precalculus

[unique hill]

and the angle between the ground and the top of radio tower is 47, which is given

[slate oracle]

We know the height of the building which is ~41.

Ok

[unique hill]

so you repeat the same process you did for finding the height of just the building

what would be the appropriate trig ratio to use here?

[slate oracle]

Since we know the adjacent and oppo, we would use we would use tangent again

[unique hill]

yes, tangent, you are given the 47 degree angle and the adjacent, which is 59

[slate oracle]

Then we repeat the process

[unique hill]

yup

[slate oracle]

So then, it would be tan35=47/59

Wait.

[unique hill]

look at what you are given

and what you are trying to find that's not given

[slate oracle]

The height of the tower!

So tanx=47/59

Right?

[unique hill]

tower + building to be exact

[slate oracle]

Or.. yeah

[unique hill]

so first construct the tan ratio

tan 47=?

and u know that tan is opp/adj

[slate oracle]

1.07

toa, yes

[unique hill]

that's incorrect, you have one unknown side

ok first, tan 47 = opp/adj

and you know that adj is 59

[slate oracle]

wE DON'T KNOW OPPO

[unique hill]

yes we dont

[slate oracle]

so tan47= x/59

[unique hill]

correct

[slate oracle]

tan47(59)

[unique hill]

mhm

[slate oracle]

63.26

So it's ~63

[unique hill]

63.27 (rounded to 2dp)

[slate oracle]

ohok

[unique hill]

ok so now you've found out the height of the building and the height of building + radio tower

[slate oracle]

So we would subtract the previous answer with the current answer we got

[unique hill]

what do u need to do to find just the height of radio tower

[slate oracle]

Which is..

63-41

is..

uhh

22!!

[unique hill]

you might want to round the numbers after the calculation

[slate oracle]

oh

[unique hill]

for more accuracy

https://cdn.discordapp.com/attachments/363224154469826562/694711932040511560/trig4.PNG

[slate oracle]

SO then

It's basically (tan35)(59)-(tan47)(59)

Which is

~22

[unique hill]

other way round

[slate oracle]

Again.

Yeah because it'll come out negative.

[unique hill]

59tan47 - 59tan35

[slate oracle]

Does it matter if 59 is in the front btw?

No.. right?

[unique hill]

the height of the building plus the radio tower is 59tan47
just the height of the building is 59tan35

[slate oracle]

Since we need to round it to be nearest hundredth place

Ok

[unique hill]

the 59 matters alot...

[slate oracle]

Yeah but

59tan47=tan47(59)

right?

[unique hill]

whut

[slate oracle]

Ok OIk

Nothing

[unique hill]

yes

internet lag

soz

[slate oracle]

But the final answer is 21.96

Since rounding to nearest hundredth

[unique hill]

yep you're correct

and that's the height of the radio tower excluding the height of building

[slate oracle]

Thanks again

So many thanks

[gaunt bay]

may i get some help now?

[unique hill]

np, happy to help

yes u may

[gaunt bay]

gimme a sec

[unique hill]

https://cdn.discordapp.com/attachments/363224154469826562/694706405969494076/c8c93c4307db6f586f2dc4521e3adb7f.png

i believe that's ur question

[gaunt bay]

yes

[unique hill]

you're trying to find the height of the airplane, but you're missing one side to do that
so you're trying to find that missing side

[gaunt bay]

im confused cause i try to use tan but i forget how to properly use it and when i tried searching couldnt finjd something good

[unique hill]

in the pic, you see how AE=150ft?

it's given

[gaunt bay]

yes

so i tried Tan(8.9)=150/x

[unique hill]

that seems correct

yes

[gaunt bay]

i keep getting a negative number

[unique hill]

you're trying to find x, in this case it's BE

so reaarange the tan(8.9)=150/x

[gaunt bay]

i got x=150-tan(8.9)

[unique hill]

???

no you dont do that

you're trying to find x

in tan(8.9)=150/x

$tan(8.9)=\frac{150}{x}$

[gaunt bay]

do i not need to single out x?

[unique hill]

yes u have to single out the x

you're trying to find x here

$tan(8.9)=\frac{150}{x}$

why is texit not working

[gaunt bay]

so i multiply the x to both side making it Xtan(8.9)=150 is this right?

[unique hill]

it would take an extra step if u do that

tan(8.9)=150/x

but yes you're right

[gaunt bay]

whats the easier step?

[unique hill]

just go straight to x=150/tan(8.9)

[gaunt bay]

ohhhhhhhhhhhh. wow i didnt think about deviding. thats where my mistake it. ughh

[unique hill]

it's like saying 6/2 = 3 and rearrange it so that it's 6/3 = 2

[gaunt bay]

yea you right. i think that was it. i can do the rest on my own. help out gabe if you can. that was literally the one thing i missed xD

[unique hill]

aight good job

[gaunt bay]

i was going mad cause i kept getting negative. but now i get why

ty again

[unique hill]

np

@viscid thistle

https://cdn.discordapp.com/attachments/363224154469826562/694715313811357786/unknown.png

[viscid thistle]

yea

[unique hill]

and to your answer, yes there is a quick way

look at the question and see what's common

[slate oracle]

Are.. you.. finished?

[unique hill]

one eighty u need help?

[slate oracle]

Yeah

[unique hill]

not sure when gabe will come back

[slate oracle]

I got 50 for the first part

[unique hill]

@viscid thistle @ me when u come back

[slate oracle]

Now I don't know what to do

sin12/50=sinb/79

[unique hill]

ok wait lemme read the question

[viscid thistle]

hey

[slate oracle]

Is the answer 50?

That's what I got.

[unique hill]

@slate oracle i'll deal with gabe first if u dont mind

[slate oracle]

Yeah

[unique hill]

ok gabe, do u see what is common in the question?

https://cdn.discordapp.com/attachments/363224154469826562/694715313811357786/unknown.png

@viscid thistle u there?

[viscid thistle]

yes

x-1

[unique hill]

yep, since x-1 is common, you can substitute sth (like a pronumeral) as x-1

@slate oracle im doing yours atm

[slate oracle]

Thank you

[unique hill]

wait holup

brb

[slate oracle]

All good

[unique hill]

i got the same step as you but somehow i got the wrong answer...

well my answer seems wrong

[slate oracle]

Lol

It's fine, thanks for everything

[unique hill]

i got 19.6 degrees

which seems wrong...

[slate oracle]

That's all the questions I had trouble with

Wahh

Welp, it's fine.

@unique hill Thank you so much for helping me with the 3 questions I was struggling with!

[unique hill]

that 49 is correct tho

for distance

np

[slate oracle]

mkmk

[unique hill]

@slate oracle

[slate oracle]

?

It's wrong

[unique hill]

apparently i got the ambiguous case for sine rule...that's my mistake

[slate oracle]

Ain't it

Oh

[unique hill]

anyways the distance is 49, which u got right?

[slate oracle]

Yeah

[unique hill]

now try drawing a triangle with the known distance and angles

[slate oracle]

Wait why?

Is that even necessary in this?

[unique hill]

so that you can interpret it better

and yes it's necessary

[slate oracle]

So 49 isn't the final answer..?

[unique hill]

49 is the distance

the question asks for distance and the change of angle

[slate oracle]

omg

[unique hill]

https://cdn.discordapp.com/attachments/363224154469826562/694723880060977212/foif.PNG

miles away from Peoria = the distance u found

[slate oracle]

So

in12/50=sinb/79

sin12/50=sinb/79

Let's do it without the triangle for now.

[unique hill]

btw, i believe it's 49

not 50 when rounded

cuz the value is 49.102...

[slate oracle]

ok

sin12/49*=sinb/79

[unique hill]

yep thats correct

you're trying to find the change of angle, which is b in ur eqn

[slate oracle]

Ok

x=0.33

ish

or 0.34~

[uncut mulch]

what's x?

[unique hill]

that looks incorrect

you mean b

[slate oracle]

sorry

I always confuse myself like this

lol

So then what's after?

There's no way b=0.33

[unique hill]

uh find b 🙂

no your b is incorrect

[slate oracle]

erhiuer

[unique hill]

that's why

[uncut mulch]

sin(b) approx 0.3345

[unique hill]

^

[slate oracle]

So we divide by sin?

[unique hill]

nono

[uncut mulch]

that would be a big fat NO

[unique hill]

you inverse sine the answer u got

@uncut mulch is texit down again?

[uncut mulch]

sin isn't a magical object being multiplied to b

[slate oracle]

LOL

[unique hill]

$sinx=\frac{79sin(12)}{distance}$

[obsidian monolith]

Nightingale:

[unique hill]

oh nvm

@slate oracle btw keep the exact values for the distance

[slate oracle]

So

We're trying to find distance right?

[unique hill]

the exact value part is crucial

it's approx 49

[slate oracle]

rIGHT

[unique hill]

u need the exact value cuz when i did it, the rounding bit gets rly rly close between rounding up or rounding down

[slate oracle]

Sorry

My dad keep calling me

[unique hill]

no worries

[slate oracle]

So are we separating distance from everything?

Or.. x?

No we're separating x

[unique hill]

ok we're trying to find the exact value of the distance

[slate oracle]

mb

x=sin^(-1)79sin(12)/distance

await

[unique hill]

distance

[slate oracle]

We don't know the x though

[unique hill]

yeah but to find x, we need to find the distance

exact value of distance pls

[slate oracle]

Sorry sorry

sinx/12=79sin/distance

[unique hill]

'sin' ?

[slate oracle]

Uh what?

[unique hill]

the 'sin' by itself...

[slate oracle]

Ok ok.

[unique hill]

and also, we need to draw a triangle to find exact value of the distance

[slate oracle]

What

Thought we didn't..

[unique hill]

ok this is the part that's crucial:

if it's rounded normally without the exact value, the answer ends up differently than u did if it's in exact values

cuz rounding in tenth place

so dont round the distance to 49

[uncut mulch]

when evaluating the distance you should've had something like:

,w sqrt(79^2+31^2-23179*cos(12 deg))

fkn wolfram

[unique hill]

this is why u need the triangle

[slate oracle]

Ok..

[uncut mulch]

triangle helps visualise things

[unique hill]

$sqrt(79^2+31^2-27931*cos(12))$

tf

[slate oracle]

lol

[obsidian monolith]

Nightingale:

[uncut mulch]

$d = \sqrt{79^2+31^2-2\cdot31\cdot79\cdot\cos(12\deg)}$

[unique hill]

cos

[obsidian monolith]

ramonov:

[unique hill]

@slate oracle that's ur exact distance

[slate oracle]

ok yeah

[unique hill]

do u how to get that exact form?

[slate oracle]

SO then the equation is sinx=79sin(12)/distance(which is approx. 49)

[uncut mulch]

and the value evaluated from that could be stored into your calculator

[unique hill]

^ important

[uncut mulch]

or be a parentheses god and enter it all in the calc at once

[unique hill]

i did it once without storing and when rounded to tenth place it's a different answer

[slate oracle]

oh

[unique hill]

so are u aware that we used the cosine rule to find the exact area?

[slate oracle]

Yes

[unique hill]

ok now u know the distance

[slate oracle]

So can't we write it as sinx=79sin(12)/sqrt(79^2+31^2-27931*cos(12))

?

[unique hill]

$sinx=\frac{79sin{12}}{\sqrt{79^2+31^2-27931*cos{12}}}$

looks horrendous without the sqrt command

[uncut mulch]

you can let
d = sqrt(stuff)
in an earlier line to save space

[unique hill]

also it's wrong

wait

[uncut mulch]

which you should've done when approximating the distance anyway

{}

\cdot

[unique hill]

...

[stuck lark]

night you may want to follow ramonov's latex to improve

[uncut mulch]

\sqrt

[obsidian monolith]

Nightingale:

[unique hill]

god

there we go

[slate oracle]

So...

[uncut mulch]

close enough

[slate oracle]

can't we write it as that

[fleet yew]

simplify that

[slate oracle]

ok

[stuck lark]

$\sin(x)=\tfrac{79\sin(12^\circ)}{\sqrt{79^2+31^2-2(79)(31)\cos(12^\circ)}}$

[uncut mulch]

what's tfrac

[obsidian monolith]

RokettoJanpu:

$\sin(x)=\tfrac{79\sin(12^\circ)}{\sqrt{79^2+31^2-2(79)(31)\cos(12^\circ)}}$

[stuck lark]

$\tfrac12\frac12$

[obsidian monolith]

RokettoJanpu:

[uncut mulch]

mmk

[unique hill]

@slate oracle u with us

[slate oracle]

yeah

so i simplify that to find x?

[unique hill]

yep, inverse sine the whole thing

[uncut mulch]

apply arcsin to both sides to find x

[stuck lark]

i use tfrac a lot more often after learning how useful it is to nicely cram all those identities on my trig sheet

[unique hill]

@uncut mulch i dont think he knows what arcsin is

[slate oracle]

so then it's csc(0.33)

[unique hill]

uhhhh

[slate oracle]

or..

[unique hill]

exact form

[slate oracle]

3.04

[uncut mulch]

no

[unique hill]

so did u store the values into the calculator?

[uncut mulch]

that's why we use arcsin

[slate oracle]

ok

19.54

[uncut mulch]

instead of sin^(-1) which is bad notation

[slate oracle]

but they're the same thing

no?

[unique hill]

nope

[slate oracle]

bjklgyjghhknl

[stuck lark]

as long as you clearly understand that $\sin\inv$ refers to $\arcsin$ instead of $\tfrac1\sin$ then you're fine

[obsidian monolith]

RokettoJanpu:

[unique hill]

arcsin doesnt not equal to csc

[uncut mulch]

$\sin^{-1}(x) \not\equiv \frac{1}{\sin(x)}$

[slate oracle]

yeye

[obsidian monolith]

ramonov:

[unique hill]

lmfao that equivalent sign

[stuck lark]

it's needed

[slate oracle]

=+-

[uncut mulch]

oh btw how do you get texit to display emotes

[slate oracle]

SO then 19.54 is my answer?

[unique hill]

180-that

cuz u found the ambiguous angle by using sine rule

it has two different sol

[stuck lark]

texit devs only handpicked a select few emotes that you can actually render

off the top of my head there's $\thonk\catthink$

[obsidian monolith]

RokettoJanpu:

[unique hill]

$\thonk$

[stuck lark]

$\mniip$

[obsidian monolith]

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

[stuck lark]

nope

[obsidian monolith]

Nightingale:

[unique hill]

@slate oracle uh u with us?

[slate oracle]

yeah

[unique hill]

180- the answer u got

[slate oracle]

im deceased

[unique hill]

180-(stored answer in calculator)

[slate oracle]

yes

im going to commit suicide

[uncut mulch]

its better to have applied the cosine rule again

[slate oracle]

NO NO STOP

IM DONE

I GOT THE ANSWER

[uncut mulch]

so you don't have to bother checking for ambiguity

[unique hill]

okay what is it

@uncut mulch true

rounded to tenth decimal place bc question asks for it

[slate oracle]

160.45?

[unique hill]

tenth

[slate oracle]

46*

[unique hill]

place

no that's hundredth place

[slate oracle]

5*

LMAO

[unique hill]

160.5 should be ur answer

[slate oracle]

omg

what math are u in btw

u sound like hawking to me

[unique hill]

wdym by what math?

hawking????

[slate oracle]

like do u study calculus ab

or something

[unique hill]

im in 11th grade

[slate oracle]

.

.

.

wow

i look so stupid now

[unique hill]

...

but u get the answer right?

[slate oracle]

yeah

im sure i did

[unique hill]

nice good job

[slate oracle]

all u lmfao

[unique hill]

np

[harsh cipher]

nightingale you are not in 11th grade

proof please

[fleet yew]

!cough @harsh cipher

[unique hill]

Mirrion’s still not infected

@harsh cipher what proof?

[fleet yew]

!cough @harsh cipher

[unique hill]

!sneeze @harsh cipher

[harsh cipher]

lolol

that you're not in 11th grade

sorry I was talking to dad

[unique hill]

what makes u think im not

[harsh cipher]

it was just an assumption

[unique hill]

alrighty

[harsh cipher]

yep 🙂

[daring yarrow]

aight dumb question

but if i do 3*-∞

does that make it +∞

is this -∞

[willow bear]

yes

[carmine elbow]

I want to see if this is right

[lilac pier]

, rotate

[unkempt stirrup]

,rotate 180

[lilac pier]

@carmine elbow What's up with the numerator after plugigng x=0 in (gof)(x)

[carmine elbow]

That’s f(x) inside of g(x)

[lilac pier]

Yes but what did you do after that

[carmine elbow]

They asked me to plug in zero @lilac pier

Oooh, I see what you mean. The numerator is supposed to be zero

[tardy ridge]

@viscid thistle limits

x^4 is the largest power so it'll impact it almost 100% when x-> infinity

[sharp marsh]

How do I do this?

[fleet yew]

How do you do what?

[sharp marsh]

lol sorry the message didn't go through

Give me a moment

[lilac pier]

@sharp marsh part (b) is straight forward, you just have to plug in the values

[sharp marsh]

Yep

I'm just wondering about c

[lilac pier]

oh

part (a) also done?

[sharp marsh]

uhh

[lilac pier]

part c is basically what happens as t approaches +infinity

[sharp marsh]

yeah

Idk how to do A and C

[lilac pier]

have you studied derivatives?

@sharp marsh If you have studied derivatives, you could find the derivative of the function and see if it is > 0 or < 0

[sharp marsh]

uhh

no

I'll just do another problem for now

f(x) = 2(x + 5)/x^2 + x − 20

Is the horizontal asymptote y=0
and is the vertical asymptote x =-5,4?

And when does the vertical asympote not exist*

[willow bear]

you're missing some parentheses here

around the denom, to be more specific.

ASSUMING for a moment that you meant $f(x) = \frac{2(x+5)}{x^2 + x - 20}$, factoring the denominator gives you $(x+5)(x-4)$, and the $(x+5)$ factors in the num and denom cancel out. so here you have a hole, rather than an asymptote, at $x=-5$.

[obsidian monolith]

Ann:

[sharp marsh]

So it's only x = 4?

because of the hole?

For the vertical?

[willow bear]

the only vertical asymptote here occurs at x=4, because of the factor of (x-4) that did not cancel out with anything else

[sharp marsh]

ok

[sharp marsh]

So when does the vertical asymtote not have one?

[willow bear]

i think you aren't wording your question properly

[sharp marsh]

Why is it not -5...

[uncut mulch]

consider what you determined in part a)

and what's written above

[sharp marsh]

OHH

that makes so much more sense

[slate echo]

Does anyone know how to express tan(-315°) as a trigonometric function of an angle in quadrant 1? I'm pretty sure the answer is but idk what the steps are

[uncut mulch]

consider the periodicity of the tangent function

[tardy ridge]

whenever you add 2pi the value doesn't change

[carmine elbow]

Can someone explain to me how to find values on a graph? Problem number 57

[fleet yew]

@carmine elbow this is a composite function

it is asking for $(f ∘ g)x$

[obsidian monolith]

AMD:

[fleet yew]

which is the same thing as $f(g(x))$

[obsidian monolith]

AMD:

[fleet yew]

so you must find f(g(3))

to do that

you find g(3)

what is g(3)

[carmine elbow]

let me see, hold on

0?

[fleet yew]

yep

you know g(3) is 0

and you must find f(g(3))

so plug in 0 for g(3)

and find f(0)

[carmine elbow]

is that 4?

[fleet yew]

yeah you got it

[carmine elbow]

are there any videos on this? I want to make sure I don't forget how to find values on a graph

[fleet yew]

here's a good one

https://www.khanacademy.org/math/precalculus/x9e81a4f98389efdf:composite/x9e81a4f98389efdf:composing/v/evaluating-composite-functions

[carmine elbow]

Thank you for the help!

[sharp marsh]

log8 x + log8(x + 3) = log8(x + 15)

How do I do this

I am at the part where it is log8(x^2+3x) = log8(x+15)

[unique hill]

is that base 8 or no?

[sharp marsh]

yep

Canceling it would be x^2+3x=x+15 right?

[unique hill]

btw u might want to type sth like log_8 x or use texit

[fleet yew]

yes

[sharp marsh]

Ok

Then what do I do after?

[fleet yew]

then move all the terms to one side

[sharp marsh]

I see I see

I know what I did wrong

thanks

[unique hill]

that was quick

[sharp marsh]

yeah

So like for this problem

log_6 x + log_6(x + 6) = log_6(x + 24)

Is the answer 8?

Or is it both 8 and -3?

Or did I just do it wrong?

or 3

nvm it's 3

thx

[fleet yew]

you have to plug it back into the original equation

see if it still works

[sharp marsh]

o h

[daring yarrow]

AMD has covid

[hollow prawn]

https://cdn.discordapp.com/attachments/326138737606262786/695097217857486988/chrome_UMhjYPeeIG.png

help

[unique hill]

u already posted this somewhere

[hollow prawn]

ik but people don't answer it

[unique hill]

during 1:28

[hollow prawn]

ok thx

[unique hill]

not getting the solution for u

[hollow prawn]

whyy

[unique hill]

we help u get to the solution

[hollow prawn]

ok

[unique hill]

what's the point of learning if u just have the solution but not the steps

[hollow prawn]

because I can just figure out how u got da solution

[unique hill]

so you're good now?

[hollow prawn]

no

I still don't know

[unique hill]

okay first

$\frac{1}{3}+\frac{1}{2}$

[obsidian monolith]

Nightingale:

[unique hill]

how would u approach that

[hollow prawn]

lcd

2/6

1/3

[unique hill]

and what do u do if you have the LCD?

[hollow prawn]

add the top

[unique hill]

https://cdn.discordapp.com/attachments/326138737606262786/695097217857486988/chrome_UMhjYPeeIG.png

so what should be your first step here?

[hollow prawn]

i get -vw+v^4/20

which I don't think is right tbh

[unique hill]

i dont think that's correct

ok lets go back to the fraction

[hollow prawn]

ok

[unique hill]

$\frac{1}{3}+\frac{1}{2}$

[obsidian monolith]

Nightingale:

[unique hill]

can u provide me the answer? guide me through each step pls

[hollow prawn]

LCD first

3*2

6

now add the numerator

1+1

2

2/6

simplify

/2

1/3

[unique hill]

can u type it out in a proper format

[hollow prawn]

what do you mean

[unique hill]

like how u would write ur working out

to solve that question

[hollow prawn]

3 * 2

6

1+1

2

2/6

2/6 = 1/3

[unique hill]

is that what your working out looks like when doing fractions?

[hollow prawn]

yea

[unique hill]

ok are you sure the final answer is 1/3

so are u saying that 1/3 + 1/2 = 1/3

your working out is crucial here

[hollow prawn]

nvm

so 5/6

[obsidian monolith]

Nightingale:

[hollow prawn]

no

5/6

[unique hill]

yes

now can u list the working out required to get that answer?

[hollow prawn]

1/3 = 2/6

1/2 = 3/6

2/6 + 3/6

5/6

[unique hill]

yes that looks more correct

but....

anyways

https://cdn.discordapp.com/attachments/326138737606262786/695097217857486988/chrome_UMhjYPeeIG.png

[hollow prawn]

yea

[unique hill]

guide me through the steps for this one

[hollow prawn]

vw-v^4

-v^4+vw

[unique hill]

steps

[hollow prawn]

what u mean

4*5

20

-v^4+vw/20

[unique hill]

your first step should make them the same denominator

[timid lagoon]

this is precalc

[unique hill]

yes ik what u did but the working out...

[hollow prawn]

ok so 5vw/20 - 4v^4/20

[unique hill]

yes that's better

wait it's wrong

[timid lagoon]

um

[unique hill]

second term

[hollow prawn]

y

oh

4v^4/20

[unique hill]

yes

now that you have the same denominator, what do u need to do next?

[hollow prawn]

subtract numerator

[unique hill]

yes

[hollow prawn]

(5vw-4v^4)/(20)

[unique hill]

parenthesis

plz

$\frac{5vw-4v^4}{20}$

[obsidian monolith]

Nightingale:

[hollow prawn]

that the answer?

[unique hill]

no your parenthesis is still wrong

the numerator is inside the parenthesis

ok there we go

[hollow prawn]

ok thx

for ur help

[unique hill]

np

[daring yarrow]

,rotate

[daring yarrow]

nvm

[unique hill]

which question

[daring yarrow]

solved it myself the second i sent

sorry

[unique hill]

aighty

[daring yarrow]

was confused on f but miswrote the question

so ofc it didnt work out

[unique hill]

wait so do u need help or not?

[daring yarrow]

nono

[unique hill]

okie

[daring yarrow]

thx anyway

[unique hill]

O.o

@vestal birch

here

[vestal birch]

#1 anyone?

Thanks @unique hill

[lilac pier]

@vestal birch You can start from RHS, you should know tan = sin/cos

tan^2 will give you sin^2 / cos^2

simplify then

[vestal birch]

Oh ok thanks 🙂

[novel dirge]

Hello, does the binomial formula belong here?

I need some help with understanding how it works

Because I don't know what to do with the coefficient in the parantheses

[willow bear]

$\binom{n}{k}$ is just another notation for $n$ choose $k$, or in other words $\frac{n!}{k!(n-k)!}$

[obsidian monolith]

Ann:

[willow bear]

sometimes it's denoted $C_n^k$ iirc

[obsidian monolith]

Ann:

[novel dirge]

so what am I supposed to do with it?

do I just leave it there or what?

I understand the other part of the formula

but not how to connect it with the parantheses and get the result

[willow bear]

can you show the problem you're doing

[novel dirge]

It is the upper left corner

Just started doing it

these were sent by my prof.

and she didn't even explain how to do this, just sent us the formula and practice problems

[willow bear]

uh

okay wait hang on

problem 1?

[novel dirge]

yes, starting from that

[willow bear]

can you transcribe or translate that? i'm having a bit of trouble reading the directions

[novel dirge]

define the fifth member in the expanded form of the binomial

[willow bear]

fifth term

okay

and that's... $(x^{1/2} + x^{2/3})^{42}$ then?

[obsidian monolith]

Ann:

[novel dirge]

$(x^{1/2} + x^{2/3})^{12}$

[obsidian monolith]

Mr.Pancake:

[willow bear]

oh 12

okay

well

expanding this out via the binomial theorem you get $$\sum_{k=0}^{12} \binom{12}{k} (x^{1/2})^{12-k} (x^{2/3})^k$$

[obsidian monolith]

Ann:

[willow bear]

for the fifth term, take the term corresponding to k=4 - since k starts at zero

$\binom{12}{4} (x^{1/2})^8 (x^{2/3})^4$

[obsidian monolith]

Ann:

[willow bear]

then simplify

,w 12 choose 4

[obsidian monolith]

Display results online and refine query

Requested by Ann#0413

[novel dirge]

I get that part

how to get it to that

but what I am confused about is what to do with the thing in the parantheses

how do I actually simplify that

[harsh cipher]

Good afternoon everyone

checking in

[novel dirge]

hello

[willow bear]

as i said

$\binom{n}{k} = \frac{n!}{k!(n-k)!}$

[obsidian monolith]

Ann:

[willow bear]

this is just the defn

[novel dirge]

so do I also have to calculate that and multiplay it with the second part?

that seems like a ton of work to do

[willow bear]

$\frac{12}{4} = \frac{12 \times 11 \times 10 \times 9}{4!}$

[obsidian monolith]

Ann:

[willow bear]

the other part is just x raised to some powers

it's not that bad

[novel dirge]

so it is just that?

calculating it and multiplaying with everything?

that is much more simple than I thought

thank you

[willow bear]

multiplying* but yes

[novel dirge]

is there a way to find which term doesn't contain x, without expanding the entire problem?

[willow bear]

take the general term, calculate the exponent on x in terms of k, and set that to zero

[novel dirge]

can you explain that a bit more?

[willow bear]

so your thing now is (x + x^-2)^12 yes?

[novel dirge]

yes

[willow bear]

$\sum_{k=0}^{12} \binom{12}{k} x^{12-k} x^{-2k}$ is what you get upon expanding it with the binomial theorem

[obsidian monolith]

Ann:

[willow bear]

$x^{12-k}{-2k} = x^{12-3k}$

[novel dirge]

yes

[obsidian monolith]

Ann:

[willow bear]

you want 12-3k=0

so k=4

and your term is $\binom{12}{4}$

[obsidian monolith]

Ann:

[novel dirge]

Ann:
@obsidian monolith is there supposed to be an X there?

...

[willow bear]

no

lmao

the point is that we made it so that there wouldn't be an x

as the problem asks

[novel dirge]

just found out that quotin teXit doesn't work

[willow bear]

x^0, if you insist

[novel dirge]

but how do we keep one X, but not the other?

[willow bear]

what do you mean

[novel dirge]

$x^{12-k}{-2k} = x^{12-3k}$

[obsidian monolith]

Mr.Pancake:

[willow bear]

no

x^(12-k) * x^(-2k) = x^(12-k - 2k)

[novel dirge]

that one is correct?

[willow bear]

our terms are $\binom{12}{k} x^{12-3k}$ for $k = 0, 1, ..., 12$. we want the one that doesn't contain an $x$. for it to not contain an $x$, the exponent on the $x$ has to be zero. so it's the one with $k=4$.

[obsidian monolith]

Ann:

[novel dirge]

yes, I understand that part

but not this $x^{12-k}{-2k}$

[obsidian monolith]

Mr.Pancake:

[novel dirge]

where did the second X go?

[willow bear]

IT'S NOT $x^{12-k} - 2k$ !!!!!!!!!!!!

[obsidian monolith]

Ann:

[willow bear]

oh god.

oh god i typoed.

sorry

x^(12-k) * x^(-2k) = x^(12-k - 2k)

[novel dirge]

...

[willow bear]

this is what it was meant to be

[novel dirge]

That is what I was asking all along

Now I do understand how to do it, or at least I think

I will try it

[unique hill]

IT'S NOT $x^{12-k} - 2k$ !!!!!!!!!!!!
lmao

[obsidian monolith]

Nightingale:

[novel dirge]

I am stupid

how do I get the $x^{12-3k}$

[obsidian monolith]

Mr.Pancake:

[willow bear]

$x^{12-k} \times x^{-2k} = x^{(12-k) + (-2k)}$

[obsidian monolith]

Ann:

[novel dirge]

ohh

I messed up the problem

I thought it was - instead of x

so couldn't figure out how to the it with subtraction

do know how to do it with multiplication though

thank you

[unique basin]

Is this allowed in math, to use diffrent unit scales on diffrent axis?

There should be a minus, for 4 and 8. Forgot that

[willow bear]

sure

[unique basin]

Would this be allowed aswell?

[willow bear]

if you want

[daring yarrow]

could someone solve this please

i solved it and got -2

so did my friend

but the online correction key it should be something completely different

LL = +oo and LL = -oo

[willow bear]

LL?

@daring yarrow can you show how you got -2? i'm curious to see where you fucked up

[daring yarrow]

left limit and right limit

dont bully please

[willow bear]

$x^3 + x^2 - x - 1 \neq (x-1)(x+1)$

[obsidian monolith]

Ann:

[daring yarrow]

(x-1)(x+1)²

does that fix anything

[willow bear]

uh huh

and it also radically changes what you're left with

[daring yarrow]

but i had that

wait

no i didnt

bruuuh thanks

one day i wont make small brain mistakes anymore

@willow bear the (0/0) part is correct tho right?

the book says that the top part is equal to -2

so its -2/0

[willow bear]

once you cancel off one pair of (x+1) factors.

that's when it becomes -2/0

[daring yarrow]

oh right

cant solve rn cuz there's a live class rn

will do after tho

thanks alot

[viscid thistle]

Yeah just replace x with -1

You find -2/0

But

+0 or -0?

[willow bear]

it's a two-sided limit

[viscid thistle]

Yes

[daring yarrow]

+0 -0

thats a thing?

[viscid thistle]

Yes

For exemple

3/-0 = -infinity

And 3/+0 = +infinity

Sorry I don't know if the notation is correct

If you search the limit of 1/x you will find that

x -> 0

[daring yarrow]

oh

idk i just make a grid-like thingy to see what the value of the bottom part is when it comes from the left and whe it comes from the right

doing that i guess

feels weird writing -0 tho

[viscid thistle]

Ah ok

[daring yarrow]

how does one stop making dumb mistakes in math tho?

[viscid thistle]

Do a lot of exercices

If you want I have a funny exercise

[daring yarrow]

sure

[viscid thistle]

Ok

[harsh condor]

-0 and +0 exists in programming, idk about maths

[viscid thistle]

Find two way to prove that
4^n-1 can be divided by 3
n is a natural integer

[daring yarrow]

proof by idk allowed?

if so :
idk
Q.E.D

i'd probs go for an induction proof or something but im too lazy

[viscid thistle]

Sorry I'm a little French guy I don't understand your post

[daring yarrow]

idk = i dont know
Q.E.D = what u put at the end of a proof

its an attempt at being funny

[viscid thistle]

Induction is when you prove the hereditary?

[harsh condor]

By induction, you assume the base case is true.

[viscid thistle]

Ok so yes induction works

[harsh condor]

Assuming 4^(0)-1 is supposed to be divisible by 3

[viscid thistle]

Yes yes

There are another method that is way more simple though

[harsh condor]

Although I think induction is unnecessary

[viscid thistle]

Yes

[harsh condor]

I think the key is to know the definition of Odd and Even numbers:
Even: 2x (2 times any number is even)
Odd: 2x - 1 (even numbers - 1 are all odd)

[viscid thistle]

More simple

In which class are you?

[daring yarrow]

would induction work?

[harsh condor]

Yes induction should work

[viscid thistle]

Yeah induction work

But it's unnecessary

Hint : ||congruence||

[harsh condor]

That makes me think of triangles

So the statement was 4^n - 1

[daring yarrow]

this isnt true tho is it cuz if u take n = 2

[viscid thistle]

Solution :
||
4 = 1 mod 3
4^n = 1^n mod 3 = 1 mod 3
4^n - 1 = 0 mod 3
||

[daring yarrow]

4 is divisble by 3?

[viscid thistle]

No

[harsh condor]

4^2 = 16

16 -1

=15

[daring yarrow]

oh wait

-1 isnt in the power

xddd

[harsh condor]

which is divisible by 3

[viscid thistle]

No 1 is after

[daring yarrow]

my bad

[viscid thistle]

You should do that with congruence

[harsh condor]

You can do that with mod?

lol didn't realize

[viscid thistle]

Yes

Mod work pretty well

[harsh condor]

(1^n mod 3) feels slightly illegal

[viscid thistle]

Congruence preserve power

Integer power*

[harsh condor]

So given X^n
if X is congruent to mod(a, b)
then mod(a^n, b) is congruent to X^n

[viscid thistle]

What is mod(a, b)

[harsh condor]

a mod b

[viscid thistle]

X = a mod b
X^n = a^n mod b

[harsh condor]

since a^n mod b, I wasnt sure if u mean (a mod b)^n, or (a^n) mod b

So given X^n
if X is congruent to "a mod b"
then "(a^n) mod b" is congruent to X^n
e.g
4 is congruent to "1 mod 3"
then "(4^n) mod 3" is congruent to 4^n

[viscid thistle]

Sorry I don't understand duh

(4^n) mod 3 is congruent to 4^n

You mean

4^n = 4^n mod 3?

[harsh condor]

So you said:
4 = 1 mod 3
4^n = 1^n mod 3 = 1 mod 3

[viscid thistle]

Yeah

Because 1^n = 1

[harsh condor]

But say:
9 = 4 mod 5

9^n = 4^n mod 5 = 4 mod 5

wouldn't work

[viscid thistle]

Nobo

It's is

Because 1^n always equals 1

[harsh condor]

Right so only with A = 1 mod B

So if you have A = 1 mod B,
then A^n = 1^n mod B = 1 mod B.

[viscid thistle]

Yes

Because 1^n = 1

[harsh condor]

Yep that makes sense

[viscid thistle]

Post avant swarm

Toujours le même exo t'as vu ?

[harsh condor]

post avant swarm sounds like some band name or genre

[viscid thistle]

No it's French

[harsh condor]

I noticed

[earnest flint]

4^n -1 = 0 mod 3
4^n = 1 mod 3
Since 4 = 1 mod 3
4^n = 1^n mod 3 = 1 mod 3

[viscid thistle]

Impressive swarm

You have learn your lesson very well

[earnest flint]

^^

[viscid thistle]

:nose:

[harsh condor]

For all real x's where x != 2, there is a unique y where y=-x/(x-2) (which i assume means 1-1)

Any idea how I'd do this?

[willow bear]

For all real x's where x != 2, there is a unique y where y=-x/(x-2) (which i assume means 1-1)
uh,,, no? as stated, this is not the same as saying "the function f(x) = -x/(x-2) is injective"

would you mind showing the exact statement of the problem you're doing

[harsh condor]

https://d2vlcm61l7u1fs.cloudfront.net/media%2F6a4%2F6a47a432-dd2b-4110-86a4-8426e083c5f8%2FphpE6LmFa.png

Random question from google actually. Just googled for proving math question examples

I assumed "for all real x's ..., there is a unique y", to mean one-to-one.

[willow bear]

no

wrong assumption

you don't know what one-to-one actually means

[harsh condor]

I believe that 1-1, means the domain maps to codomain

or something along those lines.

[willow bear]

yeah no

[harsh condor]

What does it mean?

that maps distinct elements of its domain to distinct elements of its codomain

So basically any function which has a unique output for every unique input?

Or am i wrong here?

[willow bear]

you're using the word "unique" sloppily

we say that a function is one-to-one if it never maps two different inputs to the same output. that's one way of putting it which doesn't go into imprecision.

[harsh condor]

Sure, I guess that works.

What would be a case where "a unique outptu for every unique input" would be incorrect?

[willow bear]

there's a difference between incorrect and sloppy

[harsh condor]

I see:

For all real x's where x != 2, there is a unique y where y=-x/(x-2)
The assumption that it is 1-1 is incorrect in this case.

"As two different inputs" could "map to the same output?"

[willow bear]

no it's not that

the statement just does not say anything about any function being one to one or not

[harsh condor]

Right, so I cannot assume it means it is 1-1

Because they didn't state it.

[viscid thistle]

Where you have to prove that

You must prove y exists then y is unique right?

I think you should take y and y' and find that there are equals

[harsh condor]

For all real x's where x != 2, there is a unique y where y=-x/(x-2)
I think I have to prove that there is a unique y, for all real x's

[willow bear]

Right, so I cannot assume it means it is 1-1
not only are you being sloppy again, you seem not to understand my point at all

[viscid thistle]

yn= x/(2-x)

So y exists

[harsh condor]

Sorry, I seem to have missed it, I interpreted that you were saying if "the statement does not say anything about the function being one to one" it cannot be assumed to be one to one.

[viscid thistle]

Now let's search if y is unique

[harsh condor]

Sure, how would I search?

[viscid thistle]

To prove that y is unique cannot you just write that "x/(2-x) have only one image when x=/=2"?

[harsh condor]

Not entirely sure tbh, it was a random question from the internet:
https://images-ext-1.discordapp.net/external/HziZOpVHsh52_q_WMjlQQf1N0NIpSnkavpW-BGX46LQ/https/d2vlcm61l7u1fs.cloudfront.net/media%252F6a4%252F6a47a432-dd2b-4110-86a4-8426e083c5f8%252FphpE6LmFa.png?width=400&height=102

[viscid thistle]

I try a method

[harsh condor]

And I got up to: y=-x/(x-2)

[viscid thistle]

-x

Crap I don't find the same value I retry my calcul

I find the same expression

Hmmm

I'm right

I will verify with your expression

[harsh condor]

I was thinking, if I add a 'a' and 'b', I could try prove for equality.

y = -a/(a - 2) = -b/(b - 2)
implies a = b

[viscid thistle]

y0 =...
y1 =...

Yes try that

Ok your expression is wrong

[harsh condor]

probably

[viscid thistle]

Show your calculs

I will tell you where you made a mistake

[eternal thunder]

To prove that y is unique cannot you just write that "x/(2-x) have only one image when x=/=2"?
@viscid thistle you'll have To justify this

[viscid thistle]

Yes

Bon je te parle en français

En gros j'ai y0 = x0/(2-x0)

[eternal thunder]

In english please

[viscid thistle]

Et y1 = x1/(2-x1)

[harsh condor]

x = 2y/(y+1)
(y+1)x = 2y [multiply both (y+1)]
yx + x = 2y [open brackets on (y+1)]
yx = 2y - x [minus -x on both sides]
yx - 2y = -x [minus 2y on both sides]
y(x-2) = -x [factor out the y]
y = -x/(x-2) [divide (x-2)]

[viscid thistle]

Et si je prouve

Bizarre

Je trouve pas son erreur

[harsh condor]

Bonjour!

[viscid thistle]

Heu bonjour

Ah yes

-x/(x-2) = x/(2-x)

Ok si we are both right

[harsh condor]

Noice
So does that mean proving
y = -a/(a - 2)
z = -b/(b - 2)
y = z implies a = b
Should count as proving it?

[viscid thistle]

Yes I think

Done

I have prove it with this method

[harsh condor]

I think so too.
-a/(a-2) = -b/(b-2)
-a/(a-2)*(b-2) = -b [multiply both sides with b-2]
-a(b-2)/(a-2) = -b
-ab+2a = -b(a-2) [multiply both sides with a-2]
-ab+2a = -ab-2b
2a = -ab+ab-2b [add +ab to both sides]
2a = 2b
a = b [divide by 2]

Does this count as a proof though?