precalculus

[median moss]

Sorry, are you using this channel

[neon mural]

nope

go for it

[median moss]

Thank you

Question about identity and self inverse functions. Here I have the definitions of both functions;
**Identity function:**A function which does not change the output
Self inverse function: A function in which its inverse, is its original function

Would it be correct if I wanted to prove that any function is an identity/self if I did

$$f(x)=x$$

[obsidian monolith]

Mist

[median moss]

$$f(x)=f^{-1}(x)$$

[neon mural]

{-1}

[obsidian monolith]

Mist

[neon mural]

uh

never heard of self inverse

even function maybe?

not sure

[median moss]

But for even functions isn't it like f(x)=f(-x) or something

or is that odd

[neon mural]

but yeah f(x)=f^-1(x)

oh wait yeah you right

fx=f-x

-fx=f-x is odd

[median moss]

thats right

I need to review those actually

[neon mural]

havent done that in ages

my days have been filled with calculus :(

[median moss]

I don't know if I should be scared for cal or not

Self teaching maths is so hard

[neon mural]

eh

it's not too bad

optimisation was kinda difficult

but

[median moss]

I just have a terrible teacher so I have to self teach pre cal, cal 1 and half of cal 2 in like 4 months

[neon mural]

other than that, not too bad

[median moss]

its doomed for me

[neon mural]

yeah idk what that means i don't do american syllabus

[median moss]

what country r u form

[neon mural]

aus

but i reckon you could get that done?

i got diff done in a week

[median moss]

It's methods + specialist plus first maths at uni

[neon mural]

dk

specialist is vic maybe?

not too sure

[median moss]

well atar

Are you doing HSC?

[neon mural]

yeah

[median moss]

So its the maths you take for the highest maths in hsc plus the mandatory one if you take the highest one, plus first year uni cal kinda

hsc is like nsw maths right

[neon mural]

it is

but nsw maths is harder than vic maths

so idk if you can compare it like that

but sure maybe

[median moss]

I do an international program so I just compare it with the us system lol

[neon mural]

i see

well

shouldn't be extremely difficult until the uni part

[median moss]

i shall try my best sir

[neon mural]

havent had experience with uni maths

[median moss]

wait how does maths work in nsw

[neon mural]

alright so

you have general maths

so like the non-calc courses

so basic stuff

[median moss]

r we gonna get banned for talking about this

[neon mural]

so you have genny 1 and genny 2

[median moss]

in this channel

[neon mural]

mmmm

[median moss]

XD

[neon mural]

yeah take it to chill

[grave tartan]

Oh yeah just another form, the standard makes it easier to define as a function

[bronze canyon]

Ramming my head against a wall here. Anyone have a quick proof of this for a,b,c,d suitable reals

[bronze canyon]

Oh it’s not true that’s why 💀

[bronze canyon]

Denominators need to be positive it turns out

[viscid thistle]

Hello, can anyone show me the how to get from A to B by factoring and simplifying? Appreciate it

,rccw

[pliant locust]

Let $$t=(49-x^2)$$
Also we know,
$$t^{1/2} = t^1 * t^{-1/2}$$
$$t^1 = t^{3/2} *t^{-1/2}$$
Then $$\frac{(49-x^2)^\frac{1}{2} + 3x^2 (49-x^2)^{-\frac{1}{2}}}{49-x^2}=$$
$$\frac{t^{1/2} + 3x^2 t^{-1/2}}{t}= $$
$$\frac{t^{-1/2}(t + 3x^2)}{t^{-1/2}*t^{3/2}}=\frac{(t + 3x^2)}{t^{3/2}}=$$
$$\frac{49-x^2+3x^2}{(49-x^2)^{3/2}}$$

[obsidian monolith]

ds.b16rts

[viscid thistle]

Broooo thanks ! 🙏

[robust nest]

need help with this problem and several others like it

[sick steppe]

Start on RHS then use double angles to simplify num and denom, then simplify, then re-expand with double angle

[robust nest]

@sick steppe have no idea what any of those are. We are supposed to do this purely with trig identities

[sick steppe]

ok then LHS and write everything in terms of sine and cosine

[robust nest]

sorry, what is LHS

left hand side?

ok if thats what you mean then I turned tan on top into sin/cos, then 1 became sin^2+cos^2 and -tan^2 became -sin^2/cos^2

[grave tartan]

then change the one

into an equivalent fraction with cos^2 as the denominator

[robust nest]

so 1 would just be cos^2/cos^2

and you take away sin^2/cos^2 from it

now im lost

[grave tartan]

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

[obsidian monolith]

AyeWaddup

[grave tartan]

thank god no errors

[robust nest]

ah so multiply now? i can flip the bottom right

[grave tartan]

see how it almost looks like your RHS

well you can move the cos^2 to the top

[robust nest]

i got it

tysm

[grave tartan]

dwai

[robust nest]

i have another one that i may need help on if you dont mind

[grave tartan]

send

[grave tartan]

do you know any double angle formulas

[robust nest]

nope

[grave tartan]

like sin(2a)

or cos(2a)

[robust nest]

nope we arent supposed to know them for this

[grave tartan]

hmmm

[robust nest]

only trig identities and thats it

[grave tartan]

those are identities lmao

but I wont use them

yeah okay

write the numerator as one fraction

and the denominator as one fraction

lmk if you have trouble

[robust nest]

ok one second

[robust nest]

kk sorry had to eat

so @grave tartan I got sin^2-tan * 1/cos^2-cot

[grave tartan]

oh sorry I meant like

one fraction

ok first

split tan and cot into their respective sin and cos representations

[robust nest]

ok so -tan becomes -sin/cos, -cot becomes -cos/sin

[grave tartan]

$\frac{\sin^2{\theta}-\frac{\sin{\theta}}{\cos{\theta}}}{\cos^2{\theta}-\frac{\cos{\theta}}{\sin{\theta}}}$

[obsidian monolith]

AyeWaddup

[grave tartan]

is that what you got

[robust nest]

yes

[grave tartan]

now right the squared trig functions as fractions with common denominators as the other fractions on their respective line

sorry if that doesn't make sense

I meant just express sin^2 as an equivalent fraction with denominator cos

and same for the bottom portion

lmk when youre good

[robust nest]

kk so on top i got

(sin^2*cos)-sin/cos

on bottom (cos^2*sin)-cos/sin

@grave tartan

[grave tartan]

Ok so

The final sin on the bottom

You can bring to the very top of the fraction

And the denominator below that

You can bring to the new bottom

Because of how fractions work

I can explain more in depth later but i dont have the time rn lol

[robust nest]

alright all good

kk so i got

sin^3*cos-sin^2/cos^3*sin-cos^2

[grave tartan]

its basically just $\frac{a}{\frac{b}{c}} = \frac{ac}{b}$

[obsidian monolith]

AyeWaddup

[robust nest]

yes

[grave tartan]

wait imma read

[robust nest]

i moved sin to the top and cos to the bottom and distributed them

[grave tartan]

im gonna copy and paste rq

$\frac{\sin^2{\theta}-\frac{\sin{\theta}}{\cos{\theta}}}{\cos^2{\theta}-\frac{\cos{\theta}}{\sin{\theta}}}$

[obsidian monolith]

AyeWaddup

[grave tartan]

$\frac{\frac{\sin^2{\theta}\cos{\theta}-\sin{\theta}}{\cos{\theta}}}{\frac{\cos^2{\theta}\sin{\theta}-\cos{\theta}}{\sin{\theta}}}$

[obsidian monolith]

AyeWaddup

[grave tartan]

so then

$\frac{\sin^3{\theta}\cos{\theta}-\sin^2{\theta}}}{\cos^3{\theta}\sin{\theta}-\cos^2{\theta}}$

[obsidian monolith]

AyeWaddup

$\frac{\sin^3{\theta}\cos{\theta}-\sin^2{\theta}}}{\cos^3{\theta}\sin{\theta}-\cos^2{\theta}}$
```Compilation error:```! Argument of \frac  has an extra }.
<inserted text> 
                \par 
l.55 ...sin^3{\theta}\cos{\theta}-\sin^2{\theta}}}
                                                  {\cos^3{\theta}\sin{\theta...
I've run across a `}' that doesn't seem to match anything.
For example, `\def\a#1{...}' and `\a}' would produce
this error. If you simply proceed now, the `\par' that
I've just inserted will cause me to report a runaway
argument that might be the root of the problem. But if
your `}' was spurious, just type `2' and it will go away.```

[robust nest]

what happened here lol

[grave tartan]

no clue I can see it rn lmao

[robust nest]

sin^3*cos-sin^2/cos^3*sin-cos^2 was this correct?

[grave tartan]

yeah

factor out sin^2 on top and cos^2 on bottom

[robust nest]

ll

got it

tan^2

[grave tartan]

good!

[robust nest]

thanks

[grave tartan]

np man

just work with manipulating the expression using the identities you know

and you'll be good for future problems

in this one we took apart tan and cot

and then made equivalent fractions

[robust nest]

hm alright its kinda hard for me to understand where to go and what direction but I think by practicing it itll eventually be natural

[grave tartan]

my best advice is be suspicious

always be suspicious

that tan and cot messes everything up here right

if they weren't there we'd have a nice expression

I just mess around with it in my head to see if it goes anywhere

[mild swan]

If you want to prove this identity, I'd start with clearing out the embedded complex fraction

[grave tartan]

oh no we just did it

[robust nest]

lmaoo

[grave tartan]

I sent it again for reference

[mild swan]

Ah alright

📈

[robust nest]

real quick question

[grave tartan]

yeah wassup

[robust nest]

what does - do to a natural log again

[grave tartan]

like subtraction?

[robust nest]

like a negative natural log

[grave tartan]

oh so the same log rules apply

[robust nest]

wait so

[grave tartan]

but saying $log_{\frac{1}{e}}{x}$ is weird

[obsidian monolith]

AyeWaddup

[robust nest]

like for example -ln|sec| what would that be simplified to

would that be ln|cos|?

[grave tartan]

yeah pretty much

if you are using the | | as parenthesis yeah

[robust nest]

yeah

alright that makes sense

[grave tartan]

if absolute value it should be the same but ill text rq

test*

[robust nest]

nah its just parenthesis

[grave tartan]

yeah it works both ways lmao

[robust nest]

interesting

[grave tartan]

anyhoo gl man

hope you do well

[robust nest]

thanks hope you have a nice rest of your day

[grave tartan]

Np and ty will do

[clever pecan]

hi

[bold osprey]

did i do this correct?

[bold meadow]

yep

[bold osprey]

thanks!

[blissful kayak]

I literally had to do that exact same expression in my Precalc class....

It's a messy process though

But try rewriting everything in terms of just sine and cosine

That's how I was able to do it

But again, it's messy

[grave tartan]

Oh yeah no I got it eventually it just took me till 3:00 cause I was already tired

[leaden stratus]

The protocol of a medical treatment says that a particular medicine should be in infusion for a long time. The model that describes the concentration C of the medicine in blood (μmol/L) in function of time t (in hours) is this expression: see picture.

To get an efficient result, the concentration should be stationary at circa 16 μmol/L. Can this treatment be efficient? When will the concentration goes past 15 μmol/L? The book says that yes, it's efficient and that goes past 15 μmol/L after 28 hours circa.

[pliant locust]

Picture is to to get time for when c(t) is 15.

As to answering the question about efficiency I just determine the limit of e^-t/10 is 0 when t approaches infinity so the infusion is stable.

[leaden stratus]

@pliant locust thanks a lot. Why is it stable when the limit is 0?

[pliant locust]

$$e^x \to 0 \text{when} x\to -\infty$$

I cba to latex, maybe there is a way to show it is stabile without using limits but ull explain

[obsidian monolith]

ds.b16rts

[leaden stratus]

Actually, if you plug -infty, gives infty @pliant locust

But if you plug +infty is 0

But of course, we choose +infty

So this means 16(1-0)

And it's always stable to 16

[pliant locust]

Yes!

I know you got it but might aswell upload the pic since I already took it

,r

[leaden stratus]

Ok, thanks again!

[lost onyx]

Can anyone explain finding the vertical asymptote of some equation like this

[neon mural]

can you factorise the denominator for me? @lost onyx

basically, once you've factorised the denominator, you let that equal to zero

so for example, if i had

[obsidian monolith]

lofi hip hop radio

[neon mural]

i could factorise that to

[obsidian monolith]

lofi hip hop radio

lofi hip hop radio

[neon mural]

you arrive at the conclusion that x=-1 and -2

but since that's the denominator, and the denominator cannot be zero, your vertical asymptotes are -1 and -2

@lost onyx

[errant nest]

?

[worn scaffold]

Factoring the inequality gives us (2x+1)(x-4), meaning we use -1/2 and 4 as our intervals. To figure out when the inequality is true you can just plug in values of x around the intervals. For example plugging in 0 would not work, as that would yield 0 is greater than or equal to four, but a values like -1/2 or 4 would work. Hopefully this is all the proof you need.

[bold meadow]

@errant nest I think I saw that solved in another thread

[errant nest]

yeah it was

how about this?

I dont get the second part

[lost onyx]

@neon mural thanks i appreciate it

[neon mural]

:)

[pliant locust]

If f(x) approaches a limit when x approaches x=4 then you have a hole. If f(x) approaches infinity either from left, right or both then you have a vertical asymptote.

In other words, define you function so that its limit is not infinity when x approaches x=4.

[pine turtle]

Can u get the answer?

[jagged sun]

What does the ! mean?

Sorry if its a dumb question

[past meadow]

factorial

[jagged sun]

What does it do?

[past meadow]

$n!=n \times (n-1) \times (n-2)... \times 2 \times 1$

[obsidian monolith]

Sneaky

[past meadow]

so for instance $5!=54321=120$

[obsidian monolith]

Sneaky

[jagged sun]

Oh

So root ten factorial

is

root 10 * root 9 * root 8 * root 7 * root 6 * root 5 * root 4 * root 3 * root2 *root 1

[past meadow]

well that ends up being correct, but its not actually (sqrt(10))! its sqrt(10!)

(sqrt(10))! doesnt make any sense, factorial is only defined for whole numbers

[jagged sun]

Oh I see

[past meadow]

so its $\sqrt{109...21}$

[obsidian monolith]

Sneaky

[past meadow]

which is actually the same as you said since you can split the sqrt up

[jagged sun]

Oh ok

Thanks for the help

[past meadow]

no worries

[iron acorn]

the factorial can be represented by the gamma or pi functions so not just whole numbers but yeah

[past meadow]

yeah ur right

i should teach the gamma function to the precalculus student

i understand the gripe lol thats a joke

but the gamma function is more of an extension of the factorial, the general definition i'm gonna give to a kid in precalc is just the one for naturals lol

[iron acorn]

haha aight true good point i must say

oops

[candid prawn]

Would anyone be able explain this to me?

[worn scaffold]

Switch the places of x and y. For example, y=x+1 would be x=y+1 or y=x-1 in the inverse.

[bright grail]

i get that

but i have to also find f-(3)

so i need to factor this

no idea on how to factor

[worn scaffold]

I'm not too sure about how to go about factoring that but since a point (a,b) in inverse would be (b,a) finding what value of x makes y=3 in the orginal equation would give you the answer, and looking at it x=1 works. This probably isn't the right way to go about this, because its just guessing.

[round fossil]

how to solve for x in thsi case

ping in respond

[worn scaffold]

Is there any other information?

[round fossil]

no

@worn scaffold

I'm very confused

its trigomonetry

this is supposed to be a right triangle

[worn scaffold]

oh

[latent palm]

plx

@worn scaffold

did you get an answer

@round fossil needs it

[worn scaffold]

I am very confused

I think it would be a range

like

[latent palm]

@round fossil he is confused

[worn scaffold]

-infinity to 0

[round fossil]

yeah

imma just put something like that

[brazen tundra]

how do u solve this?

[brazen tundra]

these two

@helper

<@&286206848099549185>

[fading token]

Find a system of linear equations that describe the problem

[brazen tundra]

like this?

@fading token

[fading token]

Yeah

[brazen tundra]

is this correct

wait hold on i messed up

i got the answer

[brazen tundra]

how do i answer bottom question

<@&286206848099549185>

[lucid owl]

Can anyone help with the steps for condensing
1/2 log (9x) + 3/2 log (4x^3)

[ancient vale]

can anyone help me solve these two problems

[subtle ruin]

Is there a way to go from vertex form to standard

[pliant locust]

Assuming x is always positive

[robust nest]

i need some help with this question

dont know where to start

[uncut mulch]

factorisation

[robust nest]

hmmm

should I factor out sin^2x?

[uncut mulch]

that would be the most efficient route

[ancient vale]

@pliant locust awesome thanks!!

[pliant locust]

I made a correction, uploading

[robust nest]

need some help with this problem

[uncut mulch]

consider expressing everything in terms of sin and cosine, then simply by getting rid of the nested fractions

[robust nest]

i got secx @uncut mulch

does that sound right to to you?

[uncut mulch]

yeh

[mild swan]

@robust nest you can also graph it on desmos.com

[thorn moat]

I need help interpeting compression and stretch of graphs, for this graph the functions is f(x)=x^2. i know that graph pictured is -(x-2)-1 but im not sure how to interpet the strech and compression part

[neat iron]

i guess you meant $-(x-2)^2-1$?

[obsidian monolith]

Uchigawa

[vivid comet]

can somebody help me do this

not sure how to convert it

[sick steppe]

find sin then find csc

[vivid comet]

ye sin is just 25/7 how do i convert that to csc tho

[sick steppe]

how did you get sine?

[pliant locust]

Draw the unit circle, find sine using geometry and then you have it.

[lime sequoia]

Or just a triangle

Cosine is adj/hypotenuse

Find the opposite side by using pythagorean theorem

Then find cosecant = hypotenuse/opposite

[bright grail]

without trial and error

and is it correct that i crossed out the lns?

[velvet granite]

@bright grail so your top 2 sections look correct

but when you added in e

you didn't get the correct factors

adding in an e takes away the ln which is what you did correctly

so now all you should have is x^2+1/x^3=2

[bright grail]

i didnt use an e

i just moved x^3 to the other side

and then let it equal 0

to find the roots

[velvet granite]

then how did u cancel out the lns?

[bright grail]

i thought if both sides have ln u can cross them out lmao

whats the rule for ln?

[velvet granite]

thats true, but only with an e

you have the correct idea tho

anyway i think your final answer is correct so good job

👍

[bright grail]

but how do u factor

i know its x =1 because that'll let it equal 0

but i want to know how u do it algebraically

[velvet granite]

youd have to multiply a and c then find the factors that give your the value of b

[bright grail]

can you show me

[sick steppe]

x=1 -> the function = 0 means x-1 is a factor

so long/synthetic divide to get a quadratic factor * (x-1)

[bright grail]

i realize that

but how do u factor properly

w/o realizing that x-1 will make it equal to 0

[sick steppe]

RZT requires a slight bit of guess and check to find p(a)=0

once you find a factor then you just long/synthetic divide

[bright grail]

i see, thanks homie

[ancient vale]

i got the first blank right but what do i type in the calculator for the (pi/2)+0.1

[sick steppe]

,rotate

,rotate 180

[sick steppe]

just plug in x = pi/2 + .1

[lunar axle]

someone plz help!!

[astral mantle]

l'hopitals

Since you have 0/0

[velvet granite]

@lunar axle you need to get rid of the fraction

[inner sand]

If a function is continuous, it does not have limits right? so the lim of x^2 as x approaches 2 does not exist

[serene heath]

no that's not what continous means

[grave tartan]

No I think they know what continuous means, they must have the wrong understanding of limits

[mild swan]

@lunar axle did you get it?

[pliant locust]

Write the sum in numerator as same fraction.
Turn the numerator into 2(x+2) and then cancel out x+2.

[rare zephyr]

ay for b does it mean

this?

t is 4

[simple urchin]

I need help understanding the binomial theorem. How do you compute the () columns? Like is it a 2X1 matrix?

I understand that it is a power expansion, but the vertical columns, I just don't see the use for them.

[novel creek]

@simple urchin

hope that helps

[simple urchin]

LOLOLOL, yeah i know that from probability. so you are saying that these two are the same?

This is the same formula from probability and also the same formula in the binomial theorem. WOW

I never thought is was the same thing.

[rich saddle]

Yeah, it's typically referred to as the binomial coefficient.

That () notation is read "n choose k" just like the nCk you've probably seen in counting/probability.

[clever pecan]

,rotate 360

[hoary adder]

,rotate 1080

[fiery ridge]

ooh how to do

[obsidian monolith]

( ͡° ͜ʖ ͡°)

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

[viscid thistle]

,rotate 129600

[onyx gazelle]

can anyone help me with system of equations no matter how much i try to figure this out i always end up on square one

[echo wagon]

Well, what is it and what have you tried?

[onyx gazelle]

I have an idea of where to start

You multiply one of the equations so you can cancel out a variable

but then thats where it gets confusing for me

[echo wagon]

Well, do that and tell me what you get

[onyx gazelle]

Well I multiply the second equation by negative 2

then get -5x-2z=-13

[echo wagon]

Check the second term

[onyx gazelle]

oh wait

-5x-4z=-13

[echo wagon]

Yes

And now?

[onyx gazelle]

You need a second equation right?

[echo wagon]

No

Because you have 2 equations and 3 unknowns, you don't have one solution but infinitely many

So you will have a free variable

[onyx gazelle]

Right

So you solve for z?

[echo wagon]

Sure, go for it

[onyx gazelle]

z= -5/2x +13/2

[echo wagon]

No

[onyx gazelle]

yea

i see what i did

change 2s to 4s

[echo wagon]

Yes

Then sub back and solve for y

[onyx gazelle]

3x+y-5/2(x)+13/4=8

[echo wagon]

5/4

[onyx gazelle]

Right

7/4(x)-19/4=-y

and than inverse all positives/negts

and then you get y

y=-7/4(x)+19/4

[echo wagon]

Looks right

[onyx gazelle]

Than do the same for x?

[echo wagon]

What do you want to do for x?

[onyx gazelle]

solve for x

[echo wagon]

.

Isn't x your free variable?

[onyx gazelle]

In terms of z?

Im not sure

[echo wagon]

I don't know what you mean by in terms of z

[onyx gazelle]

Im just restating something to myself

[echo wagon]

Have you heard the term free variable before?

[onyx gazelle]

I have an idea

but not a solid foundation

[echo wagon]

There are infinitely many solutions. So you say one of your variables is free and can take on any value, say x. So x can be any number. And for each value of x, we get one value of y and z. So x in free, and y and z depend on the value of x.

[onyx gazelle]

Makes sense

[echo wagon]

x is free, y = -7/4 x + 19/4, z= -5/4 x + 13/2

That's your solution

[onyx gazelle]

So whats the in terms of z in the question than?

[echo wagon]

If you choose a specific value of x, then you can find the corresponding y and z values

Oh, didn't read that

You were supposed to make z the free variable, and find x and y in terms of z

[onyx gazelle]

so i cancel out the z first

instea

d

and than repeat the process

[echo wagon]

No. If you cancel out the z, then you can't solve for the other variables in terms of z.

You need to cancel out one of the others, say x, then you have y and z left. Then you solve for y in terms of z.

[onyx gazelle]

So instead of solving for z you solve forx?

[echo wagon]

Yes

x and y need to be in terms of z

[onyx gazelle]

-5x-2z=-13

ok

-2/5(Z)+13/5

that should be the x equation

Than you plug it in for y?

[echo wagon]

Why is it -2z again?

[onyx gazelle]

thats the original equation we got when we did elimination

[echo wagon]

No

[onyx gazelle]

is it 4z

That 2 i swear man

[echo wagon]

Yes

[onyx gazelle]

-4/5(Z)+13/5

so that should be the equation you plug in than correct

[echo wagon]

Yes

[onyx gazelle]

Plug it in for y?

no no

x

-7/5(z)+39/5+y=8

-1/5

-7/5(z)-1/5=-y

y=7/5+1/5

7/5(z)

So then, z is free

y=7/5(z)+1/5

and x=-4/5(Z)+13/5

[echo wagon]

👍

[onyx gazelle]

So thats good?

than with this would you skip the elimination

and just use the -4x+9y=7 equation?

[echo wagon]

Yes

[onyx gazelle]

Alright sounds easy enough

so then you would want to solve for y with the equation because its asking for x to be the free variable

[echo wagon]

Yes

[onyx gazelle]

y=4/9y +7/9

[echo wagon]

x*

[onyx gazelle]

and than you plug that in for z

yep

[echo wagon]

Yes

[onyx gazelle]

x-3y+4/9z +7/9

3y=x+4/9(x)+7/9

13/9x+7/9

/3

13/27(x)+7/27

z=13/27(x)+7/27
and y=4/9y +7/9

so than x is free

Easier now that I know what to do but thats all of math is it not

Well thank you @echo wagon

[echo wagon]

No problem!

[viscid thistle]

https://cdn.discordapp.com/attachments/407563152100818954/802471500737150976/20210123_163451.jpg

hey guys, what step am I making a mistake in? Trying to use the property of equality for complex numbers however it's not working out for me

[lilac pier]

@viscid thistle Last line is wrong

[viscid thistle]

Yeah, that was dumb

Thanks

[lunar axle]

could someone verify if these are correct for me??

[bright grail]

and if so, what rule is that

[viscid thistle]

defn of e

$e^{\ln( {\color{blue}{x}} )}={\color{blue}{x}}$

[obsidian monolith]

Al𝟛dium

[bright grail]

been stuck here for a while

the inverses become 1/cos or whatever

but how do u evaluate

[sick steppe]

cos inverse is not 1/cos

$\cos{x} = c \iff x = \cos^{-1}{c}$

[obsidian monolith]

moshill1

[bright grail]

so what is it equivalent too

[sick steppe]

cos(x) = -1, what's x?

[pliant locust]

If you have a hard time remembering trig values I can recommend memorizing the following picture. All you have to picture is an equilateral triangle which has all angles of pi/3 and all sides are of equal length. Just knowing drawing an equilateral triangle as a start and pythagoras you can derive all trig values, for 45 degree is the exception which is just a right triangle with side length 1.
If you know the values but still can’t solve cos(x)=-1 or cos(x)=-1/2 then I suggest you do more exercises regarding the unit circle in trigonometry.

[lilac storm]

can anyone please help me with this question.

[novel cargo]

what is it you have problem with?

[lilac storm]

so i have typed in the eq N(x) = 1.42x(t) into geogebra, but i don't understand the next instruction under the N(x) = 1.42x(t) eq

and have i typed in the first eq N(x) = 1.42x(t), is this correct?

are you there?

[novel cargo]

yeah

was testing

[lilac storm]

oh, ok

[novel cargo]

I tried to put in X(t) first and then N(x) but geogebra won't accept

this is maybe bc they changed their interface and your instructions are obsolete

what you can do tho is to put in X(t) first

that's the longer equations

then put in N(x), which references X(t)

the problem right now is that when you put in N(x) first, it references X(t) and since you haven't entered it yet, the program doesn't accept

you got that?

[lilac storm]

i think so.

[novel cargo]

N(x) references X(t), therefore it comes after it

[lilac storm]

sorry, i'm not getting what is happening, so i can see the eqs but how did you get each eq after the first one in the picture?

[novel cargo]

you see the arrow on the bottom part of the second and third equation?

[lilac storm]

yes

[novel cargo]

what comes after that arrow is what geogebra calculates

what comes before the arrow is what I typed in

geogebra puts those together

[lilac storm]

oh, ok

[novel cargo]

when I say N(x) = 1.42 * X(x)

it will plug in X(x) from the previous cell and write it out

[lilac storm]

so for (a), it says find an expression for the number of jobs created (in millions) per year in the next t months, so is the answer the computed geogebra eq (the arrow one)?

[novel cargo]

yes

[lilac storm]

so basically your just multiplying the 1.42 of the N(x) eq with the expression in the X(t) eq, right?

[novel cargo]

yes, basically each housing construction creates 1.42 jobs

[lilac storm]

ok

so for part (b), when it asks finding how many jobs will be created in 6 and 12 months so do i just sub in 6 and 12 into the eq that we just got?

[novel cargo]

yes, I think so

[lilac storm]

yeah so i get it till here and then after that i am confused again that what i have to do?

[novel cargo]

are in graphing mode or CAS mode?

[lilac storm]

cas mode

[novel cargo]

Ahh, OK, I just realized that your homework functions in CAS mode

the steps that they gave

it actually works if you do the step by step they gave

define N2(t) = N(x)

Then write Substitute... like instructed

[lilac storm]

so i subbed in 6 and 12 to find the answers so when i sub in 6, the answer i got is 2.24 and for 12 i got 2.48, is this correct?

[novel cargo]

yeah, for 6 is correct

12 also correct, IG

[lilac storm]

and that's it for the question, right?

[novel cargo]

I think so

[lilac storm]

ok, thank you so much.

[novel cargo]

you welcome

[dark acorn]

I need some help with this (pls let me know if i sent this to the wrong channel)

[novel cargo]

the denominator must not become zero

[dark acorn]

i got it, its A, but thx

[dark acorn]

i need help

[serene heath]

what have u tried

[dark acorn]

idk where to start

[green kernel]

square both sides

[dark acorn]

i did that

to get -5=x

but its wrong

for some reason

[green kernel]

show working

[green kernel]

that’s not the working but aight

[dark acorn]

no

[green kernel]

you need to keep in mind that when you square both sides, it can generate incorrect solutions

[dark acorn]

i just wanted to show my answer sheet

[green kernel]

so if you solve the thing and it gives you a solution which is wrong then you need to discard it

[dark acorn]

x+3 = 2x+8
-5=x
this is what i did after sqauring it

[green kernel]

yeah

but if you plug in -5 to the original equation

[dark acorn]

where?

[green kernel]

you get sqrt(-5 + 3) on the left hand side which is not defined (square rooot of negative number)

[dark acorn]

oh

so i replug it in

and if it doesnt work

theres no solution??

[green kernel]

yes, because squaring both sides can introduce solutions which don’t solve the original equation

[dark acorn]

so to solve that issue

[green kernel]

so you need to check that the values you get are correct for the original eqn

yeah

[dark acorn]

you just plug the answer into x and redo it?

[viscid thistle]

@dark acorn there's nothing more to do in the reals

You just noticed how the solution was extraneous, there's no x in R that satisfies the eqn.

[kindred grove]

Yo is this bitch correct?

1

[uncut mulch]

first two are wrong,
you didn't screenshot the full answer for the last one

combination of poor notation and errors in signs for the first one,
you chucked a freshman's dream in the second one

[viscid thistle]

isn't that calculus?

[onyx gazelle]

Can anyone tell me what endpoints are

I feel stupid for forgetting

[strong ermine]

@onyx gazelle In a graph?

[onyx gazelle]

I figured it out

[strong ermine]

Awesome.

[onyx gazelle]

I had a massive brain fart

[strong ermine]

It happens to the best of us.

Just recently I couldn't remember that the cube root of 1 is 1.

[viscid thistle]

i'm not sure if this is precalc but, i have no clue how to this, can someone help please

[onyx gazelle]

Im pretty sure thats physics

[strong ermine]

This is a physics question lmao.

Go to #old-network if you need Physics help.

[viscid thistle]

i knoww, but i got given this question for integration homework so idek what's going on

[onyx gazelle]

#old-network message @viscid thistle

wait shit

https://discord.sg/physics

there

join that

[viscid thistle]

thanks sm

[onyx gazelle]

np

[kindred grove]

hey guys i snored cocaine and now i am programming the next Facebook

who wants to hop on the project ?

[strong ermine]

[strong ermine]

https://www.youtube.com/watch?v=9BozYg5nM-8

[onyx gazelle]

anyone able to help me with this

i have an idea of what to do

[strong ermine]

When I look at this I think system of equations.

[onyx gazelle]

That would make sense

thats what we are going over

I just cant figure out the equation

[strong ermine]

Hm.

I suck at figuring out equations too.

[onyx gazelle]

I was thinking of ratios but I dont think that would work

than I thought a Linear equation

and graph the equations to figure out where the two would intersect

[strong ermine]

Yes.

Take a look at Linear Programming.

https://www.youtube.com/watch?v=Bzzqx1F23a8

[onyx gazelle]

Got a shorter video?

or only this

[strong ermine]

There are several examples they show.

Just take a look at one.

Skip to 7:50

I think this one might help you.

[onyx gazelle]

a;right

[onyx gazelle]

@strong ermine I have found three equations so far and not sure if i need more

I got x+y less than or equal to 40

5x+15y= points

and 5y+x less than or equal to 60

60 for an hour

and 40 for the amount of questions

[strong ermine]

😅

I am bad at systems.

[onyx gazelle]

Anyone else able to help i got like 3 questions i need to do that i cant figure out

that involve systems

[pliant locust]

[strong ermine]

[pliant locust]

Ops, this was apparently a math channel sorry abt that

[strong ermine]

[onyx gazelle]

anyone able to help me with this

It just doesnt make sense to me to begin with

are you looking for the gradual climb of amount of time spent

[somber beacon]

this makes no sense to me

[sick steppe]

complementary angles add up to 90 degrees = pi/2 radians

[somber beacon]

okay, so how would I do this

90-65 for the first one, right?

[sick steppe]

yes

since you need angle x st x+65 = 90

[somber beacon]

what

[sick steppe]

what

[somber beacon]

is the first one 25?

[sick steppe]

yes.. 25+65=90

[somber beacon]

and how would I go about doing the second one

let me try it first

is this anywhere near correct @sick steppe

and then for supplementary that means the same thing except it's 180 degrees or pi radians

right?

[sick steppe]

yes

[somber beacon]

did I do it right

[sick steppe]

yes

[somber beacon]

oh okay it makes sense now

thank you

wait I have two more questions for you @sick steppe

how do I do these

[somber beacon]

why is it adding 2pi radians in the top one but subtracting in the bottom one?

[sick steppe]

cause they're both co-terminal

[somber beacon]

oh, I can do either

[velvet granite]

yeah

[sick steppe]

it's just showing that +/- a rotation is co-terminal

[somber beacon]

I see

so my questions are asking for both

[sick steppe]

no it asks for a co-terminal

[somber beacon]

it says find a positive and a negative coterminal angle

for each given angle

[sick steppe]

then yeah do both

[somber beacon]

@sick steppe @velvet granite how do I do the angle drawing? like how would I do 5 and 7

[velvet granite]

do you know how to convert radians to degrees?

[somber beacon]

yes

[velvet granite]

then convert it for the necessary ones

[somber beacon]

thats it?

and where would that be

[velvet granite]

sorry sorry sorry

ok

so you know the unit circle correct?

[somber beacon]

no that comes in the next unit

[velvet granite]

shucks

ok back to where we were at before

5pi/3 *180/pi

= 300 degrees

[somber beacon]

yes

I just googled a pic of the unit circle i'll just use that for reference

but where would -185 degrees be @velvet granite

[velvet granite]

you start from your x side and go clockwise

because positive angles are counterclockwise

[somber beacon]

so right here?

[velvet granite]

yes

[somber beacon]

oh okay thats simple

[velvet granite]

since Pi is equal to 180 on the unit circle

[somber beacon]

yes

[velvet granite]

yep

[somber beacon]

wait

one more question

how do I do these?

I have no idea what degrees-minutes-seconds are

[velvet granite]

https://www.youtube.com/watch?v=W9kousU6AI0

https://www.youtube.com/watch?v=pt-anJytCJ0

@somber beacon

[somber beacon]

oh so its just the number + the next number /60 + the next number /3600

[ripe adder]

I dont see how I am wrong

[grave tartan]

Is there a vertical shift on that graph?

[ancient rampart]

is a vector with 0 magnitude still a vector

[serene heath]

the zero vector is a vector yes

[ancient rampart]

but why

isn't it just a point

[sick steppe]

no, it's a vector which each entry being 0

$\vec{0} = [0,0,...,0]$

[obsidian monolith]

moshill1

[ancient rampart]

o

[wispy aurora]

Sub x = y^2

[lilac storm]

i have started this question and i know how to solve it but i've just gotten stuck in one of the first steps so if anyone can help me

so i started by rationalizing the numerator and denominator

[wispy aurora]

ghost ping not appreciated 😦

Oh yeah ig that works too?

BRUH

Who's ghost pinging me

[viscid thistle]

idk

[wispy aurora]

hey hobo

wassup

[viscid thistle]

yoo

[wispy aurora]

Did you catch who the ghost pinger was?

I wanna kill him

and unfriend him if I'm friends with him

And never talk to him agian

XD

[viscid thistle]

I think the ghost pinger is a genius

god

[wispy aurora]

So stealthy eh?

[wispy aurora]

Where did that get you

[lilac storm]

so in the denomiator i am left with x-25

[wispy aurora]

right

[lilac storm]

but the little confusion that i have is that how do i multiply out the numerator

[wispy aurora]

Well what did you multiply the denominator by?

[lilac storm]

(sqrt x - 5) times (sqrt x + 5)

[wispy aurora]

Right

[lilac storm]

which leaves me with x-25

[wispy aurora]

so you get (x-25)(sqrt(x)+5)/(x-25) correct?

[lilac storm]

exactly

[wispy aurora]

Can you see the next step?

[lilac storm]

what i am thinking is now i have to first multiply out the numerator as well

[wispy aurora]

but you don't

[lilac storm]

oh, ok

[wispy aurora]

there's a cancellation you can do

You don't need to multiply things out usually

leaving them as is is usually pretty good

[lilac storm]

i can't believe it, i didn't even see that

[wispy aurora]

F

It'll come with time/practice

[lilac storm]

so the answer i got is 10

[wispy aurora]

yup

[lilac storm]

is this correct

[wispy aurora]

I mean

technically

[lilac storm]

ok thank you so much

[wispy aurora]

np

rip idk what happens with limits and perfect squares

time for my own question

[ripe adder]

I am trying to find the general form of a sine wave

this is the wave

how do I use the 5pi/8 to find the phase shift

[sick steppe]

where does the 1st valley to the right of x=0 occur on the regular cos(x) graph?

actually scratch that, do you have the horizontal dilation?

[ripe adder]

@sick steppe I have solved this problem

but I have a similar one

if you could help

[sick steppe]

maybe

[ripe adder]

you said horizontal dialation

which in this one, I need to shift it pi/3 to the right

and have it end at 7pi/3

[sick steppe]

dilation is compression / stretch
phase shift is translation

[ripe adder]

So I need to translate this pi/3 to the right, and then after that make it 7pi/3 wide?

or vice versa

[sick steppe]

but you can note that the distance along the x-axis is 7pi/3 - pi/3 = 2pi so there isnt a dilation

[ripe adder]

ah

[sick steppe]

so it's just been moved pi/3 to the right

(and other things obviously)

[ripe adder]

my problem with this one right now is

in desmos

I inputted 2pix/7pi/3

[sick steppe]

dilation isnt translation

the 7pi/3 isnt anything to do with the period

pi/3 and 7pi/3 are "like" points due to the periodicity, so the distance b/w them will be the period

pi(7/3 - 1/3) = 6pi/3 = 2pi

[ripe adder]

ah I see

ok

[sick steppe]

so the period is 2pi which is typical

[ripe adder]

and its been shifted upwards 1

[sick steppe]

yep since the axis of symmetry is at y=1

[ripe adder]

so would I just say 2sin(x-pi/3)+1 ?

[sick steppe]

you want to move pi/3 to the right

[ripe adder]

I see

[sick steppe]

and horizontal changes have the sign reversed from what you get for vertical

yeah that looks right

,w graph y=2sin(x - pi/3) + 1 on the interval [pi/3,7pi/3]

[ripe adder]

on desmos the points are off though

[obsidian monolith]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by moshill1#9771

[ripe adder]

hold on

how did you set the interval

[sick steppe]

if you write the domain you want after the eqn in desmos

f(x) {pi/3 < x < 7pi/3}

[ripe adder]

Ok

so is my current equation correct?

[sick steppe]

yes it's right

[ripe adder]

oh I see

I hate doing these where the midline isnt 0

makes it so hard to see if my points are correct

thanks @sick steppe

[wicked tendon]

so if there are question channels already. Is this for funneling them?

[strong ermine]

This channel is used to discuss precalculus/get help

But using questions is better

[sick steppe]

More for targeted questions; pre-calc questions go here or the general question channels

[wicked tendon]

Thanks for clarifying because i was seeing if any question channel was open so someone can explain rational number problems

[sick steppe]

just post the question

and no clue why you pinged me

$\omega$ is how fast it takes to rotate

[obsidian monolith]

moshill1

[sick steppe]

so it will cover a full rotation (2pi) in 60 seconds

ok so you agree $\omega = \frac{\pi}{30} \frac{rad}{s}$?

[obsidian monolith]

moshill1

[sick steppe]

so angular speed is the change of the angle, linear speed is the change of the distance. so it'll go a distance of 1 revolution (the circumference, think of it like a track) in 60 seconds

$v = \frac{2 \pi r}{60} = \frac{\pi}{30} r = \omega r$

[obsidian monolith]

moshill1

[sick steppe]

so what's v in inches/s?

wr is omega (angular speed) times r (radius)

8 inches, yes

what's omega?

pi/30 yes

what's r?

so what's $\omega r$ ?

yes

$v = \frac{4 \pi}{15} \frac{inches}{s}$

[obsidian monolith]

moshill1

[sick steppe]

v is linear speed

since its units are distance over time

also, if something doesnt make sense, say so.. dont say "that makes sense" to just ask about what it means later

yes that's what i just said

$v = \frac{2 \pi r}{t}$

[obsidian monolith]

moshill1

[sick steppe]

distance / time

yes, and units

what?

v = 10pi meters?

what are units of speed?

,calc 16/40

[obsidian monolith]

Result:

0.4

[sick steppe]

well 40 seconds is 2/3 of a minute, so it'll have moved 2/3 of the circumference

so $d = \frac{2}{3} \cdot 2 \pi r$

[obsidian monolith]

moshill1

[sick steppe]

idfk you keep changing the question

you also havent finished 1b

no, you found v in inches/s

it asks for it in miles/h

this is why units are important

.05 mph yes

your engineer friend should be able to help if they're an engineer

ok well you know the speed of the hand (v), so use regular distance formula

$v = \frac{d}{t}$

[obsidian monolith]

moshill1

[sick steppe]

yes, or times v by t, making sure the units match

cause that's the definition of linear speed.. distance / time (3rd time saying it btw)

apply what we went through

$\omega$ is the time to do 1 revolution

[obsidian monolith]

moshill1

[sick steppe]

30 rev / minute = 1 rev / (1/30) min

1min = 60 sec -> (1/30)min = 60/30 sec = 2 sec

I'm not going to hold someone's hand through multiple questions... use the previous question as scaffold

[viscid thistle]

is the answer to this 0.99 (2 dp)

because theta = acos(-1/4) so

cos(2theta) = 1 - 2sin^2(theta)

=> cos(2theta) = 1 - 2sin^2[acos(-1/4)]

[bright grail]

i got this for the first one, not sure if its correct

the second one im lost

[sick steppe]

use a different version of the double angle

[jolly gust]

How would I simplify (sin x+ cos x)^2/(1+2 sin x*cos x)

i need help concept wise

[cunning obsidian]

this is the first step, can you see what to do next

the green is a hint

[jolly gust]

make that equal to 1

[cunning obsidian]

yes, and then its the same on the top and bottom, so it's equal to...

[jolly gust]

AHhhhhhh

1

[cunning obsidian]

AHHHHHHHH

YES

[jolly gust]

thanks my guy

[cunning obsidian]

no problem

[strange brook]

can someone please help me i've been to 4 different hw servers

[lilac pier]

@strange brook what's the question asking for? value of alpha?

[sick steppe]

use compound angle on the LHS

[strange brook]

its asking to find the sum or difference identity for sine to prove each identity

[sick steppe]

yeah.. so what's the issue?

[strange brook]

umm you have to get the left side to equal the right side????

[sick steppe]

yeah

have you used the compound angle formula..?