precalculus

[lilac pier]

yeah ann

it is everybody's method!

yeah

you have to know where it's negative and positive

do you know the ASTC trick?

i can show you

idk about that, but ill give you an easier one

[willow bear]

"the ASTC trick"

no

no

x = cos, y = sin is a way better way to think about it

[lilac pier]

it's just a way to memorize, but ann is right, you do have to know how it happens

[willow bear]

yeah it does

first quadrant, x and y are both positive

second quadrant, x is negative, y is positive

third quadrant, both negative

fourth quadrant, x is positive, y is negative

[lilac pier]

A = All, S = Sin, T = Tan, C = Cos

[willow bear]

yeah like. meh.

that tells you the what without the why

[lilac pier]

yeah i know

[willow bear]

so it sucks

[lilac pier]

forget it then

use the x and the y

what do you mean besides the degree ones

yeah so they're easier for you

it'll be positive

tangent is positive in the third quadrant. even if you dont know that, both the negative signs give you a positive sign

[lilac pier]

dont divide that way

1 sec

[willow bear]

there is a lot wrong with this

first off

$\divisionsymbol$ bad

[obsidian monolith]

Ann:

[willow bear]

second, your use of $\implies$ is inappropriate

[obsidian monolith]

Ann:

[willow bear]

that seems to be a poorly drawn 9

or

no

a poorly drawn 2

sup is doing $\frac ab \big/ \frac cd = \frac ab \times \frac dc$

[obsidian monolith]

Ann:

[willow bear]

how do you multiply fractions

do you

not know how to multiply fractions

[uncut mulch]

$\frac a1 \times \frac 1b = \ ? $

[obsidian monolith]

ramonov:

[uncut mulch]

can you multiply those?

[willow bear]

you either can or you can't

go review multiplying fractions

[uncut mulch]

that's the basics of multiplying fractions

[willow bear]

bc this is what you are missing

[uncut mulch]

what about: $\frac ab \times \frac dc = \ ?$

[obsidian monolith]

ramonov:

[uncut mulch]

(yes, a/1 * 1/b = a/b)

yes ok good

well you just demonstrated that you know how to multiply fractions

a/b * d/c = ad/bc

can you apply that here?

[willow bear]

no.

first off

do not use x for multiplication, especially capital X

second, no, -1*2 is not 1.

[uncut mulch]

nowhere did he write: -1 * 2 = 1
he got that after simplyfying

[native sequoia]

$\frac ab\cdot\frac bc=\frac{ab}{bc}=\frac{ba}{bc}=\frac bb\cdot\frac ac=\frac ac$

[obsidian monolith]

EpicGuy4227:

[native sequoia]

idk if that'll help

now let a=-1, b=2, c=-sqrt(3)

see if you understand that Bunny

[uncut mulch]

$\frac {-1}{2} \cdot \frac {2}{-\sqrt{3}} = \frac {\cancel{-}1}{\cancel{2}} \cdot \frac {\cancel{2}}{\cancel{-}\sqrt{3}} = \frac {1}{\sqrt{3}}$

[obsidian monolith]

ramonov:

[hoary valley]

I'm asked to find vertical asymptotes,

My answer is x=3 , x=0 , x=-3

But apparently it's only x=-3 , x=0

why?

[willow bear]

the numerator has a factor of (x-3)

there's a hole at x=3 rather than a VA

[hoary valley]

Thanks !!

[viscid thistle]

How do you figure out the exact value of
csc(-pi/3)

[serene heath]

Can u figure out csc(pi/3)

[viscid thistle]

Not really

All I can think of is sin(3/pi)

But I dont know where to go from there

[serene heath]

Erm

Csc(pi/3)=1/sin(pi/3)

[viscid thistle]

Oh

I think I got it now

Square root of 3 over 2?

All under 1?

[stuck lark]

$\sin(-x)\not\equiv\sin(x)$

[obsidian monolith]

RokettoJanpu:

[viscid thistle]

Im sorry I still dont follow

[stuck lark]

csc(pi/3) != csc(-pi/3)

[viscid thistle]

Would it be

-((2 x squarerootof3)/3)

[harsh cipher]

Hi

how do i know if

(2a^3b^-2)^3 is power of a power or power of a product?

[hexed ermine]

What

[harsh cipher]

reviewing exponent laws

[hexed ermine]

I dont understand power of a power or power of a product

[harsh cipher]

(X^a)^b is power of a power

[fallen cloud]

If you write it that way then it's x^(ab)

You mean

X^(a^b)?

[harsh cipher]

(xy)^a is x^ay^a

which is power of a product

[fallen cloud]

Yeah

[harsh cipher]

my question: how do I distinguish between the two

[fallen cloud]

Parenthesis

[harsh cipher]

so...(2a^3b^-2)^3

is it because we're powering base "a" ?

-_-

[fallen cloud]

Uh

[hexed ermine]

Yeat.exe has stopped working

[green zenith]

can cos^(4)x-sin^(4)x=1-2sin^(2)x be verified?

[hexed ermine]

Sure

Its not the same however

[green zenith]

so its not identity?

[hexed ermine]

No

its cos^2-sin^2

Not ^4

[green zenith]

oh my god

just wasted like 3 hours figuring that out

that is what i noticed

thanks for the response

[native sequoia]

no, cos^4(x)-sin^4(x)=cos^2(x)-sin^2(x)....

@green zenith

[green zenith]

What is that? @native sequoia

[native sequoia]

a true statement

[green zenith]

Hmmm

How is it equal?

[native sequoia]

apply difference of squares on LHS

[green zenith]

Oh my god

😲

[odd helm]

nvm

Wait

[green zenith]

@native sequoia that is insane guy

Thanks

[native sequoia]

np

[final mauve]

Huh? Lemme read the question carefully

[odd helm]

For the problem in yellow

I did (x-1)(x-1) and

Got -4x^3 + x^2 -2x + 1(x+5)

And I thought it’s end behavior was negative and odd but

The problem here gets all of the x degree terms and adds them to get x^6

And idk why

[uncut mulch]

for end behaviour of a polynomial, you only need to focus on the leading term. (which you can get from the product of the leading terms of its factors)

-4x^3 + x^2 -2x + 1(x+5)
you are ignoring the multiplication

-4x^3 (x-1)^2(x+5) = -4x^3 (x^2 - 2x + 1)(x+5)

[odd helm]

Mk got -4x^6

Thank you

[green zenith]

how do i go about solving this one (1/2)sin(2arcsin(x/6))?

how do i multiply arcsin by 2 and sin by 1/2?

[uncut mulch]

use the double angle identity

[green zenith]

oh

i haven't gotten to that i guess

thanks

[uncut mulch]

by solve you meant simplify right?

[green zenith]

i don't know what you meant by simplify but it's to find the the value of the expression

[merry sphinx]

Simplify or evaluate mean the same thing as finding the value of an expression

[green zenith]

ah

that is good to know

i thought simplify was to make expressions less complicated

[uncut mulch]

solve involves finding the value of an unknown variable,
and since you don't have an equation, solve doesn't apply here

[viscid thistle]

I need help with the following problem

A rain gutter is to be constructed of aluminum sheets of 15 inches wide. After marking off a length of 5 inches from each edge, the sides are bent up at an angle of theta.

Express the area A of the opening as a function of Theta.

[uncut mulch]

is this question split into multiple parts?

[rigid sun]

Tf

How we get area if we only have 1 dimension

Am I sped ?

[uncut mulch]

do you have a diagram?
what formulae do you know for calculating the area of a triangle?
@viscid thistle

[viscid thistle]

We can find sin and cos with opposite/hyp and adjacent/hyp no?

[harsh cipher]

hi guys

[uncut mulch]

sin and cos of what?

[harsh cipher]

why do I have to change 8^x to numerator if its already a positive exponent?

[viscid thistle]

the triangles that are formed

[rigid sun]

You don’t

[willow bear]

what 8^x

[uncut mulch]

can you show us your pic?

[harsh cipher]

last part it changed to -8x

[stuck lark]

$a^{-b}=\frac{1}{a^b}$

[obsidian monolith]

RokettoJanpu:

[harsh cipher]

sorry typo

[uncut mulch]

both forms are acceptable

whoops i misinterpreted the shape.

[harsh cipher]

this is first lesson of 8 lesson in the unit

i'm in trouble 🙂

[uncut mulch]

yeh use trig to find the sides

[harsh cipher]

ok I need to look at integral exponent rule again

[uncut mulch]

in terms of \theta

[viscid thistle]

we just need to find the rectangle I think I just did 5 * B

first at least

since its A= Rectangle + 2 Triangles

[uncut mulch]

what expression did you get when you did that?

[viscid thistle]

5*(5sintheta)=25sintheta?

for rectangle

is that correct?

[uncut mulch]

yeh

[viscid thistle]

is that it for the rectangle?

now I just have to find the area of the rec

triangle*

[uncut mulch]

yeh

[viscid thistle]

area of t = 1/2 a*b so can just get cos =a/5 -> a=5costheta

wait what so its just 25/2 sinthetacostheta

I just add after?

since thats just one triangle I would need to make that times 2 right?

[uncut mulch]

yes

[viscid thistle]

so I'd have 25sintheta + 2(25/2 sintheta*costheta) and 2's would cancel out

[uncut mulch]

yeh

[viscid thistle]

what would be next step after

do I factor out the 25 or something

[uncut mulch]

depends on the question,
you can if you want but not needed

[viscid thistle]

it would be 25sintheta + 25sintheta*costheta no?

[uncut mulch]

$A = 25\sin(\theta) + 25\sin(\theta)\cos(\theta)$

[obsidian monolith]

ramonov:

[viscid thistle]

yeahhh not sure what next step would be here I mean like would I have to use identity here

on the 25sin and cos?

[uncut mulch]

this is an expression of area in terms of theta.
unless they want you to write it in a certain form, its sufficient

[viscid thistle]

yeah I think they would just want me to factor out the 25sin theta

and just have it be 25sintheta(1+costheta) i guess

[glass ledge]

aihgt

aight*

zeta

@rare cloud

[rare cloud]

Oh dear.

[vague yew]

aight im here

[glass ledge]

its quick

[rare cloud]

Please save me when I do inevitably fail.

[glass ledge]

you learn about partial fraction decom yet?

[vague yew]

YOU DO THAT IN PRECALC?

[rare cloud]

Oh gosh.

[vague yew]

Oh fuck

[rare cloud]

Save me, Elliott.

[glass ledge]

is that supposed to be a algebra thing

or

[rare cloud]

I don't know crap.

[vague yew]

Pfd comes up in a lot of higher stuff

[glass ledge]

yeah but its simple

[rigid sun]

It’s just reverse common denominator

[glass ledge]

so i don't know the name of it

but you know the A/x + B/x

thing right

[rare cloud]

I don't even know what you're supposed to do, mate. I literally started Precalculus today.

[vague yew]

so you want one of x^3 and 27?

[glass ledge]

well

I thought it was

like

since thats just (x-3)^3

[vague yew]

I'm actually really rusty at this fuck zeta dont look

[glass ledge]

i thought it was A/(x-3) + B/(x-3) + C/(x-3)

[rigid sun]

[uncut mulch]

uh, no. it isnt (x-3)^3

[rigid sun]

[glass ledge]

wait

o yeah

im retarded

lmaoo

[rigid sun]

difference of cubes

[glass ledge]

that explains it

thanks for the help guys

gn

[willow bear]

you killed a puppy

[vague yew]

all in a days work

[rigid sun]

[glass ledge]

[willow bear]

every time someone rewrites $(x+y)^p$ as $x^p + y^p$, a puppy dies

[obsidian monolith]

Ann:

[uncut mulch]

at least one person has that accident every day

[glass ledge]

lmao

all that hype for nothing huh

[vague yew]

shhh

[rigid sun]

Evertime $\sqrt{a^2+b^2}= a+b$ happens, I get turned on

Sheet

F me

[willow bear]

\

not /

[obsidian monolith]

maleb1964:

[rare cloud]

I've done that more than once.

[willow bear]

ew

[rigid sun]

My calc teacher would yell at you if you did this

and make fun of you

[rare cloud]

Good thing I'm not at Calculus yet. Still have some time to fix my mistakes.

[rigid sun]

That’s what we all thought

[glass ledge]

what are you learning rn zeta

o btw the teacher speeds it up so that the juniors/seniors finish earlier so they have time to study for SAT/AP test

[rare cloud]

I'm starting to self-study Precalculus. As in I just started today, so I'm super sorry that I can't help.

[glass ledge]

wait

what are you in school

ib4 you say 7th grade or some shit

[rare cloud]

10th. My school curriculum is messed up so we don't really do the whole Algebra to Precalculus to Calculus stuff.

I'm also pretty dumb, so there's no way could've done that in 7th.

[glass ledge]

ay

im 10th too

my school (for regular track) is Algebra - Geometry - Precalc-calc

prolly same huh

[rare cloud]

I suppose so.

[languid crane]

we dont even have precalc

just algebra stuff and then geometry and then calc

[rare cloud]

The nth root of a^n is |a| if n is even. Does this hold if we, for instance take the nth root of a^m, if m and n are both positive, even integers?

[willow bear]

The nth root of a^n is |a|.
only for n even

[vague yew]

yeah

[rare cloud]

Right, that's what I meant. My bad.

Let me fix that.

Oh wait, never mind. I just answered myself by looking at the next page of my book. My bad.

[vague yew]

is -5^2 pos or neg, and is (-5)^3 pos or negative?

[rare cloud]

Both are negative.

[vague yew]

ye

[hoary valley]

So I solved the problem in green circle with the "difference of cubes formula" that I posted beneath it..
I did the same to the problem in the red circle but It didn't work.. Why?

[willow bear]

$x^3 - 9x$ isn't a difference of two cubes.

[obsidian monolith]

Ann:

[willow bear]

$x^3 - 9x = x(x^2 - 9) = x(x-3)(x+3)$

[obsidian monolith]

Ann:

[willow bear]

@hoary valley

[hoary valley]

@willow bear But how did you know that it's not a difference of two cubes? $x^3 - 9x$ looks like $x^3 - 27$

[obsidian monolith]

♥ Joshwa ♥:

[willow bear]

no it doesn't.

i mean

sure you can write $x^3 - 9x$ as $x^3 - (\sqrt[3]{9x})^3$ but who does that

[obsidian monolith]

Ann:

[rare cloud]

9x isn't a perfect cube so I don't see why you'd want to factor it as such.

[willow bear]

it's just pointless

[rare cloud]

Yeah.

[hoary valley]

Ok thanks guys!

[hoary valley]

Guys, I replaced X with 2 without simplifying anything, since the denominator isn't 0 , Is that ok?

Or is that the wrong way to solve it?

[willow bear]

no that's exactly the right way to do it

[hoary valley]

Thanks!

[viscid thistle]

@rare cloud https://www.youtube.com/watch?v=9OOrhA2iKak&list=PLDesaqWTN6ESsmwELdrzhcGiRhk5DjwLP
These can be incredibly helpful for precalc, and he's still adding videos to that playlist. He also has complete playlists of Calc 1, 2 and 3 lectures. Wonderful lecturer. Hope this helps. :)

[rare cloud]

I'll have a look later. Thanks!

[viscid thistle]

Good luck!

[rare cloud]

Thank you! I'll certainly need it.

[tame wedge]

I want to graph |x(x-2)|, I tried to find where the function is negative by setting x(x-2) < 0. When you solve you get x<0, x<2. Why is this wrong?

[serene heath]

By definition your function is never negative

[tame wedge]

Yes I understand, but if we let it be negative I can find where it is so to be able to graph it, no?

[serene heath]

No clue what you mean

[tame wedge]

Well, if I know where f(x) is negative, I figured I could graph it.

[willow bear]

|x(x-2)| is never negative

[serene heath]

By definition your function is never negative

[tame wedge]

So should I instead set x(x-2) >= 0?

[serene heath]

Are you trying to graph the original function?

[tame wedge]

Yeah, |x(x-2)|\

[serene heath]

Ok yes u can find where x(x-2) is negative

Then reflect across the x axis

[tame wedge]

Can I use null factor law with inequalities?

Because its a law when ab = 0 not otherwise

[serene heath]

U can split it into cases yes

[rare cloud]

x(x - 2) = 0 when x = 0 and x = 2. For x(x - 2) < 0, roughly sketch the shape of a quadratic and notice that 0 < x < 2 gives a negative value of x(x - 2).

[tame wedge]

I want to solve x(x-2) >= 0, the answers are x>=2, x<=0

I dont understand why x<=0, the sign gets flipped.

[serene heath]

If x is less than 0 then so is x-2

Negative times negative is postive

[rare cloud]

I was taught this way. Since the equation is an upward-facing quadratic equation, it has this shape, and is negative from 0 < x < 2.

[tame wedge]

Thanks guys

[rare cloud]

Why is this (3a + 1)√a and not (3|a| + 1)√a?

[willow bear]

because the presence of that sqrt(a) term requires a >= 0

[rare cloud]

Ah. Yes. You're right. Thanks.

[odd helm]

For this I get how you get the equation for the graph because you use the x coordinate when y=0

But why is it (x+1)^2 and not just (x+1)?

[undone drift]

there is thing for odd and even powers of the roots

https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:poly-graphs/x2ec2f6f830c9fb89:poly-intervals/v/polynomial-zero-multiplicity

for even powers, the graph will just touch the x axis. For odd powers the graph will cross through...

[blissful wadi]

any help with this limit

$\lim _{x\to \infty } (\sqrt{1+x^2} +x) \cdot arctan (\sqrt{1+x^2}-x ) $

[obsidian monolith]

dvaix:

[willow bear]

\arctan

so what have you tried so far

[blissful wadi]

multiplying by the conjugant

getting a minus out from the arctan

but it only worked on x--> -inf

[willow bear]

$x \to -\infty$

[obsidian monolith]

Ann:

[willow bear]

anyway

[blissful wadi]

( oh thanks just started lreaning maths latex )

[willow bear]

the inside of the arctan is 1/(x + sqrt(1+x^2))

so you can substitute $t := x + \sqrt{1+x^2}$ and be left with $\lim_{t \to +\infty} t \arctan(1/t)$

[obsidian monolith]

Ann:

[blissful wadi]

ooooh

got it got it thankks

[long mason]

,w plot f(x) = |x(x-2)|

[obsidian monolith]

Display results online and refine query

Requested by Goëtia#9527

[long mason]

https://cdn.discordapp.com/attachments/363224154469826562/641966142868422656/20191107_194337.jpg

[rigid sun]

,w plot f(x) = 1+x^2

,w plot f(x) = 1+x^2

[obsidian monolith]

Display results online and refine query

Requested by maleb1964#6739

[rigid sun]

,w plot f(x) = sqrt(1+x^2)

[obsidian monolith]

Display results online and refine query

Requested by maleb1964#6739

[rigid sun]

,w plot f(x) = sqrt(1+x^2)-x

[obsidian monolith]

Display results online and refine query

Requested by maleb1964#6739

[hoary valley]

What's a hole?

[lilac pier]

X cant be -3, otherwise the function will be undefined. There's no hole, it's just gonna be empty

[uncut mulch]

when simplified you get 1/(x+3)
however due to the form of the original expression, it is undefined when x=5

i.e. there will be a hole / removeable discontinuity at (5, 1/(5+3))

[rigid sun]

?

What

[odd helm]

So someone sent me a video yesterday explaining why it’s (x+1)^2 and not just (x+1)

It’s because this graph shows that x=-1 has an even multiplicity

Since the y values don’t switch their sign

It continues to be positive until x=2

But how do you know it will be (x+1)^2? Why couldn’t it be like (x+1)^4 or (x+1)^6

Those are even numbers right?

[patent beacon]

Those are possibilities, yeah

[uncut mulch]

"a possible equation"

^

[patent beacon]

I would mark those correct if I were marking

[odd helm]

Alright

Makes things easier

Thank you

[viscid thistle]

Anyone know how to solve find the zero and inequalities problems?

[wise kelp]

Do you have an example?

[viscid thistle]

Here's inequalities

[rare cloud]

Here's the method I follow to do those.

You just write each factor out and check their signs within each interval. This is because the sign of the factors don't change within each interval.

[scenic musk]

,w plot f(x)=abs(x)

[obsidian monolith]

Display results online and refine query

Requested by mike0x81#3484

[rigid sun]

,w plot f(x)= 69.420

[obsidian monolith]

Display results online and refine query

Requested by maleb1964#6739

[viscid thistle]

nice

[rare cloud]

Is there some specific character for this, or is it two brackets squished together?

[serene heath]

$\floor{x}$

[obsidian monolith]

the one n only:

[rare cloud]

Wait, what? Aren't those two totally different-looking things?

Does that also basically mean to round down to the nearest integer?

[serene heath]

Yea

Its called the floor function

[rare cloud]

Oh. Should I follow the textbook, or that floor notation?

I suppose that makes more sense. The bar at the bottom tells me to round down.

[uncut mulch]

its more intuitive and probably more recognisable

use whatever you'remost comfortable with / what your teacher accepts

[viscid thistle]

hello guys, i beg for help, there's this exercise that seem simple but i still can't grasp it, anyone who can help would appreciate that. here's how it sounds: The equation x^2 -5x -7 = 0 has roots alpha and beta. The equation x^2 +px +q = 0 has roots alpha + 1 and beta + 1. Find the value of p and the value of q.

wanna add that I don't need the answer, i need the step-by-step process to be shown, or at least 'shortly explained'

[patent beacon]

Find the roots of x² - 5x - 7. That's α and β.
Use α + 1 and β + 1 to construct the other quadratic

[viscid thistle]

that is wrong, there's something else i have to do

idk is it really right to use abc formula here?

i'm confused, cuz it says that i'm already given roots which is alpha and beta

and the other equation has roots alpha+1 and beta+1

so when i have the value of alpha and beta, how do i then go kind of backwards to find the equation

[patent beacon]

x² - 5x - 7 has roots that you can find numerically. I don't see why you'd want to avoid doing that, just because they called those roots α,β

[viscid thistle]

if i were to use abc formula

lets say i have the numbers for alpha and beta

what do i do then

[patent beacon]

A quadratic with roots r1 and r2 is (x - r1)(x - r2)

You can multiply that by any real number, but you won't want to here

[viscid thistle]

does (x - (alpha in my case))(x - (beta in my case) apply to all quadratic equations?

(x-r1)(x-r2) ......... - why is it minus in the brackets

[patent beacon]

Let's say we have the quadratic
(x - r1)(x - r2)
What do you get if you sub r1 into it?

[viscid thistle]

x^2 +r2x -r1x +r1r2

damn did it wrong

[patent beacon]

That's what you get if you expand the quadratic, I'm not asking for that. Instead, what do you get if you sub r1 into it?

[viscid thistle]

sub r1 into, wdym?

[patent beacon]

"sub r1 into it" is a common way to say "replace all x with r1"

[viscid thistle]

so like (x - (alpha+1))

[patent beacon]

Are you okay with "why is it minus in the brackets"? I was trying to answer that

[viscid thistle]

yes of course i'm okay with that, the reason why i'm asking this is that i have no idea what i would do if there was another similar exercise like this one, but anyway (x - (alpha+1)) (x - (beta+1)), i sub the roots alpha and beta that i got from abc formula, what do i do then

[patent beacon]

That's perfect. Expand it, you'll get a regular quadratic

You can just match up p and q at that point

[viscid thistle]

is that really all i have to do? let me summarize it up again, so the equation x^2 +px +q = 0 is equal to (x - (alpha+1)) (x - (beta+1)?

that will give me the values of p and q when i get there

and signs have to be the same right

which are all positive

[patent beacon]

I don't know anything about the signs, I don't believe there's a restriction there

[viscid thistle]

i mean "the equation x^2 +px +q = 0 has roots alpha+1 and beta+1

so it means that equation should have plus signs and all that etc. no?

[patent beacon]

Yes, the equation will have plus signs

[viscid thistle]

okay, i'll try it all now, i'll see if it works 8)

as i said, really nothing hard, but still complicated at the same time

[patent beacon]

There's another way to go about this. The second polynomial's roots are just the first, but shifted 1 to the right.

That must mean the second polynomial is the same as the first, but shifted 1 to the right

[viscid thistle]

makes sence somehow ^

i'll see that as well

[patent beacon]

The second polynomial is then (x - 1)² - 5(x - 1) - 7. Simplify that, p and q follow

[viscid thistle]

yes! that's kind of the way my classmate solved it, i saw exactly something like this, although memory is vague so i dont remember much, i just quickly overlooked his paper

but that should make sence, wait, i'll try both ways anyway

[patent beacon]

Good luck, let me know if you need anything else

[viscid thistle]

sure! this was already a big help, i'll definatelly ask more, thank you vm for help!

it works like spreading butter on bread, thanks a lot once again!

okay it's kind of awkward but got one more that i'm stuck on

7x^2 -8x +p = 0, (p in the set of Q), one root is three times the other root. find the value of p

[patent beacon]

@viscid thistle
Despite the setback of not knowing p, you can still use the abc formula to get information about the roots

[viscid thistle]

don't know how to solve it since p is x here, how do i sub that into abc formula

i mean yeah, i just say that my (C) is (X) unknown number

i get the answer but

[vernal rapids]

Treat p as a constant

Like you would to with c

[viscid thistle]

so can i just say my p is 1?

[vernal rapids]

Nope

Leave it as p

[viscid thistle]

as a constant"

i'm not sure i understand

leave it as p?

[vernal rapids]

What is the general (or abc) formula?

[viscid thistle]

(-b)+-sqrt(b^2-4(a)(c)) all that over / 2(a)

[patent beacon]

Just put p into the formula, in place of the number

[vernal rapids]

a=7
B=-8
C=p

[viscid thistle]

yeah i get the answer where sqrt has 4(a)(x)

uhm okay kind of hard to imagine this, lemme do it like this

since the equation is 7x^2 -8x + p = 0 (i plug values in abc formula) i get 4+sqrt(16 - 7x) / 7

^ 7x?

is that right then

x is p

[patent beacon]

So the roots are
8 ± √[64 - 28p] all over 14

[viscid thistle]

yes^!

[patent beacon]

Please don't let x be p, seeing as x is the variable not the root. You'll get confused in no time

[viscid thistle]

oh its only because i use calculator, and there's no p really

i write it as p of course

so then, moving to the main point, one root is three times the other root

or... one root is 3x times the other root

does this really work with abc formula

[patent beacon]

This is really messy to write

[viscid thistle]

: D

i understand

[patent beacon]

I was hoping the roots were simpler lol. But, that's both roots. One is 3x larger than the other. That's enough to make an equation

Where p is the only unknown

I don't like the logic. Let's try this instead. Factored, it will look like:
7(x - a)(x - b)

[viscid thistle]

yeah roots look messy

7(x-a)(x-b) : / wait i'm thinking

[patent beacon]

7x² + 7(-a - b)x + 7ab

[viscid thistle]

a and b are roots?

[patent beacon]

Yes

[viscid thistle]

why are they both minus

wait

factor only 7x^2?

oh no

factor the whole equation

right

[vernal rapids]

They are minus because the product will be 0 if one of the terms are 0

So (x-a) or (x-b) must be zero, in other words
X=a or x=b

If they were (x+a), then x would have to be -a

[viscid thistle]

ooowh

and you say just (x-a)(x-b) because you dont have the value, of course : P so i will get some values a and b from factoring that equation right?

[patent beacon]

That's already factored. We want to expand it

I say 7(x - a)(x - b)
Because the leading coefficient will have to be 7

[viscid thistle]

but it's 7x^2 - 8x +p = 0 it does have values

ooowh

i got you

i understand now

[patent beacon]

After expanding my form:
7x² + 7(-a - b)x + 7ab

This implies
-7a - 7b = 8

Because we want that middle term to match up

[viscid thistle]

just a sec, i'm simplifying myself for the sake of practice

so the "7(-a -b)x" is = 8

[patent beacon]

It would have to be, if the factored form we made up is equal to the quadratic the question is giving us

[viscid thistle]

yes okay

and this is again, two roots?

or just one root

"-7a -7b" are two roots?

because now i'm confused again, the main point was one root 3x the other, i still dont know what to do

hmm, 7x^2 -7ax -7bx + 7ab, is that how it looks now

[patent beacon]

a and b are each a root, yeah

Yes, you have that correctly expanded. See how you can get:
7x² + 7(-a - b)x + 7ab
From there?

[viscid thistle]

so i looked back again, and thought; don't we have use 7(x-r)(x-3r)? 3r because one root is 3 times the other?

[patent beacon]

Oh true. That's essentially the way I was about to go

By letting b = 3a

Fair, your way is cleaner

[viscid thistle]

owh okay, yeah sorry : D i mean your way isn't bad either if you're gonna do that step later

lets continue your way now, so it would be 7x^2 + 7(-a -3b)x + 7ab

[patent beacon]

Nah yours is better. After expanding
7x² - 56rx + 21r²

[viscid thistle]

oh damn, well i see, calculator kind of couldn't find it, more like; showed nonsense

so maybe it was wrong to go your way?

[patent beacon]

I don't know what you're putting into your calculator, or even how to enter this into a calculator?

[viscid thistle]

owh right, it's an app

"photomath"

that's what im using

it shows everystep, you can do almost everything there, from sums to integrations to derivatives

etc. but nevermind that now

[patent beacon]

Well, we want to make the middle term 8

So that our quadratic matches the question's quadratic

[viscid thistle]

sorry, idk if thats true, but i got 7x^2 -28rx + 21r^2

[patent beacon]

,w expand 7(x - r)(x - 3r)

[obsidian monolith]

Display results online and refine query

Requested by Kaynex#7520

[patent beacon]

Oops, good call

That's why you don't do this in your head

[viscid thistle]

calculator showed it, but for the sake of practice : D gimme a sec

: D

yes okay, actually did it on paper, what an achievement well anyway

okay. we got this weird "non-quadratic" now?

[patent beacon]

We built a quadratic that has roots r, and 3r.

And a leading coefficient of 7

[viscid thistle]

yeah

[patent beacon]

The question is asking about the polynomial
7x² - 8x + p

[viscid thistle]

yeah

[patent beacon]

Our polynomial is
7x² - 28rx + 21r²

[viscid thistle]

no wait, the value of p

[patent beacon]

We want them to be the same polynomial

For that to be true,
-28r = -8

That logic make sense?

[viscid thistle]

we get r as 2/7

[patent beacon]

Yus. Therefore, in order to make these polynomials the same, we're forced to set r. We know the roots!

[viscid thistle]

8/28 reduced.

[patent beacon]

Finally, since these are the same polynomial,
p = 21r²

[viscid thistle]

oooo

so i test if mid-value of the new pylonomial i've got is the same as from the original one

if it is, then 21r^2 is p, as the last coeffecient

from this new equation

[patent beacon]

Yes. All of the coefficients have to be the same, if they're the same polynomial

So most of our work is fitting the coefficients

[viscid thistle]

oookay so that's how you test it

okay, interesting, that was fun though

not gonna lie, learned here more than in school

[patent beacon]

These are pretty tough questions for someone learning about quadratics

[viscid thistle]

i'm 11th grader

[patent beacon]

Good luck oop

[viscid thistle]

thanks sir, or mam : )

no, mate fits here if i dont know the gender

^ not math related stuff

xD

[patent beacon]

Np at all, feel free to ask if you have anything else

[viscid thistle]

sure!

[viscid thistle]

okay looks like i got one more

consider the function f(x) = (1-k)x^2 + x + k, x in the set of reals. Find the value of k for which f(x) has two equal real roots

looks tough, i can't hold it anymore i'm gonna go to sleep, gonna check tomorrow if anybody have answered this

[rigid sun]

Uhhh

@viscid thistle

Remember the vertex form of a quadratic, we will simply use that form, but the k component of that is 0

Aka, the quadratic in vertex form will look something like this

$(gx +n)^2$

[obsidian monolith]

maleb1964:

[rigid sun]

Now, since our quadratic is reduced to vertex form, we can conclude that g =

$\sqrt{1-k}$

[obsidian monolith]

maleb1964:

[rigid sun]

This is justified because if g is the sqrt(1-k) then (gx)^2= (1-k)x^2

now, we’ll need to establish the relationship between n and k, similar to how we made g=sqrt(1-k)

notice that for any given quadratic, say (x+3)^2, the last term of the quadratic is n^2,

$(x+3)^2=x^2+6x+9$

[obsidian monolith]

maleb1964:

[rigid sun]

in this case n=3

n^2=9

and the last term is 9

similarly, we can take (gn +n)^2 and establish that k, the last term of the quadratic has k=n^2

the final thing we need to do is relate everything back to the equation

for a vertex quadratic , b always equals 2ng*x

We can deduce this from unfailing the vertex form to standard form

(gx+n)(gx+n)

(gx)^2+ 2gxn+n^2

notice the b term, the term with x to the power of 1, is 2xgn

in the equation in the problem, the b term is just x by itself

that means, 2xgn=x

Therefore, 2gn=1

g=1/2n

and now, we can begin to substitute

$g=\sqrt{1-k}$

[obsidian monolith]

maleb1964:
Compile Error! Click the reaction for details. (You may edit your message)

[rigid sun]

$\sqrt{1-k}=1/(2n)$

[obsidian monolith]

maleb1964:

[rigid sun]

$k=n^2$

[obsidian monolith]

maleb1964:

[rigid sun]

$\sqrt{1-n^2}=1/(2n)$

[obsidian monolith]

maleb1964:

[rigid sun]

square both sides, then multiply the denominator of the right to both sides

$1-n^2=1/(4n^2)$

[obsidian monolith]

maleb1964:

[rigid sun]

$4n^2-4n^4=1$

[obsidian monolith]

maleb1964:

[rigid sun]

Rearrange, the resubstitute k=n^2, move everything to the left so it is all equal to zero

$-4k^2+4k-1=0$

[obsidian monolith]

maleb1964:

[rigid sun]

Using factor methods, or the quadratic formula, k=1/2

Personally, I factored

I used the split method to split 4K into 2k+2k

$-4k^2+2k+2k-1=0$

[obsidian monolith]

maleb1964:

[rigid sun]

Group

$2k(-2k+1)+ -1(-2k+1)=0$

[obsidian monolith]

maleb1964:

[rigid sun]

$(2k-1)(-2k+1)=0$

[obsidian monolith]

maleb1964:

[rigid sun]

Both linear factors will lead to the same answer

$2k-1=0$

[obsidian monolith]

maleb1964:

[rigid sun]

$2k=1$

[obsidian monolith]

maleb1964:

[rigid sun]

$k=(1/2)$

[obsidian monolith]

maleb1964:

[rigid sun]

You can graph the function y= (1-.5)x^2+x+.5 and see it only touches the x axis once, aka 2 real roots that are equal

,w plot f(x)= .5x^2+x+.5

[obsidian monolith]

Display results online and refine query

Requested by maleb1964#6739

[rigid sun]

@viscid thistle you’re gonna have to scroll up a long way

[rare cloud]

Am I understanding the following correctly?

f(a) is a local maximum value if f(a) is greater than or equal to f(x) for all x in some open interval (b, c), that contains a.

[patent beacon]

I don't know if you're understanding it, but the statement seems correct

In fact, the statement seems advanced. Where's that from?

[rare cloud]

From my Precalculus text. I think I understand it as because if it were a closed interval, if we have [a, b] and we take f(a), like on the edge of a graph from a to b, then it wouldn't mean it's a maximum, as there could be something directly to the left of a that produces a greater value of f. Whereas if a is in an open interval, where there is no single minimum or maximum value, it has to be the highest point in its surrounding?

[patent beacon]

@rare cloud
Yeah that's a good understanding

Almost as if that came out of an analysis textbook

[rare cloud]

Ah. I suppose it's intuitive enough.

[rare cloud]

Since g(4) is undefined, is the answer for (d) is [-4, -2] U [2, 3] (what I think), or [-4, -2] U [2, 4]?

Ah. Thanks. Just wanted to check.

[rare cloud]

I'm trying to find the minimum value here, which apparently is 7.5mi/h, but I seem to just get a cubic graph when I plot this function. Am I doing something wrong?

The graph keeps decreasing towards negative infinity, so I have no idea how I'm supposed to pick a minimum.

[willow bear]

are there any constraints on v

[rare cloud]

"A fish swims at a speed v relative to the water, against a current of 5mi/h. Using a mathematical..." is all that's on the previous page. Nothing else.

I'm assuming v > 5, but even then, there's no minimum.

Is it just 5 × 150%? If so, that's dumb.

[willow bear]

actually no

there is a minimum if you constrain this to v > 5

[rare cloud]

Wait, really? Hold on.

[willow bear]

$\dv{E}{v} = 27.3\frac{3v^2(v-5) - v^3}{(v-5)^2}$

[obsidian monolith]

Ann:

[rare cloud]

This is supposed to be Chapter 2 of Precalculus though.

[willow bear]

h

wait how tf are you supposed to even find the minimum if you don't have access to calculus

[rare cloud]

Graphically. Maybe I just need to find the right viewing window. One moment.

[rare cloud]

Oh my! So sorry.

I didn't think to check upwards of 4,000 to find the other piece of it.

My bad!

Thank you so much!

I guess I just need to be more careful next time. So sorry for wasting your time.

[viscid thistle]

@rigid sun thanks a lot!

[obsidian monolith]

Kanjee:

@PhD in Quick Maff still i dont understand how did you know or get that it's $\sqrt{1-k}$

[viscid thistle]

you simply said and verified that it's

$\sqrt{1-k}$

[obsidian monolith]

Kanjee:

[viscid thistle]

but how did you get that

cuz i personally wouldn't think of it myself

(g)^2 = 1-k, then g = sqrt(1-k)... am i stupid or this is something that i'm supposed to know everytime when i bump into these kind of problems

[willow bear]

i mean...

what do you think sqrt() even is?

[viscid thistle]

ehh i know what it is: the number multiplied by the same number gives me what's inside the sqrt

but these cases are extreme for me, when in other hand it's easy for you

but anyway

not blaming anyone

[willow bear]

sqrt(x) is the (positive) number that when SQUARED gives x

(sqrt(x))^2 = x

is the definition

[rare cloud]

Sorry if off-topic, but I'm curious - why are you not using the discriminant to solve that question?

[viscid thistle]

because someone like @rigid sun guy gives me something more complicated xD

[rare cloud]

Ah. Seems like discriminant is an easier way of tackling that question.

[viscid thistle]

and i just figured out i can just use the discriminant instead of all this fuss he gave me, but i consider it valuable as well because he's doing something more complicated

but i dont get how he's thinking through all that

he/she

is discriminant all we need to know

for these kind of problems

[rare cloud]

Yes.

Because in $\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, if $b^2 - 4ac = 0$, then both roots have to be equal.

[obsidian monolith]

Zetanoxic:

[viscid thistle]

didn't know this fact... D: i lack practice i guess

but thank you for that

[rare cloud]

Practice definitely helps.

I just learnt about function composition today. Pretty sick stuff.

[viscid thistle]

oh yes, you are talking about those f(g(x)) right

[rare cloud]

Yeah. Never learned that in school yet. Pretty cool.

[viscid thistle]

and when it comes to complex numbers, which we are gonna use composition functions with

and here i can't solve vertex xD

oh hilarious. but yeah pretty sick stuff

[rare cloud]

I barely got started with complex numbers. Only know how to solve quadratics and general arithmetic with complex numbers.

[viscid thistle]

i mean, there's only imaginary part (i) and and real part

(i) is just sqrt -1

[rare cloud]

There's still a few components of complex numbers I haven't covered.

Especially De Moivre's Theorem. Still have 6 chapters before I reach it though.

[viscid thistle]

what grade r u?

if not personal

cuz i'm 11th grader

i wanna know the level of other countries

in math

cuz a country like china, knows this stuff what i learn now in like 6th grade xD

[rare cloud]

I'm in 10th.

Trying to self-study a bit of Precalculus.

[viscid thistle]

ohh you see. and you did better what i couldn't in 11th grade, truly there's inequality in all countries

i dont think im mistaken

[rare cloud]

Not really. It just happens to be something I've learnt in the previous chapter of this book.

[viscid thistle]

self study is what i always did at here, books just doesn't help, they're complicated to understand : P

or because i lack mindset

could be both idk but you get the point

[rare cloud]

Do you use Khan Academy? That saved my life back in the day.

I do like books, but it's usually easier to understand when someone explains it verbally.

[viscid thistle]

oh yes, i agree with verbal explaining part

i used to use khan academy, but know i watch "organic chemistry tutor" instead

he covers everything as well, if not more than khan academy

that guy makes 2-3 hour videos of only one topic with lots of exercises

the problem is then with your own memorization

[rare cloud]

Ah. Khan Academy does have pretty basic exercises.

[viscid thistle]

yeah i think i'm gonna go through his exercises as well, i just can't move on not knowing basic stuff first

[rare cloud]

Right. Math tends to build on previous knowledge, so having strong foundations is definitely important.

[viscid thistle]

^yes

[rare cloud]

I'm hoping to finish Chapter 2 of my Precalculus text today. Might not have time to finish the test at the end though.

[viscid thistle]

i wish you best luck though. : )

[rare cloud]

Thanks! Hope everything goes well for you as well.

[viscid thistle]

Thanks : ) !

i'll do my best

[rare cloud]

I think I'm being too ambitious trying to finish this 1108-page text by the end of the year, but I'll try.

Man, so many graphs that I need to sketch. Not a huge fan of those.

[viscid thistle]

i never really liked functions in general xD

but they're like most important in math, ah well too bad.

i'm more interested in derivatives and integration, limits etc.

which is linked to functions as well, but it's more fun lol

[rare cloud]

At least you know what you're interested in. I don't have a damn clue.

All I know is that it's not sketching a million graphs of transformations.

[viscid thistle]

haha, true ^ doing Sums, or at least trignometry, is much more fun than functions

always better

at least in my opinion

[rare cloud]

Series are in Chapter 12 for my book. It'll be a while before I get there.

[viscid thistle]

oh you haven't learned series yet?

yeah they will be a biiiiiit hard at the beginning

but always easy to grasp if you practice

[rare cloud]

I certainly haven't. Only covered Fundamentals and Functions chapters so far.

[viscid thistle]

yeah well you'll see it, it's interesting i can tell that much : D

[rare cloud]

Yeah. Looks fun!

[languid crane]

@viscid thistle functions are fun

pun intended

but yeah I like calculus more though'

[rare cloud]

Aha!

I see what you did there.

[viscid thistle]

if you have good skills and understand a lot, i agree they're hella fun : )

[languid crane]

lol i dont understand a lot prollly

but theyre still fun

same with calculus

and trig

[rare cloud]

I probably could've learnt all this cool stuff if I weren't such a useless, lazy person. Better start now. Have loads of cool stuff ahead.

[languid crane]

true

[viscid thistle]

true

indeed

[languid crane]

I cant wait to get into exploring advanced math

but first ofc need to have the basics down

[viscid thistle]

yeah yeah

me too, i wanna jump straight into calculus 2

xD is that weird

[languid crane]

its not weird

[rare cloud]

I'm really tempted to jump in, but I need to learn sums, trigonometry and polar coordinates, so I'm just going to be patient and learn this prerequisite stuff first.

[languid crane]

I wanna move to calc 3 or linear algebra or heck even diffy geo or pdes but i need to study calc 2 first

and other stuff

[rare cloud]

I think I'll go for Calculus, then Linear Algebra, then who knows.

[viscid thistle]

calculus 1, there's nothing more than integration, derivatives, limits, functions, sums and trigonometry right

[languid crane]

does anyone have a practice test for precalc which only includes things like trig, polar coordinates etc (so no log functions etc)

cuz id like to revise those

[viscid thistle]

nope sry

[languid crane]

@viscid thistle should be right

[rare cloud]

I haven't even gotten there yet. Sorry.

[languid crane]

not really sums either

[viscid thistle]

i've gotten into sums, but i have no idea where i put the previous test from 3 months ago

xD

[languid crane]

what do u mean by sums

[viscid thistle]

test with sums

[languid crane]

... show me an example

[viscid thistle]

yeah... no idea where it is : P

ehm

[pale kettle]

you mean like convergence tests right

[languid crane]

are those calc 1??

i thought those were calc 2

[rare cloud]

Can someone help me with the intuition behind exchanging x and y with inverse functions?

Like, for example, f(x) = x³ - 1.
To find the inverse, we let y = f(x).
y = x³ - 1
x³ = y + 1
x = (y + 1)^(1/3)
[•] Exchanging x with y,
y = (x + 1)^(1/3)
f^(-1)(x) = (x + 1)^(1/3)
I'm questioning the step marked with [•].

Correct.

Oh! The inverse function of f plugs in a value of x to give its inverse!

Ah! I think I got it.

Thank you! My brain needed a kick there.

[languid crane]

what

I thought the x = (y+1)^1/3 would be the inverse

why not??

tf does f^(-1)(x) mean

[rare cloud]

The inverse function of f?

[languid crane]

huh

[obsidian monolith]

JY1853:

[rare cloud]

Yeah. That.

[languid crane]

but isnt the inverse f(y) thing

[rare cloud]

I have no clue. That's how I'm supposed to do it in my textbook though.

[languid crane]

f(x) aint x

its y

so f^(-1)(y) = x which is (y+1)^(1/3)=x

you can try it out for yourself

[rare cloud]

Yes. It's a reflection in the line y = x, right?

[obsidian monolith]

JY1853:

[languid crane]

yeah ik

but the graph is the same

its supposed to be

on the f(x) = y and f^(-1)(y)=x

that aint false right

idk why ur talking about graphs when it asked the inverse function

yeah ik

so this aint right?

[rare cloud]

That's correct. Just that I'm trying to find f^(-1)(x) and not f^(-1)(y) because it's more useful or something.

[languid crane]

oh ok

sorry

[rare cloud]

Man. This whole inverse function idea is throwing me for a loop.

[scenic musk]

https://cdn.discordapp.com/attachments/326138772477575180/642845954512257049/360640011638669316.png

[stuck lark]

@scenic musk no diff. pick what looks good to you

[harsh cipher]

I think there is a typo

10^7

is 100,000,000?

[rare cloud]

Pretty sure it's 10,000,000.

[hexed ermine]

1 with 7 zeros

[rare cloud]

That.

[harsh cipher]

thx i made a mistake in the first step

🙂

hi again

queston c)

3^3 on the right. Then 3^ 1/2

He described that's what radical looks like.

[rare cloud]

Not sure what you mean?

But $27\sqrt{3} = 27 \cdot \sqrt{3} = 3^3 \cdot 3^{1/2} = 3^{3 + \frac{1}{2}} = 3^{7/2}$?

[obsidian monolith]

Zetanoxic:

[harsh cipher]

how to get 3 1/2

cube root 3 is 1.44

[rich flint]

@harsh cipher when you convert the radical of a square root to a power it would be squareroot(3) to 3^(1/2)

[rare cloud]

Yeah. $\sqrt[n]{a} = a^{1/n}$.

[obsidian monolith]

Zetanoxic:

[rich flint]

Way better than my attempt zeta I need to learn that

[harsh cipher]

ok that's awesome thank you!

[rich flint]

I tried to explain it but no clue where the at symbol comes in

[rare cloud]

LaTeX is certainly useful. Whenever I try to type fractions without LaTeX, it gets messy very quickly.

[harsh cipher]

lol

[rare cloud]

Just learnt synthetic division for the first time today. Interesting stuff.

[viscid thistle]

Nice

[rare cloud]

Sorry for the long wall of text ahead, but I'd like to check my knowledge:$\$

The Rational Zeros Theorem basically just states that the rational zeros of a polynomial $P(x) = a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0$ have to be in the form $\frac{p}{q}$, where p is a factor of $a_0$ and q is a factor of $a_n.\$

Descartes' Rule of Signs states that in P(x), the number of positive real roots is either the number of variations in signs (ignoring terms with a coefficient of 0) of P(x), or less by an even whole number and the number of negative real roots is either the number of variations in signs in P(-x), or less by an even whole number.$\$

The Upper and Lower Bounds Theorem states that if P(x) is divided by x - b (b $>$ 0) using synthetic division such that the row of quotients and remainder does not have negative coefficients and if P(x) is divided by x - a (a $<$ 0) using synthetic division such that the row of quotients and remainder alternates between nonpositive and nonnegative coefficients, then every real root of P(x), c, must be so that $a \le c \le b.\$

Did I understand correctly?

[obsidian monolith]

Zetanoxic:

[tall merlin]

For those people out there who are beginners or even experts on calculus, I made a program for you that visualizes calculus graphs. Its on scratch so you can view it in your browser. Play around with the variables and have fun! https://scratch.mit.edu/projects/343428185

[rare cloud]

Where does the $\overline{a_i}$ come from in the second line?

[obsidian monolith]

Zetanoxic:

[rare cloud]

Oh wait. Is it that the complex conjugate of a real number is itself?

Ah! Got it. My apologies.

[hexed ermine]

@rare cloud yeah your understandings of the theorems are correct

just worded a bit oddly

but im able to understand it

[rare cloud]

Ah. Thanks!

[tame wedge]

How do you simplify the following:

Into: sqrt(2)-1

[willow bear]

$\sqrt{\frac{1 - \frac{\sqrt{2}}{2}}{1 + \frac{\sqrt{2}}{2}}} = \sqrt{\frac{\sqrt{2} - 1}{\sqrt{2} + 1}} = \sqrt{\frac{(\sqrt{2}-1)(\sqrt{2}-1)}{(\sqrt{2}+1)(\sqrt{2}-1)}}$

[obsidian monolith]

Ann:

[willow bear]

@tame wedge

[distant nova]

how do I solve it

if I divide by 5^ x

then it's (2/5)^x + (3/5)^x + (4/5)^x=1

LHS is a decreasing function right?

how do I go about from here

[rare cloud]

I have no idea how to solve it, but would writing 4^x as 2^(2x) help?

[willow bear]

no

[rare cloud]

Oh. Interesting.

[distant nova]

Thanks Ann

[willow bear]

please don't post the same thing across multiple channels

[distant nova]

Sorry

I won’t next time

[distant nova]

How is 2x+sin 3x not one-one in “its respective domain”

differentiating gives 2-cos 3x right?

[willow bear]

no

2 + 3cos(3x)

[distant nova]

Oh

Yeah that makes it wrong then

Sorry for the dumb AF question

[rare cloud]

In what situations would the Upper and Lower Bounds Theorem be useful in practice? Just when there are a lot of rational zeros to try?

[willow bear]

what theorem

can you state it

i don't know any theorem by that name but maybe it's only a consequence of not having heard that name

[rare cloud]

I don't see the point in chasing the bounds if I only have a handful of results to try.

[rigid sun]

Not for every polynomial

Sometimes you have a massive list of # to try

so if you don’t know Newton’s method, you’re completely donezo

[rare cloud]

Ah. Interesting.

Have no idea what that is, but I'm sure I'll come across it in due time.

[trim fable]

can someone help me with these

i did 9a how would u do 9b?

[summer sierra]

Can someone help me with this question?

[trim fable]

oh wow I wish i could

[candid onyx]

Did I do this right ?

[viscid thistle]

@trim fable Well they told you that ‘t’ was the time in seconds

And there’s 60 seconds in a minute

I think you get it now right?

[trim fable]

is t the period?

@viscid thistle

but like the period is 6/5 and i did multiply it by 60 but

im getting 72 and the answer is 50

[viscid thistle]

@trim fable can you show me how you calculated the period?

[trim fable]

ok

2pi/k
= 2pi / 5pi/3
= 2pi * 3/5pi
=6/5

@viscid thistle Like that 😄

oh i got it people!!

with @viscid thistle 's help

[plush matrix]

hey hey people

I have a question

is the End Behavior of a poly is determined by the leading term? or the overall multiplicity of the poly??

[torn swift]

by the sign and power of the leading term

[plush matrix]

thanks bbe

[hexed geyser]

yo guys

0 = 12x^2 - 29x + 15

how do i solve this

without using the quadratic formula because im taking calculus and we arent allowed to use a calculator and i cba to multiply and divide big numbers

i dont see a common factor or anything

[plush matrix]

plug 0 into x?

maybe idk

[hexed geyser]

what

[rigid sun]

ooof no calculator

[hexed geyser]

i mean i can multiply big numbers

(lattice ftw)

just didnt know if there was another way

[rigid sun]

Why can’t u use the quadratic formula?

[hexed geyser]

i can

just seems out of place that she would put this on a homework problem

[rigid sun]

Then do it

[hexed geyser]

so didnt know if there was another way

[rigid sun]

What

Lol no