precalculus

[proud raven]

and sometimes you cant see the difference

[gaunt mural]

Ok

So if i have a horizontal strech by two its a vertical compression by one half?

[proud raven]

if you look at something thats not a quadratic you'll see it though

no

certainly not always

[gaunt mural]

but like for like quadratics and shit

[proud raven]

i can give you examples where that is not true at the very least

id say for all intents and purposes they are two different things that just sometimes happen to be the same

[gaunt mural]

Ok

[uncut mulch]

@vague zephyr ask your teacher why they marked it wrong.
someone else can dbl check whether there's an issue

[vague zephyr]

he wont tell me, he just says to use my readily available resources which aren't really helpful

[gaunt mural]

Also if a function has an inverse you use the horizontal line test but what do you use the vertical line test for?

[proud raven]

iirc the vertical line test is for functions right

its how you tell if something is a function of some variable or not

[gaunt mural]

k

[proud raven]

i guess its clearer to say its how you confirm that a single input doesnt map to multiple outputs

when you have f(x)=y

[gaunt mural]

One more question rational functions are usually infinitely discontinuous. I know not always but usually compared to like quadratics or cubic

[vague zephyr]

inflame do you know anything about sets and intervals? unions and intersections?

[gaunt mural]

kinda

[uncut mulch]

you reasoning was fine, A is a subset of C so you don't need to include that
you're left with B U C which is what you had

[vague zephyr]

thats what I thought but somehow it is incorrect

[gaunt mural]

Idk why its wrong thats what I would've done

[vague zephyr]

thank you all for your help, im going to send an email to my teacher explaining my methodology and issues with the problem

[gaunt mural]

What are you using if its khan you can ask for a hint

[proud raven]

looks like ww

[vague zephyr]

i used youtube, khan, and mathisfun

[proud raven]

i dont see what is wrong with your answer either 🤷

fwiw

[gaunt mural]

Are you taking honors or on level pre cal

[uncut mulch]

btw for a) it is preferable to put the lower set on the left

[vague zephyr]

this is an equation on webworks, its a HW page he set up

im in college

[gaunt mural]

I am in Hs

[proud raven]

should be $(- \infty,\ -8)\cup[-5,\ \infty)$ right

that was your first answer

[obsidian monolith]

jan Niku:

[vague zephyr]

yes

[proud raven]

😅

so yea

🤷

[viscid thistle]

Yo

[vague zephyr]

again, thank you all for the help

[uncut mulch]

b) and c) also missing = signs in the question

[viscid thistle]

Do you guys have any website to practice Math?

[uncut mulch]

khan

[proud raven]

pauls

reddit has some great problem sets if youre looking to drill too

or uni webpages

if you really want to drill wolfram has an infinite problem generator for more basic stuff

[gaunt mural]

Kuta is really good cause they have ws and answers and if u get them wrong u can go on youtube where people do the sheet and explain each problem

https://www.kutasoftware.com/

[proud raven]

video-worked examples

[gaunt mural]

Lol yea

Also Online Chem tutor is really good at explaining math stuff aswell

[chilly jackal]

are you talking about the organic chemistry tutor?

[proud raven]

martin van beizen is great, wish id found him earlier

[gaunt mural]

Yes

On youtube

[proud raven]

but you really need to focus on his material and understand every video, so it can be a lot of work

[gaunt mural]

PatrickJMT is good too

[chilly jackal]

he's a godsend, he's carrying me through all of my classes

[gaunt mural]

IKR

[proud raven]

JMT yea

organic chem tutor starter pack

[gaunt mural]

lolool

[viscid thistle]

Hi guys

How would I solve this...

|x+3|=|2x+1|

I understand that I would have to write out four equations

1. x+3 = 2x+1
2. x+3 = -(2x+1)
3. -(x+3) = 2x+1
4. -(x+3) = -(2x+1)

I'm unsure which one to use

[proud raven]

all of them right

although some are the same

[viscid thistle]

Yes.... all four equations

[proud raven]

4 and 1 are equivalent

[viscid thistle]

I know 4. is exactly the same as 1.

I don't know about 2 and 3 though

[proud raven]

i cant imagine youll get 4 solutions

what happens when you solve them

[viscid thistle]

I think it's better to use two

for 1.
i got x = 2

[proud raven]

you know all of them are valid so you need to solve all of them unless you can show they're equivalent to another solution anyways

[tribal sun]

@proud raven my phone died sorry

[proud raven]

oh i was wondering lol

i was hoping you didnt give up

😄

[tribal sun]

i had to drive back home and im on my desktop lol

haha

[viscid thistle]

2 and 3 are equivalent too

[proud raven]

@viscid thistle i got two solutions

so that makes sense

[tribal sun]

i wanted to go somewhere quiet where i could study but library was closed idk any other place

lol

[viscid thistle]

x=2, and x = -4/3

[proud raven]

yup

[viscid thistle]

those are my solutions

[proud raven]

@tribal sun I had asked you if you understood where f(x+h)=4(x+h)+3 came from

[tribal sun]

oh ok so every time some thing says f(x)= ax+b, you replace the x in that thing with (x+h)??

[proud raven]

yea, like f(x) = 2x, then f( 😄 ) = 2 😄

[tribal sun]

OMG

my mind has been given information :0

ok now what about the spoopy equations such as f(x)=X^3+5

[proud raven]

I mean its the same you just need to be careful about parens

f(x+a) = (x+a)^3 + 5

and remember all your algebra rules, etc

[tribal sun]

f(x)=X^3+5 ; f(+ )

[proud raven]

lol

[patent beacon]

f() = ³ + 5

[proud raven]

we were literally just doing this in ODE too

for integrating factor for first order linear

[tribal sun]

f(+ )^3+5

is that right

[proud raven]

yea

[tribal sun]

yes

[patent beacon]

Nop, the notation is getting a little mixed

[tribal sun]

🙂

0us

[proud raven]

so

[tribal sun]

:0

[proud raven]

oh

yea

no f

[patent beacon]

f is the name for your function, it doesn't appear in the function

[tribal sun]

ohhh ok

(+ )^3+5

🙂

[patent beacon]

Is f( + ), exactly

[tribal sun]

ye

have any of you gone to starbucks to study befoe

[proud raven]

i usually just study wherever i am unless theres a problem

travel eats time and motivation

[tribal sun]

Ken Eckert, English Professor, South Korea (Canadian)
Answered May 1 2017 · Author has 5.3k answers and 9.4m answer views
Originally Answered: Why students go to Starbucks to study instead of study at home?
Confession: When I was a college student, and even now when I’m doing work as a professor, I love studying in a coffee shop or a pub. Your home can be surprisingly hard to get things done in, with the temptation of your computer, the phone ringing, and siblings or parents bothering you. I like the consistent white noise of a cafe/pub, with background music, people talking, pool balls clinking, and so on. This cushion of sound works for me. Plus, there’s coffee or beer.

183.4k views · View 4.3k Upvoters

true tho: Your home can be surprisingly hard to get things done in, with the temptation of your computer, the phone ringing, and ** siblings or parents bothering you.**

[proud raven]

dont let yourself get too sidetracked you were almost done with that problem

[tribal sun]

ye

f(x+h)=4(x+h)+3 back to it

f(x+h)=4(x+h)+3 - 4X+3

what does one do after that

[proud raven]

heres the steps i would follow

write out what f(x) and f(x+h) equal

then plug them into your quotient without simplifying

then perform one algebraic step on each line with a small note about what youre doing

dont try to simplify in your head or cram too much on one line, paper is still cheap

etc

[tribal sun]

what happend to the f(x+h)=

[proud raven]

oh you mean why didnt i write it

i was skipping steps 😔

only to demonstrate thats as much work as i can possible do for you really and thats probably too much 😔

[tribal sun]

OH fok

wait

my gears are turning

4h/h

cancel h

BOOM 4

[proud raven]

wheres your steps

[tribal sun]

on my ppr

[proud raven]

i mean i dont need to see them but your teacher will

[tribal sun]

lemme send

[proud raven]

also i guess worth saying im not sure the question asked you to simplify lol

no its okay

i believe you

just make sure not to skip steps when youre working through them

[tribal sun]

my i phone xr is charging slow

[proud raven]

even if you feel like you see where its going

[tribal sun]

u need to see it

i need help to make sure im doing the steps correctly lol

u are my teacher now

[supple nimbus]

hi folks, can you suggest me a book for precalculus 🙏🏽 I am learning for myself

[tribal sun]

i got a book

[proud raven]

@supple nimbus #books-old

[supple nimbus]

thank you 🙏🏽

@tribal sun which one

[proud raven]

theres an open stax precalculus book i think

@supple nimbus https://openstax.org/subjects

[tribal sun]

https://www.slader.com/textbook/9780134119281-precalculus-enhanced-with-graphing-utilities-7th-edition/

[supple nimbus]

wow

[tribal sun]

not the answers part

[supple nimbus]

what is this amazing thing openstax

[tribal sun]

wut is open stax

[proud raven]

free introductory math books

some schools use them now

[supple nimbus]

thanks a lot man, this is so so good

[tribal sun]

GLIGGA

its FREE??

[viscid thistle]

Dang.

you have to collect like terms

This is actually so helpful.

4h / h

lim is 4

[tribal sun]

yes i got that solution @viscid thistle

[viscid thistle]

Okay, need any more help?

[tribal sun]

i might be good for now ill see what pops up along the wat

way

[proud raven]

you want good practice

?

find difference quotient of $f(x)=\frac{1}{x}$

[obsidian monolith]

jan Niku:

[tribal sun]

i got a shit load for homework lol

ok lemme try

[proud raven]

youll inevitably have one with a fraction in homework or exam im guessing so no stress if you dont have time

[tribal sun]

i mean i have time and hmmmmm the confusion is setting in again

f(x+h) - 1/(x+h) / h

[proud raven]

that f is creeping in again

[tribal sun]

;O

[fallow oxide]

Hey I have a question that I'm not sure how to start:

The latitude of Miami, is 25˚47'16". The latitude of Cleveland is 41˚28'56". Find the velocity of each city in space due to the rotation of the earth on its axis. Which one is traveling faster? By how much?
Radius of the Earth is 3960 miles.

[proud raven]

you need to know if you have tilt in there too im guessing

[fallow oxide]

if i do, it's not provided

[proud raven]

no

[tribal sun]

[proud raven]

,rotate -90

[proud raven]

theres your worksheet

i dont understand what that f in a box means

[tribal sun]

oh lol

f box i didnt erase it

[proud raven]

you know how to do these problems so this is more a test of your ability to not confuse yourself 😄

[tribal sun]

with out f

[proud raven]

im sorry i dont follow

when you substitute you do not have any f's remaining

here

substitute

but importantly here note that the output of f(x) and the output of f(x+h) do not have any f's in them

[proud raven]

for simplifying id suggest doing one thing on each line with a note about what you did, you dont have to be that severe but if you cannot follow what you did to get from step to step you are going to be in trouble

you may want to practice just setting some of these up if you feel that is a better use of your time

you will eventually abandon this process entirely, but its a good way to make sure you can set things up and follow your own thought process, so dont stress too much over the particulars

[tribal sun]

Is that right tho

[proud raven]

yes

[main vessel]

Having trouble finding a method to get the x intercept 😓

[viscid thistle]

I can’t remember those 5 step to get a 1 problem

[fresh kelp]

nvm i got it

[vague zephyr]

Deku?

[drifting helm]

what does it take to make .030005 = 1

Trying to simplify a fraction for this kinda problem

[tribal sun]

So like is that the answer for my question

[viscid thistle]

Yoo is precalculus hard?

[stuck lark]

not if your algebra skills are solid

[viscid thistle]

Not really good at algebra but gotta learn

[vague zephyr]

this is the email my instructor sent me

im still not sure how to go about solving this

@uncut mulch would you be able to help me with this?

[uncut mulch]

A and C overlap, what's your teacher on about

e-mail a pic to your teacher showing that they overlap

if they didn't then there wouldn't be a solution for part c

was that in response to your old answer, or updated answer?

[vague zephyr]

my old answer was (-inf,-8)U[-5,inf)

it was in response to the old answer

this updated answer isnt really an answer it was just me messing around

[uncut mulch]

his response was a bit dodgy

[vague zephyr]

im not sure anymore

well thank you again for your time

[uncut mulch]

ask them in person

or show them how A and C clearly overlap

(-inf,-8)U[-5,inf) was completely fine

[vague zephyr]

yeah i can ask him tomorrow

[tribal sun]

@viscid thistle the only way you’ll get batter is actually take time to study

I’m not good at algebra either

[viscid thistle]

It’s boring to study bruh @tribal sun

I want to but it’s boring

[weary raft]

yo can someone help me

[viscid thistle]

Sure.

[weary raft]

what am i supposed to do here?

[viscid thistle]

Do you know how functions work?

[weary raft]

yes

im just not sure what its asking

[viscid thistle]

Oki.

Let's make the problem easier.

Given $f(x)=2x+1$, find $f(2)$.

How would you go about solving this?

[obsidian monolith]

[TD] Aurora:

[weary raft]

plug in the 2 and solve for y

[viscid thistle]

So given $f(t)=\frac{2t^2+3}{t^2}$, find $f(y+3)$

[obsidian monolith]

[TD] Aurora:

[viscid thistle]

What would you do here?

[weary raft]

just plug in the y+3 and solve

wow ok

[viscid thistle]

Precisely.

Yes.

[weary raft]

thanks sm

[viscid thistle]

Mhm.

[sharp pagoda]

anyone got some sort of video i could watch, that goes over more or less the basic concepts of precal. ive got a dynamics test coming up this week and havent done anyting math related for several months now, i dont have any worksheets or textbooks that could help me through. if anyone has a link to something like this itll be very appreciated

[weary raft]

oof

[sharp pagoda]

lmfao

[weary raft]

ill try to find one

[sharp pagoda]

thanks in advance

sorry for my profile picture btw i know its disturbing

[viscid thistle]

https://www.youtube.com/watch?v=ubDIa9m3_T4

Skip like a minute and the last minute.

It's him advertising himself.

He runs through everything Precalc goes through.

[sharp pagoda]

thank you kindly

just a quick question since im here

[viscid thistle]

Anything you don't know I suggest you refer to https://openstax.org/subjects https://www.khanacademy.org/ or https://brilliant.org/

[sharp pagoda]

assuming you taken calc obviously , anything related to combs and perms , does that come up anywhere ?

[viscid thistle]

Combs and Perms is something that I was taught in Alg II.

Just refer to one of the websites I suggest for anything you don't understand.

Concepts you don't understand after researching it, then you can come back here to ask questions.

We help people not teach people entire concepts here :/

[sharp pagoda]

oh i know that you guys have a life aswell! well thank you for this information, ive copied everything 🙂

[viscid thistle]

Quick question y=g(x+10) lets say I have point (5,-7) we subtract 10 from 5 I know that, so we get -5, what would happen if I changed it to the following
y=g(5x+10)?

I keep getting (-1,-7)

[chilly jackal]

i'm lost, what are you trying to achieve from subtracting these?

[viscid thistle]

Suppose (5,-7) is a point on the graph of y=g(x), what point is on the graph if it goes through y=g(5x+10).

What do I have to do to the 5 next to the x? I know that if it was (x+10) alone I just have to subtract 10 from 5, and I would get -5

[chilly jackal]

divide by 5, i think.

so then x = -1

you're trying to solve for X right? @viscid thistle

[viscid thistle]

yeah

[chilly jackal]

is that right?

[viscid thistle]

Yes

and if the point was (-5,-1) and the equation is (5x+5)?

[chilly jackal]

i'm not sure about that one, because i end up getting y = g(5)

but we aren't given g(5)

[viscid thistle]

it was (-2,-1)

hm I'm having trouble seeing what step

or what to do

[tribal sun]

Please explain

How to do

[chilly jackal]

@tribal sun

are you trying to find the inverse function?

@viscid thistle weird, and there are no other missing components in the problem?

[tribal sun]

Difference of quotient

@chilly jackal

[viscid thistle]

I got it already you multiple by the inverse of the 5

[chilly jackal]

oooooooooh, sorry i wasn't any help @viscid thistle

[viscid thistle]

you got me going on the idea thanks though

[patent beacon]

@tribal sun
Don't forget to divide everything by h

[pseudo sonnet]

how is my calculator getting 1.5

tfffff

[uncut mulch]

not enough parentheses

[pseudo sonnet]

?

[uncut mulch]

it was entered as
1/(-2) * (-3) which is (-3)/(-2) = 1.5

[viscid thistle]

Your calculator is reading it like this: $\frac{1}{2(0)-2}(0-3)$

[obsidian monolith]

leviosa:

[viscid thistle]

More parantheses.

[pseudo sonnet]

see

same thing 1.5

[willow bear]

it doesn't know that the (x-3) is supposed to be in the denom.

[viscid thistle]

(x+1)/[(2x-2)(x-3)]

[pseudo sonnet]

oh

i see

🤦

ty

[viscid thistle]

👍

[pseudo sonnet]

can someone check my answer for the first question under deeper understanding

this is what i came up with for my equation

[wintry jewel]

@pseudo sonnet Hey man that looks good , should be correct for what the question is asking

[pseudo sonnet]

Construct a polynomial inequality that has only 1 solution. Explain your answer and why it works.

can someone tell me what 1 solution means in this case?

is it like (x, x2) would be an example of one solution

like (1, 4)

im confused lol

[patent beacon]

Something like x² + 1 ≤ 0, that's a polynomial inequality

However, mine doesn't only have one solution

@pseudo sonnet

[pseudo sonnet]

your's doesnt have any solutions

😂

[rigid sun]

What

If a polynomial inequality only had one solution

Would that even be an inequality?

[gaunt mural]

How do you know if a function is odd?

[stuck lark]

If a function f is odd, f(-x) = -f(x)

[willow bear]

by checking whether it satisfies the definition of "odd function" @gaunt mural

[patent beacon]

@rigid sun
Yeah, for example x² ≤ 0

That's an easy one. Hopefully Vici made their own

[rigid sun]

im probably misunderstanding what "one solution" is

[serene heath]

which ones true for all x?

the one kaynex posted defo isnt

[rigid sun]

ooof

i read that wrong

[viscid thistle]

Why is this not working

Wait, maybe I have a typo

[viscid thistle]

I don't understand this questions

What do they mean? All together? Then I need the last number to get the sum

What am I supposed to do??

[viscid thistle]

gl :)

im just as clueless

:(

I see...

[uncut mulch]

are you familiar with arithmetic sequences?

[proud raven]

if a rotational problem with constant acceleration is asking for the magnitude of the acceleration vector

that just menas the rate of acceleration right

im looking at slader but i still cant quite tell if ive gotten the right answer

does the magnitude not have rads?

i got the same number but different units and i cant quite tell if thats the final answer

[stuck lark]

@proud raven radian is dimensionless

[proud raven]

oh right

does magnitude have dimensions?

should there be a unit on this at all if theyre asking for just magnitude

or the speed is the magnitude

[stuck lark]

the magnitude of a quantity is irrelevant to what dimensions the quantity has @proud raven

if my velocity is "2m/s east" then the magnitude of my velocity is "2m/s". magnitude just means take away the direction. taking the magnitude has no influence on the units

[proud raven]

okay 😄

thank you

[stuck lark]

no prob

[rose locust]

I feel really dumb for asking this but why is sin90 equal to 1? When sin is y/r it would be 1 divided by the square root of 2?

[stuck lark]

what's y/r?

[rose locust]

y/r replaces o/h which is 1 divided by the square root of 2

[ruby frigate]

sine is Opposite / Hypotenuse, here both are the same number

[native timber]

we assume that (x, y) is a point on the unit circle (therefore for all such points, r = 1)

r isn't the square root of two

[rose locust]

okay, but then cos90 will be x/r which is 1/1 , but the textbook says 0.

[native timber]

if the angle is 90deg, x is 0

[ruby frigate]

This should make things clearer

Consider the x value cosine, and the y value as sine

Would highly recommend memorizing the first quadrant and the other 3 can be figured out with a few tricks if you know the first quadrant

[rigid beacon]

yea basically you memorize the first quadrant

and then learn to place the triangles on the unit circle in the right direction

and since the unit circle lies on the xy-plane

if the line goes a certain direction, you know the sign

[viscid thistle]

New question 😢

I know how to describe if a sequence is quadratic, linear or exponential but not a serie!!!

[ruby frigate]

Is there more information to the questions

[upper flint]

find domain

I am confuzzled

[uncut mulch]

4 or 7?

[upper flint]

both

[rigid sun]

for number 4, the domain is all real numbers

[upper flint]

how do i write work for that

[rigid sun]

for number 7, 4 to infinity including 4

[upper flint]

uhh

explain

[rigid sun]

ok

well there's to main things you need to look out for in domain

[uncut mulch]

are there any values of x where f(x) is undefined

[ruby frigate]

Well if you put anything <4 you will have a negative in the square root

[rigid sun]

remember,

the demoniator cannot be zero

[upper flint]

x^2 + 1 is not equal to 0

[rigid sun]

and the radical cannot be negative

[upper flint]

then just write that out

is that right for domain of 4

[warped crane]

How do the arrows work? If it's

I have (x - 1) (x + 3)

(x + 3) = 0 etc etc for X but how do you work arrow notation into this?

[stuck lark]

what arrows are you talking about?

[warped crane]

Like hmm

The solution is X < -3 or -3 < x < 1 or x > 1

How did they get the arrow notations?

[willow bear]

arrow notation?

[warped crane]

I'm definitely using the wrong term

[willow bear]

can you show what you're referring to as arrow notation

[warped crane]

Sorry I mean greater than, less than.

Like how do you work out which way it'll point?

[willow bear]

what are you even asked to do

write the domain of f?

[warped crane]

Yeah

I have this

[willow bear]

the domain of f consists of all real numbers except for -3 and 1

[warped crane]

I want to know how they get < >

[willow bear]

the domain of f consists of all real numbers except for -3 and 1

your question is kinda indicative of overthinking ngl

draw a number line

mark off the points -3 and 1 on it

see that the number line gets broken into three intervals

[warped crane]

Right I have this

[willow bear]

each of which can be described either in interval notation or with an ineq

no

no no no

i didn't say to graph the function.

the function itself is irrelevant for now.

[warped crane]

Oh just the line

[willow bear]

do what i told you to do.

[warped crane]

ok

[willow bear]

draw a number line
mark off the points -3 and 1 on it
see that the number line gets broken into three intervals
each of which can be described either in interval notation or with an ineq

[warped crane]

So we have -3 and 1 on the line

So it goes x < -3 because if it goes x > -3 then 1 would be in it?

[willow bear]

...no

[warped crane]

And the domain of f consists of all real numbers except for -3 and 1

[willow bear]

look

just look at the number line

and quit overthinking it

[warped crane]

Ah ok

So pretty much just two blank spot on the number line because the domain of f is everything but -3 and 1

So the < > is just an indication of that gotcha

Thanks @willow bear

[lyric ether]

I need help factoring this out correctly

ended up just expanding the entire thing but would like to avoid that in the future...

[obsidian monolith]

Namington:

[short sorrel]

@lyric ether

this will allow you to make a few simplifications

[lyric ether]

I ended up with something like (t1-t0)(v0+a0t0+1/2a0(t1-t0)) while attempting to do so

Which I figured out that the a0t0 does not belong there after I expanded everything out

But not entirely sure why this is the case

@short sorrel

[short sorrel]

what do you mean, "does not belong there"?

so we have

$(t_1 - t_0)(v_0 + a_0 t_0 + \frac12 a_0(t_1 - t_0))$

[obsidian monolith]

Namington:

[short sorrel]

now, we do have to expand a bit here, but it's much less complicated than it would've originally been

distribute the 1/2 a_0 into (t_1 - t_0)

$(t_1 - t_0)(v_0 + a_0 t_0 + \frac12 a_0t_1 - \frac12 a_0t_0)$

[obsidian monolith]

Namington:

[short sorrel]

now we note that $a_0 t_0 - \frac{1}{2} a_0 t_0 = \frac12 a_0 t_0$

[obsidian monolith]

Namington:

[short sorrel]

$(t_1 - t_0)(v_0 + \frac12 a_0 t_0 + \frac12 a_0t_1)$

[obsidian monolith]

Namington:

[short sorrel]

now, depending on the convention your class uses

this might be enough

or they might want an additional factoring step

with the 1/2 a_0

in that case, you'd get \
$(t_1 - t_0)(v_0 + \frac12 a_0 (t_0 + t_1))$

[obsidian monolith]

Namington:

[short sorrel]

again, depends on the convention of your class

[lyric ether]

Thank you, I just thoght I did something wrong rather than requiring more steps to arrive at the answer.

And now that I look into it... it really just flew past my head...

Seems like summer break really threw me off

[lyric ether]

@short sorrel sorry to bother you again but could you take a look at this?

Because it seems like it is not t1+t0 but rather t1-t0?

[short sorrel]

,w factor (t_1 - t_0)(v_0 + a_0 t_0) + (1/2) (t_1 - t_0)a_0(t_1 - t_0)

[obsidian monolith]

Display results online and refine query

Requested by Namington#1983

[short sorrel]

bleh so many alternate forms

but uh lemme see

,w (t_1 - t_0)(v_0 + a_0 t_0) + (1/2) (t_1 - t_0)a_0(t_1 - t_0) = (t_1 - t_0)(v_0 + (1/2)a_0 (t_1 - t_0))

[obsidian monolith]

Display results online and refine query

Requested by Namington#1983

[short sorrel]

yeah, you made a mistake somewhere

looks like

ill check

[lyric ether]

Just came to revisit as my prof had this ans

[short sorrel]

oh hold on

I see the issue

you somehow factored $(t_1^2 - t_0^2)$ into $(t_1 - t_0)^2$, it looks like

[obsidian monolith]

Namington:

[short sorrel]

which is not correct

im referring to this step

if you factor out $(t_1 - t_0)$ from $(t_1^2 - t_0^2)$, you should be left with $(t_1 + t_0)$

[obsidian monolith]

Namington:

[short sorrel]

since $(t_1^2 - t_0^2) = (t_1 - t_0)(t_1 + t_0)$

[lyric ether]

Oh I see

[obsidian monolith]

Namington:

[lyric ether]

Thanks again

Probably shouldn't blindly follow my math prof but 1st yr anxieties....

[long crown]

How to square a binomial Useful to grasp this for #precalculus , hope you find some value 🙂 https://youtu.be/cMLzSCAX_2g

[willow bear]

holy fuck, 16+ minutes?!

[long crown]

@willow bear Got quite a few questions in there! 😉

And I go very slow and in depth.. Got some critique for being slow, but many people seem to enjoy it aswell... I rather go to slow than to fast I feel like...

Theres a speed up function after all 😉

[heady jewel]

stop with the winks lmao

[willow bear]

i mean idk maybe i shouldn't really be complaining, what with me not being in the target audience and all

[pseudo sonnet]

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

Notice that when we plug and into the expression we get the positive number: 2166, and when we plug and we get the negative number -13,254. State all values of x and y for which the above expression is zero or positive. Explain how you got your answer.

edit: I found something out, im currently editing my solution. ill post whatever i end up finding in a few seconds

can anyone help me

not really sure what to do next

[uncut mulch]

don't forget the y

[pseudo sonnet]

oh right

good catch

[pseudo sonnet]

nvm G's

got it solved

thanx~

[low musk]

Can I do this? Kinda dumb question but I duck at math

[proud sparrow]

@low musk you mean $\frac{\frac{-h}{x^2+xh}}{h}=\frac{-1}{x^2+xh}$?

[obsidian monolith]

Element118:

[proud sparrow]

If so, yes, think of it as dividing both the numerator and denominator by h

[low musk]

oh i get it now thanks @proud sparrow

[long crown]

@heady jewel 😉 😘

[heady jewel]

?!

[rancid flame]

Could anyone recommend a good resource for studying and practicing trigonometric identities? I've been using Khan Academy, but it just seems to skim the basics of them.

[marble raven]

I got 80% on my first test

It dropped my grade from 100% to 85%

Super discouraging to have a B right now

FUCK

[shrewd turtle]

@rancid flame there are honestly too many identities in trig so you just have to learn the ones the class says to learn

tan= sin/cos

sin^2 + cos^2 = 1

cot= cos/sin

tan^2 +1 = sec^2

learn the unit circle it doesnt go away

those ones are pretty common off the top of my head

[ruby frigate]

You need to know how they relate to each other, the pythagoren identities, the double angle, are very common

[shrewd turtle]

double angle shows up in calc a lot but not in precalc

[ruby frigate]

The rest not so much but it is still good to know them

well if he is taking precalc, good practice now

[shrewd turtle]

exactly

[rancid flame]

I'm aware of those three, and I know the basics of the unit circle. I just get overwhelmed when equations start popping up with the different trig functions.

[ruby frigate]

Its all practice

[shrewd turtle]

yep

[ruby frigate]

Lets say you have to check the identity list for every question in your homework, eventually youll see there is a pattern

and wont have to check as often

[shrewd turtle]

trig more sense the more of it you learn

generally

[rancid flame]

That's why I was wondering if there's somewhere online I can do practice tests or something. My book is no help and Khan Academy seems to have a huge gap in that area.

My book is written as if it's trying to teach someone that already knows the material.

[ruby frigate]

What kind of practice are you looking for

like identity proofs

?

[rancid flame]

I guess? I'm not really sure honestly. After learning the unit circle, I just kind of got lost.

My book throws formulas around a lot and doesn't really walk you through anything.

[shrewd turtle]

so what are you trying to do with the problems?

[rancid flame]

Well, for instance, I got a problem like this on my practice exam:

Find the exact value of sin(2x), cos(2x) given that tan x = -4/23, -pi/2 < x < 0.

I have no idea how to even begin approaching that problem.

[shrewd turtle]

well, know that tan= sin/cos

[ruby frigate]

What Quadrant is tangent negative

[rigid sun]

[rancid flame]

Not sure off the top of my head. I know the positives and negatives for sin and cos, but I'm not sure about tangent.

[ruby frigate]

think about it

since tan=sin/cos

[lucid crow]

when is sin negative?

ask urself that

[rancid flame]

So, cos is negative in quadrant 2 and sin is positive in quadrant 2. So tangent is negative in quadrant 2?

[rigid sun]

👌

[ruby frigate]

Correct

You need another quadrant

[rancid flame]

And the opposite in quadrant 4. sin is negative and cos is positive.

[ruby frigate]

Exactly

in q1 both are positive so tan is positive

in q3 both are negative and since negative /negative is positive , tan is positive

[rancid flame]

Okay.

[ruby frigate]

So which quadrant do you think its in

2 or 4

[rancid flame]

Hmm...

4?

[ruby frigate]

Sorry something just came up, can some one else help him out I gtg

[rigid sun]

ok

so to continue

[rancid flame]

I think it's quadrant 4 because x < 0 and greater than -pi/2. So, would that be between 270 and 360 degrees on the unit circle, or am I in the wrong line of thinking?

[rigid sun]

very good

ok so you know it is in quadrant 4

[rancid flame]

I'm not sure if I'm thinking right about it or not.

[rigid sun]

you are

0=360 degrees= 2pi

in trig

[rancid flame]

Right.

[rigid sun]

ok

so you know that sin is y and cos is x

[rancid flame]

Yes.

[rigid sun]

but, sin=opposite/hypotenuse

[rancid flame]

Yeah.

[lucid crow]

SOH CAH TOA gang what up

[rigid sun]

so we can use pythagorean's theorem to solve for the hypotenuse

remember, tan= height/length

and height^2+length^2= Hypotenuse^2

what is our hypotenuse?

[rancid flame]

Okay, so the equation would be sin(2x)^2 + cos(2x)^2 = (-4/23)^2?

[rigid sun]

not quite

(-4/23)= tan (x)

tan(x) is not the hypotenuse

[rancid flame]

Would the hypotenuse just be 1 then?

[rigid sun]

remember, sin will give you the height (y) and cos will give you length (x)

not in this case

[rancid flame]

Hmm..

[rigid sun]

-4/26= height/length=tan(x)

so...

(-4)^2 + 26^2=?

?= hypotenuse

[rancid flame]

Okay, it was -4/23, but it should be sqrt(545) I think.

[rigid sun]

ok

so we have an opposite (height)

and a hypotenuse

and opposite/hypotenuse= sin(x)

so what is sin(x)?

[rancid flame]

Hmm...

-4/sqrt(545)?

[rigid sun]

yeah

now here's the no brain part

did your teacher give you the double angle formula?

[rancid flame]

No, I'm studying online independently mostly. Preparing for university entry.

[rigid sun]

ok

this is important to know

you see, we're trying to find sin(2x) and cos (2x)

so we're going to use the formulas on the left

i'd recommend memorizing these formulas, because they'll come back later

for now, we're going to use the left hand side, the double angle formulas

[rancid flame]

That sounds like a nightmare to memorize.

[rigid sun]

Quizlet helps

anyway

so we just found out that sin= -4/sqrt 545

You see any formulas for sin2x or cos 2x that only have sinx in them?

[rancid flame]

No?

[rigid sun]

i mean, they'll have other numbers, but there will be know unknown values

since we know the value of sin x

[rancid flame]

Oh, so the first one should only have cos as an unknown.

[rigid sun]

yes

we have not found cosine

the second one also has cosine

so does the third

but, the 4th one does not

the 4th formula tells us that we can find cos2x by using only 1 and sin

so lets use that one

go ahead and calculate cos2x by using the 4th formula on the left side

[rancid flame]

So, sin x^2 = 16/545

[rigid sun]

yes

[rancid flame]

Multiply by -2 = -32/545

[rigid sun]

continue👌

[rancid flame]

So, cos x = 513/545

[rigid sun]

cos 2x= 513/545

[rancid flame]

Oh, right.

[rigid sun]

now

we need to find sin2x

but there is only one formula and that uses cosx

so lets find cosx real quick

recall that tanx =-4/23 and the hypotenuse was sqrt545

[rancid flame]

-4/23

But yes.

[rigid sun]

cos x =?

[rancid flame]

Hmm...

If cos 2x = 513/545, then cos x = 513/1090?

[rigid sun]

that would be a mistake

cos(2x) does not equal 2cos(x)

[rancid flame]

Okay.

Not sure then.

[rigid sun]

its the same way you found sin(x)

the ratio -4/23 gave you the height and length

and we found out the hypotenuse

you took the height over hypotenuse and found sin(x)

-4/sqrt545

now do the same thing you were able to do earlier except use the adjacent(the length) over the hypotenuse

cos x=?

[rancid flame]

Hmm, not sure. Losing my train of thought a bit.

[rigid sun]

that's ok

just remember tan= opposite/adjacent

adjacent =23

hypotenuse=sqrt545 as we found earlier

[rancid flame]

Okay, so cos x = 23 / sqrt(545)

[rigid sun]

yes

you see it's very similar to finding sinx

the only difference is you used the adjacent (23) instead of the opposite (-4)

remember you got -4/sqrt545= sinx but this time you got 23/sqrt545=cosx

[rancid flame]

Okay, I think I get that, but the 2x part is still throwing me off some.

[rigid sun]

the 2x part simply means double the angle

[rancid flame]

Hmm, okay.

[rigid sun]

so right now, the sin of angle x= -4/sqrt545

luckily mathematicians already solved for you how to find the sin of double that angle

so sin(2x) doesn't mean double the answer of sinx, it means that we are trying to find the sin of an angle double the quantity of our original angle x

an example would be like sin(20) to sin(40)

[rancid flame]

Okay.

[rigid sun]

ok

so back on track

now going back to the formulas sheet

we have to find sin2x

but the formula involves quantities we just solved for!

sin = -4/sqrt545

cos= 23/sqrt545

all we need to do now to find sin2x is plug those values into the formula

[rancid flame]

So, sin(2x) = (-8/sqrt(545)) * (23/sqrt(545))

[rigid sun]

yes

[rancid flame]

so sin(2x) = -184/545.

[rigid sun]

yes

[rancid flame]

and cos(2x) = 513/545.

[rigid sun]

you've solved the problem

[rancid flame]

Okay, and that matches the answer on the test. What a headache...

Thanks for walking me through it though.

[rigid sun]

yes

don't forget to study this chart over the next few days

sin(x/2) is half of the angle

i.e. 20 degrees and 10 degrees

[rancid flame]

Is there anyway to derive these in a reasonable time? I'm not sure I could possibly memorize all those.

[rigid sun]

oooooooooo no

ive derived these myself

and it took about an hour

[rancid flame]

Geez.

[rigid sun]

there are multiple proofs that go into these

let me see if i can find you a quizlet

[rancid flame]

I don't know how someone can cram those all in their head. They all seem arbitrary to me.

[rigid sun]

you may not need all of them

depends on what you're dong

doing

if you just want to get ready for college, just keep the chart handy

practice problems with trig and use your best memory, then look at the chart if you need to

[rancid flame]

I doubt professional mathematicians have all these memorized. Why do we need to memorize them for tests? -_-

[rigid sun]

tests won't give you the formulas

are you still in HS?

[short sorrel]

professional mathematicians use very little trig

[rigid sun]

boi

[short sorrel]

trig is mostly used by engineers

[rancid flame]

No. Taking prerequisite classes for university.

[rigid sun]

ok

[short sorrel]

and some physicists

[rigid sun]

ok

well

[short sorrel]

trig still provides useful examples though, so it's relevant in the teaching of mathematicians

[rigid sun]

keep the chart handy

khan academy will probably give you more problems using those formulas

[rancid flame]

I just mean that someone in a profession wouldn't need to know them all because they could just use a reference. So why should someone have to memorize them for a test?

[rigid sun]

i mean

you could say that about anything

[rancid flame]

You would still have to know how to use them.

[rigid sun]

i mean you

[rancid flame]

Sorry, I'm just frustrated and ranting.

[rigid sun]

won't have to use them all the time

i forgot those formulas

i only remembered them because of what i'm doing now

you'll know when to start studying them

[rancid flame]

That's what I mean. Why do you have to remember them for a test when you'll forget them soon after?

[rigid sun]

to pass the test

[rancid flame]

That's my point. XD

You learn them to pass a test, you forget them after, so what have you gained?

[pale kettle]

Why were you taught the multiplication table?

Why don't you just look it up every single time you have to multiply numbers?

[rigid sun]

thats what i do

[pale kettle]

Yeah no one does

[rigid sun]

there are things in education you'll never want to do or even never do again

like my 8th grade dance PE unit

[pale kettle]

learning how to stay active and fit is an important lesson for life I'd say

[rigid sun]

in this case, you struggled through something you didn't know and did it

that's an important life skill

to be open to learning

most of the time

[pale kettle]

Most of the actual topics taught people will never remember, but people use the skills and concepts from all their classes all the time

[rigid sun]

Most of my friends can breeze through subjects like math, but I respect someone who works their tail off to get it right

[rancid flame]

True, but things like this just seem like they're encouraging you to memorize an arbitrary algorithm rather than actually learn why it works.

[rigid sun]

i mean

you can learn why it works

but it's such a long proof

[pale kettle]

Learning why it works is how you understand material

[rigid sun]

i know i hate plug and chug math too

[pale kettle]

I doubt anything that would be done in high school for you guys would be that long of a proof

[rigid sun]

nah dude

proof is long

its aids

never doing that again

[pale kettle]

I'd be surprised but

[rancid flame]

Yeah, it just seems like that was what I dealt with in high school. "Nevermind how it works, just remember it works."

[rigid sun]

my school is tryhard

[pale kettle]

Even if you can't memorize a full technical proof, you can always memorize the general ideas of how the proof goes

[rigid sun]

yes

a lot of other subjects are like that oo

like biology

there's no way they can teach you exactly how photosynthesis works in HS bio or how proteins are synthesized

its just the general concept

well

good job and peace out

[rancid flame]

Thanks for walking me through that.

[viscid thistle]

Hello guys I think there's something wrong here

I can't cancel that two next to n^2

[short sorrel]

whered the brackets go?

[lucid crow]

yo

missing bracket

[short sorrel]

thats better, but now the next step is invalid

$(a+b)c = ac + bc$

[obsidian monolith]

Namington:

[short sorrel]

you need to distribute the n/2 in

alternatively, you could factor a 2 out of (2+2n)

either works

[viscid thistle]

?

[short sorrel]

thats better, yeah

so we have $600 = n + n^2$

[obsidian monolith]

Namington:

[short sorrel]

you can solve this as a quadratic

[viscid thistle]

Yeah

[short sorrel]

$n^2 + n - 600 = 0$

[obsidian monolith]

Namington:

[viscid thistle]

Noup, didn't teach me how to solve quadratics

[short sorrel]

thats... a bit weird then, but thats ok

i guess you can just guess-and-check this

just try inputting some values in $n^2 + n$

[obsidian monolith]

Namington:

[short sorrel]

and you should be able to get 600

with a specific value

[viscid thistle]

Can't I put 2n^2, then cancel the exponent for then divide by 2?

[short sorrel]

I'm not sure what you mean

$n^2 + n$ is not the same thing as $2n^2$

[obsidian monolith]

Namington:

[viscid thistle]

Oh

Well

...

I don't know how to do quadratics

I changed school so maybe that's why

[short sorrel]

again, for this if you're not familiar with quadratics

i'd approach it with guess-and-check

we have $n^2 + n$ and want it to equal $600$

[obsidian monolith]

Namington:

[short sorrel]

so lets try some values

say, n = 20

,calc 20^2 + 20

[obsidian monolith]

Result:

420

[short sorrel]

nope, thats short

30 maybe?

,calc 30^2 + 30

[obsidian monolith]

Result:

930

[short sorrel]

nope, too big

what about 25?

[viscid thistle]

25!

[short sorrel]

,calc 25^2 + 25

[obsidian monolith]

Result:

650

[short sorrel]

thats just a bit too much

maybe 24

,calc 24^2 + 24

[obsidian monolith]

Result:

600

[short sorrel]

there we go.

n = 24.

[viscid thistle]

:00

[short sorrel]

alternatively, if you're at least familiar with factoring quadratics

we rearrange this to $n^2 + n - 600 = 0$

[obsidian monolith]

Namington:

[short sorrel]

then factor the trinomial

$(n + 25)(n - 24) = 0$

[obsidian monolith]

Namington:

[short sorrel]

recall that anything multiplied by 0 is 0

so if either of those factors are 0

the whole thing is 0

so we have two solutions, given by:

n + 25 = 0
n - 24 = 0

solving n + 25 = 0 gives us n = -25

and solving n - 24 = 0 gives us n = 24.

now, we can disregard the negative solution, as I'm assuming this is a sequences & series problem

and it doesn't make sense to have a negative number of terms

so the remaining solution is n = 24

[viscid thistle]

You made a correct assumption

What's a trinomial factor?

I would really want to understand what you did :)

[short sorrel]

ah, so you're not familiar with factoring?

essentially, factoring quadratics is a way to "break them" into smaller factors

if we have:

$(x + a)(x + b)$

[obsidian monolith]

Namington:

[short sorrel]

this expands out to

$x^2 + xb + xa + ab$

[obsidian monolith]

Namington:

[viscid thistle]

Oh I think I remember this!

[short sorrel]

so we want two numbers a and b that sum to the coefficient of x

and multiply to the constant ab

in x^2 + x - 600, the coefficient of x is just 1

(so a and b have to add to 1)

and the ab term is -600

so they have to multiply to -600

the numbers that multiply to -600, and add to 1, are 25 and -24

so a = 25, b = -24 (or vice versa, doesn't matter)

thus we get

$x^2 + x - 600 = (x+25)(x-24)$

[obsidian monolith]

Namington:

[short sorrel]

this approach works specifically when the coefficient of the x^2 term is 1; otherwise, we need a couple more steps, but its the same idea

and we can check this:

$(x+25)(x-24) = x^2 - 24x + 25x - 600$

[obsidian monolith]

Namington:

[short sorrel]

and of course, -24x + 25x = 1x (or just x)

so thats $x^2 + x - 600$

[obsidian monolith]

Namington:

[viscid thistle]

But how did you know it was 25—24?

[short sorrel]

again, we needed numbers that satisfied these properties:

a + b = 1

a * b = -600

25 and -24 are the only numbers that fit this

[viscid thistle]

Oh okay

[lucid crow]

whats going on u gotta solve that quadratic eq?

[viscid thistle]

Thank you gift

[lucid crow]

shiet

it doesnt have to be solved

[viscid thistle]

I'm paying attention

🤔

Well

The problem is this ^^

[rose locust]

How do I start off with answering this ? I’m trying to find theta

[short sorrel]

solve it as a quadratic

rearrange it to $6\sin^2\theta -5\sin\theta + 1 = 0$

[obsidian monolith]

Namington:

[short sorrel]

and you'll note that this is a quadratic equation, but instead of "x" or whatever as a variable

we have sin theta

still, you can treat it the same way for now

start by figuring out what values sin(theta) must take

and then figure out theta from there

[if you like, you can do a substitution, like u = sintheta, to write 6u^2 - 5u + 1 = 0 instead; this may look more familiar]

[undone tide]

Y’all real quick

Would any circle equation

Let’s say tangent to the X axis or something

Just be the distance of the Y coordinate of the center of the circle from the X axis?

By just be I meant the radius

[short sorrel]

if i'm understanding what you're saying correctly, yeah

more precisely, it'd be the absolute value of the y coordinate

assuming this is what you have in mind

[undone tide]

Yes pretty much,

Thank you for the affirmation

Affirmation? Assurance?

One of those big words

[short sorrel]

confirmation?

[undone tide]

Yes thank you

For that also