear]

did you mean csc(pi x)

also there is not one asymptote there are an infinite number

and they happen precisely at the integer coordinates

i.e. x = n, for n ∈ Z

[toxic coral]

Yeah csc(pi x)

Sooo it's x=pi n?

Or simply just x=n since pi is already in there

[cobalt storm]

Please tell me I’m not tripping here

None of these answers are true

[blissful ridge]

Yeah, all of them look wrong
(Though don't ask for helps on tests, you'll get banned)

[cobalt storm]

Oh this is just a homework that we submit my bad

[viscid thistle]

Anyone know how to solve a systems of log equations with an uncommon base

[north island]

convert to a common base

[viscid thistle]

i tried that

its where i mess up

like i tried doing 6log8(y) into 6log2^3(y)

and then where from there you know

[cobalt storm]

Use product law

For the left side

To simplify

Then make both side 4^

[north island]

nahhh just think about it intuitively

https://www.youtube.com/watch?v=N-7tcTIrers

[cobalt storm]

Oh nvm those are uncommon bases

[viscid thistle]

Wait nvm

i found it

i think i convert to log10

[north island]

noooo

[viscid thistle]

no?

not a base 10

why not

[north island]

because it's not neat

okay i highly recommend watching the linked video

but i'll summarise

logs are like multiplicative steps

log2(x) is the number of 'steps' it takes to get to x, from 1, with the step size 2

and 8 is 3 steps of 2

[viscid thistle]

mhm yes

[north island]

so log8(y) = 3log2(y)

...

gonna be really embarrassing if this is wrong

[viscid thistle]

so that becomes 18log2(y)?

[north island]

think so

[viscid thistle]

for my problem i meaqn

because its 6*3

[north island]

yes

[viscid thistle]

thats what i did

but i think its not correct

like if we use a value for y

[north island]

why not

ok

[viscid thistle]

like lets say u have 3=y

they dont match yup

up

[north island]

why not

y = 3 is a bad value

but ok

[viscid thistle]

and also makes sense because it would need to be y^3 to be multiplied

instead of 2^#

2^3

[north island]

what

[viscid thistle]

like if its log8(y) right

[north island]

yes

[viscid thistle]

or

6log8(y) rather

ok

[north island]

yes

[viscid thistle]

in order for it to become 18

it has to be 6log8(y)^3

[north island]

no

[viscid thistle]

the y has to be cubed to multiply

not the 2

[north island]

6log8(y) = 18log2(y)

log8(y) = 3log2(y)

[viscid thistle]

plug any number in tho

and its untrue

[north island]

give me a counterexample

[viscid thistle]

wym

[north island]

try one out

[viscid thistle]

like a different example or number

[north island]

try it

y = 64, let's say

[viscid thistle]

sure

[north island]

...

[viscid thistle]

so

18log2(64) = 6log8(64) we're testing then?

[north island]

do 3log2(64) = log8(64)

it's equivalent

just dividing through by 6

[viscid thistle]

but why isnt it true with 18 and 6

[north island]

it is true

[viscid thistle]

theyre supposed to be the same scale factor

[north island]

it's just simpler to consider 3 and 1

[viscid thistle]

18log2(64)

is 108

[north island]

yes

[viscid thistle]

and 6log8(64) is 12

[north island]

shit what have i done

[viscid thistle]

lol its fine

i had the same issue

[north island]

waaaait

i'm a foool

[viscid thistle]

?

oh

[north island]

it's the other way around

[viscid thistle]

wym

[north island]

6log2(64) = 18log8(64)

[viscid thistle]

huh

[north island]

i am a moron

[viscid thistle]

oh

do u divide by 3 instead multiply???

[north island]

log2(64) = 3log8(64)

[viscid thistle]

you're a lifesaver if thats true

[north island]

because you're taking 1/3 the steps

[viscid thistle]

ohhhh

yes

ye

yes

you're the best my g

[north island]

i'm a buffoon

[viscid thistle]

lol nah just confused like me

ok so ill do that conversion now

thx sm!!!!

[obtuse kite]

Where tf did the 38 go?!

Never mind it’s the sh*tty website being a troll

[calm grove]

probs wanna multiply the 2 into (x-14), then synthetically divide the resulting equation

[heady field]

does anyone know this and can explain it to me?

also what does the U mean?

[small forge]

@calm grove That doesn't work

[gaunt mason]

U is the union. It means the solution is both the interval on the left and the interval on the right of "U"

@heady field

[heady field]

wdym?

[gaunt mason]

What is unclear?

[heady field]

wait am i supposed to answer it based off of what i see in problems and not the graph?

[patent beacon]

A and B are two different sets (or intervals)

A U B means "whatever is in either A or B"
U is called the "Union" of the sets

[gaunt mason]

a,b,c,d are not the problems, but the solutions.

you are supposed to answer based on the graph

[heady field]

ok i think i understand

what it be b then?

[gaunt mason]

Yes.

The function rises from -1 to 1, and also from 3 to infinity

So both these two intervals are the solution, and you "unite" them with U

[heady field]

thank you!

i am studying for test tomorrow

i thought it was (x,y) U (x, y) so that's why i was confused

[gaunt mason]

cool

[obtuse kite]

i am studying for test tomorrow
@heady field good luck on the test. I took a similar one recently

[heady field]

ty

i think i prepared

[knotty echo]

Can anyone help me?

[heady field]

i am guessing it would be the top degree so x^7/x^4 = x^3, so odd

take that with a grain of salt though because i am not done precalc

sadly the practice quiz didn't give me results to see if it was right or wrong, does anyone know the answer to this? i guessed it was a polynomial

[velvet granite]

@heady field its a poly cuz the x^3 correct?

[heady field]

i guess so

just had me confused

thanks

[velvet granite]

try plugging it in

to a calculator

[heady field]

that makes more sense

i was just scared that some technical way it wasn't

and i assume that'd mean \sqrt[4]{x^5-x^2+3} would be one as well

[viscid thistle]

Hi, can someone help explain why there is an extraneous constraint?

And what an extraneous constraint is ? Thank you

[gaunt mason]

Can anyone help me?
@knotty echo you need to apply the definitions of even and odd.

[hard hedge]

Can someone explain how to do 3 and 4 please?

[velvet granite]

@hard hedge where your unit circle?

For #1 its asking where sec is positive and tang is negative

sec is just 1/cos right, so we need to find where cos is positive and where tang is negative

cos is positive in quadrants 1 and 4 and tang is negative in quadrants 2 and 4

so your answer would be quadrant 4 because it matches the needs of both identities

[jolly gust]

How would I prove that $f(x) = 5*3^x and prove that f(x+2)-f(x) is divisible by 10$

[obsidian monolith]

Big Chungus:

[jolly gust]

So far I have

$(53^x )(53^2) - (5*3^x )$

[obsidian monolith]

Big Chungus:

[jolly gust]

I dont know how to prove that this is divisible by 10

[willow bear]

wording

also $f(x+2) \neq f(x) f(2)$ just sayin'

[obsidian monolith]

Ann:

[willow bear]

also bad tex earlier

$f(x+2) - f(x) = 5 \cdot 3^{x+2} - 5 \cdot 3^x \ = 5 \cdot 9 \cdot 3^{x} - 5 \cdot 3^x \ = 45 \cdot 3^x - 5 \cdot 3^x \ = 40 \cdot 3^x$

[obsidian monolith]

Ann:

[willow bear]

which is not divisible by ten for all real values of x, so you'll have to be more specific as to what values x is supposed to range over

unless you want me to maliciously comply with your underspecification

[jolly gust]

I dont know my teacher just said to prove if it is 10

ill just write down it isnt

[willow bear]

...

do you have the exact wording

like, without any omissions or alterations

[jolly gust]

Yeah hold yo

up

If f(x) = 5 * 3^x, show that f(x+2) - f(x) is divisible by 10

[willow bear]

that's it?

no indication of what x is?

[jolly gust]

nope

[willow bear]

then it's false as stated

f(-98) - f(-100) is not even an integer let alone one divisible by 10

[jolly gust]

ok thamks

[south bear]

can someone explain the unit circle to me?

My teacher taught the class it but he didn't really explain in depth wut it actually is

[grand berry]

the unit circle is a circle of radius 1, and the various points along the circle have certain angles with respect to the positive x axis

and those points have coordinates that are given on the diagram

[south bear]

hmm, ok. Is this related to like sin, cosine, and tangent?

[grand berry]

Yep

The x coordinate is the cos of the angle

[south bear]

The professor mentioned something about that

[grand berry]

The y coordinate is the sin of the angle

[south bear]

why tho? I don't understnad

[grand berry]

tbh I forgot

sorry

[south bear]

hmm ok, do you know anything else about it?

It's ok

[grand berry]

pretty sure it has something to do with sin^2(theta)+cos^(theta) = 1

[south bear]

woah that sounds confusing

is that some sort of theorem?

[grand berry]

well it’s just the Pythagorean theorem

[south bear]

thetha is just the degree right?

[grand berry]

Yes

The angle

[south bear]

oh ok, so another way to write the pythagorean theorem?

[grand berry]

you know how the Pythagorean theorem is a^2 + b^2 = c^2

[south bear]

yes

how does the unit circle related to this

[grand berry]

the coordinates on the unit circle are (cos(theta), sin(theta))

And the radius is 1

[south bear]

hmmm ok

[grand berry]

So then you use the Pythagorean theorem and get sin^2(theta) + cos^2(theta) = 1

[south bear]

ok tysm

That means a whole lot of sense

why is unit circle important in all this tho?

If it's just another way of doing pythagorean theorem?

[grand berry]

unit circle helps you memorize cos and sin of a bunch of angles

[south bear]

ok that makes sense

Tysm again

[grand berry]

[potent imp]

unit circle helps define sin(x) and cos(x) for any real number

if u used the right triangle definition sin 240 wouldnt make sense because u cant have a 240 degree angle in a right triangle

[topaz flint]

The accerleration of a shuttlecock is given by a(t) = 34.6t-52.1 where t is time in seconds since shuttlecock was hit. initial velocity = 40 ms^-1. Calculate the velocity after 2 seconds.

[violet jetty]

differentiation of logs

would highly appreciate help

[jagged glade]

Chain rule?

[violet jetty]

yeah

[topaz flint]

i got the answer 7.2ms^-1

can someone confirm?

[jagged glade]

Right, alchemist was here first, didn't notice

,calc 34.6/24-52.12 + 40

[obsidian monolith]

Result:

5

[jagged glade]

@topaz flint
I'm not sure, I got 5

[topaz flint]

oh

[violet jetty]

struggling with the working backwards

[jagged glade]

do you know how to apply chain rule in this question?

[violet jetty]

not particularly sure

[topaz flint]

i integreted the acceleration andd got v(t) = 17.3t^2 - 52.1t + 40

then t = 2

[jagged glade]

Yea

,calc 17.34-52.12+40

[obsidian monolith]

Result:

5

[topaz flint]

well

i calculated wrong lol

[jagged glade]

Lol

[topaz flint]

thanks for the help

[jagged glade]

Yw

@violet jetty
Recall: (f(g(x)))' = (f'(g(x))(g'(x))

[violet jetty]

yeah

product rule

[jagged glade]

This is chain rule

[violet jetty]

...

[jagged glade]

Anyways

(ln(f(x)))'= d(ln(f(x))/df(x) * f'(x)

So what is $\frac{\dd ln (f(x))}{\dd f(x)}$

[obsidian monolith]

Biscuit:

[jagged glade]

It's similar to $\frac{\dd ln (x)}{\dd x}$

[obsidian monolith]

Biscuit:

[violet jetty]

am i suppoed to answer?

[jagged glade]

Oh, I forgot the question mark

Yea

[violet jetty]

I'm gonna be honest, not sure

[jagged glade]

Okay... Have you learn the derivative of ln(x)

[violet jetty]

derivative of ln(x) is 1/x?

[jagged glade]

Yea

[violet jetty]

ok formatting is weird apparently

[jagged glade]

So what is $\frac{\dd ln (f(x))}{\dd f(x)} ?

[violet jetty]

(1/x)/1?

[jagged glade]

Since it's with respect to f(x), we don't have to change it. And since bot is dead, let's let f(x)=y

Then we will have ln(f(x))=ln (y)

Derivative of ln(x) w.r.t. x is 1/x
Then what is the derivative of ln(y) w.r.t y ?

[violet jetty]

y

[jagged glade]

1/y

Therefore we have ln(f(x)) = 1/y times f'(x)

Hence we plug the f(x) back in and get 1/f(x) times f'(x) equals

f'(x)/f(x)

[violet jetty]

true

you're very smart

[jagged glade]

That's how we get d (ln (f(x)))/dx= f'(x)/f(x)

It's just practice, and practice makes perfect :)

[violet jetty]

thanks for the help

[stuck lark]

@jagged glade @violet jetty (ln o f)'=f'/f

$(\ln\circ f)'=(\ln'\circ f)\cdot f'=\frac1f\cdot f'=\frac{f'}f$

[obsidian monolith]

RokettoJanpu:

[dawn depot]

Find the gradient of the tangent at each function at the given value of x. How to solve this? i get the derivative then i plug 1 and -3?

[uncut mulch]

if the derivative is only in terms of x, you only need to use the x-coordinate of the point i.e. 1

[dawn depot]

the derivative here is 2+5x^-2?

[violet jetty]

2x^2+5/x^2

[uncut mulch]

2+5x^-2?
yes

[violet jetty]

use quotient rule

[uncut mulch]

quotient rule is overkill

[dawn depot]

so now to find the gradient i plug in 1 then what?

[uncut mulch]

that's it

plug in 1

[dawn depot]

ur saying answer is 7?

[uncut mulch]

yes

[dawn depot]

book says 13

[uncut mulch]

weird, can you show the full question

as printed

error with the book ig

[dawn depot]

oh ok

[jagged glade]

ty for pointing out typo😂 RokettoJanpu

[cedar pawn]

What's the position of these oblique asymptotes? y=2x for +inf and y=-2x for - inf. Function is sqrt(4x^2-1)

For +inf= sqrt(4x^2-1)-2x=0

Don't you do this afterwards (sqrt(4x^2-1))^2=(2x)^2

Then your become 4x^2-1=4x^2 and then u will become - 1=0 but that's wrong I think

How do you do it then

[viscid thistle]

??

[cedar pawn]

?

[viscid thistle]

the function is even so you might aswell study it for x > 0

ok I got it

you are trying to get q

$q=\lim_{x\to\infty}f(x)-mx$

[obsidian monolith]

HoboSas:

[cedar pawn]

I already have q

It's 0

I just want to find the position

[viscid thistle]

wym position

y = 2x

[cedar pawn]

This v(x)

Difference function

[viscid thistle]

can you explain yourself

[cedar pawn]

Translated

[viscid thistle]

I just want to find the position
of what?

[cedar pawn]

The formula is v(x) = f(x) - mx- q

Location

[viscid thistle]

never heard of it

anyway you need to find it?

[cedar pawn]

Yeah but slip it. If a function has 2 horizontal asymptotes then it can't have an oblique right?

Skip*

[viscid thistle]

no

it already has 2

[cedar pawn]

What

[viscid thistle]

asymptotes

[cedar pawn]

A function with 2 horizontal asymptotes doesn't have an oblique asymptote

[viscid thistle]

yes

[cedar pawn]

A function with a horizontal asymptote for +inf can have an oblique asymptote for - inf right?

[viscid thistle]

yes

[cedar pawn]

Okie

[proper hornet]

I was thinking 60y = 1.5x

I got that by setting both equations equal to zero and then setting them equal to each other

[sick steppe]

Just find slope of the line then plug it into slope-point formula

[round quest]

how do i find the sum of all integers between two numbers

[viscid thistle]

area beneath a curve of y=x from n to m I think?

i might be wrong

[blissful ridge]

how do i find the sum of all integers between two numbers

The interval you want to sum will form an AP

[viscid thistle]

triangular numbers apparently @round quest

[proper hornet]

@sick steppe to find the intercept do I minus y to get them both equal to zero and thrn set them equal to each other?

Slope*

[sick steppe]

For a slope m and point (a,b): $y-b=m(x-a)$

[obsidian monolith]

moshill1:

[viscid thistle]

(n(n+1))/2 - (m(m+1))/2 = all integers summed from m to n

[sick steppe]

that's point slope @proper hornet

[viscid thistle]

@round quest

[round quest]

its not working

its giving me a different result

i need to find the sum of all integers from -10 to 50

[north island]

what?

[proper hornet]

@sick steppe ok how do i find the poiny?

Point*

[north island]

so what you can do is

[sick steppe]

you're given 2 points, pick one

[north island]

you can pair them up

[proper hornet]

Ok

[north island]

-10 + 50 = -9 + 49 = -8 + 48 etc.

so you get (a-b)/2 terms of (a+b)?

30 * 40 = 1200

[viscid thistle]

@round quest sorry it's (n(n+1))/2-(m(m-1))/2

try that

[round quest]

i found out why

i didnt include n0 as a term

[viscid thistle]

this is inclusive

so sum of 4 through 6

of upper and lower bound

[north island]

oh wait for my one you need to add one more for the middle term?

oops

[proper hornet]

@sick steppe if i solve for m I get m= y/x -70

[sick steppe]

what?

slope uses the 2 points

[proper hornet]

So (a,b) and (x,y)?

[sick steppe]

no what?

you have 2 points given to you

[proper hornet]

I mean eith thebformula

[sick steppe]

(2.5,175) and (4,235)

im not talking about the formula, im talking about the points you have

cause you said 2 arbitrary points

[proper hornet]

Oh then (y2-y1)/(x2-x1)?

[sick steppe]

yes that's slope

[proper hornet]

Ok

After finding the slope what would i plug in for x and y to find b?

Just the x and y of a point?

[north island]

yes

[viscid thistle]

@proper hornet

m = rise/run

[proper hornet]

Yea

I got y=40x+105

[viscid thistle]

if it hits both points you're probably good

[proper hornet]

Nope

It's wrong darn

[north island]

dy/dx = 60/1.5 = 90

[proper hornet]

Oh my calulatir lied

[viscid thistle]

(235-175)/(4-2.5) = m

[proper hornet]

I got 40

[viscid thistle]

y-235=m(x-4)

y=40x+75

[north island]

what

m = 90

[proper hornet]

How does 60/1.5 = 90 lol

[viscid thistle]

yeah...

[proper hornet]

Its not a differential

He was thinking a differential

I was once at calc 2 level then i forgot everything :(

[round quest]

how do i do this

[north island]

sum from 1 to 200 of (3n-1) = 3 x sum from 1 to 200 of n + sum from 1 to 200 of -1

[round quest]

is there a formula for that

[echo wagon]

$\sum_{n=1}^m n = \frac{m(m+1)}{2}$

[north island]

let's set 10 as an upper bound

[echo wagon]

But also, it's an arithmetic series, so you can use the partial sum formula for an arithmetic sequence @round quest

[obsidian monolith]

Lunasong:

[round quest]

thanks

what happens when n is not 1

[echo wagon]

n is an index/variable, it's not a single value

[round quest]

wait

that formula gave me a different answer

it gave me 20100

the answer is 60100

[north island]

divide by 2

and then multiply by 3

and then do the -1s

[echo wagon]

That would be because you just worked out the sum of n

But you have 3n-1

Not the same thing

Have you done arithmetic series? You can just use the partial sum formula for that, it's probably better

[misty ocean]

sin x=3/5, x lies in quadrant 1, and sin y=5/13, y lies in quadrant 2. Find sin(x+y)

[echo wagon]

Draw the plane to find cos x and cos y too

And then expand sin(x+y) using the compound angle formula

[misty ocean]

oh wait i completely forgot those existed lol

thanks

[velvet granite]

@topaz obsidian so something like x(x-3)^2(x+2)

im just not sure if that goes through 5,70

Dont listen to me im dumb

[uncut mulch]

introduce a scaling factor to be determined

if you perform a vertical shift like that, you no longer have the specified roots

[velvet granite]

@uncut mulch ah so what you are saying is the the a value must change

[sick steppe]

yes

[cobalt storm]

Wouldn’t this be false since x = 4 is not a point in the inverse of f(x)

Since this is f o f^-1 we can’t use an x value that isn’t apart of the inverse function right

[north island]

it isn't? thought f^-1(4) = 2

[cobalt storm]

Oh wait bvm

Yeah you’re right I just saw the (2,4)

In the original function

So it’s inverse would be (4,2) making 4 an x value

Yeah thanks

[idle moat]

not sure where i gone wrong

[patent beacon]

There's a π/3 in it, but the expression isn't just π/3

[north island]

it's sin(pi/3), which is sqrt(3)/2

[idle moat]

how do i find out what quadrant csc theta > 0 is in?

[velvet granite]

@idle moat well, csctheta, is the same as 1/sin theta

and sin is positive in quadrants I and II

[remote veldt]

Can anyone think of a way to maximize and minimize $f(x) = \frac{3x + 1}{x^2 + 3}$ without calculus

[obsidian monolith]

Nicholas:

[remote veldt]

a high schooler I know asked me this question and I'm like 90% sure that the teacher means for them to just use graphing software

[viscid thistle]

heyo

i need a tad of help with law of sines

[radiant notch]

hello can someone help me with composing functions with functions. I don’t understand the order. So like for example if we had f(x) = 2 and g(x) = 4
(f o g)(x)

What would the order be ??

[uncut mulch]

(fog)(x) = f(g(x))

[radiant notch]

So would it be, 2(4(x))

[quick mirage]

Note that whatever we have as x f(x) always equals 2

So f(g(x)) = 2

But if we had something like f(x) = 2x

Then we would substitute the x values with whatever x ends up being

[radiant notch]

ohhhhh

[quick mirage]

For example if g(x) = 2

And we use f(x) from the example above

(f o g)(x) = f(g(x)) = f(2) = 2(2) = 4

[uncut mulch]

wrong dot

solid dot means something different

[quick mirage]

sorry I can’t find the right dot

[uncut mulch]

just use o

[quick mirage]

Okay, will do in the future

Thank you

[radiant notch]

@quick mirage ok thank you !! :)

[round quest]

is this right

[steep plume]

Been doing introduction to Combinations and the teacher pulls up wit this, and says research. Asking for help

[willow bear]

what is your definition of nCr? @steep plume

[steep plume]

As in the formulae?

[willow bear]

as in, how did your teacher introduce nCr to you

since it clearly wasn't by means of this formula

[steep plume]

nCr=n!/(n-r)!r!

[willow bear]

did your teacher DEFINE nCr to be that?

[steep plume]

Yes

[willow bear]

then the proof will consist of two words

"By definition."

[steep plume]

Sent the wrong image

That’s the one

[willow bear]

oh

well this is a simple matter of writing out the RHS with your formula & doing some simplification

have you done any work in that regard?

[steep plume]

This is all me and my friend have worked out. All we could think of 💀

[willow bear]

keep going

you have $\frac{(n-1)!}{(n-r-1)! r!} + \frac{(n-1)!}{(n-r)! (r-1)!}$

[obsidian monolith]

Ann:

[willow bear]

you can use $(n-r)! r!$ as a common denominator; this will require multiplying the first fraction by $\frac{n-r}{n-r}$ and the second by $\frac{r}{r}$

[obsidian monolith]

Ann:

[trim cave]

Use derivatives

If you're talking about whether the function is increasing or decreasing

[limber compass]

but how did they go from

How did the go from the first line to the second line?

[willow bear]

replace sqrt(p^2 - 4q) with something like R for convenience

$-p + R = n(-p-R) \ -p + R = -np - nR \ nR + R = p - np \ (n+1)R = (1-n)p$

[obsidian monolith]

Ann:

[willow bear]

$R = \frac{1-n}{1+n}p$

[obsidian monolith]

Ann:

[harsh rampart]

https://i.imgur.com/gLVIbl3.png c is -2r^-2 - (3/2) x^-1/2 right?

[viscid thistle]

no

d/dx

[harsh rampart]

what?

[viscid thistle]

2/r is just a constant

[harsh rampart]

so ur saying derivative of 2/r is 0?

[viscid thistle]

yes

[harsh rampart]

then i have to find derivative of -3 sqrt x?

[viscid thistle]

we are differentiating with respect to x

not to r

[harsh rampart]

so the answer is - (3/2) x^-1/2 ?

without -2r^-2

[viscid thistle]

yes

[harsh rampart]

how to know which variable we are solving with respect to?

not to r

because it f(x)?

[viscid thistle]

if you have f'(x) its gonna be with respect to x

otherwise d(f)/dr

[harsh rampart]

oh ok ok

[viscid thistle]

can someone help me

[sour hemlock]

poorly worded question

so, if I understand the question, increasing the rent on every tenant by 100 will decrease the total number of tenants by 10?

and we are only allowed to increase rent in increments of 100?

let x be the number of 100 increments made to the rent
(50 - 10x)(850 + 100x)

so I guess you could find the maxima of the above

[gaunt mason]

At 850$ the building is full with 50 apartment units.

So less than that is not optimal, and not needed to consider.

When increasing the rent from this point the appartment ocupancy decreases. The decrease is likely linearly, at a rate 10 units/$100.

let x be the number of 100 increments made to the rent
(50 - 10x)(850 + 100x)
@sour hemlock I think it is sligtly more clear to let x be the increase in $.
(850 + x)(50 - 10/100 * x)

[mystic umbra]

you can increase the rent in any amount. the 100 per 10 is a rate

and yes this is a find the maximum

O = occupied units

U = units

V = vacant units

R = revenue

P = price of rent

R = P(U-V)

or you can say R = PO

we knot that V = (P-850)/10

because 100/10 = 10

so also O = U-(P-850)/10

R = P(U-((P-850)/10))

U = 50

maximize R = P(50-((P-850)/10))

so you find the zeros of the derivative. look at the graph. that's it.

[viscid thistle]

sigma notation

[willow bear]

did you mean to say "i am asked to write these summations in sigma notation and i am struggling, please help me!"

[serene heath]

simple typo

[willow bear]

@viscid thistle

[viscid thistle]

lmao

very simple typo

kk

lemme fix dat

i am asked to write these summations in sigma notation, and i'm not struggling im just challenging other people in this math server good luck.

[willow bear]

lmao

[viscid thistle]

@willow bear

[quick mirage]

im just challenging other people
@viscid thistle good one

[sand harbor]

LOL

[viscid thistle]

im not capping

small brained potatoes

[willow bear]

sounds like you're trying to get ppl to do your hw for you

these are not challenge problems by any means lol

[viscid thistle]

smh

[quick mirage]

Okay, then why don't you tell me what these are in sigma notation

[viscid thistle]

ok

how do i write math stuff here?

really @willow bear

if their not challenging do them

eeeeeeeeeeeeeeeeeeexxxxxxxxxxxxxxxacccccccccccccctlyyyyyyyyy u have NO IDEA

[quick mirage]

If you need help man just say so

[willow bear]

$\sum_{k=-26}^{-18} \frac{k+27}{k+28}$ and $\sum_{j=100}^{119} \frac{1}{2(j-99)}$ respectively lol

[obsidian monolith]

Ann:

[quick mirage]

LOL

[viscid thistle]

ur wrong

[serene heath]

based

[willow bear]

i'm not lol

[viscid thistle]

L

[willow bear]

these sigma notations are entirely valid for the sums you gave

[viscid thistle]

OOP

i sent the wrong ones

[willow bear]

you did huh

[quick mirage]

Oh yeah just take those answers and hand them in

see what happens

[viscid thistle]

its not even hw lmfao

its not due i just have it as practice

[quick mirage]

Okay, so you want us to help guide you through the practice correct?

[willow bear]

i mean idk about you but i don't think it's a good move on your part to attack someone who, yknow

knows her shit pretty well in this branch of math

[quick mirage]

Respect goes both ways indeed

[willow bear]

i intentionally gave notations which are not the "canonical" or simplest ones, by the way.

be grateful i didn't use $\thonk$ or $\catthink$ as variables, cause i very well could've

[obsidian monolith]

Ann:

[quick mirage]

Emojibra

[viscid thistle]

sry lol

[somber yew]

Catthink in preamble

[viscid thistle]

Can anyone help me with this Pre calc problem?

[uncut mulch]

are you familiar with function notation?

[viscid thistle]

Ik its the way a function is written

[uncut mulch]

where are you stuck exactly?

[viscid thistle]

Well I have math dyscalculia and I dont remember how to do this specific problem unfortunatly

[uncut mulch]

if h(x) = x
h(🍌) = ?

[viscid thistle]

[uncut mulch]

if h(x) = x
h(1) = ?
h(2) = ?
h(a) = ?

this is supposed to test your basic understanding of the notation

[viscid thistle]

i dont know this, Im sorry man. I cannot rmbr for the life of me. I dont know what to plug in to what you just stated /:

[uncut mulch]

well if you start with
h(x) = x
to determine h(1) you could simply sub x=1

hence h(1) = 1

similar idea if you wanted to find h(2) or h evaluated at anyother value

h(🍎) = 🍎

h(🍌 ) = 🍌

as for your problem,

$f(g(x))$ could be obtained by replacing instances of $x$ in your $f(x)$ with $g(x)$ giving you:
$$f(g(x)) = \frac{g(x)-8}{g(x)+7}$$

[obsidian monolith]

ramonov:

[uncut mulch]

and then use algebra to show that that simplifies to x

[viscid thistle]

Okay, I will work that out, then I will send what I have if that is okay

Can I solve for f?

@uncut mulch

[uncut mulch]

wdym?

[viscid thistle]

Hold on

Here it is @uncut mulch

[uncut mulch]

no

[viscid thistle]

sigh

[uncut mulch]

that is complete nonsense

g(x) is the function g evaluated at x,
and you are explicitly given the value of that

g(x) is NOT the product of g and x

[viscid thistle]

Sorry

[uncut mulch]

f(g(x)) is NOT the product of f g and x

[north island]

madre de dios

[uncut mulch]

this is FUNCTION notation

[north island]

cool it lol

[uncut mulch]

due to this poor understanding you should look up some youtube videos on it

[viscid thistle]

alright

[velvet blade]

r's degree can be 1/2 also right?

[viscid thistle]

I think this is about integer degree

so deg(r) has to be 0

[velvet blade]

So in polynomials, integer degrees are a must?

[viscid thistle]

that's how they are defined

[north island]

not just integer, gotta be natural number degrees i think

[velvet blade]

Why not rational degrees? Aren't we able to define it?

Or is it just out of the scope of this book?

[viscid thistle]

that's not a polynomial

[north island]

not all polynomials would have certain nice properties then

[viscid thistle]

not just integer, gotta be natural number degrees i think
yes correct my bad

[north island]

roots would be completely different

not as useful

[velvet blade]

Okay, so are there more general definitions of functions with multiple terms with different degrees?

[north island]

y'know, i've never really heard about them

i don't know what they're called, if there's a name for them

[velvet blade]

Okay. Another doubt: Is there a proof for when if you add one term to the other, it's degree doesn't increase?

Or is it trivial?

[viscid thistle]

one term?

[velvet blade]

Like, ax³+bx². Adding those two terms won't increase the degree right?

The degree of ax³

[viscid thistle]

the degree of the polynomial is the highest power in it

[velvet blade]

3•2³+6•2² = 3•2⁴ right? I was just thinking of some numbers that do this

Because there's this proof remark about how the leading coefficient doesn't change if you factor out one of a common factor in the terms

[north island]

well that doesn't hold for all x

is the point

[velvet blade]

It holds for very few of the pairs, but the book said it can't happen, so this remark holds about polynomials

[gaunt mason]

The book said it can't happen what?

@velvet blade

[velvet blade]

Wait, let me send a screenshot

[gaunt mason]

@velvet blade I don't get it. It said what

[quaint mason]

stuck on both of these problems.

[velvet granite]

5%

is the same at 5/100 @quaint mason

so its for 100 feet, the road goes down 5 feet

[quaint mason]

how do u know its 100 feet tho

[velvet granite]

you kinda have to assume because they dont give you any other information

[gaunt mason]

100 feet is defined precisely.

The first sentence says 5% slope means that elevation_change = 5% * horizontal_distance_change

That is elevation_change = 0.05 * horizontal_distance_change @quaint mason

And in the next sentence the elevation_change (the descent) is said to be 5 feet.

So you put that into the formula from the previous line and you get the horizontal_distance_change

[quaint mason]

elevation change??

[gaunt mason]

What is unclear about these words?

elevation change is increase or decrease in elevation (height)

[velvet blade]

How do I do this?

[gaunt mason]

That's x^3

Or A*x^3 + B where A and B are any numbers.

[velvet blade]

fuck that was simple

better snap

instead of x-c, would x+c have worked to prove this?

if f(c)=0, won't a_0 =0?

<@&286206848099549185>

[fading token]

... no?

Where'd you get that conclusion from

[velvet blade]

f(c)=0?

[fading token]

How does that lead to a_0 = 0?

Show me

I would like to see your thinking

[velvet blade]

a_0 doesn't have x as a factor, so there's no way 'c' is multiplied by 'a' in that constant term

[fading token]

To be clear that we're on the same page, you do realize that we have $f(c)=0=c^n+a_{n-1}c^{n-1}+\ldots+a_1c+a_0$, right?

is the bot ded

[obsidian monolith]

Rijinaru:

[velvet blade]

yes

So a_0= 0 right?

[fringe stream]

a_0 would be 0 only if c = 0

[velvet blade]

wait

[fringe stream]

is this the case?

[velvet blade]

Ok I misunderstood

Ok, so since all the coefficients are integers, and f(c)-a_0 = -a_0, that means LHS is divisible by a_0 and c is one of it's factors?

Or did I get it wrong?

[fading token]

$c^n+a_{n-1}c^{n-1}+\ldots+a_1c+a_0=0\implies -a_0=c(c^{n-1}+a_{n-1}c^{n-2}+\ldots+a_2c+a_1)$ i.e. $a_0=ck$ with integer $k=-(c^{n-1}+a_{n-1}c^{n-2}+\ldots+a_2c+a_1)$, and hence $c|a_0$

[obsidian monolith]

Rijinaru:

[velvet blade]

Ok so that's how you write it...

Thanks

Was there anything wrong with my answer?

[fading token]

Yes, you were trying to divide by $a_0$ whereas you want to prove that $c$ is a divisor

[velvet blade]

Okayy

[obsidian monolith]

Rijinaru:

[viscid thistle]

How do I do this?

What have you tried so far

I tried making them into their own triangles and making a proportion

[viscid thistle]

Consider one specific trig ratio on the small one @viscid thistle

To get an specific side length so that we can use that length plus x, to use again another specific trig ratio to finally solve for x

@viscid thistle So would it be 93/53.7=93/x?

Why didn't you use the trig ratio

Idk what you did

Since the smaller triangle is a 60 30 one I thought that the smaller leg would be 93 divided by root 3

It is

But i mean

How did you got that eqn

53.7 is indeed the base of the small triangle correct

I didn't know how to do the trig ratio here

Now you can consider solving using
tan(30)=93/(x+53.7)

Do you understand why? The adj side of the big triangle is basically the 53.7 and the x together

I get it

From there it is just a point of algebra, i have to go, if you have any doubts ask away

Good luck.

thank you

I got it correct this time

[autumn marsh]

26! ≡ 29 (mod n) I need help finding n

[willow bear]

presumably, n would have to be a divisor of 26! - 29

[limber compass]

@willow bear Thank you for helping me out yesterday 🙂

[viscid thistle]

Bro

im challenging u guys

what is the

good luck my fellows

theres no way any of u guys can solve this

if any of you guys do

u get a free pass to my belly

[blissful ridge]

How did you get the value of a_n??

[viscid thistle]

yes

I'm confuzzled

[blissful ridge]

nth term of GP is given by

T_n=ar^{n-1}

[viscid thistle]

well no shit

[willow bear]

oh it's you again

[viscid thistle]

rude!

[willow bear]

??

[amber linden]

I can't seem to figure this one out...

[obsidian monolith]

GTX1080Ti:

[willow bear]

f(x) = (x+2)(x-5)Q(x) + ax + b

ax+b is your remainder

f(-2) = 4, f(5) = -3

[amber linden]

Ahhh, I got the right answer now. Thanks bro!

[willow bear]

please don't bro me