precalculus

[rigid sun]

just do it

[hexed geyser]

ye ik how to do it

wtf

so i end up with 29 + or - sqrt(144-720)

that aint right is it

like you cant have a negative number up in there

oh nevermind

[rigid sun]

?

[hexed geyser]

sec

[rigid sun]

lol

[plush matrix]

I'm having trouble with this polynomial: P(x)= (x+3)^2(x-3+2i)(x-3-2i) anyone know how to solve this and get rid of the complex #?

[rigid sun]

Try u sub

U=x-3

[plush matrix]

oh sorry, I was working on some other problems

so you were saying

[torn swift]

you want to find where it equals zero?

[plush matrix]

the question says: Find a polynomial of degree 4 which has the zeroes -3 and -3+2i. You may leave your answer in factored form, but do not leave any complex numbers in your final answer.

[torn swift]

almost there, just multiply out the above polynomial then

[plush matrix]

so that would be (x+3)(x+3)(x-3+2i)*(x-3-2i)?

[torn swift]

yea, you will have to just do the dirty work of multiplying out the entire thing

[rare cloud]

(x - 3 + 2i)(x - 3 - 2i) = [(x - 3) - 2i)][(x - 3) + 2i] = (x - 3)² - (2i)², which becomes a real number.

[plush matrix]

oh shit u rite

[rigid sun]

Lol get it

U right

[plush matrix]

WHHEEZEE

[rare cloud]

You OK? Sounds like you require medical attention.

[rigid sun]

Sounds like u need medical attention

U

[rare cloud]

I guess "U" need mental help.

[plush matrix]

we killed it

you can stop now

[rigid sun]

[plush matrix]

ln e^(pi/2)

what even is this

[rigid sun]

What’s what

,w calc e^(pi/2)

[plush matrix]

it says evaluate

[torn swift]

well, use exponent laws

[obsidian monolith]

Display results online and refine query

Requested by maleb1964#6739

[torn swift]

you don't need a calc to know what ln(e^(pi/2)) is

[plush matrix]

would it just be 1^(pi/2)

[torn swift]

we can already see the exact value of it

[rigid sun]

Ik

[torn swift]

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

[obsidian monolith]

⚡Amphy⚡:

[torn swift]

apply this law first

[rigid sun]

^

[torn swift]

then you apply what ln(e) is

[plush matrix]

oh yee

third law

or whatever

[stuck lark]

imagine log rules having a defined order

[plush matrix]

reeeeeeee

so ln(e) cancel out and that leave out pi/2?

[stuck lark]

sure

[plush matrix]

would you do the same thing for b^(log base b square root of 7)?

[rare cloud]

$b^{log_{b}\sqrt{7}}?$

[plush matrix]

ye

[stuck lark]

latex gore

[rare cloud]

God dammit.

[torn swift]

$b^{\log_{b}(\sqrt{7})}$

[obsidian monolith]

⚡Amphy⚡:

Zetanoxic:

[rare cloud]

Phew.

[torn swift]

that is in fact just sqrt(7), just like you said

[plush matrix]

the one without the ( ) for sqrt7

[rare cloud]

Same thing.

[plush matrix]

$\log _{10}\left(\frac{\left(\left(y+z\right)x^3\right)}{\sqrt{y^3z}}\right)$

[obsidian monolith]

The Belliest of Bells:

[plush matrix]

this makes me creamy in my panty

[hexed ermine]

o.O

[rare cloud]

Oh.

[plush matrix]

and there's another one but in reverse

idk how but I got log(y+2)+2logx-3/2logy-3/2logz

[hexed ermine]

whats the question

[plush matrix]

use laws of log to rewrite

as sum or differences of log

[rare cloud]

Ah.

No idea where you got the (y + 2).

[plush matrix]

oh

my Z looks like a 2

my dumbass

[hexed ermine]

close

the 2nd and 4th terms are wrong

[rare cloud]

Yep.

[plush matrix]

2nd as in logx^2?

[rare cloud]

It's a cube though.

[plush matrix]

oh god

[rare cloud]

Or is it?

[plush matrix]

my handwriting is working against me

SOS

send help

and the 4th one

[rare cloud]

z is to power ½. How did the 3/2 get there?

[plush matrix]

oh I thought because 1/2log(y^3z)

would become 1/2log(3logy+3logz)

:<

so it's log(y+z)+3logx-3/2logy-1/2logz?

[rare cloud]

I believe you would be right.

[hexed ermine]

ya

[plush matrix]

$\frac{1}{2}\left(4lnx-2ln\left(y\right)-5ln\left(x^2\right)+2ln\left(y^3\right)\right)$

[obsidian monolith]

The Belliest of Bells:

[plush matrix]

last one before I can break out of this mortal shell

[rare cloud]

You trying to convert into a single log or what?

[plush matrix]

ye

[rare cloud]

Maybe expand out and write in terms of powers first?

[plush matrix]

I can't do the other way around for some ducked up reason

how do you rewrite -5lnx^2

[hexed ermine]

bring the 2 out front

also, is it (ln(x))^2

or ln(x^2)

[plush matrix]

it's ln(x^2)

[hexed ermine]

ok

so move the 2 out front

[plush matrix]

I meant 5lnx^2

do I move lnx^2^5

[rare cloud]

5 ln (x²) = ln (x²)⁵ = ln (x¹⁰)?

[plush matrix]

idk man

[plush matrix]

is that really correct?

[torn swift]

yes

[plush matrix]

$ln\sqrt((x^(4))/((y^(2))/(x^(10)*y^(6))))$ yell at me if I'm wrong

[obsidian monolith]

The Belliest of Bells:

[plush matrix]

$ln\sqrt((x^(4))/((y^(2))/(x^(10)*y^(6))))$

[obsidian monolith]

The Belliest of Bells:

[plush matrix]

fuck

$ln\left(\sqrt{\frac{x^4}{\frac{y^2}{x^{10}\cdot y^6}}}\right)$

[obsidian monolith]

The Belliest of Bells:

[torn swift]

try to simplify the things inside the square root first

[plush matrix]

how would I do that

and that's what I got from this btw

$\frac{1}{2}\left(4lnx-2ln\left(y\right)-5ln\left(x^2\right)+2ln\left(y^3\right)\right)$

[obsidian monolith]

The Belliest of Bells:

[torn swift]

oh, you had to start there and go to simplified, I see then

[plush matrix]

yes yes

[plush matrix]

time to play shadow kee[

wait wrong chat

my b

[torn swift]

lol

[tame wedge]

If I have some polynomial p(x) with a set of real or complex roots; if I take the same polynomial but multiply it all by some factor 'k', do the roots change? Like does k*p(x) have the same roots as p(x)? For any degree?

[willow bear]

what do you think

do the equations p(x) = 0 and kp(x) = 0 have the same or different solutions

[tame wedge]

Im guessing they do because desmos paints a fair picture

Is it because you can use nfl on any factoried poly and k never equals zero?

[willow bear]

if "nfl" is supopsed to mean "null factor law"

then that doesn't even depend on the lhs of your equation being a polynomial or not

[tame wedge]

Yeah sure but if it is a polynomial...

[harsh cipher]

Hi

why do I have to take the inverse on the right side of the equation?

((2/5))^-3

[willow bear]

:?

[knotty spear]

its so that you have 2/5 as base

[harsh cipher]

125/8

[knotty spear]

on both sides

[harsh cipher]

is it because of that purpose only?

[knotty spear]

which means that you can say that the exponents must be equal

yes

[harsh cipher]

I understand that first thing I have to do is make base the same on both sides

@knotty spear thanks!

[knotty spear]

[harsh cipher]

this channel deserves more 🙂

[velvet elm]

I am having trouble with this and the book doesn't cover any similar examples.

If I expand it using the hint, I get 2sin(tan^-1 6)cos(tan^-1 6), but i don't know what the inverse tan of 6 is

Oh there's a double angle formula for sin2\theta that turns into tan...that might work

[rugged nimbus]

Anyone can give me example on how arc length in cartesian coordianates dy question example like?

[rugged nimbus]

Ive search it eveeywherw but only found arc length in y=f(x) form

[viscid thistle]

If i wanted to factor a polynomial function, how would i do so without using my teacher's stupid guess and check method

currently, we are just using every single possibility and checking if it is divisible

[lucid forum]

That's the same way we do in the textbook. I don't believe there's an alternate method, unless the trinomial is obviously a perfect square trinomial. I'd love be proven wrong though!

[torn swift]

going through the synthetic division isn't all that time consuming

you can usually tell if it's going to end up as 0 a few terms in

[fleet yew]

lamo

[odd helm]

Idk why the flow chart starts at positive

I'm supposed to plug in values before the VA right?

But when I do I get a negative value for y

Or is that not what I'm supposed to do?

[uncut mulch]

the zeroes of the numerator are: x= -3, 0, 1
the asymptote is at x= -2
those are your values of interest

when x<-3, f(x) will be positive

[pseudo sonnet]

have i answered this question correctly?

first i showed with sums identity

[willow bear]

well

[pseudo sonnet]

then product

identity

[willow bear]

you could have just said sin(2x) = 2sin(x)cos(x) right away

you could also have given an example of a value of x for which sin(2x) != 2sin(x)

[pseudo sonnet]

like if I set x = pi/4

[willow bear]

sure

uhhh wait.

that one might not work?

nvm it works

1 is not sqrt(2)

[pseudo sonnet]

idk if she wants to prove it by plugging in

or by explaining the sin(2x) identity

🤷‍♂️😂

can my final answer simplify further

[willow bear]

no

[pseudo sonnet]

did i set up correctly for my part c using the sum trig identity

[stuck lark]

👍🏽

[lucid forum]

Hey! I was wondering how to solve this equation. I thought of multiplying by the conjugate, but that seems like it will give me a radical on the right side, so not too sure how to go about solving it.

√(x + 8) - √(x - 4) = 2

[willow bear]

isolate one of the radicals
square both sides
isolate the radical that remains
square both sides again
you'll end up with a polynomial equation
solve it
check each solution for validity

[lucid forum]

Ah! I'll try that. Thanks!

Oh wow! Got the answer of x = 8. Thank you! Should've tried it before asking for help. My bad!

[pseudo sonnet]

can someone tell me where i went wrong

nvm i got it

[wraith idol]

Nothing went wrong

[pseudo sonnet]

i shouldve used a different identity

[wraith idol]

Yes

[pseudo sonnet]

sec^2(x) = tan^2(x) + 1

[wraith idol]

Yes that

Exactly

[pseudo sonnet]

then the 1 - 1 cancel

gg

thanks

can anyone help me with this?

[uncut mulch]

combine like terms

[pseudo sonnet]

like so?

[uncut mulch]

yeh

[pseudo sonnet]

ty

that was hard to spot

hiding in plain sight

[pseudo sonnet]

am i doing this right?

[uncut mulch]

yeh continue

[pseudo sonnet]

how

wait i think ik

nvm lol

got nowhere

[uncut mulch]

you can't split fractions like that

can any trig identities be used here?

[pseudo sonnet]

i have no idea

[uncut mulch]

your goal might give you a hint

[pseudo sonnet]

oh dude

the bottom part looks like the similar sums trig identities

wait

omfg

i know

1 secondddd

like so

[uncut mulch]

yeh

[pseudo sonnet]

thanks g

[uncut mulch]

mention you're starting from the RHS at the start

[pseudo sonnet]

in that way?

[uncut mulch]

yeh

[pseudo sonnet]

thanks

[pseudo sonnet]

https://i.imgur.com/HtwgI3M.png

Can anyone help me with this

is this suppose to be turned into tan^2(x)

i dont see how that would help me

[willow bear]

numerator is sin^2(45°) + sin^2(90°) + sin^2(1°) + sin^2(89°) + sin^2(2°) + sin^2(88°) + sin^2(3°) + sin^2(87°) + ... + sin^2(44°) + sin^2(46°)

[pseudo sonnet]

did u rewrite that but putting the 45 and 90 in front

b/c those are angles we can measure easier

[willow bear]

,,, it doesn't matter where they go

they just aren't paired up w anything

[pseudo sonnet]

wdym paired up

[willow bear]

look at how i wrote the thing down...

let me be even more explicit

sin^2(45°) + sin^2(90°) + [sin^2(1°) + sin^2(89°)] + [sin^2(2°) + sin^2(88°)] + [sin^2(3°) + sin^2(87°)] + ... + [sin^2(44°) + sin^2(46°)]

[pseudo sonnet]

OH

because 45 is the middle term

and 90 would go with 0

but sin(0) is 0 so it doesnt matter anyways

do I have to rewrite every single sin term with sum identities? 😦

[willow bear]

,,,,,,,,, what

no

sin^2(x) + sin^2(90°-x) = 1

[pseudo sonnet]

is that an identity or something

<@&286206848099549185>

[plain lake]

cos(x)=sin(90-x)

[tardy crescent]

Just think of a right triangle, and how the cosine of one of the acute angles is the sine of the other acute angle

So sin(90 deg - x) = cos (x), therefore sin^2(x) + sin^2(90 deg - x) = sin^2 (x) + cos^2 (x), which is the Pythagorean identity and equal to 1

[pseudo sonnet]

https://i.imgur.com/HtwgI3M.png

how does it help me with my question

[serene heath]

Read what Ann said again

[pseudo sonnet]

OH

she wrote them in terms of all of the angles adding to 90

so the numerator is root(2)/2 + 1?

[plain lake]

the beauty of telescoping

well i guess this technically isn't telescoping

but the spirit is there

[pseudo sonnet]

does the numerator equal root(2)/2?

[serene heath]

No

[pseudo sonnet]

i mean

root(2)/2

[uncut mulch]

No, how are you getting sqrt(2)/2?

[viscid thistle]

it telling me this is false

😦

[pseudo sonnet]

im not sure im confused

[uncut mulch]

how are you applying

sin^2(x) + sin^2(90°-x) = 1

[pseudo sonnet]

OH

she rewrote all those terms

and they equal to 1

so is the numerator

44 + sin^2(45)

[uncut mulch]

why did you change it from 45 to 44?

[pseudo sonnet]

i thought i miscounted

after adding all the rewritten terms

[uncut mulch]

how many pairs of numbers are there that add to 90?

[pseudo sonnet]

45

[uncut mulch]

is that meant to include the sin(90°)?

[pseudo sonnet]

yeah

[uncut mulch]

what term is left over?

[pseudo sonnet]

sin^2(45)

[uncut mulch]

so should it be 44 +sin^2(45°) or 45 + sin^2(45°)?

[pseudo sonnet]

44 +sin^2(45°)

[uncut mulch]

why 44?
you just said that all your other stuff adds to 45

[pseudo sonnet]

because 44 terms + 1 more term = 45

if its 45 terms + sin^2(45) = 46 total terms

?

[uncut mulch]

you can think of this in two ways
introduce a +sin(0) term
and you'll have 45 (pairs adding to 90) + sin^2(45)
0+...+44 (45 terms)
90+...+46 (45 terms)
45

OR
you'll have 44 pairs
+sin^2(90°)
+sin^2(45°)

1 + ... + 44
89 + ... 46
90
45

[pseudo sonnet]

ahh i see now

[vocal pivot]

been stuck on this all day 🤦🏽‍♂️

[pseudo sonnet]

89-46 = 43

43 + the other 2 terms (45) and (90) = 45

[uncut mulch]

what no
the terms between 89 to 46 inclusive is 89 - 46 + 1 = 44
sin^2(90) = 1
and you also have sin^2(45)
so your total sum is 44 + 1 + sin^2(45) (and the sin^2(45) can be simplified)

sin^2(1) + sin^2(89) = 1 right?
sin^2(2) + sin^2(88) = 1 right?

...
sin^2(44) + sin^2(46) = 1

and there are 44 of these pairs and these will add up to 44

you also have sin^2(45) and sin^2(90)

so numerator = 44 + sin^2(90) + sin^2(45)

[pseudo sonnet]

i was right the first time when I said 45 + 1/2

[uncut mulch]

yes

[vocal pivot]

any help?

[uncut mulch]

do something similar for the denominator and then simplify the fraction

where are you stuck?

[vocal pivot]

who me?

[uncut mulch]

yes

[vocal pivot]

the entire thing

i dont understand it at all 🥺

[uncut mulch]

have you ever solved any equations before? eg, quadratics?

[vocal pivot]

i have however i keep getting this wrong

[uncut mulch]

what answers were you getting?

[vocal pivot]

ill send a pic of it since i cant type it

[uncut mulch]

describe your steps

[pseudo sonnet]

in regards to my question, the identity sin^2(x) + sin^2(90°-x) = 1 doesnt apply for the denominator right

[uncut mulch]

similar idea for cosine
all revolves around sin^2(x) + cos^2(x) = 1

[pseudo sonnet]

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

OH

[vocal pivot]

factor on the left side of the equation is equal to 0, the entire expression will be equal to 0

then

Set the first factor equal to 0
and solve

[uncut mulch]

ok, what were your solutions to
2sin(x) = sqrt(3) ?

[pseudo sonnet]

recalling what Ann said

sin^2(45°) + sin^2(90°) + [sin^2(1°) + sin^2(89°)] + [sin^2(2°) + sin^2(88°)] + [sin^2(3°) + sin^2(87°)] + ... + [sin^2(44°) + sin^2(46°)]

and that cos^2(x) = 1 - sin^2(x)

https://i.imgur.com/HtwgI3M.png

could I rewrite the denominator terms

as 1 - sin^2(x)

for each term?

[uncut mulch]

same idea as in
cos^2(1) + cos^2(89) = 1
etc

[pseudo sonnet]

cos^2(45°) + cos^2(90°) + [cos^2(1°) + cos^2(89°)] + [cos^2(2°) + cos^2(88°)] + [cos^2(3°) + cos^2(87°)] + ... + [cos^2(44°) + cos^2(46°)]

like that?

[uncut mulch]

yeh

[pseudo sonnet]

sin^2(90) = 1

cos^2(90) = 0

so all those terms just equal 0?

except cos^2(45)

[uncut mulch]

cos^2(1) + cos^2(89) = 1
etc
only the cos^2(90) = 0

[pseudo sonnet]

for example

[cos^2(1°) + cos^2(89°)]

does that expression equal 0

[uncut mulch]

cos^2(1) + cos^2(89) = 1

[pseudo sonnet]

wait what how

[uncut mulch]

cos^2(90-x) + cos^2(x) = 1

[pseudo sonnet]

i thought the whole point was the degrees add to 90

and cos^2(90) = 0

[uncut mulch]

the idea is that (90 - x) + x = 90
cos(90-x) = sin(x)
sin^2(x) + cos^2(x) = 1

[pseudo sonnet]

then..

isnt it just the same outcome

45 + 1/2

[uncut mulch]

no.

because you have 44 pairs adding to 1 → 44
cos^2(90) = 0
and cos^2(45)

[pseudo sonnet]

wait what dude

cos(0) = 1

cos(0)cos(0) = 1 * 1 = 1

[uncut mulch]

whoops

typo

meant to be cos(90) = 0

[pseudo sonnet]

cos^2(90-x) + cos^2(x) = 1

is this also true for

sin^2(90-x) + sin^2(x) = 1

[uncut mulch]

yeh

[pseudo sonnet]

so the expression in the end turns into (45+ 1/2) / (44 + 1/2)

which is

91/89

@uncut mulch

[uncut mulch]

yep

[vague zephyr]

im not sure where to go

i have found all of the zeroes and i know which multiplicities are odd and which are even however i dont know what each multiplicity is

[tardy crescent]

If you know the zeroes, you also know the multiplicities

[vague zephyr]

how are they related

[tardy crescent]

odd multiplicity is when it goes through the x-axis, even multiplicity is when it just touches the x-axis

[vague zephyr]

i know that much but how do i determine which number to use for the multiplicity

[tardy crescent]

So look for what the graph does at each zero, if it goes through the x-axis then there's odd multiplicity

[vague zephyr]

odd meaning 1, 3, 5, 7, ...

[tardy crescent]

yes

[vague zephyr]

even meaning 2, 4, 6, 8

[tardy crescent]

Yes

[vague zephyr]

so how do i determine which number to use?

for 8 would the multiplicity be 8

[tardy crescent]

The way I'm reading, you could say "the zero x = -8 has even multiplicity"

Do you know the degree of the function?

[vague zephyr]

the degree is odd so 7?

[tardy crescent]

No, quadratics can only have 2 zeroes. If it has 4 zeroes it must be at least a quartic, but it looks like it's odd degree

[vague zephyr]

or 5

[tardy crescent]

Has to be 7 or higher

Because you have 3 roots with even mutliplicity

And one with odd multiplicity

[vague zephyr]

the odd root would be 6 correct?

[tardy crescent]

Oh hang on

[vague zephyr]

so would the multiplicity of 6 be 7

[tardy crescent]

Read the bottom of the question

They're only asking you to answer "even" or "odd" for multiplicity

[vague zephyr]

god f**king damn it

[tardy crescent]

There's no way we can tell the exact number from the graph

I missed it too, small text

[vague zephyr]

that makes this a whole lot easier

[tardy crescent]

Oh yeah, otherwise the multiplicities could be almost any number

[vague zephyr]

got it right now thank you

that is true

[tardy crescent]

you're welcome

[vague zephyr]

this is the last question i have, could you help me out with this one too?

[tardy crescent]

Okay, so that graph gives us two zeroes for the function

[vague zephyr]

-2 and 1

[tardy crescent]

yep

[vague zephyr]

1 being the odd and -2 being even

[tardy crescent]

Yes,

[vague zephyr]

how would i generate a polynomial from this graph

[tardy crescent]

You know the factors of the function now, since you know the zeroes and it's a cubic

[vague zephyr]

(x-1)^3(x+2)^2=y?

[tardy crescent]

In this case, the (x - 1) only appears once

[vague zephyr]

oh ok

[tardy crescent]

so you have (x -1)(x+2)(x+2), but there's a bit of a trick

remember that it has to go through point 0, 4

If you plug in 0 for x, you get -4

[vague zephyr]

so 4=(x-1)(x+2)(x+2)?

since in our situation the 4 is the y intercept

[tardy crescent]

When x is zero, yes, but we're missing one thing from our equation: the correct sign of the leading coefficient

So the zeroes are all the same, but since the y-intercept of (x-1)(x+2)(x+2) is (0, -4), we need to flip the whole graph across the x-axis for it to hit (0, 4), by multiplying the whole thing by -1

The factors of your polynomial are definitely (x-1)(x+2)(x+2), but if you want the function to have a y-intercept of (0, 4), you need to multiply by -1:

[vague zephyr]

so solve this equation and flip the sign?

[tardy crescent]

-1 * (x-1)(x+2)(x+2)

[vague zephyr]

or just flip the sign

because I ended up with -x^3-3x+4

[tardy crescent]

When I multiplied it out I got -x^3 -3x^2 +4

[vague zephyr]

yeah i forgot the ^2

[plucky storm]

Are there any online calculaotrs I can use to check my answers for something like this:

[tardy crescent]

ah ok, then you got it

[vague zephyr]

but I answered it and it says its wrong

[tardy crescent]

Hmm, let's see

[vague zephyr]

nope nevermind i forgot to submit lol

[tardy crescent]

Oh okay

[plucky storm]

Here's what I got for 12 if someone can glance at it

[tardy crescent]

Looks good to me

nice job

[plucky storm]

Thanks, I missed the last two classes so I'm trying to catch up on my own

[vague zephyr]

im back with more homework

my instructor posted more problems

[viscid thistle]

Hello! I hope you didn't miss me. I need help with this

Oh, sorry Hak!

[vague zephyr]

It's all good lol, college precalc is never fun especially when you're professor refuses to help you in class with the material. He goes over a few examples on the board and then has people do group work, this is usually left to the one person who knows there stuff (usually me because of tutoring). Then when you have questions he asks you to refer to your group for help... No one knows what they are doing.

[viscid thistle]

:( that sucks

[plucky storm]

That's how my summer class was that I dropped for pre-cal

He did a "flipped" type deal

Which if you don't have a basic understanding of the material prior, you develop bad thought processes while trying to teach yourself, and then the develop the muscle memory using the wrong methods.

It might work in a higher level class, but not intro.

How does this look?

Theta terms in q3, not cos 😂

Oh, I Screwed that up, I wrote it down all wrong

[rich flint]

@viscid thistle you do composition of functions. Plug p(x) into q(x) and you should get x if they are inverses of each other

[tardy crescent]

Hey, sorry I was afk. @vague zephyr , you got the factors of the polynomial correct, but the leading coefficient needs to be different to make the function go through (-1, 32)

[vague zephyr]

Its all good, and thank you

so i solve the equation and I get y=0

not sure where to go from there

[plucky storm]

Alright, I'm pretty sure I messed this one up.

[tardy crescent]

You can write an equation: A*(1)(-4)(-4)(-1) = 32, where A is your leading coefficient, and those parentheses are where you substitute -1 for x in your factors: (x+2)(x-3)(x-3)(x)

So if you plug in -1 for x in your factors, you get a result of -16,

It's kind of tricky because the zeroes are correct, but the actual function is flipped across the y-axis and stretched by a factor of 2

So to get the right function, you can just multiply by -2, to get: -2(x +2)(x - 3)(x - 3)(x). If you plug in -1 for x there, you'll get the 32 that you're looking for, and all the zeroes remain unchanged and correct

[vague zephyr]

that makes much more sense than the approach of trial and error my teacher alludes to

[tardy crescent]

Ah yeah, in cases like those, where you need to get a function through a point, remember that you can change the leading coefficient or multiply the whole function by a constant, and still keep the correct zeroes.

That's usually quicker than trial/error, because multiplying by a constant A is the only thing you can do to it without changing the zeroes

Hi @plucky storm , your answer looks right except for that last part. The inverse sine function won't ever give you a value outside of the 1st or 4th quadrants, so you just have to take the value it gave you and relate it to the 3rd quadrant.

sin (x) = sin (Pi - x)

[plucky storm]

Are you looking at the one that's oriented correct?

[tardy crescent]

Yes

[plucky storm]

The one above I wrote down the question wrong

Alright

Oh, I shared the wrong image. Ugh.

It was asking for Sin and cos theta

Not theta itself... let me take another picture

[tardy crescent]

oh okay

[tardy crescent]

Oh okay, so in this case you're in the third quadrant, where cosine is negative

[trim fable]

oh is this trig

im doing trig too

[tardy crescent]

And sine is also negative

[trim fable]

i have a test tom..

[tardy crescent]

Yes, this is trig

[trim fable]

on it

;-;

[tardy crescent]

Good luck with your test, Star2825. So Xodus, since your angle is in quadrant 3, the values you circled for sine and cosine are actually negative

[plucky storm]

That's correct, I'm not sure why I was thinking Cos was positive

[trim fable]

hey I have a question

[plucky storm]

it would have to term in q2 with sin negative, cos positive

[trim fable]

ill ask after ur done answering theirs so

[plucky storm]

Go ahead, I'm working on other problems while talking and working this one out as well

[trim fable]

ook

whenever i draw the graphs i end up messing up where it starts

and i need help with the equation for that

[tardy crescent]

Okay, so if it reaches a minimum height of 0.25 m above the floor, that would be your lowest value in the cycle, and 2.75 m is your highest value in the cycle, and 1.5 m is your y-intercept

[trim fable]

yeah

i got that

its just i started the graph at its min

then goes up to the Equation of the axis then the max

but the answers started it the opposite like max to min

so like why?

also the period isn't given so how do i solve for that

[tardy crescent]

Oh okay, I think you might need to calculate the circumference of the wheel, since the minimum height is 0.25 and max is 2.75 m,

The circumference is 2.5*Pi

So two full cycles should give you 5*Pi in total distance moved

The period is 2.5pi, since that's one full cycle

[trim fable]

oh

oh

waitt

how did u find that

did u subtract them?

[tardy crescent]

yes

[trim fable]

oh

why

[tardy crescent]

2.75 - 0.25 = 2.5 meters

[trim fable]

yeah but why

[tardy crescent]

Picture the wheel, for it's a circle

[trim fable]

u got the radius

oh so thats

[tardy crescent]

The minimum is the bottom of the circle, the max is the top, and the difference is the diameter

[trim fable]

uh wait idk

ohh

OHH

so the radius is half of that

[tardy crescent]

Yes

[trim fable]

yeah so

why dont u use 2.5/2

for the period

and then 2pi/k=2.5/2

to find k

[tardy crescent]

The period is 2.5pi, because you're measuring the distance traveled, which gives you sin ((2/2.5)x), not accounting for the vertical offset that you'll need to start at 1.5 m

[trim fable]

oh

well thank u

[tardy crescent]

it's not done yet, you'll still need to vertically shift it

[trim fable]

oh

[tardy crescent]

You can shift it vertically to get you to 1.5 m at the y-intercept, and then multiply the sin ((2/2.5)x) by a constant because otherwise the amplitude would be wrong, and you'd only vary between 0.5 and 2.5

[trim fable]

oh ok

[lyric echo]

is precalc hard

[stuck lark]

if your algebra is good then you have a chance

[lyric echo]

hm, dunno if it is

im trying to study this by myself, but i gotta catch up on everything first

my high school math teacher was a literal fraud who couldnt teach math

im stressed thinking about all of this, though

i dont recognise anything

[stuck lark]

try not to overwork yourself. take breaks as needed. feel free to ask around here if you got questions

[lyric echo]

thank you

i hope my math journey will be a consistent and happy one

[stuck lark]

it won't be happy all the time but if you're consistent it may work out

[plucky storm]

For #50, how would i use the half angle ID to find it?

What makes sense would to do sin(180+15), then use the half angle 30/2 for the 15?

Is that possibly what it wants?

Or, maybe not now that I look at the identities

The two examples in the book are useless for this and they both are easy numbers, they use 15 and 22.5

Oh that was obvious actually....

Sin (195) = 1/2sin(390) = 1/2 sin(30)

Now I can use the half angle ID to solve it

[plucky storm]

So Cot(x/2), with tan x = -sqrt(5)/2, given 90<x<180

That would place it in 90/2<x/2<180/2 | 45<x<90. In Q1

Then how could Tan be negative if it's Q1

That's what I got disregarding the signs, but tan would still need to be positive?

[dry dagger]

Dead chat

[viscid thistle]

Is sin is positive and cos is positive at the same time then u'd be in quad 1 thus making tan pos

Ohhh hold up

@plucky storm

there's a rule

[plucky storm]

Based off mine though, what q would it be?

Because I'm pretty sure it's in q1 based off of the given info

[viscid thistle]

whats your question

[plucky storm]

Oh no I got it now

[viscid thistle]

https://tenor.com/view/thumb-thumbs-up-thumbsup-gif-7585355

[plucky storm]

Theta is in q2, but theta/2 is in q1

Since tan(x) would be sqrt(5)/-2

[trim fable]

can someone help me

what would the domain and range be for

y=-0.5sin3x

[viscid thistle]

alright so

what can x be?

what numbers can we put in for x

does 99.4221 work?

[trim fable]

sure

😛

[viscid thistle]

how bout 999999999999999999999999999999999999994221

[trim fable]

lol

YEAH!!

[viscid thistle]

well so we know domain isn't restricted

[trim fable]

yeahh

[viscid thistle]

-inf<x<inf

[trim fable]

oh

[viscid thistle]

so how much can y be

[trim fable]

restricted to

[viscid thistle]

what is the maximum value sin(x) can be

[trim fable]

one cycle actually

[viscid thistle]

y=-0.5sin3x

[trim fable]

max is 0.5

min -0.5

[viscid thistle]

Yea

[trim fable]

ik the range

soo

well

[viscid thistle]

aight

[trim fable]

how about for this

one

wait so

for the one i just asked domain iis: {x|0<=x<=2pi,X(element of the real)}

right

question the only thing that affects the domain is the period and horizontal translation?

[viscid thistle]

im not familiar with that notation but one cycle is not 2pi

consider setting y=0

0=-.5sin(3x)

sin(3x)=0

sin^-1 (0)=3x

[trim fable]

and the only thing that affects range is amplitude and equation of the axis?

[viscid thistle]

sin is 0 when x = 0, or x=pi

but we got 3x

so
0/3 and pi/3

[trim fable]

oh

question the only thing that affects the domain is the period and horizontal translation?
question the only thing that affects the domain is the period and horizontal translation?

lol^^

right?

[viscid thistle]

that's hard to answer as a yes or a no

if you move the stuff to the right of left, yeah

assuming it is restricted and not -inf to inf

[trim fable]

oh

well assuming that

[viscid thistle]

but if it is restricted, horizontal shifts would indeed affect domain

[trim fable]

its always restricted

to one cycle

[viscid thistle]

then the answer is simply yes

[trim fable]

360 = 2pi

YEAHH

ok

I have a test tom for trig ahh

im stressed and ye..

[viscid thistle]

nah bruh you doin the work right

relax

[trim fable]

yeah

[viscid thistle]

you ready

[trim fable]

i hope ;c

[viscid thistle]

ez pz

[trim fable]

sorry lol

my math mark already dropped 6% now im stressed

im good at math but when im stressed i do dumb things

lol

[viscid thistle]

I learned one thing about grades

[trim fable]

?

[viscid thistle]

dont worry about them

[trim fable]

they dont define u

oh ok

lol

are u in uni?

[viscid thistle]

If you tried your best, and you learned something, the numbers aren't going to change

[trim fable]

or high school

oh

[viscid thistle]

uni

[trim fable]

oh

coool

[viscid thistle]

Stress will not help you do better

get rid of it

[trim fable]

yeah

ok

ill try

tho its quite hard

how about for this equation

1.5sin2(x-pi/4)-3 restricted to one cycle

[viscid thistle]

brutal

[trim fable]

period is 180 or pi?

but like with the shift it would be at

[viscid thistle]

,w graph (1.5)sin^2(x-pi/4)-3

[obsidian monolith]

Display results online and refine query

Requested by Babowwz#1394

[trim fable]

135

[viscid thistle]

So you know domain not restricted

[trim fable]

right?

huh

[viscid thistle]

so if it's 180, but loses pi/4, thats pi-pi/4, 3pi/4

suppose y=1.5sin2(x-pi/4)-3

we wanna know how long a cycle is

set it equal to 0

1.5sin2(x-pi/4)-3=0

[trim fable]

yeah

[viscid thistle]

3=1.5sin2(x-pi/4)

[trim fable]

180-45

135

[viscid thistle]

,calc 3/1.5

[obsidian monolith]

Result:

2

[trim fable]

oh

[viscid thistle]

2=sin^2(x-pi/4)

[trim fable]

hmm

yeah help them first

ill try to do the domain

then ask u 😛

if im right

LOL

[viscid thistle]

,calc (10^-(4.1))/(7.9(10^-5))

[obsidian monolith]

Result:

1.005478778132

[viscid thistle]

pretty much equal

so

,calc 10^-4.1

[obsidian monolith]

Result:

7.9432823472428e-5

[trim fable]

ok soooo

[viscid thistle]

wat happen

[trim fable]

domain= {x| +pi/4 <=x<= 13pi/4}

@viscid thistle

[viscid thistle]

me go bed, i'm not sure bout that though

[trim fable]

oh ok

[viscid thistle]

,w period of 1.5sin2(x-pi/4)-3=y

[obsidian monolith]

Display results online and refine query

Requested by Babowwz#1394

[viscid thistle]

wat de fek

anyhow if you need more help you can ping the helpers with @ helpers

@trim fable

but only after 15 mins of wait time

[trim fable]

ok

[warm crescent]

$ S={x|2^x-1$ is prime$}$

[obsidian monolith]

Raftaar:

[warm crescent]

,calc 2^7

[obsidian monolith]

Result:

128

[warm crescent]

Isn't the set of primes and The S equal

[native sequoia]

,calc (2^11-1)/23

[obsidian monolith]

Result:

89

[willow bear]

,calc (2^11 - 1) - 23 * 89

[obsidian monolith]

Result:

0

[serene heath]

Nice

[willow bear]

the tex is painful to look at though

[warm crescent]

Nice dp . Ann

Lol

[willow bear]

dp?

[serene heath]

Pfp

Display picture =pfp

[willow bear]

oh

[viscid thistle]

dp = double penetration. please stick to using pfp when referring to somebody's display picture, friends. thank you

[willow bear]

i tend to say "avatar"

[viscid thistle]

i used that term a lot until i realized its still kinda ambiguous, you dont want those freaking kids to confuse it with the movie avatar haha you know how silly those gosh darn kids can get

[willow bear]

ok boomer

[viscid thistle]

wtf

[serene heath]

lol

[heady jewel]

LOL

[pale kettle]

what do you think?

[hoary valley]

Guys what is " diverges " is it like " doesn't exist " ?

in limits.

[serene heath]

Negation of the definition basically

[rigid sun]

Diverge means converge to infinity

[pale kettle]

It honestly depends how people use it

Some people would say that the limit of sin(x) as x goes to infinity diverges

[hoary valley]

It's the same as Doesn't exist. thanks for trying to confuse me.

[rigid sun]

Could also mean oscillation

Sequence sin(pi/2*n) diverges

[pale kettle]

I mean, some people say that the limit of x as x goes to infinity exists

and the value is infinity

So

[serene heath]

Some people are weird

[willow bear]

i am of the opinion that a sequence approaching infinity has a limit but diverges

or rather, i adhere to that convention

[gusty igloo]

Please help, I’m stumped

[fading token]

!15m

Hmm the bot doesn't work

Don't just ping helpers

#❓how-to-get-help

[hazy hull]

anyone up to help me with this https://documentcloud.adobe.com/link/track?uri=urn%3Aaaid%3Ascds%3AUS%3A3d9745da-ff64-4792-a3a7-a06de39af6eb hopefully in VC?

[willow bear]

@gusty igloo what have you tried and where are you stuck

[viscid thistle]

Can anyone help with this question: A) A team of four is to be chosen from a group of four boys and four girls.
Find the number of different possible teams that could be chosen.
B) Find the number of different possible teams that could be chosen, given that the team must include at least one girl and at least one boy.

I'm not sure if i should use either combination formula or permutation

in A

and B is not clear for me...

i'm guessing that in A) it's combination formula: 8C4 - (C (8,4))

formulas look like this

[hoary valley]

I can't use direct substitution here right?

I have to solve it like this right?

[willow bear]

why would you not be able to use direct substitution on the first limit

last i checked, the real numbers 11 and 0 weren't equal

[hoary valley]

-1

But thanks

[willow bear]

5 - 3(-2) isn't -1.

[hoary valley]

the numerator is -1

[willow bear]

i was talking about the denominator

since that is the only thing that could have caused issues

[hoary valley]

Guys, I think it's True, but Chegg says it's False.. what's going on??

I say it's True due to the denominator equals 0

[willow bear]

consider f(x) = g(x) = x-5

then the limit of f(x)/g(x) clearly DOES exist, and is equal to 1

0/0 is indeterminate

[hoary valley]

Sure, but 0 causes issues in the denominator..

[willow bear]

yes but that DOESN'T MEAN the limit didn't exist.

0/0 is indeterminate

"you can't tell what the limit is or if it exists at all" is not the same thing is "the limit doesn't exist"

"issues" should NOT be taken to mean "nonexistence of limit"

not here

[hoary valley]

This is a True statement.
What's the difference between this and the one above?

I don't see any difference. this might help

[willow bear]

you don't see any difference?

[hoary valley]

No

[willow bear]

not even between the 0 and the 2?

[hoary valley]

Both have 0 in the denominator

[willow bear]

are you blind or just inattentive?

0/0 is a very different form from 2/0.

[hoary valley]

How so?

[willow bear]

the quotient of a function approaching 2 and a function approaching 0 is unbounded.

so at best it goes to positive or negative infinity, and at worst it oscillates with varying degrees of wildness

i guess your book considers an infinite limit as "not existing"

if this isn't obvious then i am not going to be able to explain this to you, especially not past midnight my time.

[hoary valley]

Figure it out..

[willow bear]

...

you are

COMPLETELY

missing the point

[serene heath]

That's not how limits work

[hoary valley]

I guess If it said Undefined it would be True.

[serene heath]

With limits you're concerned with what happens as x approaches a certain point

Key word is approaches

[hoary valley]

Yeah dude

I know

[serene heath]

U dont necessarily care what happens at that point

So u cant just plug in, get an undetermined form and say its DNE

[hoary valley]

Sure, I get what you saying, but then why this is True?

[serene heath]

The indeterminate form doesnt tell u much, the limit may still exist

Or it may not

Is 2/0 indeterminate form?

Ok look at it this way

The top approaches a certain constant value

But the bottom gets smaller and smaller

So the whole thing keeps getting larger and larger

Indefinitely

[hoary valley]

Ok so it exists..

[serene heath]

No cuz here it depends which side u approach 0 from

From the left u get -inf

Right u get inf

Either case its dne

[hoary valley]

Amazing, now why we can't do the same to the other one

[serene heath]

Refer back to Ann's example

[hoary valley]

Wait so if we approached this from the right it will be +infty , and from the left it will be +infty still ? that's why it exists?

[harsh cipher]

Why do we divide by log 5?

[frank willow]

Anyone need tutoring?

[brisk frigate]

Why is the phase shift wrong? I thought I was supposed to set: ((2x)+(pi/5))=0 and that was the answer, maybe the teacher put the wrong symbol? Because if I answer with pi/5 the answer is shown as correct but I think it should be negative like the other examples

Or is the negative symbol from the -15cos affect it?

[eternal lotus]

I think it just wants the magnitude it's shifted, not including the direction

Your direction was indicated from the phase shift

The dilation factor at the front should not have an effect of how much it is shifted left/right

[harsh cipher]

can anyone help me with above question?

[valid violet]

@harsh cipher $\log_b(a) = \frac{\log a}{\log b}$

[obsidian monolith]

gfauxpas:

[valid violet]

where log means natural log but the equation holds for any log on the rhs

any type of logarithm

[rigid sun]

,w plot y=(x^2-2x+1)/x^2

[obsidian monolith]

Display results online and refine query

Requested by maleb1964#6739

[spring garden]

Hey lads, can someone see if I'm Rationalizing the Denominator correctly?

[willow bear]

yeah, that'll work

[spring garden]

Yay thanks

[astral escarp]

Hello, can anyone explain what's happened here

Hello can anyone tell me what happened here?

[serene heath]

they multiplied top n bottom by that long term

[native sequoia]

difference of cubes

then difference of squares

[astral escarp]

thanks!

[spring garden]

Yooo, nothing severe, but can a function's graph have a Horizontal Asymptote And Still have a part of the graph pass through that Asymptote?

Like parts of the graph clearly approch but never pass through, but maybe there's a Tangent-like portion that passes through only once?

[willow bear]

yes a graph can cross its own asymptote

as many times as you want

[spring garden]

Okay cool, heh

[serene heath]

Horizontal?yes vertical? Na

[willow bear]

i can give you an example of a rational function with y=0 as an asymptote which it crosses n times for any given natural number n

$y = \frac{(x-1)(x-2)\cdots(x-n)}{x^{n+1} + 1}$

[obsidian monolith]

Ann:

[spring garden]

Big thanks!

[compact cosmos]

Example where f(x) = x and f2(x)=sin(x) ; f(x) if x >= 0 otherwise f2(x)
https://www.wolframalpha.com/input/?i=((cos(0)x+-+1)+((x+%2B+Abs[x])%2F(2+Abs[x]))+%2B+1)+((sin(x)+-+1)+((-x+%2B+Abs[-x])%2F(2+Abs[-x]))+%2B+1)

A 'nice' way of making piecewise functions / combining sub-functions with different domains into one (if you don't want to use the neat-looking notation. You can replace |x| with sqrt(x^2) if you don't want to use abs.value (for real numbers)
One can shift the function below to add more sub-functions. It is also possible to take a derivative of the whole piecewise using the chain rule (real numbers)

((f(x)-1)(x + abs(x))/(2 abs(x))+1)*((f2(x)-1)(-x + abs(-x))/(2 abs(-x))+1)

[hoary valley]

Why the answer is not +infty?

[pale kettle]

why do you think it is

[compact cosmos]

http://prntscr.com/px1gi2

[hoary valley]

Why you did that ? it didn't have parenthesis... @compact cosmos

It's only x^2

[compact cosmos]

Binomial Expansion

http://prntscr.com/px1hwp

http://prntscr.com/px1i9j

its equal meaning that you can replace it

[hoary valley]

Nice man, they should give you Honorable* role, you are better than 90% of the helpers here @compact cosmos

[viscid thistle]

yo

I need some help

tan 2 theta - tan theta = 0

[compact cosmos]

[tan(theta)]^2=tan(theta)
assume tan(theta) = t
t^2 = t
t^2-t=0
t(t-1)=0; factorization
t=0 v t=1
therefore tan(theta)=0
tan(theta)=1
0
pi/4 +pi*n [real numbers])

[willow bear]

tarp might've meant tan(2θ), rather than tan^2(θ)

[viscid thistle]

^

im also doing sum dif formulas

finding two unit circle points that add or sub to -11pi/12 is hard

[willow bear]

... why would you even need that

[viscid thistle]

using add sub formula

?

[willow bear]

,,,,

why not rewrite tan(2θ) as 2tan(θ)/(1 - tan^2(θ))

then substitute x := tan(θ)

get a rational equation in x

solve it

then go back to θ

[viscid thistle]

oh i was talking about two different problems

i will try that

i got 2x/1-x^2 = x

[willow bear]

no, you got 2x/(1-x^2) = x.

[viscid thistle]

what the difference

[obsidian monolith]

Ann:

[viscid thistle]

ah i see

Im solving it

i think i got somethign odd

-x^2 = 2?

-2

tanθ = sqr2?

[willow bear]

missed at least two solutions

uh

actually i'm not even sure if sqrt(2) is a solution anymore

can you show your work

[viscid thistle]

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

[willow bear]

what happened in that last like

what did you do

[viscid thistle]

is it -x^2+x+2?

[willow bear]

wh

no that's not what i'm asking you

how did you go from 2x = x - x^3 to 2 = x - x^2

[viscid thistle]

lol

[willow bear]

ack

stupid autocorrect

[viscid thistle]

x-x^3/x

oh i see

[willow bear]

no you can't just divide one term on the rhs by x and claim to have divided the rhs in its entirety by x

[viscid thistle]

bc its subtracting

[willow bear]

and you also have to take note of the solution x=0

which you may miss otherwise

[viscid thistle]

so ( (x-x^3)/ x) - 2 = 0

[willow bear]

,,,,,,,,,,,,,,,,,,,

FRACTIONS REQUIRE PARENTHESES IN PLAINTEXT

[viscid thistle]

``

[willow bear]

no not like that

you don't mean $x - \frac{x^3}{x}$ do you

[obsidian monolith]

Ann:

[viscid thistle]

NopE

so
( (x-x^3)/ x) - 2 = 0

[odd helm]

Zeros at 0 and -4

VA at -3 and 2

So I did y=x(x+4)/(x+3)(x-2)

But that doesn’t work

Could someone explain? Thank you

[stuck lark]

think HA

[odd helm]

Ah it’s at 3

But like a part of the graph hits 3

I heard sometimes graphs can touch HA so is that what’s happening here?

[stuck lark]

HA gives end behavior, ie says what y value the graph approaches for big positive x or big negative x

[odd helm]

Alright so it’s not the same as VA it just says what the graph approaches

Thank you

[stuck lark]

no prob

[odd helm]

So if it’s going to be at 3 I just add a 3 to the x right so it would be 3x(x+4)/(x+3)(x-2)

Because 3/1=3