precalculus

[willow bear]

fine

can you repost it

i cbf to scroll all the way up

[viscid thistle]

I used the same method and it worked

I think our teacher isn’t gonna give us a fuck ton of extraneous solutions

[coarse storm]

You are missing the other solutions

Is there a restricted domain?

[willow bear]

I think our teacher isn’t gonna give us a fuck ton of extraneous solutions
bad practice, bad practice, bad practice

[viscid thistle]

I feel like that’s harder

Yea I know this is gonna nite my ass in calc

bite*

@coarse storm yea domain is restricted

[willow bear]

so you're willingly subjecting yourself to months if not years of struggle further on

okay

imma dip then

[viscid thistle]

LOL

But ur method confuses me in the end

@willow bear

[coarse storm]

@willow bear Does have a point

[viscid thistle]

But I’m confused at the end

[coarse storm]

What are you confused about?

[viscid thistle]

Tf do I do after this

Like I can’t solve it normally

[willow bear]

yes you can

look

[coarse storm]

What question is this?

[willow bear]

you've FACTORED the damn thing

[viscid thistle]

Or can I just find all my solutions then divide by 2

[willow bear]

do you remember algebra 1

[viscid thistle]

Yea room 7

6th period

[willow bear]

do you remember why you sought to factor quadratics in the first place

[viscid thistle]

Yea

To find solutions

[willow bear]

the EXACT SAME PRINCIPLE applies here

solve the two smaller equations cos(t/2) = 0 and 2sin(t/2) - 1 = 0 individually

then take the union of their sol sets

it really fucking isn't as hard as you're making it out to be

[viscid thistle]

But like cost/2 is a half angle

[willow bear]

so WHAT

if you can solve cos(t) = 0

then you can also solve cos(t/2) = 0

you just gotta stop OVERTHINKING the fuck out of it

so WHAT if it's a half angle. i could give you a problem like cos(2495.874586973495863485678345768437658t) = 0 and it would be not a SINGLE BIT harder to solve than cos(t) = 0

[willow bear]

don't just write "pi/2 3pi/2"

of COURSE if all you do is blurt out numbers you'll never get anything

like that's

absolutely no secret

cos(t) = 0 has solutions t = pi/2 and t = 3pi/2... on [0, 2pi) at least

cos(t/2) has solutions t/2 = pi/2 and t/2 = 3pi/2...

seriously bruh

[viscid thistle]

So it’s the same for both?

But it’s pi and 3pi?

For t/2 one

[coarse storm]

@viscid thistle

Is this Q36?

[viscid thistle]

35

[coarse storm]

What are you confused with?

[willow bear]

lachlan can you take over

[coarse storm]

Sure

[willow bear]

ok it's all yours then i'm out

[coarse storm]

Enjoy

[viscid thistle]

Uhhh

Lachlan can I call u

To make it easier 😢

[coarse storm]

Erhh, not really, give me a few minutes

Let me write something out

[viscid thistle]

It’s almost 1am

And I still gotta remember formulas

Fuck me

[coarse storm]

@viscid thistle, let me know if you are confused with anything.

[viscid thistle]

Ty

:))

[coarse storm]

Did that clear up everything?

[maiden furnace]

need some more help so the problem is this

https://gyazo.com/29acbd50b608c50b2a9997743ae46999

i pull out gcf

but after that im stuck

[lilac pier]

What's the question asking for?

[maiden furnace]

factor completely find all zeros and state the multplicity

[lilac pier]

$x(x^{4} + 6x^{2} + 9) = x[(x^{2})^{2} + 2(x^{2})(3) + 3^{2}]$

So you have $x(x^{2} + 3)^{2} = 0$

[obsidian monolith]

Sup?:

[lilac pier]

Now you can solve for x @maiden furnace

[stone temple]

Does he here say

x² = -3
x=√-3

So x can be an imaginary number? Or does that make no sense

[maiden furnace]

but once u pull out an gcf of x doesnt 6x^3 go ti 6x^2

[lilac pier]

Yeah it does.

Oh my bad.

[obsidian monolith]

Sup?:

[lilac pier]

All you had to realize was that it was (a+b)^2 in disguise

[stone temple]

Sup what I said was correct?

[lilac pier]

x can be imaginary yeah.

Unless it was previously stated that x is a real number in the question.

@maiden furnace Did you understand how I factorized?

[maiden furnace]

No im trying i understand that u rewrote it as ^2 instead ^4

but after that

im lost

[lilac pier]

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

[obsidian monolith]

Sup?:

[lilac pier]

So what I did was follow this.

To complete the square you need a^2, I got that by (x^2)^2

You need 2a(b), so I separated 2 and then a.

You're automatically left with b.

There is also another way but I think this is more easy but I'll show.

$x^{4} + 6x^{2} + 9 = x^{4} + 3x^{2} + 3x^{2} + 9 = x^{2}(x^{2} + 3) + 3(x^{2} + 3) = (x^{2} + 3)(x^{2} + 3) = (x^{2} + 3)^{2}$

[obsidian monolith]

Sup?:

[maiden furnace]

ahh i see it now

and then the quadratic formula?

[lilac pier]

No need to use quadratic formula.

$x(x^{2} + 3)^{2} = 0$

[obsidian monolith]

Sup?:

[lilac pier]

From this, we have x=0 or x^2 + 3 = 0

[valid violet]

or

[lilac pier]

Yeah I was confused on that. Thanks

You can use the quadratic formula on x^2 + 3 = 0 or you can just get x^2 = -3, x = +-sqrt(-3) which gives us x = +-sqrt(3) i

[maiden furnace]

ahh i see got it

but when u go to square it become + or - because u pull an i out correct?

[lilac pier]

When we take square root of both sides, then we get a + -. It hasn't got anything to do with i.

For example $\sqrt{4} = \pm 2$

[obsidian monolith]

Sup?:

[maiden furnace]

oh ok

[lilac pier]

Because square of +2 and -2 = 4

[maiden furnace]

and no negative under the square root sign so then u pull and i out correct?

[lilac pier]

Do you know what "i" is?

[remote veldt]

Wait wait wait. When we say $\sqrt{4}$, we mean the positive square root. $\sqrt{4}$ has only one value. What has two solutions is $x^2 = 4$. There, the solutions are $\pm\sqrt{4}$

[obsidian monolith]

Nicholas:

[remote veldt]

If someone writes $\sqrt{a}$ they mean the positive square root

[obsidian monolith]

Nicholas:

[lilac pier]

My bad.

[maiden furnace]

i is an imagery number

[lilac pier]

First you should read what Nicholas said.

Then come back to i.

[maiden furnace]

yeah i understand that

[remote veldt]

That’s just a very important notational thing - if, on a test, you miss the $\pm$ you’ll lose points, so I thought it was worth clarifying :)

[obsidian monolith]

Nicholas:

[maiden furnace]

ty for that

[lilac pier]

So we have $\sqrt{-3}$

[obsidian monolith]

Sup?:

[lilac pier]

What you do is just $\sqrt{3} \cdot \sqrt{-1} = i \cdot \sqrt{3}$

[obsidian monolith]

Sup?:

[lilac pier]

i = sqrt(-1)

[maiden furnace]

oh so i was thinking of i wrong

i thought of i as take out negative sign and put an i outside of the radical

[lilac pier]

Almost the same thing. That negative sign has a coefficient of 1.

anyways I g2g so good luck

[maiden furnace]

thank you for ur help\

[still yew]

Demoivres theorem is easy

[maiden furnace]

https://gyazo.com/823c3a886c2c96dc44de4138c9631797 need help with this one as well

its ask to find all zeros

wait

wait

i think i got it nvm

[limber compass]

I need help with this

7b.

I am really new to calc

[remote veldt]

@limber compass it goes 1 revolution per 1/10 of a second. 1 revolution is a full circle - how many radians are in a full circle?

[limber compass]

2pi radians

@remote veldt

[remote veldt]

yeah

so if a revolution is a full circle

you're going 2pi radians per 1/10 of a second right?

[limber compass]

Yes

wow you are explaining really well

[remote veldt]

if you're going 2pi per 1/10 of a second, how long does it take you to go exactly 1 radian?

aw thanks!

[limber compass]

divide by 2pi?

[remote veldt]

yeah!!

so 1/(20pi) of a second

[limber compass]

Omg! Thank you.

thats it right?

[remote veldt]

yep

if it asks for a decimal value, you could punch that into your calculator

[limber compass]

Yes.

@remote veldt need help for d

i got the answer.

but I am having trouble explaining it to myself.

so circum = 2piR
= 0.8pi

then ... ? I am confused.

Oh wait

[remote veldt]

in one revolution of the tire, the car travels as far as the circumference of the tire

[limber compass]

i think I got it

don't tell me anything I think I got it.

YAS

I think i got

so basically, the circume ference is 8pi

and if u like make the circle into a flat line. it will travel that much.

and speed = d/t

so d = 0.8pi

and t = 0.1
as 1 rev = 0.1secs

so d = 0.8pi/0.1
which shud give u 25m/s

sorry i gtg in hurry so didn't explain properly.

will be back in 10 minutes

[harsh cipher]

Question

Is the graph on the right incorrect? Because there is no vertical shift up 3 units

It's displayed properly on the left though...

confused if he just forgot about it or...?

[clever inlet]

Probably

[harsh cipher]

what do you mean probably?

[hallow lava]

definitely

[harsh cipher]

it's either he forgot to draw the shift or I misunderstood something

[hallow lava]

you can just plop it into wolfram

https://www.wolframalpha.com/input/?i=pi%2F4+*+tan(x%2B2)+%2B+3

[harsh cipher]

this guy has been frustrating me with these type of mistakes for 5 units

I drew it desmos

https://www.desmos.com/calculator

and I see that he did

it on *

y=\tan\left(\frac{\pi}{4}\left(x+2\right)\right)+3

oh no

$\y=\tan\left(\frac{\pi}{4}\left(x+2\right)\right)+3$

[obsidian monolith]

אewb64:

$\y=\tan\left(\frac{\pi}{4}\left(x+2\right)\right)+3$
```Compile error! Output:

! Undefined control sequence.
<recently read> \y

l.54 $\y
=\tan\left(\frac{\pi}{4}\left(x+2\right)\right)+3$
The control sequence at the end of the top line
of your error message was never \def'ed. If you have
misspelled it (e.g., \hobx'), type I' and the correct
spelling (e.g., `I\hbox'). Otherwise just continue,
and I'll forget about whatever was undefined.

[stuck lark]

$y=\tan\left(\frac{\pi}{4}\left(x+2\right)\right)+3$

[obsidian monolith]

RokettoJanpu:

$y=\tan\left(\frac{\pi}{4}\left(x+2\right)\right)+3$

[harsh cipher]

$y=\tan\left(\frac{\pi}{4}\left(x+2\right)\right)+3$

[obsidian monolith]

אewb64:

[harsh cipher]

there 🙂

[kindred valve]

Why cant you sqrt all of a^2 + b^2 = c^2 to make it a+b=c

Is there a rule or something

[clever inlet]

You can square root both sides

[kindred valve]

But a+b=c isnt true

[hallow lava]

you can square root both sides, but not a and b individually

[kindred valve]

So what would the alternative be

To the pythagorean theroem is you sqrt it

[hallow lava]

sqrt(a^2 + b^2) = c

lol

[clever inlet]

^

[neon garden]

For this problem I got the quotient correct but the remainder is wrong can anyone see where I went wrong?

Nvm, figured it out dumb mistake ^

[full garden]

hey guys can someone please clarify for me if it's possible to find the square root of f(-1) or is this a typo

[clever inlet]

It's possible

[stuck lark]

f(-1)=??

[full garden]

howwww

so the f(-1) inside the radical

?

[stuck lark]

what is f(-1)

[full garden]

like for this f(x) thing

[stuck lark]

what. is. f(-1)?

[full garden]

@clever inlet please tell me is it supposed to be under the radical

what do you mean

f(-1)?

[stuck lark]

plug in x=-1 into f(x)

[hallow lava]

evaluate f at x = -1

[full garden]

but it's under a squre root

[clever inlet]

So?

[stuck lark]

idc about the square root now!

tell me

[full garden]

and the f is also under the squre root

[stuck lark]

f(-1) is what???

[full garden]

9

[stuck lark]

now sqrt that

[full garden]

3

[stuck lark]

gj

[clever inlet]

There

[full garden]

wiat that's it

[clever inlet]

Done

[full garden]

?

[clever inlet]

Yes

[full garden]

oh why am I so retarded

thank you guys

what hte hell

I can't believe I messed up on that

[stuck lark]

i don't need peanut gallery comments, ok thank you

[clever inlet]

Just think of it simply first

Don't overcomplicate things

[full garden]

thank you guys so much

thank you guys

[stuck lark]

you're welcome

[fluid shore]

peanut gallery?

[gloomy vigil]

yo when it comes to questions regarding deposits and shit, when do you add a +1 to the n value and when do you not add it

Ex: Suppose that you invest $1,000 on Dec 31 of every year, in a bank account which pays 3.5% interest, compounded annually. You make a separate $1,000 deposit on the last day of each of 2015, 2016, 2017, and 2018, and then tally up the total amount of money in the account right after making your deposit on Dec. 31, 2018.

FV = (PMT(1+i)^n -1)/i FV = ($1,000(1+0.035)^4 -1)/0.035

even tho the dif in time is 3 years, it's gotta be raised to the 4th cuz that's how it be (i think)

[fleet yew]

do you know how to generate a line given points

[sharp marsh]

(x,y)

y being the number on the right

y + 7 = 9/7(x+6)

y + 7 = 9/7x + 54/7

y = 9/7x + 5/7?

?

[full garden]

hey guys I have a very silly question

the answer to what's inside the bracket is obviously the squre root of (x-1)

but what about that restriction

like if I put square root of (0-1)

wouldn't that be like i

[stuck lark]

seems like we should instead say $x\in(1,\infty)$

[obsidian monolith]

RokettoJanpu:

[full garden]

like this restriction?

[stuck lark]

btw that says x is a real number between 1 & infty. that's what i think the dom restriction should be

[full garden]

yeah bro but like this restriction came with the question

[stuck lark]

poorly written then

[full garden]

oh alright bro I will tell the teacher

thank you bro

[stuck lark]

yw

[winter isle]

how can you tell whether to use law of sines or law of cosines to solve an oblique triangle

[soft night]

Idk

[uncut mulch]

depends on what you're given

[soft night]

just use the criteria

[uncut mulch]

in some cases you may be able to use either

[soft night]

gug

[viscid thistle]

use the sine law when ur given 2 lengths and an angle or 2 angles and one length

[uncut mulch]

dunno sure why they needed to split AAS

[soft night]

These are actually really easy to program solutions to on the calculator

[fleet yew]

you don't even need to program it yourself

just look it up

[soft night]

You can’t look stuff up during the test

[fleet yew]

Look up the program

And install it on your calc

Depending on what model you got you can even store notes on it

Just straight up write down all the concepts

[harsh cipher]

Question

[rigid sun]

[harsh cipher]

[hexed ermine]

make a triangle

[harsh cipher]

why is the answer - 7pi/4?

I got root 7 /4

[hexed ermine]

it shouldnt have pi

[harsh cipher]

i know

typo

sin theta is opp vs hypotenuse

[hexed ermine]

well lets see, sintheta=sqrt(7)/4

[harsh cipher]

and theta lies between pi/2 and 0

[hexed ermine]

ok so sin(pi+theta) is in the 3rd quadrant

so its negative first of all

[harsh cipher]

hmmm

i didn't get that that must be it

[hexed ermine]

yep

and the angle is made the same with the horizontal

[harsh cipher]

so from the beginning I had to solve pythagoras in quadrant 3

that makes sense because, this time in quadrant 3 hypotenuse is negative.

YES! (this is how I sound if i get the problem correct)

[fleet yew]

NO!

[harsh cipher]

wrong

ok then

y value is -root 7

🙂

is that correct or still wrong

sigh i just rewrote my question and realized something was wrong.

[fleet yew]

-sqrt7/4

[harsh cipher]

explain to me please

oh that's what i said

in quad 3

radius isn't negative

[rigid sun]

sin(pi+x)=-sin(x)

[harsh cipher]

Question

how did the answer become 8 root 3/ 3

i couldn't get the denominator.

my answer is just 8 root 3

hi ramonov 🙂

[uncut mulch]

how are you getting: $8 \sqrt{3}$?

[obsidian monolith]

ramonov:

[harsh cipher]

well.

the angle is in quadrant 4

11pi/6 has a reference angle of 30 degrees

30 degree has a ratio of 1, root 3 and 2

cos is adjacent/hypotenuse

so that's root3/2

1/ cos is

1/1/root3/2

[uncut mulch]

ambiguous fractions

[harsh cipher]

lol

ok

$(1/1)/sqrt{3}/2$

[obsidian monolith]

אewb64:

[uncut mulch]

still ambiguous

[harsh cipher]

if you multiply by 4 you get 8 root 3

[uncut mulch]

$ 1 \big/ \big(\frac{\sqrt{3}}{2}\big)$

[harsh cipher]

Yea

[obsidian monolith]

ramonov:

[harsh cipher]

because secant is $1/cos

[uncut mulch]

first simplify that

[harsh cipher]

okay

$2/sqrt{3}

[uncut mulch]

yes

[harsh cipher]

$2/sqrt{3}$

[obsidian monolith]

אewb64:

[uncut mulch]

\sqrt

[harsh cipher]

$2\sqrt{3}$

[obsidian monolith]

אewb64:

[harsh cipher]

ahh

[uncut mulch]

not that

$\frac{2}{\sqrt{3}}$

[obsidian monolith]

ramonov:

[harsh cipher]

$\frace{2}{\sqrt{3}}$

[obsidian monolith]

אewb64:
Compile Error! Click the reaction for details. (You may edit your message)

[harsh cipher]

sigh

[uncut mulch]

\frace

[harsh cipher]

$\frac{2}{\sqrt{3}}$

[obsidian monolith]

אewb64:

[uncut mulch]

yes

[harsh cipher]

finally i need to practice a lot more

well that's what I got

that isn't.

[uncut mulch]

multiply that by 4 to get something equivalent to what the answer says

and then rationalise it

my multiplying by sqrt(3)/sqrt(3)

[harsh cipher]

awesome

well that's doing algebra.

is it the proper way of solving 4sec 11pi/6?

it made sense root 9 = 3

most of my test questions are general concepts doesn't make you think very hard.

so I'm not too worried.

[uncut mulch]

$4 \sec( \frac{11\pi}{6}) = \frac{4}{\cos(-\frac{\pi}{6})} = \frac{4}{\sqrt{3}/2} \
= 4 \cdot \frac{2}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{8\sqrt{3}}{3}$

[obsidian monolith]

ramonov:

[harsh cipher]

genius.

Thanks a lot for your help.

[harsh cipher]

Question

i made my graph as 3 square is = pi/3

the above picture is the answer answer key

for the equation y=-4sin 2 (x-pi/3)

can my starting point be at (0,4) for a sine function?

nope

it'll be at -pi/3 and 0

and diving the 12 square into 3 square = pi/3. At (0,4) that'll be pi/2

[mortal parcel]

If i need help this is the right place

Correct

[full garden]

hey guys I have a very silly question here it would be really nice if you guys can just clarify it for me

so right here I think that the answer should be b because it's 4 units to the right

but it says that the correct answer is 2 units to the right

can someone please clarify which answer is correct

[clever inlet]

2x - 4 = 2(x - 2)

[full garden]

bro but I never like seen anyone do this before

like i my textbook they never factor out the 2 bro

they just say like horizontal compression by 1/2 and it like never has an affect on the horizontal translation value

[clever inlet]

🤔

[dense kayak]

can anyone help me with this? sin(2x + 1/4 * pi) + sin(1/4 * pi - 2x) > square root(6) * sin(2x). Solve for x

[fleet yew]

wolfram alpha won't even do it

sorry bud

[harsh cipher]

Question

The answer is phase shift of $+pi/6$

$pi\6$

$frac{pi/6}$

$pi/6$

[obsidian monolith]

אewb64:

[harsh cipher]

sigh...

y= 5sin 2(x+ pi/6) + 8

ok...

sine curve...

if it had a maximum at pi/12... and the distance between is 6pi/12...

distance between next minimum at 7pi/12..

our starting point must be on the left side of the y-axis. The smallest phase shift is -pi/6?

[harsh cipher]

Hi

Can someone help me with the question above

[patent beacon]

@harsh cipher
The phase shift of y = 5sin 2(x + π/6) + 8
is -π/6

[harsh cipher]

I can't get that answer

can you explain how you arrived at that answer?

[patent beacon]

I didn't lol just clarifying the above

Alright, so your curve has a maximum at π/12, the regular sine curve has a maximum at π/2

[harsh cipher]

oh okay i'm still trying to figure it out 🙂

yep

that's correct

[patent beacon]

There's a minimum at 7π/12, which should be at 3π/2

[harsh cipher]

really what i have to do is pi/12 divided by pi/2

[patent beacon]

The difference between these two is π/2, when it is normally π. Ergo, horizonal compression by 1/2

[harsh cipher]

between these two what do you mean by these two

[patent beacon]

I mean between the x-values of the min and max. They're π/2 apart, which is half the distance apart they should be

So that's asin 2(x - p) + b

a and b are pretty easy to figure out, need those two?

[harsh cipher]

no

I have amplitutde and displacement figured out

but I don't think we need those

[patent beacon]

5sin 2(x + p) + 8

Yeah you're right they don't really come up, but the horizonal compression is important

[harsh cipher]

there is a point on the y axis and a point between pi/2 and 7pi/12

which are on the vertical displacement

and the next point to the left is the answer

but i need to see that equation not in my imagination

lol

[patent beacon]

Let's say you took the compression out of the curve. The new max would be x = π/6, the new min would be x = 7π/6

But that max should be π/2

[harsh cipher]

maybe this will help

if you understand it

[patent beacon]

So they translated to the left by π/2 - π/6 units

[harsh cipher]

no sine curve has 5 points starting at 0

0 1 0 -1 to 0

[patent beacon]

,w graph 5sin(2(x + π/3)) + 8

[obsidian monolith]

Display results online and refine query

Requested by Kaynex#7520

[harsh cipher]

cool

[patent beacon]

Damn. Not the one. It's been a while oop

,w graph 5sin(2(x)) + 8

[obsidian monolith]

Display results online and refine query

Requested by Kaynex#7520

[harsh cipher]

this is what is says in the answer

what it says

[patent beacon]

That's the graph we have. How far to shift it so that max hits π/12?

[harsh cipher]

,w graph 5sin((x+pi/6))+8

[obsidian monolith]

Display results online and refine query

Requested by אewb64#6887

[patent beacon]

The max happens when x = π/4 on that graph, so you'd need to go left π/6

[harsh cipher]

I still don't understand

🤣

[patent beacon]

,w graph 5sin(2x) + 8

[obsidian monolith]

Display results online and refine query

Requested by Kaynex#7520

[patent beacon]

This is the correct graph, but without the phase shift

The max of this graph is at π/4

We want to phase shift it so that the max is at π/12

So that's a shift by π/4 - π/12 = π/6

[harsh cipher]

okay got it

@patent beacon thank you

[patent beacon]

Hopefully it helps oop

I guess my hint is to find everything except the shift, then compare that graph to the one you want

[harsh cipher]

well the graph definitely helped and your explanation was simple

[slow wharf]

$log_{\frac{1}{2}}(2x+3) \ge 0$

[obsidian monolith]

<3 Hoodie <3:

[slow wharf]

How do you solve this inequality?

I have a brain freeze...

[serene heath]

what have u tried

[slow wharf]

Sec

$(\frac{1}{2})^{log_{1/2}(2x+3)} \ge (\frac{1}{2})^0$

[obsidian monolith]

<3 Hoodie <3:

[slow wharf]

$2x+3 \ge 1$

[obsidian monolith]

<3 Hoodie <3:

[slow wharf]

And that's already incorrect

Is there some log rule I am missing?

[hardy abyss]

if a < b is it true that (1/2)^a < (1/2)^b?

[slow wharf]

What is a what is b?

Oh sec

No, it's not true

[hardy abyss]

right

but you used this false rule

is there a similar, correct, rule you can use?

[slow wharf]

I don't know...

[hardy abyss]

you can look at the graph of f(x) = (1/2)^x to try and figure it out

[slow wharf]

Okay

I am guessing

if the base of a logarithm is less than 0 then I switch the inequality?

[hardy abyss]

you mean less than 1?

[slow wharf]

Yes

[hardy abyss]

you shouldn't have to guess, but yes

you can graph (1/2)^x and just see that it's decreasing

[slow wharf]

But it is still not correct

[hardy abyss]

well what do you end up with?

[slow wharf]

I get

$2x+3 \le 1$

[obsidian monolith]

<3 Hoodie <3:

[slow wharf]

$x \le -1$

[obsidian monolith]

<3 Hoodie <3:

[slow wharf]

$x \in (-3/2, -1)$

[obsidian monolith]

<3 Hoodie <3:

[slow wharf]

Is the solution

Ohhhh

[hardy abyss]

don't forget you need to be able to evaluate the logarithm at 2x + 3

[slow wharf]

Yeahhhh

Thank you

[hardy abyss]

also pay attention to whether you want (-3/2, -1) or (-3/2, -1], I'm not sure if the original problem had a strict inequality or not

[slow wharf]

Yes, it had a strict

[hardy abyss]

good job then

[slow wharf]

Waiit

$[-3/2, -1)$

[obsidian monolith]

<3 Hoodie <3:

[slow wharf]

?

Sorry, it wasn't strict

I meant

Larger or equal

[hardy abyss]

no, if you allow x = -3/2 then you will be trying to evalulate the log at x = 0 which doesn't make sense

but yes, if you have less than or equal to, as you wrote it originally, then you should include x = -1

[slow wharf]

Yes, you're right

For some reason desmos doesn't graph it correctly

It says that -3/2 is part of the inequality but it isn't

Thank you though

[serene heath]

desmos gives u an asymptote there

[slow wharf]

It doesn't make dotted line

[slow wharf]

What about something like this

$sin^{-1}{x} > 0$

[obsidian monolith]

<3 Hoodie <3:

[slow wharf]

How do you solve this inequality?

Hmm

This is more of a using your brain

Or I can just graph the function

[harsh cipher]

Question

How does the period change to 20?

[clever inlet]

well

what's the usual period of the sin function

[harsh cipher]

4?

2pi

[clever inlet]

yes

so you're dividing your argument by 10

it should make sense that multiplies the period by 10

and multiply by pi

it should make sense that divides the period by pi

[harsh cipher]

still don't get it

I'm so worried if i dont understand this am I going to make it through cs program

[clever inlet]

which part don't you understand?

[harsh cipher]

if you divide the argument by 10

[clever inlet]

your function will go slower

[harsh cipher]

okay.

[clever inlet]

so it takes longer to complete a period

10 times longer in fact

[harsh cipher]

how do you write it as an equation?

to figure out period is 20

because when writing sinusoidal functions with rational period

[clever inlet]

i mean

[harsh cipher]

its y= cos2pix

[clever inlet]

it should make intuitive sense

but

in general

[harsh cipher]

well it didn't

😦

[clever inlet]

🤔

[harsh cipher]

lol

[clever inlet]

think about it carefully

[harsh cipher]

okay let me think about it with seriously

[clever inlet]

say you have cos(2pi * x)

2pi is the usual period

of the cosine function

but we are multiplying our argument by 2pi

so the function completes a period much faster

2pi times faster in fact

[harsh cipher]

in the question isn't it pi

[clever inlet]

i'm using the example of cos(2pi * x)

the concepts are the same

you should be able to apply the concepts from one trig function to another

[harsh cipher]

but in the question we are dividing by pi not 2 pi

[clever inlet]

i'm using the example of cos(2pi * x)

[harsh cipher]

okay

then if'were multiplying by pi it takes longer

[clever inlet]

compared to?

[harsh cipher]

2pi?

[clever inlet]

i guess you can think of it like that

[harsh cipher]

lol

[clever inlet]

cos(pi * x) is not as "fast" as cos(2pi * x) and hence has a longer period

in general

whatever multiplier you have on x

denote it by T

the period is given by 2pi/T

for sine and cosine functions

[harsh cipher]

I mean I drew the graph as if period was 10

and I look at the answer and the period was 20

the period*

so I have to rewrite my equation as 2pi*x and then graph it

[clever inlet]

[harsh cipher]

hahaha

[clever inlet]

just think it through

test points if you need

[summer monolith]

How do they simplify it like this? Why only the parentheses got multiplied by A in the denominator?

[willow bear]

the num and the denom were both multiplied by a

multiplication doesn't... distribute over itself

[summer monolith]

a*(b*c)...
This simple truth can seem to sink in for some reason

[leaden stratus]

The result is [3(sin alpha + 1)]

But I can't figure out how to get this result

[uncut mulch]

what have you tried?

@leaden stratus

[uncut mulch]

why did your $\cos^2(\alpha)$ change to $-\cos(2\alpha)$

[obsidian monolith]

ramonov:

[leaden stratus]

I think it's a refuse

I forgot to write "+"

And it's not cos(2a) but cos^2a

[uncut mulch]

write the 2 higher,
if its on the same level, its going to be interpretted as 2a

anyways

try do something with
2 - 2sin^2(a)

[leaden stratus]

2-2sin^2(a) is a relationship but ×2?

[uncut mulch]

or alternatively write
cos^2(a) in terms of sin(a)

would probably be more efficient

[leaden stratus]

Like cos^2a = 1 - sin^2a?

[uncut mulch]

yes

[leaden stratus]

So now the numerator is -3sin^2(a) + 3

[uncut mulch]

good. try factorising that

[leaden stratus]

3 (-sin^2a + 1)

[uncut mulch]

can be factorised further

[leaden stratus]

3cos^2a?

[uncut mulch]

factorise, (not write in terms of cos)

note that 1 = 1**^2**

[leaden stratus]

🤔

[uncut mulch]

and sin^2(a) = (sin(a))^2

3 (-sin^2a + 1) = 3( 1**^2** - (sin(a))^2)

i've bolded the ^2 to indicate that they are both: ___

[leaden stratus]

Ok, now...

(1+sina^2)(1-sina^2)

@uncut mulch ?

[uncut mulch]

parentheses

[leaden stratus]

Oh true

[uncut mulch]

(sin(a))^2 or sin^2(a)

[leaden stratus]

First one you wrote

[uncut mulch]

also the 3 in front

[willow bear]

also the uh

[leaden stratus]

?

[willow bear]

you basically just said a^2 - b^2 is the same thing as (a^2 - b^2)(a^2 + b^2)

and also have not filled in ramonov's blank

[uncut mulch]

whoops missed that

error

[leaden stratus]

Wait

We were that 3(1^2 - (sin(a))^2)

[uncut mulch]

yes. and the current goal is to factorise that

i've bolded the ^2 to indicate that they are both: ___

?

[leaden stratus]

🤔

I'm figuring out

They're bot squared

[full garden]

hey guys I have a question

I am trying to graph the rational function but I need to know how to find the vertical and horizontal asymptotes, but whenever click on a youtube video they are using limits and I didn't do that in school yet, so is there a way to find the vertical and horizontal asymptote without using limites

[serene heath]

uhh

sticky one

what's your function?

[serene heath]

you can find the vertical asymptote right?

[full garden]

yeah bro

[noble fulcrum]

Hello guys quick question: for f(x) = 3+x+(e^x) is f^-1(x) possible to find

I get stuck at ln(x-3 ) = ln(y+e^y)

[clever inlet]

I doubt it's possible in terms of nice functions

[neon garden]

Can anyone help with 49 b?

[clever inlet]

what have you tried

[neon garden]

Honestly I don’t even get what it’s asking

if a rectangle has a base of 2 and height 3 then it’s area it’s 6

it’s asking to find the base of a second rectangle with the same area

So wouldn’t 3 work?

but I feel like that’s way too simple

[clever inlet]

3 for what?

[neon garden]

The base of a second rectangle

and 2 for the height

because it’s asking for a rectangle with the same area

[clever inlet]

does it fit under your parabola though?

base of 3 implies x = 1.5

4 - (1.5)^2 = 1.75

so the height can't be 2

[neon garden]

oh I didn’t know it had to be under the parabola

That makes much more sense

[clever inlet]

it has to have 2 corners on the parabola

[neon garden]

Yeah I still don’t know how to start

[clever inlet]

well

consider a point on the parabola

for any given x

what's the corresponding y?

[neon garden]

f(x)?

[clever inlet]

sure

which is?

[neon garden]

4-x^2

[clever inlet]

yes

hmm

this approach leads to a cubic equation

do you know how to solve those?

[neon garden]

factor?

[clever inlet]

it's technically factorable

but not in any obvious way

long/synthetic division is the other way

[neon garden]

Yeah I know how to do long and synthetic division

[clever inlet]

ok cool

so for any given x

the corresponding y is 4 - x^2

[neon garden]

yes

[clever inlet]

would you agree

that x is half the length of the base of that rectangle?

[neon garden]

yes

[clever inlet]

and the height is 4 - x^2

actually you should have already done this for the last question i realise

either way

A = 2x(4 - x^2)

we want to solve 2x(4 - x^2) = 6

[neon garden]

Yeah I got -2x^3 + 8x for part a

[clever inlet]

-2x^3 + 8x - 6 = 0

or

2x^3 - 8x + 6 = 0

[neon garden]

Ok I’ll try that

[clever inlet]

you know x = 1 is a solution

so (x - 1) is a factor of the LHS

divide it out

[viscid thistle]

Whats the difference between [f x g] and (f x g)

[clever inlet]

🤔

[viscid thistle]

Theres none?

[clever inlet]

i don't believe so

[neon garden]

Ok I did all that but my answer isn’t the same as the key

Not sure where I went wrong

wait

Figured it out

[clever inlet]

solve for x

double it

to get the base

[neon garden]

I got about 2.6

Yeah that matches the key

alright great thanks

[buoyant hound]

are these equivalent

[dim jungle]

no

i dont think so

[rigid beacon]

nah that looks good @buoyant hound

[buoyant hound]

okay nice

[full garden]

hey guys just a quick question please

here I just plug the x inside of the p(r) and x^2+5

like that's what I did in the previous exercises but this this one seems kids odd because it's like p(x)

[rigid beacon]

nah that's right

what did you get as the end answeer @full garden

[full garden]

for some reason that question doesn't get answered at the back of the book

do we do this the same way as we do the other ones

[rigid beacon]

I'm asking what did you get as you end answer

and what "other ones" are you referring to for your 2nd image

[full garden]

sorry that was a stupid question

[rigid beacon]

you're good

[full garden]

sorry that was very silly of me to ask

[rigid beacon]

nah don't sweat it

[full garden]

I was just asking because it looks about weriod plugging in p(x) in a p(r)

[rigid beacon]

completely normal

[full garden]

oh really

so do I just continue and plug in x instead of r?

[rigid beacon]

yup

so your end answer will be some function of x plus a constant

[full garden]

thank you so much man I really appreciate it 💙

thank you

[sharp marsh]

Do I do (4(-4)^2 ) * -4-4

[tardy crescent]

No, you do (4(-4)^2 + 1 ) * (-4 - 4)

[sharp marsh]

65 * -8 right?

thx

[tardy crescent]

yep np

[clever inlet]

Which of your coordinates contains 7?

[harsh cipher]

Hi

Question

Can an asymptote be on the y axis of the grid for a tangent function?

[clever inlet]

sure

[harsh cipher]

it can?

[pale bison]

grid for a tangent function
what does this mean

[harsh cipher]

can we draw an asymptote of a tangent function at 0

(0,0)

ill show you

[pale bison]

oh you mean tan(x).?

hmm

what?

[harsh cipher]

yes

[pale bison]

tan(x) does not have an vert. asym. at y-axis

[empty fulcrum]

You can just vertically shift it to make one

[pale bison]

they're talking about tan(x) precisely? of course you can shift it

[empty fulcrum]

He said a tangent function, so I took it as any sort of tangent no matter what x is shifted by

[pale bison]

oh you mean tan(x).?

yes
but ok whatever

i suppose we can say tan(x) = tan(x-c) for any c, why not

[uncut mulch]

*horizontal shift

[pale bison]

...

[harsh cipher]

lol

x cannot equal (n(pi)/2 + pi/4

that means our asymptote to the left is at 0?

[uncut mulch]

how are you getting 0?

[patent beacon]

You can't have 2x = nπ/2 for any odd n

Which you've likely calculated

[harsh cipher]

that's what I did, yes

that's our arguement

[patent beacon]

Note x can be 0

[harsh cipher]

why can't you have odd "n"?

[patent beacon]

You can't have a vertical terminal arm. That's at π/2, 3π/2...

[empty fulcrum]

Think about the tan of 3pi/2

Ye ye

[harsh cipher]

well i guess the y axis is -3pi/4

[patent beacon]

A better way to say what I'm saying is
2x ≠ π/2 + nπ

[harsh cipher]

too hard..

sigh

[empty fulcrum]

What's the sin of pi/2

[harsh cipher]

1?

[empty fulcrum]

Okay

What's the cos

[harsh cipher]

0

for pi/2

[empty fulcrum]

And tan=sin/cos right

[harsh cipher]

yes

y/x

[empty fulcrum]

Therefore, tan pi/2= 1/0

It's undefined

What's what we're saying

So any odd pi/2 value is going to have an undefined tangent

pi/2, 3pi/2, and so on

[patent beacon]

I like to think of tanθ as the slope of the terminal arm. Vertical slopes don't work

[empty fulcrum]

Ye ye

An intuition like that will actually benefit you once you get to calc

[rose locust]

how do I turn a y = square root of x graph into a y = mx+b style graph?

[willow bear]

you can't because y = sqrt(x) isn't a linear function and its graph is not a line

[rose locust]

if i square y it will give me a straight line, thats all i know

[willow bear]

if you square a linear function it'll no longer be linear

what led you to ask this question in the first place?

[rose locust]

ill send you a picture of what i am working on

I think, what i wrote down in my notes is if you have a y = x^2 graph you can make a y vs x^2 graph if that makes sense

[harsh cipher]

Ann you were here the whole time?

[willow bear]

define "the whole time"

[pale bison]

integer time units

[serene heath]

since the first human was born

[fleet yew]

Polynomial time

[harsh cipher]

like online but busy?

I have a question again

$-2sin frac{3pi/4}+cos frac{3pi/4}$

[obsidian monolith]

אewb64:

[harsh cipher]

omg

[uncut mulch]

\

and ()

[stuck lark]

$-2\sin\br{\frac{3\pi}4}+\cos\br{\frac{3\pi}4}$

[obsidian monolith]

RokettoJanpu:

$-2\sin\br{\frac{3\pi}4}+\cos\br{\frac{3\pi}4}$

[harsh cipher]

yes that's right

$-2\sin\br{\frac{3\pi}4}+ \cos\br{\frac{3\pi}4}$

[obsidian monolith]

אewb64:
Compile Error! Click the reaction for details. (You may edit your message)

[harsh cipher]

errr

[stuck lark]

i predefined custom commands

[harsh cipher]

for background color

?

[stuck lark]

no for big

[harsh cipher]

well anyways...I couldn't figure out the question

[stuck lark]

\br is not default

[harsh cipher]

cool

[stuck lark]

what'd you try in evaluating these trig things

[harsh cipher]

ok

what i got was sin 45 and cos 45 is

1/ root 2

-2 x1 root/2 + 1 root/2

opps

[fluid shore]

Not quite

[harsh cipher]

-2x 1/ root 2 + 1/root 2

[stuck lark]

abhi i don't want to dogpile in here

please and thanks

(that means more than 2 people camping in 1 channel)

what quadrant is the angle 3pi/4 in?

[harsh cipher]

its in quadrant 2

[stuck lark]

what's the sign of cos there?

[harsh cipher]

neg

ative

[stuck lark]

cos(3pi/4)=?

[harsh cipher]

-1 /root2

[stuck lark]

-2 x1 root/2 + 1 root/2
what should you have instead?

[harsh cipher]

-2?

I'd like to say I don't know

[stuck lark]

you evaluated -2sin(3pi/4) fine but messed up cos(3pi/4)

[harsh cipher]

okay but the answer is something way off

it says -3 root2/2

[stuck lark]

don't read the ans yet

[harsh cipher]

ok

is that a bad way to study math?

[stuck lark]

not always

[harsh cipher]

ok could you show me what we'd get if we added -1 /root 2 + 1/root 2?

[stuck lark]

you're off track

[harsh cipher]

okay

[stuck lark]

we're evaluating -2sin(3pi/4)+cos(3pi/4), we can evaluate each term then add em up

what's -2sin(3pi/4)

[harsh cipher]

2/root2

-2 /root2

[stuck lark]

-2 /root2
ok. put that aside. cos(3pi/4)=?

[harsh cipher]

-1/root 2

[stuck lark]

put em together. what sum are you computing?

[harsh cipher]

-3/root2

[stuck lark]

done

[harsh cipher]

answer says -3root2/2

[stuck lark]

equal

[harsh cipher]

equal?

[stuck lark]

same thing

[harsh cipher]

how so?

[stuck lark]

try figuring out yourself

[harsh cipher]

okay

[harsh cipher]

okay all i had to was rationalize the denominator.

do*

@stuck lark thank you for your help.

[summer monolith]

multiplication doesn't... distribute over itself
Does this mean I can write 2*(2a(b^2-4ac)^(1/2)) as 4a((b^2-4ac)^(1/2)
?

[willow bear]

that's got nothing to do what i said, and also your second expression has unbalanced parentheses

[summer monolith]

2*(2a(b^2-4ac)^(1/2)) as 4a((b^2-4ac)^(1/2))

fixed

[willow bear]

that's got nothing to do what i said

[summer monolith]

What about a(bc)=abc
?

[willow bear]

that has nothing to do with me saying "multiplication doesn't distribute over itself"

what i said is that, in general, $a(bc) \neq (ab)(ac)$

[obsidian monolith]

Ann:

[summer monolith]

Ok. I'll refrase
Is 2*(2a(b^2-4ac)^(1/2)) equal to 4a((b^2-4ac)^(1/2))
?

[willow bear]

yes

[summer monolith]

Thks

[stuck lark]

@harsh cipher very naisu. and no prob

[sturdy haven]

If you look by 2pi on the right

What does the zero signify

[willow bear]

wdym

do you have no qualms with the same angle in degrees being labeled both as 0° and as 360°?

[sturdy haven]

No

Not 0 degreesp

Just the zero

Right above 2pi

[valid violet]

@sturdy haven 0 radians

[sturdy haven]

Thanks

[full garden]

hey guys can someone please help me with this very easy question that I am too stupid to answer

I tried plugging in 65 and solving for t but I got the question wrong

I keep on getting the question wrong

like for b) shouldn't I use the second equation and do 65=22t+1

[serene heath]

uh