precalculus

[cold jackal]

x=9, y=-2

Based on what I said what would the new slope be?

[lament garden]

I accidentally left the window of the work and it changed to another question

[cold jackal]

Oh ahaha

same type of question or?

[lament garden]

I was trying to work on the question and it disappeared

Yeah just different numbers

[slender zinc]

can somebody tell me the difference between precalc and algebra 2...lol

[cold jackal]

Okok

Precalc uses trig a lot

And it’s a prerequisite to calculus so it introduces the concepts

[slender zinc]

like limits?

or

[cold jackal]

Yeah and continuity etc

[lament garden]

Ok so the x=2 and y=-5?

[cold jackal]

Yes

And based on what I told you what would the new perpendicular slope be

Negative reciprocal

[lament garden]

-1/2x?

[cold jackal]

Closeeee

It’s negative

Reciprocal means you basically switch the numerator and denominator

I didn’t clarify what that meant sorry

[lament garden]

-4?

[cold jackal]

yessss

[lament garden]

Ohhh

[cold jackal]

good good

so we have x, y and the slope (m)

[lament garden]

Yeah

[cold jackal]

Now plug in x and y into the point slope form

Do you know it?

[lament garden]

Yeah i do

[cold jackal]

Okay

So there’s your answer

I hope that helped

Lmk if you need clarification

[lament garden]

Is the answer y-2=-4(x-2)?

I did the work and it fave me that

[cold jackal]

Y=-5

But yeah

On a side note if it asks you to make it parallel, the slope stays the same

[lament garden]

Oh ok

Then it asker for the slope intercept form and i got y=-4x-13

[cold jackal]

I did not get that

[lament garden]

Theres a second part

[cold jackal]

I mean that’s not the answer I got

[lament garden]

Oh

[cold jackal]

y=mx+b

-5=-4(2)+b

-5=(-8)+b

Add 8 to both sides

b=3

[viscid thistle]

This isn’t a poly right

Is something with the power of 1 a poly?

[serene heath]

yea u need integer powers

*pos

[viscid thistle]

Anyway to factor polynomials

ez

[viscid thistle]

,rotate

[viscid thistle]

You can have a poly w/ deg 0

1*

@viscid thistle no

$f(x) = 0$

[obsidian monolith]

rudy:

[viscid thistle]

$y=x-5\sqrt{x}$

https://en.m.wikipedia.org/wiki/Zero_polynomial

[obsidian monolith]

Gryff:

[viscid thistle]

Some define it as undefined

Take your cup of tee

I think of it to be logical for it to be zero

2 is more logical than 0.

Otherwise, f(x) = 100 = f(x) = 0x^2 + 0x + 100

and so on

You realize I'm talking about this...?

I said “You can have a polynomial with degree 0”

Honestly don't know if polynomials can be just a constant.

Also saying that you can have a polynomial with a degree of 0 for that problem can throw people off.

[sonic rivet]

you certainly can have a constant polynomial

f(x) = Cx^0

for whatever constant C you please

[native sequoia]

except 0

uh for degree 0

[viscid thistle]

Yeah that's what I thought ^

[hardy abyss]

what's wrong with constant polynomials?

[willow bear]

polynomials of degree 0 are still polynomials... and so is the zero polynomial... otherwise you don't even have fucking closure under addition for crying out loud

[cedar blaze]

hey, so for 1 - cube root of x / x - 1

if i'm going to use diff of cubes

for the 1 - cube root of x

how would that work out? x to the 1/9 ?

[proud sparrow]

no, more like x-1 is the difference of cubes

@cedar blaze

[rigid beacon]

Constants are part of polynomials yes

Hell we even say (x-1)² is a binomial squared

Which yields a trinomial

[viscid thistle]

Long dividing polynomials is hard af

[willow bear]

is it really

it's nothing but a matter of practice

[languid crane]

its not hard imo

[willow bear]

@viscid thistle are you just gonna complain or are you gonna actually post a problem that's giving you trouble if there is one

[proud sparrow]

It's just tedious at most

[viscid thistle]

Nah I’m just gonna complain @willow bear

[willow bear]

take that elsewhere then

[viscid thistle]

@willow bear nah

[willow bear]

...

complaints like yours don't belong here.

[lime perch]

A sum of 600 dollars is invested for 6 years and interest is compounded quarterly. There is 1200 dollars at the end of 6 years what is the nominal annual rate? I know I'm solving for r in A=P(1+r/4)^24, but I can't figure out how to find r. I get stuck when I raise it to the reciprocal cause I get the answer -3.34 which is clearly not right

[fossil kiln]

If you still need help

Did you put in the 600 for the Principle value?

[lime perch]

Sorry I didn't set it up fully, I did

I talked to my buddy about it and I guess I didn't divide 1200 by 600 initially so it threw off my answer.. I didn't realize I had to do that first

[graceful wing]

In order to find the amplitude of an equation, you take the absolute value of the coefficient right?

so y=−4sinx, the amplitude is 4 right?

[wild turtle]

I am in Pre Cal/Trig and have 3 online assignments to take before I'm done with the course for good. The section is Trig identities. I am extremely lost on this section if someone wouldn't mind personal messaging me for some 1on1 help.

[rigid beacon]

@graceful wing yes

[plain rose]

Does anyone know what type of function a sum-product,product-sum, and product-product is?

And if you can omit values that go against the pattern?

[plain rose]

<@&286206848099549185>

[toxic jasper]

Who needs help

[proud wraith]

Could someone explain where I would start this?

[frozen needle]

$\frac gf!\prn{\frac 12}=\frac{g!\prn{\frac 12}}{f!\prn{\frac 12}}$

[obsidian monolith]

Tuong:

[proud wraith]

o

that makes a lot more sense.

ty

[frozen needle]

You're welcome

[proud wraith]

Quick question, do you guys quickly type out those codes?

or is there a website you guys use to copy and paste to get those images?

[frozen needle]

a discord bot does it

you type in some LaTeX and it turns the code into an image

$1+2+3$

[obsidian monolith]

Tuong:

[proud wraith]

Oh, so you guys memorized the symbols represented?

[frozen needle]

pretty much

I'd say LaTeX is largely widespread

[short sorrel]

it's the standard for academic writing in mathematics

(as well as related fields, like physics and statistics and CS)

[proud wraith]

Good to know. Thanks guys.

[serene heath]

Np

[short sorrel]

if you ever find a bunch of math papers/modern math textbooks and notice they all "kind of look the same"

that's why

(see: arxiv)

[proud wraith]

Do you foil this? How would you foil a fraction?

[short sorrel]

"foil" is used when multiplying together multiple terms

i.e. of the form (a + b)(c + d)

that's not what you have here

do you know what $(f \circ g)(x)$ means? are you able to write it?

[obsidian monolith]

Namington:

[proud wraith]

No

[short sorrel]

$(f \circ g)(x) = f(g(x))$

[obsidian monolith]

Namington:

[short sorrel]

does that help?

[proud wraith]

Yes!

[viscid thistle]

so i have an answer for this but i'm not sure if it's correct

is it -5+8a+4h

Lemme see.

ok thanks

@viscid thistle Yes it's correct.

ok thanks bro

👍

[green zenith]

hello. i need some help looking for the x intercept of y=-3x^2+18x-11.
I already ended up on this equation. 0=-3(x^2-6x+11/3).
My question is how do i factor this one to find the x intercept?

[proud sparrow]

x intercept, i.e. roots

if the x intercept are rational, you can use rational root theorem

otherwise, probably just complete the square

just divide by -3, it wouldn't affect the value of x that works

[green zenith]

i dont know the first half of what you just said
im gonna have to look it up

but if im going to complete the square, how am i going to do it?
i'm sorry if i sound like a noob but the fraction is just bamboozling me.

[patent beacon]

@green zenith
I wouldn't pull out the 3, as you get an ugly 11/3

[proud sparrow]

It's quite similar to (x-3)^2

fractions are numbers too

[green zenith]

so i include 3 when factoring?

@proud sparrow yeah. im just not used to factoring with fractions. its much more confusing.

[proud sparrow]

you can divide by -3

but if you are just doing factoring, just look at the things inside the brackets

[green zenith]

okay. ill try and ill get back to you.

this is really complicated

is there an equation for this kind of factoring?

[sullen pendant]

I had a question on a quiz today and Idk if i got it right

x^4 + x^2 + a
Find a if 4i is a root

[proud sparrow]

okay

Try plugging in 4i, what do you get?

[proud wraith]

Could someone give me some pointers for B and C?

[stuck lark]

starting with part B, what have you tried?

[proud wraith]

I don't know what they're asking for

I know the domain for (a) would be [-4,3]?

[stuck lark]

do you understand the notation of "g(2)"?

[proud wraith]

No

[stuck lark]

the question has defined some function g that takes all inputs x on the interval x=[-4, 3]. g(x) is a way to write g as a function of x

knowing this, what does g(2) mean?

[proud wraith]

x=2?

[stuck lark]

that's part of it. part B wants you to think about when the input x is equal to 2

if it helps, think about g(x) as the y value on the graph of g(x) at some x value

[proud wraith]

So 3?

Is it also asking for domain?

[stuck lark]

part A is asking for the domain of g. part B is asking for something different

So 3?
explain what you mean by this

[proud wraith]

When the y-axis reaches 2, it takes 3 points to the right.

So 3?

and -4 for c

[stuck lark]

@proud wraith hmm that seems a little backwards

remember, x is the x value while g(x) is the y value on the graph

[proud wraith]

Thanks, I get it know.

[stuck lark]

awesome

[pseudo sonnet]

can someone check my work for those 2 problems

[pseudo sonnet]

and that one as well 🙏

[viscid thistle]

@pseudo sonnet , think about what happens for

$(-2)^n, n>1$

[obsidian monolith]

RyanE:

[viscid thistle]

For the first problem

[pseudo sonnet]

oh i see

so

would it be

2 * (-1)^n

on the numerator

[viscid thistle]

Correct

[viscid thistle]

Can anyone help me with a?

Nevermind solved

[pseudo sonnet]

@viscid thistle are my other problems/solutions correct

[viscid thistle]

@pseudo sonnet , as far as I can tell, the other 3 are correct

[pseudo sonnet]

ty

[velvet elm]

Hi guys, my homework threw a problem at me and I have no idea where to begin, it's not covered in the chapter.

Find the x-coordinate of the vertex of the parabola
y = (x − a)(x − b).
(Your answer will be in terms of the constants a and b.) Hint: It's easier here to rely on symmetry than on completing the square.'

answer should be in the form of x=?, it says not to complete, but I'm not sure what i'm supposed to do after I distribute.

[pale kettle]

You don't need to distribute

Just draw a picture of the situation and think about the symmetry

[velvet elm]

not sure how I draw a graph of that.

[pale kettle]

What information does y = (x-a)(x-b) tell you about the parabola

Think about the roots of the quadratic

[velvet elm]

well it's positive

oh you mean like x-a=0, x-b=0, so x=a, x=b

[pale kettle]

What does that tell you about the quadratic or parabola

[velvet elm]

that it has two x-intercepts?

the brain is churning slowly right now..I've been at this for hours today 😩

[pale kettle]

Yeah it has two x-intercepts

What are those two x-intercepts?

[velvet elm]

well, a and b

[pale kettle]

Now draw the picture

You don't know what a and b are, but just put them somewhere

And see where the vertex is in relation to a and b

[velvet elm]

Not getting it, i'm probably overthinking it. I don't know what the value for a and b are, but they're on the x-axis, the vertex would be halfway between a and b

[pale kettle]

That's just the answer

[velvet elm]

how do I write that in the form of x=?

there's only one input for an answer, i'm assuming X= means the vertex

[pale kettle]

What's the number halfway between a and b/

[velvet elm]

need to use the distance formula for that don't i

[pale kettle]

No you don't

Both a and b are just normal numbers

[velvet elm]

i'm missing something painfully obvious. going to take a couple minutes break, think i've been overworking myself

(a+b)/2

midpoint formula 🙃

or that thinking got me there at least lol

Thanks Zopherus!

[pale kettle]

Nice work

[velvet elm]

Sometimes a quick break is all you need..Helps the brain crunch things

[odd badge]

@velvet elm you could also have expanded the quadratic out

and then used the formula for the x-coordinate of the vertex, -b/2a

where a is the coefficient of the x^2 term and b is the coefficient of the x-term

check: (x-a)(x-b) = x^2 -ax - bx + ab

so a = 1

and b = -(a + b)

so you get (a+b)/2

which is what you had ✅

[velvet elm]

b=-(a+b) from -ax-bx >> -x(a+b)?

[odd badge]

haha

same thing whatever

i use b because the coeff notation using a b and c is standard

it's kinda confusing ig

[velvet elm]

Yeah i was following you, I was just trying to see where you got the coeff for the x-term.

cause there was -ax and -bx

[odd badge]

yup

yeah you can prob just look at it and see it

skip a few steps

[velvet elm]

When I first tried the problem I did initially expand it, but I just wasn't sure what it was asking and how to get there. took me a while to think of the midpoint

[odd badge]

yeah

the vertex's x-coord is always going to be -b/2a tho

[velvet elm]

It doesn't help I forget some of those smaller formulas.

I had done that last semester but this trig/precalc class is just jumping straight through things, and the book is freakin horrendous

[odd badge]

worst case scenario just derive the formula -b/2a from the derivative

[velvet elm]

don't know derivatives yet

[odd badge]

meh they're not too hard

but yeah just remember the formula then

[velvet elm]

probably not, I think we're coming up on it soon.

[odd badge]

also yeah for a second order fct the midpt b/w zeros will be the vertex

so you were on the right track by intuitive thinking anyway

[velvet elm]

I skipped to some easier work for now. I've got a ludicrous amount of hw to get through this weekend cause I was on vacation, need to catch up 😩

[odd badge]

rip

[velvet elm]

8^(z + 1 )= 32(sqrt2), I am equating exponents. Does 8^(z+1) become 2^[3(z+1)] or just 3z+1

[obsidian monolith]

euler/fermat:

[velvet elm]

Ok that's what i thought, it's just been a while since i've these.

[obsidian monolith]

euler/fermat:

euler/fermat:

[velvet elm]

dang that's handy, i've been meaning to learn latex

[odd badge]

u left a space in there

rip

yeah i think it's a package

that latex doesn't have by itself

[obsidian monolith]

euler/fermat:

euler/fermat:

[velvet elm]

Thanks! I was pretty sure 3 would distirbute to the z+1, the rest of it i can do

Yeah

appreciate it.

Should I just attack google, or is there a preferred learn latex guide?

[obsidian monolith]

euler/fermat:

[proud sparrow]

@velvet elm There's a latex guide here

[velvet elm]

I have lyx installed on my laptop, just haven't used it yet.

not enough time since I just have so much to get through

[proud sparrow]

https://discordapp.com/channels/268882317391429632/488106178094694400/556173635195240488

I think just try it out

[velvet elm]

Cool thanks i'll save that.

[proud sparrow]

and then we can guide you with it

[velvet elm]

let me just say, I'm so happy I found this place 😄 haha

[odd badge]

rip lmao

[velvet elm]

I saw someone do that actually, I thought it was interesting way to do things.

[odd badge]

$\LaTeX$

[obsidian monolith]

thief:

[odd badge]

there it is

[velvet elm]

lol

[velvet elm]

I'm having trouble finding the X-Intercept for this exponential function.

I can't equate the -1=9^-x

oh wait..I'm confused, just not wrong! There is no x-int,

there's a y-int which is what i thought i was doing, and was wondering why i can't find it

[willow bear]

32 is equivalent to 2^5
it's not EQUIVALENT to 2^5, it IS 2^5

[echo plaza]

thanks ann

[obsidian monolith]

Display results online and refine query

Requested by euler/fermat#9844

[willow bear]

don't "blyat" at me and then try to hit me with a dictionary definition

you definitely weren't talking about some sort of equivalence relation on R other than equality

thank you for notifying me of a fact i was already aware of

[astral star]

can someone help me on how to go about #71, just @ me

[obsidian monolith]

Display results online and refine query

Requested by euler/fermat#9844

[uncut mulch]

foil it like any other binomial

[willow bear]

foil

[astral star]

ookay thanks

[shrewd urchin]

Ann are u gifted ?

[willow bear]

what's your definition of gifted

[serene heath]

Christmas gift

[shrewd urchin]

I am gifted.

Actually I am smarter than all of you. But I am an unexperienced problem solver.

[vernal rapids]

I am smarter than all of you

[shrewd urchin]

Yes

[vernal rapids]

Its brave to say something like this here

[next willow]

It's still to vague

https://en.wikipedia.org/wiki/Theory_of_multiple_intelligences

[shrewd urchin]

Lol i was jk

[next willow]

Lol me too

rhythmic intelligence

It wasn't that bad to be deleted lol

[shrewd urchin]

Ok let me rewrite

I am different person to different people
For one i am talented
For the second: i fucking suck and i am good for nothing
For the third : you need a therapist u mentally ill .
For the fourth: you are pragmatic u will make it to far.

[heady jewel]

yes you really need therapy

youre just sick

[shrewd urchin]

The third person spotted

[odd helm]

Why is it imaginary over real

I thought it should be tan(1/-2)

Since 1 is opposite and -2 adjacent in the complex plane

[sonic rivet]

do you know euler's identity?

[velvet elm]

For the x-inputs in this example, does the base (10) go to the denom and the exponent (1) go to the x-exponent around outside of the fraction? or does 10^1, 10^2, so on, go in each X

[odd helm]

No

[velvet elm]

sorry, no to which part?

[desert socket]

Hey so I’m working on finding the inverse of the following:

(X-4)/(2x+2)

When I stop at this point:

Do I just multiply by -1 to get 2(x+2)/(1-2x)

I get -2y-4

And factored the -2

[velvet elm]

Nvm I'd have to write it out lol

[desert socket]

Oh okay lol

The image was where I stopped and Mathway says 2(x+2)/(1-2x) is the answer

So is it just a multiply by -1 situation

[velvet elm]

if you factor a -1 from 2x-1, it would cancel the - on the fraction

and you'x get 1-2x

[desert socket]

Brilliant

Thank you

[velvet elm]

How do you find the horiztonal asymptote?

The degrees of the num and denom are the same, but the leading coeff is 1, but the horiz asymptote isn't 1/1=1,

also I tried dividing the polynomials but I have to refresh how to do that and got stuck, do I need to continue doing that?

[stuck lark]

but the horiz asymptote isn't 1/1=1
really?

[velvet elm]

y=1 isn't accepted and this is the graph tho

so ya I can "see" that it wouldn't be 1..But the math says it should?

[stuck lark]

the math says it should, and i can also actually see graphically that the right-end & left-end behavior involves a horizontal asymptote y=1

[velvet elm]

so desmos says the left line turns at -0.056

how........

[stuck lark]

do you know what end behavior is?

[velvet elm]

vaguely. I'm playing catch-up with this chapter

as the line appraoches zero, unbounded positive or unbounded negative, thing?

[stuck lark]

nah

you got any notes on end behavior?

[velvet elm]

Yes, classmate sent me notes. Odd degrees and even degrees?

[stuck lark]

@velvet elm more generally, how the function behaves as x approaches infinity and -infinity

you guys worked with limits yet?

[velvet elm]

No.

So the asymptote deals with the behavior of the line as it tends towards inifinty? Not, for a lack of a better term, near the origin/

in this example, where it crosses the x-axis?

[stuck lark]

right end behavior tells you how a function f behaves as x approaches infinity

if you find any horiz asymptote related to the right end behavior, that tells you the particular y value that f(x) approaches as x approaches infinity

[velvet elm]

Ok, following.

[stuck lark]

examining right-end behavior, you correctly said that the horiz asymptote is y=1. and the graph indeed shows that the function approaches 1 as x heads off to +inf

[velvet elm]

Ok.

You keep saying right end, does left-end deal with -infinity and describe the behavior of the other line?

[stuck lark]

left-end is about how f(x) behaves as x heads off to -inf

[velvet elm]

How do I find the left end asymptote?

I presumed that wasn't a thing since it's the same function.

[stuck lark]

how did you find the asymptote for right-end?

[velvet elm]

According to my instructions, because the numator and the denominator have the same degree for the leading coeff, You divide the coeffes (after distributing it becomes (x^2+11x+28)/(x^2-2x+1), so the leading coeff of x^2 and x^2 is 1/1

but it's not accepting that answer.

sorry don't know latex yet so I can't write it out nicely

[stuck lark]

your hw website isn't accepting y=1 as the horiz asymptote for left end?

[velvet elm]

Correct.

[stuck lark]

can you screenshot the hw q?

[velvet elm]

WTF!!!!! I just re-entered it AGAIN and it accepted it this time.

I swear i tried typing it in multiple times

[stuck lark]

[velvet elm]

https://tenor.com/view/cry-morning2-brown-cony-gif-13009332

Time to take a break I think! Thanks for the help though, and allowing me to understand the material more thoroughly.

[stuck lark]

no prob man

[loud quartz]

How do you do this

[merry sphinx]

Alright so the one-to-one property just means that the stuff inside ln() are equal

So it then becomes a regular quadratic equation that you solve

[loud quartz]

I get -1 but it says incorrect

@merry sphinx

[merry sphinx]

Sometimes quadratic equations have false roots

So plug it back into the original equation to check if it is incorrect

[loud quartz]

-1 works on the equation but not in the quiz

[merry sphinx]

give me a second

You should get 2 roots

[loud quartz]

-1 and 0 but 0 doesn't work in the equation

Is 0 extraneous

[merry sphinx]

0 is not the root

[loud quartz]

-4x=x^2-3x.

[merry sphinx]

Ignore what I said

I messed up

[loud quartz]

X^2+1x

X(x+1)

X:0 x:-1

[merry sphinx]

yeah

so 0 and -1 works right?

[loud quartz]

-1 works in the equation but 0 doesn't

[merry sphinx]

yeah cause log

[loud quartz]

Is -1 and 0 solutions and 0 an extraneous solution

[merry sphinx]

I think -1 would be the solution and 0 would be the extraneous because ln(0) is undefined

[loud quartz]

It says that -1 is incorrect is there another solution?

[merry sphinx]

It says -1 is not a solution?

Or does it need another

[loud quartz]

No just that there could be more

It's either wrong or there is another

[merry sphinx]

I don't get why cause -1 should be the only sol

I ran it through a couple websites like wolfram

[loud quartz]

NVM the program was bugged you had to get both parts correct

-1 was rogt

Thanks

[merry sphinx]

Ok i was so confused for a second

np

[velvet elm]

Can anyone explain how I do this?

how do i solve it as f(x)=k

[patent beacon]

What is the asymptote?

[velvet elm]

y=3

horizontal asymptote

[patent beacon]

Very nice! So then, when is f(x) = 3?

[velvet elm]

....y=3?

or set the function=3

my brain is a little mush at the moment, i've been at this for 2 days :*(

[patent beacon]

Forget the y = k thing. You want to know where the function crosses it's asymptote. You already know the asymptote is at y = 3

So when does the function cross y = 3?

[velvet elm]

I don't know.

when it nears 0?

No imm thinking of somehting else.

Sorry i'm kind of stumped. I'm looking at the graph and i can see it crosses 3 at (-27,3) for the left end, but i'm not sure what i'm supposed to do with it, what the next step is

My textbooks is overkill when it comes to its explanations

[cedar minnow]

Hello. Can someone please help me do #8?

The function is y = 5root(32x) btw.

If someone can help with this I would really appreciate it.

[uncut mulch]

would it help if it was written as

$\lim_{h\to 0} \frac{\sqrt[5]{32(1+h)} -\sqrt[5]{32(1)}}{h}$

[obsidian monolith]

ramonov:

[uncut mulch]

*also not 5 root but 5th root of

[cedar minnow]

Right.

My bad.

Btw that’s the limit I’m asked to evaluate. That corresponds to the function:

[obsidian monolith]

Eh Canada:

[cedar minnow]

$\sqrt[5]{32(x)}}$

[obsidian monolith]

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

[rigid sun]

well

it looks like a derivative limit no?

so use the graph

derivative=slope of tangent line

[cedar minnow]

Yeah so do I literally just draw a tangent to the point (1,2) and take the slope?

@rigid sun

[rigid sun]

well you could

actually yeah do that

cus it asks you to estimate not exact

[stuck ivy]

Yeah I need to check my answer for this question

I'm not sure if 5pie/3 is in the correct interval of 0 < x < 2pie

[uncut mulch]

fix the 2pi/2 first

[stuck ivy]

Oh yeah mistyped there

good catch

[uncut mulch]

also just pi not pie. pie is food

[stuck ivy]

Haha that's true

[uncut mulch]

is 5/3 less than 2?

[stuck ivy]

is 5pi/3 in my interval?

No idea haha

I think it's 300 degrees

So i'm allowed to keep 5pi/3 in my answer right?

[uncut mulch]

think about my previous question
is 5/3 less than 2
its simple division

[stuck ivy]

yes

For this question are the solutions pi/6

cause it says type any angle in radians between 0 and pi instead of 2pi

[uncut mulch]

in which quadrants are tan negative?

[willow bear]

is*

[azure sundial]

quadrants ii and iv

[flint river]

''Find a so that f(x) is continuous...''

how do you evaluate f(0) ? dunno what to do with that sin angle

I mean I guess it oscillates infinitely near zero

is it possible to get a value for it?

[safe maple]

if f is continuous then f(0) is the limit of f(x) when x -> 0 and x <= 0

it oscillates but it is getting smol

it tends to 0 so you need a to be -4

i think

and your exercise is messed up, it should be < above and >= below

otherwise you divide by 0

[flint river]

right that's what I thought

wait what if that's made on purpose :p

[safe maple]

nah

who gave you this thing?

[flint river]

given that, it means f(0) can be calculated but how? I thought some trig trickery would do the work

(which I suck at, trig I mean)

school, why??

[safe maple]

it sucks hard

[short sorrel]

you cant take $\sin(\frac{1}{0})$

[obsidian monolith]

Namington:

[short sorrel]

theres no magic property of trig that makes 1/0 a valid thing

[flint river]

OK, thats good to know

[safe maple]

the 7 has no purpose, they make you solve a+4 = 0 for no other reason than to annoy you

[flint river]

typos are everywhere, tho

[safe maple]

and they can't even get the inequalities right

[flint river]

lmao

[safe maple]

it's about continuous continuation right?

you have that theorem?

Prolongement continu

[flint river]

yep, haven't memorized it tho

[safe maple]

ok, donc une fonction definie sur un intervalle auquel il manque un point a, (comme R* par exemple) se prolonge en une fonction continue sur R lorseque les limites epointées en a a gauche et a droite existent et sont égales
et il y a UNICITE de ce prolongement

avec l'unicité tu peux conclure

[flint river]

I ain't french

[safe maple]

you're french, right?

oh shit

[flint river]

lmao

[safe maple]

si is a french word

[flint river]

it's spanish too

[safe maple]

so i tought you were

[flint river]

np

[safe maple]

fak

so if you take some interval (IR here)

and take out one point (0 here)

and take some continuous func on that set

if the limits from the left and from the right are the same,
then your func can be continuated (?) continuously and in ONE AND ONLY ONE way,
meaning here there's a func on R that equals your func on R* and is continuous

so you need to prove it exists by calculating the limits, you'll get a func g
g(0) will be your common limit
then it will be unique so you know that f continuous if and only if f = g

[flint river]

func g? another function? how come?

[safe maple]

it's the continuous prolongation of (R*->R : x -> f(x)) onto R

you prove it exists, and you know its unique, so it must be equal to f if and only if f is continuous

[flint river]

oh, didn't know that, yeah our classes aren't that rigorous

oh btw I didn't notice the error on that inequality, that made me think some crap like doing:

[safe maple]

it's the best redaction i find

[flint river]

$$\lim_{x\to 0} 7xsin\left(\frac{1}{x}\right)\
\lim_{x\to 0} 7xsin\left(\frac{1}{x}\right)\frac{\frac{1}{x}}{\frac{1}{x}}\
\lim_{x\to 0} 7x\frac{sin\left(\frac{1}{x}\right)}{\frac{1}{x}}\frac{1}{x}\$$

[obsidian monolith]

ike98:

[safe maple]

XD no

[flint river]

oh how do you even make a new line here in latex?

[safe maple]

you post another thing?

[flint river]

cool lets try

I mean on a single message

well w/e

[safe maple]

you just need to calc the limit going to zero FROM THE LEFT of 7xsin(1/x)

[flint river]

oh right

[safe maple]

from the left and not taking 0 ofc

so you know how to do that?

[flint river]

hmm

[safe maple]

how does it look, you think?

xsin(1/x)

[flint river]

lemme see

oh 0 multiplied by mmm what's the word

[safe maple]

you mean that the x part tends to 0?

[flint river]

right and sin is... narrowed?

dunno what's the proper word

[safe maple]

bounded

[flint river]

yeah that

[safe maple]

between -1 and 1

so -1 <= sin(1/x) <= 1

and if you multiply by x

...

[flint river]

why multiply?

oh

right

[safe maple]

to get xsin(1/x) in the middle

[flint river]

nevermind

yep

cool!

[safe maple]

so you got some funnyly named theorem to conclude

here it's the policemen theorem

but there's a lot of names to it

[flint river]

so the fact that the function is being multiplied by x is what makes it continuos on x = 0 I guess right?

I mean the sin func

[safe maple]

is what makes its continuous prolongation exist and be of value 0 at 0 you mean

but yeah essentially

[flint river]

yea

[safe maple]

and then they took the right side of R, and changed the func there to a + 4

[flint river]

I was being confused by that 1/x argument :S

[safe maple]

should be just about applying the continuous prologation theorem now

yeah it's wiggly but it gets crushed by the x

[flint river]

so why that latex crap I pasted didn't make sense?

[safe maple]

first of all if you write lim x -> 0 before proving it exists you got an issue

and the whole point was that you needed to prove it exists

and idk why you did the 1/x stuff

[flint river]

hmm, lim x->0 sin(x)/x isn't it 1?

[safe maple]

nvm

the limits didn't even exist so there ain't much to say

anyway, imma sleep hf

[flint river]

thanks! 😄

[safe maple]

Np

[viscid thistle]

Hello guys, I have this situation with my teacher, he wants me to explain how do I know that adding the first and last number of a sequence for then multiply it by the number of pair in the sequence will actually give the correct sum of all the terms in the sequence

how in the frickiry frick I explain that???

[patent beacon]

Do you mean Gauss' method?

@viscid thistle

Nvm, googling that just gives a ton of gaussian elimination stuff.

Sum = Average × Number of terms
And the average of an arithmetic sequence is the first and last number divided by 2

[proud wraith]

Is this right? If so, why do I remove the three at the top?

[short sorrel]

why would the 3 affect the domain?

[patent beacon]

Actually this does better than I ever could https://math.stackexchange.com/questions/2260/proof-for-formula-for-sum-of-sequence-123-ldotsn

[proud wraith]

I wish I could tell you @short sorrel lol

[short sorrel]

that's my point

it doesnt

[viscid thistle]

Here I am!

Sorry

[proud wraith]

Oh. Ok thanks 🙂

[short sorrel]

but yeah your answer seems fine

[viscid thistle]

@patent beacon

I figured something out with a classmate

[merry sphinx]

What about for odd number of terms? Try to prove that

[viscid thistle]

Me?

[short sorrel]

uh

[merry sphinx]

Nvm, i thought you were doing sum of 1 to n, nice job on figuring that out though

[short sorrel]

the odd case is identical

[merry sphinx]

Yeah that is my point, but that reasoning on the paper only seems to consider even number of terms

unless I am missing something

[viscid thistle]

If it's odd is going to work the same way

I've done it in the past

Math magic

[merry sphinx]

Yeah math is magical

[viscid thistle]

Okay, I need help

Okay so I have this so far

But when I plugged my numbers the result is 2 numbers incorrect

So I did this but I don't think this is mathematically correct, to just put a +2 for convenience

Okay I don't have it anymore

<@&286206848099549185>

It doesn't work for the next one

I really doubt this is a sequence at all

I don't understand it

[proud wraith]

Sorry to cut you off Elara. Am I suppose to get numbers with this to plug in? Any pointers where to start?

[pseudo sonnet]

yeah

just pick 3 values from each condition

and sketch it

id also do x = 0

and find the y value for that

[proud wraith]

Thank you 😄

[viscid thistle]

I have to be blind

Maybe I'm not seeing something

[patent beacon]

It's very hard to read off the paper. What are you trying to do?

[pseudo sonnet]

yeah fr

[viscid thistle]

A recursive rule for the sequence 2,8,26,80

[pseudo sonnet]

well first

is it geometric or arithmetic

[viscid thistle]

Is the result of a geometric series

It's just a sequence

[patent beacon]

It's neither, sadly

[pseudo sonnet]

oh gg

what do u do in that case

[viscid thistle]

Ignore that recursive rule part

I don't have the less idea

I'm blocked

I hardly know what am I supposed to look for

[short sorrel]

sorry, can you give the full question?

there's infinitely many "rules" that would generate that sequence

such as $3^n - 1$

[obsidian monolith]

Namington:

[short sorrel]

or less trivially, something like OEIS http://oeis.org/A124721

[viscid thistle]

This is all I have/can give you

Just find recursive rules if it's a sequence

[short sorrel]

well, as you observed, each term increases by 2, then 6, then 18, then 54...

this is a geometric sequence

[viscid thistle]

It is

[short sorrel]

and the formula for that geometric sequence is 3n - 1 (if we take n starting at 1)

er wait

2*3^{n-1}

[viscid thistle]

I checked from pass classworks and I think 3n -1

[short sorrel]

$2 \cdot 3^{n-1}$

[obsidian monolith]

Namington:

[short sorrel]

this is the formula for the $n$ term of that underlying geometric series

[obsidian monolith]

Namington:

[short sorrel]

therefore, the $n$th term of the sequence itself is $\sum_{i=1}^{n} 2\cdot3^{i-1}$

[obsidian monolith]

Namington:

[short sorrel]

or recursively, the first term of the sequence is 2, and the term $a_{n+1}$ is $a_n + 2\cdot3^{n}$

[obsidian monolith]

Namington:

[viscid thistle]

Oh my, I'm really blocked I don't understand anything

Where did the 2 came from?

[short sorrel]

its the starting term of the geometric series

[viscid thistle]

Okay okay

[short sorrel]

again, the geometric series is 2, 6, 18, 54; so we generate each term by multiplying 2 by the common ratio (3) a certain amount of times

and to generate the "main" sequence recursively

we add the next term of the geometric series

[viscid thistle]

I didn't catch that last part

[short sorrel]

you noted that the "main" sequence is generate dby

2 + 6 + 18 + 54 + ...

[viscid thistle]

Yeah

[short sorrel]

to "generate a sequence recursively" means to figure out the next term from the previous term

[viscid thistle]

Okay okay

[short sorrel]

so if I gave you, say, 18, the recursive formula lets you figure out that the next term is 54

[viscid thistle]

Got it

[short sorrel]

er

rather

if i gave you 26

you should be able to figure out the next term is 80*

and you can use that underlying geometric series to figure that out

26 is 2 + 6 + 18

so we add on the next member of the geometric series

which is 54

so the next term is 26 + 54

which is 80

[viscid thistle]

Got it

[short sorrel]

the $n$th term of the geometric series is generated by $2 \cdot 3^{n-1}$

[obsidian monolith]

Namington:

[short sorrel]

so the $n$th term of the larger sequence, defined recursively, is given by $a_n = a_{n-1} + 2\cdot3^{n-1}$

[obsidian monolith]

Namington:

[short sorrel]

a_{n-1} is the previous term of the sequence

and 2 * 3^(n-1) is the new term of the geometric series

so for example, if we had 26, and wanted to find the next term

[viscid thistle]

You lost me

[short sorrel]

we know 26 is the third term

so $a_4$, the 4th term, is equal to $a_{4-1} + 2\cdot3^{4-1} = a_3 + 2\cdot3^3$

[obsidian monolith]

Namington:

[short sorrel]

and again, the third term, $a_3$, is $26$; so this is $26 + 2\cdot3^3$

[obsidian monolith]

Namington:

[short sorrel]

i.e. $26 + 54 = 80$

[obsidian monolith]

Namington:

[viscid thistle]

I'm going to eat dinner

I going to calm down

[serene heath]

Mmmm dinner

[pseudo sonnet]

can someone check if i did my work correctly

ignore the upper half of the paper

the sequence is 1,3,9,27,81

and i was asked to express first 10 terms using summation notation

and find the sum using the summation theorem

this one too

[void quiver]

IS ANYONE

AVALBILE

AHHH

f x x2 and gx 1x are shown below in the viewing window
0, 5 by 0, 5. Sketch the graph of the sum f  gx by adding
the y-coordinates directly from the graphs. Then graph the sum on
your calculator and see how close you came.

[viscid thistle]

What the heck are those boxes

[pseudo sonnet]

can someone help me with my solutions

[viscid thistle]

@pseudo sonnet Yeah I checked you're all good.

[pseudo sonnet]

https://i.imgur.com/7e7HPUh.png

@viscid thistle is that right?

the second part of the page

after S = 5

[dusk skiff]

Can someone explain why lim x->infinity ln(x) = infinity?

Anyone?

[viscid thistle]

@pseudo sonnet Why can't n be greater than or equal to 1?

If $n=1$ then $\frac{(-1)^{1-1}}{4^{-1}}=\frac{1}{4^{-1}}=4$

[obsidian monolith]

leviosa:

[pseudo sonnet]

oh shit

ur right

idk why i put n >= 3

[viscid thistle]

Lol.

Other than that it's all good.

[pseudo sonnet]

what about the explicit formula

it doesnt work

for n = 5

[viscid thistle]

Oh lemme check.

$a_n=4 \cdot (\frac{-1}{4})^{n-1}$

This?

[pseudo sonnet]

the 1 should be negative

but yeah

[viscid thistle]

I mean A.) You're forgetting a negative.

Oh.

[pseudo sonnet]

nah just my hand writing lol

[obsidian monolith]

leviosa:

[viscid thistle]

Wait it should work for 5...?

You probs forgot to count the n-1 in the power.

$a_5=4 \cdot \frac{1}{4^4} \ a_5= \frac{4}{4^4} \ a_5=\frac{1}{4^3}=\frac{1}{64}$

[obsidian monolith]

leviosa:

[pseudo sonnet]

lmao what

thats wrong bro

4^5-1 = 4^4

= 256

[viscid thistle]

Uhh there's a constant 4 in the beginning.

[pseudo sonnet]

idk where u subtracted that extra 1 from

[viscid thistle]

$\frac{x^1}{x^4}=x^{1-4}=x^{-3}=\frac{1}{x^3}$

[pseudo sonnet]

what?

[obsidian monolith]

leviosa:

[pseudo sonnet]

but dude

i used constants

not variables

its

(-1) / 4

[viscid thistle]

Okay fine.

[pseudo sonnet]

not x / x

[viscid thistle]

Wait for a second.

$a_5=4 \cdot (\frac{-1}{4})^{5-1} \ a_5=4 \cdot (\frac{-1}{4})^{4} \a_5=4 \cdot \frac{1}{4^4} \ a_5= \frac{4}{4^4} \ a_5=\frac{1}{4^3}=\frac{1}{64}$

[obsidian monolith]

leviosa:

[viscid thistle]

$\frac{4}{4^4}=\frac{4}{4 \cdot 4 \cdot 4 \cdot 4}$

[obsidian monolith]

leviosa:

[pseudo sonnet]

oh shit

im dumb

when i checked my work

i forgot to multiply 1/256 * 4

[viscid thistle]

Lol yeah.

[pseudo sonnet]

sorry my bad

[viscid thistle]

Nah it's all good.

[pseudo sonnet]

https://i.imgur.com/kA0DPen.png

for this one

i know from 1 to 3 , r = 3

and from 3 to 12, r = 4

and 12 to 60, r = 5

[viscid thistle]

The increases by 1 for every interval.

[pseudo sonnet]

so im guessing some sort of factorial

[viscid thistle]

Starting from 2.

Recursive is similar to what you did before.

[pseudo sonnet]

dont i use a factorial here

because the ratio increases by 1 every time

[viscid thistle]

Oh wait yeah.

Soz mb.

[pseudo sonnet]

any help anyone

?

[proud wraith]

I got 3x+1 and -1/3 for the domain. The answer key says 3x+1 and 1/3 for the domain. Why don't you make it negative?

[viscid thistle]

Check out my Youtube, I post daily Statistics content to help you out with AP Stats, and math in general. Please subscribe, as I am a 13 year old who wants my content to be recognized. My yt is https://www.youtube.com/channel/UCG4WyfGQk3MYMqdX9xJ0rmA?view_as=subscriber

Why is there no AP stats section in this discord?

oh nm sory

[vague zephyr]

this question has been a pain in my behind for about an hour now

does anyone know what it is?

[blazing parrot]

what is g(4)?

[vague zephyr]

i dont know

ive just been given the graph

wait

g(4)=2

[blazing parrot]

try again

[vague zephyr]

1.5

[blazing parrot]

what is g(1.5)

[vague zephyr]

.5?

-.5

[blazing parrot]

almost there

[vague zephyr]

-.75

[blazing parrot]

one section down is -.5

how many sections down is g(1.5

[vague zephyr]

-.25

[blazing parrot]

yes

[vague zephyr]

i thought i already tried that, ill try it again though

no, it is incorrect

[blazing parrot]

hmm

oh i see

[vague zephyr]

so what seems to be the issue?

[blazing parrot]

okay the graph slopes up uniform from x=0 to x=4 right

by how much does it increase over this period

[pseudo sonnet]

https://i.imgur.com/kA0DPen.png

can anyone please help me

[vague zephyr]

i am not sure

[blazing parrot]

what is g(4)-g(0)

[vague zephyr]

0

[blazing parrot]

no

[vague zephyr]

g(4) is 1.5 and g(0) is -1.5

[blazing parrot]

subtract not add

[vague zephyr]

3

[blazing parrot]

so y increases by 3 when x increases by 4

so how much does y increase when x increases by .5

[vague zephyr]

.15?

[blazing parrot]

no

you can treat it like a ratio

so y:x is 3 : 4

[vague zephyr]

.375

[blazing parrot]

yes

how about a 1.5 increase in x

[vague zephyr]

so would the answer be -.375?

[blazing parrot]

no

[vague zephyr]

so .5 now becomes 2?

[blazing parrot]

what do you mean

[vague zephyr]

you said there was a 1.5 increase in x

.5625?

[blazing parrot]

i was asking how much does y increase by when x increases by 1.5

no not .5625

[vague zephyr]

3

[blazing parrot]

no

[vague zephyr]

im not sure where to go with this

[blazing parrot]

since y increase by .375 when x increases by .5

then you treat it like a ratio and multiply both sides by 3

then when x increases by 1.5 y increases by 0.375*3

so when x increases by 1.5 from 0 to 1.5 y increased by 1.125

so g(1.5) is g(0)+1.125

[vague zephyr]

this becomes -.375

which was what i answered above

[blazing parrot]

oops sorry i read wrongly

[vague zephyr]

its okay lol

[blazing parrot]

heh

@pseudo sonnet since it’s a recursive formula what is the relation between each term and the next

[hexed bolt]

Could I have help on b and c

[gleaming lagoon]

i need help on 14 & 15

their easier but idk if i have it right

[willow bear]

what are your answers for 14 and 15?

[gleaming lagoon]

i have no way to check so im really stressed i have 0 clue whats going on

sry

uploading img rn

[willow bear]

ok first off

calm down

as best you can

[willow bear]

man this is a mess

do you not have a separate piece of paper onto which you could copy each problem and write out your working immediately below

[pseudo sonnet]

https://i.imgur.com/kA0DPen.png

can someone help

the ratio multiplier increases by factorial

thats all i could see

[viscid thistle]

Hint.

You have the first term.

The 2nd term is a_1 times 2!.

3rd term is a_1 times 3!.

Etc.

@pseudo sonnet

[willow bear]

https://cdn.discordapp.com/attachments/359052604149465088/615491471684009984/unknown-143.png

[hoary valley]

What is the angle between 2 unit vectors?

[willow bear]

[hoary valley]

Unit vectors are vectors with 1 in magnitude " i , j and K "

+they have direction

[willow bear]

why don't you post your problem exactly as stated instead of posting vague shit like you just did

[hoary valley]

I don't have a problem, I'm just asking

[pseudo sonnet]

@viscid thistle

[hoary valley]

I wanna know what's the angle between each unit vectors

[pseudo sonnet]

does that work as my explicit formula

[willow bear]

@hoary valley the angle between two unit vectors is equal to the arccos of their dot product.

if you so insist on an answer this particular.

[hoary valley]

@willow bear My teacher literally said " The angle between two units vector = 90 degrees" <<< I couldn't swallow it..

[willow bear]

...unless "unit vector" somehow only refers to i, j and k, and not the entire literal sphere of unit vectors

that is false

the angle between i and (i+j)/sqrt(2) is 45°, for example. and they are both vectors of length 1.

[short sorrel]

i've definitely seen "unit vector" used to refer to the standard basis vectors, particularly in high school/intro physics texts

this is "incorrect", but go with what your class uses.

if you're talking only about the standard basis vectors\textemdash that is, the vectors $\hat{i}, \hat{j}, \hat{k}$ that look like $\begin{pmatrix}1\0\0\end{pmatrix} \begin{pmatrix}0\1\0\end{pmatrix} \begin{pmatrix}0\0\1\end{pmatrix}$

[obsidian monolith]

Namington:

[short sorrel]

then they are, in fact, all orthogonal

if you're talking about the unit vectors as a whole - i.e. every vector of length 1 - then of course they can have any angle <= 180

[hoary valley]

Unit vectors = i hat, j hat and k hat.

[short sorrel]

yes, in that case

the angle between each pair is 90 degrees.

you should be able to check this intuitively by visualizing it

[hoary valley]

So she's right.. each pair of unit vectors are perpendicular to each others..

[short sorrel]

or you can check this explicitly: $\theta = \arccos \left(\frac{\hat{\imath} \cdot \hat{\jmath}}{||\hat{\imath}|| \cdot ||\hat{\jmath}||} \right) = \arccos \left(\frac{0}{||1|| \cdot ||1||}\right) = \arccos(0) = 90^{\circ}$

[obsidian monolith]

Namington:

[willow bear]

is... English not your native language?

[short sorrel]

but really, this should be intuitive.

it should be clear that vectors i, j, and k here are 90 degrees from each other

if not, you need to review your geometry

severely

[hoary valley]

It makes perfect sense actually now that I think about it, Thanks a lot 🖤

[obsidian monolith]

Lionel:

[prisma prairie]

does this look right?

[proud sparrow]

as expected? It jumps to 19 and stays there

domain can be simplified

you can combine the last 2 intervals

range is argh

please make sure your intervals are (small number, big number)

@prisma prairie

[prisma prairie]

oh

[uncut mulch]

where are you getting the (5,inf) (and (inf,5)) in the range?

[prisma prairie]

yeah i was thinking of the domain there lol

is that better?

[uncut mulch]

better than before, but are you able to combine those intervals in the range?

[uncut mulch]

it can still be simplified further