precalculus

[timid dock]

so we use cos(2π - x) = cos(x) when the value of arccos is not in 0 to pi

more questions incoming

y=1/3 sin(-5x)

does the -5 determine the flip

and what happens if it's y=-1/3sin(-5x)

[steel venture]

sin is odd so a negative number multiplied to the x leads to a flip

the negative on the outside acts the same way

[timid dock]

ok so the signs determine it

got it

[wise bison]

How do I prove (ab)^n = a^n • b^n using induction? Not sure how to do induction with 2 variables.

[willow bear]

n natural, a and b real?

@wise bison

if so, then you're inducting on n only. a and b are real-valued and hence impossible to induct on.

[wise bison]

Oh! I'm so stupid. I'm supposed to use n. Not reading properly. So sorry.

For n = 1: (ab)^1 = ab = a^1 • b^1
Assuming (ab)^k = a^k • b^k,
For n = k + 1: (ab)^(k + 1) = (ab)^k • (ab)^1 = (ab)(ab)^k = a^k • a^1 • b^k • b^1 = a^(k + 1) • b^(k + 1)
Is this good enough? Sorry, first time doing induction.

[willow bear]

(ab)^(k + 1) = (ab)^k • (ab)^1 =* (ab)(ab)^k =** a^k • a^1 • b^k • b^1 = a^(k + 1) • b^(k + 1)

the equalities i marked with * and ** need to be elaborated on

you silently used the commutative law

which you probably should indicate

[wise bison]

Oh. I'm not allowed to assume commutativity? I don't see any of this stuff in the examples.

[willow bear]

you are allowed to assume commutativity, you just need to be explicit about using it at this level

[wise bison]

Ah. Do I also need to state if I use associativity and distributivity?

[willow bear]

well assoc tends to kinda bog you down a lot

more so than commutativity

[wise bison]

Oh. Thanks a lot! I'll try to practise more, I guess.

[dusky wave]

How would I do this?

[uncut mulch]

what have you tried?

[dusky wave]

Taking the log of both sides

[uncut mulch]

good. continue

[dusky wave]

What I get doesnt compare at all to the answer so

[uncut mulch]

what do you get?
the answer has several forms depending on the log base you used

[dusky wave]

So I get 1-6xlog5=xlog4

The actual answer is the x I wrote down

Idk whee the second log 5 comes from

[uncut mulch]

distribution/parentheses

(1-6x)

[dusky wave]

Of what

[uncut mulch]

never forget

[dusky wave]

So how do you get past this?

[uncut mulch]

$\log 5^{1-6x} \neq 1 - 6x \log 5 \
\log 5^{1-6x} = (1 - 6x) \log 5$

[obsidian monolith]

ramonov:

[dusky wave]

Wouldn’t it be x+x

[uncut mulch]

wdym?

apply the log rule properly, then isolate x from there

[dusky wave]

I got to log5= 6log5x+Log4x

Idk what to do with the x’s

[uncut mulch]

try not to write the x like that

it can be misinterpreted as log(5x) and log(4x)

[dusky wave]

Kk

So you can factor it out right

[uncut mulch]

which isn't what you have

yes

[dusky wave]

Kk

Alright thanks a bunch man

[carmine birch]

Hey

at 43

how do i get z1 to be in trigonometric form

[valid violet]

Isnt it already in trig form @carmine birch

[willow bear]

no it's not

[carmine birch]

nah

i have to make it be smth like r(cosa +isina)

in order to use moivre’s formula and the other

stuff

[viscid thistle]

So I'm asked to find the length of the radius of a circle on a graph with the extremities of the diameter being 2,3 and 6,1

The answer is 2,2 but I'm not entirely sure how

[viscid thistle]

<@&286206848099549185>

@viscid thistle can you elaborate on what you were asked?

Unfortunately I don't think I can

On a graph to scale,

I could take a picture but it's all in French

There are quite a few people here who speak french (I think)

if its a graphic, it doesn't matter regardless

The graphic isn't in the question

But it is implied that we would need to use the same formula

That we would if a graph were given

Wait

Ah, well, I'll try to know what you mean then. It gives you the coordinates of 2 "ends" of the diameter, right?

It's asking for the radius so I could just get the distance between the two points right then decide that by two?

eyup, what I was thinking

I'm dumb idk y I didn't just think of that it should've came to mind

since the radius is just D/2

I'm just scared

I had to look up radius in english

[odd helm]

My answer was 2 triangles

The answer was supposed to be no solution

Could someone explain where I messed up

[warped willow]

which question

[uncut mulch]

number 5?

i have no idea what you're attempting to do in your work

[warped willow]

if it is number 5, its no sol. because the angles dont work

or my math is off

[uncut mulch]

the longest side is opposite the largest angle (in a triangle)

from that, it should be trivial

[broken jungle]

can someone explain to me why we use parametric equations rather than expressing y in terms ofx?
also
why do we have restrictions on the x and y of parametric equations?
like
x = 3t +4
y = t -5
why is x only restricted to x >= 4?

[uncut mulch]

i'm assuming t is some sort of variable representing time?

[broken jungle]

so t can't be negative?

in any parametric equation

[uncut mulch]

depends on the parameter.

[warped willow]

have you heard of negative time?

[uncut mulch]

but if it IS time, then it can't be negative

[broken jungle]

idk if it's time

it's just an equation being model by x(t) and y(t)

if it's not time

can it be negative then?

[uncut mulch]

yes

[broken jungle]

wait

anyone here with a ti84 know how to graph parametric equations

if I graph

x = t^2 + t
y = t^2 - t

it gives me a line

but it's supposed to give me this:

[valid violet]

Try extending the t interval

[broken jungle]

I extended t to the same as the one I had on desmos

-500 < t < 500

but it still gives me a line

[valid violet]

Viewing window?

Zoom fit?

[broken jungle]

@valid violet I think I fixed it

I just updated my calculator

god

but im like

still so worried

bc idk why the updating fixed it

or if it'll ever stop working again

[broken jungle]

got question

fuck

it's rotated

but the question is pretty much

The area bound by the relation abs(x) + abs(y) = 2 is ?

Why is it 8 instead of 4?

[uncut mulch]

,rotate 90

[broken jungle]

if I graph abs(y) = 2 - abs(x)

I find the area using bh/2; which is = 4

[uncut mulch]

show me your graph

[broken jungle]

@uncut mulch

what

wait

,rotate 90

[broken jungle]

ignore the f(x) part a thte top

[uncut mulch]

that isn't the graph of
abs(x) + abs(y) = 2

[broken jungle]

but

how would I graph abs(x) + abs(y)?

ugh

[uncut mulch]

graph
y = 2 - |x| and
y = - (2 - |x|)
with the domain of x in [-2, 2]

[broken jungle]

I don't get it

[uncut mulch]

which will get you a square

[broken jungle]

how did you come up with that

[uncut mulch]

well from
|y| + |x| = 2
rearranged gets
|y| = 2 - |x| right?

[broken jungle]

yup

[uncut mulch]

from properties of abs vals
y = 2 - |x| OR
-y = 2 - |x| → y = - (2 - |x|) = |x| - 2

also from |x| + |y| = 2
the maximum value of |x| is 2 which means x is in [-2,2]

[broken jungle]

so |y| could be either y or -y?

[uncut mulch]

yes

[broken jungle]

I thought absolute value means that the value has to be positive

[uncut mulch]

for y>= 0 |y| = y
for y <0, |y| = -y

eg, if y = -2
|y| = |-2| = 2 = - (-2) = -y

[broken jungle]

ah okie

wait

when you said

the maximum value of |x| is 2

how did you get [-2,2]

[uncut mulch]

solving the inequality |x| <= 2
but it should be intuitive

[broken jungle]

x has to be less than 2

I get that

but

why does it have to be greater than -2

[uncut mulch]

what happens when x is <-2?

what will |x| be?

[broken jungle]

OH

yeah

sorry

I get it now

[wild wadi]

Hello everyone

Do any of y’all know how to verify double angle identies

[dim jungle]

https://en.wikipedia.org/wiki/Proofs_of_trigonometric_identities#Angle_sum_identities

double angle identities is just a type of the angle sum identities

[wild wadi]

Ay thx bro 👌

[obsidian frost]

hello

can someone help me with factoring polynomials

[valid violet]

[obsidian frost]

how to factor 5 terms to its simplest form?

[patent beacon]

Guess rational roots

[willow bear]

"5 terms"

kinda worrying that "5 terms" and not "4th degree" is your first description of it

[obsidian frost]

roots?

like it has to equal zero?

[valid violet]

[obsidian frost]

is it not 5 different terms

[willow bear]

this polynomial DOES have five terms but that,,, isn't as important as its degree being 4

[patent beacon]

So a root of a polynomial is any place it equals zero.
If a polynomial has a root at x = r, then (x - r) is a factor

We describe a polynomial with its highest degree, not number of terms. This is a 4th degree polynomial. We sometimes call this a quartic.

[obsidian frost]

i see

and what formula am i supposed to use exactly?

quadratic formula?

no two numbers multiply to get 60 but add to negative 43 as well

[patent beacon]

Quadratic formula works on quadratics

This is not one of those

[obsidian frost]

what formula would you use

[willow bear]

i mean first off do you understand why factoring your polynomial is equivalent to solving the equation (your polynomial) = 0

[valid violet]

There is a quartic formula but it's not practical, no one uses it

[obsidian frost]

to get the simplest form right

[uncut mulch]

simplest form is subjective.
do you know about the factor theorem?

[obsidian frost]

no never learned that

[patent beacon]

What happened right after you took that picture?

[obsidian frost]

i took it right before class ended

[patent beacon]

Like, have you never seen a polynomial of degree higher than 2 get solved?

Or factored

[obsidian frost]

i have

[uncut mulch]

although sometimes not explicitly stated, you should have applied it in some form when doing quadratics
^

[obsidian frost]

but those are like one step equations

[patent beacon]

"those"?

[obsidian frost]

polynomials with degrees of 2

[patent beacon]

Yeah. Larger ones are a bit different

[obsidian frost]

i dont know how to solve larger one outside of 2nd degree

[uncut mulch]

do you agree that if
a is a root of P(x), then P(a) = 0?

[obsidian frost]

yes

[valid violet]

No

[obsidian frost]

you use synthetic division to check that

[uncut mulch]

whoops my mistake

^ and from that (x-a) is a factor of P(x)?

missed a line

[obsidian frost]

ummm

[uncut mulch]

if a is a root of a polynomial P(x), then
P(x) = (x-a)Q(x)
P(a) = (a-a)Q(a) = 0

[valid violet]

this is a conseqence of Bezout's identity

[uncut mulch]

sry about those typos

to find possible "nice" roots, you want to apply the rational root theorem

which involve testing: $\pm \frac{ \text{factors of constant term}}{\text{factors of coefficient of leading term}}$

[obsidian frost]

never used rational root

[obsidian monolith]

ramonov:

[uncut mulch]

here the factors of the coefficient of the leading term is 1, so you would test +- factors of 60

[obsidian frost]

all through synthetic division correct?

if it equals 0, then its correct

[uncut mulch]

more efficient to test if its actually a root first

from factor theorem

and then do the division if it is

[obsidian frost]

but idk how to factor that in the first place

i cant factor -43x + 60

[uncut mulch]

you don't factor like that

[obsidian frost]

no possible values

foiling?

[uncut mulch]

eg if you test whether x=1 is a root.
P(1) = 1 + 3 -21 -43 +60 = 0

then (x-1) is a factor

then you do the division

which makes it easier to find your other factors

[obsidian frost]

wait isnt the division that same thing as inputting the value into x to get 0

[valid violet]

P(a)=the remainder of dividing P(x) by (x-a), i think that's your question

[upbeat prairie]

I can't seem the find the equation of the asymtotes anywhere

Could someone tell me what the formula was

I remember it was something like

y = a/bx

[hollow bluff]

what would this middle number be

[viscid thistle]

did someone say

"r o w e ch e l o n f o r m"

[patent beacon]

Thoughts so far? Have you got a)?

[vague zephyr]

2

sorry about the late response, I was busy

i cannot figure out the horizontal asymptote

[stark trout]

where do you think they're at?

[viscid thistle]

I am having trouble with sin(6theta)cos(2theta)

I know the formulas to seperate them

I know sin(6theta) can be turned to

sin(2theta+4theta)

can some one help me with this problem

I have been stuck with this problem for a while now

@viscid thistle

do you think you can help me with this?

I worked with my professor on this question

how ever i seem to have forgotten the next step after this

[fluid shore]

?

You're having trouble with an expression?

What's the question? You want to simplify sin(6theta)cos(2theta)

[viscid thistle]

yes

@fluid shore well no

I need to express

it as a sum

that contains only sines or cosines

i am confused as to where to continue

I managed to get sin(6x)

down to sin(2x)cos(4x)+cos(2x)sin(4x)

[fluid shore]

Do you know the addition formulae?

Okay

Then?

[viscid thistle]

I am confused after that part

[fluid shore]

Keep reducing and simplifying

[viscid thistle]

do i reduce the cos(4x)?

wuldnt they just be equal to

like 1?

[fluid shore]

Well, yes you do reduce it

?

[viscid thistle]

so since

[fluid shore]

Since when would cos(4x) be equal to 1?

[viscid thistle]

sin(2x)sin(2x)-cos(2x)cos(2x)

would be wha it would be reduced

wait no that's if its plus i think

yeah

so sin^2(2x)

sin^2(2x)-cos^2(2x)

if i am not mistaken?

(sin(2x))^2-(cos(2x))^2

[fluid shore]

Wot

[viscid thistle]

is that not what it is reduced to?

[fluid shore]

Okay okay, write it down on a piece of paper and show me

[viscid thistle]

cos(4x)

Oh okay

[fluid shore]

No. cos(4x) = (cos(2x))^2-(sin(2x))^2

[viscid thistle]

oh right sin is after

yeah that's what i tried to get to]

i think this is the part where i am totally lost

after i get to this section

[fluid shore]

,rotate

[viscid thistle]

that's neat

okay

[fluid shore]

Can you simplify further?

[viscid thistle]

lets see

[fluid shore]

Use identities. Basically, just spam the identities as much as possible but be smart about their use as well

[viscid thistle]

right

wat no that's the wrong

thing

its sin(6x)sin(2x)

thatdoesnt matter

all that much

from where you are at with that but i wonder if i have to use that someway

also isn't it

wait nvm

don't you have to reduce sin(4x) aswell?

oh right

[fluid shore]

Ah I'm sorry, you mentioned that you wanted to simplify sin(6x)cos(2x)

My bad

[viscid thistle]

its okay

also

https://gyazo.com/8c77eae22ff6cab42c535a3a231418b1

i am confused at this spot

so sin(4x)cos(2x)

how does that convert to

2sin(2x)cos^2(2x)

?

[fluid shore]

Well,

Sin(4x) = 2sin(2x)cos(2x)

[viscid thistle]

oh wow

that's neat

right

ohhh right

you did the

formula

ahhh isee that makes sense

and that's why its cos^2(2x)

ahhokay

[fluid shore]

🙂

Yeap, I just dabbed my way through it really fast

[viscid thistle]

then you took out the factor

common factor

sin

wait

no

you distributed?

interesting

[fluid shore]

No. I added the first term in the bracket to the third time

Now, you can simplify further by removing the cos^2(2x) using a particular identity

[viscid thistle]

I know the final form is something like 2cos or 2sin(theta)sin(2theta)

...

i for got

that's what i had when i was with my professor

hmm

cos^2(2x)

that's

hmm i am like not sure at all to be honest at the moment

all i know is

oh wait

i know

cos^2-1

= sin^2(2x)

cos^2(2x)-1

is that it?

ah no

i mean

sin^2(2x)-1

[fluid shore]

Sin^2(2x) + Cos^2(2x) = 1

:/

[viscid thistle]

right

[fluid shore]

Simplify further lol

[viscid thistle]

hmm

[fluid shore]

Don't give up, it's a lot of work but you can do it

[viscid thistle]

I don't wanna ever give up

on math

I am a programmer so i need to get my math credits

despite having a 4.0 i still struggle a lot and in the end after the hard work i manage

my final is in a few weeks so i am going over all the things i am having trouble remembering

this is the sole problem that i am pooped with haha

[fluid shore]

Having a 4.0 means absolutely nothing if you don't have an intuition for what you're doing

[viscid thistle]

i appreciate your help

[fluid shore]

Forget about remembering things. Focus on understanding.

[viscid thistle]

yeah I agree

Right

lets see

so I was thinking

3sin(2x)(sin^2(2x)-1)

do i distribute this?

hmm

[fluid shore]

Okay first of all

Is that the correct formula?

[viscid thistle]

lets see

[fluid shore]

Is that the correct substitution for cos^2(2x)

[viscid thistle]

give me 2 seconds

i need to revisit

oh is it 1-

[fluid shore]

Think, think, think

[viscid thistle]

1-sin^2(2x)?

yeah i am pretty sure that is it

[fluid shore]

Idk, you tell me? Is it the correct formula?

[viscid thistle]

yes

[fluid shore]

Good. Have some more confidence in yourself

Now, what's the next step?

[viscid thistle]

hmm

I should then be able to distribute 3sin(2x)

3sin^3(2x)

oh

wait

then

3sin^3(2x)-3sin(2x)

(3sin^3(2x)-3sin(2x)) but then i should be able to subtract an exponent from this

3sin^2(2x)

making it that

is that correct?

wait

[fluid shore]

Wait, are you sure you're doing your algebra correctly?

Okay to simplify things

Let sin(2x) = u

[viscid thistle]

hmm okay

[fluid shore]

Now, you have 3u(1-u^2) - u^3

So, simplify that

[viscid thistle]

got it

okay i got 3u-2u^3

2u^3-3u

[fluid shore]

Are you sure that's the correct simplification?

[viscid thistle]

i am 80% sure

lets see

[fluid shore]

Look very closely.

[viscid thistle]

so first distributing

3u

to 1

would be 3u

then to u^2

would be 3u^2

so 3u-3u^3

then i subtract u^3

oops

si teag

so i sub u^3

to 3u^3

making it 2u^3?

2u^3-3u

hmm

[fluid shore]

No .

3u(1-u^2) -u ^3 = 3u -3u^3 - u^3 = 3u -4u^3

[viscid thistle]

aihiflf

I am sooooooooo stupid!

AHHHH!

ITS A NEGATIVE AHHHHHH

[fluid shore]

No you're not. You just needed a nudge in the right direction

Take it as a learning experience and move forward

[viscid thistle]

No i am actually stupid right now haha

I had 3 hours of sleep

last night

I apparently got 15 extra points on my final exam for no reason

for my computer science class

apparently colleges don't even look at the extra points

gotten in classes

despite me having 132.35% lol

and wasted my sleep

over that which was useless

[fluid shore]

Then, you are not at your best state of mind to solve problems. You can't call yourself stupid if you're not in the best mental state to solve a problem.

[viscid thistle]

https://gyazo.com/f8c2ed19e64032b1636acdb90b4cf505

Haha thanks for cheering me up

I was proud i got the extra credit though

[fluid shore]

That's a job well done to you

[viscid thistle]

But yeah

I will have 2 days of good sleep

on my day of the final exam

exams are on the 12th

so 7 more days yikes

I am not nearly prepared yet considering ic ant even do simple algebra

eek

okay lets continue

so now that we have

3u-4u^3

[fluid shore]

That's not fair to yourself. You're tired. Everyone makes mistakes when they're tired.

[viscid thistle]

Haha true.

[fluid shore]

Yeap, let us continue

Now, do you have a way of simplifying sin(2x)

[viscid thistle]

hmm

oh

isn't it

2sin(x)cos(x)?

[fluid shore]

Are you sure?

[viscid thistle]

yes

i am 100% sure lol

[fluid shore]

Good.

Continue

[viscid thistle]

hmm

3(2sin(x)cos(x)) -4(2sin(x)cos(x)) )^3

6sin(x)3cos(x)-(8sin(x)-4cos(x))^3

is this the correct path?

hmm

[fluid shore]

Hmmm are you sure about that?

[viscid thistle]

no honestly I am not

hmm

so is

3u -4u^3

and u = sin(2x)

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

3(2sin(x)cos(x))-4(2sin(x)cos(x))^3

[fluid shore]

Yes. If there are too many terms, then just evaluate each term one by one

[viscid thistle]

hmm that looks right

[fluid shore]

Go on

[viscid thistle]

I think 6sin(x)3cos(x)-(8sin(x)-4cos(x))^3 is correct

hmm lets see

[fluid shore]

Nope

[viscid thistle]

oof

lets see

[fluid shore]

Where did that minus come in suddenly?

[viscid thistle]

3u -4u^3

[fluid shore]

With your second term, are you sure you're doing the algebra properly?

[viscid thistle]

hmm lets see

-4u^3

-4(2sin(x)cos(x))^3

(-8sin(x)-4cos(x))^3

[fluid shore]

,rotate

[viscid thistle]

ohh the root 3

i thought that was after

wait

8^3

[fluid shore]

Root 3? You mean the cube?

[viscid thistle]

oh right

yeah cube sorry

8*8=64

so 4*4

= 16

[fluid shore]

Wait hold on, you're trying to simplify sin(6x)sin(2x), yes?

[viscid thistle]

yes

wait no

I am trying to

express as cosine or sine

prompt is

:Express the given product as a sum containing only sines or cosines.

[fluid shore]

Right. So express it as a sum of sin(x) & cos(x)

[viscid thistle]

hmm

What do you mean by that

oh frick

x3

not x2

222 = 32

right

umm

e.e

[fluid shore]

Like, basically, the thing inside the brackets cannot be anything other than an x?

HUH

[viscid thistle]

?

[fluid shore]

222 = 32?

[viscid thistle]

2*2=4

Idjlkaidf

Umm

UMMMM

[fluid shore]

Lol

[viscid thistle]

8*4

e.e

[fluid shore]

4*2^3 = 32

[viscid thistle]

32

[fluid shore]

Lol

[viscid thistle]

e.e

[fluid shore]

You can't write it as 222 =32

People will think you're mental

[viscid thistle]

XD

I am sorry eeeek

so when you distribute it doesn't get sent to cos?

that's interesting

i guess it looks different

Ill have to remember it is a polynomial and sin(x) is basically x

or sin(2x) is basically x

hmm okay

6sin(x)cos(x)-32sin^3(x)cos^3(x)

so we need to get rid of

cos or sin

is the next step i believe

so i can subtract those

no i cant

umm

hmm

lets see

[fluid shore]

Remember, you need to simplify it as much as possible

[viscid thistle]

hmm factor?

factor out a 2

3sin(x)cos(x)-16sin^3(x)cos^3(x)

[viscid thistle]

https://gyazo.com/5461905a37e8b05b7cacd5b3cdbbb125

,rotate

[obsidian monolith]

Couldn't find an attached image in the last 10 messages

[viscid thistle]

,rotate

[pale bison]

did your dog ate all the pixels

[signal mountain]

Hi. If I have f(x)=2x^2+kx-4k how do I find values of k for which f(x)=0 has real solutions?

[narrow marsh]

Show what uve tried

[signal mountain]

i solved the discriminant and got k^2+32k >= 0

[narrow marsh]

ok

[signal mountain]

but that didn't give me the correct answers according to the book

[narrow marsh]

What are the answers in the book

[signal mountain]

k <= -8, k>= 0

[narrow marsh]

Well Im mystified

[signal mountain]

printing error?

[narrow marsh]

cant see anything wrong with your working at a glance

b^2-4ac has to be bigger than 0

[fluid shore]

Did you put the correct quadratic equation above?

[narrow marsh]

thats

k^2 + 32k surely

[fluid shore]

Is the question that you typed in exactly the thing stated in the book?

Just a bit of a difference could result in a different answer. That's why i'm asking.

[narrow marsh]

Pretty sure its wrong. One way to check is to plug in something like -8

[fluid shore]

Yea the book seems to be wrong.

[narrow marsh]

The book forgot the 4 in 4ac most lilely

[signal mountain]

ok, thanks guys ❤️

[iron spoke]

hey, can someone help me out?

[fading token]

[iron spoke]

if you have a fraction like 2x-1 over 2x, you can not cancel out the 2's right?

[uncut mulch]

be more specific, depends what you are doing overal

$\frac{2x-1}{2x}$?

[obsidian monolith]

ramonov:

[iron spoke]

trig identities

yeah, that one

[uncut mulch]

and what do you think it could equal to?

[iron spoke]

1x over 2x

[uncut mulch]

that wouldn't be allowed, i don't know what you even did to get that

[iron spoke]

yeah i got confused

[uncut mulch]

an alternate form would be:

[narrow marsh]

@iron spoke

$$2\times 3 + 1\over 2$$

[obsidian monolith]

Shuri2060:

[uncut mulch]

$\frac{2x-1}{2x} = \frac{2x}{2x} - \frac{1}{2x} = 1 - \frac{1}{2x}$

[obsidian monolith]

ramonov:

[iron spoke]

so in terms of 2sin(x)-1 over 2cos(x) those are the steps i would do?

[uncut mulch]

can you post the whole question and what its asking for?

[iron spoke]

okay

let me find an example if thats okay with u

[uncut mulch]

that is very different from what you were asking

[iron spoke]

i could not find an example, my bad

But there is situations where i come across 2sinx+cosx over 2sin^2x and tried to cancel out the coefficients of 2 which gave me different answers

thats why i was wondering if u can or can not cancel the 2's out

[uncut mulch]

$\frac {ab + c}{ad} \neq \frac{ \cancel{a}b + c}{ \cancel{a}d}$

[obsidian monolith]

ramonov:

[uncut mulch]

are you attempting to cancel like that?

[iron spoke]

yeahh, thats what I mean

[uncut mulch]

that would not be allowed

[iron spoke]

oh i see

[uncut mulch]

you are attempting to divide both the numerator and denominator by "a" to cancel it
however that would ignore the "c"

[narrow marsh]

https://cdn.discordapp.com/attachments/363224154469826562/652504999737426007/321445119171756033.png

[iron spoke]

oh alright, so whenver its + or - that would not be allowed?

[narrow marsh]

Try cancelling the 2s in that

3.5 = 4?

[iron spoke]

yeah i get it now aha

thanks alot you two!

[signal mountain]

Hey, can anyone help me with this problem?

[narrow marsh]

whatve u done

[signal mountain]

not much

i know if x-2 is a factor then f(2)=0 but i'm not sure if that's relevant

[fluid shore]

What does it mean for the given polynomial to be divisible by some linear factors.

Of course it is. It gives you the possibility of forming two equations with a & b. Forming two linear equations in a & b will surely allow you to retrieve them.

lol shuri

nice

[narrow marsh]

what i was about to say probably woulda worked

ehh

or maybe not

[signal mountain]

so i have f(2)=18-4a+2b

[narrow marsh]

anyway ^ is easiest

[fluid shore]

where R(x) is the remainder polynomial

Of course, I'm being very sloppy but I don't quite care tbh.

Let me know if I've made any arithmetic errors lol

I'll probably want to write out proper derivations of all these results when I wake up.

[signal mountain]

ah i think i get it now

thanks

[fluid shore]

🙂

[viscid thistle]

are my calculations are correct?

[patent beacon]

1 + 0i

Is a 4th root of unity, because if you take the 4th power, you get 1

[stark willow]

so i guess use
$r^n*(e^{in\theta})$

[obsidian monolith]

tryingtolearn:

[patent beacon]

You get very far with that, yeah. Just gotta know how to use it

[valid violet]

the correct way to take nth roots of unity in general is to write

where this is to be understood as an equality of multiple-valued expressions

if you don't know what I mean by multiple-valued I can explain

oh, typo

sec

$1^{1/n} = e^{2 \pi i k/n}$

[obsidian monolith]

gfauxpas:

gfauxpas:

[valid violet]

where k is an integer

for k = 0, 1, 2, ..., n-1 you'll get n answers and then you'll start repeating answers

for k >= n

latex bot broke D:

oh we're in the precalc chat

maybe you don't know enough to use that method then

[broken jungle]

guys

i ned help

So you got the question

A salad bar has 7 ingredients. How many different salads are possible where two salads are different if they don't include identical ingredients?

[rich ocean]

wouldnt it be 2^7 - 1

i assume each of the 7 ingredients can either be present or not present in the salad, and that you need at least 1 ingredient (meaning the only invalid combination is no ingredients)

but the "if they don't include identical ingredients" throws me off a lil cuz isnt that obvious

[red niche]

Okay what's the question

[obtuse stirrup]

on the interval [-7,-5]

is it wrong to say there are multiple global max

on that interval

[red niche]

What is the question asking you?

[obtuse stirrup]

or is that just completely wrong because global extrema refer to the entire function

theres no question, im just trying to understand this for myself

[red niche]

Okay

So first and foremost, you know what a local maximum/minimum is right?

[obtuse stirrup]

yes

[red niche]

Okay, so if the question asks something like:

In which interval are there local maximum/minimums for the function, f(X)?

A) [-7,-5]
B) [-2,4]
Etc.

Then your answer would be fine.

[obtuse stirrup]

Ok im just trying to understand the rigorous definitions of local and global extrema properly

like for example the function sin(x) has max value of 1 recurring

and min value of -1

All those points are local max and local min on its domain but does it have an absolute max and min?

[red niche]

That is the absolute max and min

[obtuse stirrup]

So there CAN be multiple occurence of the absolute max and min?

[red niche]

@obtuse stirrup https://math.stackexchange.com/questions/2108731/local-max-and-global-max-of-sin

does not have to be unique. It just has to be greater than or equal to all other values

[obtuse stirrup]

Ok I found the rigorous definition

so yea there can be multiple counts of local max and min

thanks

@red niche Also the answer sheet for the previous question says it's only A

I dont know what function says otherwise though

[red niche]

@obtuse stirrup The answer sheet is wrong then

There is a local maximum at X= 5.

Then f''(X) <= 0 will always follow

[obtuse stirrup]

Oh wait I think you're right

because the slope is positive approaching x and negative past x

wait no hmmmm

wait yea would it have to cross the x axis like this

to some degree

forgive my terrible drawing

therefore the second derivative will always have to positive

Nvm I lied thats the exact reason why its only A,
B and C say less than 0

[hexed geyser]

hey

ayone on right now?

how do i do this

like i have no idea how to "work backwards" with the quotient rule

like im trying to think of, what equation can be derived to give -13/y^14

and its frying my brain

ill try rewriting it as -13*y^-14 one sec

[hexed geyser]

i got it im good guys

[harsh cipher]

I need help setting the appropriate windows with this question.

I can't get the graph on my calculator to look like the solution. 😦

[harsh cipher]

nm figured it out

[harsh cipher]

how do I solve the second line with TI83 calculator

y= log base 1/3 (2) +1

[willow bear]

what

1. why is this in this channel
2. why are you pinging me
3. what does this have to do with math

blink blink

what the fuck are you on about

you have not answered even one of my 3 questions

what feet pics

#chill

[undone pawn]

lmao

[viscid thistle]

wtf happened here lmao

[fluid shore]

Ah

Some guy asked for feet pics

& Ann felt displeased. So that guy was kicked

To be fair, it was really creepy

[willow bear]

i didn't feel displeased, i felt utterly disgusted

[fluid shore]

Ah yes, that ^

[heady jewel]

Bruh

[fluid shore]

d(Bruh^2)/d(Bruh) = 2*Bruh

[reef moth]

woah

[fluid shore]

Bruh

[undone pawn]

woah

[fluid shore]

bruh

[heady jewel]

hurb

[nova dew]

$\lim_{x\to\infty} \frac{2x + 1}{2x^2 + 3x + 5}$

[obsidian monolith]

CoolShot:

[nova dew]

so

[fluid shore]

Okay, what have you thought of?

[nova dew]

my reasoning is

that

the denom is quadratic

So increases at a faster rate than the linear numer

so it should be =0?

[willow bear]

yes, if $\deg(P) < \deg(Q)$, then $\frac{P(x)}{Q(x)} \to 0$ as $x \to +\infty$.

[obsidian monolith]

Ann:

[nova dew]

what happens if $x\to -\infty$

[obsidian monolith]

CoolShot:

[willow bear]

think about it yourself

[fluid shore]

What do you think happens? @nova dew Make arguments 😄

[nova dew]

It would pretty much be the same I assume

since you know -0=0

[willow bear]

not quite

[nova dew]

hmm well the denominator still does increase at a faster rate than the numerator right?

[willow bear]

P(-x) and Q(-x) have the same degree as P and Q respectively

[nova dew]

ok so it would still be zero then

[willow bear]

yes

[nova dew]

is this how you actually calculate limits? like look at how things change or is there a "formal method"?

[willow bear]

...

[nova dew]

why the ellipsis

[fluid shore]

Well, there is a formal definition of a limit. But calculating limits themselves often requires a bit of creativity

There are some limits that require a few more techniques that you'll learn down the road

Like L'Hopital's Rule, for example.

But yea, it's just something you need to get practice with.

[obsidian monolith]

Lionel:

[nova dew]

I see

[trim fable]

how would

u do this

1a.)

nvm

i got it

[agile steppe]

Hi

How do you do this?

I know area of speed time graphs gives distance, but what do I do if it is curved?

[serene heath]

Do what it says

3 rectangular strips

U only need an estimate

[ripe dust]

How did they simplified the input to 8,

the only thing I think that I can do is bringing the 2 in front

ah nvm lmfao exponent rule oof

the fact that it's written as ln^2(stuff) instead of (ln(stuff))^2 fucked me up there

[trim fable]

how would u do

3a

[serene heath]

Log properties

[trim fable]

ik but

im not

getting the

right answer

[uncut mulch]

show your work

[trim fable]

ok

i did

log8(24x4) over log8(3)

=32

=2^6

lol

the answer is 5/3

lol

@uncut mulch there

[uncut mulch]

uh... recheck your log rule for subtraction

[trim fable]

uh

u divide

[uncut mulch]

you divide the argument

[trim fable]

oh

?

[uncut mulch]

=log_8(24*4 / 3)

[trim fable]

yeah

which is log8(36)

right?

32*

[uncut mulch]

log_8(32)

[trim fable]

yeah

[uncut mulch]

and then apply change of base

[trim fable]

log8(2^6)
6log8(2)

[uncut mulch]

2^6 \neq 32

[trim fable]

6log(-1/3)

right?

the answer is 5/3

so im confused

lol

[uncut mulch]

fix your power first

also how did the argument of your log become negative?

2^6 \neq 32

[trim fable]

huh

why

that?

[uncut mulch]

,w calc 2^6

[obsidian monolith]

Display results online and refine query

Requested by ramonov#0921

[trim fable]

64/32 =2

[uncut mulch]

how does that change the fact that 32 isn't the same as 2^6

[trim fable]

oh

[uncut mulch]

and the next issue,
log8(2) is just 1/3

[trim fable]

ye

[uncut mulch]

NOT log( - 1/3)

[trim fable]

oops -3?

[uncut mulch]

no

[trim fable]

no?

coz

[uncut mulch]

1/3 is not the same as -3

[trim fable]

uh

isn't

-3 just 1/3

tho

wait

NO

i meant

3^-1

[uncut mulch]

i mean you could write it like that if you want

[trim fable]

ye

[uncut mulch]

and overall you should have
= log_8(2^5) = 5log_8(2) = 5 * 1/3 = 5/3

[trim fable]

ohhhhh

okkkkkkk

thanks

[uncut mulch]

there are major issues with your notation

[trim fable]

oh

rip

@uncut mulch

i need more help if thats fine

with d

[uncut mulch]

what have you tried?

this looks less complicated than the previous one

[trim fable]

oh really

log5(10x(25/4)^1/2)
=log5((250/4)^1/2)

and now im stuck

[uncut mulch]

oh no, you can't multiply the 10 into the square root like that

firstly, do you think you can simplify the sqrt(25/4)?

[trim fable]

lollllllllll

5/2

oh..

ok i see

ik what i did now

ok makes sense

thanks 😛

[trim fable]

oh

uh

for e

when u divide

square root 5 by square root 40

do the square roots cancel?

@uncut mulch

[uncut mulch]

you can divide the terms inside the root, but the sqrt remains

[trim fable]

oh so

then

ok

so how wouldi

simplify

square root of

1/8

2-3 (so -3 x1/2) = -3/2?

2^-3/2?

or 2^1/6

[uncut mulch]

2-3 (so -3 x1/2) ?

[trim fable]

2^-3

so the -3 times the 1/2 from the

square root

but the square root

[uncut mulch]

1 sec

[trim fable]

to get rid of it

ok

[uncut mulch]

2^(-3/2) looks ok

[trim fable]

ok

yay

oh

i needed help with this

the

6 square root 3

is confusing me idk what to do with it

[uncut mulch]

combine your logs and simplify first, worry about the radical later

[trim fable]

but

its

log 32 (24/16x6 square root 3)

then

[uncut mulch]

*log_3

[trim fable]

log 32 (1/4square root3)?

yeah

oops

[uncut mulch]

no, the 6sqrt(3) multiplies to the numerator

[trim fable]

?

why

[uncut mulch]

because you are "adding" log_3(6sqrt(3))

so multiplication, not division

[trim fable]

ye but

the first part was

division so

doesn't the division go with the

wait

ok

[uncut mulch]

you are subtracting log_3(16) so you divide by 16

[trim fable]

yea

ok

so