precalculus

[elfin dirge]

Ok so now that we know a = -1/2 let's solve for b

[narrow peak]

how do u know where the abs apply

applies

[elfin dirge]

where the | symbols are

What it surrounds

[narrow peak]

like as in

where does the mod apply

[elfin dirge]

Oh I see

It's the left side that's being affected by it

[narrow peak]

how do u know

[elfin dirge]

because in the equation |x+b|, it's only different from (x+b) when x < b

Aka the left side

[narrow peak]

let me think

[elfin dirge]

Oh actually I'm wrong about that

[narrow peak]

then what is correct

[elfin dirge]

It's usually the left side, but because it's c minus the absolute value it's the opposite

So the slope is 1/2

[narrow peak]

i formed 2 eqns anyway

[elfin dirge]

Ok

[elfin dirge]

Ok let's find b now

So since 12 = -b/a, we can solve for b

[narrow peak]

sure

let me try

[elfin dirge]

Sorry got the formula wrong

Fixed it

It's negative because usually it's x minus the horizontal offset, not plus it

[narrow peak]

what if i just subbed a=1/2 into my eqn

[elfin dirge]

Oh that'd work too

[narrow peak]

and solve simultaneously

aight

[shadow plaza]

@narrow peak a=1/2, b=-6, c=2??

[elfin dirge]

Don't spoil

[narrow peak]

idk

thats what im gonna find out

lets pretend i never saw that

[elfin dirge]

Btw hmm it doesn't actually matter what side you choose a to be the slope of when it's inside the absolute value brackets. Because if you choose the negative version, then b will also be negative. And you can factor out a negative sign, And because it's inside the absolute value it goes positive

[narrow peak]

nope subbing in wont work

need a 3rd eqn

[elfin dirge]

Ok so just solve it the other way then, it's faster anyway

12 = -b/a

[narrow peak]

ill do 12=-b/a right

yes

[elfin dirge]

Yeah

[narrow peak]

let me just re read above so i can understand

cuz this is practice exam paper and i need to present properly

[elfin dirge]

Ah I see

[narrow peak]

done thx

xd

[elfin dirge]

Alright

[narrow peak]

oop not done

part 2

,rccw

[elfin dirge]

Whatd you get for b?

[narrow peak]

-6

[elfin dirge]

Dope

Alright do you know how to solve this one

[narrow peak]

nope thats why im asking

[elfin dirge]

Alright

[narrow peak]

ok wait sub a b c

solve for m?

[elfin dirge]

?

[narrow peak]

• some use of discriminant

[elfin dirge]

Discriminant?

This isn't a quadratic

[narrow peak]

nvm its not quadratic

yep

dis my first line

[elfin dirge]

Let's start off with some brainstorming first

It'll make our lives actually easier

[narrow peak]

bet

[elfin dirge]

So there's going to be two regions where it will intersect it only once

Either by just touching the tip of the vertex

Or by being low enough to only hit one side of the graph

[narrow peak]

one side of the graph wym

[elfin dirge]

Let me show you

This is when it just touches the vertex

[narrow peak]

i understand that one yes

[narrow peak]

ohh

[elfin dirge]

And that is when the slope of m is less than the slope of -a

And we know -a=-1/2

so there's our first region done

[narrow peak]

what is what now

[elfin dirge]

What is your question

[narrow peak]

oh

ok i am braindead

why m<-a

[elfin dirge]

If the slope is too high, it'll intersect on the right side

[narrow peak]

ohhh/

[elfin dirge]

Alright

Now let's find m such that our line passes through the vertex

[narrow peak]

sub 12,2?

[elfin dirge]

Yes

[narrow peak]

m=-1/6

[elfin dirge]

You got it

So our m can either be equal to -1/6 or less than -1/2

[narrow peak]

ohhhhhhh

ok done

[elfin dirge]

Hope this helped!

[narrow peak]

yes ofc it did

if it didnt help i wouldve abandoned ship

[elfin dirge]

Lol

[vague aurora]

For combinatorics with probability. I did a question that said there was a free throw percentage of 80% and we wanted to find out the probability of scoring exactly 3 of 5 baskets. Broadly speaking, I understand that each combination would have a percent chance, and that each one also has a frequency. Multiplying each percent by its frequency and adding them for all of the combinations would result in 1. Algebraically, why do the percent chance for each combination work out for this? I think it has something to do with the percent probabilities summing to one (80% free throw plus 20% miss = 100%), but I'm not sure where to go from here.

[unborn yoke]

hi

im doing series rn

and im confused about a formula

Fack

should a be on the outside

or multiply with the numerator first like the first pic shows

[weak needle]

For combinatorics with probability. I did a question that said there was a free throw percentage of 80% and we wanted to find out the probability of scoring exactly 3 of 5 baskets. Broadly speaking, I understand that each combination would have a percent chance, and that each one also has a frequency. Multiplying each percent by its frequency and adding them for all of the combinations would result in 1. Algebraically, why do the percent chance for each combination work out for this? I think it has something to do with the percent probabilities summing to one (80% free throw plus 20% miss = 100%), but I'm not sure where to go from here.
@vague aurora I advice you to check #probability-statistics

[viscid thistle]

@unborn yoke same thing

Doesn't matter

[unborn yoke]

what

im getting different answers for those 2 formulas though

[viscid thistle]

No way

[weak needle]

im getting different answers for those 2 formulas though
@unborn yoke No you don't

The 2 formulas are the same

[unborn yoke]

order of operations??

the 2nd one makes sense to me

but if you look at the first one wouldnt you multiply a with (1-r^n)

[weak needle]

order of operations??
the 2nd one makes sense to me
but if you look at the first one wouldnt you multiply a with (1-r^n)
@unborn yoke They are both the same

[obsidian monolith]

HoboSas:

[viscid thistle]

Do you agree?

[unborn yoke]

🤦‍♂️

long day guys

sorry for the brain fart lmao

thanks for the explanation ❤️

[viscid thistle]

No problem

[unborn yoke]

not sure why im getting different answers though lemme check again

W looks good i had a typo

[viscid thistle]

Probably

[wide lynx]

what would the normal f graph look like?

[elder junco]

Easy

Take the anti derivative

F(x) = ∫ f’(x)dx

So, let’s see...

f’(x) has roots -3, -2, 1 and 4

Um

It probably has an imaginary root too

Can’t tell from the picture alone though.

Do you know F’(x) exactly @wide lynx ?

[wide lynx]

no

its just from the graph

i do need a pretty good idea cause i need the maxs and mins etc

[elder junco]

Oh

If you want to find the minimums and maximums of f(x), set f’(x) equal to 0.

[wide lynx]

i have no equation for f'(x)

[elder junco]

Oh

Oh

[wide lynx]

its just drawing the graph that i think is kinda

i just never thought it was necessary

[elder junco]

Well it’s impossible without the actual f(x) function

Hmmm

Oh well.

[wide lynx]

wtf

[elder junco]

Ikr

[wide lynx]

so how do u answer something like this

cause you need to draw the graph precisely but i never really learned how

[uncut mulch]

you can still get the general shape (just not the vertical shift/intercepts)

[elder junco]

Example

[wide lynx]

i just need to find the max's and mins

my graph doesnt look all that different from thishttps://cdn.discordapp.com/attachments/363224154469826562/742866983099432960/image0.png

[uncut mulch]

identify turning, inflection points

[wide lynx]

inflection points are when concavity changes sign or when derivative does?

or norma;

.....

[uncut mulch]

look up second derivative test if you're not sure

[elder junco]

f’’(x)=0 will give you inflection points of f(x)

[wide lynx]

i have no equation guys

im going off a graph here

[elder junco]

Do you want to sketch?

[wide lynx]

yeah

[elder junco]

You can approximate a sketch

[wide lynx]

just find the flat spots?

yeah approximate

[elder junco]

Uhhhhh

[wide lynx]

like draw the f graph from f'

[elder junco]

See how f’(x)=0 touches the maximums and minimums of f(x)

Yeah

My best guess

[wide lynx]

wait i dont need an equation or anything

[elder junco]

But it’s an example

You just need to look at the graph

Erm

Blue is f(x)

[wide lynx]

im lost

[elder junco]

Me too

[wide lynx]

i just wanted to draw a graph maan

[elder junco]

I’m sorry.

I tried.

[wide lynx]

it is ok

[wide lynx]

what does the critical at x=-2

do

because i do something like this

critical points can only be max and min right? or can it be a flat spot

[elder junco]

Well critical points are when the derivative of the function = 0

Um how can I explain

[wide lynx]

nah i know

[elder junco]

So to answer your question yes

Because the slope is 0

[wide lynx]

im just thinking like critical points are always maxs or mins of the function

[elder junco]

The zeroes of the function that is a derivative of the original function, are the x value that pertain to the maximums and minimums of the original function

[wide lynx]

english

[elder junco]

Example here

See how the zeroes of the prime function penetrate the x value of the critical points of the original function?

That’s a rough idea

[wide lynx]

yeah i still dont get any like knowledge on how to do this

i feel like im getting the x values wrong

cause how do u somewhat accurately determine the steepness of the graph

[elder junco]

Well we don’t know f’(x)

[wide lynx]

man idk

yeah we're just estimating

[elder junco]

So that is the estimate

[wide lynx]

but it asks me for specific x values where there is max and min

so if my estimate is off im wrong? i dont get it

[elder junco]

The max and min of f(x)?

Sorry

(Lol)

[wide lynx]

yeah

it asks for inflection points maxs and mins

[elder junco]

Well the max and mins of f(x) are the zeroes of f’(x)

[wide lynx]

but which is which

its like

[elder junco]

-3, -2, 1 and 4 are zeroes

[wide lynx]

1. what are the maxs
2. what are the mins
3. what are the inflection points

[elder junco]

And are the max and mins of f(x)

[wide lynx]

.

but which is which

ill just guess maaan

supported guess

[elder junco]

Well I gave you 50% I think you can find out the rest

You see, we don’t give out direct answers here

We just help

[wide lynx]

what

[elder junco]

oof

I’m starting to get tired

[wide lynx]

ok

ty for help

[elder junco]

I Tried

Inflection points of f(x) ⇒ f’’(x)=0 (two derivatives)

[wide lynx]

yea

i looked it up tyty

[elder junco]

And those inflection points happen to be the mins and max of f’(x)

Np

[raven compass]

hello, can someone please explain this answer? i'm confused about what happened in line three, where they merge the two fractions together

it uses the binomial theorem combination formula

[patent beacon]

(n - k - 1)! = (n - k)!/(n - k)
With me on that one?

[raven compass]

not really 😖

does that mean (n-k) is 1

[patent beacon]

That's an identity, true for any n and k

Aight so let's take a look at the factorial function. There's really only one thing that defines it. You've got to know that:
(n + 1)! = (n + 1)n!

[raven compass]

ohh i get it now

you divide the (n-k) so the factorial starts from the next term

(n-k-1)

[patent beacon]

Yeah yeah you can make that identity from the definition

So good, now if you apply that, both terms will have the same denominator. You can combine the fractions after that

Oh you'll have to do the same thing with (k - 1)!

In general if you find you have constants in your factorial, try to get them out

[raven compass]

ah my brain is fried

do i multiply each side with a different number

[patent beacon]

I want to get the "- 1" out of (k - 1)!

So I use
(k - 1)! = k! / k

[raven compass]

o ok and the k! stays on the bottom

solved it thank you!

[vague aurora]

can someone point me to a source to learn parametric equations?

it isnt on khan academy precalc

[steel venture]

what exactly about parametric equations

https://www.youtube.com/watch?v=97pe-QlSGqA

this dude is great for a lot of things

[opaque olive]

$\frac{q_3+q_2-2q_1}{p_3+p_2-2p_1}(x-p_1)+q_1=\frac{q_3+q_1-2q_2}{p_3+p_1-2p_2}(x-p_2)+q_2$

[obsidian monolith]

Yes:

[opaque olive]

best way to solve this for x?

[steel venture]

start by isolating the x terms to one side

[opaque olive]

ive tried that and got a very ugly fraction

[steel venture]

can you show your work

[opaque olive]

yep

$a=q_3+q_1-2q_2\b=p_3+p_1-2p_2\c=q_3+q_2-2q_1\d=p_3+p_2-2p_1$

[obsidian monolith]

Yes:

[opaque olive]

i wrote them as a b c d

[steel venture]

the x(a/b - c/d) bit is right, and i can't really see it simplifying too well so i'm assuming that fraction looks right

is there a reason why you think the fraction should be simple

[opaque olive]

yep cuz i have the answer lol

the answer should be $\frac{p_3+p_2+p_1}{3}$

[obsidian monolith]

Yes:

[opaque olive]

i dont see it simplifying into that

so im thinking i missed a factor or something early on

any idea?

[steel venture]

right side of top should be - p2(c/d)

• q2 - q1

[opaque olive]

wait wdym

[steel venture]

so you distributed

$\frac{q_3+q_2-2q_1}{p_3+p_2-2p_1}(x-p_1)+q_1=\frac{q_3+q_1-2q_2}{p_3+p_1-2p_2}(x-p_2)+q_2$

to

[obsidian monolith]

maximo:

[steel venture]

$x\frac{q_3+q_2-2q_1}{p_3+p_2-2p_1}-p_1\frac{q_3+q_2-2q_1}{p_3+p_2-2p_1}+q_1=x\frac{q_3+q_1-2q_2}{p_3+p_1-2p_2}-p_2\frac{q_3+q_1-2q_2}{p_3+p_1-2p_2}+q_2$

[obsidian monolith]

maximo:

[steel venture]

which we can simplify to

$x\frac{a}{b}-p_1\frac{a}{b}+q_1=x\frac{c}{d}-p_2\frac{c}{d}+q_2$

[obsidian monolith]

maximo:

[steel venture]

we good so far?

[opaque olive]

ok

[steel venture]

then you move around the terms, so the x terms to the left and the p1 term to the right

$x\frac{a}{b}-x\frac{c}{d}+q_1=p_1\frac{a}{b}-p_2\frac{c}{d}+q_2$

[obsidian monolith]

maximo:

[opaque olive]

yep

[steel venture]

then the q1

$x\frac{a}{b}-x\frac{c}{d}=p_1\frac{a}{b}-p_2\frac{c}{d}+q_2-q_1$

[obsidian monolith]

maximo:

[steel venture]

do you see the difference to what you wrote on the paper

[opaque olive]

i do lol

my bad

[steel venture]

all good

try it from here

[opaque olive]

yep gonna give it a try

$x(ad-bc)=p_1ad-p_2cb+bd(q_2-q_1)$

[obsidian monolith]

Yes:

[steel venture]

yup

[opaque olive]

shall i put back the original values now

since i dont see anything else cancelling

[steel venture]

yeah i'd go ahead and do that

but it still seems like it would be kinda wild

[opaque olive]

seems like alot of work

[steel venture]

did you copy the problem right cause that seems like a lot of work

[opaque olive]

https://gyazo.com/761fe0245e807dfb73cf896d5a66f6a0

https://gyazo.com/7bd48fda7ced6376bf8ea2e784217240

[steel venture]

,w $\frac{q_3+q_2-2q_1}{p_3+p_2-2p_1}(x-p_1)+q_1=\frac{q_3+q_1-2q_2}{p_3+p_1-2p_2}(x-p_2)+q_2$

[obsidian monolith]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by maximo#5992

[opaque olive]

is everything good?

[steel venture]

$\frac{q_3+q_2-2q_1}{p_3+p_2-2p_1}(x-p_1)+q_1=\frac{q_3+q_1-2q_2}{p_3+p_1-2p_2}(x-p_2)+q_2$

[obsidian monolith]

maximo:

[steel venture]

yeah everyhting from the given problem to what is written down looks good

[opaque olive]

i feel like this is too much work

[steel venture]

same

[obsidian monolith]

maximo:

[steel venture]

i mean

if even wolfram doesn't feel like doing it

then i don't think they'd make you do it either

[opaque olive]

lol

has it got something to do with centroids? @steel venture

[steel venture]

i couldn't say

ive never really dealt with centroids during geometry so im not sure about the math behind it

is this for geometry class

[opaque olive]

no

from a random test

[sudden mauve]

random test??

that ur currently doing?

@opaque olive

[steel venture]

he/she has the answers

[opaque olive]

i mean its a test from a previous year

[steel venture]

what class is it for

[opaque olive]

not for school

[steel venture]

but what class took this test

[opaque olive]

its a pre university test

[steel venture]

but like, i want to know what was covered in the class that might be used during the test

so maybe there's some calculus that was used regarding the slopes of the lines

[opaque olive]

oh gimme a sec

[steel venture]

or maybe its some centroid work like you talked about

yeah for sure

[opaque olive]

yeah i was told that doing the algebra would work

an alternative method is recognising centroid or something (idk)

[steel venture]

so if we go from

$x(ad-bc)=p_1ad-p_2cb+bd(q_2-q_1)$

[obsidian monolith]

maximo:

[steel venture]

and reintroduce the terms

[opaque olive]

yeah doing the algebra should work

[steel venture]

maybe it all works out?

[opaque olive]

yeah it should work out

[steel venture]

i mean it doesn't look like it would at a glance but maybe so

[opaque olive]

yeah

https://www.admissionstesting.org/Images/47831-step-specification-2020.pdf

thats all the content that they expect the candidate to know

its a question from 2004 so i guess that its just poorly written

making the person do that kinda algebra

im gonna bash out the algebra lol

[steel venture]

that's what i'd do at this point

unless you can confidently take the centorid path

[opaque olive]

i cant lol, the question probably expects people to just do the algebra

or use ceva's thereom

[vague aurora]

sin 2x times sin 2x is sin squared 2x right?

The 2x doesn’t change?

[upbeat bone]

Ya

You mean like this right? $\sin(2x)\cdot \sin(2x)=\sin^2(2x)$

[obsidian monolith]

Wilston Lynx:

[vague aurora]

Yeah

Thanks

[upbeat bone]

Yeah, that's correct

You're welcome!

[vague aurora]

@upbeat bone yaknow how the half angle identities have the plus minus? When would you actually change the sign of your answer?

Like when do they apply?

[upbeat bone]

Sorry I was playing some games

Do you know the the cos(2x) formula?

Which comes in 3 different forms?

The half angle are derived from that, which while you're deriving it, you can get sin^2(x)=... and cos^2(x)=... , and when you take the roots of both sides, you can get that the sin(x)=... or cos(x)=... will be in the form of plus minus

Ohh wait, I think I misinterpret what you're asking there--

[harsh smelt]

sign prolly depends on the concrete problem

like if I know that angle is in (pi/2, 3pi/2) range i would take cos negative

[uncut mulch]

depends on quadrant

[graceful ledge]

Can someone please help me with this one?

[willow bear]

what is giving you trouble here?

[graceful ledge]

The um... h(g(0))

[willow bear]

ok that doesn't help me help you at all

fine

can you tell me what g(0) would be?

[graceful ledge]

g(0) = -5/1

-5

[willow bear]

simplify.

yes, g(0) = -5.

so given that g(0) = -5, what must h(g(0)) be?

[graceful ledge]

I honestly don’t know.

[willow bear]

are you unable to evaluate h(-5)?

[graceful ledge]

No

[willow bear]

why not?

[graceful ledge]

...

[willow bear]

what's stopping you from calculating h(-5)?

[graceful ledge]

You asked me if I was unable

But I’m able

[willow bear]

oh, my bad

anyway, is there any other issue here that needs addressing?

[graceful ledge]

No, thanks.....

I get it now

❤️

[prime yoke]

How do you convert x^2 + (y-1)^2 = 1 to polar form? I know the answer is r = 2sin(theta), and i got as far as r^2 - 2sin(theta) = 0, but idk where to go from there

[echo wagon]

Why do you say the answer is r = 2sin(θ)?@prime yoke

[prime yoke]

oh wait nvm

i figured it out haha i dropped an r

[echo wagon]

Lol

[prime yoke]

with the 2rsin(theta)

[echo wagon]

Okay, good

[wide lynx]

can someone help me graph f from this

[elder junco]

You had this question yesterday.

@wide lynx

[wide lynx]

yeah

ik

i want to learn how to graph it though

u told me roots and to get on with it

[elder junco]

Hmm

[wide lynx]

@elder junco also yesterday did you say the inflection points are the roots

cause ithought thats when f'' changes sign

[elder junco]

I said that f’’(x)=0 are the inflection points of f(x)

You misunderstood.

[vague aurora]

How would u be able to tell that the x-4 root is to the power of 3 or 5 etc?

Is it possible?

[wide lynx]

what

[elder junco]

I think he needs more answers than questions.

[wide lynx]

wait f''=0

are the f' = 0

no?

[elder junco]

What?!

What are you asking placement?

[vague aurora]

@wide lynx do u know the order of the function?

[wide lynx]

ia m

levitating irl

[vague aurora]

huh

[elder junco]

Try to factor the function just by knowing the roots as shown.

f’(x)=0 are the Minimums and Maximums of f(x).

f’’(x)=0 are the inflection points of f(x).

Au revoir.

[wide lynx]

but f''=0 when f'=0

[elder junco]

The critical points.

[wide lynx]

but

i dont see why its not like

x=1 is inflection point

or x= -1

[elder junco]

Hmmm

I wonder if there is anyone knowledgeable to be able to recognize the equation of a graph by sight...

Actually a lot of info is missing.

It is impossible.

[viscid thistle]

You just need to sketch it

[elder junco]

You can only get a rough idea on the inflection and extremes of f(x).

Right

But that’s all

[viscid thistle]

Uhm no?

[elder junco]

How would you do it Hobo??

[wide lynx]

yeah

sketch it

i need an accurate sketch and im just wondering how you guys would do it

[viscid thistle]

You know the stationary points

Inflection points

And when the function is increasing / decreasing

[elder junco]

Exactly.

The zeroes of the f’(x) are the extremes of f(x).

[viscid thistle]

Yep

[elder junco]

So Um...

You can only have a rough idea

[viscid thistle]

Yes because of the y-shift

[elder junco]

Just take this as an example.

[wide lynx]

this is what i wrote

ignore inflection points

didnt edit

[elder junco]

Well it could look better

[viscid thistle]

Looks pretty good

[elder junco]

You’re getting the idea.

[viscid thistle]

yes

[elder junco]

Doesn’t look too good for x=-2

Needs an extreme there for f(x)

[viscid thistle]

That part needs to be fixed

[elder junco]

Example

I recommend putting putting points where you know there are extremes of f(x) first

And then you can plot where you know there are inflection points with a different color

[vague aurora]

Are these limits literally just plugging in the limit value for x?

At least for the first one

So -4/3 for the first one

Does the arrow mean going from the positive end?

Like for 54 it’s -2.999999999 etc?

[wide lynx]

wait

why does it go down @ x=-2

[vague aurora]

For 54 the limit doesn’t exist right?

Someone please confirm anything I’m saying

[elder junco]

Still having trouble ): @wide lynx ?

[wide lynx]

uh

i dont think so?

i understand what youre saying

[elder junco]

Ok good

[wide lynx]

it goes down

so its a max

[elder junco]

Yay

❤️

[wide lynx]

[elder junco]

Wow! Nice

[wide lynx]

awesome

thanks for the help

[elder junco]

You’re welcome

[wide lynx]

i had a problem on x=-2 because i thought when f'>0 its increasing

[viscid thistle]

@vague aurora

$\lim_{x\to a^+}$ means the limit as x approaches \verb|a from the right side| and $\lim_{x\to a^-}$ means the limit as x approaches \verb|a from the left side|

[obsidian monolith]

Al𝟛dium:

[vague aurora]

@viscid thistle so right side means slightly more positive and left means slightly more negative? And they need to have the plus or minus in the top right? I also read that if they are different and it isnt specified you write no limit right?

[viscid thistle]

$$3^+\approx 3.0000001 $$ $$3^-\approx 2.999999$$

[obsidian monolith]

HoboSas:

[viscid thistle]

I wouldn't word it as slightly more positive or more negative, it's a vague and bad definition. Look at this chart for clarification of the function: f(x)=1/(x-3) as we approach say, x=3 from the right at a vertical asymptote:

``f(x) | x

-10 | 2.99
-1000 | 2.9999
-10000| 2.999999999
10 | 3.01
1000 | 3.0001
10000 | 3.00000001
``

,w plot 1/(x-3)

[obsidian monolith]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Al𝟛dium#4084

[viscid thistle]

You can see from the chart i just made above, how where the function approaches 3 from the left side, at a vertical asymptote, the values of x get extremely close to 3 without touching 3. And as it is a vertical asymptote, the values of f(x) will increase rapidly. Similiar stuff from the right side of 3, values of x close to 3 without actually touching 3.

@vague aurora

[wide lynx]

just a quick question

does in terms of w mean f(w)= w variables or f(w) = l variables

[viscid thistle]

Occupied channel still, waiting for their response.

[vague aurora]

@viscid thistle Ah I see, thanks. So if it is not specified what side it is approaching from and both sides have two different values, you would write that there is no limit right?

[viscid thistle]

@viscid thistle Ah I see, thanks. So if it is not specified what side it is approaching from and both sides have two different values, you would write that there is no limit right?
@vague aurora yes. The limit would not exist

But i wouldn't say: "it is not specified what side it is approaching from", $\lim_{x\to 3}$ is just the limit as x approaches 3. Not from the right nor left

[obsidian monolith]

Al𝟛dium:

[vague aurora]

Thanks

[kindred whale]

Can someone please help me with Sigma Notation

[sudden mauve]

post your question @kindred whale

[kindred whale]

If im understanding it right a should be correct? but unsure

[sudden mauve]

,rotate

[viscid thistle]

The start is good

Sure but that sigma looks ugh.

[kindred whale]

lol

fixed it to look better first time ever seeing the symbol so my bad

[viscid thistle]

$\sum$ so that you have inspiration

[obsidian monolith]

Al𝟛dium:

[kindred whale]

idk where to go after here I am confuseed

[viscid thistle]

Look at the 3 terms, tell me which term varies after each sum

[kindred whale]

the second term on numerator

but do I add those?

[uncut mulch]

write the i-th term in terms of i

[kindred whale]

so im missing 2i+1 for the sum

[uncut mulch]

wdym by missing 2i +1?

[kindred whale]

Idk im lost as I said

[uncut mulch]

sry I used the wrong variable above.

anyway to express it in sigma notation you need to express your summand in terms of the index in this case: i

$ a_\red{1} = \frac{1 + \red{1}}{100} \ \
a_\red{2} = \frac{1 + \red{2}}{100} \ \
a_\red{3}= \frac{1 + \red{3}}{100} \ \
a _\red{i }= ?$

[kindred whale]

yes

so for the sum of the notation it is 135/100 simplified down to 27/20 or 1.35

[obsidian monolith]

ramonov:

[uncut mulch]

the sum for part b) is indeed 27/20

I'm referring to what you're missing in part a)

$\sum_{i = 1 }^{15}$, by itself doesn't represent the summation of anything

[obsidian monolith]

ramonov:

[kindred whale]

got it

thanks its 1+i/100

[uncut mulch]

()

[ionic timber]

is this right?

for summer hw

[obsidian monolith]

I have just restarted and am loading myself, please wait!

[viscid thistle]

I think it should be like this.

[harsh smelt]

yep

colored points are correct

[ember remnant]

solve over the interval [0,2pie] : 2sin(3x)+1

How do I even start? @viscid void if you know how to solve this equation

[uncut mulch]

you don't have an equation...

🥧 is not the same as

[rapid lance]

@ember remnant there is no equation but generally, since sin(15) = sin(375) they dont want "trivial" solutions

or in radian pi/4 has the same terminal arm as pi/4 + 2 pi

[uncut mulch]

but if you were actually asked to solve:
2sin(3x)+1 = 0 over that interval,
first find the general solution for 3x, hence x
and then generate from your general solution the values in that interval

[lime bolt]

if k is an integer, solve: k+3 @ember remnant

[rapid lance]

youd solve solve sin(k+3) in the integers

[lime bolt]

🤨

[uncut mulch]

solve: 0

[ember remnant]

Does the interval matter?

[rapid lance]

@ember remnant yes

[ember remnant]

Is 2pi the period then?

[lime bolt]

it is multiplied by 3

[rapid lance]

@ember remnant the 0 to 2pi is to limit solution so that you dont need to write +2k*pi, k in the Integers at the end

[uncut mulch]

consider the steps I've outlined above

[ember remnant]

But it asks me to solve for x

[uncut mulch]

yes

[ember remnant]

I can find the period but not x

[uncut mulch]

what's your current issue with solving for x?

if you were actually asked to solve
2sin(3x) +1 = 0,
first find the general solution for 3x

[ember remnant]

@uncut mulch I simplified it to x= sin^-1(1/6)

But now I need to solve for x over the interval [0, 2pi]

how do I do that?

[simple edge]

how do i do this?

[viscid thistle]

Well I mean that’s just (sqrt(2-2cos(10*))/2

Maybe multiply by conjugate?

Idk ‘simplify’ is quite vague. That operation is pretty relatively simple in and of itself, if there’s a smaller simplification it’s not obvious to me.

[lost pawn]

Use half angle identity for sin

[tribal wasp]

How I factor this into 3 equivalent expressions

[vague aurora]

If I have ln (x+2)^2 is that the square of the log or the log of the square?

@tribal wasp r u asking for the cube root?

[tribal wasp]

Nah I have to factor it with i or something

I got (x^2 + 2)(x^2 + 1)

But can I factor x^2 + 2 into (x + i sqrt(2))(x - i sqrt(2))?

[ember remnant]

sin^-1(sqrt(3)/2) = pi/3

How am I supposed to know that? Is there a trick or something?

@viscid void

[steady palm]

@ember remnant are you asking why the sin of root3 over 2 is pi/3?

i'm assuming thats what it is. arcsin is inverse so opposite. sin of pi/3 is sqrt(3)/2 so just work backwards. if you're not sure why any of that is true i'm guessing you're not visualizing it.

or maybe this helps, on how you arrive at that:

pythagorean theorem to get the sides, then you should know what sin/cos/tan/csc/sec/cot are as far as what sides they are

if you're about to take calculus know this shit because it had a lot of trig functions in the integrals

[uncut mulch]

I simplified it to x= sin^-1(1/6)
@ember remnant
trig functions don't work like that

sin(ax) = b doesnt imply sin(x) = -b/a (nor sin(x) = b/a)

sin(100x) = 100 does not mean sin(x) = 100/100 = 1

consider that it was A instead.
what would be the general solution to
2sin(A) + 1 = 0

[steady palm]

oh i see the confusion. the parenthesis for the trig function is an argument, not a variable you manipulate with algebra.

I think when I first started trig I was making the same mistake

[uncut mulch]

what are you struggling with?

[gilded mirage]

Idk where to start

[viscid thistle]

**rapidly **

[uncut mulch]

well firstly, where do vertical asymptotes come from?

[gilded mirage]

Infinite discontinuities?

[uncut mulch]

not quite what i was expecting.

[gilded mirage]

[uncut mulch]

can you determine the vertical asymptote for $y= \frac{1}{x-9}$?

[obsidian monolith]

ramonov:

[uncut mulch]

i probably didn't phrase the original question well

[gilded mirage]

X=9?

[uncut mulch]

yes.

what about:

$y = \frac{1}{(x-7)(x+4)}$

[obsidian monolith]

ramonov:

[gilded mirage]

X=7 and x=-4

[uncut mulch]

now what about going in the other direction

if you were given the vert asymptotes, can you get something similar to the above examples I presented?

[gilded mirage]

Hm

Something like this i think

[uncut mulch]

ok. that's a decent start for the denominator.
now we want to satisfy the condition for the horizontal asymptote

the easiest way to get a (non-0) horizontal asymptote is if the numerator and the denominator has the same degree

[gilded mirage]

So 2nd degree in this case?

[uncut mulch]

yes

and you also want the asymptote to be y=10,
so you simply need to have a coefficient of 10 on x^2 term on the numerator

[uncut mulch]

you could introduce linear and constant terms if you want to be fancy, but you'd need to make sure that the numerator doesn't have a factor of (x-2) or (x+2) otherwise you'll have a hole

yep that will work

[gilded mirage]

Thanks so much ❤️

[uncut mulch]

should've mentioned non-0 asymptote earlier (edited now)

[kind cave]

i am taking precalc honors this year what would be a good way to prepare myself?

[willow bear]

algebra

make sure you have a solid grasp of algebra

[earnest jungle]

Any help?

[viscid thistle]

Nice try

[acoustic harbor]

don’t @

[proven marten]

Nice try
@viscid thistle But it says Sample questions?

[viscid thistle]

They did @ everyone

That's why I said that

[earnest jungle]

JC Denton

Help???

Please.

[earnest jungle]

?

[proven marten]

I'd like to but I dont have enough knowledge

[earnest jungle]

😦

[fossil crown]

From

#❓how-to-get-help

Rule 1

The help channels are solely for help with math, so feel free to post your question. Asking whether you can ask a question or if anyone knows about some specific topic is unnecessary, so please try to avoid questions of that nature.

[willow bear]

@earnest jungle do you still need help

and if so, what with

[ornate wolf]

Can someone point me in the right direction?

I end up at :

-1 + (1/x+iy) = 1 + i

Assuming it's right so far, how do I proceed?

Ok, so I went further and made it:

(x-iy / x^2 + y^2) = (i+2) / 1 

[echo wagon]

I end up at :

-1 + (1/x+iy) = 1 + i

@ornate wolf From here, take the - 1 across then take the reciprocal on both sides

[ornate wolf]

@ornate wolf From here, take the - 1 across then take the reciprocal on both sides
@echo wagon I end up with:

1/(x+iy) = 2+i

[echo wagon]

Did you finish reading halfway through?

[ornate wolf]

Ok, so I went further and made it:

(x-iy / x^2 + y^2) = (i+2) / 1 

@ornate wolf is this wrong?

[echo wagon]

Not wrong

[ornate wolf]

Did you finish reading halfway through?
@echo wagon Sorry I thought I did the reciprocal

[echo wagon]

Taking the reciprocal means, go from $\frac{1} {x+iy} = 2 + i $ to $x+iy = \frac{1} {2+i} $

[obsidian monolith]

Lunasong:

[ornate wolf]

ah, thank you

@ornate wolf is this wrong?
@ornate wolf then i shold not have done this

[echo wagon]

If you did that, you would get two equations with two variables when equating the real and imaginary parts. Probably still solvable, but unnecessarily long

[ornate wolf]

thanks so much

do I multiply the RHS with conjugate?

such that:

x+iy = (2-i )/ 5

[jade heron]

Yh

[ornate wolf]

how should i proceed?

[echo wagon]

Equate real and imaginary parts, since x and y are real

[ornate wolf]

Equate real and imaginary parts, since x and y are real
@echo wagon sorry what does that mean?

[echo wagon]

If a, b, c and d are real then $a+ib = c+id$ means $a = c$ and $b=d$

[obsidian monolith]

Lunasong:

[ornate wolf]

sorry i can't solve it till now @@

[echo wagon]

If a, b, c and d are real then $a+ib = c+id$ means $a = c$ and $b=d$
@echo wagon Do you understand this and agree with it?

[obsidian monolith]

Lunasong:

[ornate wolf]

Yup

Complex numbers a+bi and c+di are equal if and only if a=c and b=d.```

Is x = 2/5 , iy = - i/ 5 ?

y = -1/5?

edit: look slike it

Thank you so much everyone!!!!!

[jade heron]

Np

[novel cargo]

how can you show that y = 10^x is the inverse of y = log10(x)?

[echo wagon]

How did you define log_10?

[novel cargo]

it means that to what power do we need to take 10 to get x

is that what you mean?

[echo wagon]

Okay, yeah

To show that f and g are inverses, show that f composed with g and the reverse is the identity

[novel cargo]

ahah

so if I put the output of one into the other I get the original input of the first

[echo wagon]

Yes

[novel cargo]

hahah

thanks Luna

[echo wagon]

Np

[novel cargo]

also, what does it mean "is the identity"?

[willow bear]

the identity function

f(x)=x

[novel cargo]

ahh, thanks

[novel cargo]

$x^6 = 3^{18}$

[obsidian monolith]

az:

[novel cargo]

looking at something like this, is there a better way to solve other than taking to the power of 1/6?

[blissful ridge]

Well, that's the simplest way to do it

[willow bear]

why is raising both sides to the power of 1/6 unacceptable

[novel cargo]

no, it is

I'm learning and had difficulties to find how to do this at first

then came up with this

[viscid thistle]

I mean you can do it by another way i think

Wait a sec

[novel cargo]

wanted to see what you guys think and all

[viscid thistle]

Okay

I have watched many competition maths video and one method that i see a lot is this one $$x⁶=3^{18}$$ we want to make both sides similar so that we can make a solution for x, you'll see what i mean: remember exponent laws $(a^b)^c=a^{b\cdot c}$ $$x⁶=3^{3\cdot 6}$$ $$x⁶=(3^3)^6$$ and finally see that by "similarity" $${\color{green}{x}}{\color{blue}{⁶}}=({\color{green}{3^3}}){\color{blue}{⁶}}$$ so $$x=3³=27$$

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

@novel cargo another method

[uncut mulch]

ew. missing (real) solutions

[viscid thistle]

Oh wait a sec

[novel cargo]

this is cool @viscid thistle

[viscid thistle]

I forgot the negative one

[novel cargo]

also, using colors, nice

[viscid thistle]

x=-27 is also a solution, forgot to add

[novel cargo]

$\log_{8}{x} = \log_{8}{12}$

[obsidian monolith]

az:

[novel cargo]

when we remove the log8 from both sides, what operation are we exactly doing?

[willow bear]

we're taking 8^(both sides)

raising 8 to the power of both sides

however you wanna phrase it

[novel cargo]

substituting them for x in the exponential function 8^x

lol

thanks

[violet knoll]

Looking to understand this. Thanks!

[novel cargo]

to do this you have to complete the square

then you will find the shift on the x and y axis and also the constant factor a

[violet knoll]

Where do I find the h and K?

and thank you for helping az

[novel cargo]

look at it like this first

-2 * (x^2 -2x - 1/2)

can you now complete the square inside the parenteses?

it must become something in the form of (x+h)^2 + k

[violet knoll]

let me try real quick (im a little slow in mathimatics)

[novel cargo]

take your time

I'm learning myself

[violet knoll]

so you divided the equation in half pretty much... then do you square what you divided?

[novel cargo]

I haven't divided it

I've factored out -2 but it's still there outside of the parenthesis

let's continue from -2 * (x^2 -2x - 1/2)

you need +1 to complete the square so we write

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

so we add and subtract 1

now we have x^2 -2x + 1 inside the parenthesis

this equals (x - 1)^2

so we get:

-2 * ((x-1)^2 -1 -1/2)

-2 * ((x-1)^2 - 3/2)

-2(x-1)^2 +3

now that's in the form a(x -h)^2 - k

you see what happened?

[violet knoll]

I'm trying to, but I don't completely understand what a completed square means. I'm going to watch another video and see if i can get it.

[novel cargo]

$\frac{1}{\log_{4}{x}} + \frac{1}{\log_{5}{x}} = 1$

[obsidian monolith]

az:

[novel cargo]

so, I solved this and I'm happy

yeey

[proven marten]

gj baby yoda :)

[novel cargo]

$2 \cdot \frac{\log_{2}{25}}{\log_{2}{5}} \cdot \frac{\log_{2}{9}}{\log_{2}{3}}$

[obsidian monolith]

az:

[novel cargo]

how would I solve this?

[acoustic harbor]

change 25 to 5^2

and use log properties to move the 2 in front

same with the 9

[novel cargo]

oh man

why didn't I see this?

thanks

[novel cargo]

$\log_{4}{(x)} + \log_{4}{(x + 6)} = 2$

[obsidian monolith]

az:

[novel cargo]

has two solutions: 2, -8

-8 isn't in the domain of logarithmic functions

so I remove it from solutions?

[viscid thistle]

Yes

[novel cargo]

thanks

[tall herald]

Hey, I got a problem. Prove without calculating why "xy-1=(ax+by+c)(Ax+By+C)" can't be factorised, where a,b,c,A,B,C are real numbers.

I think I see a few implications here, but I might be wrong

[patent beacon]

Why that factorization can't happen? Hmm.

[tall herald]

So there's no way to get the constants x and y out of the parenthesis, since it will sort of "add up" on eachother I guess

But we're told to think geometrically

Sorry my english is a bit rusty when it comes to using it for math D:

But I hope you guys understand what I'm trying to say lol

[patent beacon]

No no that's clear, but I'm not sure how to answer haha. I'll think about it or someone smarter than I might know

[tall herald]

Ahh ok, thanks 🙂

[willow bear]

if such a factorization were possible then the graph of xy-1=0 would look like two intersecting straight lines

namely they'd be the lines with equations ax+by+c=0 and Ax+By+C=0

but the graph of xy-1=0 is a hyperbola

@tall herald @patent beacon

[viscid thistle]

Hey

[novel cargo]

a log-log graph of f(x) = x is a straight 45 degree line

is that correct?

[proven marten]

not sure what you mean

[novel cargo]

f(x) = x is a straight 45 deg line in a cartesian plane, right?

now, the scales on both axes are logarithmic

[willow bear]

yes it'll look the same on a log log plot

[novel cargo]

thanks

also, does anyone know a good tool to test this?

I looked at geogebra but it doesn't seem to have such a scaling option

[willow bear]

test what

[novel cargo]

scale the axes

and plot

not plot, write functions

[willow bear]

you can make a log-log plot of y = f(x) by graphing e^y = f(e^x) i guess

[proven marten]

Desmos?

[willow bear]

or 10^y = f(10^x)

[novel cargo]

Desmos?
@proven marten will look at it

or 10^y = f(10^x)
@willow bear will try this out. Haven't still grasped it fully.

[fleet yew]

,w loglogplot y=x

[obsidian monolith]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by AMD#4434

[fleet yew]

It is 45 deg but the scaling on the axes is not the same

[willow bear]

that's an artifact of WA

[onyx venture]

how do I show that $f(x) = \left(1 + \frac{1}{x} \right)^x$, $x > 0$ is a monotonically increasing function?

[obsidian monolith]

Abrar:

[onyx venture]

my attempt is taking me to conclude that it's monotonically decreasing but the graph shows clearly that it is increasing 😔

Here is what I tried

for $h>0$,
$$(1 + \frac{1}{x+h})^x < (1 + \frac{1}{x})^x$$

[obsidian monolith]

Abrar:

[onyx venture]

and since $(1 + \frac{1}{x+h})^h > 0$

[obsidian monolith]

Abrar:

[onyx venture]

actually nvm

I think I figured it out

[novel cargo]

the problem asks how many circles can be made that pass through all three fixed points

intuitively you can see that only one solution is possible, but the algebraic solution was very fascinating to me

[willow bear]

the algebraic solution is rly just a bunch of heavy alge-bruh

[novel cargo]

the solution I saw wasn't that computation heavy or anything, it was rather smart

[lime bolt]

every three points has precisely one circle which passes through all of them

[novel cargo]

not every three point

[lime bolt]

yes every three

[novel cargo]

what do you do if all the points agree in their x or y coords?

[lime bolt]

ok obviously other than a degenerate one

[willow bear]

every three points not all on the same straight line

uniquely determine a circle

...in euclidean geometry anyway

[novel cargo]

Ann, we have an obv case, like the one I had a few days ago

so, anyway

suppose, we have points (h1, k1), (h2, k2), and (h3, k3)

[lime bolt]

tbh u can construct an infinte radius circle which passes through 3 collinear points

[novel cargo]

ohh, that's probably true altho my math level isn't that advanced yet

[willow bear]

"infinite radius circle"

that's a big thonker right there

[novel cargo]

we approaching Star Trek territory then?

anyway, you can setup a system of three circle equations with those three points

now, if you think about it, with two points, you gonna have a free variable which is the reason for infinite solutions

three points, gives you a unique solution, so only one circle possible

also, three collinear points, must result in a system with no solution I assume

the connection between these ideas is actually really wonderful

I wish I understood more

[novel cargo]

Will every light ray approaching a parabolic mirror bounce towards the focus?

Answered "Yes", and turned out to be wrong.

Every light ray parallel to the the optical axis bounces back towards the focus

Also learned that this is how amplifier dishes that you see in spy movies work

[gloomy path]

if youre in the extended complex plane, maybe not that much of a big thonker

[novel cargo]

how far should the microphone be from the bottom of the reflector

basically asks for the y coord of the focus point

I assumed vertex will be at origin so b and c will be zero

and setup 4 * p * 8 = 20^2

the focus is 12.5 cm above bottom of reflector

appreciate if somebody can confirm and/or give me other input

[harsh smelt]

x^2 = 4py where p is focus-vertex distance

@novel cargo what is equation of parabola then?

[novel cargo]

would be x^2 = 50y

(x^2)/50 = y

[harsh smelt]

ok i seem at x = \pm 20 you have y = 8

20^2 = 4p(8)

400 = 32p

,calc 400/32

[obsidian monolith]

Result:

12.5

[harsh smelt]

yep correct

so focus is 0, 12.5

[novel cargo]

I'm struggling with this since yesterday

made me feel really bad

I was confusing everything

[harsh smelt]

this is process of learning

[novel cargo]

I was thinking: what about the the x^1 and its coeff?

[harsh smelt]

wdym

[novel cargo]

the general form of a parabola ax^2 + bx + c

[harsh smelt]

complete the square and you wil be able to get form (x-h)^2=4p(y-k)

[novel cargo]

I didn't see the connection yesterday

that when vertex is at origin, you don't have any x^1 because h and k are zero

and was also afraid of using just formula

[viscid thistle]

Can someone help

[sudden mauve]

@viscid thistle I don't think this is the right channel to post that type of advanced math question

[viscid thistle]

Ok how advanced is it

#point-set-topology probably

[rapid lance]

how can i prove that x^3 + y^4 = 19^19 has no solutions in the integers?

[viscid thistle]

where is this problem from?

has potential to be extremely hard

[rapid lance]

my teacher gave it to me, im in grade 8

so ive tried using the reflective sum of points

that didnt work for me

[willow bear]

consider it modulo something maybe?

[rapid lance]

ok

[willow bear]

mod 3 or mod 9 or something that makes the rhs 1 perhaps

[rapid lance]

thx

[viscid thistle]

try mod 13

[rapid lance]

both mod 13 and mod 9 dont work 😦

[viscid thistle]

what did you do with mod 13?

[rapid lance]

i brought 19^19 to 7 mod 13

then made a table with x^3 and y^4, in which i got 8, -1

which gives me 7

[willow bear]

,w 19^19 mod 13

[obsidian monolith]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Ann#0413

[viscid thistle]

hm I skimmed a solution which used mod 13 so that's why I suggested it, but you are right and it has an error

and mod 13 doesn't work

[rapid lance]

maybe mod 6???

i think bringing the rhs to 1 wont work since then we can always get 0, 1

,w 19^19 mod 11

[obsidian monolith]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by ohNoiAmHere#6879

[rapid lance]

,w 19^19 mod 7

[obsidian monolith]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by ohNoiAmHere#6879

[viscid thistle]

wait a second, I think mod 13 does work

[rapid lance]

y = 5, x = 5 i think that should give you 7

[viscid thistle]

,w (5^4+5^3) mod 13

[obsidian monolith]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Botn#8995

[rapid lance]

it doesnt! wow, ill try that again then

[viscid thistle]

what led to -1 being a possible residue for y^4?

[rapid lance]

me doing this by hand 😫

[viscid thistle]

recheck whatever led to that

[rapid lance]

i will

x : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10

x^{3}: 0, 1, 8, 1, -1, 8, 8, 5, 5, 1, -1

x^{4}: 0, 1, 3, 3, 9, 1, 9, 9, 1, 9, 3

this is my new chart

[viscid thistle]

why does it only go up to 10?

[rapid lance]

mb, 11, 12 i did, but forgot to type

they work too

[viscid thistle]

work as in, they are redundant?

[rapid lance]

yea, i got 11: 5, 3 and 12: 1, -1

[viscid thistle]

-1, 1 would probably be a better way to put that

[rapid lance]

yea

[viscid thistle]

so this reduces to {0,±1,±5} and {0,1,3,9}

[rapid lance]

yes

[viscid thistle]

and no combinations are 7?