precalculus

[harsh cipher]

you're correct.

[undone pawn]

asymptotes arent part of the function at all

[viscid thistle]

^^

[undone pawn]

they're just lines towards which the function approaches, but never intersects it

[viscid thistle]

That’s true.

[willow bear]

not entirely true

a function may cross its own asymptote

[undone pawn]

oh

[viscid thistle]

That’s also true

[undone pawn]

ohh

right

[harsh cipher]

good morning Ann

🙂

[willow bear]

for example, (x-1)(x-2)(x-3)/x^7

[undone pawn]

well one part of the function approaches but never intersects its respective asymptote?

[willow bear]

this one crosses its own horizontal asymptote at y=0, thrice

[undone pawn]

hmm

[willow bear]

no, you can even have infinitely many intersections

sin(x)/x

[undone pawn]

lmao

[viscid thistle]

eHh

[willow bear]

this cannot happen if all you're considering is rational functions

[viscid thistle]

It’s 1:26 AM. Lemme go grab a soda and then we talk asymptotes

[willow bear]

but rational functions are not the only kind of function out there

[undone pawn]

fair enough

so how are they defined

besides just 'denominator goes to 0'

[harsh cipher]

so....for my question....

since I cannot factor the denominator I need to use quadratic formula

[viscid thistle]

True

[harsh cipher]

and the horizontal asymptote is y=0

[willow bear]

In analytic geometry, an asymptote of a curve is a line such that the distance between the curve and the line approaches zero as one or both of the x or y coordinates tends to infinity.

[harsh cipher]

that's the best description

🙂

[undone pawn]

nice

[willow bear]

i ripped this one off wikipedia.

what

why the sully

[viscid thistle]

sUL.Ly

[viscid thistle]

for polar coordinates, how can i learn to do these

https://cdn.discordapp.com/attachments/683852688009723909/692261465112969256/unknown.png

[noble thistle]

what "k" means ? for an example, X=K(pi) or (pi) : 2 + 2 (pi) k.

we are using this with sin and cos stuff

[stuck lark]

in that case, k most likely denotes an integer

[noble thistle]

is it important to have it or can i just skip it ?

[stuck lark]

very important

[noble thistle]

i still cant find it usefull

[stuck lark]

find all real x that satisfy sin(x)=0

[noble thistle]

for an example ?

[stuck lark]

yes

[noble thistle]

so it can be: x0 = 0 + 𝜋 ∙ 𝑘 or x0 = 𝜋 + 2𝜋 ∙ 𝑘

[stuck lark]

x=pi*k is enough, and remember to state that k is an integer

[noble thistle]

also im really confused about this: the highest point 1 took: x= 𝜋/2 + 2k𝜋.

i mean how, if 2𝜋 is on 0

SIN line

and its like 3/4 of periode diffrence

wait

if k is there mean that im using difference between max and min points without starting point 0k

[fleet yew]

@noble thistle the extrema (maxes and mins) of the sin function are the zeroes of the cos function

Think about where cos is zero

[noble thistle]

i know that

but was i wrong?

[sturdy haven]

helbiz prease

nuner 13

[jagged glade]

Weird, How could x-coordinate be positive for a point on Quadrant II?

This is what I picture

[uncut mulch]

yeh, the question doesn't make sense

[sturdy haven]

tahts what is thinking

[fleet yew]

Maybe they just mean the magnitude

That's what i assumed

[oblique thicket]

That means that the editor of that textbook wasn't caring at all.

[viscid thistle]

Sin(a) + cos(3a) = ??
I am supposed to make it into a product and simplify. The theme is sum and difference into product.

[harsh cipher]

Hi!

can anyone help me find the name of this book

it's for my ap stats 12 course

seems like pearson something but can't find it

[patent beacon]

@harsh cipher
https://www.chino.k12.ca.us/Page/14454

[harsh cipher]

Thank you.

you're a legendary googler!

[patent beacon]

Lol yes I've done some googling

[lethal oracle]

It wants me to solve this system of equations

I’m not sure where to start

Do I use something like substitution

[rich charm]

<@&286206848099549185> does ayone know this wtf

[viscid thistle]

tan(x) = -v/u

cosx(u-vtanx)=secx

cosx(u+v^2/u)=secx

(u+v^2/u)=sec^2x

u+v^2/u=1+tan^x

u+v^2/u=1+v^2/u^2

u(1+v^2/u^2)=1+v^2/u^2

u=1

v=-tan(x)

if you plug this in into the first equation, everything checks out

sin(x)-tan(x)cos(x)=0

for the second equation

cos(x)-tan(x)sin(x)=secx

cos^2(x)-sin^2(x)=0

cos(2x)=0

2x=pi/2+pi*n

x=pi/4+pi/2*n

u=1 and v=tan(pi/4+pi/2*n)

if you look at the unit circle, tan(pi/4+pi/2*n) is equal to either 1 or -1 depending on n

[harsh cipher]

hey guys

can anyone try youtube.com

their bgc is white?

wtf?

nm

ip reset and flush dns

[harsh cipher]

hi

if a pie chart does not display any numbers.

What word would I use to describe that the information is inaccurate?

[uncut mulch]

what are you referring to when you say "information"

[harsh cipher]

the pie chart

say the pie chart is divided into 5

[uncut mulch]

a par chart without numbers doesn't mean that chart itself is inaccurate

[viscid thistle]

you can try estimating the percentage of each part and comparing to any other information

[harsh cipher]

the question is is this an appropriate display for the genres

why/why not

[viscid thistle]

are you able to provide us with a visual?

[harsh cipher]

sure

[uncut mulch]

^

[viscid thistle]

I think an image might help us understand it better

[harsh cipher]

question 1

[viscid thistle]

the question is asking if the diagram is well-suited for displaying the data

like some diagrams won't work with certain data

[harsh cipher]

oh

my answer was no because pie chart doesn't display any number

s

[viscid thistle]

I think you can say that

[harsh cipher]

awesome

[uncut mulch]

depends on what you actually want

the chart it self would have to label "first 120..."

[viscid thistle]

a way I could see a person answering "yes" is that you can still compare the sizes of each part and determine which is most/least common

[uncut mulch]

and it does display relative amounts

which genres are more common etc

if you had extremely good measurement tools, you could obtain the numbers

but ofc it would be better if they were given on the diagram

[harsh cipher]

thank you!

one more question

when writing difference of squares equation

[willow bear]

this isn't a difference of squares

[harsh cipher]

oh yes

you're right

I need to sleep haha

difference of cubes

bonjour Ann

[willow bear]

$(2x)^3 - (3y)^3 = (2x - 3y)[(2x)^2 + (2x)(3y) + (3y)^2]$

[obsidian monolith]

Ann:

[harsh cipher]

yes but why did symbolab write it differently?

I can see from the refined answer that its (x)(y)

which is 6xy

[willow bear]

i don't know! symbolab is crap!

[harsh cipher]

okay

[willow bear]

for some reason it chose to do it that way!

i don't know!

[harsh cipher]

thank you Ann

passe une bonne journée

[limpid rampart]

guys

i am having some minor difficulty

[willow bear]

mhm?

[pale bison]

shh they're still typing

[limpid rampart]

$f(z) = z^3 + pz^2 + qz - 15 , f(z) = 0 has three roots which are a , 5/a , (a + 5/a -1 )$

[pale bison]

tex gore

[limpid rampart]

lol

hold on

[pale bison]

$f(z) = z^3 + pz^2 + qz - 15 , f(z) = 0$ has three roots which are $a , 5/a , (a + \frac5a -1 )$

[limpid rampart]

yes

thank you

[willow bear]

are you sure you didn't mean $\frac{a+5}{a-1}$

[obsidian monolith]

Ann:

[willow bear]

for that last root

[limpid rampart]

no

[obsidian monolith]

Publius:

[pale bison]

this?

[limpid rampart]

yes

[willow bear]

ok so what are you asked for

[limpid rampart]

to solve for p

[willow bear]

p only?

[limpid rampart]

and solve the equation f(z) = 0

completely

[pale bison]

as i see it, system of equations

[willow bear]

ok so what is your difficulty

[limpid rampart]

not sure where to start

[willow bear]

vieta's formulas

[limpid rampart]

should i rearrange to make each variable the subject

alright

ill go with that

[pale bison]

i knew you're gonna hit me with a better method... time to google lol

[willow bear]

denoting the roots as $r_1, r_2, r_3$, you have $\begin{cases} -p = r_1 + r_2 + r_3 \ q = r_1r_2 + r_2r_3 + r_3r_1 \ -(-15) = r_1r_2r_3 \end{cases}$

[obsidian monolith]

Ann:

[limpid rampart]

oh shit

thats a lot easier

lol

thanks

[sturdy haven]

There were other answers

But I am confused about the question

[pale bison]

1 radian is roughly 57.2958 degrees

[willow bear]

compare 2 to 0, pi/2, pi, 3pi/2 and 2pi

[solemn sluice]

what exactly is a fundamental vector?

[willow bear]

uhhh

context?

[solemn sluice]

uhhh iodk LOL

i was just wonderign

is thetre like a definiton

[carmine elbow]

I need help converting this problem into standard form

the standard form looks like a(x-h)^2 + k

[torn oriole]

do you mean vertex form?

@carmine elbow

[carmine elbow]

yes

they're the same thing?

[torn oriole]

iirc, standard form is ax^2+bx+c

whereas vertex is the a(x-h) blahblahblah

[carmine elbow]

oooh ok

I assume you know how to help me then?

[torn oriole]

yis

just looked it up, heres what i found

but lets get onto the helping

so you have to find h and k here

$h=\frac{-b}{2a}$

[obsidian monolith]

Jaboi:

[torn oriole]

and $k=f(h)$

[obsidian monolith]

Jaboi:

[carmine elbow]

ok

so h is 1?

[torn oriole]

yes

so now you can plug in the value of h for x in your original equation to get k

[carmine elbow]

so, I'm plugging in 1 to get k, is that right?

[torn oriole]

yes

then you can plug in your found values to the vertex form

[carmine elbow]

got it

[torn oriole]

[carmine elbow]

k = 5?

[torn oriole]

yes

[carmine elbow]

what's next?

[torn oriole]

just put in the values to get it into the form

we have a, h, and k

so just input

[carmine elbow]

ok. Thank you for helping me!

[torn oriole]

np!

[carmine elbow]

How would you find the axis of symmetry for the function?

[slender bay]

@carmine elbow if its quadratic in x

Then axis of symetry is given by x=-b/2a..where a and b denotes the things they generally denote in a quadratic

[fleet yew]

@slender bay parantheses

[carmine elbow]

It is, but the equation is in standard form

[unique jewel]

Hello

-90x -10² + 5 ? @carmine elbow

[iron pike]

https://gyazo.com/994ff86baf28f4d1f6538b01dfcf502c

Could anyone help me w/ this

[unique jewel]

Sure

[carmine elbow]

No, -9(x-1)^2 +5 @unique jewel

[unique jewel]

What steps do you usually follow to draw a function's graph ?

[iron pike]

a b c

find a b c

[unique jewel]

-9 (x-1)² +5 , k = 5 and h =1

Exactly, a = 2, b = 8, and c = 5

[iron pike]

then i do -b/2a

i forgot what for but i think thats what i do

[unique jewel]

Great, now we're trying to find a remarquable point !

-b/2a = -8/4 = -2

Now your turn, replace x by -2 and find the value of y

[iron pike]

-2/1 right

[unique jewel]

Oh yeah oops 🤣 sorry sorry !

[carmine elbow]

I’m confused, are you helping me, him, or both of us?

[iron pike]

both of us

[carmine elbow]

Ok

[unique jewel]

-9 (x-1)² +5 , k = 5 and h =1
@unique jewel @carmine elbow

[carmine elbow]

Ok

[iron pike]

oh and then -2(x) us part of the vertix

and i find the y by plugging in

[carmine elbow]

Where did the -8 come from when you were explaining it earlier?

[iron pike]

and then find 4 more by moving the left and to the right

so like this right ?

[narrow peak]

,rotate

[narrow peak]

@iron pike if u want a graph just plug in random values of x

like 12345

[carmine elbow]

Got it

[iron pike]

like in the picture right where that like chart is ?]

@narrow peak

May someone solve the question, and i feel i would follow better

don't need to draw the chart thou

[unique jewel]

@carmine elbow what -8 exactly

Oh ! I replaced (-b) by its value (-8)

Alright I'll do

[narrow peak]

https://gyazo.com/994ff86baf28f4d1f6538b01dfcf502c

this one

@iron pike

[iron pike]

is this the answer ?

@narrow peak @unique jewel

https://gyazo.com/d7a0fc779d2bd0ebe57140871390d125

is the next question

lol and thats it

[narrow peak]

is ur y intercept @ 5

[unique jewel]

I found different results, lemme check your answer

[narrow peak]

,w graph 2x²+8x+5

[obsidian monolith]

Display results online and refine query

Requested by Hmm#2554

[narrow peak]

epic

[unique jewel]

Yep that's what I found

You solved the equation y = 0 right ?

[narrow peak]

,w factor 2x²+8x+5

[obsidian monolith]

Display results online and refine query

Requested by Hmm#2554

[narrow peak]

just plug in x values of 1,2,3,4,5

get the y value for each

plot those points

and draw the curve

[unique jewel]

,rotate

[unique jewel]

You solved the equation 2x²+8x+3 in order to find the two points where the curve meets the x axis

[carmine elbow]

Nvm, the negative 8 was for the other person’s problem @unique jewel

[atomic talon]

can someone help me find the solution for

hold on

[unique jewel]

@atomic talon

You let X be x² and then u get an easy equation that you can solve

But don't forget that you've found the values for X not for small x

If X = a then x² = a x = √a or x = -√a etc ...

[atomic talon]

thanks so much!!!!

i really appreciate it

[unique jewel]

Anytime

[harsh cipher]

question

Determine the full expanded equation of a polynomial if it has a root at x=2 with multiplicity 1, a root at x= -3 with multiplicity 2, and a y -intercept (-36).

[fleet yew]

@harsh cipher write it out

You have the factors

And the y int

And an unknown leading coefficient

[harsh cipher]

so far I have

y= a((x-2)^1)((x+3)^2) -36

[fleet yew]

No you cant have that minus 36 there

[harsh cipher]

oh

[fleet yew]

Because that changes the polynomial

[harsh cipher]

then I must substitute y = -36?

[fleet yew]

Yes

And x=0

Because it is at the point (0, -36)

[harsh cipher]

I see

one minute please

okay then

(2)(x-2)((x+3)^2)

(2x^3)+(8x^2)-(6x)-(36)

🙂

@fleet yew ty!

[fleet yew]

Nice

👍

[limber compass]

I am brain dead ?

[rose trellis]

what r some good pre calc textbooks

[harsh cipher]

i've came across another problem

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

wow Ann is not around today

[viscid thistle]

what's your question? @harsh cipher

[harsh cipher]

simplify and answer with positive exponents only

[viscid thistle]

$\frac{{(3x-2)}^{-2}}{{(2x^{2})}^2}$

[obsidian monolith]

Silver Hunter:

[viscid thistle]

well, what can you do to get the top term to have a positive exponent?

[harsh cipher]

we can divide the coefficients?

no

[viscid thistle]

no

do you know how to make negative exponents into positive ones?

[harsh cipher]

yes we write 1/(x)

[viscid thistle]

then what can you do to $(3x-2)^{-2}$

[harsh cipher]

integral exponent rule

[obsidian monolith]

Silver Hunter:

[viscid thistle]

yes

[harsh cipher]

hmmm

1/9

[viscid thistle]

? how did you get that?

[harsh cipher]

it's wrong

[viscid thistle]

$x^{-n}=\frac{1}{x^n}$

[obsidian monolith]

Silver Hunter:

[harsh cipher]

yes I understand that

[viscid thistle]

let's say (3x-2)=a

and now the equation is $a^{-2}$

[obsidian monolith]

Silver Hunter:

[viscid thistle]

how can we change that to a positive exponent

[harsh cipher]

1/a2

1/a^2

[viscid thistle]

yup

and since (3x-2)=a

[harsh cipher]

okay but the orginal question is (3x)^-2

[viscid thistle]

we can say $(3x-2)^{-2}=\frac{1}{(3x-2)^2}$

[obsidian monolith]

Silver Hunter:

[viscid thistle]

((3x-2)^-2)/((2x^2)^2))
was this not your expression?

[harsh cipher]

wrote it wrong

[viscid thistle]

oh ok

[harsh cipher]

its (3x)^-2

sorry about that

[viscid thistle]

well we can still take the same approach

it's fine

[harsh cipher]

okay

[viscid thistle]

if a=(3x)

[harsh cipher]

so then its 1/(3x)^2?

[viscid thistle]

yup

[harsh cipher]

sorry parentheses

we have a fraction in the numerator

[willow bear]

in all seriousness, isn't this algebra and not precalc

[harsh cipher]

it's logs

the first lesson goes over exponent laws

[willow bear]

the first lesson goes over the exponent laws that everyone somehow forgot

everyone taking the class, that is

[harsh cipher]

it's just me? my math skills are far from average

😛

[viscid thistle]

you don't necessarily need to have a fraction in the numerator. if you have a negative exponent in the numerator, you know then that you have to move it from a numerator to a denominator

[harsh cipher]

but we don't a negative exponent in the denom

[willow bear]

you know then that you have to move it from a numerator to a denominator
meh

this whole "have to" mindset

you never """have to""" do anything

[viscid thistle]

I'm trying to make it simpler

¯_(ツ)_/¯

they don't necessarily need to, but they need to have positive exponents so they need to in this case

[willow bear]

there's a difference between a requirement you see right there on the paper and a nebulous "have to" that makes the student think negative exponents are actually illegal or something

[harsh cipher]

let me show you the picture of the equation

[willow bear]

also, y'all have been misusing the word "equation"

there's a difference between an equation and an expression: the former consists of two of the latter joined by an equals sign

[harsh cipher]

okay!

[viscid thistle]

well, moving on

[harsh cipher]

awesome

[viscid thistle]

if you want to have positive exponents, you want to move any negative exponents in the numerator to the denominator, right?

because essentially $\frac{x^{-n}}{1}=\frac{1}{x^n}$

[obsidian monolith]

Silver Hunter:

[willow bear]

that's much better now as far as explanations go

bc this time instead of imposing arbitrary rules, you motivate

[harsh cipher]

yes

[viscid thistle]

so if you moved $(3x)^{-2}$ to the denominator

[obsidian monolith]

Silver Hunter:

[viscid thistle]

it would be same as $\frac{1}{(3x)^{2}(2x^2)^2}$

[obsidian monolith]

Silver Hunter:

[harsh cipher]

uh yes

[viscid thistle]

are you confused?

[harsh cipher]

no

Im not

[viscid thistle]

I can give you another way to look at it

ok

so you don't always need to think the numerator and denominator as separate expressions

[harsh cipher]

I agree

[viscid thistle]

so for simplifying

[harsh cipher]

(9x^2)(4x^4)

add exponents

[viscid thistle]

and multiply the coefficients

[harsh cipher]

yes

[viscid thistle]

and you should be done

[harsh cipher]

that's it

you made it easy for me to understand

thank you!

[viscid thistle]

no problem!

glad to help :)

[harsh cipher]

🙂

[atomic talon]

Does anyone know how to do this?

[willow bear]

do you know the relationship between a polynomial's degree and the number of roots it can have

[viscid thistle]

no bro

just give us the answer

thx x

[willow bear]

ok first off

i'm not your bro. please don't call me that

second, we don't give out answers here.

@viscid thistle @atomic talon

[viscid thistle]

ok help us

its either d or e

we cant figure it out

[atomic talon]

help me understand then

i think its asking how many numbers could fit in x right?

or does it mean what actual value fits in the x

bc if its asking how many solutions it would be 1 right? bc only 2 would work there

[steel tulip]

Even roots of a positive number yield two solutions.

That’s for every even root

[serene heath]

look up ''the fundamental theorem of algebra''

[willow bear]

@steel tulip that's not what this is about.

and it's also wrong

[steel tulip]

My bad then

[atomic talon]

@willow bear can u help me

thx for trying tiessie

i appreciate it

[willow bear]

i think its asking how many numbers could fit in x right?
this is a bad way of putting it

[atomic talon]

ok sorry?

im just trying to understand

[willow bear]

a polynomial equation can have at most as many distinct roots as its degree

you know what the degree of a polynomial is, right

[atomic talon]

ua

ya

[willow bear]

so can you tell me the degree of the polynomial x^6 - 64

[atomic talon]

2

WAI

no

wait

ait ya

ya 2

[willow bear]

no

[atomic talon]

then what

[willow bear]

the degree of x^6 - 64 is 6, not 2.
you told me you knew what the degree of a polynomial was, but either you were confused about it or you lied.

[atomic talon]

why would i lie

im confused then

obviousl

why do u think im asking for help

[willow bear]

some people have actually lied to me about this sort of stuff. dunno why they do it.

anyway

that's beside the point

forgetting about this problem for a moment

what do you think the DEGREE of a polynomial is

[atomic talon]

exponet

[viscid thistle]

exponent?

[willow bear]

no

give me something longer than a one word reply

[viscid thistle]

LOL MIA

[atomic talon]

so its x to the 6th so is that why 6 is the degree

WHY?

[viscid thistle]

omg

LMFAO

[willow bear]

why what

[atomic talon]

why are u questioning me

[steel tulip]

The degree of a polynomial is the highest power in the polynomial

[atomic talon]

like im confused

[willow bear]

i'm trying to get you to phrase things clearly, which is the first step to actually understanding things.

[atomic talon]

ok well-

anyways

thanks ig

[willow bear]

the degree of a polynomial is the highest exponent of x appearing in the polynomial (with a nonzero coefficient)

[atomic talon]

is that right tho

is that why 6 was the degree

bc it was the exponet

the only exponet

but if there were more it would be the highest

like u said

[willow bear]

6 is the exponent on x^6 and there are no higher exponents on any x's around

[atomic talon]

oh okay i understand that part now

thx

[viscid thistle]

hi can someome please help me

[slender bay]

Do u knoe the formula for r+1 th term

If not then

3rd option is coreect

[torn oriole]

Dont just tell them

[leaden stratus]

I can't figure out how to do cos (4a)

[serene heath]

use double angle

[viscid thistle]

thx @slender bay

[leaden stratus]

@serene heath cos 2a = cos^2(a) - sin^2(a)

So that cos 4a? 🤔

[serene heath]

yes

but instead of a its 2a

cos(4a)=cos(2a+2a)

[leaden stratus]

@serene heath so sin 4a is sin (2*2a)?

[serene heath]

yes

[leaden stratus]

Uhm thanks. I'll try it

[viscid thistle]

@mia#0666 the answer its 2 so D, if you are still looking for it. Basically because:

2^6 - 64 = 0
64 - 64 = 0
0 = 0
Proven

how do i find the fundamental frequency of the superposition of two sinusoids
and btw they arent harmonics

[viscid thistle]

pls help

my time is gonna run out

😭

[unique jewel]

@viscid thistle

[viscid thistle]

not an exam

it was homework

❤️

[dawn void]

Who would use discord during an exam..?

[viscid thistle]

fr

[fleet yew]

@viscid thistle ur time ran out already

Lmao

[viscid thistle]

i know

😭

[leaden stratus]

Hi guys, I need to find cos B and tan B

I found cos B

But I'm unsure about tan B

[soft mica]

Can you tell me your answer for cos B?

Sorry the working is a bit messy

@leaden stratus

Ping me when u reply:)

[slender bay]

@viscid thistle option 3 as at x=2 deno becomes 0

[viscid thistle]

@viscid thistle (next time try looking for values that X cannot equal, or look for the vertical asymptotes of the graph.)

[analog relic]

Since $\sin{\alpha}=\frac{1}{2}$, $\alpha=30$. Since $\alpha+\beta=90, \beta=60 \implies \cos{\beta}=\cos{60}=\frac{1}{2}$ and $\tan{60}=\frac{\sqrt{3}}{3}$

[obsidian monolith]

ιχιση_тαкαgι:

[leaden stratus]

@soft mica 1/2

[soft mica]

yep that's correct

when you were working out cos beta what did you use for the value of beta?

@leaden stratus

[analog relic]

60 degrees

Because alpha + beta = 90

[leaden stratus]

@soft mica 360 - 2alpha

Because we studied the double angles

Actually tan B result is sqrt3 according to my book

[soft mica]

yep so beta is 60 degrees

[leaden stratus]

We could find cos B by doing cos (alpha - beta)

[soft mica]

and tan 60 is an exact value that is equal to sqrt3

[leaden stratus]

Yup, but you should find it only using trig formulas

[soft mica]

you mean cos ((a+b)-a)?

[leaden stratus]

No

cos (360 - 2a)

Then cosacosb + sinasinb

[soft mica]

hmm are you asked to solve using trig like this?

much easier to use exact values..

[leaden stratus]

Yes, we need to use trig

I know it's easier that way

[soft mica]

ahh ok ok

are you all good with the answer now?

[carmine elbow]

Hello, I need help finding the minimum/maximum for an absolute value

[slender bay]

Ueah what is the prob

U see the modulus..break it

U will find two linear functions at the breaking point whose graph is a straight line...and then u can either plot to solve or u can solve them algebraically as well

The breaking point is x=-1/2

[lost sphinx]

Break it into seperate functions like if x-a>0 then replace modulous with a bracket only and if x-a<0 replace it with a bracket with a negative sign outside

[desert crystal]

Hey guys, this is probably a rlly easy question, I know I just have to use substitution but I dont know what to do next ;-;

[slender bay]

Solve them

Put that valuue of y in the second equation

And get a quadrstic

[desert crystal]

so expand it?

[slender bay]

Yeah

[desert crystal]

well Ill try it again

[harsh condor]

heres the pic

just thought it looked nice

[desert crystal]

haha ok but I need to solve it using the equations

[slender bay]

U are giving her/him the ans directly

Against policy

[lost sphinx]

Ban

[harsh condor]

tbh, there is no answer there lol

[lost sphinx]

Sadlyf but ban

[harsh condor]

unless u go do pixel measurements

[lost sphinx]

Bruh (1,-3)

[desert crystal]

no there should be to on the two places it crosses over

[lost sphinx]

I can clearly see it

[desert crystal]

two answers

[slender bay]

Me also

[harsh condor]

well i can't :/

[slender bay]

@desert crystal dont u see the pic

[harsh condor]

and neither has the person

[lost sphinx]

Ban plep you crossed the line dude it's end of the line now

[harsh condor]

nooo

[lost sphinx]

😂

[harsh condor]

rip me

[desert crystal]

I shouldnt use the pic to solve it

[harsh condor]

no lol

[lost sphinx]

Anyways bro solve that shit simultaneously whoever asked it and apply the quadratic formula if you can't factorise

[harsh condor]

u cant draw it irl

[slender bay]

Ueaj...good to go thwn

[desert crystal]

welp I got x=0 or 9

and thats wrong

[lost sphinx]

Yea now subs it in the linear obtain y and check calculations

[harsh condor]

show workign

[slender bay]

@desert crystal maybe u screwd

[desert crystal]

ok I did it on paper hang on a sec

maybe

[lost sphinx]

@slender bay are you a nihilist

If you're girl you'll have a nice handwriting

Unlike me

[harsh condor]

so u have:
(x - 4)^2 + (y-2)^2 = 25
and y = x -3

so you replace y, with x - 3
(x - 4)^2 + (x - 3 - 2)^2 = 25
which is:
(x - 4)^2 + (x - 5)^2 = 25
then you can expand both using the rule
(a - b)^2 = a^2 - 2ab + b^2

[slender bay]

@lost sphinx how did u know

[desert crystal]

Well fricl

[lost sphinx]

Your pfp dude

[harsh condor]

It should be:
2x^2 - 8x + 16 + 2x^2 - 10x + 25 = 25

[slender bay]

Wjere did ur 16 go

[lost sphinx]

I can spot a nihilist in the darkness of the night

[desert crystal]

lol oh

[harsh condor]

nihilists don't exist

[lost sphinx]

We do

[desert crystal]

hang on ill do it again

[harsh condor]

because its unnihilistic to be a nihilist

[desert crystal]

ugh im so dumb

[slender bay]

Hey i am present here.

[lost sphinx]

Plep I know your dirty secret now

[slender bay]

@desert crystal yeah..everybody is

[harsh condor]

lol

[slender bay]

Except TAO

[lost sphinx]

I am not dumb you're offending me now pmg

[desert crystal]

im asking this chat cuz im too scared to ask my nerdy friends

[slender bay]

Well it depends..u r dumb w.r.t einstien

😂

[lost sphinx]

I don't know but in India there's this saying "jisne kari sharam uske phootey karam" so like

[slender bay]

Everyrhing here is relative

[lost sphinx]

Which means

Well I can't explain

[slender bay]

Which means if u ger shy ..u are to suffer

[desert crystal]

guys I got it

thank youuu

[lost sphinx]

It means if you're shy you probably are pretending to be something you are not

[harsh condor]

gg

[lost sphinx]

Or you have that hidden regret

[desert crystal]

im not pretending im not dumb, Im just in the top maths class and dont want to drag people down

like bother them

[slender bay]

Who cares if u do

[desert crystal]

I care

[lost sphinx]

Bruv I was not saying you're dumb

[harsh condor]

wtf is going on

[lost sphinx]

I'm not translating it properly

So calm down

[harsh condor]

@desert crystal u did the question good job, ur not dumb

[slender bay]

As i said..its relative

[lost sphinx]

You're obviously mot dumb

*not

[desert crystal]

awww thats so sweet but I do rlly need to improve my maths

[slender bay]

Why are u into maths

[lost sphinx]

Ping @slender bay any time he's two times jee topper

[desert crystal]

bruh im not into maths

[harsh condor]

anyway guys i have a precalc question for ya'll
why the hell does x^2 + y^2 = r^2 draw a circle

[desert crystal]

it just does

[lost sphinx]

It doesn't have the xy term

[harsh condor]

hint: its pythagoras

[slender bay]

Dude...dosnt matter

[lost sphinx]

Hold up plep you earned massive respect by asking this question

Yea I know

[harsh condor]

yea?

[lost sphinx]

Hold up

[slender bay]

The amswer ks coming

[lost sphinx]

I am getting a feel by the pythagoras thingy

[slender bay]

Distsnce formual is indeed pythagoras

And why should we evn be bothering bt that😂

[lost sphinx]

Dude distance formulae, complex numbers,

😂

[harsh condor]

lol wut

it isn't supposed to be that complicated

[desert crystal]

Im pretty sure complex numbers aren't involved with this

[willow bear]

what even are y'all doing rn

what's the original problem

[harsh condor]

they already completed it

[lost sphinx]

Complex numbers can be used to derive this @harsh condor

|z-z0|=r

z0=0

Take modulous

We know that this equation is supposed to represent a complex number that rotates about a fixed point

I was off to some work

[willow bear]

We know that this equation is supposed to represent a complex number that rotates about a fixed point
bad desc

[karmic wave]

what does that even mean

[harsh condor]

we were talking about the formula for a circle

x^2 + y^2 = r^2

[slender bay]

Thats how you make a mess out of nowhere😂

[viscid thistle]

Is this channel open

Nvm

[slender bay]

Yeaj

[carmine elbow]

Hello, I need help finding the minimum/maximum for an absolute value

[slender bay]

I think i told u

If u lack the basics then i suggest u to do some related exercises

[fleet yew]

@carmine elbow what exactly are you having trouble with

[slender bay]

The exact problem is she/ he posts the problem and within 10 minutes she disappears again reappear after 1 hr, post it again and again show the same periodic behavior..thats all

[carmine elbow]

I'm still here. That

I just need to find the minimum/maximum for this function

[fleet yew]

Are you trying to find the maximum? Or are you trying to solve for that inequality?

Two different problems

[slender bay]

Are u still here with us

[carmine elbow]

yes

the internet cut out

[slender bay]

Ok...do u know how to break modulus

[carmine elbow]

?

[slender bay]

I mesn do u know thr definition of modulus function

[pale bison]

til these functions deserves a name

[slender bay]

I am out..of it

[heady jewel]

someone asked me this q and can someone tell me what this question means

the question is :- out of the statements x^2+y^2=r^2 and y=√(r^2-x^2) which options states that y is a function of x

[fleet yew]

Ok

Do you know what the definition of a function is

[heady jewel]

a map

what is he even asking

[fleet yew]

What kind of map?

[heady jewel]

,w function

here

[obsidian monolith]

Display results online and refine query

Requested by Lionel#5497

[fleet yew]

Each element of the domain (x) is associated with ONLY one value of the codomain (y)

For every x there can be only one y

Make sense?

[heady jewel]

you mean there can be exactly one value of f(x)?

[fleet yew]

Yes

For each value of x

If you want to think of it graphically

There is something called the "vertical line test"

A vertical line (any vertical line) may only pass through the function ONCE

[heady jewel]

but what does his question mean

[nocturne stream]

It means that which one says

That y depends on x and follows the vertical line test.

[fleet yew]

When we say that y is a function of f

y = f(x)

It means that y is solely dependent on the value of x

So y = (some collection of terms of x)

[heady jewel]

both are the same

the second one is simplified

[fleet yew]

No they are not

Square roots are multivalued

[carmine elbow]

What is a modulus function?

[nocturne stream]

Square roots don't have negative inputs. x^2 does accept negative input.

[fleet yew]

@carmine elbow modulus of a number z is its distance from zero

So the modulus of 5 is 5

The modulus of -5 is 5

It's the absolute value

@heady jewel which equation says y = something else

[carmine elbow]

ok. I think I get it

I ask because I don't know how to find the minimum/maximum value for an absolute value function. My professor is saying that it's possible even though Mathway is saying that only quadratic functions can have max/min

[fleet yew]

Absolute value functions do have a max and a min

Do you have a specific problem i can use as an example to show you?

Ping me if you still need help

[viscid thistle]

Can someone tell me how to solve 4th?

[fleet yew]

@viscid thistle find the integral

[viscid thistle]

Isn't that the problem?

[sharp marsh]

How do I do this again?

log4(256)^−4

[viscid thistle]

Uh

-16?

[sharp marsh]

Yeah but how do I do it

to get answer

without calculator

[viscid thistle]

I didn't use a calculator

[sharp marsh]

No i'm saying how do I do it

without a calculator

[viscid thistle]

And uh you just take the power of -4 outside the log and solve -4*(log4[256])

256 is 4 raised to the power 4

There you have it

[sharp marsh]

oh

Um that's awkward

Thanks

[harsh cipher]

Hello

[carmine elbow]

@fleet yew -8|2x+1|+6

[fleet yew]

an absolute value function is a function of the form

$a |b(x-h)| +k$

[obsidian monolith]

AMD:

[fleet yew]

where the coordinates of the vertex (the max or min) are (h, k)

so in this example, you will need to factor out what's inside the absolute value sign

$-8|2x+1|+6 = -8|2(x+\frac{1}{2})| + 6$

[obsidian monolith]

AMD:

[fleet yew]

from there you can see the h and k coordinates

so the vertex is at

(-1/2, 6)

[sharp marsh]

So this problem log8 x + log8(x + 3) = log8(x + 15)

What do I do

I got x^2 * 3x = x + 15

[fleet yew]

$\log_{8}(x) + \log_{8}(x+3) = \log_{8}(x+15)$

[obsidian monolith]

AMD:

[fleet yew]

$x^2 + 3x = x + 15$

[obsidian monolith]

AMD:

[fleet yew]

$x^2 + 2x - 15 = 0$

[obsidian monolith]

AMD:

[fleet yew]

@sharp marsh

i pretty much did it for you there

[sharp marsh]

o

Solve the logarithmic equation algebraically. Approximate the result to three decimal places.Solve the logarithmic equation algebraically. Approximate the result to three decimal places.

Why decimal though

Wouldn't the answer just be 5 and -3?

[fleet yew]

@sharp marsh you factored wrong

(x + 5) (x-3) = 0

x = -5, 3

got your signs switched up

[sharp marsh]

oops mb

[fleet yew]

-5.000

3.000

lmao

[sharp marsh]

It says it's wrong

How do I do it where I get decimal places

[uncut mulch]

watch out for extraneous solutions

[carmine elbow]

Thank you @fleet yew!

the equation was a bit confusing.

[leaden stratus]

A terrible identity to verify!

Where can I start?

Would it be okay to start from tan and write it as sin/cos?

[lilac pier]

Yes since everything on RHS is all sin and cos

[leaden stratus]

What more, now? @lilac pier

[lilac pier]

you get sin^2 (a/2)

you should know sin^2 (x) + cos^2 (x) = 1, so sin^2 (a/2) + cos^2 (a/2) = 1. This way you get sin^2 (a/2) = 1 - cos^2 (a/2)

[leaden stratus]

Ok

Where do I get that?

@lilac pier

[lilac pier]

get what?

[leaden stratus]

you get sin^2 (a/2)
@lilac pier

[lilac pier]

You have tan(a/2)

That's = sin(a/2) / cos(a/2)

Oh

Actually yeah I realize my mistake lol sorry

You have to use the double angle identity of sin. sin(A) = 2Sin(A/2)Cos(A/2)

Tan(A/2) = Sin(A/2) / Cos(A/2)

[leaden stratus]

Oh cool

So that I get 2sin^2(a/2) @lilac pier

Where cos (a/2)? 🤔

[lilac pier]

yeah mb

cos(a/2) in the denomintor from tan

cancels out the cos(a/2) which came in the double angle identity of sin

[leaden stratus]

2sin(a/2)cos(a/2) * sin(a/2)/cos(a/2) cos (a/2) cancels out

[lilac pier]

so all you have now is 2sin(a/2) sin(a/2)

Yes

[leaden stratus]

Oh it was your mistake

[lilac pier]

Yeah.

[leaden stratus]

Ok, so 2sin^2(a/2)

Should we now do something on the right?

[lilac pier]

Yeah now use sin^2 ( a/2 ) = 1 - cos^2 (a/2)

[leaden stratus]

So 2(1-cos^2(a/2))

[sharp marsh]

e^2x − 9e^x + 20 = 0

How do I do this?

[slender bay]

Sub e^x= t

Reduces to a quadratic in "t"

Then use quadratic formula..or u can factorise..

@sharp marsh

[leaden stratus]

@lilac pier

[slender bay]

Its still not solved or wht

[lilac pier]

@leaden stratus I'm really sorry I was gone

[slender bay]

Now he's gone...an eye for an eye

[leaden stratus]

I'm back! @lilac pier

Don't worry

[slender bay]

Now u are up : he s gone..an eye for an eye

[odd helm]

I’m struggling on this problem

Because I used rational root theorem on this polynomial

But when I graph the polynomial on my calculator it says the answer is in between 4 and 5 but using rational root theorem none of my possible zeroes are in between 4 and 5

So the answer is irrational but if I plug in 4.372 it says it is wrong

[willow bear]

no

you are not asked to actually solve the equation

you are only asked to analyze it using the rational root theorem

you wrote down the POSSIBLE rational roots as ±(1,2,3,4,6,8,12,24)/(1,2,3,6)

among the rational numbers whose numerator is one of 1, 2, 3, 4, 6, 8, 12, 24 and whose denominator is one of 1, 2, 3, 6

which one is the largest?

@odd helm

[slender bay]

#Being ignored😂

[willow bear]

3 minutes isn't quite as big of a delay to be associated with ghosting

[slender bay]

#Time is passing

[leaden stratus]

@lilac pier I'm giving up

Can't do it 😂

[lilac pier]

@leaden stratus It's better to take RHS first

[leaden stratus]

What can we do?

[lilac pier]

you can make cos(A) = 2Cos^2 (A/2) - 1

[leaden stratus]

Sure?

[lilac pier]

multiply that with sin^2 (A/2)

Yeah

[leaden stratus]

Uhm, do I have to rewrite sin^2(a/2)?

[lilac pier]

no

[leaden stratus]

It is -2cos(a) so -2(2cos^2(a/2)-1)???

[lilac pier]

Yeah

And multiply that with sin^2 (a/2)

see what u get

[leaden stratus]

2sin^2(a/4)

[lilac pier]

?

[leaden stratus]

Yeah

[lilac pier]

did u use an identity again

[leaden stratus]

Hmm I used Photomath since I've closed my exercise book

It's late here

[odd helm]

Sorry I wasn’t ignoring Ann I just went offline so the answer would be 24 I guess

Thank u

[harsh cipher]

Hi

question

question b

second part of the equation

why is the answer log base 2 8y in the numerator

sorry about the parentheses

i get that the denom is root(x)

[unique hill]

log base 2 8y comes from the 3

cuz 3 can be expressed as base 2 log form

[harsh cipher]

I don't understand

[uncut mulch]

$3 = \log_2(8)$

[obsidian monolith]

ramonov:

[harsh cipher]

yes

that makes sense to me

I need thinking time hold on please 🙂

[unique hill]

take your time

[harsh cipher]

hahahaha

so am I supposed to recognize that

the 3 can be written as log base 2 8?

[uncut mulch]

basic application of a log law

[harsh cipher]

in order to express the original equation as single logarithm

I do need to express the 3 as log base 2?

[uncut mulch]

for this question, yes

[unique hill]

you need them to be the same base

[harsh cipher]

yes

I read the +3 as log base 2 (y+3)

sigh

[harsh cipher]

I'm confused again

-1/2 log

-1/2 log base 2 x

nm

[steel venture]

you got it?

or still need help?

[willow bear]

"log base 2 x"

log_2(x)

@harsh cipher

[harsh cipher]

thank you

I will write it like that from now on.

@willow bear merci

@steel venture i got it

[steel venture]

ok 😄

[novel dirge]

Do combinatorics belong here?

[steel venture]

yeah i guess

[novel dirge]

This is how I started

but Idk how to continue

we never did the !

I know what it means but no special operations or rules on what to do with it

and our teacher only gave us the basic formulas

now, I know this is comparing 2 combinations

but I got stuck here and have no idea what to do

[steel venture]

what are you trying to solve

like whats the end goal

[novel dirge]

the first line

that is what I got from the teacher

probably to find x

[steel venture]

ah ok