precalculus

[acoustic laurel]

g(x) = sqrt(x) + 69*420

what happened from f to g

[viscid thistle]

Big number begone

[acoustic laurel]

anndd what to do for big number begone?

from f to g?

[viscid thistle]

Cancelled it?

It just disappeared

[acoustic laurel]

how does a division cancelled out?

what operation would have to be done?

[viscid thistle]

Multiplication

Oooh :0

Wait

[acoustic laurel]

and that would mean f to g change would be?

heheheh

[viscid thistle]

It was dividing only the x and not the whole equation...

It was a vertical compression???

[acoustic laurel]

horizontal compression

because from f to g the /something was cancelled out by multiplying

and this was done on the variable

so it is horizontal

[viscid thistle]

Okay, but I have this, and the teacher didn't say it was wrong, how can I apply this?

Am going to copy what you wrote

[acoustic laurel]

lol rip copying is technically against the rules

we're trying to encourage conceptual understanding

but this time, it is what it is

so I think it'll pass

[viscid thistle]

It's not going to copy copy

[acoustic laurel]

🙂

[viscid thistle]

Just you know

Your ideas into my words

[acoustic laurel]

vertically stretching is multiplying the whole function

lol i think im so burned out lol

[viscid thistle]

So, only x being affected + a multiplication being involved = horizontal stretch/compression

[acoustic laurel]

if it is a fraction less than one that is being multiplied with, then it is a compression

but if only x, always horizontal

[viscid thistle]

So many combinations >~<

Another question, when horizontal, how does stretch/ compress look?

[acoustic laurel]

however, keep in mind that the multiplication with a fraction less than one is technically dividing

and that only applies for vertical

[viscid thistle]

Ooh

Okay okay

Wait

[acoustic laurel]

omg im getting disorganized with my thoughts

[fringe wadi]

I have a question about function multiplicity:

I don't understand why in order to determine the end behavior we have to add the exponents of the factors

[viscid thistle]

Okay let's get this chairman

We got this

Dimple question, how is the graph affected when horizontal stretch and horizontal compression

[acoustic laurel]

it becomes horizontally stretch and compressed lmao

[viscid thistle]

:V

Visually

[acoustic laurel]

just think of the graph shrinking (when being compressed) or growing (when being stretched) horizontally

[viscid thistle]

But I can't see that in a horizontal way

[acoustic laurel]

@fringe wadi I can address you

[viscid thistle]

My brain cannot make an image of it

[fringe wadi]

thank you!

i've got an example that i've been trying to work with

[obsidian monolith]

Cameron:

[acoustic laurel]

I think adding them up and checking whether they are odd or even leads back to the graph of x^2 vs x^3

[fringe wadi]

for this function, the sum of the exponents would be 6

but like i dont get how you can just add them together when the bases of the exponents are not the same

[acoustic laurel]

you don't add them up to work with them in a manner where you would have to fondle with x(x-3)^(2)(x-1)^3

it adds to six

so what do we know about even functions

[obsidian monolith]

Cameron:

[fringe wadi]

that would be the expression expanded

[acoustic laurel]

their end behavior has both lines going off the same direction

they point upwards

both upwards

like

,w graph of x^2

[obsidian monolith]

Display results online and refine query

Requested by Chairman Rovic#4063

[acoustic laurel]

because 6 is an even number

[fringe wadi]

i get the end behavior portion of it

[acoustic laurel]

@viscid thistle also, don't stress on making a mental picture

just think of the values of the function for every x

[fringe wadi]

but the actual multiplicity of the function is determined by the sum of the exponents

at least thats what i've been told

[acoustic laurel]

@viscid thistle think about how they get affected with horizontal or vertical stretches or compressions

mhmm

[fringe wadi]

i just dont understand why that works

the summing of the exponents

[acoustic laurel]

you can get the multiplicity of some repeating binomial

but I don't think you can for a function

you just determine odd and even multiplicity for a repeating binomial to check whether it crosses or bounces off the x axis

[fringe wadi]

yes

but you also have to initially find out what side of the x axis the function starts on

and as it was explained to me that's why you have to sum the exponents to find the highest multiplicity

[acoustic laurel]

mhmm lol idk

I think it has more to do with whether the leading coefficient is negative or positive

but with that rule, i'll look for it

it would be more intuitive since you wouldn't need to expand haha

@fringe wadi again, im sorry for not being informed enough. but thank you for bringing this to my attention

i'll post here once i've come across something pertaining to that method

[fringe wadi]

no worries! i just found it strange

in all cases we've practiced in class it worked fine

and even expanding the expression completely reveals that the highest present degree is indeed 6

so my thought was that in all cases the sum of every factors exponent will be the highest degree in the expanded expression?

i havent found an example where that isnt the case, but i wasnt sure if it was some widely proved thing

[desert socket]

Hello so I’m working on finding the zero and the example I’m following has me stumped in regards to what happens to the remainder 12.

We start with the polynomial at the top and know that x=2 is a zero so I started doing the synth division.

The example immediately factors the (x-2)(x^2 +4x+3) but doesn’t explain what happens to the remainder.

Well distributing and adding the +12 at the end of the distribution gives me the original polynomial but they removed it to factor for some reason.

[proud raven]

why do you think that 2 is a zero

are you sure you havent accidentally a sign somewhere

[desert socket]

That was part of the example.

[proud raven]

2 is not a zero of that polynomial

[desert socket]

It specified that 2 was a zero...

[proud raven]

its possible im stupid

1s

im pretty sure

ill double check with a different tool

but

thats calculator and synth division lol

if you divide a poly by a zero, there is no remainder

[desert socket]

Let me capture my screen on my laptop 1 sec.

[proud raven]

[desert socket]

it expected us to do long division which i did

long or synth

i did synth

[desert socket]

and then the guy got this..

[proud raven]

theres 2 imaginary and 1 real root

[desert socket]

I’m sorry can you explain

[proud raven]

sure, but you can just graph it too

hmm

if you make the 2 a negative instead of positive you have all real roots

2 isnt one of them though

[desert socket]

guess this guys just weird then

[proud raven]

lol one second

[desert socket]

this was the 'end result'

[proud raven]

he just fucked up a sign

try it with a negative 6

[desert socket]

christ

[proud raven]

youll see theres no remainder

[desert socket]

Yup I see that

You’d think they would review the maths before uploading this stuff on the official textbook video guides lol

Thanks dude

[proud raven]

[desert escarp]

gamer log equation?

[hybrid charm]

what does that mean 🙀

but it's not an equation

[desert escarp]

=?

[uncut mulch]

you posted an expression with little to no context. did you need help finding its value or something?

[dim jungle]

its just -3/5

log4(1)-log4(64^1/5)

[wind cedar]

I think I have this set up right, but I'm not really sure what to do next. I thought you set x to be a value that would make A or B = 0 so you can solve for the other one? But how to do that isn't jumping out at me

[hybrid charm]

you didn't factor it right

also either write x or t

[dim jungle]

Yes

J choose x to be one that cancels one of em

[wind cedar]

I did factor it right

[dim jungle]

So x=1/5 to solve for B or x=3/2 to solve for A

[hybrid charm]

also you need to find B you can't just say make them 0
instead make their coefficients 0

[wind cedar]

Yes

So how I was shown to set it up is to multiply both sides by the factored denominator. Which is how I got 8-3t = A(5t-1)+B(2t-3)

But I dont know what value to use, the leading coefficients are throwing me off

Nvm, I think I got it

[dim jungle]

I said above

[wind cedar]

yeah I saw that, but I was trying to figure out how, but I got it

thanks

[hybrid charm]

,w factor 10t^2+13t-3

[obsidian monolith]

Display results online and refine query

Requested by Nick ( 11111010011 )#3263

[hybrid charm]

😸

[lethal oracle]

It wants me to solve for exact solutions in the interval 0,2pi

[uncut mulch]

what have you tried?

[somber echo]

is anyone good at rational equations ?

[lethal oracle]

@uncut mulch

[uncut mulch]

good start.
do you have any ideas on what to do next?

[lethal oracle]

Factor it

[uncut mulch]

yep

[lethal oracle]

Ya I got it now thanks

[uncut mulch]

I didn't really do anything 😀

[lethal oracle]

Same instructions for this one

[willow bear]

ok first off

the sin(2x) was best left unexpanded

second

[lethal oracle]

How do I know to leave it unexpaneded

[willow bear]

sin(4x) could have been expanded to 2sin(2x)cos(2x)

third, your handwriting is a bit wack

[lethal oracle]

Sorry let me rewrite it

[willow bear]

and you are missing the =0

anyway the thing is the lhs is essentially a function of 2x rather than of x alone

[lethal oracle]

What would be the next step

[willow bear]

after you get to $\sin(2x) + 2\sin(2x)\cos(2x)=0$?

[obsidian monolith]

Ann:

[willow bear]

why don't you give it some thought yourself

[lethal oracle]

Should I factor out sin2x?

[willow bear]

should you?

[lethal oracle]

I think so, cause then I can start figuring out my solutions

[willow bear]

then do it

[lethal oracle]

Would x=pi/2k be the right equation to figure out my solutions for sin2x=0

[willow bear]

do you mean $x = \frac\pi2k$ or $x = \frac{\pi}{2k}$?

[obsidian monolith]

Ann:

[lethal oracle]

The first one

[willow bear]

then yes those are indeed the sols of sin(2x)=0

[lethal oracle]

For cos2x=-1/2

Is it x=4pi/6+pi(k)

[willow bear]

is it?

sounds like there's something you're missing

[lethal oracle]

Hmmm

I’m not sure

I got all the solutions except one

[willow bear]

$\cos(2x) = -\frac12$ gives two families of solutions: $x = \frac23\pi + k\pi$ and $x = -\frac23\pi + k\pi$

[obsidian monolith]

Ann:

[hoary valley]

These are the answers..

I chose (d) , cuz 2 multiplied by the highest value cosθ can get will equal 2.. and e > 2 , is that a valid solution ?

[willow bear]

yes, e > 2 ≥ 2cos(θ) for all real θ

[hoary valley]

Awesome

[hoary valley]

How to shift this function to the right and left ?

[rigid beacon]

Why does it say f(y) what

Anyways, for a given f(x)

f(x+a) shifts it left a, and f(x-a) shifts it right a

@hoary valley

[hoary valley]

@rigid beacon I miss typed that, Are you saying that f(x)=1/(x^2-a) shifts it to the right ?

[proud sparrow]

@hoary valley

by replacing x with x+a (a positive)

now 0 behaves like a

so the curve moves to the left by a

[rigid beacon]

1/(x+a)^2 shifts it left by a

And then -a to shift it to the right

[hoary valley]

@rigid beacon Uh Thanks

[rigid beacon]

Does that make sense?

[hoary valley]

Yes, I have to use parentheses and square the whole thing

[hoary valley]

I'm asked to find the domain of this function... Help

I know I should make the base on both sides similar, but I don't know how to do that with e

[willow bear]

solve e^2x = 1

[hoary valley]

I can't get rid of e

2e^2x = e

[willow bear]

what

where the f did you get that from

where did the extra 2 come from

[hoary valley]

I'm trying different things

[willow bear]

ever heard of this thing called log

[hoary valley]

I can't solve it

It's like anti exponent

[willow bear]

you can't not have heard of logarithms

the natural logarithm

have you ever used the natural logarithm in any capacity

at all

[hoary valley]

ln x = e^ln or something

[willow bear]

no!

ugh

no

look up "natural logarithm"

[hoary valley]

Ok

[hoary valley]

Thanks Ann, I solved it.

[green zenith]

i have a question about composite functions

is the domains of f(g(x)) the domain of g(x) AND the not simplified form of f(g(x))?

[patent beacon]

f(g(x)) is itself a single function, and as such it has a single domain

You can't put in anything that g can't accept.

And you also can't put in anything that, after being mapped by g, f can't accept

[green zenith]

so by that, f(x) is also considered the in the domain of f(g(x))

?\

ah i understand

so f(x) isnt considered

but the value of g(x) onto f(x)

[jaunty mason]

hi, ineed help with this question

Prove by mathematical induction that, for all positive integers n, 10^n + 3 * 4^(n+2) + 5 is divisible by 9

,$ 10^n + 3 * 4^{n+2} + 5 is divisible by 9

[willow bear]

$10^n + 3 \cdot 4^{n+2} + 5$ is divisible by 9

[obsidian monolith]

Tāhā:

Ann:

[willow bear]

ok so what is giving you trouble

[jaunty mason]

well, i can put k + 1 into it but not sure what i need to do next

[willow bear]

do you know in general how induction proofs go

[jaunty mason]

uh-huh

but i'm really only comfortable with proving sums by induction

not really good with divisibility

[dim jungle]

3*4^(n+2) will always be a certain number

🙂

Mod 9 that is

For pos n

That's ur hint

[jaunty mason]

@dim jungle ahh i don't get it

3*4^(n+2) mod 9 isn't equal to zero

🤔

[dim jungle]

Yes

But

It's always equal to a certain number

Mod9

[willow bear]

is it tho

[dim jungle]

Yes it is think binomial theorem

@willow bear

[willow bear]

is 4^n constant mod 9 tho

[dim jungle]

3*4^(n+2) is

[willow bear]

oh

yeah you're right

[green zenith]

hello i need help with exponential equation

i want to know how to solve 9^x=3^x+12

[stuck lark]

12 in the exponent or not?

[green zenith]

no

so what i did was isolate 9^x and 3^x

so its 9^x-3^x=12

and log both sides?

is this correct?

[willow bear]

well you CAN log both sides but it's not gonna be very useful

[green zenith]

so i cant log 9^x and 3^x?

oh it must be the quadratics

it is the quadratics

i already forgot

sorry for bothering you guys

[stuck lark]

it is the quadratics
elaborate

[jaunty mason]

I'm still not sure what we established when we found that mod 9 was eqaul to a certain number @dim jungle

[dim jungle]

If it's equal to a certain number mod 9, substitute it in

[jaunty mason]

umm what

substitute what where?

[dim jungle]

Nvm

J

Do u know wat number it always equals

[jaunty mason]

1

[dim jungle]

The middle part always equals 3

[jaunty mason]

wouldn't we need to prove that as well?

[dim jungle]

$3\cdot 4^{n+2}$ is 3 mod 9

[stuck lark]

rip texit

[dim jungle]

You can prove that through binomial theorem easily

[obsidian monolith]

Jbao:

[dim jungle]

Then add other stuff

[jaunty mason]

what

idk full context

what am i adding together?

why am I adding them together?

@dim jungle Sry for the ping spam but I kind of still need help

also i'm fairly certain we're not supposed to do it this way anyways

something similar to this probably

[trim fable]

can someone help me

with this

[patent beacon]

How far have you gotten?

[trim fable]

oh

i was starting with a stretch

so I have the asymptotes drawn out

but like idk how i would

include the 3 points on the right with these specific asymptotes

@patent beacon

like i just drew a stretch of what i did on paper

[ruby frigate]

if given a function do you know how to find the VA and the HA?

[trim fable]

ye

but the question is above this

@ruby frigate

i have to write an equation but i was drawing a sketch to help

[ruby frigate]

yeah I see that

[trim fable]

oh ok

so ix if i were to just make a basic equation that

f(x)= 6x/3x-12

that would give me a ha of 3

and va at 4

[ruby frigate]

no it wouldnt

play around on desmos and see for yourself

VA is when the denominator =0

HA has some rules but a tip would be just to make the highest power of top and bottom the same

and then just divide by their coefficients

[trim fable]

oh right

i meant

x^2-16

@ruby frigate

how would i make an equation with the properties where f(x) has to be greater than or less than 1/2

[green zenith]

hello i need help forming equations of circles

in the equation Ax^2+By^2+ax+by+c=0, what do i do with A and B when A and B > 1?

[urban vigil]

for the remainder thereom

let’s say that you’re looking for the remainder of a polynomial equation divided by TWO (x+/-n)

would you add together the remainders of each

[green zenith]

uhhh

no?

sir

[green zenith]

hello!

i need help solving this equation (log(x^2)+log(x^3))/log(60x)=7

what i use the product rule. Then the change of base. Then I change it in exponential form. Is that correct?

[green zenith]

halp

please?

[dim jungle]

just use log properties

you dont need to change base at all

combine all the log to one single log

[green zenith]

can you make it into one single log without using change of base?

[dim jungle]

yes

search log properties

[green zenith]

okay

i dont understand

i thought log(a)/log(b) is not equal to log(a-b)?\

[green zenith]

😧

[dim jungle]

it is

[harsh cipher]

Hi guys

and girls

I've previously asked this question but wanted to ask again

video explanation said it can foil out the line above the equation with arrows.

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

using pascals triangle. I'm not sure what that is.

[dim jungle]

J google online 🙂

When u get numbers forming a triangle u know it's right

There r a lot of cool concepts surrounding it so u can look into that if interested

Rly not necessary for a problem like this tho

[noble halo]

a little lost on this

i know i could just search it up but I’d like to know an explanation

[dim jungle]

(Sqrt(x))^2 isn't empty

It's j x but only when x is pos

[noble halo]

oh i just left it empty cus i didn’t know the answer lol

wdym pos

[acoustic laurel]

No, it’s not just f(x) = x. I think the domain of the square root function prevents you from using x-values that are negative to avoid a non-real answer. However, it is just f(x) = x for x-values of zero towards positive infinity.

Unless what you meant by “when is x pos” would mean that you cannot have negative position, which you can. It would be much more proper terminology to say “distance” since that would be the appropriate scalar term for some magnitude you travel in space.

[hexed bolt]

Just a quick one

What would I put?

[uncut mulch]

csc(x), cot(x) ig

[hexed bolt]

Okay

That didnt work

Oh I see

It's x = 0 not y = 0

[acoustic laurel]

Yeah, but I don’t see how you would have to change your answer. I cannot think of any other trigonometric function that would be undefined for x = 0 other than csc(x) and cot(x) which both would be undefined due to the sin(0) in the denominator that would yield a zero which would result in an undefined answer due to forcing the whole function to divide by zero.

[hoary valley]

1^x can I call this an exponential function or is it called constant ?

[willow bear]

depends on context but usually 1^x is not considered an exponential function

[haughty fern]

ive totally forgotten how this actually works

Without the two oldest children being chosen, where does 6c2 come from?

[hoary valley]

What's the range of this function? I tried graphing it, but it shows a blank graph..

[willow bear]

did you just put it into desmos as is

[hoary valley]

I used Mathway

[willow bear]

did mathway not say anything about k

[hoary valley]

Nope

[willow bear]

mathway is garbage then

bc if it assumes k=0 it should at least say so outright instead of just doing it silently...

[hoary valley]

Ok, I used Desmos apparently its range never hits 0, so it has a HA at 0 , So the range is (0,+infty), right?

[willow bear]

... no? its domain is (-k, k)

it can't have any horizontal asymptotes

by your logic, y = x^2 has a horizontal asymptote at -1

[heady jewel]

i thought log(a)/log(b) is not equal to log(a-b)?\

@dim jungle stop saying wrong things

[hoary valley]

I still can't find its range, Help.. I know its domain is
(-k,k) , but I'm unable to find its range.

@blazing monolith I'm not allowed to replace K with a negative number right?

@blazing monolith this with it being as is, the range will be [0,k]

but if it's on the denominator, I have no idea.

[uncut mulch]

|| sqrt(k^2) != k||

[green zenith]

hello!

i need help solving this equation 8x^(2)e^(-5x)-5x^(3)e^(-5x)=0

[uncut mulch]

$8x^2e^{-5x}-5x^3e^{-5x} = 0$

[obsidian monolith]

ramonov:

[uncut mulch]

?

[green zenith]

yes

i did try to solve it by moving 5x^3e^-5x to the other side and ln both sides but im not sure if thats the correct solution

[uncut mulch]

notice that e^(-5x) cannot be zero

[green zenith]

okay

yeah i can see that

[uncut mulch]

does that give you any ideas on what to do?

[green zenith]

not really

i dont see how that would help

[uncut mulch]

generally if it was another term, you would factor it out
but since you know e^(-5x) can't be 0, you can divide everything by it when solving for x.

and you'll be left with solving
8x^2 - 5x^3 = 0

[green zenith]

wait what happened to e^(-5x)?

oh we divided it

thats right

[uncut mulch]

if you want to go through the intermediate step
e^(-5x) (8x^2 - 5x^3) = 0
there are no real solutions to e^(-5x) = 0

[green zenith]

oh okay now i see

well that means then that every exponential term cant be 0?

[uncut mulch]

(excluding 0^x which isn't really considered an exponential anyway)

[green zenith]

that is interesting

good to know

thank you

i think i can take it from here

thanks

[rigid sun]

No problem

[viscid thistle]

hello

how do I solve
sec^2t(1 - sin^2t)
using fundamental identies?

[serene heath]

What did u fry

Try

[viscid thistle]

i tried to multiply by 1/cos

but i didnt really know what to do after

[serene heath]

What is 1-sin^2

[viscid thistle]

i dont know

[serene heath]

$\sin^2+\cos^2=1$

[obsidian monolith]

the one n only:

[serene heath]

Yes?

[viscid thistle]

isnt that a pythogrean identity?

[serene heath]

Sure

Now what's 1-sin^2

[viscid thistle]

im sorry i still dont get it

i dont really get the basics either

[willow bear]

if a+b=1 then what is 1-a

[viscid thistle]

1 - a = b?

[willow bear]

now look at lemon's question again

[viscid thistle]

1 - sin^2 = cos?

[willow bear]

no

[trail badge]

Look at the equation ;-;

[vague zephyr]

would someone be able to help me with these?

[gaunt spoke]

hey @vague zephyr you still there/need help

[vague zephyr]

i finished the first 2, just need help on the last one

[gaunt spoke]

ok so

you see that the amplitude is 6

so a =6

[vague zephyr]

what i have so far is 6 sin ((4pi/5)x)+6

[gaunt spoke]

the period is 2/5

2pi/period=b

[vague zephyr]

oh

[gaunt spoke]

2/2/5

[vague zephyr]

my period is incorrect then

[gaunt spoke]

do you see why the period is 2/5?

[vague zephyr]

yes because the first interval begins at 0 and ends at 2/5

[gaunt spoke]

the horizontal shift is 0 so c=0

yes

the avg value is 6

so

6sin(2pix)

+6

[vague zephyr]

ooohh

[gaunt spoke]

oops

wrong

hold up

5pix

because it is 2pi/2/5

which is 10pi/2

[vague zephyr]

2pi/2/5x,

[gaunt spoke]

which is 5pi

[vague zephyr]

oh ok

i get it now

thank you so much

[gaunt spoke]

just flip the 2/5 to 5/2 and multiply the 2pi by it

np

[vague zephyr]

yep, forgot to multiply by the recip

[gaunt spoke]

yeah

think about it like this to not forget that

2/5 is less than one

so it is going to get bigger

the numerator will be larger/if it is negative, smaller

and if you are dividing by .4

well that is the same as multiplying by 2.5

because 1/.4=2.5

[pulsar viper]

How would u solve ax^2 + 10x + c = 0 without using the quadratic formula

My answer has to be in terms of a and c

[tepid cedar]

what is the correct formula for figuring out compounded monthly interest?

is this one it?

so 1/12 would be a month for n right?

ok i get it

[pulsar viper]

How do u do 1 and 2??

[sleek wolf]

Is anyone online able to help with chapter 5.3 tangent function graphing

[viscid thistle]

What would be the reciprocal identity of (sin^2x + cos^2x)/sinxcosx

[uncut mulch]

what have you tried?

[lethal oracle]

Well I’m not sure where to start is the thing

I know I gotta get i wanna get it to sin2x

[uncut mulch]

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

[lethal oracle]

Okay so I got (2)(2sinxcosx)-1=0

[rigid sun]

stop

Go back

[uncut mulch]

^

[rigid sun]

Do you know what 2sin(x)cos(x) is?

[lethal oracle]

Sin2x

[rigid sun]

ok

Substitute

[lethal oracle]

So (2)(sin2x)-1=0

[uncut mulch]

leave the one on the rhs

[rigid sun]

put the one back on the other side

[lethal oracle]

Okay

[rigid sun]

Now solve

[lethal oracle]

Even though the 1 is on the rhs?

[rigid sun]

Yes

[lethal oracle]

So I’m solving 2sin2x?

[rigid sun]

Yes

[uncut mulch]

solve: $2\sin(2x) = 1$

[lethal oracle]

Oh alright

[obsidian monolith]

ramonov:

[rigid sun]

[lethal oracle]

Is the equation I use to solve 2x=4pi/6+pi/2k

[rigid sun]

Don’t do that

Sin2x=1/2

Inverse sin both sides

[lethal oracle]

My instructor has us do it that way

[rigid sun]

That’s aids

[lethal oracle]

He requires it me for this take home quiz

[rigid sun]

Aids

just use brain

No formulas

[lethal oracle]

Would that be the right equation?

[rigid sun]

Idk I don’t do that

I just use brain

Sin(2x) has a period of pi

Which fits twice into 2pi

Sin(x) passes 1/2 twice per period

So 2*2

Four solutions

You can find these four solutions by doubling your period

2pi*2= 4pi

Find all values of x under the 4pi where sin(x)=1/2

then half all those values

[lethal oracle]

So the equation isn’t pi/3+pi/2k=x

I need to show this equation in my work

[rigid sun]

Aids

That’s more force*distance

[uncut mulch]

ambiguous fractions

[lethal oracle]

More work, yes

Sorry pi/3+(pi/2)k=x

[uncut mulch]

if you wanted to do it like that
2x = pi/6 + 2kpi → x = pi/12 + kpi
2x = 5pi/6 + 2kpi → x = 5pi/12 + kpi

[lethal oracle]

Why is it 2pi

[uncut mulch]

period of sin is 2pi

general solution is split into two sets

[lethal oracle]

So how do I know when to use pi/2, pi, or 2pi?

That’s something I’m confused on

[uncut mulch]

add the period while its still in the form 2x =

[lethal oracle]

So I got sin2x=1/2

Sin is 1/2 at pi/6 and 5pi/6

So I get my period by looking at these two values, right?

[uncut mulch]

you get your period from the function

[lethal oracle]

Oh

[uncut mulch]

which is sin which has a period of 2pi

[lethal oracle]

How would I do that?

Does it always have a period of 2pi?

[uncut mulch]

the parent function has a period of 2pi

[lethal oracle]

Sinx has a period of 2pi

Right?

[uncut mulch]

yes

[lethal oracle]

Is the period for cosx, pi?

[uncut mulch]

period of cos(x) is also 2pi

[lethal oracle]

So when would I have pi/2 as my period?

[uncut mulch]

depends on the question / transitional shifts

[lethal oracle]

Oh

Do you know an example of when it would be

[uncut mulch]

cos( 2pi/(pi/2) * x) = cos(4x) would have a period of pi/2

(relative to x)

[lethal oracle]

Is it because it’s cos4x

?

[uncut mulch]

wdym?

[lethal oracle]

Is it pi/2 because it’s a function of cos4x?

[uncut mulch]

period is pi/2 because 2pi/(coeff of x) = 2pi/4 = pi/2

did you understand how to get

2x = pi/6 + 2kpi → x = pi/12 + kpi
2x = 5pi/6 + 2kpi → x = 5pi/12 + kpi

[lethal oracle]

You divided both sides by 2

[uncut mulch]

did you understand the first part?

[lethal oracle]

You got pi/6 or 5pi/6 by looking at where sin is equal to 1/2

[uncut mulch]

specifically the + 2kpi

[lethal oracle]

Cause the period is 2pi?

[uncut mulch]

yeh (relative to 2x)

what would your 4 solutions be?

[lethal oracle]

pi/12, pi/6, 5pi/6, 13pi/12

[uncut mulch]

some of those arent right

[lethal oracle]

The last one?

[uncut mulch]

pi/12 and 13pi/12 are fine

[lethal oracle]

I’m not sure what I did wrong

[uncut mulch]

x = 5pi/12 + kpi

[lethal oracle]

Oh shoot

So my solutions are 5pi/12, 5pi/6, 13pi/12, 17pi/12

[uncut mulch]

where are you getting 5pi/6?
and where did your pi/12 go?

reminder that

x = pi/12 + kpi
x = 5pi/12 + kpi

[lethal oracle]

Okay I sent them wrong it’s: pi/12, 13pi/12, 5pi/12, 17pi/12

[uncut mulch]

yeh thats better

[lethal oracle]

I have this other one, it’s cosx+cos3x=0

Would my solutions be pi/4, pi/2, 5pi/4, 3pi/2, 7pi/4, 3pi/4

[uncut mulch]

yeh.

[lethal oracle]

How do i determine how many triangles a set of measurements make

For example the measurements in 17

[viscid thistle]

Is homework not allowed to be answered by others here?

[ruby frigate]

people wont solve it for you, if thats what you're asking

[green zenith]

What happens when i have 2 log exponents in one number?

For example 10^((log1)+(log2))

Will i just simply add the value of the logs?

I think in this case, i would you use the product rule, right?

But what if it log1 and log2 is multiplied instead of added?

[serene heath]

Well log1 is just 0

So it simplifies nicely

[green zenith]

Oh yeah

But what if it was a different value

Its now 10^((log50)(log9))

[serene heath]

Cant do much

[proud sparrow]

@green zenith What base is the log? If base 10, you can change it to 50^log(9), but that's somewhat ugly

[green zenith]

Yes it is base 10

Can i also do it 9^log(50)?

And so it cant be simplified to a number?

[proud sparrow]

yeah that would be okay

9^log(50) can be 9^(1+log 5)=9*9^(log 5)

[green zenith]

I didnt know that last part

[willow bear]

50=10*5

[viscid thistle]

Hello, why is sin(-Ax) * cos (-Bx) equal to -sin(Ax)*cos(Bx)?

[green zenith]

Yeah but how did 1+log5 happen?

[willow bear]

log(10)=1

presumably

[green zenith]

Oh its log10+log5

Oh okay

[willow bear]

@viscid thistle sin(-t) = -sin(t), cos(-t) = cos(t)

[green zenith]

Thanks guys

[viscid thistle]

Thank you, I completely forgot about that

[green zenith]

Is it possible that a graph have 2 horizonntal asymptotes?

[uncut mulch]

yes

[green zenith]

I only saw the three rules of finding a horizontal asymptote. Are there other rules?

[proud sparrow]

what 3 rules?

[green zenith]

The 3 rules based on the degree of the polynomial in the rational function

[proud sparrow]

for rational functions, that's all you need to know

y=e^x though

[green zenith]

But based on the rules, i can only find 1 horizontal asymptote

So in e^x, there are multiple asymptotes?

Oh yeah e approaches a certain amount of number

So in a rational function, there is always only one horizontal asymptote?

[willow bear]

Oh yeah e approaches a certain amount of number
what the fuck is that supposed to mean

[green zenith]

I think im confused with something else

[proud sparrow]

So in a rational function, there is always only one horizontal asymptote?
At most one

[green zenith]

Ahhh

[viscid thistle]

In a rational function if there is a zero at a certain point but there is also a horizontal asymtope on the x-axis can the function go through the x-axis to reach the zero

[uncut mulch]

wdym by reach the zero?

[viscid thistle]

Like

If there's a zero there it means the function needs to pass that point that is on the x-axis right, but there's a horizontal asymtope on the x-axis, so it would cross through the x-axis?

[uncut mulch]

ah. the horizontal, oblique asymptotes describe the end behaviour

it is possible for the function to pass through it multiple times

[viscid thistle]

Ok

Thanks cuz I realized it would be impossible for there to be a zero there if it couldn't pass through the asymtope

[nocturne tangle]

hi i need help

I already have mu and Vmax
i just need to solve for theta and radius with this information and this formula

<@&286206848099549185>

[uneven tendon]

@nocturne tangle square both sides, divide the the numerator and the denominator of that fraction by $\cos\theta$ and solve for $\tan\theta$

[obsidian monolith]

Nuke:

[uneven tendon]

Also,

[merry sphinx]

Bot is broken

[uneven tendon]

Again? rip

[green zenith]

if sin = negative number, shouldn't csc also be = negative number?

[ruby frigate]

Yes

[green zenith]

and that works with cos and sec, and tan and cot?

[rigid beacon]

yup

[green zenith]

okay thanks. i just want to make sure. the book is making me doubt myself.

[clear glade]

why is pi and 5pi over 3 there, I don't get what I did wrong

[stuck lark]

$\cos(\pi +2\pi n)=-1, n \in \bZ$

[obsidian monolith]

RokettoJanpu:

$\cos(\pi +2\pi n)=-1, n \in \bZ$

[clear glade]

i dont get it

[stuck lark]

cos(pi)=-1, what about cos(3pi)?

[clear glade]

-1

[stuck lark]

cos(5pi)?

[clear glade]

-1

[stuck lark]

notice for each multiple of 2pi i add, cos remains -1

what i typed above says the same, where n is any integer

[clear glade]

oh interesting

I still don't get where the 5pi/3 comes from

[stuck lark]

solve the problem again, but remember that there are infinitely many values of theta that satisfy the equation

infinitely many until you apply the usual restriction that theta lies between 0 and 2pi (which you yourself never specified but i'm assuming it here)

[clear glade]

ok I will give it a shot

so yea it just keeps going

oh wait I mightve messed up

gonna try again

yea i dont get it

[stuck lark]

cos(WHAT) = -1?

[clear glade]

lol what?

[stuck lark]

cosine of WHAT equals -1?

[clear glade]

pi

[stuck lark]

not just pi

https://i.gyazo.com/79bfdb16d810ed211ba322afe1183b2a.png

[clear glade]

ok I get that

I don't get where we get 5pi/3 from

like I know pi is where cos is -1

and when we solved the equation we got pi/3

[stuck lark]

is there a way to neatly state all solutions of x to cos(x) = -1?

[clear glade]

pi/3 + 2pi(n) I'm guessing

where n is an integer

[stuck lark]

i let n = 0, so that's pi/3+0. cos(pi/3) is not -1

[clear glade]

so you got 1/2 for the cosine

then you looked where else cos is 1/2

and that was the answers along with what we solved basically

?

[stuck lark]

x = pi/3 + 2pi*n is not the solution to cos(x) = -1

[clear glade]

i know

but it's the other places where cos=1/2?

[stuck lark]

didn't ask you to solve cos(x) = 1/2

[clear glade]

oh

this is too hard

well not hard but confusing

[stuck lark]

x = pi is ONE solution to cos(x) = -1 but it's not the ONLY solution

[clear glade]

right

I understand that

[stuck lark]

so state the other solutions using n for an arbitrary integer

[clear glade]

thanks for the help but I give up lol

[clear glade]

ahh

I finally get it @stuck lark

Arigatou Roketto-senpai!

I was doing the equations wrong basically

sorry to bother u for u so long lol

[stuck lark]

@clear glade very naisu!

[clear glade]

[trim fable]

hey guys

I have a question

is someone able to help me

[rigid sun]

K

[trim fable]

yay

this

x^2+3x+3 cant be factored tho :C

so my question was can i move the top and bottom to the other side and simplify then factor?

@rigid sun

[rigid sun]

Sure

You can try that

[trim fable]

oof

im so confused with this

coz i got x^2-3x-3/-2x+3

by doing that

u know what nvm

ill ask my teacher tom

[hexed bolt]

How would I do the last one?

[limber bone]

tan(theta) = sin(theta)sec(theta)

[brisk frigate]

Can I ask for math help here as long as its related to Pre-calc? or does it have to be on one of the questions chats

I feel like if its precalc this would be the place for it right?

[grizzled orchid]

yes

[brisk frigate]

k thnx

[summer sierra]

−5e^7x − 4 = −52

how to solve for x?

[uncut mulch]

what have you tried?

have you learned logs?

[summer sierra]

nvm i solved it

but i have a new problem

The population of a small town is modeled by the equation
P = 1650e0.5t
where t is measured in years. In approximately how many years will the town's population reach 20,000? (Round your answer to two decimal places.)

[uncut mulch]

exponential growth function?

[dim jungle]

Should it be e^ stuff

[summer sierra]

This is what I tried

[uncut mulch]

$P(t) = 1650e^{\frac{t}{2}}$

[obsidian monolith]

ramonov:

[summer sierra]

Yes exponential growth I suppose

[uncut mulch]

how did you get from the 3rd line to the 4th line?

your right side of the equation just changed

[summer sierra]

thats what 20,000/1650 is

oh i mean

the power rule lets u do that

like bring the exponent to make it the coefficient

[uncut mulch]

$e^{0.5t} \neq 0.5 t e$

[obsidian monolith]

ramonov:

[summer sierra]

??

[uncut mulch]

that's what i'm seeing

and you can't do that

[summer sierra]

what should i do instead?

[uncut mulch]

use logs

specifically the natural log

[summer sierra]

So like this right

Then t equals that

[uncut mulch]

which parts are you dividing by 0.5? just the left or both the right and the left?

[summer sierra]

Both

To get t by itself

[uncut mulch]

then write division on both

[summer sierra]

thanks i managed to solve it

[opal swallow]

guys

how would you derive this

[rigid sun]

There’s nothing here

d/dx []= []

[pseudo sonnet]

why does my calculator say pi/4, y is 0

i got y is 1

[stuck lark]

$y=\tan(2x-\frac{\pi}{2})$ not $y=\tan(x)$

[obsidian monolith]

RokettoJanpu:

[pseudo sonnet]

gotcha

i got 0 thanks

so is there a quicker way to finding these Y output values

[stuck lark]

desmos

[pseudo sonnet]

or i have to plug in x, simplify, evaluate for sin/cos everytime

what

[stuck lark]

https://www.desmos.com/calculator

[pseudo sonnet]

what about it?

[stuck lark]

https://www.desmos.com/calculator/op1velke2f as an example

[pseudo sonnet]

what the hell

i got positive one for pj/8

ik it should be negative

i plugged in pi/8

then i evaluated pi/4

[stuck lark]

you forgot how the function's defined

[stuck lark]

you forgot how the function's defined

[pseudo sonnet]

?

[stuck lark]

read the problem again and tell me how y is defined

[pseudo sonnet]

wdym

y = tab(2x-pi/2)

tan*

[stuck lark]

plug in x=pi/8

[pseudo sonnet]

oh my god...

i forgot to subtract the pi/2 when i first did it

🤦

[young goblet]

When doing transformations, you have inverse, reflection, expansion/compression, and shifts right

what's the order you do them in?

[pseudo sonnet]

i get amp, period, phase, vertical shift

wait are u talking about trig lol

[young goblet]

no lol

we're not there yet

like i"m just talking about transforming a function in general

like say y = -3f(2(x-5))+4

there's vertical stretch, horizontal compression, reflection across X axis shift right by 5 units and up by 4 units right

but then in waht order do I do it?

cuz I know that some orders matter like stretch before shift

and then what if I throw in an inverse

[radiant night]

I keep messing up on factoring that f(x) equation and I cant move forward without simplifying it

[fading token]

,w 2x^2-12x+16

[obsidian monolith]

Display results online and refine query

Requested by Rijinaru#5601

[fading token]

There you go, under "Alternate forms"

[radiant night]

Thank you

[young goblet]

So does inverse come before everything else or after again?

[edgy sun]

part c

[serene heath]

x=x-1+1

[edgy sun]

explanation please?

🤔 nice joke i guess

[uncut mulch]

oh that would've been a hint to part a)

for part c)
integrate using the recommended substitution

double angles. draw triangle to find cos(theta),
back substitute and you should reach the answer

[opal swallow]

nayone here can help me ?

what are some good ways to practice conjugates

[brisk frigate]

Anyone know how to solve this? My school is bilingual so we just started math in english this month, i dont really understand what is asking me to do

At first i thought it was synthetic division but idk what they mean by "with multiplicity 2"

[serene heath]

Mutltiplicty refers to how many times the root occurs

So (x-1) 1 has mult of 1

(X-1)^2 is 2

[brisk frigate]

oh i see

but since there is a comma after the "x = -3" idk if the multiplicity refers to that one or another one that i have to find?

[serene heath]

Think they mean -3

[brisk frigate]

okok thnx

I havent really done any "hard" math like this in a while, I think I had to check for factors of "C" and then plugged those values into "X" to check which ones made the equation be "0"

But 81 prob has a lot of factors so, is there a way to just get the answers without that trial and error on my calculator?

I was also thinking about synthetic division but i dont really remember much about it and i think in order to do that i need another set of numbers to divide by or something like that?

[serene heath]

81?

[brisk frigate]

well i was thinking 81 bc thats the value of "C" in the first one

1. f(x) =x^4 + 6x^3 + 18x^2 +54x + 81

[serene heath]

Oh

I'd solve 4 first

Since it's the easiest one

[brisk frigate]

oh okok, i'll get the values for x on that one

i factored it to: (x^2-9) (x^2-9)

(I always get confused with algebra and functions n stuff like that so im writing what i did, in case the answer is wrong)

so i sqrt both sides

[serene heath]

Yea that's fine

[brisk frigate]

and got: sqrt(-3)

[serene heath]

Erm

[brisk frigate]

as the answer on both

[serene heath]

Not quite

Factor out x^2-9

[brisk frigate]

oh

(x-3)(x+3)

so my answers are: x=-3, x=-3, x=3, x=3

[serene heath]

Yes

[brisk frigate]

OH i get it now

its d bc of what you said before about multiplicity

oh wait

[serene heath]

No

[brisk frigate]

nvm

c

[serene heath]

Yes

[brisk frigate]

read it wrong

xD

[serene heath]

Lol

[brisk frigate]

k now ill just look up how to get the values of polinomial equations bc i forgot about it and i should be good to go

tysm for the help

[paper temple]

https://cdn.discordapp.com/attachments/634219944967405571/637073628055207937/JPEG_20191024_194246.jpg

I'm in college precalc, but here's my notecard for the test tomorrow

[thorny island]

the fuck

[viscid thistle]

lool

[remote maple]

Can anyone explain how I do this

[viscid thistle]

ok lets expand that first expression

[uncut mulch]

double angle identities

[viscid thistle]

yea

actually thats a bad idea

[uncut mulch]

(in reverse)

[viscid thistle]

yeah

actually why not plug some values in

cos(0) = (1+cos(0))/2

what ?

oh yeah

never mind

anyway

cos^2(pi/2) = (1+cos(k*pi/4))/2

bruh

[uncut mulch]

do you know your double angle identities for cos? @remote maple

[viscid thistle]

0 = (1+cos(k*pi/4))/2

cos(k*pi/4) = -1

k * pi/4 = pi

[remote maple]

@uncut mulch i think so

[viscid thistle]

k = 4

oops

ok listen to ramonav

im right but listen to him

[uncut mulch]

it has 3 forms. what are they?

[remote maple]

Cos^2A - Sin^2A, 2Cos^2A-1, 1-2Sin^2A

[uncut mulch]

parentheses would be a bit better but ok

since you have cos^2(2x) in you question, which one should you apply here?

[remote maple]

2nd one

[uncut mulch]

whoops, left out a 2
yeh. what equation would you start with?

[remote maple]

2Cos^2A-1

[uncut mulch]

equation, not expression

[remote maple]

So, "2Cos^2A-1 = 1+cos(kx)/2" ?

[uncut mulch]

no. it's simply just your double angle Identity

2cos^2(A) - 1 = cos(2A)

[remote maple]

oh

[viscid thistle]

Need halp. How to do dis?

[uncut mulch]

and then substitute A for 2x and then rearrange

and equate terms

[viscid thistle]

@viscid thistle you need to zoom dude

[remote maple]

Think I got it, thanks you guys

[uncut mulch]

fow much do you know about functions
(Inputs/outputs)
have you done compositions before?
do you understand fog notation

[viscid thistle]

I cant figure out how to simplify

(1 - sin^2x) / (csc^2x - 1)

All I know is that (1 - sin^2x) turns into (cos^2x) but I can't figure out what to do with the bottom