precalculus

[hasty wadi]

ur right

lol

wait, so do we totally disregard arcsec (0) ?

cuz there is no answer for it?

[patent beacon]

There's no number x that would make sec(x) = 0.

So while sec(x) = 0 would solve our equation, we just can't make that happen.

[hasty wadi]

so we disregard it

and we stick to sec(x) = 1

we cross out sec (x) = 0

right?

[patent beacon]

Sure, that works

[hasty wadi]

so arcsec (x) = 1

same thing as arccos (x) = 1

?

[patent beacon]

You're backwards.
arcsec(1) = x

[hasty wadi]

yes sorry

my answer is 2(pi)(n)

but

were looking for [0,2pi)

sooooooo the answer is only 0

?

[patent beacon]

Then arcsec(1) = arccos(1/1) = 0

And yeah, the only answer that solves is indeed 0

[hasty wadi]

yes, it's the reciprocal (which ends up being 1 anyways)

alright, i'll put 0

let's see if its right B)

@patent beacon thank you so much dude, your a real help

i just got confused with the arcsec (0) = arccos (1/0)

my fault on that

lmao

[patent beacon]

No problem at all. I make dumb mistakes like that all the time, lel

Did it work?

[hasty wadi]

i think im really understanding it now lol

yes

it did 😃

thanks lots dude, i really appreciate it

👌

[patent beacon]

Np. Feel free to ask if you have anything else

[hasty wadi]

thanks 😄

[viscid thistle]

How can I find the turning points of this function without calculus?

If I need calculus, how can I do this without a calculator?

[severe verge]

this isn't a function

[viscid thistle]

ok

but if I graph this, the turning points, how can I find them algebraically?

\

[dense terrace]

u have to use implicit differentiation

[viscid thistle]

I can't simply squareroot both sides?

[dense terrace]

but what do u get after square rooting both sides

[viscid thistle]

y=\sqrt{x}\left(x-4\right)

wait soz

[dense terrace]

ya but

what does that tell u

[viscid thistle]

It tells me the intercepts

[dense terrace]

thats it right

[viscid thistle]

Yea

[dense terrace]

hmm

[viscid thistle]

but +-

[dense terrace]

u cant find the stationary points

[viscid thistle]

Yeah, with desmos I've checked them to be at 1.3333

but I don't know the method to arrive at that value of x

[quartz garnet]

Differentiate both sides

[viscid thistle]

I have and it's pretty ugly

0 = 0.5 + 1/sqrt(x) - 2/x

[dense terrace]

nop

=tex 2y\frac{\rm dy}{\rm dx} = (x-4)^2 + 2x(x-4) \
(x-4)^2+2x^2-8x=0

[dense terrace]

x = 4 or 4/3

[viscid thistle]

ok thanks

but just wondering, is this possible without calculus?

without guess and check of course

[dense terrace]

um

[viscid thistle]

and neither a calculator

[dense terrace]

oh ya theres a way

plot the graph on a graph paper

[viscid thistle]

my problem book asks to use minimal calculus but this could be it

ah nah that's probably it then

thanks :_

😃

the differentiation not the plot

[wise umbra]

Hello My doods

I am trying to get through pre calculus xD

[clever inlet]

Nice

[wise umbra]

lol u know whats sad

I have that F+ ;)))

freshman pre cal is hard

along with chem

[clever inlet]

Rip

[wise umbra]

lol

[tall oxide]

Can someone check my answers to see if I’m right?

[viscid thistle]

Can someone help me with synthetic division? ;-;

The problem is (4x^4 +2x^3 -4x^2 +2x +2)÷(2x+1), I don't know what to do with the leading coeff of the divisor ))))::::

@ me if you a real one 💯 💯 💯 😤 🙏 💪

[dense zealot]

Ok

so

x = -1/2

So we put -1/2 on the left

Make an upside down division bar

Put the coefficients in

Hm

Make the top polynomial

Have a leading coefficient of 1

If u wanna make it simple

If u don't...

Then

Ud make the leading coefficient 2

So u get 2x^3-2x+2

[viscid thistle]

@dense zealot oh wait so when you factor out the 2 from (2x+1) do you ÷ the whole poly. by 2? So after factoring it'll be (2x^4 +x^3 -2x^2 +x +1)÷(x+1/2)???? If so you either made it make complete sense to me or I'm more lost than ever ❤

[dense zealot]

Hi

1 sec I'm eating

just divide it by -1/2

what

Here

[viscid thistle]

You're fine take your time

But I thought the formula was x-k and we take k

Binomial or sommmm

[dense zealot]

So

U just divide the polynomial by 2

And the divisor too

so u have x+1/2 = 2x^4+....

Cuz u want the divisor to have leading coefficient 1

So u divide it by 2

And u do the same for the polynomial

Then u can synthetically divide easily

[viscid thistle]

So basically factor out the leading coeff of the divisor for both?

[dense zealot]

?

U want 2x+1

To have a leading coefficient of 1

So u divide everything in it by 2

To get x+1/2

And since u divided that by 2

U divide the polynomial by 2 too

So u get 2x^4 + x^3 - 2x^2 + x +1

[viscid thistle]

I understand it now but I guess I'm over complicating it lmao

Would we ever multiply the whole thingy by two again? Like once we find q(x)?

[dense zealot]

No

Since u divide both by 2 it's balanced

[viscid thistle]

Oh alright, thank you so much ❤ Sorry I'm super slow ;-;

[dense zealot]

Mhm

[viscid thistle]

Can someone help me super quicj

The question is as follows

"Suppose a function f(x) has an inverse function of f^-1(x). Then determine f^-1(5) if f(17) = 5"

Number 13

[high cliff]

...

its 17

[viscid thistle]

Care to explain how

[high cliff]

the function f maps 17 to 5

therefore the inverse of f must map 5 to 17

[viscid thistle]

ahhh got it

thank you thank you

[viscid thistle]

in the second problem, the HA would be equal to zero since the degree on the bottom is larger than the one on top right?

[dense zealot]

Hi

second problem?

Yes

[viscid thistle]

Yeah it would @🚀✈#8003

[viscid thistle]

Yo 😂 how 😂 do 😂 I 😂 find 😂 the inverse function of x/(x+2) ;-; 😂

[wise umbra]

lol idk

would it be (-x) - (x times 2)

Basically the exact opposite

I have no clue

probably 100% wrong

[viscid thistle]

It really be like that 😞 💯

Nvm I'm good ^:^

[dense zealot]

Hi

u first want to see if it passes the horizontal line test

Then switch y and x

And solve for y

@viscid thistle

[viscid thistle]

Thanks fammy ❤

[hoary yoke]

its hard doe

=tex y = \frac{x}{x+2} \Rightarrow yx+2y = x \Rightarrow\2y = x(1-y) \Rightarrow y=\frac{2x}{1-x}

[elder smelt]

any help?

number 6

[calm thicket]

Unless I'm missing something, you just subtract them, no?

=tex \int_5^7\cos(x)+7-\ln(x-3){\rm~d}x

[elder smelt]

oh

wait that's it?

wtf

oh ok

LOL

thanks for the help

[dapper oar]

Yep, though always make sure if it's asking to find an area that you subtract the larger one from the smaller one

If the curves cross then you'll need to break it up (though they don't here)

[past jay]

Not complaining or anything, but are integrals in precalc for some schools....?

[true vigil]

no it's calculus

[past jay]

I didn't think so but I've got some back-woods hs so

[viscid thistle]

https://cdn.discordapp.com/attachments/387447746791079957/415700668674736128/image.jpg

please help me

[dense zealot]

hi

35 m/s

u can use the function for that

for v_0

x = 35 * costheta * t

so we want a minimum theta

70/35 = costheta * t

[dense zealot]

3 = (-1/2)(9.8)(t^2) + 35(sintheta)(t) + h

so 2 = costheta * t

2/t = costheta

actually u migt wanna isolate t

t = 2/costheta

so 3 = -4.9(4/cos^2(theta)) + 35(sintheta)(costheta) + h

@fringe stream wouldnt h = 0 cuz u want to height to be lowest .-.

uwu

[fringe stream]

what's h for?

[dense zealot]

the height its being shot from

[fringe stream]

yes.

[dense zealot]

so we're minimizing -4.9(4/cos^2(theta)) + 35(sintheta)(costheta) -3 = 0?

[fringe stream]

=pup x^2 - 70x + 14.6 = 0 solve for x

[granite stirrup]

Query made by @fringe stream
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=x^2+-+70x+%2B+14.6+%3D+0+solve+for+x
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/

[fringe stream]

eeeek.

o i see.

=pup 11.6x^2 - 70x + 14.6 = 0 solve for x

[granite stirrup]

Query made by @fringe stream
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=11.6x^2+-+70x+%2B+14.6+%3D+0+solve+for+x
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/

[fringe stream]

even worse. 🤢

[dense zealot]

worse

.-.

minimization of theta.

minimum is

[fringe stream]

=pup 19.6x^2 - 70x + 22.6 = 0 solve for x

[dense zealot]

15.93....

[granite stirrup]

Query made by @fringe stream
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=19.6x^2+-+70x+%2B+22.6+%3D+0+solve+for+x
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/

[fringe stream]

okay not that bad.

[dense zealot]

so

factor out 2

4sin(2(theta - pi/2))

and so

phase is 2pi/2

= pi

amplitude is 4

phase shift is to the right pi/2

because

asin(bx+c)

u wanna factor it

asin(b(x+c/b))

and 2pi/b will be period

c/b will be to right if its negative and vice versa the shift

and a is amplitude

mhm

yw!

can I have a hint for a

Not the answer plz

Ik roots of unity

But I don't get how to apply it

Ohhh it's 0 right?

wait

it's re(roots of unity of z^m -1) Right

So 0

[viscid thistle]

ty @dense zealot @fringe stream

[dense zealot]

7.32?

Halp

hint

[quartz garnet]

Jeez

Try to think about it in quadratics

Assume that the tenth degree polynomial is factorized to five quadratics

[dense zealot]

@quartz garnet is there a faster way to do it using roots of unity

I mean the topic of the section is called that .-.

<@&286206848099549185> ??????????

[past jay]

👀

[hexed ermine]

LOL

[past jay]

What math did you say you were in?

[dense zealot]

no math

math is bad!

[quartz garnet]

Uuh

[past jay]

Just for kicks?

[dense zealot]

?

just for PARENTS

[brisk zealot]

?

where

[dense zealot]

above

-_-

https://cdn.discordapp.com/attachments/363224154469826562/415717764246929418/JPEG_20180220_225349.jpg

7.32

hint

[brisk zealot]

oh

so

[dense zealot]

with roots of unity

[brisk zealot]

think about what multiplying conjugate does

[dense zealot]

ik

a^2 + b^2

that doesnt help

[brisk zealot]

well

how do you get the real and imaginary parts

[dense zealot]

by using roots of unity

to solutions to the deca-ith

whatever theyre called

so i have to factor it

some way

but idk how

@pine kindle ;-;

[brisk zealot]

errr

nah

im los

all i can think is that

[viscid thistle]

cute hands

[brisk zealot]

the bottom is u^2

• a number

[dense zealot]

;-;

hodgeee helppp

ur smart right .-.

k i gotz it

(x/13x-1)^10 = -1

let u sub x/13x -1

use roots of unity

solve for sub

[hasty wadi]

@dense zealot can i please get assistance?

[dense zealot]

hi

use sine laws

[hasty wadi]

the sum and difference laws?

[dense zealot]

sin(180-x) = sin(x)

[hasty wadi]

?

[dense zealot]

test a few values

ull see a pattern

also note

sqrt3/2 = sin(pi/3)

[drowsy plaza]

@dense zealot NICE!!!!

aops precalc book 😄

I like what I see

I'll be going through that next year ;D

[dense zealot]

so we set the expression to sin(pi/3)

@drowsy plaza dont, itll save u some worthless time .-.

its torture

[drowsy plaza]

yes but good torture

I'm excited

[dense zealot]

no, not good torture

[drowsy plaza]

are you taking their online course?

[dense zealot]

it will lead to permanent brain damage

[drowsy plaza]

I'm going through intro to algebra right now

[dense zealot]

ofc not

[drowsy plaza]

with their online course

[dense zealot]

sin(x + pi/6) - sin(x - 7pi/6) = sin(pi/3)

[drowsy plaza]

@dense zealot what other aops have you gone through?

[dense zealot]

all

except calc

[drowsy plaza]

nice

do you self study them?

or what?

how many chapters do you try to do a week?

[dense zealot]

what

3 hrs a day over summer break

idrc during school time

[hasty wadi]

im lost on how to approach this

https://cdn.discordapp.com/attachments/363224154469826562/415723961608044546/unknown.png

[grizzled hull]

Alrighty

[dense zealot]

No need to expand it like that

-sin(x-7pi/6) = -sin(5pi/6)

sin(pi/6) - sin(5pi/6) = sin(pi/3)

they're equal

wait I messed up

idk don't ask me

uwu

[dense zealot]

why?

Did u do that

Why would u use sum identities -_-

WHY

;-;

[patent beacon]

sin(7π/6) = - 1/2

[hasty wadi]

lmao, i don't know any other way

[patent beacon]

And I don't see any easier way either

[grizzled hull]

Distribute negatives, combine like terms, isolate the trig functions with the variables.

[hasty wadi]

can you help me with the next step tho cuz im really lost now lmao

[patent beacon]

You made a mistake. sin(7π/6) = -1/2

[hasty wadi]

oh

[patent beacon]

Wait, just do x + π/6 = u

[patent beacon]

At that point, you can just drop the brackets since they're not doing anything

[hasty wadi]

yes

[patent beacon]

And the problem becomes much simpler

[hasty wadi]

but combining like terms with trig functions here confuses me

[patent beacon]

Imagine sin(x) = u
cos(x) = v

[patent beacon]

Combine like they're each an independent variable

[hasty wadi]

the cosines here cancel out

[patent beacon]

Yup yup!
sqrt(3)sin(x) = sqrt(3)/2

sin(x) = 1/2

[hasty wadi]

🤔

should i just divide both sides by (square root of 3/2)

[patent beacon]

Divide both sides by sqrt(3)

[hasty wadi]

but wouldn't (sqrt(3)/2) get rid of the whole frac?

[dense zealot]

Or be a nerd and multiply by two and square both sides then check the extraneous solution

[patent beacon]

We don't wanna! sin(x) = 1/2 should be ringing your bells like Christmas morning

[hasty wadi]

i don't get that concept of sin(x) = 1/2

wat

[dense zealot]

U have to memorize them

0, 30, 45, 60, 90...

degrees

U did that in the beginning of precalc

[hasty wadi]

lmao, i've delved into this shit, sin (x) = 1/2 doesn't ring a bell at all

[dense zealot]

let's see

U know rule for 30, 60, 90

Right

Triangle

[patent beacon]

sin(π/6), go!

[dense zealot]

2x , x , xroot3

[hasty wadi]

yes

[patent beacon]

You know the answer, you did it above

[hasty wadi]

@patent beacon now that's making sense to me

[dense zealot]

so 1/2, 1/2 root3 , 1

That's 30,60,90

[hasty wadi]

oh

[dense zealot]

1/2 is smaller side

[hasty wadi]

ohhhhhhhhhhhhhhhhhhh

[dense zealot]

So it's 30°

[hasty wadi]

wait so how would i write it out?

[patent beacon]

sin(π/6) = 1/2

So if sin(x) = 1/2, x = π/6

However, x also equals 5π/6

Because unit circle

[hasty wadi]

but what about the square roots? how did you get rid of dem?

[dense zealot]

Square

or divide

[patent beacon]

They're on both sides. I divided by sqrt(3)

[hasty wadi]

ahh

?

is this what you did

?

like, is that remotely right lmao?

@patent beacon

[patent beacon]

Exactly

Divide everything by a number

[hasty wadi]

you should end up getting "2sinx = 1" by the end of it no?

[patent beacon]

Yeah

[hasty wadi]

it's right 😃

thank you so much for the assistance dude

@dense zealot @patent beacon thank youuuuuuu 😃

[dense zealot]

I didn't do anything

[patent beacon]

No problem at all. Feel free to ask if you need anything else

[deft mauve]

https://cdn.discordapp.com/attachments/269573202018041856/415728329522413568/image.jpg

Anyone can help?

I'm lost with this q

[dense zealot]

what does it want

Prove that the remainder is?

[deft mauve]

prove that it's that mess on the bottom

[dense zealot]

ok

[deft mauve]

All I know is that the remainder is a linear of form ex + f

[dense zealot]

try using long division

On a third degree

With that quadratic

[deft mauve]

Is it possible to do it without doing division

[dense zealot]

Ya

[deft mauve]

and just using the remainder theorem

[dense zealot]

But I'm too lazy

.-.

Mmm

So it's divisible by that

meaning

It's divisible by (x-a) and (x-b)

So we know

When we plug in

[deft mauve]

If I let the remainder be ex+f I get P(a)=ae+f & P(b)=be+f

Since if you divide by quadratic it results in linear?

[dense zealot]

I guess

But we need a more generalized form sooo

So if u use bezout

Wait what

I messed up

e.e

[deft mauve]

If we let P(x)=Q(x)(x-a)(x-b)+R

[dense zealot]

U might wanna ask someone else since it's midnight and I'm braindead

[deft mauve]

lol np

[dense zealot]

@deft mauve look in #help-2

[viscid thistle]

Thank you all for helping me pass my last two precal tests v..v ❤

[deft mauve]

Anyone that can help me with b)

[deft granite]

@deft mauve what have you tried?

[gritty blaze]

z = 2

[hasty wadi]

can i please get sum assistance for this problem?

here's some work i did

i get a weird decimal answer for sin and i don't know why

[severe verge]

is this a test 👀

[hasty wadi]

no

home work assignment

[severe verge]

is this your​ last attempt 👀

[hasty wadi]

no

lmao, not last attempt

[severe verge]

ok i like this emoji tho 👀

[hasty wadi]

👀 same

[dense zealot]

Half angles!

sqrt(1+cosx/4)

I think

Don't trust me

[hasty wadi]

it's over 2

[dense zealot]

For sin

It can be

[dense zealot]

But I put everything in the sqrt

[fringe stream]

what you were doing is right.

[dense zealot]

uwu

My memory is failing me

[fringe stream]

just simplify what you got.

[dense zealot]

As always

[hasty wadi]

but im getting a weird decimal answer

i plugged it into my calculator

[fringe stream]

don't approximate.

just leave it in radicals.

[hasty wadi]

that's gonna be one weird radical lmao

i'll try it

broo im getting funky asnwers

[fringe stream]

looks good.

[hasty wadi]

i tried sin

i mean tangent

but it's not simplified to a fraction

also

sin is wrong

IT WAS PLUS???

WAT

okayyyyy???

tangent is fucking me

i can't find an answer in radical form

[severe verge]

just divide em

[hasty wadi]

i divided both of them

[severe verge]

cancel the 130

and ur done

[fringe stream]

$$-\sqrt{129 - 16\sqrt{65}}$$

[hasty wadi]

THANK YOU ANYONE WHO HELPED ME TODAY

MUCH APPRECIATED 😃

[fringe stream]

😄

[dense zealot]

Precalculus is a weird and useless radical

[viscid thistle]

lmao

[dense zealot]

@viscid thistle hai

Old person ^^

[viscid thistle]

hello @dense zealot

lol

[dense zealot]

:p

[proper flicker]

Can someone explain logs to me please?

I've been trying to understand them for a while now

But don't seem to quite get it

[clever inlet]

what in particular?

[proper flicker]

Just the basics

Like how common log, ln log, and e work

[clever inlet]

well, you input some number into log

and it finds 10 to what power gives that input

[proper flicker]

10 to?

[clever inlet]

e.g.

log(100)

10^2 = 100

so log(100) = 2

[proper flicker]

Okay

So what if the base isn't 10?

[clever inlet]

so ln(x) is logarithm base e

and you can really have any positive number as the base

[proper flicker]

what does ln(5) mean

[clever inlet]

e to what power gives 5

[proper flicker]

the interval?

[clever inlet]

what do you mean?

[proper flicker]

how would I solve ln(5)^(x)

[clever inlet]

is that the full question?

cause that's just some expression

[proper flicker]

that's just an example I thought

[clever inlet]

there's nothing really to solve there

[proper flicker]

let me find a question in my textbook real quick

something like 4ln(5)

[clever inlet]

to evaluate that?

[proper flicker]

yes

[clever inlet]

well

calculator

[proper flicker]

i can't use a calculator in my class

[clever inlet]

then there's really nothing we can do here

[proper flicker]

i see

im guessing on the test its going to have more complicated ones

and I'd have to simplify them to that

[clever inlet]

4ln(5) is equal to ln(5^4)

but i'd say the first is simpler

[proper flicker]

log is always base 10 right?

[clever inlet]

ehh

depends

but in a high school or middle school math class

probably

[proper flicker]

so log(1,000,000)

is 10^x=(1,000,000)

and I solve for x?

[clever inlet]

you can do that

[proper flicker]

is there an easier way?

[clever inlet]

it's fine

[proper flicker]

okay

can something like ln10 be simplified?

[clever inlet]

not really

[proper flicker]

or ln1,000,000?

[clever inlet]

that can be

[proper flicker]

how would you do that

[clever inlet]

so 1,000,000 is just 10^6 right?

[proper flicker]

yes

[clever inlet]

you can throw exponents in front

which is one of the cool things with logs i guess

[proper flicker]

so 6ln(10)?

[clever inlet]

yep

[proper flicker]

would 10(ln)6 be the same?

[clever inlet]

hmm?

[proper flicker]

are those two equal?

6ln(10) and 10(ln)6

[clever inlet]

well, () for inputs

do you mean 10ln(6)?

[proper flicker]

yeah i meant that sorry

[clever inlet]

it's fine

they aren't equal

[proper flicker]

aren't they all being multiplied?

[clever inlet]

no

ln(x)

ln is the function

x is what you are inputting into the function

it's just like how you can have a function

say f(x) = x^2

f(3) = 9

f is the function, we are inputting 3 into it, and we get 9 back

[proper flicker]

and then you multiply by the number outside after inputing?

[clever inlet]

yeah

so 6ln(10) is really 6*ln(10)

evaluate ln(10) first, then multiply that by 6

[proper flicker]

ahh okay I get it now

[clever inlet]

nice

[proper flicker]

for e

that's just the opposite of ln right?

[clever inlet]

e^x is the inverse of ln(x)

and vice versa

[proper flicker]

can e be solved with number too?

like e100

[clever inlet]

what do you mean?

[proper flicker]

is e always with ln? or can be alone with a number too?

[clever inlet]

it's just some constant

that has nice properties

[proper flicker]

i see

[clever inlet]

you can do all sorts of arithmetic with e if you want

[proper flicker]

since im not using a calculator

i might not need to worry about that part

[clever inlet]

yeah

if you aren't using a calculator

you'll probably just leave everything exact

[proper flicker]

how do logs work with square roots?

[clever inlet]

example

?

[proper flicker]

log5(√ x+5)

[clever inlet]

the entire x+5 is being square rooted?

[proper flicker]

yes

[clever inlet]

well

sqrt(x+5)

that's (x+5)^1/2 right?

[proper flicker]

yes

[clever inlet]

then you can just throw that in front

and i guess it's a slightly more simplified expression

[proper flicker]

so log 1/2(x+5) ?

[clever inlet]

before the log

1/2 log_5(x+5)

[proper flicker]

okay that makes sense now

if the number was cube rooted, would it be 1/3log?

[clever inlet]

log properties can be more intuitively understood by thinking of exponent laws

yep

[proper flicker]

log cube root 1000

does that equal 1?

[clever inlet]

yep

[dapper oar]

if your log is base 10, yes

[proper flicker]

what happens if the base is a variable?

[clever inlet]

depends what the question asks for

it could be something like, convert to exponential form

[proper flicker]

i think it wants to simplify it

[clever inlet]

what's the question in particular?

though, in that case

yeah

just apply general log properties

[proper flicker]

would it be the same with ln?

[clever inlet]

yep

[merry ocean]

hi i need help with thsi question

heres the answer key to it

but shouldnt "du" be "4dx" instead of "4xdx"

[patent beacon]

Yes. Note they didn't actually carry that mistake through

[merry ocean]

hmm so that answer key is wrong?

[dense zealot]

this isnt precalc???....

[merry ocean]

oh sry

[proper flicker]

@clever inlet thanks you for helping me out earlier 😄

[tidal cypress]

Does Sat math fall under this category?

[true vigil]

yes

[tidal cypress]

@true vigil would you be willing to help me later with Sat questions?

Math isn't my strong suit

I also don't do well on exams

[true vigil]

just ask here

bunch of people ready to help

[fringe stream]

[fierce coral]

hey question how do we choose values to prove something has at least one real root

[patent beacon]

Show the something is continuous in an interval, and show that it must cross the x-axis in that interval

Have an example?

[fierce coral]

cos^-1(x)=e^x-2

show this has a real root

i just chose 0 and 1 and got a positive nad a negative number but i was wondering how to choose those numbers

[severe verge]

intermediate value theorem

oh

it's kind of arbitrary i guess

kind of guessing

[patent beacon]

Having an idea where the root might be is helpful

[severe verge]

u just pick numbers that are easy to compute with your functions and see where they land

until u have one positive one negative, then apply intermediate value theorem

[patent beacon]

Note in this case you're looking for roots of
f(x) = cos¯¹(x) - e^x + 2

[severe verge]

fancy

[patent beacon]

Any root of that is a solution to the original problem.

[fierce coral]

humm yea i guess so so its a guess and check type problem

[viscid thistle]

Hi can someone pls help me for a sec

I dont quite get the problem

The first one idk why i did it wrong. The 2nd one i dont understand

[grim yarrow]

It's

i/(i+2)

ar^(n-1)

[viscid thistle]

@grim yarrowcan u tell me how

For the first one

Bruh i forgot to put in one "(" for the second one

:GWqlabsYaoLUL:

Other times i didnt even put the ()

But can u tell me how u got i/(i+2)?

[grim yarrow]

Just looking at it lol

[viscid thistle]

:/

Ill try to figure it out why then

Thank you so much

It was right answer

[dense zealot]

ol

lol

[quartz garnet]

"Too hot for girls"

[grizzled hull]

誰がそれと言ったかな。。。

[stuck sapphire]

https://cdn.discordapp.com/attachments/367271259278409730/413682877306503188/image.jpg

thats cool

[gritty pewter]

Anyone mind giving me a hand? last problem of the set and i'm terrified to go backwards again

both cases the solution is >1

top comes out to 10/9 and bottom comes out 4 even

just feels weird for both to be no sum

[patent beacon]

That's a geometric sum
Σ 3(1/4)ⁿ

= 3 Σ (1/4)ⁿ

= 3 / (1 - 1/4)

= 4

[gritty pewter]

the rule though is |R| < 1...

its R not the whole thing

[patent beacon]

In general,
Σ rⁿ = 1 / (1 - r)

For... Yeah, unitary r

[gritty pewter]

I see where I goofed

bottom one is 4

reworking the top pone

and i get 10/9

[clever inlet]

hmm

[dapper oar]

what part is the question

[gritty pewter]

the left of =

[dapper oar]

oh wait i should scroll up

this isn't the first comment

[gritty pewter]

first one comes out to 3/5 / 1 - 3/5

[dapper oar]

i need to go to bed tbh

[gritty pewter]

3/2

[clever inlet]

10/9 looks right for the first one

[gritty pewter]

top should be 3/2, rather than 10/9, I believe

[dapper oar]

I'm getting 10/9 for the first one too

[gritty pewter]

all work, no play today making me crazy

[clever inlet]

2/(1--4/5)

10/9

[gritty pewter]

10/9 is correct

its the top one

it is not 4

[clever inlet]

3/4 is first term right?

and r is 1/4

since i starts at 1

[gritty pewter]

first term is 3, R is 1/4

[patent beacon]

2 + 2/(1 + 5/4)
= 2 + 2/(9/4)
= 2 + 8/9

[clever inlet]

look at i

i starts at 1

so first term is 3/4

[patent beacon]

It's the real value isn't it

[gritty pewter]

yes

at 3/4, the bottom comes out to 1

[patent beacon]

Try 26/9

[gritty pewter]

I get more practice

[clever inlet]

yeah

[gritty pewter]

doh

[clever inlet]

when you have sigma notation i guess

make sure to check where it starts

[patent beacon]

Because that's 2 + 2 Σ (-4/5)ⁿ

[gritty pewter]

Kanga, when you say to check where it starts

[clever inlet]

mainly for the second question

notice how i starts at 1

and not 0

so first term is -2

wait no

[gritty pewter]

ok, that didnt even register to me

[clever inlet]

if i started at 0, then yes

first term would be 3

[patent beacon]

That one's different.

[gritty pewter]

at 1, its -3

yeah, new problem set

its not 9, its -9

had i been -1, it would have been 9

[clever inlet]

i got some fractional answer for the second one i think

[gritty pewter]

yeah, im reworking it again

-5/3 is not the top

nor the bttom 9

bottom seems to be proper -2

[patent beacon]

3 × 1 / (1 + 2/3)

= 3 × 3/5

= 9/5

[patent beacon]

The top one is very much non-existent

[clever inlet]

i did -2/(1--2/3)

[dense zealot]

hi

[clever inlet]

hi

[dense zealot]

Why is complex number yeomtry weird

How do I do et

;-;

all i see is "bleh maps to weirder bleh so we bleh image and find the angle using the bleh super weird theorem to find bleh^1098 bleh."

Like I don't get Im((cos12° + isin12° + cos48° + isin48°)^6)

Here is their solution

But I don't get it ;-;

someone?

<@&286206848099549185> .-. u have been summoned

Ug u worthless helpers

-_-

cri

Someone?

[main flume]

I too am confused by this terrible textbook writer

Have you posted question 8.2?

[dense zealot]

That was it

@main flume

The question right above the picture

[main flume]

Ah

I see what the book is trying to explain

You've learnt vectors before, I'm assuming?

[dense zealot]

Nope

Vectors r next chapter

Everything in precalc but vectors and limits

[main flume]

Ah, okay

Odd

Vectors should ideally go before complex numbers, but meh

The main thing you should focus on here is the diagram

Ignore the wall of text first

On the diagram, W is w, Z is z

Do you understand that part?

[dense zealot]

Yes

[main flume]

S is w+z, does that also make sense?

[dense zealot]

Idk .-.

[main flume]

This is why vectors should be taught before complex numbers .-.

[dense zealot]

cos(48) + cos(12) = cos(30)cos(18) right

over 2

Nvm

so S is 1/2 cos(18), root3/2 sin(18)

????

[main flume]

The purpose of this approach is for you not to focus on the numbers first, but look at the question visually on an argand diagram

I've got something to tend to rn so can someone else take over

[dense zealot]

everyone else died

[main flume]

I recommend you read up on complex number addition on argand diagrams, or alternatively start reading up on vector addition

The reason why vectors should be taught before complex numbers is that complex numbers can be thought of as 2D vectors which in some cases can greatly simplify things

[dense zealot]

I get everything up to w+z= 30°

Why can u make that thing

Just from soa = 30

Nvm

I get it now

[lost pawn]

Ok so

=tex \lim_{x\rightarrow0}\frac{\sin(x)}{x}=1

[lost pawn]

Is the following limit true because

=tex \lim_{x\rightarrow0}\sin(x)=0

[lost pawn]

And then we are dividing by that small number?

Which will cause it to approach 1?

[viscid thistle]

\give Almax {🖐}

[grim yarrow]

hello~

[lost pawn]

Hello

[grim yarrow]

I love dogs

[viscid thistle]

it is true

I think

right?

[grim yarrow]

So it's a bit annoying depending on what definition of sine you're using

the limit is the derivative of sin(x) at x=0

[lost pawn]

Yes

[viscid thistle]

cos(0)

[lost pawn]

cos(0) is 1 though?

[grim yarrow]

using the unit circle definition of sin

you can use the squeeze theorem to show the limit is 1

but if you use the series definition of sine

it's pretty obvious why

[lost pawn]

@grim yarrow

The question was to show how sin(x) approaching 0 as x approaches 0 can prove sin(x)/x approaches 1 as x approaches 0

And since sin(x) is divided by its reciprocal it will cause it to approach 1

Or it is divided by something closer and closer to its reciprocal

Which would be 0

Since sin(0) = 0

This is coorect, no?

[grim yarrow]

it's not correct reasoning

sin(2x) also approaches 0 as x approaches 0

but sin(2x)/x does not approach 1 as x approaches 0

[lost pawn]

Yes

Because in that case x does not approach the reciprocal of sin(2x)

[viscid thistle]

Are you familiar with squeeze theorem?

[lost pawn]

No

I'll look it up

[grim yarrow]

http://ime.math.arizona.edu/g-teams/Profiles/JS/Calc/SqueezeTheorem.pdf

this is a good write-up

[lost pawn]

Thanks

[rocky bison]

I'm pretty sure khan academy did a video on that exact proof

[lost pawn]

Oh yes

I have seen this

Just did not know the name

But okay

This is the principle?

=tex \cos(x)\leq \frac{\sin(x)}{x}\leq 1

[grim yarrow]

yes

[lost pawn]

Ok

Thanks for the help :>

[viscid thistle]

@grim yarrow don't you just plug what x is approaching and see if it is defined before doing anything?

[grim yarrow]

it's 0/0 indeterminate

[summer beacon]

haha

again , so u dont have to jump

but yea

[summer beacon]

no

the issue is that

i have 2 functions

2 aprox. polynomials

and this function

being R^3 -> R

and i have a point x0 = (0,-1,1)

Then i have to find the approximated P1 and P2 in that point

ye

Finding the gradient

[summer beacon]

yea but the function is f(x)

thats my issue

yup

by my lecturer

ye

like gradient(x,y,z) = (f ' x, f'y , f'z)

?

i get this

Maple

well it's easier and helps making fancy plots

[granite stirrup]

Failed to make the request.
Maybe you should try again?

If this error keeps recurring, you should report it to DXsmiley on the official MathBot server: https://discord.gg/JbJbRZS

[summer beacon]

kewl

aaah i see

the function is yzexp(1-x^2-(1/2)*y^2-(1/2)*z^2) btw

[summer beacon]

why -2 in front?

[summer beacon]

=pup gradient of yz e^(1 -x^2 -y^2/2-z^2/2)

[granite stirrup]

Query made by @summer beacon
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=gradient+of+yz+e^(1+-x^2+-y^2%2F2-z^2%2F2)
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/

[summer beacon]

wtf

[granite stirrup]

Query made by @pine kindle
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=hessian+of+-2*y*z*e^(1-x^2+-y^2%2F2-z^2%2F2)
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/

[summer beacon]

haha

yup

[summer beacon]

why u keep puttin -2 in front

haha

[summer beacon]

aah yes

thx

mathbot on a 300% workload

haha

thx dude @pine kindle

[lost pawn]

=tex f_1(x)=\left ( \frac{x}{x-1} \right )^3

[lost pawn]

=tex f_2(x)=\left ( \frac{x}{x-1} \right )^4

[lost pawn]

The question is which of these functions satisfies

=tex \lim_{x\rightarrow 1}f(x)=\infty

[lost pawn]

I have concluded that they both grow rapidly, but f_2(x) grows faster

But both approach infinity?

<@&286206848099549185>

[vital tartan]

Nope

it's actually equal to 1

[viscid thistle]

wait what

[hollow obsidian]

wat

[lost pawn]

The denominator at x=1 is 0 in both cases

[viscid thistle]

the first one does not approach infinity

the second one does

[vital tartan]

wait nvm

didn't see the right limit

is it

[viscid thistle]

oh you thought it was the limit to infinity lol

[vital tartan]

=tex \lim_{x \to 1} \frac{x}{x - 1}