precalculus

[rare pagoda]

but yes completing the square does exactly the same thing

[muted steeple]

let's see. So i have a problem $h=-16t^2+24t$ and the goal is $h_{max}$

[obsidian monolith]

꧁༺Vocal༻꧂:

[muted steeple]

so to complete the square i just do $h + \left(\frac{24}{2}\right)^2=-16t^2 + 24t + \left(\frac{24}{2}\right)^2$

?

[obsidian monolith]

꧁༺Vocal༻꧂:

[muted steeple]

i honestly don't know where this is getting at.

[rare pagoda]

what you are doing is correct

[muted steeple]

i'm doing, but not understanding, which is not good.

in my book

[rare pagoda]

you are going to get an equation of the form $h(t) = -a(t-t_0)^2 + h_{max}$

[obsidian monolith]

magnusChadson:

[rare pagoda]

where the max height will be h_{max} and the time associated with this height is t_0

[muted steeple]

i don't see how the above distills to that

[rare pagoda]

also, I think you forgot to divide through by -16 before starting to complete the square

[muted steeple]

it's gonna result in fractions, if i do that

[rare pagoda]

that's okay

if you find this method too tedious, you can do what I told you before of using the formula I gave you or finding the average between the two roots

[muted steeple]

$\frac{h}{-16}+12^2=t^2+\frac{3t}{-2}+12^2$

is what i get so far, i think it's correct so far right?

it's not tedious, is just, im confused about my goals here.

[rare pagoda]

I don't think that you divided the by -16 properly

[obsidian monolith]

꧁༺Vocal༻꧂:

[muted steeple]

is this correct now?

[rare pagoda]

also isn't v_0 40?

[muted steeple]

you are right, i was looking at the wrong initial value, sorry.

let me change taht.

[rare pagoda]

also I meant to divide through by -16 to get the equation $\frac{-h}{16} = t^2 - \frac{v_0}{16}t$

[obsidian monolith]

magnusChadson:

[rare pagoda]

and then complete the square

[muted steeple]

why shouldn't i plug $v_0$?

[obsidian monolith]

꧁༺Vocal༻꧂:

[muted steeple]

it's a given value

[rare pagoda]

you can

[muted steeple]

'kay so then i have $\frac{-h}{16} + \left(\frac{\frac{5}{2}}{2}\right)^2 = t^2 - \frac{5t}{2} + \left(\frac{\frac{5}{2}}{2}\right)^2$

[obsidian monolith]

꧁༺Vocal༻꧂:

[muted steeple]

$\therefore \frac{-h}{16}+\frac{25}{16} = t^2 - \frac{5t}{2} + \frac{25}{16}$

[obsidian monolith]

꧁༺Vocal༻꧂:

[rare pagoda]

the rhs can be simplified to (t- 5/4)^2

[muted steeple]

ohh

[rare pagoda]

aka there is your t max

and now isolate for h

and you'll get h_max is 25

[muted steeple]

$\therefore \frac{-h}{16}+\frac{25}{16} = \left(t-\frac{5}{4}\right)^2$

[obsidian monolith]

꧁༺Vocal༻꧂:

[muted steeple]

how does this gives me t tho?

[rare pagoda]

isolate for h first

[muted steeple]

i guess multiply by the LCD 16?

[rare pagoda]

ye

[muted steeple]

$-h + 25 = 16\left(t - \frac{5}{4}\right)^2 \ therefore -h = 16\left(t - \frac{5}{4}\right)^2-25 \therefore h = -16\left(t - \frac{5}{4}\right)^2+25$?

[obsidian monolith]

꧁༺Vocal༻꧂:

[rare pagoda]

great

[muted steeple]

so now i move the h to rhs and just solve like a quadratic right?

actually

what

[rare pagoda]

noooo

[muted steeple]

that does not make sense

like i said, im not understanding what im doing here

[rare pagoda]

so you now have a quadratic in vertex form

[muted steeple]

what is a vertex form?

[rare pagoda]

one sec

so you know the square of any real number > 0

?

[muted steeple]

yes.

[rare pagoda]

= 0 acutally

so -16* some positive number will be <= 0

[muted steeple]

yep.

[rare pagoda]

so in order to maximize -16(t-5/4)^2 + 25, we want -16(t-5/4)^2 to be as small as possible

[muted steeple]

sorry, what do you mean when you say maximise?

[rare pagoda]

to make it as big as possible

so we do that by setting t = 5/4

and then we get h = 25

[leaden stratus]

Hello. Today we tried to solve this, but even our teacher was unsure about the steps. How would you solve this? Thanks.
@leaden stratus up

[acoustic harbor]

what’s the problem

@leaden stratus

[leaden stratus]

@acoustic harbor use the search option and find the messages sent by me. You'll find the post...

[proud jetty]

what does sin^2 become?

does it have a variable?

[acoustic harbor]

what do you mean

it’s sin^2(pi)?

the variable is theta

and you plug in pi into theta

[jaunty mason]

i think, maybe, he's confused by the notation?

[proud jetty]

yes

[jaunty mason]

sin^2 (x) = sin (x) * sin (x)

[proud jetty]

like is it gona be 3pie^2-2pie?

[acoustic harbor]

sin^2 is the same as (sinx)^2

no

sin pi times sin pi

[jaunty mason]

se-senpai?? 👀

[proud jetty]

sin basically pie when solving right?

[acoustic harbor]

no

theta is pi

[proud jetty]

so sin is sin?

[acoustic harbor]

yes

[proud jetty]

no number variable?

[acoustic harbor]

no

the variable is theta

and it tell you that theta is pi

[proud jetty]

could ya tell me how i would go about putting it in a calculator?

[acoustic harbor]

you don’t need a calculator but here

3(sin(pi))^2-2sin(pi)

should give you 0

[proud jetty]

oh yea i was putting it wrong

ic thx alot

[viscid thistle]

May I get help with this please?

[smoky needle]

Can someone help me with this problem?

[acoustic harbor]

is this a calculator problem

[smoky needle]

If you’re referring to mine, I don’t think it is.😅

[crimson merlin]

,help

[obsidian monolith]

A brief description and guide on how to use me was sent to your DMs! Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[crimson merlin]

,help cmd

[obsidian monolith]

You really shouldn't take it literally. Please type ,help rolemod for example. Full command list can be found with ,list.

[viscid thistle]

what is the difference between tan^-1 and arctan

nothing right?

[sweet wyvern]

Nothing

[viscid thistle]

ok

thats what i though

thought

[sweet wyvern]

But there is no cot button allowed?

[viscid thistle]

here

Find, with proof, a sequence of buttons that will transform $x$ into $\frac{1}{x}$.

[obsidian monolith]

Disabled_Skooter:

[viscid thistle]

First we have to get x on our screen, then we press [tan-1] then we press [tan].

@sweet wyvern

no wait

it doesnt

[sweet wyvern]

that would still yield x

[viscid thistle]

it just cancels out

yea ok

hmm

ok

so

[sweet wyvern]

oh nvm @viscid thistle your original proof worked

[full garden]

hey guys I feel very embarrassed for asking this but how do i solve this...

[sweet wyvern]

I'm blind

[viscid thistle]

ok

[sweet wyvern]

the one with cot

[viscid thistle]

isnt it 12?

ah

[full garden]

no bro

[uncut mulch]

don't overthink it

[full garden]

is it just 4?

[viscid thistle]

lmao

wait can i get some help with my problem

[full garden]

me first :(

[rare pagoda]

@viscid thistle draw a right triangle with sides 1, x and sqrt(1+x^2)

[viscid thistle]

nvm

im a clown

and i got it

[full garden]

hey guys can someone please help me with this question

can I just say that a must be an even number ?

[rare pagoda]

not exactly

for a function, (x,y), each x needs to have a unique y

aka when you have (a, x) and (a, y) then x = y

so basically in your case a cannot equal 2, 4, or 6

[full garden]

oh

sorry what?

yes bro reading over it now I understand what you were saying

thank you bro makes sense

[hexed cairn]

hi, does anyone know how to do this?

[fleet yew]

@hexed cairn do you know how to work with vectors

This is pretty much just asking you to draw a triangle when it comes down to it

Same basic trig you've been doing since preschool

[hexed cairn]

kinda? i’m certainly not the sharpest tool in the shed

[fleet yew]

Do you know what the complex plane is at least

[formal iris]

can i get help on 24

[viscid thistle]

How did u get the answer for tan(2x)

[smoky needle]

Can someone help me on 4? I did the work prior to it in #3.

[viscid thistle]

Ok i think I can help

So in the sqaure root you have 36- x^2

[smoky needle]

Thank you!!

Yes

[viscid thistle]

You can't have a negative number in the sqaure root

[smoky needle]

Oh did I do #3 wrong?

[viscid thistle]

So your x has to be a value that, When sqaured, gives you a value less than 36

No u did 3 correct

So u have to have an x value such that x^2 < 36

[smoky needle]

Okay! I will try that out!

So will I just have to plug anything for x?

[viscid thistle]

You have to plug in a value that, when squared, gets you a number less than 36

[smoky needle]

So any number as long as it’s less than when squared, got it!

[viscid thistle]

Yes

[smoky needle]

To both Xs right?

[viscid thistle]

Yah

[smoky needle]

👍🏽👍🏽

I used 2...lol I don’t know if that’s right

[viscid thistle]

It's not

I'll give you a hint: there's a number that, when you sqaure it, you get the number you're subtracting it with.

X has to be less than that number

[smoky needle]

Can it be 6^2?

[viscid thistle]

No because then the value inside your sqaure root would be 0

It's gotta be less than that

[smoky needle]

Hmm okay, I’ll redo it again!

So when squaring it, it has to equal what I’m subtracting it with, right?

Sorry I’m kinda slow

[viscid thistle]

It has to be less than what you're subtracting it with

No worries

You don't need to apologize :)

[smoky needle]

Thank you for your help! I appreciate it!

[viscid thistle]

So x^2 has to be less than 36

[smoky needle]

I used 5^2

[viscid thistle]

That works

[smoky needle]

But that would get sqrt of 11

So far now I have 5 sqrt of 11

[viscid thistle]

Can you use numbers wirh decimals in them?

[smoky needle]

I am actually not sure, our teacher never really said anything about this problem..😕

[viscid thistle]

Ah. It seems kinda vague

[smoky needle]

Yeah..my friends and I don’t know how to solve it

[viscid thistle]

x has to be the closest value to 6 without actually being 6

[smoky needle]

Would 5.4^2 work since that = 34.81. It is the closest

Sorry I Meant 5.9^2

[viscid thistle]

Yah

That's what I would put

[smoky needle]

I now have 5.9 sqrt of 1.19

[viscid thistle]

Oh wait it says largest possible value

My bad lol

Your number should be furthest away from 6

[smoky needle]

Haha no worries!

[viscid thistle]

Uh yah so x has to be 1 I think

[smoky needle]

Oh wait 1?

[viscid thistle]

Uh yah I think 1

[smoky needle]

So it would be 1 sqrt of 35?

[viscid thistle]

Oh wait but you have to multiply by 1 after...

After doing trial and error I found that 3 is the correct answer

[smoky needle]

Okay, so is the final answer or the value that I replace x with? 😅

[viscid thistle]

Yah so x is 3

That gives the maximum area

I realized why this question was tricky

If x = 3, the area is 81

[smoky needle]

Yeah it’s tough. Lol
And ohh okay, I’ll plug in 3 for x and see what I get

I ended up getting this😅

[viscid thistle]

Do that but with x = 1 thru 5

[smoky needle]

So I do 1,2,3,4,5?

[viscid thistle]

Yah

Cuz then you'll get the answer

[smoky needle]

Awesome, I will do that

[viscid thistle]

Ok thx

[smoky needle]

Hmm I think I am doing it incorrectly

I am getting negatives inside the sqrt and isn’t that wrong?

[viscid thistle]

Can you show me what u have

This is what I have if x= 1

@smoky needle

This is what I have when x = 2

[smoky needle]

Sorry I had it sending but my WiFi is acting up

I’ll resend it

[viscid thistle]

No worries

[viscid thistle]

I got 16.58 for 5 × sqrt(11)

[smoky needle]

Oh okay, I’ll do that for the rest. I just left it with the sqrt!

[viscid thistle]

So looks the answer is when x = 5 then After All

[smoky needle]

Can you explain why x=5 is the answer?

It wouldn’t be 4? Because that’s the largest?

[viscid thistle]

Actually you're right it's 4

[vernal moon]

It’s technically sqrt(18) I reckon

[viscid thistle]

What's sqrt(18)

[vernal moon]

It’s about 4.23

[smoky needle]

🤯

Wait what is sqrt(18)? Sorry

[willow bear]

sqrt(18) is sqrt(18)

it can be simplified as 3 sqrt(2) if one so desires

[viscid thistle]

Ok ann

[willow bear]

what

[vernal moon]

Ok Ann means ok Ann

It can be simplified to k if one so desires

[viscid thistle]

🤣🤣🤣

[willow bear]

fuck both of you

[vernal moon]

I’m sorry ann

[viscid thistle]

Ann stop being mad pls Ann im begging you

[smoky needle]

Wait I think I got the answer, it is sqrt(18)...I think

[vernal moon]

it is

[smoky needle]

I’ve been at this question for a while 💀💀

Do we need to do the derivative for this problem?

[vernal moon]

well this is in precalc

so no calc allowed zone

smh

my head

[pulsar spindle]

maybe move the x inside the square root

then find the maximum peak

[viscid thistle]

Oh crap I can't believe I didn't think of the derivative

[vernal moon]

this is precalc

its cheating ;-;

jk

[pulsar spindle]

matches the root 18 answer

[viscid thistle]

@smoky needle u need the deravitive

[smoky needle]

I’ve done derivatives before lol, but how can I do it with this problem?😅😅

[viscid thistle]

Search up the product rule

Ight so first rewrite sqrt(36-×^2) as (36- x^2) ^ 1/2

Then u gotta take the derivative of that and multiply it with x

Next you take the deraivitve of x and multiply it with (36 × x ^ 2) ^ 1/2

Then you add your products

And set a =0

Then solve for x

[pulsar spindle]

its cheating ;-;
@vernal moon Cheating would feel like going on desmos and graphing the function

I hate calculator papers man

[vernal moon]

i was jokin lmao

personally

without calc

i would just

square it

then notice

its a quadratic

-u^2+36u

where

u=x^2

then use quadratic turning point

then square root

and u get sqrt(-36/(-2))

which is sqrt(18)

[smoky needle]

Can I show my work? I got this far

My work is sending right now, it just takes a while for it to send

Wait Nevermind, I think I did it wrong

[vernal moon]

i mean

u could just

use the quadratic turning point formual

[smoky needle]

What is that?

It’s been a while

Sorry

[vernal moon]

let p(x) be a quadratic polynomial ax^2+bx+c

the turning point of p(x)

is

the x cord of the turning point

will be

-b/2a

[smoky needle]

Would it be okay if I just do the derivative since I’m more familiar with that?

[vernal moon]

i mean

its basically a derivative in disguise

cause

[smoky needle]

I’m just stuck where I’m at 😅

Ohhh okay

[vernal moon]

d/dx(ax^2+bx+c)=2ax+b

-b=2ax

-b/2a =x

the turning point has a derivative of 0

so this works

[smoky needle]

Okay I will try it!

[smoky needle]

It can be simplified to 3sqrt (2)?

[native sequoia]

you can derive -b/(2a) from noticing that the turning point doesn't change respect to c and then averaging the 2 x-intercepts that we can force to exist by changing c

[native sequoia]

actually slightly better to just notice ax^2+bx+c=c has solutions x=0 and x=-b/a and taking average

[gray marten]

Hey, would it possible if anybody could give me some tips for solving this limits question: lim x->+∞ ln(x^2)/ln(2x) Any help would be amazing!

[willow bear]

log laws

[gray marten]

So, if I use log laws, it would become 2 ⋅ lim ln(x)/ln2x

but im not sure what to do after that

[willow bear]

there is still another log law you can use

[gray marten]

sorry i'm looking through all my notes and I'm still not sure which log law i can use 😅

[pulsar spindle]

@gray marten have you tried L'hopital rule?

[gray marten]

I don't think my teacher has ever mentioned it

so im assuming im not supposed to use it yet in the homework?

[pulsar spindle]

ah... then ignore that suggestion

[gray marten]

is it necessary to solve this question???

[willow bear]

no

[pulsar spindle]

No, probably should be another way

[willow bear]

no it's not

log(2x) = log(2) + log(x) lmfao

[pulsar spindle]

than factor out ln (x)?

[gray marten]

BRUH how did i miss that

lmao

[pulsar spindle]

oh wait... I got there but forgot to sub in the limiit

I am an idiot

[gray marten]

thank you for the help guys!

much appreciated 🙂

[sullen canyon]

lads anyone here

is g(X) = 10 sin (30x) + 20

[willow bear]

no

[sullen canyon]

its not 30x is it

or what did i do wrong

[willow bear]

3 * 2x is far from 30x

[sullen canyon]

so you don't multiply it by 5?

cuz i did 5 * 3 * 2x

[willow bear]

why would you multiply it by 5?

$f(2x) = 2\sin(6x) + 4$

[obsidian monolith]

Ann:

[willow bear]

$5f(2x) \neq 5 \cdot (2 \sin({\color{red}5} \cdot 6x) + 4)$

[obsidian monolith]

Ann:

[sullen canyon]

so its 10 sin (6x) +20

[willow bear]

yes

[sullen canyon]

thanks

[gray marten]

Hey, does anyone know how to do this question? Given f(x) = ((e^2x)-1)/((e^x)-1)) Name the x value(s) at which the function isn't continuous. For each discontinuity, determine whether the discontinuity is removable. If removable, indicate the point that would make the function continuous at the particular value of x.

I'm thinking that I have to use the derivative? but im not exactly sure

[serene heath]

where is your function defined?

[gray marten]

my worksheet doesn't specifically say anything

regarding where the function is defined

[patent beacon]

A real function is defined everywhere except where "a definition doesn't make sense". Your worksheet will never generally tell you where functions are defined as it's something you can find.

In this case, there's an x-value that causes a division by 0. A function is never defined at such an x-value

@gray marten

[fleet harbor]

Can anyone please help with this question?

I just need to figure out where to start

[patent beacon]

Can you express (2z + 1)/(4z - 4) in rectangular form?

Or actually, split each z into x + yi first

[fleet harbor]

I did that and afterwards kept only the imaginary parts, but I have no idea where to go from there

Would I need to multiply the denominator by the complex conjugate?

[patent beacon]

It sounds like you've done something off, after doing this, your simplified expression should be clear

You can always divide a complex number by another by multiplying by the complex conjugate, but we may not need to

Actually nvm I think you do need to perform the division first

[fleet harbor]

Alright thanks

Actually, wouldn't this make the denominator purely real?

[patent beacon]

That's the point haha

We want i out of the denominator

[fleet harbor]

but would I be dividing by 0 if I do the Im?

Soz, I don't fully understand how the Im works

[patent beacon]

Im(a + bi) = b

So you want a rectangular form in those brackets

,w simplify (2x + 2yi + 1)/(4x + 4yi - 4)

[obsidian monolith]

Display results online and refine query

Requested by Kaynex#7520

[patent beacon]

Wolfram why

[viscid thistle]

Lol

[fleet harbor]

I got (16xyi - 4yi + 8x^2 - 4x - 4)/(16y + 16x^2 - 32x + 16) after multiplying

[gray marten]

Sorry for the late response, thank you that makes more sense. @patent beacon

[fleet harbor]

Nvm I multiplied wrong

[muted steeple]

Hi, guys. How can the determinant be used to find the highest point in a quadratic?

More specifically, i have a question which asks me "How fast would a ball have to be thrown upward to reach a maximum height of 100 ft?" and I'm given the following formula $16t^2-v_0t+h=0$ and the hint suggests using the discriminant.

[obsidian monolith]

꧁༺Vocal༻꧂:

[muted steeple]

So based on what i know so far, i need to find $v_0$ and $h = 100$ so the equation can be written as $16t^2+v_0t+100=0$

[obsidian monolith]

꧁༺Vocal༻꧂:

[muted steeple]

not sure how a determinant would apply here given that $v_0$ is an unknown coefficient, which is required for a determinant.

[obsidian monolith]

꧁༺Vocal༻꧂:

[viscid thistle]

The display initially shows 0. Prove that there is a sequence of buttons that will produce $\frac{3}{\sqrt{5}}$.

[obsidian monolith]

Disabled_Skooter:

[viscid thistle]

Can I get help with this really quick?

[serene heath]

context?

[viscid thistle]

its question C

@serene heath

[viscid thistle]

guys

[viscid thistle]

@severe verge do you think you could help me?

[severe verge]

just check a bunch of things til you get one that does the job

well, probably a good idea to draw a unit circle diagram on paper and just try to get to 3/rt5 there first

[full garden]

hey guys I have a very fast question please

so i have to find the inverse and i got to this point

and I was wondering of I can make the last part like the equation in the question

like can I break the donamirator of a + b into a - a

i feel like the question might have a mistake because i don't see a mistake in my steps 😅

can someone please clarify

or can i even do it that way if the equation is not set to y -_-

can someone please help me

[viscid thistle]

you have to swap all x's for y's in the first step

you didn't do that

@full garden

How would I do this problem?

[full garden]

@viscid thistle i did bro

[viscid thistle]

no u didn't

you missed a y

[full garden]

by = will equal yy?

[viscid thistle]

no

[full garden]

i missed a y?

[viscid thistle]

one of your y's should be an x

[past meadow]

the second term, you didnt swap the y

[viscid thistle]

^

bingo

[full garden]

ohh i seee

[viscid thistle]

you should have bx not by

[full garden]

omg i see

thank u so much @viscid thistle @short geyser 🖤 thanku

[viscid thistle]

np

❤️

[viscid thistle]

Did i do this right?

[viscid thistle]

How did u calculate the hypotenuse

[daring yarrow]

does this look right?

i messed up on the left limit from that fraction with the squrae brackets, its supposed to be -4-4-4

aka -12

but apart from that i didnt make any writing errors so question still stands

[pseudo lance]

to calculate the height of centre of mass of this shape, can i do (centre of mass triangle +centre of mass rectangle) /2

[fading token]

only if the triangle and rectangle have the same mass

[pseudo lance]

ugh sad

if there was a y-axis there, how would u be able to calculate the a of the triangle

the distance to the centre point

is there a formula or something?

[hollow eagle]

hmm. I gotta think about this. I believe you take the center of mass of the triangle (which will not be halfway up the triangle), and then use CoM formula Xc = ( x1m1 + x2m2 ) / (m1+m2) alongside the center of mass of the rectangle to get the center of mass of the Y coordinate

note that the com of x coord will always be halfway along because it's symmetrical

[pseudo lance]

alright thanks

[sharp marsh]

3^(x + 6) = e^(3 − x)

Help

[viscid thistle]

@sharp marsh solve for x?

@sharp marsh send the exponential x+6 to right side

It would become 3= e^(3-x)/x+6

No just use log

Yes I'm coming there

[acoustic harbor]

yeah you can’t send the (x+6) to the right side

[viscid thistle]

log3 = 3-x/x+6

Yeah wtf

WTF

[acoustic harbor]

just ln both sides

[viscid thistle]

You can

just ln both sides
@acoustic harbor yeah

It would be reciprocal.on the opp side

Even if you use log on both sides you'll still get log3 = 3-x/x+6

[acoustic harbor]

no

[viscid thistle]

?

[acoustic harbor]

it’s in terms of ln

not log3

[viscid thistle]

Yes

I forgot about base notations solving for x

Yes it would be ln

[acoustic harbor]

you don’t want to use log base 3

like ever

[viscid thistle]

Using In makes it easier

Base 3?

[acoustic harbor]

idk what you did lol

[viscid thistle]

Log3 meaning log base e (3)

Log subscript(e) 3

[acoustic harbor]

that’s ln

[viscid thistle]

Yes

I thought you misunderstood the denoted 3 for base

[acoustic harbor]

you wouldn’t get that answer

[viscid thistle]

ln3 = 3-x/x+6

[acoustic harbor]

no you can’t bring the (x+6) to the right

[pale bison]

[viscid thistle]

sigh

Then why did I get same one as application of log on both sides

Like, its wayyyyy easier to do it.
$In(3^{x+6})=In(e^{3-x})$

$(x+6)In3=3-x$

If a^m = b then a = b^1/m

[obsidian monolith]

Al3dium:

[pale bison]

It would become 3= e^(3-x)/x+6
That's incorrect @viscid thistle

[viscid thistle]

You will then take log because of e

e will become in base.for ln3

But thats ¡In!!!!!!

Its what i did before

Denotion wrong

Process right

Ln yes ln

[pale bison]

log is the same as ln in many contexts

They're both correct

[viscid thistle]

If a = e^b then ln(a) = b

Apply that @pale bison

[pale bison]

[viscid thistle]

Lol

yo big brain

But not on this channel lol

[fierce slate]

how would one solve this?https://prnt.sc/sc5cd9

[opaque mountain]

@fierce slate Where's the initial equation?

[fierce slate]

thats the whole question

https://prnt.sc/sc5hwo @opaque mountain

[opaque mountain]

I'm not sure then..

[viscid thistle]

none of those answers look correct

you'd end up with 12 if you plug in every answer

[serene heath]

16 works

[viscid thistle]

how?

[serene heath]

wym how

[viscid thistle]

16 doesn't work

[serene heath]

why not

[viscid thistle]

because like i said earlier, you'll just get 12

[serene heath]

what

you'll end up with (x-4)^2-4

difference of 2 squares

[viscid thistle]

you'll end up with (x^2 - 8x + 16) + 12 -16 which simplifies to x^2 + 8x + 12 when u remove the brackets

[serene heath]

-8

not +8x

[viscid thistle]

yes i know

[serene heath]

ok

x^2-8x+16 is just (x-4)^2

[viscid thistle]

hmmm i think i get what you mean now

[serene heath]

they put those brackets there for a reason

[viscid thistle]

i thought you had to simplify it after

[opaque mountain]

Can I at least get an explanation on what's that question is asking?

[serene heath]

do you know what a difference of 2 squares is?

basically asking you to pick a number so u get something of the form

a^2-b^2

[viscid thistle]

How would I do 19

[serene heath]

what have u tried

[viscid thistle]

I have done log(sinx*cosx) but I’m stuck there

[misty heath]

what is the definition of a log?

[viscid thistle]

Ok then use the def of log

Yeah lol

@viscid thistle if you have $\log_{\frac{1}{2}}{(\sin{x}\cos{x})}=1$

Use the definition for log

$\log_a{b}=c$

$a^c=b$

[misty heath]

then you can either consider the double angle formula for sine or use the fact that sine is an odd function and solve the equation that way

first line has an extra } I think

cos{x}})}

[obsidian monolith]

Al3dium:

[viscid thistle]

True

Alright, so I have 1/2=sinx*cosx

Ok

So $\sin{2x}=2\sin{x}\cos{x}$

[obsidian monolith]

Al3dium:

[viscid thistle]

@viscid thistle doesn't it seem familiar? Like if we could apply this to your problem ...

Ohhh

Thank you very much

Np

Another question sorry

How would I do this

Oh my

Wait lol

@viscid thistle do they ask you to simplify

It says solve for x

can i simplify u, big brain ?

@viscid thistle multiply both sides by 100

Okay

So I would have 100x^log(x)=x^3

can i simplify u, big brain ?
@viscid thistle bet

What would I do from there

take the log of the exponent or something

Yeah lol i answered the same on #help-1

@viscid thistle dont post it twice.

Alright

So lets keep the one here @viscid thistle

Take the log of both sides

There is smth weird on line 3

[tardy ridge]

substitute x=10^n then solve for n

then solve for x after solving for n

[viscid thistle]

@tardy ridge he can take out the 3 as it is an exponent of log

But anyways, the 3rd line looks wrong

Just dont take the 100 to the left side

I’m confused

,rccw

[viscid thistle]

Srry for the overloaded ink on the middle, i had to make it clearer that it wasn't the base of the log lol

@viscid thistle

All the logs are base 10 btw, in some, i just didnt put them (on the last lines)

@tardy ridge could you verify this lol im a lil rusty on logs

Okay

Now try to continue

[tardy ridge]

everything is right

except 2log10 is unnecessary

just write 2

[viscid thistle]

Oh true yeah

@viscid thistle so yeah try to continue from last line but remember that $log_{10} (10)=1$

So you can simplify that

So I have log(x)^2=3logx-2

[obsidian monolith]

Al3dium:

[viscid thistle]

@viscid thistle what would you try here

Move the right to the left and factor

[charred latch]

How would I complete the square for this problem x = y^2 + 6y + 5

[viscid thistle]

@viscid thistle yeah try that

(Srry for too much tags lol now im realising)

[viscid thistle]

@viscid thistle

is 3logx-2

as

3log(x) - 2

or

3log(x-2)

theres a difference

[tardy ridge]

...

[viscid thistle]

wdym ... u idiot

whats wrong with clarification u fucking monkey

[gilded stirrup]

[viscid thistle]

what

im literally right

[limber compass]

What do i do?

Do i make denominator same?

I am just brain dead sorry.

[next kernel]

Recursively defined sequences?

[limber compass]

?

right?

[next kernel]

a1 = -2; an = n + a(n-1)

[tardy ridge]

yes you can make the denominator into x^2(x-1)

and then put it as a polynomial in terms of x

[limber compass]

now I am stuck

@tardy ridge ^ : D

[tardy ridge]

you can get rid of the x^2(x-1)

[limber compass]

I did.

[tardy ridge]

why did you replace everything with p and q

[limber compass]

bceause its easier.

[tardy ridge]

ok well I have x^2(A+C) + x(B-A)-(B+1)=0

and then I thought wouldn't it be nice if everything equaled 0

[limber compass]

Hmm I am thinking of something like this...

[tardy ridge]

that seems fine too

[limber compass]

hmm, so what were you talking about.

[tardy ridge]

wait what are you stuck on

do you know what p,q, and s are

[limber compass]

haha got it sorry.

had a lightbulb moment.

[native shadow]

precalculus be like

but thats easy precalc lmao

jk

[viscid thistle]

(3+4i)(-2+5i)
convert the rectangular form to polar form

how do I do that?

[willow bear]

well you might consider first computing the product while still in rectangular form

and then converting the result to polar

[dire raptor]

@viscid thistle I will try, you first expand the brackets.

-6 + 15i - 8i -20

= -26 + 7i

now you convert

[willow bear]

that's exactly what i said, but they chose to ignore me as it seems

except that i didn't straight up do it for them

[dire raptor]

Yeah i know

But I wanted to straight up do it

[viscid thistle]

Aight sir

So I figured it out

botth fo you were kinda wrong

I have to coonvert each individually

and then multiply them

@willow bear I wanna learn :pp I need this for calc

[dire raptor]

Both ways work.

though with your method you gotta know some extra rules.

such as cis(a) * cis(b) = cis(a+b)

etc etc

[willow bear]

if the actual directions in the problem were not what you posted, then it's kind of on you for misleading us, and not on us for not following directions neither of us knew.

[dire raptor]

I have to a gree with Ann here.

Since you didn't specify Ann assumed any method is allowed. Which means logically she chose the most efficient one

[sullen canyon]

if i have log (base 9) (cos 2x+2), how could i make that into log (base 3)

do i use the change of base formula?

[willow bear]

$\log_9(z) = \frac{\log_3(z)}{\log_3(9)}$

[obsidian monolith]

Ann:

[sullen canyon]

@willow bear after that i need to make it into log (base 3) sqrt cos 2x + 2

how do i do that

[willow bear]

oh god there are so many parentheses you're omitting there

but $\log_a(z^p) = p \log_a(z)$

[obsidian monolith]

Ann:

[sullen canyon]

idk how i can go from the change of base formula to what i need to get

[pale bison]

$\log_9(\cdots)=\frac{\log_3(\cdots)}{\log_3(\cdots)}$

[obsidian monolith]

Publius:

[pale bison]

consider something like that

[sullen canyon]

itd be this right?

i'd have log (cos 2x+2) / log (9)

but how do i go from there

[pale bison]

what is $\log_3(9)$?

[obsidian monolith]

Publius:

[sullen canyon]

2 log (3) right?

so then i do the same thing with the top

take out the 2

[willow bear]

log_3(9) is 2.

[sullen canyon]

ye

so i have the 2 now

[willow bear]

so you have $\log_9(\cos(2x) + 2) = \frac{1}{2}\log_3(\cos(2x) + 2)$

[obsidian monolith]

Ann:

[sullen canyon]

yes

[past meadow]

log law ::

[sullen canyon]

bruh

i'm dumb

[past meadow]

as ann said above, $\log_a(z^p) = p \log_a(z)$

[obsidian monolith]

Sneaky:

[sullen canyon]

cheers everyone 👍

ye i just realized

i'm not very good with logarithms as you can see

[past meadow]

it just takes practice. you'll get better

[pale bison]

its ok we forgive you

[sullen canyon]

its just when there are a lot of different topics, and we haven't done logs in months and we get a question on logs, i kinda forget how to do it

[harsh cipher]

Hello

question about graphing hyperbola

(x^2)-(y^2) = -36

trying to come up with the restriction under the radical sign after writing y= root x^2-36

[hexed ermine]

Is it root x^2-36?

[harsh cipher]

x^2 is greater than or equal to -36

yes

root -36 is where I'm stuck

[hexed ermine]

Are you sure its y=sqrt(x^2-36) ?

[harsh cipher]

original question

x^2- y^2 = -36

y^2 = x^2+36

y= root x^2+36

[hexed ermine]

Right

+36 not -

[harsh cipher]

now x^2+36 is greater than or equal to 0

x^2 is greater than or equal to -36

yes

oh maybe I need to isolate x^2

[viscid thistle]

Whats the matter

@harsh cipher

[harsh cipher]

matter : graph hyperbola x^2-y^2 = -36

show asymptotes

i think im right

need to isolate x^2 this time

since radius is negative

[muted steeple]

can someone explain the highlighted part?

does it just mean d = x^2+2xc+c?

also not sure what it means when it says "we can also obtain the coefficients from the solutions".

when i say not sure, i mean i don't understand how

[hexed ermine]

Well you can plug in your solutions into a quadratic with arbitrary coefficients

[muted steeple]

okay, what does that achieve or tell me?

so far i understand this sentence as; if you obtain d from some quadratic with some coefficients, you can use the d to find the coefficients of the said quadratic.

which sounds useful, but i dunno how.

[pure dagger]

is this screenshot skipping a page or something

the sentence just makes no sense lol

[muted steeple]

sorry, let me provide more context.

so for a i determined solutions x=5 and x=4

and it is indeed that 5*4=20 and 5+4=9 which is middle coeff

[pure dagger]

i think exercise 137 has nothing to do with what's written on the second page

it rather looks like there's an exercise 140 to the right of it, which continues onto that page

[hexed ermine]

^

[pure dagger]

this is the kinda math help i was born to do

yes there we go

then, your highlighted part just refers to what follows in the parts of the exercise a, b, and c

namely, you're supposed to verify that given some numbers, and someone tells you "these are the solutions to a quadratic equation", you can reconstruct the quadratic equation from the solutions

[muted steeple]

so (a) now tells me that i can reconsturct const and middle coeff

[pure dagger]

ye

[muted steeple]

not sure about the case where the leading term has coeff of >1

[pure dagger]

yeah, you don't get that one; the problem is, that if you have a quadratic equation (e.g. x^2 - 1 = 0), then all multiples of that equation have the same solutions (e.g. 50 x^2 - 50 = 0)

so you can only reconstruct the coefficients from the solutions if you fix the leading coefficient

so the correct statement is: "if you have the two solutions of a quadratic equation, you can uniquely reconstruct the quadratic equation from it, up to a scaling factor"

[muted steeple]

so it has to explicitly tell me the leading coeff if it's non 1?

[pure dagger]

yup!

[muted steeple]

okay, i understand now.

[pure dagger]

cool! glad to help

[muted steeple]

so whatever solutions i got now, say i have coeff of 2, then i'd need to scale them both by factor of 2?

[pure dagger]

mhm i think it's more complicated

you might understand it better if you do the exercise

[muted steeple]

cuz i know that first and last term scale exponentially

[pure dagger]

once you've done c), try to see what happens if you use leading coefficient 2 or 3

[muted steeple]

okay, then, i'll do that.

thanks for clearing up.

[pure dagger]

np!

[muted steeple]

Hey @pure dagger mind clearing up few more things for me concerning the the (c)?

let me write out what i've done so far for c.

So to prove $r_1 \cdot r_2 = c$ using the quadratic formula, i set

$r_1 = \frac{-b+\sqrt[]{b^2 -4ac}}{2a}$ and

$r_2 = \frac{-b-\sqrt[]{b^2 -4ac}}{2a}$.

Since i need to prove $r_1 \cdot r_2 = c$

i plug these two equations

$\frac{-b+\sqrt[]{b^2 -4ac}}{2a} \cdot \frac{-b+\sqrt[]{b^2 -4ac}}{2a} = c$
Which then using the difference of squares rule means
$\frac{b^2-b^2+4ac}{4a^2}=c \therefore \frac{c}{a} = c$ which seems to be wrong.

[obsidian monolith]

꧁༺Vocal༻꧂:

[muted steeple]

it should cancel out to be to c = c right?

[sand skiff]

It works if a = 1

[muted steeple]

so it only works if a = 1?

[sand skiff]

Seems so

[muted steeple]

is that the conclusion i should get from this?

oke

[sand skiff]

If there is a constant in front of (x-r1)(x-r2), c will not be r1*r2 unless that coefficient is 1

If that makes sense

[muted steeple]

(1-r1)(1-r2) = 1^2 -r2-r1+r1*r2)?

[sand skiff]

a(x-r1)(x-r2)

[muted steeple]

oh, so this form you mean when factoring

[sand skiff]

= ax^2 -a(r1 + r2) + ar1r2

Yes

Exactly

[muted steeple]

i was confused for a second about how you got these solutions in that form

😄

[sand skiff]

I understand 😄

Hope it makes sense

[muted steeple]

yep, makes sense

thanks.

i just realised that also proves the $r_1 + r_2$ part 😄

[obsidian monolith]

꧁༺Vocal༻꧂:

[orchid pier]

hi im stuck on a homework question says compute ((1/2) - ((sqrt3)/2)i)^20 i get two answers -1/2 + (sqrt3/2)i or -1/2-((sqrt3)/2)i)

are either of these right?

express the answer in form a + bi

[willow bear]

the second one is right

the first one is wrong

idk how you got two answers

but probably worth looking into

[orchid pier]

thank you very much, i thought so too

[patent beacon]

Do you know what it means for a graph to be "increasing"? Just means it has a positive slope

[hexed ermine]

Ok

So A(x+2)^2(x-3) right?

A here is a coefficient needing to be determined

[hexed ermine]

Yeah

[winged ether]

can someone help me how to do difference quotients? I am really stuck on this one problem where you find teh difference quotient of f(x) = x^1/2

so difference quotient of square root of x

[serene heath]

try finding a unit vector in that direction first

[full garden]

guys can someone please clarify what this question is asking

like am i supposed to derive the x = -b/2a formula ?

I've just never seen anything like this formula before

[tardy ridge]

the vertex is always the midpoint of the two roots

[full garden]

yes

so is this question asking to derive that formula

[tardy ridge]

I didn't use the axis of symetry, I just foiled that into y=a(x^2 -(r+s)x+rs) then completed the square and get that answer

[full garden]

so u completed the sqaure from that strange formula ?

can u elaborate on that u just said please

so u completed the square after expanding the factored form ?

so we're supposed to get x = -b/2a as the answer ?

because if that is the answer then u helped me a lot

[tardy ridge]

what no I proved -a((r-s)/2)^2 is the max or min

[full garden]

what is the question even asking

[tardy ridge]

by expanding a(x-r)(x-s)

[full garden]

is the question asking for that ?

[tardy ridge]

it's asking you to prove the max or min of a(x-r)(x-s) is -a((r-s)/2)^2

[full garden]

and how do u complete the square when u get y = a(x^2-rx-sx+rs)

[tardy ridge]

ok lets ignore the a for now

so you would have x^2-(r+s)x+rs

[full garden]

yes

[tardy ridge]

now you can complete the square

[full garden]

okay

so after i complete the square it will give me the formula for min and max?

isn't min and max the same thing as the vertex

and the vertex can get from -b/2a?

[tardy ridge]

the min and max is the y value at -b/2a

[full garden]

yes yes

[tardy ridge]

when you complete the square the a(x-k)^2+ N the n is the max or min

[full garden]

so if i find the vertex

yes yes

[tardy ridge]

yes yes

[full garden]

haha

: )

[tardy ridge]

wait what you just have to find N that's all-

[full garden]

okay will try thank u so much bro thank u

hey guys I am supposed to complete the square in order to get the answer that I circled on the page but for some reason i didn't get that answer

becasue I still have to add + rs and multiply by -1

can someone just please clarify the solution for me please

guys please..

[reef crest]

1 sec lemme look at it

is that a(x-r)(x-s)?

[full garden]

love u bro yes please

so the question was

[reef crest]

ok ill see if I get that

[full garden]

thank u bro

[reef crest]

this is confusing lol i think im almost there tho

[full garden]

i hate my teacher

but yeah im waiting bro

[tardy ridge]

rs=4rs/4

r^2-2rs+s^2=(r-s)^2

and also you need brackets

[full garden]

yes but I have to still add the + rs

and multiply by the a

[tardy ridge]

your thing is wrong

the rs should be inside the brackets

[reef crest]

or equal to ars

thats the only thing that's wrong that ive seen though

[tardy ridge]

you should have $(-r^2-s^2-2rs)/4+rs$ inside the brackets

[reef crest]

math is hard to format smh

[obsidian monolith]

Fishraider:

[tardy ridge]

then you can get -(r-s)^2/4

which is -((r-s)/2)^2 and you add the a later and then you're done

[full garden]

bro is this equal to -(r-s)^2/4

where did the + rs go

[viscid thistle]

then you can get -(r-s)^2/4
@tardy ridge its -(r+s)^2/4

[full garden]

thank u @tardy ridge @reef crest @viscid thistle thank u guys a lot

[reef crest]

i didnt really help LOL sorry

[viscid thistle]

How would I do 23

isn't the answer given?

No?

log 18 = a?
4

isn't that the answer

I don’t think so

I did that and my teacher said it was wrong

[tardy ridge]

@tardy ridge its -(r+s)^2/4
@viscid thistle no it isn't, I was talking about the final answer.

[viscid thistle]

Oh srry lol

[reef crest]

if 2^a is equal to 3, then that means 2^a^2^2 would give you 18

so then 2^4a

i think

[tardy ridge]

separate log 18 into log 2's and log 3's

[reef crest]

idt u need any log 3s right?

[tardy ridge]

if 2^a is equal to 3, then that means 2^a^2^2 would give you 18
3^2^2 = 18?

wut

[reef crest]

oh wait am i being dumb

wait no that's not what it is lol unless im thinking wrong

it's not (2^a)^2^2

[tardy ridge]

18 is not even a power of 3

[reef crest]

wait i think im wrong

i was thinking

oh wait yeah i am very wrong

i somehow thought

3 squared gives 9 and that squared again is 18

[viscid thistle]

This is what I have so far

[reef crest]

im dumb

[viscid thistle]

Idk if it’s correct

[reef crest]

ask fishraider lol i have no idea what's going on rn

[viscid thistle]

Anyone know that?

[tardy ridge]

log3/log2=a

log18/(2log2) is what you are trying to find

log 18 = 2log 3 + log 2

[reef crest]

i think it diverges right?

because the power of the base increases

and each term gets closer to zero

sorry not divergent

convergent

for the sum there's a formula

i cant remember it rn though

[tardy ridge]

a(r^n-1)/(r-1)

[reef crest]

is that the right one

[tardy ridge]

lim r^n as n -> infinity when |r|<1 is 0

yes

[viscid thistle]

Got it

[full garden]

hey guys I have a very quick question

I tried solving for this and I got the slope is either -4 or 0

but when I tried -4 it was a secant when I graphed it

so 0 seems to be the correct answer, but can that be right?

becaue the question says in the form of y=mx+b

[viscid thistle]

if 0 looks correct then its probably the right answer

[full garden]

so it can be correct even if the question says the line is in the form of y=mx+b?

[viscid thistle]

yah

[full garden]

❤️ thanks bro

[viscid thistle]

np

❤️

[tardy ridge]

@full garden m = 0 or 4 by setting the discrimminant to 0

[full garden]

@tardy ridge yES BRO I LOVE u but like when i checked the graph it looked like when it was -4 it touched the graph twice bro

[tardy ridge]

No?

https://www.desmos.com/calculator/97vtirqzwe

[full garden]

bro how did u get 4

i got -4

$x^2 + (-(2x-mx)+1=0$

[obsidian monolith]

iRazur:

[full garden]

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

[obsidian monolith]

iRazur:

[full garden]

$m(m+4)=0$

[obsidian monolith]

iRazur:

[full garden]

that's why i got -4 bro

are my steps wrong

[tardy ridge]

$x^2 + (-(2x-mx)+1=0$

[obsidian monolith]

Fishraider:

[full garden]

because that's the b bro

[tardy ridge]

No.

[full garden]

i tried like following this step

because there was a negative on the outside bro

bro can u please who me the steps that u took

[lilac pier]

@full garden YEs there is a negative outside but it's being squared

[full garden]

yes

[lilac pier]

negative squared means negative times negative which is just positive