precalculus

[blazing parrot]

,rotate

[blazing parrot]

wat do you mean by b and c value

[proud jetty]

the phase shift and the period

@obsidian monolith

[keen rapids]

does anyone know the restriction on this?

9√(3x−21)

[proven dove]

how would i graph a sin graph that has a period of 2

[small cedar]

each cycle is 2

[proven dove]

what if I have a pre drawn x scale like the one in @obsidian monolith

[past meadow]

Are you looking for the equation?

[tardy ridge]

@keen rapids well you can't sqrt a negative

what do you think happens to numbers lower than 7

[novel dirge]

If I have 6 guys and 7 girls, how many different ways are there to pick 6 people, of which at least 2 are girls?

this is combinatorics, but I am not sure what to use

[willow bear]

i'd count the ways to pick 6 people with at most 1 girl among them

then subtract that from 13C6 (the number of ways to pick 6 people w/o restrictions)

[novel dirge]

why so?

and why at most 1 girl?

I am preparing for an exam, so I would like to understand the approach

[willow bear]

you take the total number of ways to pick 6 people

and you subtract the ways that don't fit the restriction of "at least 2 girls"

[novel dirge]

And that I would do by just choosing 4 out of 6?

or 13?

[willow bear]

no

the number of ways to pick 6 people with at most 1 girl among them is 6C6 + (6C5)*(7C1)

to write it in a particularly formal way

6C6, a.k.a. 1, is the one (1) way to pick 6 people out of our 13 with 0 girls among them, i.e. boys only

(6C5)*(7C1) is the number of ways to pick 5 boys and 1 girl

[novel dirge]

so if I want 2 girls, it would be (6C4)*(7C2)?

[willow bear]

that's the number of ways to pick exactly 2 girls.

[novel dirge]

oh

so for at least 2?

another one is, how many 7 digit numbers, can you make from the numbers 0,0,0,7,7,8,9?

[pale bison]

what have you tried?

[novel dirge]

I am not sure what to use

was thinking about just multiplying the probabilities

but Idk if the numbers are reusable

so it would be 6x7x5x4x3x2x1

or maybe use permutation with repeatable numbers

but Idk what to do with the cases where 0 is first place

[pale bison]

did someone ghost ping me..

right, but look, 000 is the same as 000 even if i use "different" zeros first and formost

so you really only have 0, 7, 8, and 9 to choose from to fill in 7 slots
_ _ _ _ _ _ _
now think about how many choices you have for each

oh i guess not you can't reuse the numbers

hmm

[novel dirge]

yeah

that confuses me, too

[pale bison]

ok, so you still have 7 slots

and so 7 factorial

but you need to remove the same numbers

you can do that by dividing 7! by some number

so uhh think about it

[novel dirge]

hmm

but I also have repeating numbers

[pale bison]

right that's the point

you need to remove them to avoid overcounting, and you can do that by dividing

[novel dirge]

what about derivation with repeatable elements?

[pale bison]

uh yeah

so suppose you have like 0 0 0 7 7 8 9, then you can swap 0s and 7s and get the same number, so in essence, you can write every number in 3!*2! ways

[novel dirge]

?

[pale bison]

sorry broken english

[novel dirge]

np

it isn't the english I am confused with

[pale bison]

oh

well look, there are only 2 ways to rearrange those two 7s, that's why i wrote 2!

and same goes for those three 0s, there are 3! ways to arrange them

[novel dirge]

yes

so would I use permutation for that?

I mean, I would use it if there weren't the 0 there

because a number can't start with 0

but Idk how to remove them

[pale bison]

oh you aren't allowed to have 0s in front?

[novel dirge]

well, I haven't seen the number 0123 count as a 4 digit number

it would be 123

[pale bison]

ok that's fine i guess

so instead of 7!, you only have 4 choices for the first digit, and rest the same

agree?

[novel dirge]

I guess

[pale bison]

.. wdym

it's a yes or no question, what don't you get..?

[novel dirge]

no, I understand that part

but just not what to do in the end

[pale bison]

ok, you divide it by 3!*2! since every number can be written in that many ways

[stuck lark]

@pale bison rassword

[pale bison]

hush

[novel dirge]

wait

what do I divide?

[pale bison]

(4*6*5*4*3*2*1)/(3!*2!)

the top is the total number of ways to arrange those digits with no leading 0

and the bottom is the number of ways each same digit can be rewritten

[novel dirge]

ok

I get it now

thank you

[pale bison]

np

[stuck lark]

@pale bison neat sketch

[pale bison]

betcha

[viscid thistle]

I have literally no idea how to do this

Any of them

[proven marten]

Well what do you know

[viscid thistle]

Nothing

[proven marten]

Do you know your identities

[viscid thistle]

Kinda?

[analog relic]

Kinda?

List all the identities

That you know of

[viscid thistle]

I just don’t get how this all functions together

[analog relic]

Ok ok

Let's focus on the first question

What would you do first?

[viscid thistle]

That’s my problem

I have no idea

[analog relic]

Just think

Anything at all

Just throw ideas

[proven marten]

^

[analog relic]

No, not that

Try again

[viscid thistle]

Try and match it with an identity I guess?

[analog relic]

$\tan^{2}x-\tan^{2}x\sin^{2}x$

Try and match it with an identity I guess?
@viscid thistle No no, try to do something other than with the identities

Just think of it like another algebra issue

Don't focus on the identity so much

[obsidian monolith]

Lσνιηg✧Sσνєяєιgη:

[viscid thistle]

Something something 1/cotx

[proven marten]

whats an identity for tan x

another one

[viscid thistle]

Sin x/cos x

[proven marten]

yup

[analog relic]

Yes

[viscid thistle]

How do I know which one to use

[analog relic]

But I think that's just going to make things complicated

Try something else ah

[proven marten]

Huh?/

I worked it out, its fine

[analog relic]

Oh ok

[proven marten]

but

actually if its simpler another way then

we should introduce it

[analog relic]

Sure

Try factorising

$=\tan^{2}x(1-\sin^{2}x)$

[obsidian monolith]

Lσνιηg✧Sσνєяєιgη:

[analog relic]

Then try from here

[proven marten]

^

@viscid thistle Tell us if you dont get it

[analog relic]

@viscid thistle you good?

[viscid thistle]

Oh

[analog relic]

Then can you use any of the identities?

Try ah

[viscid thistle]

Uhhhhhh

I think?

[analog relic]

Tell us what you think

[viscid thistle]

Only thing I can think to do is take the last one and move them all over to the left side

[analog relic]

Menaing

Meaning?

Sorry sorry, I don't quite understand

Take the last one?

Move them to the left side?

[viscid thistle]

1 = sec^2 x - tan^2 x

Move them all to the left I think?

[analog relic]

No need

$1-\sin^{2}x=\cos^{2}x$

[viscid thistle]

Cos^2 x

[proven marten]

good

[analog relic]

Then $\tan^{2}x=?$

[obsidian monolith]

Lσνιηg✧Sσνєяєιgη:

Lσνιηg✧Sσνєяєιgη:

[viscid thistle]

Sec^2 x-1

But what do I do with all of this

I have a cos^2 x and sec^2 x-1

[analog relic]

Another identity?

Any other identities?

Let me give you a hint, $\tan^{2}x=(\tan{x})^{2}$

[obsidian monolith]

Lσνιηg✧Sσνєяєιgη:

[analog relic]

@viscid thistle

[viscid thistle]

1?

[analog relic]

1?

[viscid thistle]

Oh wait they’re not the same lmao

[analog relic]

Sin x/cos x
@viscid thistle ^^

[viscid thistle]

I’m sorry but where are all of these coming from

[analog relic]

The identities

$\tan{x}=\frac{\sin{x}}{\cos{x}}$

[obsidian monolith]

Lσνιηg✧Sσνєяєιgη:

[viscid thistle]

Ohhhhh

But what happened to the other cos^2 x and sec^2 x-1

[analog relic]

So, $=\tan^{2}x(1-\sin^{2}x)=\frac{sin^{2}x}{cos^{2}x}(cos^{2}x)=?$

[obsidian monolith]

Lσνιηg✧Sσνєяєιgη:

[viscid thistle]

Sin^2 x?

[analog relic]

Ya

That's it

Good job

Now, try the second question

$\frac{\cos^{2}x}{1-\sin{x}}$

[obsidian monolith]

Lσνιηg✧Sσνєяєιgη:

[viscid thistle]

Is this one just 1?

Oh wait it’s not sin^2 x

[analog relic]

Oh ok

So?

Try ah

Just throw out ideas

[viscid thistle]

So cos2x/1-(1/csc)

[analog relic]

No need so complicated man

Any other ideas?

[viscid thistle]

1-sin^2 x/1 - sin x

[analog relic]

Yes

$\frac{1-\sin^{2}x}{1-\sin{x}}$

[obsidian monolith]

Lσνιηg✧Sσνєяєιgη:

[analog relic]

Now, the next step, abit harder

You remember $a^{2}-b^{2}=(a-b)((a+b)$

[obsidian monolith]

Lσνιηg✧Sσνєяєιgη:

[viscid thistle]

sure?

[analog relic]

Now, think of $a=1$ and $b=\sin{x}$

[obsidian monolith]

Lσνιηg✧Sσνєяєιgη:

[analog relic]

So $1-\sin^{2}x=?$

[obsidian monolith]

Lσνιηg✧Sσνєяєιgη:

[viscid thistle]

Cos^2x

[tardy ridge]

He is not wrong

[analog relic]

Cos^2x
@viscid thistle yes it is, but not about that

$a^{2}-b^{2}=(a-b)(a+b)$

[obsidian monolith]

Lσνιηg✧Sσνєяєιgη:

[analog relic]

You know this right?

[viscid thistle]

It looks familiar?

[analog relic]

Factorisation?

No, not a clue?

[viscid thistle]

Oh oh oh

I had a brain fart there for a second

[analog relic]

Ok

So $1-\sin^{2}x=?$

[obsidian monolith]

Lσνιηg✧Sσνєяєιgη:

[viscid thistle]

(1-sin) (1+sin)

[analog relic]

What's with the x outside?

[viscid thistle]

Don’t know just thought it went there for some reason

[analog relic]

Oh ok ok

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

[obsidian monolith]

Lσνιηg✧Sσνєяєιgη:

[analog relic]

Then can alr?

[viscid thistle]

So it’s just 1+sin x

[analog relic]

Yes

[viscid thistle]

Huh

These seem a lot simpler than I though they were

[analog relic]

Because they are

You're just not that exposed, so you don't quite know how to tackle it

So, at the start, just try anything you can think of

Next question, $\frac{1-\tan^{4}x}{1-\tan^{2}x}=?$

[obsidian monolith]

Lσνιηg✧Sσνєяєιgη:

[viscid thistle]

Is this also just factoring?

[proven marten]

noice

[analog relic]

Try ah

Anything

Just go for it bro

[viscid thistle]

((1+tan^2x) (1-tan^2x))/1-tan^2x

[analog relic]

Nice

[proven marten]

looking good

[viscid thistle]

=1+tan^2x

[analog relic]

Good good

Any identities?

[viscid thistle]

=sec^2x

[analog relic]

That's great

You did it again

Look at that, 3 in a row

[viscid thistle]

Huh

[analog relic]

3 questions in a roow

[viscid thistle]

Alrighty

Thanks for your help

This makes a whole lot more sense now

[analog relic]

$\frac{\sec^{4}x-\tan^{4}x}{1+sin^{2}x}$

[obsidian monolith]

Lσνιηg✧Sσνєяєιgη:

[viscid thistle]

tryina figure out how to factor the top row

Or wait

Would this just be through perfect squares

[proven marten]

It doesnt hurt to go for it

you can always backtrack

[analog relic]

Ya

Just try man

[viscid thistle]

Ok yep I’m lost on this one

[proven marten]

try

the sec = 1/cos x

thing

Any progress?

[analog relic]

Im also trying it

Uhh what did you do?

[proven marten]

I did

give me a sec

(1/cos^4x - sin^4x/cos^4x)/1+sin^2x

[analog relic]

I solved it

[proven marten]

Oh

Wait I thought

oof my b

Did you use my method or a different one

[analog relic]

A differnet one I think

What's your answer

[proven marten]

1/1+sin^2x

[analog relic]

No

My answer just secx

[proven marten]

oof

[analog relic]

You want I sendd my soln

[proven marten]

im probably wrong

Yes plssss

[proven marten]

@viscid thistle You still here?

Ahh

nice

[analog relic]

Correct right?

Idk sia

[proven marten]

Im pretty sure youre right

[analog relic]

Idk

[proven marten]

Because I only started doing this when dile asked

[analog relic]

Same

[proven marten]

This is my second time with trig identities lol

[viscid thistle]

Yep

[analog relic]

Damn

It's ok ah

[viscid thistle]

I just got completely lost

[proven marten]

well

what part do you need help understanding

[analog relic]

This one quite difficult

I would say

[proven marten]

Yeah it is

[analog relic]

I'll break it down

Actually, no quesiton is hard or easy, it's just something you have to do

When you know how to do, it's easy. When you don't then it's hard

So, hard or easy is relative to the person

But ok ok

I'll break it down

Firstly, try to factorise the numerator

What do you get

WAit ah

Give me 10-15 mins

[viscid thistle]

It’s to the fourth

[analog relic]

Hmm?

Ok

$a^{4}=(a^{2})^{2}$

Wait ah

[obsidian monolith]

Lσνιηg✧Sσνєяєιgη:

[analog relic]

Omg

@viscid thistle

Yo

You good?

[viscid thistle]

Ah I’m sorry I had a class

[analog relic]

Yo

Since it's now like 1am

Imma sleep

Gn

[proven marten]

Gn

You were great, you helped a lot

So thank you

[proud jetty]

How about I go about finding it?

[viscid thistle]

theta = arc divided by radius

8/8 = 1

so its 1 radian

if you want that in degrees, use the conversion factor, 180/pi

[proud jetty]

@viscid thistle alright thx

[viscid thistle]

can I get some help with this? How do I find triangle ADE
I am planning to find S1 by doing -area of sector + area of triangle

[tardy ridge]

AD=sqrt(3) * a

[viscid thistle]

how do u know this?

[tardy ridge]

AN=sqrt(3)a

or is that a W

that is because it's a 30 60 90 triangle

[viscid thistle]

its N

yeah I got the AN

[tardy ridge]

AD=AN

[viscid thistle]

But N is the midpoint of BC and D is higher than B

Can someone tell me what I’m doing wrong with the first question, becuz I’m stuck there

me simplifying trig expressions

[tardy ridge]

Yeah AB=2a and AD=sqrt(3)a

[viscid thistle]

and AD = AN?

[tardy ridge]

If I am understanding the problem correctly, yes

Since A is the center of the circle

[viscid thistle]

Ohhhh

[tardy ridge]

try drawing the circle with center A and radius sqrt(3)a

[viscid thistle]

ok ok now I see

thanks!

I think I can get the area now

[tardy ridge]

Nice

[viscid thistle]

How would I do number 9?

[past meadow]

consider $log_{10}(12)=log_{10}(2 * 3 * 2 )$

[obsidian monolith]

Sneaky:

[viscid thistle]

Why would I multiply by two twice?

[past meadow]

to get 12

also consider $log_m(a)+log_m(b)=log_m(ab)$

[obsidian monolith]

Sneaky:

[past meadow]

do you see it now or nah?

[viscid thistle]

Nope

[past meadow]

$log_{10}(12)=log_{10}(2 * 3 * 2 )$

do you agree with this?

[viscid thistle]

Yes

[obsidian monolith]

Sneaky:

[past meadow]

so do you agree you could break $\log_{10}(12) $ into $ log_{10}(2)+log_{10}(3)+log_{10}(2)$

does that make sense to you?

oh its base 5

but its still the same

you can use change of base

[obsidian monolith]

Sneaky:

[past meadow]

$\log_a(b)=\frac{\log_x(b)}{\log_x(a)}$

[obsidian monolith]

Sneaky:

[viscid thistle]

This is what I have so far

[unique hill]

that doesnt seem correct

[past meadow]

I'm not sure what you've done there

[viscid thistle]

I’m still confused

[unique hill]

@viscid thistle this is occupied

[past meadow]

#help-8 would work @viscid thistle

[viscid thistle]

Thank you

Would I put log of 12 base 5 first?

[tardy ridge]

I have a question about your question

does the answer need to be purely in terms of a and b

or can you divide by some log

@viscid thistle

[past meadow]

its still in terms of a and b if its say for instance 2a+b divided by a certain log, i would think

[tardy ridge]

hmm

You are saying I can do 1/(log_12 (5)) + 0a +0b

[past meadow]

i dont see how thats equal to the question or in terms of a and b

[tardy ridge]

it's definitely equal

and it's in terms of 0a and 0b

[past meadow]

you're telling me 1/(log_12 (5))=log_12(5)

[tardy ridge]

yes

No

What

[past meadow]

oh

i see what you mean but its not what the question asks

[tardy ridge]

The question asks for log_5 (12)

[past meadow]

the question asks for log_5(12)

[tardy ridge]

log_5 (12)=1/(log_12(5))

[past meadow]

you can get it in terms of a and b. im pretty sure they dont want what you answered

[fleet yew]

Tru

[tardy ridge]

Yes.

[full garden]

hey guys i have a very quick question please

i tried solving this question on paper and it couldn't find a solution

i tried using this website to check it and it said it was false

but the question was asking to prove this identity

so does that mean that there is a solution

[full garden]

can someone please clarify

[steel venture]

if you get the left side in the same denominators you find that $\frac{(1-\cos(x)) + (1+\cos(x))}{\sin^2(x)} = \frac{2}{sin(x)}$

[obsidian monolith]

Buncho Maximos:

[steel venture]

which just simplifies to

$\frac{2}{\sin^2(x)} = \frac{2}{\sin(x)}$

[obsidian monolith]

Buncho Maximos:

[steel venture]

so i'm sure you could find a value that satisfies this (such as pi/2) but its not an identity

[full garden]

wait what

how is 2/sin^2 = 2/sin

[steel venture]

if you get them in the same base,

[full garden]

oh

[steel venture]

the base simplifies to sin^2(x)

[full garden]

@steel venture thank u so much

[steel venture]

np

[vernal spindle]

How do you find the slant asyptote

I looked it up and it says something about dividing

[fleet yew]

slant asymptote occurs when the degree of the numerator is greater than the degree of the denominator

[full garden]

@vernal spindle can i ask if this is like a public website because it looks very helpful for practicing questions

guys can someone please explain to me why these two are equal

[past meadow]

they aren't

[full garden]

oh

[past meadow]

so i'm sure you could find a value that satisfies this (such as pi/2) but its not an identity
@steel venture \

thats what maximos said

he said its not an identity

he didnt say it was true

[full garden]

alright alright thank u

[vague minnow]

hey I'm kinda having trouble with this problem:

Find all α where the vertex of the parabola y(x) = x^2 - (2 *sqrt(5) *cos(α) - 3)x - (25/4)*cos(4α) is on the y = 3x line, and the parabola intersects with the y axis in a point with a negative y coordinate

I made some progress and now I'm left with the following equation:
50t^4 - 100t^2 + 12*sqrt(5)*t + 19 = 0 where t = cos(α)
and I kinda have no idea how to solve it, so some help would be appreciated 😄

[copper vigil]

a concave up parabola with a negative y-int must have 2 real roots

so the discriminant must be > 0

[vague minnow]

hmm true, but I was thinking more about the vertex being on the y=3x line and < 0 and using the vertex formula

[glad mica]

can someone explain this one to me?

[viscid thistle]

use the pythagorean theorem

[fleet yew]

Draw treangle

[patent beacon]

sin(2x) = 2sin(x)cos(x)

But you'd need cos(x). For that, draw the triangle yeah

[glad mica]

so it should be
Sin(2x) = 2sin(5/13)cos(12/13)

[viscid thistle]

The lenghts are not necesarily 5 and 13

They can also be 10 and 26 etc

[glad mica]

but wouldnt you use the simplified form of the fraction in sin cos etc.

[viscid thistle]

Yes, but you should have it like sin(5x/13x)

Xs cancel out

I had a similar question in my exam few weeks ago and got minus points for not including x

[raven barn]

question: is induction in precalc?

[glad mica]

im currently in precal and i do not know what that is so im gonna say probably not, but it could be very late in the curriculum

[raven barn]

alright thanks i guess

[glad mica]

yeah haha i guess that didnt help like at all

[patent beacon]

@glad mica
No, sin(x) = 5/13
There's no sin(5/13) anywhere

[glad mica]

but once you draw out the triangle, it would the 5,12,13

[viscid thistle]

Could I get help with #13 please

[tardy ridge]

either choose to log 5 both sides or log 3 both sides

then use the exponent log rule

then isolate x

@viscid thistle where do you get the questions.

[smoky needle]

Would someone be able to audio call me and explain how to do some precalculus materials? I’m currently on #5 of my hw out of the 10... I’ve been stuck for 6hrs...

[unique hill]

ohh

i wont be able to video call, but i'll try to explain it by text if u want

[viscid thistle]

I can try to help too ^-^

[unique hill]

:)

[smoky needle]

Thank you, @unique hill and @viscid thistle !! I’ll send the problems right now

[unique hill]

okie

[smoky needle]

I am currently doing #5. I’ve jumped around but the last 5 seems to be the ones I’ve been stuck on..

[viscid thistle]

The first term will always be negative, so the greater value x either negatively or positively the function will go down

so you want to make the first term as small as possible

if that makes sense?

i'm bad at explaining, sorry

[smoky needle]

For -2 right? Is there like a formula I could use for it?

I’m referring to my calculus notes from high school, but I can’t seem to find a problem similar

And no you’re good! Thank you for helping me!

[viscid thistle]

You might be overthinking it, maybe try sketching the function?

[smoky needle]

Okay, I tried sketching it

[unique hill]

for question 5, you can do graph shifting for the modulus

[flint bloom]

the first term is always going to be negative since the -2 out front, so what can x be to completely get rid of that piece

[viscid thistle]

how does the sketch look?

it should be like an upside down absolute function shifted over a bit

[unique hill]

and how many units, up/down/left/right?

[smoky needle]

Oh dang....why would it be upside down? Is it because it’s a -2?

[viscid thistle]

yep

[flint bloom]

yes

[smoky needle]

I have this so far..

What does the -2 play as?

[viscid thistle]

perfect

[unique hill]

that looks correct

[viscid thistle]

that's just like squeezing or stretching

[flint bloom]

-2 flips it and stretches it

[smoky needle]

Would the maximum be 12?

[flint bloom]

yes

[smoky needle]

Ohhh! Thank@l @flint bloom

[flint bloom]

which should make sense intuitively

[unique hill]

cuz the maximum point is the vertex

[flint bloom]

you can get the absolute value piece to 0 by plugging in a -6

[smoky needle]

How would you solve it if you have to write down your work?

[flint bloom]

i would say something about how the maximum/minimumof an absolute value function occurs at the vertex

[smoky needle]

I’m referring to my calc notebook from high school and this is what I have down

Like how would you find the vertex and so forth?

[viscid thistle]

the constant dictates the vertex

[unique hill]

within the absolute value bracket

right?

[smoky needle]

So In this case, vertex (6, 12)?

[viscid thistle]

yep

[unique hill]

yes indeed

[smoky needle]

Sorry I’ve literally forgotten a lot of math

And I’m a college student now who dropped out of calc since I’ve forgotten materials before it 😓

[unique hill]

no worries, we're here to help

[smoky needle]

Thank you!

Why wouldn’t the vertex be -2?

[flint bloom]

that’s just a multiplier

[viscid thistle]

the only thing the -2 decides here is whether the function has a maximum or minimum

[unique hill]

the maximum value of the function would be -6

cuz it's the x value of the vertex

[smoky needle]

Ahhh I see I see!

[flint bloom]

the maximum is 12 but it occurs at x=-6

[smoky needle]

You guys are amazing

[unique hill]

^ so x=-6 is the max value

[viscid thistle]

no, 12 is

[smoky needle]

Wait it won’t be 12?

[viscid thistle]

it is 12

[flint bloom]

-6 is just the input that gives you the maximum value

[viscid thistle]

(-6, 12)

[flint bloom]

but the maximum itself is 12

[unique hill]

oh right ok

mb, got mixed up

[flint bloom]

another thing worth noting, the -2 in front corresponds to a maximum since its open down

[smoky needle]

No worries!!👍🏽

[flint bloom]

any negative infront of an absolute value does the same thing

[smoky needle]

Sorry, I don’t get what you mean...

[viscid thistle]

Like I said before, -2 only tells us if the function has a maximum or minimum.

Since it is -2, the function is an upside down V and has a max

[smoky needle]

If there’s wasn’t any number before the absolute value, there wouldn’t be any max or min?

[unique hill]

no it will be min

[viscid thistle]

there is always a number

[unique hill]

ya

if there's nothing in front, it will just be 1 at the front

[viscid thistle]

invisible 1 😮

[unique hill]

and 1 is positive so it has a minimum value

[smoky needle]

Ohhhh!!! Invisible 1! 😭

I see...so if it was an invisible 1. It the graph would have “V”?

But if it was -1 it would be upside down V?

[viscid thistle]

yeah

you got it

[smoky needle]

Thank you!!

Bless

[unique hill]

np

[willow bear]

@viscid thistle what's the concept you need explained

[viscid thistle]

roots of unity

i understand it a little bit

but it's very shaky

@willow bear ?

what about them is confusing to you

i am not sure how they work

work in what sense?

as in, why the n'th roots form an n-gon, for example

i just don't understand them very well

are you familiar with how complex multiplication works?

yes i am

add the angles, multiply the magnitudes

yes so

all n'th roots of unity are rotated copies of the primitive n'th root of unity xi

all rotated by the same angle and having the same magnitude

yeah so i don't understand the primitive roots of unity then lol

n'th primitive root of unity is just xi = e^(2pi i/n)

ok, but why?

because it's the only number such that x^n = 1 but x^k != 1 for any 1<=k<n

[willow bear]

okay i'm back from a phone call

right, so

sonja's thing is a bit inaccurate

so here's how i would explain why the roots of unity are what they are

@viscid thistle you here?

[viscid thistle]

yes i am

[willow bear]

alright so

there's one key thing to know

for a complex number $z = re^{i\theta}$, multiplication by $z$ as a transformation of the complex plane looks like scaling by $r$ and rotating by $\theta$

[obsidian monolith]

Ann:

[willow bear]

and multiplication by 1 is, of course, the identity transformation

i.e. the transformation that leaves everything unchanged

i am not being 101% hyper-formal here, but does this make sense?

[viscid thistle]

yes of course

this is just standard multiplication in $\mathbb{C}$

[obsidian monolith]

polynomial:

[willow bear]

yes that is what i'm talking about, nothing funky

so

$z^n = r^n e^{i n \theta}$, as i am sure you know; said in another way, this means that applying the "multiplication by $z$"\ transformation $n$ times looks like scaling by $r^n$ and rotation by $n\theta$

[obsidian monolith]

Ann:

[willow bear]

now we are interested in solving the equation z^n = 1, i.e. find those numbers z such that multiplying by z a total of n times brings you back to what you started with

so we need the transformation "scale by $r^n$ and rotate by $n\theta$"\ to be the identity transformation

[obsidian monolith]

Ann:

[willow bear]

so let's take care of the straightforward part first: the scaling factor, $r^n$, must be $1$, and hence, since $r$ is a positive real number, $r^n = 1$ implies $r = 1$.

[obsidian monolith]

Ann:

[willow bear]

i.e. in the identity transformation, there is no scaling involved.

[viscid thistle]

what do you mean by

"identity transformation"

[willow bear]

the identity transformation
i.e. the transformation that leaves everything unchanged

[viscid thistle]

where did you say that

[willow bear]

https://discordapp.com/channels/268882317391429632/363224154469826562/701041649186242580

[viscid thistle]

ah ok

yes

i didn't fixate on the identity part just the 1 part

anyway

[unique hill]

should this be in #precalculus ?

[viscid thistle]

we can go to prealgebra @willow bear

[willow bear]

i think this channel is a perfectly fine place for our discussion

let's stay

anyway

do you understand what i've said so far

[viscid thistle]

so you have z^n = 1

how do you take the magnitude of this

to get r = 1

[willow bear]

well, $|z^n| = |z|^n = r^n$

[obsidian monolith]

Ann:

[viscid thistle]

hmm idk if i agree with that

r is the magnitude no?

[willow bear]

r is the magnitude of z

r = |z| by defn

[viscid thistle]

yeah so why is |z^n| = r^n

[willow bear]

do you agree that $|zw| = |z| |w|$ for any two complex numbers $z$ and $w$?

[obsidian monolith]

Ann:

[viscid thistle]

yes of course

[willow bear]

it's a matter of rather simple proof by induction that |z^n| = |z|^n for any positive integer n

i don't want to get into it right now as i feel it'd detract from what i have to say at the moment

[viscid thistle]

ok

go on

[willow bear]

alright

so the scaling factor is 1

i.e. z = e^iθ, a pure rotation

[viscid thistle]

but r^n doesn't have to be 1?

[willow bear]

yes it does.

[viscid thistle]

[willow bear]

we want r^n e^inθ = 1

taking magnitudes on both sides leaves you with r^n = 1

[viscid thistle]

but that doesn't have to be the case

even if we "want" it to be the case

[willow bear]

yes it does, we're solving the equation z^n = 1

letting z = re^iθ, the equation becomes r^n e^inθ = 1

or are you gonna be all contrarian and disagree with that

[viscid thistle]

well here's my dilemma

so

if you let

z^n = r^n e^(i * n * theta)

then you

have r^n e^(i * n * theta) = 1, right

since z^n - 1 = 0

[willow bear]

you don't "let" z^n = r^n e^inθ

you let z = re^iθ, and z^n = r^n e^inθ follows

and yes then you have r^n e^inθ = 1, which is in fact what i have been saying all along

[viscid thistle]

ok now

when you take the magnitude of both sides

that's just r^n by definition then

?

[willow bear]

the magnitude of r^n e^inθ is r^n

yes

and the magnitude of 1 is just 1

[viscid thistle]

ok

yes

[willow bear]

so you get r^n = 1

[viscid thistle]

i agree that r^n = 1

sure

so r= 1

[willow bear]

and from that, since r is a positive real number, you have that r = 1

yes

okay, can we consider r=1 as established and move on now

[viscid thistle]

yes

[willow bear]

okay great

so now

z^n, as a transformation, looks like rotation by nθ

[viscid thistle]

not just looks

it is

[willow bear]

ok, sure, if that wording suits you better

[viscid thistle]

well is it not?

i mean that is by definition

[willow bear]

do you wanna argue pointless, detracting semantics or do you want me to continue with my explanation

[viscid thistle]

i just want to be sure

they're pointless to you because you know

i don't

i'm sorry if you think that i'm being semantic

i just want to know the details

[willow bear]

ok, fine, z^n as a transformation is a rotation by nθ.

there. happy?

[viscid thistle]

yes

[willow bear]

the point is, this rotation must bring you back to where you started

note however that it would be wrong to say nθ = 0

[viscid thistle]

so n > 1 and n in Z?

[willow bear]

n is a positive integer yes but that is not yet relevant

[viscid thistle]

= 1

i think

[willow bear]

not important

[viscid thistle]

since n = 1 is the original position

oh..

[willow bear]

n is fixed

we're solving for θ here

or will be once we have an equation

so we know that a rotation by nθ must bring you back to where you started

[viscid thistle]

this is where it gets fuzzy

why is that

[willow bear]

the transformation of rotation by nθ must be the same as the identity transformation

[viscid thistle]

why is that?

[willow bear]

z^n is rotation by nθ
1 is the identity

z^n

=

1

[viscid thistle]

what does identity mean?? i think i am misunderstanding that word

[willow bear]

the identity transformation is the transformation which leaves its input unchanged

[viscid thistle]

ok still that

AHHHH

i get it now

ok

so

we know that multiply by 1 just gives back the same thing

but because that thing is equal to 1

multiplying by that angle and its magnitude (1) is also like multiply by 1

lol

[willow bear]

ok

so

the transformation of rotation by nθ must be the same as the identity transformation

[viscid thistle]

yes

[willow bear]

now

it is true that a rotation by an angle of 0 is the identity

but 0 is not the only such angle

[viscid thistle]

we can add 2pi 🙂

[willow bear]

in fact, a rotation is the identity if and only if its angle is an integer number of full turns

AND AS SUCH, we can write nθ = 2πk, where k is an arbitrary integer

and from there, θ = 2πk/n

[viscid thistle]

yeah so z^n = 1 can be written as e^(2pi * k * i/n) = 1

[willow bear]

no

[viscid thistle]

😦

wait i think

[willow bear]

if anything, that would be $[e^{2\pi ki/n}]^n = 1$

[obsidian monolith]

Ann:

[viscid thistle]

yeah

ok

so now what

k has to be >= 1 right

in Z

[willow bear]

no, k can be any integer

however, despite the fact that $k$ can take on infinitely many values, the expression $e^{2\pi k i/n}$ will only ever return a total of $n$ values

[obsidian monolith]

Ann:

[willow bear]

this is because any two values of $k$ which differ by a multiple of $n$ will give the same value of $e^{2\pi k i/n}$, and hence $${ e^{2\pi k i/n} \mid k \in \bZ} = { e^{2\pi k i/n} \mid k \in \bZ, 0 \leq k \leq n-1 }$$

[obsidian monolith]

Ann:

[unique hill]

ann amazes me using texit every time i swear

[viscid thistle]

hmm

probably a dumb question

but

i am not sure what that last sentence means. wouldn't that apply to only even multiplies

since it'll be 2pi

for the in Z

not the non-negative one

[willow bear]

??

$e^{it} = e^{is}$ iff $(\exists m \in \bZ)(t-s = 2\pi m)$

[obsidian monolith]

Ann:

[viscid thistle]

yes exactly

so why can k be any integer here

[willow bear]

...

[viscid thistle]

?

[willow bear]

...

[viscid thistle]

i can tell its a dumb question

[willow bear]

from a purely geometric standpoint

you can spin around in place any number of full turns you want

you can spin around once, or twice, or thrice, or 1241 times, or a googol googol + 31 times, or whatever

[viscid thistle]

so.. negative values

just mean going in the opposite direction

oops

right

[rare hinge]

Can someone help me find the inverse of the function f(x) = 2(3)^x+1 -5

[steel venture]

let y = f(x)

so $y = 2(3)^x + 1 -5$

[obsidian monolith]

Buncho Maximos:

[steel venture]

first thing you can do is simplify the right

$y = 2(3)^x -4$

[obsidian monolith]

Buncho Maximos:

[steel venture]

and now you just have to switch y with x and solve for y

[rare hinge]

@steel venture the +1 is in the exponent, does that change anything?

[steel venture]

slightly

so $y = 2(3)^{x + 1} -5$

[obsidian monolith]

Buncho Maximos:

[steel venture]

just switch the y and the x from here and solve for y

[rare hinge]

I forgot how to solve when there is an exponent like that

[steel venture]

if $y=x^a$

[obsidian monolith]

Buncho Maximos:

[steel venture]

then $\log_x(y) = a$

whoops

[obsidian monolith]

Buncho Maximos:

[steel venture]

that one

[rare hinge]

Ok so convert it into a logarithmic function?

[steel venture]

you want to solve for y in $x = 2(3)^{y + 1} -5$

[obsidian monolith]

Buncho Maximos:

[steel venture]

so you'll probably use a logarithm

[rare hinge]

Ok, so what would be the solving steps?

[steel venture]

i'll ask you the same thing

what would the first step be

[rare hinge]

Add 5

[steel venture]

good

$x + 5= 2(3)^{y + 1}$

[obsidian monolith]

Buncho Maximos:

[rare hinge]

Ok then I would divide by 2

[steel venture]

$\frac{x + 5}{2}= (3)^{y + 1}$

[obsidian monolith]

Buncho Maximos:

[rare hinge]

Then this is where I dont know

[sturdy haven]

Please explain the difference between row echelon and reduced row echelon form

[steel venture]

earlten i don't think this is the best place for that

[sturdy haven]

Why it is precalculus.

[steel venture]

i guess

nvm

@rare hinge

$\frac{x + 5}{2}= (3)^{y + 1}$

from here

[obsidian monolith]

Buncho Maximos:

[steel venture]

lets just say the left side is a

and the 3 is b

and the y+1 is c

so we have

$a = b^c$

[obsidian monolith]

Buncho Maximos:

[steel venture]

how do we find c

[rare hinge]

Ok

Hm

Get it by itself

[steel venture]

ye thats what we want to do

how would we do it

hint

we have to use logarithms

[rare hinge]

Convert it to a logarithmic

So

Log(b)a= c

[steel venture]

$\log_b(a) = c$

[obsidian monolith]

Buncho Maximos:

[steel venture]

but yeah

and now lets apply the same logic to

$\frac{x + 5}{2}= (3)^{y + 1}$

[obsidian monolith]

Buncho Maximos:

[steel venture]

so if we want to trun this

into

$\log_b(a) = c$

[obsidian monolith]

Buncho Maximos:

[steel venture]

what is a going to be

[rare hinge]

Log3(x+5/2)= y+1

[steel venture]

awesome yeah

$\log_3(\frac{x + 5}{2})= y + 1$

[obsidian monolith]

Buncho Maximos:

[steel venture]

so too find y we just need to subtract 1

$\log_3(\frac{x + 5}{2}) - 1= y$

[obsidian monolith]

Buncho Maximos:

[steel venture]

$y = log_3(\frac{x + 5}{2}) - 1$

[rare hinge]

And that's it?

[obsidian monolith]

Buncho Maximos:

[steel venture]

thats the inverse

[rare hinge]

Thank you very much

[steel venture]

ofc

also

there's one more thing i wanted to show you

$\frac{x + 5}{2}= (3)^{y + 1}$

[obsidian monolith]

Buncho Maximos:

[steel venture]

when we're here

we can use the rule

$\log(a^b) = b \cdot \log(a)$

[obsidian monolith]

Buncho Maximos:

[steel venture]

and it doesn't matter what the base is

so we can just take the log of both sides

$\log(\frac{x + 5}{2})=\log((3)^{y + 1})$

[obsidian monolith]

Buncho Maximos:

[steel venture]

and move the y+1 down

$\log(\frac{x + 5}{2})=(y+1)\log(3)$

[obsidian monolith]

Buncho Maximos:

[steel venture]

and we just divide and do the minus 1 again

$\frac{\log(\frac{x + 5}{2})}{\log(3)} -1=y$

[rare hinge]

That's useful to know

[obsidian monolith]

Buncho Maximos:

[heady jewel]

dont multipost across various channels @rare hinge

[steel venture]

and the funny thing is that fraction

is just the log base change fomula

[rare hinge]

@heady jewel sorry

[rare hinge]

Here is another one I'm stuck on

I know the formula is A=Pe^rt

I'm just confused one where I get the numbers from

Ping me if you can help, I'll be trying to figure it out

[rare hinge]

This is what I got so far, just not sure how to solve for "t" which is the days it will take for the population to reach 3000 beetles

[viscid thistle]

Is the growth constant?

[rare hinge]

im assuming so

the questions just says uninhibited

i found the rate

[viscid thistle]

Im not a native speaker so i dont know that many "math words"

Just made sure

[rare hinge]

i need to show my work using the formula

but idk how

[viscid thistle]

Wouldnt it be like that?

Where r is the rate and t is days passed

Plug in you rate and solve for t

[rare hinge]

it would be 2000^rt

bc the equation is A=Pe^rt

[viscid thistle]

U gotta ask someone else

In #❓how-to-get-help

Im not ik precalc yet

[fleet yew]

@viscid thistle you cant expand that

You just leave it as the cube root of that expression

Binomial theorem only works for integer powers

[muted granite]

When I plug the 2 back in the equation does not equal. Where did I go wrong? Thanks.

[plush violet]

3^2 isnt 6

[muted granite]

lmao

thanks

[plush violet]

np lol

[muted granite]

The answer is not matching up with the calculator. Not sure what I am doing incorrectly. Thoughts? Thanks.

[wanton niche]

@muted granite how did you go from ln(6^2x) to 2xln5

[muted granite]

@wanton niche To be honest I am not sure. Just winging it.

5 is supposed to be a 6.

[wanton niche]

Yes

[muted granite]

Which step is wrong? I looked at the steps of a similar problem and it seems the same.

[wanton niche]

from the third

[muted granite]

No. I am lost.

I'll have to find a video and relearn how to do it.

[wanton niche]

I'm following on what you're doing, so here is what your third step should be:(x+2)ln(9)=2xln(6)

@muted granite this is your third step

xln(9)+2ln(9)=2xln(6)

xln(9)-2xln(6)=-2ln(9)

x(ln(9)-2ln(6))=-2ln(9)

x=-2ln(9)/(ln(9)-2ln(6))

[muted granite]

that does not get to this tho?

^correct answer.

[wanton niche]

if you put that in calculator

the answers are the same

all you have to do is simplify it

and the way you solve

[muted granite]

Thank you.

[odd helm]

I got this one right but shouldn’t the answer to be ln3 idk how it’s 3/2

[upper kelp]

Unfortunately, I think your solution to the quadratic formula:
2A² + 5A - 12 = 0
as A = 3 is incorrect.

[unique hill]

@odd helm

[glad mica]

would someone mind doing this for me, the answer i got doesnt seem to be right

I used Cos2x = 1 - sin^2x

once with the numbers plugged in

cos2x = 1 - (5/9) ?

[unique hill]

cos2x does not equal to 1 - sin^2 x

[glad mica]

whoopsie

[unique hill]

That’s cos^(2) x

[glad mica]

it should be cos2x = 1 - 2sin^2x

right?

[unique hill]

Yep

[glad mica]

alright thanks

[fleet yew]

,w cos2x = 1 - 2sin^2x

[obsidian monolith]

Display results online and refine query

Requested by AMD#4434

[unique hill]

...

[fleet yew]

!!!

[granite pine]

for this q ik the answer is 1 2 -3

but how would you calculate this?

cant seem to prove it

[viscid thistle]

look up how to do matrix multiplication

[deft spear]

they teach matrix multiplication in precalculus?

Australia must be super slow

[stuck lark]

i've seen it happen before. it never goes beyond mere matrix algebra though

[willow bear]

@granite pine do you know how matrix multiplication works

[granite pine]

oh yeah nvm im just dumb

how about this ?

[analog relic]

$9^{x+2}=6^{2x} \iff 3^{2x+4}=3^{2x} \times 2^{2x} \iff 2^{2x}=3^{4}=81 \iff 2x\lg2=\lg81 \iff x=\frac{\lg81}{2\lg2}$

[obsidian monolith]

Lσνιηg✧Sσνєяєιgη:

[autumn mesa]

@deft spear at my hs they never taught us matrices at all in the math sequence

[fleet yew]

Yeah it depends on how far your school goes

If you take calc 1 senior year then you wont get the chance

High school math is painfully slow anyway imo

Teach yourself