precalculus

[full garden]

so that makes it

[lilac pier]

think of it as (-1) * (-1) = (-1)^2 = 1

[full garden]

m^2+4m+4-4=0

so bro that means the 4 would have to be negative in order to make the equation = 0

[lilac pier]

?

[full garden]

m^2+4m = 0

so m = 0, -4

right?

[lilac pier]

it's (m+4)^2 - 4(9)

you missed the 9

do you know the formula of (a+b)^2?

It's a^2 + 2ab + b^2

[full garden]

bro i didn't have a 9 in my entire equation

[lilac pier]

So it's m^2 + 2(m)(4) + (4)^2

im following the picture

[full garden]

ill redue the entire question and hopefuly i get +4

@lilac pier @tardy ridge thank u so much I did it and I got the answers u guys gave me

[muted steeple]

Hi, I was wondering if the following is the correct way to approach this question.

if $D = b^2-4ac$ and $D < 0$ then $D = a + bi$, so the quadratic equation looks like this:

$\frac{-b \pm a+bi}{2ac} = z$

Then group the terms

$\frac{a - b \pm + bi}{2ac} = z$ Is this explicit enough to say that the two solutions must be conjugates of each other?

Not sure how to explain the why part though, it just is, no?

[willow bear]

If D = b^2 + 4ac and D < 0
then D = a + bi

[muted steeple]

how should i state it? use different variable?

or i need to prove that it's actually the case?

[willow bear]

i mean it's not the case so like good luck trying to prove a false statement

[muted steeple]

ok

[willow bear]

and you didn't even state the discriminant properly

and honestly this is kinda overkill

[muted steeple]

i state it how it's been stated to me.

in the previous chapters

[willow bear]

the discriminant of a quadratic is not b^2 + 4ac.

[muted steeple]

oh

[willow bear]

anyway

this is extraneous

[obsidian monolith]

꧁༺Vocal༻꧂:

[willow bear]

this is overcomplicated.

take a complex root z, plug it into the equation, conjugate both sides keeping in mind a, b and c are real.

[muted steeple]

then what is simplified?

[willow bear]

get that conj(z) is also a root.

so your two roots are z and conj(z).

and that's it.

[muted steeple]

uhh, how would a conjugate work for the $\pm$ part?

[obsidian monolith]

꧁༺Vocal༻꧂:

[willow bear]

what ± part

i'm suggesting a solution which does not invoke the quadratic formula at all

[muted steeple]

take a complex root z, plug it into the equation
then what equation are you talking about?

[willow bear]

the equation ax^2 + bx + c = 0, what else?

az^2 + bz + c = 0

now conjugate both sides

keeping in mind a, b and c are real and so are their own conjugates

and that, for that matter, so's zero

[muted steeple]

ah, wow. I was way too invested in the quadratic formula idea.

[willow bear]

if you do it correctly, you'll get $a(\overline{z})^2 + b\overline{z} + c = 0$

[obsidian monolith]

Ann:

[willow bear]

so $\overline{z}$ is also a root

[obsidian monolith]

Ann:

[willow bear]

so your two roots (since this is a quadratic equation and hence cannot have more than two) are $z$ and $\overline{z}$

[obsidian monolith]

Ann:

[muted steeple]

Isn't the conjugate of the real number, just the real number itself? i.e. x + 0i and its conj is just x - 0i hence x?

nvm.

[willow bear]

yes

yes that's why $\overline{az^2} = a\overline{z^2} = a(\overline{z})^2$

[obsidian monolith]

Ann:

[muted steeple]

ohh that's so smart, thanks.

[viscid thistle]

Can someone tell me how to use center and vertex as a step to solve for the equation

I know e=c/a but don’t know if it’s (c,0) or (0,c)

<@&286206848099549185>

[slow geode]

Please ping at least 15 min after posting your question next time

[viscid thistle]

Oh sorry

I didn’t know

[placid heart]

may anyone explain and proof how to find the Range of the function? Step by step, proof that subset that youve found is the Range of the function for this Domain?

[hexed geyser]

yoyo

can someone hook me up here

its been so long since i've done polar coordinate shenanigans

so for point B

r = 4 of course

and so the angle is...

uh

[viscid thistle]

look at a unit circle

[hexed geyser]

i dont wanna im stubborn

[viscid thistle]

just do it

[hexed geyser]

how can i tell fromt his pic

fine

[viscid thistle]

if u learn the unit circle you'll tell from this pic

[molten kestrel]

Anybody need help? O.o

[severe prawn]

mental help

[viscid thistle]

@viscid thistle
Yo
Is it ((x-1)^(2))/(200)+ ((y-2)^(2))/(30)=1

[dark stump]

anyone have a nice way to memorize all those, double angle formulas

half angle formulas

product to sum formulas

sum to product formulas

sum and difference formulas

[viscid thistle]

why would you memorise them

[dark stump]

uh for a test

that isnt open note

and has all of those

rip

[viscid thistle]

i mean

you can derive them all from de moivre's thm

[dark stump]

whoa really

[viscid thistle]

or just practice until you remember them

ya

[dark stump]

rip thank yuu

i just learned that formula for like next fall semester so ill probably just work on bland memorizing for now lol

thanks

[hard hornet]

as long as you know these two formulas, you can pretty much derive anything

sin(a + b)
sin^2(x) + cos^2(x) = 1

[fleet yew]

You still need half angle

Wait

[dark stump]

huh

okay thnks ill make sure to go and understand more about deriving it after my test tho cause its tomorrow rip

[patent beacon]

If you have sin(a + b)

a = b = x to get sin(2x). That's your double angle. Rearrange to get your half angle

.
If you have sin²(x) + cos²(x) = 1

Divide both sides by cos²(x) to get:
tan²(x) + 1 = sec²(x)

Divide instead by sin²(x) to get:
1 + cot²(x) = csc²(x)

[fleet yew]

@patent beacon im kind of small brain rn

How do you rearrange it to get half angle

Sin(2x)=2sinxcosx

So?

[patent beacon]

Let u/2 = x

sin(u) = 2sin(u/2)cos(u/2)

Haha, now that you mention it, that doesn't just give it outright

But with the other one I bet it can

I think they're both based off cos

[fleet yew]

Oh i see

Yeah that works

[patent beacon]

cos(2x) = cos²(x) - sin²(x)
Then use pythag and solve ya

[hard hornet]

You can compute double angle, then substitution for half angle ^^^

i lowkey forgot how to get cosine sum angle hahahahaha

[patent beacon]

The sum ones aren't very easy. I just memorize those

[hard hornet]

fair enough

[patent beacon]

Actually, if you're good with matrix multiplication they're easy

Just have to memorize the rotation matrix

Or even find a way to get that quickly, haha

[heady jewel]

if you have the pre requisites you can learn it anytime

[viscid thistle]

Mostly people learn in their mother's womb

Very trivial

[north crescent]

I learned it at 14

[fleet yew]

I learned calculus in physics class

My physics teacher was actually better at explaining it

[viscid thistle]

Lmao kinda relatable

[atomic thorn]

very quick question, does it matter if we flip f'g-g'f in the derivate ? or no ?
Quotient Rule of f/g is (f' g − g' f )/g2

in case i made it fg'-gf'/g

[patent beacon]

Yes it will give you the negative of the answer

[pale bison]

$\frac{f'g-g'f}{g^2}\neq\frac{fg'-gf'}{g^2}$

[obsidian monolith]

Publius:

[pale bison]

uh yeh what kaynex said

[patent beacon]

As
a - b = -(b - a)

[atomic thorn]

thanks a bunch. going crazy revising math with a differnet langauge

[pale bison]

i don't think that's a language problem

[atomic thorn]

i know, everyting is messed up in my head tho 😄

[muted granite]

-- f varies jointly as the square of g and inversely as the square root of h, and f=6 when g=2 and h=64. Find f when g=3 and h=36.--
is the correct formula for this problem f=(√g/√h)?

[patent beacon]

f = kg²/√h
I believe. This is why we have equations lol

Since they give you one point, you can find k exactly.

[muted granite]

When I plug in g and h, i use 3 and 36?

[patent beacon]

You can, and you'd get:
f = k(3)²/√36

Problem is, this isn't enough info for anything.

[muted granite]

right. bc there would be two variables left over. k and f.

How do I find F then? There has to be a solution.

[patent beacon]

Use the point they've given you.
f = 6, g = 2, h = 64:
6 = k(2)²/√64
k = 12

And THEN use your unknown point with k = 12

[muted granite]

oh. then you plug the k in for the second formula?

got it. thank you.

[hexed geyser]

hey

can someone help me convert this to cartesian:

so my approach so far is:

multiply both sides by r to yield

r^2 = -6rcos(theta) + 6rsin(theta)

thus r^2 = -6x + 6y

but i have a feeling this isnt right because i just hit a wall

maybe i complete the square but i doubt it

<@&286206848099549185>

[lilac pier]

r^2 = x^2 + y^2

@hexed geyser

[hexed geyser]

right

ah

so r^2 = -6(x^2) + 6(y^2)

no

but rcos(theta) = x

not x^2

[lilac pier]

?

[hexed geyser]

i feel retarded sorry i have no business being in calc II

[lilac pier]

Your working is fine

[hexed geyser]

i dont know

hmm

[lilac pier]

You have r^2 = -6x + 6y

[hexed geyser]

right

[lilac pier]

Plug in r^2 = x^2 + y^2

you get x^2 + y^2 = -6x + 6y

[hexed geyser]

OH

[lilac pier]

Now you can try to complete the square

[hexed geyser]

of course

jesus

thank you

that was obvious lol

(x+3)^2 + (y-3)^2 - 18 = 0

[lilac pier]

you can add 18 to both sides

[hexed geyser]

(x+3)^2 + (y-3)^2 = 18

[lilac pier]

and to find the radius..?

[hexed geyser]

to find radius

the

wait wut

radius for cartesian?

[lilac pier]

radius for the circle

[hexed geyser]

just square root it

no

divide by 18

no

[lilac pier]

close

square root method

close

[hexed geyser]

it needs to equal 1

[lilac pier]

no?

x^2 + y^2 = r^2 means a circle of radius r

so you have 18

can u write 18 as a square of something

[hexed geyser]

square it

[lilac pier]

how

[hexed geyser]

hmm

3sqrt(2) is what you mean?

[lilac pier]

yes

so you have a circle centered at (-3,3) with radius sqrt(18) or 3sqrt(2)

[hexed geyser]

(x+3)^2 + (y-3)^2 = 3sqrt(2)

now what

[lilac pier]

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

[hexed geyser]

ah right r^2

[lilac pier]

and you're done

[hexed geyser]

ok got it

thank you

[lilac pier]

np

[muted granite]

Partial Fraction Decomposition
-2x^3-x^2-5x+1/x^4+x^2
1)-2x^3-x^2-5x+1/x^2(x^2+1)
2)-2x^3-x^2-5x+1/x^2(x^2+1)=(a/x)+(b/x^2)+(c/x^2+1)
3)-2x^3-x^2-5x+1=a(x^2)(x^2+1)+b(x)(x^2+1)+c(x^3)

From steps 2 to 3, where am I going wrong?

[hexed geyser]

@lilac pier can you just tell me the first step for this one thats it?

[lilac pier]

same as b4

like u did

[hexed geyser]

@muted granite ill help you one sec

[muted granite]

thank you.

[lilac pier]

think of getting r to the cos and the sin

[hexed geyser]

@muted granite so you're right on step 2, just checked that

and noww

you wanna write a/x + b/x^2 + (cx + d)/(x^2 + 1)

which you did i think

then you multiply by the denominator:

[muted granite]

i did not include cx+d, whats that?

[hexed geyser]

sorry thats just the format

so d = 1

so you end up with:

then you solve the system of equations for all the a0 a1 a2 a3

which should be:

3, -2, and -5

so you end up with

[muted granite]

Yeah, I seen that on the calculator I used. But I am confused on what is (a3x+a2)

[hexed geyser]

this bit

[muted granite]

yeah

[hexed geyser]

so thats just the

hang on i have a part in my notes about that

so in this case a3 x + a2 is unknowns, you know the last one has to have an x because you multiply by the denominator

and since x^2+ 1/x^2 + 1 = 1

you know it has to be something other than that

[muted granite]

Ok. I'll review my textbook again. the process looks different. Thanks.

[hexed geyser]

hence the extra x we add

@lilac pier so how on earth do i get the r under that fraction then

[lilac pier]

divide both sides

by...?

[hexed geyser]

r of course

but then that screws up the equation doesnt it

[lilac pier]

no

the r on the other side cancels out nicely

[hexed geyser]

ah okay give me a second

ahhhh i see what you mean

3/(5cos(theta)+6sin(theta)/r

= 3/(5rcos(theta) + 6rsin(theta)

which all equals 1 because r/r

so now we can replace those with 5x and 6y

and we get y = -5x/6 + 0.5

got it

hmm

[muted granite]

Is this the correct translation for the series: 1/2 - 1/4 + 1/8 - 1/16 + ... +1/128?

[hexed geyser]

well did you try the partial sums

try plugging in 12 for n to see if it gets 1/128

[muted granite]

why would I put 12? Also, it does not get 1/128

[shut shuttle]

It's $\frac {(-1)^{n-1}}{2^n}$

[obsidian monolith]

BeautifulLie:

[hexed geyser]

conversion of y = 10x^2 to polar

rsin(theta) = 10(rcos(theta))^2

rsin(theta) = 10r^2 cos^2(theta)

how is this not right

answers apparently (1/10)sec(theta)tan(theta)?

they mustve used some random ass identity i havent seen in 3 years

[lilac pier]

both are correct

[hexed geyser]

apparently not

[lilac pier]

what they did was

isolate r

divide both sides by 10rcos^2(theta)

So that there's a "r" remaining on the RHS

[hexed geyser]

you cant isolate it from the left now?

instead?

[lilac pier]

well you can

[hexed geyser]

lemme try it again i see what you mean

[lilac pier]

but it'll cancel out anyways from the RHS

there's 2 rs on the RHS, 1 on the LHS

so there will always be one r on the RHS after simplifying

[hexed geyser]

got it thanks

[north crescent]

The highlighted area should be 1/2n

The bottom should be n=1

The top should be 7

[molten kestrel]

Hey y'all.

[upbeat moss]

This is the question i am answering

Am I on the right track?

[languid crane]

yeah u r

[fleet yew]

bro just graph it

[hexed geyser]

@lilac pier yo

whats the deal with this i thought that was right

apparently not

(its for 5,5)

[fleet yew]

read the question

lmao!

[hexed geyser]

yeah so how do i write that then

5sqrt(2), -5pi/4?

nah

hmm

[fleet yew]

lol

[hexed geyser]

maybe just -pi/4

[fleet yew]

[hexed geyser]

🤔

wasnt that either

[fleet yew]

the angle must be coterminal

[hexed geyser]

how do you mean

[fleet yew]

angle x = angle y if x=y mod 2pi

so if you add 2pi to an angle, it's the same angle

[hexed geyser]

if you minus 2pi its the same angle

?

[fleet yew]

yes

[hexed geyser]

so then

it must be

-7pi/4

ayyyyyyy

thanks @fleet yew

[fleet yew]

lmao

np

[hexed geyser]

sad part is im in calc 2

hahaha

like my professor says, "you take calculus so you can learn precalculus"

fuckin hell

[fleet yew]

bro i was thinking how you're doing polar but don't know ur unit circle

[hexed geyser]

well in high school i was a bad student and only got serious in college

so a lot of that stuff i dont know still 😛

anyway time to go have some brews cuz im done

[sharp wraith]

Hello 🙂 just asking, precalculus is not suitable for primary students right?

[north crescent]

Unless you’re really smart, precalculus probs isn’t right just yet, but it’s never too early

As long as they have a strong foundation, learning ahead is possible I started precalc in 8th, but I’m sure if you’re very driven, you could start

[remote bison]

What grade do people usually learn precalc at?

[north crescent]

My school has it at either 10th or 11th depending on which course track you take

[proven marten]

What textbook are you using?

[north crescent]

I did precalc 2 years ago and last year, and I don’t remember the name

[remote bison]

Is khanacademy's material adequate for the entire curriculum? I'm in 9th grade in an IGCSE school, we don't have US math courses

[north crescent]

For us schools, khan academy has just about everything. Not sure about yours, but I’m sure khan academy will have most if not all of what you need

[remote bison]

not for igcse math, I'm asking about khanacademy's precalculus khanacademy.org/math/precalculus

[proven marten]

Just curious are you IGCSE going into IB

[shut shuttle]

Can someone help me with this

[copper vigil]

factor out constants

like m^2/3^m is constant wrt n

[shut shuttle]

Okay after that?

[copper vigil]

use wolfram alpha

[shut shuttle]

Eh

[rocky prism]

Hello, Question for you guys

Have had this problem showing up for a while now,
I know that x ~= 66.42 but im not sure what answer I need to select, could somebody guide me through this? thanks

its an exam?

this is a question on kahn academy, this isnt from a exam

[viscid thistle]

Oh lol

Lemme see

[rocky prism]

ty

I get you take the arccos of 0.4 and that gives the value on the left (66.42) but im not sure if its you multiply by 180 or 360, and sometimes for this type of question it would say 113.58 (just as an example) would also be correct, which I dont understand

[viscid thistle]

,w cos(x)=0.4

[obsidian monolith]

Display results online and refine query

Requested by Al3dium#4084

The following error occured while calculating:
Error: Invalid left hand side of assignment operator = (char 12)

[viscid thistle]

Let me get a calculator lol

[rocky prism]

XD

[viscid thistle]

Aight so its 66.42

So its D :)

@rocky prism

[fleet yew]

you don't need a calculator if you use a bit of logic

[rocky prism]

okay, can I show you another question with the same idea?

[viscid thistle]

Yeah, i lack on logic on this topic a lil bit

So you can shine the logic to him maybe AMD

[fleet yew]

in fact there should be two answers to this question

[viscid thistle]

There are

But Khan only shows one

[rocky prism]

No it allows multiple for these types

[viscid thistle]

I mean, i dont see any other

Correct

[rocky prism]

I find this next question confusing becuase although it is the same as the one I just asked about, it still differs

[fleet yew]

think about it

cosine is an even function

so cos(x)=cos(-x)

[rocky prism]

I need a visual one second 😛

right okay

What about this though

let me double check but

I am getting the value of 8.16

But all the 8.16 values are both n * 40

why is it 40? should it not be 360? what makes the value change?

[fleet yew]

360/9=?

[rocky prism]

OH I see, okay

I didnt realize that affected it

so if cos(x) = cos(-x) does that mean cos has two solutions then?

so would it be both C and D?

[brave patrol]

so in a limit, when its lim(x-->-1^+) and lim(x-->-1^-) whats the diff

[viscid thistle]

• is from the right side, - is from the left side

[brave patrol]

ah ight thx

wait so like

lim(x-->-1^+) is the line coming from the right?

[viscid thistle]

yes

[brave patrol]

what if its just lim(x-->-1)

[fleet yew]

Thats just a normal limit

Provided it exists

[muted granite]

Where did I go wrong? I found d. But a1 seems strange.

finding d?

where are you getting 3d-8? a2 and a9 are 7 steps apart.

[viscid thistle]

Yes they are 7 steps apart, but you have to consider the y values as well

[muted granite]

so its not a9=a2+7d?

[viscid thistle]

No

As the input increases by 1 the output increases by 3

[muted granite]

whats the new equation then? a9=?

[viscid thistle]

A9 = 3d - 8

[muted granite]

where does -2 fit in all of this?

[viscid thistle]

You use it to get 3

19 - (-2) = 21

9-2 =7

21/7=3

[muted granite]

so there are two steps? The intial one I did to get 3 and then another?

[viscid thistle]

Yes

[muted granite]

idk every other problem like this doesn't have this step.

[viscid thistle]

Ok so your first part is correct

[muted granite]

right then i plug it into an=a1+(n-1)d

[viscid thistle]

I don't get why u did what u did fit the 2nd part

[muted granite]

bc im trying to find out what a1 is in the sequence.

[viscid thistle]

oh i figured out wherr you messed up

your an and n are for different inputs and output

you should have 19 = a1 + (9-1)(3)

not -2 = a1 + (9-1)(3)

@muted granite

cuz a(9) = 19

[keen sigil]

could someone explain this to me?

it's referring to an inverse function

[muted granite]

@viscid thistle thats what i first thought but it seems you must put -2 on the left side of the equation.

[patent beacon]

What is an inverse function? Like, what does f do with f^(-1)?

[keen sigil]

it's flipped on the y=x line

so every ordered pair is reversed

like (1,0) would become (0,1)

[patent beacon]

So f takes 1 and maps it to 0

And f^(-1) takes 0 and maps it to 1

[viscid thistle]

@muted granite you don't put -2 at the left side of the equation. that's your error

[patent beacon]

The answer is A.
f composed with f inverse gives back x. That's how we define the inverse function @keen sigil

[keen sigil]

alright thanks

[muted granite]

Alright. I'll take a look. thanks.

[steep geode]

did i do this question right

i did horizontal line test since that is what test we do for one to one

[viscid thistle]

yes

[keen sigil]

could someone explain this?

[remote bison]

<@&286206848099549185>

[karmic wave]

@keen sigil the bars mean "absolute value" if that's what you're asking

[keen sigil]

yea im not sure how to graph it from that

[fleet yew]

Just take the function and make it positive everywhere

[rugged linden]

So i am kinda confused about example 3

the set it {0}

*is

and 0 is an element of T

and 0 is less than or equal to 0

[willow bear]

T is not {0}.

[vague minnow]

yeah {0} is a subset of T

[willow bear]

but you aren't examining {0}

you're examining T

{0} is bounded and 0 is its maximum element, yes
but that is not what the example is saying

[true stone]

does anyone else love the feeling you get tweezing your ass hair?

i pull them out and put em in a lil bottle so i can smell them whenever i want

[north crescent]

#chill

[rugged linden]

Wait wait

I don't get what you mean

0 is the only number that can exist that is less than or equal to zero yet less than one

ohhh

shitt

yeah i forgot about fractions and whatnot oof

Since we're talking about all real and not all integer

[viscid thistle]

@Jah#5662 please never speak again

Oop my wish came true

[smoky needle]

Can someone help me on this problem?

[lilac pier]

try shifting that 2 * 3^x to the LHS

then factor out 3^x

[smoky needle]

Sorry what is LHS?

[viscid thistle]

Left hand side.

[smoky needle]

Thank you!

I will do that

Can I multiply 2(3^x) before shifting it to the other side?

[fleet yew]

How would you?

[smoky needle]

I can’t distribute it and make it 6^x?

Or is that not a thing?

[lilac pier]

@smoky needle no

[smoky needle]

Ahh I see okay.

[lilac pier]

because 3 has a power of x, we have to first evaluate 3^x and then multiply by 2, but we dont know what x is

so we let it stay 2(3^x)

[smoky needle]

I see okay! So, dumb question sorry, to clarify. I just subtract two and divide 3^x on to get it to the left side right?

[lilac pier]

No you just subtract 2(3^x) from both sides

$3^{x+1} - 2(3^{x}) + b = 2$

[obsidian monolith]

Sup?:

[smoky needle]

Oh okay thank you! I was getting confused there

This is what I have now

[tardy ridge]

3^(x+1) is 3*3^x. Anyway, how do you solve 1 equation with 2 variables? is it solve in terms of b?

[smoky needle]

Sorry, I kinda went ahead and didn’t see your message till now. I am not sure if this is correct

[tardy ridge]

??

How do you know what b is

wait is that a 6?

[viscid thistle]

lmao

[tardy ridge]

also 3^x=-4 is impossible

[smoky needle]

Yes sorry, that’s a 6😅

And yes, so I don’t think it has any real solutions?

[tardy ridge]

right

[smoky needle]

Thank you all!

For clarification, why can’t I have it as 1/3^4?

Or simplified as 1/81?

When I plug in -4 for x

[viscid thistle]

have what as 1/3^4?

[smoky needle]

3^x= -4

Is that just not possible?

[viscid thistle]

I don't know what you want to do with 1/3^4

or where you got it

[smoky needle]

Oh I was doing it wrong sorry!

Also, I was wondering if someone could check my work for this problem? I got 1/2 on it and my total score for the hw is 19/20. But I’m just really trying to end the class with A, and our hw and quizzes are worth the same percentage. I wanted to see if I did any mistake on this one and possibly ask for a regrade?

[harsh smelt]

@smoky needle what was your answer?

[smoky needle]

I got (3,9)@harsh smelt

[harsh smelt]

ok, give me a sec i ll check

[smoky needle]

Thank you. Unless I did have a mistake, I’ll take the point off lol

[harsh smelt]

i think i found some mistake but i ll finish calculations still

@smoky needle sorry, i somehow got lost in my own calculation but still

[smoky needle]

So sorry! I am barely getting your notifications. And ahh... okay. My answer just be wrong then. Lol

[harsh smelt]

i mean you even look graphically, distance will be minimal where the segment from the point to the graph is perpendicular

and at 3, 9 line from 0, 18 is not perpendicular

[smoky needle]

Thank you so much, I appreciate it.
And I see. So to make sure, we just need to see it graphically if perpendicular?

*it’s

[harsh smelt]

it is way to check the answer

[smoky needle]

Got it, thanks a lot again!

[molten kestrel]

Uhh hi.

[molten kestrel]

@viscid thistle

O.O

[viscid thistle]

hellO

[molten kestrel]

Conic sections?

O.O

[viscid thistle]

sue

Sure*

god damnit help me speak

Ok so ellipses?

[molten kestrel]

idk.

I knew this.

[viscid thistle]

Do you have any specific issues of just general...

[molten kestrel]

Uh in April.

[viscid thistle]

Yes

[molten kestrel]

Okay ik the equations okay.

I'm not dumb dumb.

[viscid thistle]

Co-vertices are the endpoints of the minor axis.

Means the shorter diameter.

[molten kestrel]

Ik

[viscid thistle]

Ok

[molten kestrel]

and focis are the almost there

ones

[viscid thistle]

sorta yea

[molten kestrel]

I'm kinda ahourhneoanf rn.

So ima be kinda dumb.

Anyways.

[viscid thistle]

There’s a formula and you can reverse engineer it to find the actual vertices I believe...

[molten kestrel]

x^2/a^2

y^2/b^2

=1

yeah the formula is

c^2=b^2+a^2

pythagorean

[viscid thistle]

mhm

Yes

I agree

Ok so

[molten kestrel]

okay so

[viscid thistle]

Standard form of an ellipse is ((x-h)^2)/a^2 + ((y-k)^2)/b^2

[molten kestrel]

center is (0,0)

No it isn't.

[viscid thistle]

=1*

[molten kestrel]

it's what I said.

Okay acc idk. I'm kinda dumb rn.

Do as you wish.

[viscid thistle]

Yeet

Essentially the same if h and k are 0

Which is where you get the center from I believe...

[molten kestrel]

mmhhm

that's why I said

[viscid thistle]

Ok then we good

Cleared that up

[molten kestrel]

x^2/a^2

y^2/b^2

=1

okay. so. uh.

I can't see clearly but.

[viscid thistle]

Any specific issues? Or...

[molten kestrel]

Well I can, but my I can see my headache. Okay so.

Now.

Center is (0,0)

and foci is

[viscid thistle]

mass confusion II, electric boogaloo

[molten kestrel]

(0,3) (0,-3)

[viscid thistle]

Yes

[molten kestrel]

and

[viscid thistle]

mhm

[molten kestrel]

covertices is

(2,0) (-2,0)

so.

what is a

what is b

and hwat isc d

what is c?

okay thanks.

[viscid thistle]

Vertices: +-a/b. (Whichever is bigger.)

[molten kestrel]

yeah.

[viscid thistle]

mmk seems like you got this mostly...

I dunno what the issue is.

[molten kestrel]

thank you, child.

[viscid thistle]

I mean A B and C yea but looks like u can solve for it

[molten kestrel]

I shall do this now.

[viscid thistle]

okie

And YW btw.

For basically nothing at all.

[molten kestrel]

[viscid thistle]

AWw cute

[molten kestrel]

Okay guys.

[viral trail]

I know vietas formula, but I cannot factorise the given expression. Help pls

[fluid shore]

You don't have to

$pq + qr + pr = 3$. Can you use that to construct the sum you want?

[obsidian monolith]

Abhijeet Vats:

[fluid shore]

@viral trail

[viral trail]

@fluid shore
I tried multiplying by p+q+r but then that gives me 3 unwanted terms like p^2r

[fluid shore]

Hmmmmmmm

[viral trail]

Squaring doesn't help either

[fleet yew]

Bro just use rational root theorem?

[viral trail]

It says real roots

They can be irrational

[fleet yew]

Have you tried

Try +-1

[viral trail]

+-1 for roots?

[fleet yew]

Nvm the roots are all irrational

Just checked

[viral trail]

Right, so how do I proceed?

I need help

[fleet yew]

Use cubic formula

[willow bear]

uh

[severe verge]

short notice im just making everry channel unusable

for the duration of the ap exams

i forgot to tell people earlier

[viral trail]

What is cubic formula?

[severe verge]

its like an hour long

[fleet yew]

Why

[olive star]

what's 16 * 3?

[tardy ridge]

probably 48

[fleet yew]

@olive star here's a cool trick

tex bot has a calculator built in

you can type in things

,calc 16*3

[obsidian monolith]

Result:

48

[fleet yew]

,calc sqrt(a^2)

[obsidian monolith]

The following error occured while calculating:
Error: Undefined symbol a

[fleet yew]

it can't do any symbolic math

but it's good at other stuff like computations

[viscid thistle]

Lol i think he was joking, wasnt he?

[fleet yew]

,calc 31*3

[obsidian monolith]

Result:

93

[tardy ridge]

what is the fourth root of 22667121

[fleet yew]

,calc fourth root of 22667121

[obsidian monolith]

The following error occured while calculating:
Error: Undefined symbol fourth

[viscid thistle]

69

+-

,calc 22667121^(1/4)

[obsidian monolith]

Result:

69

[viscid thistle]

@fleet yew

Gottem'

[fleet yew]

+-69,+-69i

[viscid thistle]

Would this be a way to solve this problem?

I tried using an equation

[gritty sandal]

which question

[viscid thistle]

number 1

[gritty sandal]

seems like you got the answer

[viscid thistle]

i did, but does my equation work?

[gritty sandal]

yeah it works, just adding the equations together and isolating x

that's how I would've done it

[viscid thistle]

ok, thanks

i also isolated the y

which seems to work as well

[gritty sandal]

u could've just subbed in x into one of the equations, a bit less work

but you did get the right answer using the right methods so good job!

[viscid thistle]

i just wanted to know how to find y as well

[gritty sandal]

once you found x=9, you can sub it into any of those equations, like x + y = 6

[viscid thistle]

yeah i know

[gritty sandal]

yeah that's kinda it

[viscid thistle]

seems good, thanks

[chilly spire]

can someone help pls

with 7 and 9

like how would i get a

bro please 😔

[viscid thistle]

is there anyway to use an equation to solve 2 a)? #precalculus message

set them equal to one another

or subtract them from one another

[gritty sandal]

the victorian curriculum has nothing about focuses, not even in year 12, so can't help you sorry

[viscid thistle]

yeah nvm

[gritty sandal]

should probably look at them tho

[viscid thistle]

im figured it out

multiply or substitute

salutations everyone, may I please get some help with my online test ? I would’ve been confident to complete it but ever since my mom came down with covid-19 i haven’t had much time to study.

i have never asked for a handout like this before I already completed my part 1 but this part 2 has been a struggle

it is only a few questions

yah go ahead

[chilly spire]

ugh :(

[viscid thistle]

try figuring it out yourself

[chilly spire]

i literally dont know the point of including the focus

[acoustic harbor]

u cant help him on a test

[chilly spire]

bc all ur missing is a

[viscid thistle]

his mom has covid

lets cut him a break

how do you even know thats true @viscid thistle

yes i live in nyc my mom is an ER nurse

I will take a picture of her ID

hold on

@viscid thistle learn the material yourself after this test

i doubt he's lying

its never good to rely on others

i know just in general

yes like I said I never do this

yeah its fine

Here is her id

I’ve been working at chipotle to provide and taking care of her so I haven’t had much time to study but I did all that I can

sorry that many people in this server don't have a heart

it’s ok i get where they’re coming from

no, its just simply me being skeptical

because its the internet

and you shouldnt just mindlessly believe people in real life as well

its fine if hes telling the truth

, which he is

idk how to ask in a mild manner, but may i please get the answers with work? it is only these last bits of questions

that are due in about an half a hour

try asking @ honorable. I havent learnt the math you have on the test yet

@viscid thistle

how do i do that? im sorry like i said i have never done this

@viral flicker could you help @viscid thistle sorry for the ping

I dont really know what to do either to be honest, I'll try to help though

you could always also tells the school your situation as well, they should be understanding

[fleet yew]

@viscid thistle tell ur mom to renew her id

expiring soon

[viscid thistle]

@viscid thistle do you still need help

i can help you on all of those if you're still here

[viscid thistle]

how do you know how to factor expressions?

like one with (c^3 + whatever + whetever)

is (c^2 +whetever) (c + whatever)

how come it cannot be (c + ) (c+) (c+_)

and then an expression with c^4

is (c^2 + ) (c^2 +)

how are you supposed to know how to factor t?

[blazing raven]

not even every two degree is factorable.

x^2 + 1

[viscid thistle]

yeah, but how do you know how to factor the ones that are factorable

also it is factorable (x-1)(x+1)

[blazing raven]

What's (x+1)(x-1)

(check your work)

[viscid thistle]

x^2 + 1

[past meadow]

you sure about that?

[viscid thistle]

oops

[past meadow]

foil (x-1)(x+1)

it wont be x^2+1

[viscid thistle]

x^2 -1...

neg•pos = neg

[muted steeple]

guys, is there some neat trick to factoring quadratics with huge numbers or everyone is just a human calculator nowadays?

[craggy dune]

an example?

[willow bear]

quadratic formula tbh

[muted steeple]

$y^2+2y-2400=0$

[obsidian monolith]

꧁༺Vocal༻꧂:

[muted steeple]

yeah, but every solution i look at just factors them mindlessly

i have no clue how they can hold all these factors in their head

[willow bear]

okay well

that 2 there is pretty small

so you can complete the square instead

and get (y+1)^2 - 2401 = 0

and then perhaps you happen to know that 2401 = 49^2

[muted steeple]

ahh, i guess i just need to git gud at mental arithmetic then.

[craggy dune]

horner method could also help

[muted steeple]

what's that?

[craggy dune]

https://en.wikipedia.org/wiki/Horner's_method

look at the examples in on the site

if you run through the method, you can use the result of the last equation to figure if you guessed to high or low

it takes a bit, but once you've gotten used to that method, polynomials that can by divided by integers is easy

[muted steeple]

Thanks, i'll try it out.

[gilded mirage]

idk what this is even asking

[sour quiver]

arithmetic progression

[tardy ridge]

#of terms * (first term + last term)/2

[wind igloo]

👀

Why ghost ping, Ann? 😛

[tardy ridge]

does the green bot store deleted messages?

[willow bear]

there was an attempted cheater here

i pung mods

cheater got b&

i deleted mod ping

[wind igloo]

Ah. Ok.

[viscid thistle]

How do I simplify 2sec^2x + 1

@viscid thistle apply $sec(x)=\frac{1}{\cos{x}}$

apply $\text{sec x}=\frac{1}{\cos x}$

[obsidian monolith]

Al3dium:

[ember crane]

hey! am I doing this right?

[serene heath]

the equation on the left has 2 solutions

[ember crane]

+2 and -2 right?

but for the left one: this is the answer

I'm so confused

[jagged timber]

Where did that k come from?

[patent beacon]

That's the answer for the right one @ember crane

[ember crane]

ops

I mean that is the answer for the right onwe

*one

my faulth

[patent beacon]

Did you want to go over that answer?

[ember crane]

yeah, because 4pi/3 is not supposed to be accepted as an answer right?

the restriction for cos inverse is 0 to 180 degree

[patent beacon]

The problem is you're using cos inverse haha

cos(4π/3) = -1/2
So it is indeed a solution

[ember crane]

wait, I'm confused

[patent beacon]

cos inverse won't give you it, but it is still one

[ember crane]

so as a solution, should I follow the restriction? or just find thee solution?

the range only in this

[patent beacon]

Any number x that satisfies
cos(x) = -1/2
Is a solution here

The inverse cos function will give you the one solution in Q2, but won't give you the solution in Q3.

Ultimately inverse cos isn't really related to your problem. The unit circle is far more valuable of a resource

[ember crane]

wait! so how about x = (17pi)/6 for cos (x) = -1/2 ?

[lusty heart]

so does anyone know how to put x^2+2y^2=16 and -5x+y^2=0 into a graphing utility?

[jagged timber]

Like Desmos?

[lusty heart]

it gives me weird decimals

[ember crane]

how?

[cosmic cargo]

Doesn’t this series diverge or is my formulation wrong?

[tardy ridge]

if |-2x|<1 it'll converge

you will need to find an x value for that

[cosmic cargo]

Just got that, thanks so much

I’m confused on c tho?

[tardy ridge]

Nice

[cosmic cargo]

How would I find the sum of the series in terms of x?

If it diverges

[tardy ridge]

you won't be able to do that if it diverges

[cosmic cargo]

Hmm okay

Thank you very much @tardy ridge 😎

[tardy ridge]

Nice

[ember crane]

how can I do this problem?

[uncut mulch]

compound angle identities

[ember crane]

I'm sorry, this one

[past meadow]

sin (a+b)=sin(a)cos(b)+cos(a)sin(b) in specific

[uncut mulch]

also compound angle identity

specifically for ||cos||

[ember crane]

I added both angles up and the number is really weird

11(pi)/21

[past meadow]

cos(a-b)

not cos(a+b) in this case

[uncut mulch]

you shouldn't be adding

[ember crane]

How should I do the question?

[uncut mulch]

pay close attention to the signs in that identity

and try applying it again

[ember crane]

which signs? I try to think but I could not come up w/ a way

[uncut mulch]

$\cos(a)\cos(b)+\sin(a)\sin(b) = \cos( a {\color{red}{-}} b)$

[obsidian monolith]

ramonov:

[ember crane]

OH!!! right, thanks you @uncut mulch

[viscid thistle]

hey can i get some clarification?

problem 7

[viscid thistle]

Find x, then plug it into sinx+cosx.

[ember crane]

I have a question how can I identify horizontal asymptotes in rational function using long division?

not the power

[viscid thistle]

you use long division

as you said

[viscid thistle]

ok

so

divide every term by x

[ember crane]

but which number indicates horizontal asymptotes

[viscid thistle]

you'll get

$\frac{2 - \tfrac{3}{x}}{1 - \tfrac{2}{x}}$

[obsidian monolith]

polynomial:

[viscid thistle]

now

$\lim_{x \to \infty} \frac{2}{x} = 0$

[obsidian monolith]

polynomial:

[viscid thistle]

and

$\lim_{x \to \infty} \frac{3}{x} = 0$

[obsidian monolith]

polynomial:

[ember crane]

I did it like this, is it correct?

[viscid thistle]

likewise for the negative infinity

so

[fleet yew]

Lc of the numerator over lc of the denominator

[viscid thistle]

the two terms become 0 essentially

[fleet yew]

Assuming that they are of equal degree

[ember crane]

my prof doesn't allow me to explain horiontal symptotes using limits, nor degree

only using number in division

[glacial coyote]

what u can do is long divide and ignore the remainder

[next kernel]

horizontal asymptote will always be y = (number)

[ember crane]

I'm struggling real hard

[viscid thistle]

but you need to use limits to explain the division

[next kernel]

and you are taught to ignore the remainder completely

[viscid thistle]

to show that the other terms will be come irrelevant

[ember crane]

@glacial coyote oh?

[viscid thistle]

@ember crane ignoring the remainder is the point, since the limit as x goes to infinity is 0

[glacial coyote]

the long division remainder being ignored is explained by limits but if you don't need to explain it just divide and ignore remainder

[viscid thistle]

hence, the remainder is 0.

[ember crane]

so in this one, which one is the horizontal asymptotes?

[viscid thistle]

for large x

[ember crane]

is it 1?

[glacial coyote]

its y=2

because your result was 2 remainder something

[ember crane]

OHHHH!!!!

got it, thnk u

[glacial coyote]

sure np

[next kernel]

@glacial coyote need help

[viscid thistle]

@ember crane but that still uses limits tho

like

literally

[next kernel]

What kind of sequence is this and how do we find the sum of the next 50 terms(if it is geometric or arithmetic)?

[viscid thistle]

you can't not use limits to explain it.

[next kernel]

{3^n/2}

[ember crane]

ohhhh! @viscid thistle thank u so much! that helps a lot

[glacial coyote]

@next kernel you can rewrite the series as (1/2)*(3^n) which is recognizable as a geometric series

[next kernel]

what is the ratio of that series? how do we know?

[glacial coyote]

the ratio is 3 try writing out a couple terms of it

you know it's 3 because 3 is raised to the nth power, and whatever number is raised to the nth power in a geometric series is the ratio

[next kernel]

so we get a number that is continuously increasing square roots of 3

we get to 3 to the 3/2 power

[glacial coyote]

wait was your series 3^(n/2)?

[next kernel]

yes

[glacial coyote]

then the ratio is sqrt(3)

i thought it was (3^n)/2 lol

[next kernel]

oooo

so it becomes (sqrt of 3)^n

[glacial coyote]

yes

[next kernel]

and the first 50 terms is just sqrt3^50

?

[glacial coyote]

(sqrt 3)^50 is the 50th term

[next kernel]

oh

would it be good to just put the equation in summation form?

[glacial coyote]

what are you looking for?

[next kernel]

sum of the first 50 terms

which appears to be a large number equivalent to 2.01E12

[glacial coyote]

do you know the formula for the sum of the first n terms of a geometric series?

[next kernel]

a1 * (1-r^n) / (1 - r)

[glacial coyote]

yeah use r=sqrt3, n=50 and that formula will give u the sum of the first fifty terms

[next kernel]

that gives me -1.47E12

[glacial coyote]

it didn't when i put it into a calculator

maybe u typed it in wrong?

[next kernel]

the correct answer is: -sqrt3 / 2 * (1 + sqrt3)(1-3^25)

[glacial coyote]

does the series start from 1 or sqrt3?

[next kernel]

the series of terms? or the equation

[glacial coyote]

is the first term 1 or sqrt3

[next kernel]

The first term is sqrt3.

[glacial coyote]

then use the formula to get (sqrt 3) * (1-((sqrt3)^50))/(1-(sqrt3))

then rationalize the denominator and (sqrt3)^50 equals 3^25

[next kernel]

aha, 25 * 2

and what do i do once i have:

sqrt3 * (1 - 3^25) / (1 - sqrt3)

?

Or is that my sum

[glacial coyote]

that is the sum but u can format it differently by rationalizing the denominator

[next kernel]

1^1/2 - 3^1/2?

I need you to walk me through it

[glacial coyote]

oh sure

first change it to (sqrt3)*(3^25 -1)/(sqrt3 -1)

because u just multiplied top and bottom by -1 there

then multiply both numerator and denominator by (sqrt3 +1)

[next kernel]

??????????
Where did the -1 come from

[glacial coyote]

just to simply multiply the top and bottom by -1

since (1- 3^25) and (1-sqrt3) are both negative, multiply them both by -1 to make them both positive

[next kernel]

Gotcha

[ember crane]

Anyone understand the question? and how to do it?

[hexed ermine]

Ok uhm first let's look at cot , what do you know about cot @ember crane

[ember crane]

cot = (cos/sin) ?

[hexed ermine]

Yep

So what is cos

Can you write it in terms of sine?

[ember crane]

so sine = cos/cot?

[hexed ermine]

No

What is cosine