precalculus

[narrow peak]

oh alright

[willow bear]

$\log_{10}(x)$

[obsidian monolith]

Ann:

[uncut mulch]

underscore{base}

[willow bear]

this is how you do subscripts longer than one character in general

[narrow peak]

b

[stuck lark]

,rccw

[willow bear]

$\log_2[(5x+3)^2] - \log_{5x+3}(2) = 1$

[obsidian monolith]

Ann:

[narrow peak]

OH

wow

[willow bear]

discord ate the underscores

[narrow peak]

shitcord

[willow bear]

aight hold up now

[uncut mulch]

i mean the notation sucks, but based on context its reasonable to assume that the exponent apples to the (5x+3) and not the log function

[willow bear]

i might have missed a two turning itself into a four somewhere.

[narrow peak]

yes u did

mr ramanov found it

i squared the entire thing cuz my algebra glitched

xd

[viscid thistle]

@narrow peak did u got answers

[narrow peak]

i got it alr

[viscid thistle]

-1/5 guess

[opaque olive]

Quick question $\sqrt{9-x^2} = 3-x?$

[obsidian monolith]

Yes:

[stuck lark]

not equivalent

[opaque olive]

ok thanks

[fleet yew]

Lmao!

I'm not trying to make fun of you but that is pretty funny

I think that's called "undergrad's dream" or something

[viscid thistle]

How is that not Equivalent?

[past meadow]

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

!=(3-x)^2

[fleet yew]

You cant distribute a power over a sum

[willow bear]

@viscid thistle (a+b)^2 ≠ a^2 + b^2 and sqrt(a+b) ≠ sqrt(a) + sqrt(b).

[viscid thistle]

https://en.wikipedia.org/wiki/Freshman's_dream

[viscid thistle]

$(a+b)^2 = a^2 + b^2$ in some rings

[obsidian monolith]

polynomial:

[harsh smelt]

i can define (a+b) be identically zero and get this but what for are you saing this

[proven marten]

Vimes, pls fix your english

[willow bear]

@viscid thistle we get it you know what a ring of characteristic 2 is now get out of this pre-uni channel

[viscid thistle]

[heady jewel]

lol euler/fermat alt maybe

who knows

executes similar behavior

[gloomy path]

yikes

[serene heath]

cringe

[proven marten]

snap

thats going in my cringe compilation

[willow bear]

@serene heath https://media.discordapp.net/attachments/359052581022203914/733247653651152906/fb52961e754540acce550b281276750a.gif

[proven marten]

Glory to Aristorska

I didnt even spell it right

[long crown]

[willow bear]

no you didn't @proven marten

it's Arstotzka

[opaque olive]

$\sum_{k=1}^{n}(\sqrt{k}-\sqrt{k-1})$

[obsidian monolith]

Yes:

[opaque olive]

can i evaluate this with the summation formula?

[lime bolt]

bruh it just telescopes

[opaque olive]

wdym

[lime bolt]

the middle bits all cancel

[opaque olive]

oh yh

the formula wouldnt work here right?

[lime bolt]

what formula do u mean

[opaque olive]

https://gyazo.com/ed9dd8ce3eeb7afb8e38e099e42a9707

[lime bolt]

this isnt an arithmetic sequence

[opaque olive]

oh ok

[sage lake]

hey hey

x^4+8x^3-x^2+2=0

help

<@&286206848099549185> ?

[fleet yew]

@sage lake try -2

[sage lake]

I don't think that's it chief

[fleet yew]

@sage lake try 2

[sage lake]

nope

[fleet yew]

Try 1

[sage lake]

nope

[fleet yew]

Try -1

[sage lake]

Should've gotten an intel chip

[fleet yew]

Bro its called the rational root test

[sage lake]

what happens if it's not rational 😭

[fleet yew]

That is probably the case

[sage lake]

So like what do I do?

I'm fucked then

[fleet yew]

Factor by grouping ig

If that works

I dont think it does

[sage lake]

sooo AMD what do I do

[fleet yew]

Is there a specific method your teacher wants you to use

Because i mean if you just need the answer you could plug it into wolfram alpha

If you really want to solve this by hand you'd have to do it numerically

[steady kernel]

$n$ is a positive integer and $p$ is a polynomial such that $p(x+1)-p(x)=x^{4n}$ for all $x.$ How do we prove $p(1-x) \ge p(x)$ for $0 \le x \le 1/2$? Suggest a better channel if polynomials do not fit here.

[obsidian monolith]

aadfg:

[sage lake]

uhhh

what is this thing

[fleet yew]

@steady kernel prove that p(x) is strictly increasing from 0 to 1

Wait actually

p(0)=p(1)

And p(0.5)=p(0.5)

[sage lake]

@fleet yew so is there a method where I can solve this by hand?

or do I just have to bring wolfram alpha to class

[steady kernel]

@fleet yew That's false, even for n=1.

[tepid cedar]

is there an easy way to do this instead of doing it manually

[viscid thistle]

manually is probably the fastest way

[tepid cedar]

this was the first problem, so i thought id just code something out so my prof can see i did the work

thanks

i havent done this in forever, i kinda forgot

[viscid thistle]

you can separate it into $\sum_{k=1}^5 k+ \sum_{k=1}^5 1$ which maybe makes it a little faster to see

[obsidian monolith]

Botn:

[tepid cedar]

smart

[viscid thistle]

easier to see it is 1+2+3+4+5+5 like that

[tepid cedar]

yeah

i guess ill do that for this too

[viscid thistle]

well actually 2+3+4+5+6 is just as easy

[tepid cedar]

wait so this one means

do this first

then this

and add them together

[viscid thistle]

the inside sum should be done first

[tepid cedar]

the second pic i posted is the inner sum right?

[viscid thistle]

$\sum_{i=1}^2\left(\sum_{j=1}^3 (i+j)\right)$

[obsidian monolith]

Botn:

[tepid cedar]

oh

ok

ahh i get it

[wide ocean]

for 9b) I'm not sure how to answer this
f of g(x) means f(x) times g(x)?
same for 9a)?

[acoustic harbor]

no it means you stick g(x) into f(x)

does not mean times

[wide ocean]

oh I see

so the second variable always goes into the first variable? @acoustic harbor

like g(x) is the "X" in f(x)

[acoustic harbor]

sure

[wide ocean]

thanks @acoustic harbor , how can we isolate g(x) in h(x) = f(g(x))

h(x) =x^2 ?

x-4 = f(x)^2

[acoustic harbor]

x-4=g(x)^2

[wide ocean]

so g(x) = root of (x-4)

?

[acoustic harbor]

ye

[pseudo cargo]

Yess

[wide ocean]

thanks @acoustic harbor

[fleet yew]

@steady kernel let x=0

Its true

[viscid thistle]

@wide ocean 😁 😁

[wide ocean]

@viscid thistle why do they ask this

isn't it solved? or they want to solve for x

[viscid thistle]

solve to find x

manipulative

[wide ocean]

I found x=3

[proven marten]

yes they want you to solve for x

[wide ocean]

-1=2/(x-5)

-x+5=2

-x=-3

x=3?

[viscid thistle]

,w 2/(x-5) + 1 = 0

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by polynomial#4757

[viscid thistle]

yes

[proven marten]

lol

no worries

Although if you kept that in #prealg-and-algebra it wouldve been better

[chilly apex]

Series

Bottom width is 50,49,48,47,46,45....0.1?

[proven marten]

which part?

[chilly apex]

a) please and b

[proven marten]

Well, what have you tried

[steady kernel]

@fleet yew No, we want to prove $f(1-x) \ge f(x)$ for all $x \in [0,1/2].$

[obsidian monolith]

aadfg:

[fleet yew]

look at the first equation. p(x+1)-p(x)=x^(4n). so when x is 0 that means that p(1)-p(0)=0, meaning that p(0)=p(1)

so there are two points in the interval [0, 1/2] where p(x) = p(1-x)

[viscid thistle]

from graph A to graph B, what would the horizontal stretch be?

is it just pi

1/2 I believe

So in the equation I would write it as 1/2 ?

or 1/2pi

[upper kelp]

I believe it should be 0.5 and not 0.5π.
If you're unsure, you can check which points are supposed to be the roots of the function, and choose accordingly.
Let's try f(x) = cos (0.5x).
In that case, we see that the roots are: f(x) = cos (0.5x) = 0 => 0.5x = 0.5π + πk, meaning x = π + 2πk which means ±π, ±3π, ... as the picture shows.
However, if we were to try f(x) = cos (0.5πx), we would get:
cos (0.5πx) = 0 => 0.5πx = 0.5π + πk, meaning x = 1 + 2k, which means ±1, ±3, which isn't the case, as the picture shows.

[opaque olive]

ive found the derivative not sure what to do next

any hints

[upper kelp]

If I may ask: Did you have troubles with proving that the inequality holds true algebraically, or something else?

[opaque olive]

well

no

i was thinking that ill need to show first that if a is bigger than or equal to 8/9

and then the derivative

but

the derivatives denominator is positive, (a squared number)

so i just need to prove that the numerator is positive now

which is what im struggling on

[upper kelp]

I hope I'm not getting rusty:
Did you get $x^4 + (a - 3)x^2 + a$ in the numerator?

[obsidian monolith]

RoiKadmon:

[opaque olive]

no

$\frac{(a-1)x^4+(2a-3)x^2+a}{(x^2+1)^2}$

thats what i got for my derivative

i used the product rule

and tidied it up alot

[obsidian monolith]

Yes:

[opaque olive]

oops the denom was squared

[upper kelp]

Okay, I got the same thing now, sorry for the delay.
May I give a few hints?

[opaque olive]

so

i need to just prove that the numerator is positive when a is bigger than or equal to 8/9 now right?

[upper kelp]

9/8, but yes, indeed. This will be the values of a for which the inequality holds true for all x values, and not specific ones.

If I may give some hints:
Hint 1:
||Notice that only the powers $1, x^2, x^4$ are present in the inequality. In that case, the substitution $t = x^2$ could help you simplify the inequality to a second-degree equality.||
Hint 2:
||Once you get a second-degree equality, you can check which values have an "intersection".||

[obsidian monolith]

RoiKadmon:

[upper kelp]

Oops.

[fleet yew]

find the determinant of that quadratic in t

[opaque olive]

ok

im gonna give it a try

[fleet yew]

glhf

[opaque olive]

i originally tried to complete the square

[upper kelp]

In a sense, the discriminant is another way of completing the square, so you were still correct. It's just that the discriminant allows you to find the values of a directly.

[opaque olive]

ok

i think ive done it

i just want to clarify my thoughts though @upper kelp

So after getting my derivate, i say that the denominator was square thus positive for all values of x

if i say that f(x) = numerator

complete the square to find the minimum y value of that function

and i found that the minimum value was bigger than zero if a was bigger than 9/8

is that fine?

[upper kelp]

Indeed.
The reason that works is that if you start with the inequality:
Ax^4 + Bx^2 + C ≥ 0
and substitute t = x², then it's sufficient to show that:
At² + Bt + C ≥ 0
holds true. This is precisely what you showed.
By completing the square, you're showing that the function At² + Bt + C is equal to:
(Pt + Q)² + K
where K is dependent on a. You've shown that if a ≥ 9/8, then K ≥ 0, which means that the function is never negative.

[opaque olive]

oh nice

I understand

thanks for the help

[thin sky]

Why is it B? Fak I forgot the most basic math of all time

I mean... I wrote D because none of the values between 0 and 2 when squared get to 4

I don't know if that is an error of the crowdsource answer or if I have gone extremely dumb right now

lol

[viscid thistle]

@thin sky Are u sure about
D to be correct.

[thin sky]

yeah! I wrote D

but the answer that a lot of people apparently wrote is B

those are "crowdsourcing answers"

answers that a lot of people agree on

but I don't see why it is not D

it should be D

[viscid thistle]

But, actually answers are all except E

[thin sky]

but the other answers also fail

because anything squared is positive

[viscid thistle]

Show me ur work

[thin sky]

I don't do work for most of these questions

for example

anything greater than 0 but less than 2

squared

[viscid thistle]

Have a try once otherwise u may forget

[thin sky]

will always be less than 4

no matter what

however, for options A, B, C

for A, (-4)^2 is 16

which is greater than 4.

for B, the same but with (-3)^2 = 9, which is more than 4

and it should be all that are less than 4

c more of the same

E is ludicrously wrong

5^2 is 25 which is way more than 4

only (0,2) -- noninclusive meets the requirements

I think I have gone mad

lol

even if options A, B, C, and E had non-inclusive values, they don't meet the requirements

[viscid thistle]

wait give me sometime i m doing

[thin sky]

okay

now that I think of it

the answer should be (-2, 2) -- non inclusive, but it does not appear

[viscid thistle]

yeah

[thin sky]

ooo maybe

maybe they are so vague that

they want us to "interpret that in (-3,3) all the values I mentioned are included

[viscid thistle]

sending one beautiful way to tackle such problems in future

[thin sky]

wow

I can't believe these dudes

like (-3,3) is so ambiguous because it might make you believe that anything between -3 and 3 should be possible.

they should have stated (-2,2)

[viscid thistle]

yeah it will not be

sorry it was fast i did mistake

[thin sky]

what will it not be?

[viscid thistle]

check options fall under the given interval in,answer

[thin sky]

thanks

the wording of the question* was awful

but that is a way of getting used to questions like that

[opaque olive]

is there any easy way to find all values of x other than considering all "critical values of x"? $|x+1|-|x|+3|x-1|-2|x-2|=x+2$

[obsidian monolith]

Yes:

[steady kernel]

@fleet yew Yes, but that doesn't imply $p(1-x) \ge p(x)$ on the whole interval, which is what we want.

[obsidian monolith]

aadfg:

[patent crow]

omg, i can't believe this is stumping me.

https://cdn.discordapp.com/attachments/570672396235833369/737504172844515368/162017091996745728.png

https://cdn.discordapp.com/attachments/570672396235833369/737508688356704270/162017091996745728.png

i calculate 216 hours, but that's not possible.

i'm doing this wrong.

you have X people doing work in 288 hours,

if you less than X people, doing the same amount of work, it should take longer than 288 hours.

<@&286206848099549185>

please save my sanity.

the answer is 384 hours, right?

[viscid thistle]

pretend work is in the form of workers*hours

therefore total work is 12*288

and this means that time taken for 9 workers is 9x = 12*288

,w 12*288/9

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by Fang#5134

[viscid thistle]

yes, it is 384 Hours

[patent crow]

https://cdn.discordapp.com/attachments/298214658714632193/737522757054496795/162017091996745728.png

okay, this is what we have to do then.

[opaque olive]

Quick question, what would be the best way of finding the derivative of $\sec(x)^2$ shall i just write it as $\sec(x)\times\sec(x)$

[obsidian monolith]

Yes:

[viscid thistle]

⛓️ 📏

[opaque olive]

yep i suspected using the chain rule

but i have 4 different products

so this is taking a while

$g(x) = e^{-ktanx}$ where $k=\sin2\alpha$

[obsidian monolith]

Yes:

[opaque olive]

i need to show that its convex between $0<x<\alpha$

[obsidian monolith]

Yes:

[opaque olive]

i think ive got it now though

[proven marten]

⛓️ 📏
@viscid thistle Iron bar scale?

[viscid thistle]

Ann made that

[upbeat bone]

lol Denton

[wide ocean]

can someone please explain why its "C"

i did f(x) -g(x)

but the equation was -root of(x+2) -x^2 +2x-7

does this combine in some way

[viscid thistle]

What do you mean combine in some way

What qualms do you have with C?

An equivalent question is, for which $x\in\bR$ is $-\sqrt{x+2}-x^2+2x-7$ also real?

[obsidian monolith]

Botn:

[opaque olive]

@viscid thistle how did you make the background white btw?

of what the TeXit bot said

[viscid thistle]

Hmm let me see

[wide ocean]

okay nvm I might have graphed it wrong initially

how can I graph this equation without a graphing calc?

$x\in\bR$ is $-\sqrt{x+2}-x^2+2x-7$ also real?

[obsidian monolith]

Cord:

[viscid thistle]

Try ,tex --color white @opaque olive

Do you need to graph it?

[wide ocean]

No but to understand what the domain is yeah

[viscid thistle]

You don’t need to graph it to recognize for which values of x that expression is real

[wide ocean]

Oh what defines real parts of the eqn

[opaque olive]

,tex --colour white

[obsidian monolith]

Your colour scheme has been changed to white

[opaque olive]

yay

[wide ocean]

like only variables?

[viscid thistle]

Look for variables in square roots and denominators

[wide ocean]

Ohh

so the root of (x+2)

[viscid thistle]

These things may give rise to complex values or undefined values

[wide ocean]

is the only thing that points to the domain not being able to equal -2

ok thanks @viscid thistle

[viscid thistle]

Sqrt(0) is fine

Sqrt(x) for negative x is not real though

[wide ocean]

Sqrt(x+2)

yes ty

[viscid thistle]

That function, ima just call it f(x), is a function who’s output is complex for all real numbers in the input less than -2.

For example

G(x) = sqrt (x-2). For all x<2 the output is complex.

For some values like 1 or 0, it is clear:
G(1) = sqrt(1-2) = sqrt(-1) = i
G(0) = sqrt(0-2) = sqrt(-2) = sqrt(2) * i
G(-2) = sqrt (-2-2) = sqrt(-4) = 2i

[gritty lintel]

Hey guys, I have a question where I need to solve the equation of sin(2x-0.35) = cos(3x) where x is greater than or equal to 0 and less than or equal to pi

I tried writing the cosine as a cofunction identity but my online course said it's incorrect, and im not sure what to do as they dont say.

[lime bolt]

just write cosx as sin(x+pi/2) and equate the inputs

[simple edge]

theres 3 solutions

[viscid thistle]

Sin(x) = cos(x-pi/2)

Cos(x) = sin(x-pi/2)

[lime bolt]

your second equation is wrong

that implies sinx=cosx

[fleet yew]

cos(x)= sin(90-x) = -sin(x-90)

[viscid thistle]

any tips to help me solve this problem would help

would the variables be the two points and time

[lime bolt]

no use the speed of the current and speed of boat

the points are the same so idk why u would use that

[viscid thistle]

and time would be the last variable @lime bolt

[lime bolt]

it isnt a variable it just 1 hour and 2.5hours

[viscid thistle]

anyone know how to do this on desmos

[past meadow]

you should just be able to click on the intersection point and it will give you the inteersection

like this

[upper kelp]

Since you don't know the coefficients of the small powers of x, it would be beneficial to look at properties of the function that relate to the coefficients of the large powers of x. (x³ in this case)
One important aspect of the coefficients of the greatest power of x is that they determine the values of the horizontal asymptotes. Consider the coefficients of x³ in this case, and the horizontal asymptotes in the graph. Which one do you think is more fitting?

[viscid thistle]

For this i found that theta = 70 but can someone help me with finding the general equation? Not sure how because there is no restriction

[pseudo cargo]

You could do $cos (2\theta) = 1 - 2sin (\theta)^2$

[obsidian monolith]

Lobo:

[pseudo cargo]

And stablish $sin \theta$ as the value you wanna find, saying $sin \theta = x$ for example

[obsidian monolith]

Lobo:

[pseudo cargo]

So you end up with the equation $-2x^2-x+1=0$

[obsidian monolith]

Lobo:

[pseudo cargo]

You solve x and then you have $sin\theta$. After that, you get $\theta$

[obsidian monolith]

Lobo:

[pseudo cargo]

Using the sine inverse

[rapid lance]

$$y^{2}=x^{3}+x+2$$ find a way to represent the number of solutions modulo p as and angle $$\theta_{p}$$ then find the frequency of each value of theta as $$\pi \left( x\right)$$ approaches infinity
i got that $$\cos \theta {p}=\dfrac{M{p}-\cdot P-1}{2\sqrt{p}}$$

[obsidian monolith]

ohNoiAmHere:

[rapid lance]

what do i do for the frequency of $$\theta_{p}$$?

[obsidian monolith]

ohNoiAmHere:

[lethal oracle]

How would I find the initial velocity required to reach the target at 300m

[viscid thistle]

You have two equations

Horizontal and vertical motion

[lethal oracle]

Can’t I use something like this?

[lime bolt]

you want to work out how long it is travelling in the air for to deduce the invariant horizontal velocity. the vertical velocity is a function of -gt, use the equation for final position=initial position +initial velocity and half acceleration times t^2. plug in the values, and work out t

[lethal oracle]

Well I used an equation that someone else gave me and I got 3.83m/s

Is that incorrect?

This is a like a base level problem btw

[viscid thistle]

Well I used an equation that someone else gave me
...

[lime bolt]

66.67 ms^-1

[lethal oracle]

So I can’t use this?

[viscid thistle]

Vertical velocity is 0

Yes you can

[lethal oracle]

It gets you 3.83m/s

[viscid thistle]

But it's too easy

[lethal oracle]

This is an entry level physics high school class I’m helping someone with

Or rather helping both of us I mean

[lime bolt]

well you at least want to understand why it is true

[viscid thistle]

^^^^

[lethal oracle]

Ya that’s true

[viscid thistle]

I would get the time from the vertical motion

[lime bolt]

yea that is what i said earlier

[lethal oracle]

Can you help me on the next part of you don’t mind?

The vertical motion is effected by gravity, right?

[lime bolt]

[viscid thistle]

Horizontal motion has constant speed

[lime bolt]

did u not understand my solution @lethal oracle ?

[lethal oracle]

No I do

Do you guys know how I would do c?

I understand there’s different ways to do this

But for the sake of a low level high school course

I think what I need to do for c is basically find the cannon balls height at 275m, right?

[lime bolt]

so we are using the velocity from b)?

[lethal oracle]

Yes

[lime bolt]

just work out the time by distance over horizontal velocity

and then use that in the vertical equation

[viscid thistle]

I think what I need to do for c is basically find the cannon balls height at 275m, right?
Yes

[lethal oracle]

So 275m/3.85m/s?

Wait not over 3.83

[lime bolt]

y postition is -half gt^2

starting from 100

plug in time, then get answer

[lethal oracle]

I’m sorry I’m having trouble

So half gt^2 that’s

9.8 squared?

[lime bolt]

yes g is gravitational constant of earth

[lethal oracle]

Yes

So when you say half gt^2

I’m sorry can you walk me through this more

[lime bolt]

use the equation for final position=initial position +initial velocity and half acceleration times t^2.
do u know why this equaiton is true

u just plug 0 into initial velovity and initial position

[lethal oracle]

Alright

I’m sorry I’m just so confused

[lime bolt]

so do u want me to explain this equation

[lethal oracle]

I understand the equation I’m just having trouble like using it

It’s vi+initial position

[lime bolt]

your finial position will be your initial plus your average velocity times how long you are moving at this velocity, note this works only for constant acceleration

[lethal oracle]

Could you show me how this would look in equation form?

[lime bolt]

like a picture of the full equation?

[lethal oracle]

Ya if you can

[lethal oracle]

So vot

Is gone right

Same with the first term too

[lime bolt]

yes as initally there is 0 vertival velocity

[lethal oracle]

So we are left with 1/2at^2

[lime bolt]

also we take the centre of the coordinate system as the starting place

that is why x_0 is 0 not 100

[lethal oracle]

Oh I see

So our a is -9.8, right?

[lime bolt]

👍

there are separate equations for horizonatal and vertical motion

[lethal oracle]

And we get t^2 from

Getting t by using d/s?

[lime bolt]

ok, as we have constant acceleration, we can regard the velocity as an average over the journey, i.e inital velocity plus final over 2

your finial position will be your initial plus your average velocity times how long you are moving at this velocity, note this works only for constant acceleration
use this

do u have paper? try and get the formula from what i have said

[lethal oracle]

Our initial is 3.83m/s right?

Are we sure that’s the right answer?

I’m using paper ya

[lime bolt]

no i got 66.67 for it

[lethal oracle]

Oh

Do you get that via this too

[lime bolt]

i told you how i got it earlier

[lethal oracle]

Ya

But our distance for the t=d/s

Is 275?

[lime bolt]

yes

[lethal oracle]

So 275/66.67

[lime bolt]

no plug into above formula, and you can see how much it has changed over the y

[lethal oracle]

So now we got (1/2)(-9.8)(4.12)^2

?

[lime bolt]

yes

[lethal oracle]

Now we got -83.17

[lime bolt]

so it is gonna be a minus, but now we just readapt our coodinate system to the question's one by adding 100 on the vertical

so now you have the answer, i suggest you try deriving the equation from the information i gave

[lethal oracle]

83m

Is the height?

Needed is what you’re saying?

[lime bolt]

no what is -83 add 100

[lethal oracle]

17

So the minimum height for it to pass over is 17 meters

Thank you so much for putting up with my confusion btw

[lime bolt]

although add the decimals to be more precise

[lethal oracle]

Could too explain though like

What adding 100 means like in terms of the math

We’re adding 100 specifically because it’s the height it was shot from?

[lime bolt]

because we set 100 as 0 to ease calculation

so we gotta add it back on

[lethal oracle]

Oh so we could have used 100 earlier and it would have yielded the same results

?

[lime bolt]

yes it would have changed the equaion a bit but same answer

[lethal oracle]

Thank you so much

[opaque olive]

i dont really know how they got line 3 from line 2

[opaque olive]

calculus?

i thought this was pre calculus

oh

i dont understand still though

how would i get rid of ln?

well multiplying be e?

oh wait

ill get ln(y+1) = ln(x) + lnC ?

oh

hmm im not too sure

hmm i think so

log_e in natural log

$log_by = x$

[obsidian monolith]

Yes:

[opaque olive]

$y=b^x$

[obsidian monolith]

Yes:

[opaque olive]

so raising both sides to the power of e?

ahh

i see

[obsidian monolith]

slimvesus:

[opaque olive]

so you mean this right? $e^{ln(y+1)} = e^{{ln(x)}+c}$

[obsidian monolith]

Yes:

[opaque olive]

so $y+1 = x + e^c$

[obsidian monolith]

Yes:

[opaque olive]

and we can right A as e^c?

oh

wdym absolute values

nah this is part of a much larger differential equation question

ok the absolute values, you mean the "critical values"

magnitude

modulus

yep

no

so y = -1 and x = 0

im not sure what youre asking about the modulus lol

hmm

the modulus was only added cuz you cant have a negative in the log

i dont know

could you just tell me the answer?

yeah i have

A = e^C

?

$|y+1| = |x| e^c$

[obsidian monolith]

Yes:

[opaque olive]

yeah thats what i sent earlier

i just forgot the modulus sign

ok

how do we go about dealing with the modulus sign

ok

it would be $|\pm (y+1)|$

?

[obsidian monolith]

Yes:

slimvesus:

[opaque olive]

ok

i agree

$\pm (y+1) = \pm (xe^c)$

[obsidian monolith]

Yes:

[opaque olive]

oke

hmm

we can just ignore those plus minus now right

ok

yep

yep

it does

question

ok so we dont need the plus minus for 2/3 cases

that could arise

but for the 3rd, we can get away with having the plus minus on one side right

@viscid thistle

but we can contain the plus minus for what ever case in the A constant rigt

yeah

[obsidian monolith]

slimvesus:

[opaque olive]

yep

makes sense

thanks

https://gyazo.com/efc7ddb859da5c91997d43034330c736

thats the actual question if you were curious

[viscid thistle]

hey anyone here

quick and easy question

how do i find the horizontal strech of a sin function

idk if this is precalc or a calc question sorru

[simple edge]

2pi/b gives the period for a given b in sin(bx)

[dark sky]

how to do dis

[viscid thistle]

Draw a triangle

And then substitute the side lengths

And then || I’m not doing this problem for you||

@dark sky still need help figuring it out

?

[lime bolt]

this is a primitive pythagorean triple

[fleet yew]

Primitive pythagorean triples are just scaled down versions of non primitive pythagorean triples

Isnt that cool?

[vale basin]

Can somebody help me here

I solved it whit lhopital rule

But our prof said we had to solve it without lhopital

But dont know how to start on this one so far

[uncut mulch]

consider conjugates

[vale basin]

I dont know what you mean like conjugates in complex numbers etc

[uncut mulch]

conjugates aren't exclusive to complex numbers

the conjugate of a+b is a-b

[vale basin]

I still dont get it to use that in this case :/

[uncut mulch]

and is a very common idea to solve these limits and prove trig identities

first combine them into a single fraction and the rationalise the numerator

by multiplying the numerator and denominator by the conjugate of the numerator

[vale basin]

So basicly multiplying with 1

[uncut mulch]

yes

[vale basin]

Oooh i see

I get a new form of that 'equation'

Thanks man !

[uncut mulch]

expression

and then apply elementary limits

[obsidian monolith]

I couldn't find a command named tex Area. Please make sure you have spelled the command correctly.

[narrow peak]

ok i need help

Without using a calculator, find the integer value of a and of b for which the solution of the
equation $2x\sqrt{5}=x\sqrt{2}+\sqrt{18}$ is $\frac{\sqrt{a}+b}{3}}$.

[obsidian monolith]

Hmm:

Without using a calculator, find the integer value of a and of b for which the solution of the
equation $2x\sqrt{5}=x\sqrt{2}+\sqrt{18}$ is $\frac{\sqrt{a}+b}{3}}$.
```Compile error! Output:

! Extra }, or forgotten $.
l.55 ...rt{2}+\sqrt{18}$ is $\frac{\sqrt{a}+b}{3}}
$.
I've deleted a group-closing symbol because it seems to be
spurious, as in $x}$'. But perhaps the } is legitimate and you forgot something else, as in \hbox{$x}'. In such cases
the way to recover is to insert both the forgotten and the
deleted material, e.g., by typing `I$}'.

Preview: Tightpage -1310720 -1310720 1310720 1310720
[1{/usr/local/texlive/2018/texmf-var/fonts/map/pdftex/updmap/pdftex.map}]

[narrow peak]

fuck

ah whatever it has what i want anyway

[viscid thistle]

Try isolating x

[narrow peak]

,w expand (2xsqrt(5)-xsqrt(2))^2

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by Hmm#2554

[viscid thistle]

I s o l a t i o n

[narrow peak]

e.e

oh

LOL

[viscid thistle]

So?

[narrow peak]

so i got it thanks

[proven marten]

Is conic section pre calc

[viscid thistle]

I think so

[rapid lance]

$$y^{2}=x^{3}+x+2$$ find a way to represent the number of solutions modulo p as and angle $$\theta{p}$$ then find the frequency of each value of theta as $$\pi \left( x\right)$$ approaches infinity,

i got that $$\cos \theta {p}=\dfrac{M{p}-P-1}{2\sqrt{p}}$$

what would i do for the frequency of values for $$\theta{p}$$?

[obsidian monolith]

ohNoiAmHere:

[rapid lance]

p is prime

M_p is the number of solutions in mod p

[opaque olive]

quick question, does $\frac{v}{2-2v^2}dv = \frac{v}{-4v}ln(2-2v^2)$

[obsidian monolith]

Yes:

[hallow thunder]

@opaque olive is this an integration?

because if so, then no

[opaque olive]

oh

its integration, whats the correct integral

[hallow thunder]

unfortunately, there is a v on top, so it is not as easy as the natural log

integration by parts should help here

or not.

@opaque olive u sub works. dont forget to add C at the end

[opaque olive]

hm

ill give it a try

[hallow thunder]

try u sub for 2-2v^2

🙂

[opaque olive]

is the answer $-\frac{1}{4}ln(2-2v^2)$

[obsidian monolith]

Yes:

[past meadow]

,w integral x/(2-2x^2) integral

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by Sneaky#0928

[past meadow]

wtf

,w integral x/(2-2x^2)

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by Sneaky#0928

[opaque olive]

howww

[upper kelp]

Hint: ||Try to notice a relationship between the numerator and the denominator in terms of differentiation.||

[opaque olive]

wasnt there like a standardised way of doing this

without parts or sub

[upper kelp]

At least from what I've learned, you can show that the numerator is the derivative of the denominator, and that gives you that the function is log (denominator).
Not sure if that answers your question.

[opaque olive]

yes

[upper kelp]

I think that's technically still a substitution of variables without explicitly doing it, but it's still correct.

[opaque olive]

okay so

[past meadow]

pretty sure your answer is correct and wa just put some of it in the constant

[opaque olive]

....

[upper kelp]

Yes, sorry. Keep in mind that:
log (2x² - 2) = log (x² - 1) + log (2)
due to the rules of logarithms.

[opaque olive]

ohhh

ok

its simplified

no wait

[upper kelp]

Oh, and a small technical note:
The integral technically needs to have an absolute value on it inside the argument of the log.
This means that you should have:
log (|x² - 1|) = log (|1 - x²|).

[fleet yew]

You can simplify it further

[opaque olive]

differentiating $ln(2-2v^2)$ gives $\frac{-4v}{2-2v^2}$ right?

[past meadow]

thats not a dollar sign

[opaque olive]

lol my bad

[fleet yew]

Remember that (x²-1) is a difference of two squares

[obsidian monolith]

Yes:

[past meadow]

yes although that expression can be simplified

[opaque olive]

so would $\frac{-1}{4}ln|2-2v^2|$ be a correct answer?

[past meadow]

\\\

[obsidian monolith]

Yes:

[opaque olive]

im used to linux

[past meadow]

yes

although not pulling out that constant may sometimes get you marked down on a test or w/e

[opaque olive]

i can now simplify yes

[past meadow]

also $\ln$

[obsidian monolith]

Sneaky:

[past meadow]

yes you can now simplify

[opaque olive]

ok ty

[graceful ledge]

1 is not prime because sin(90°) is a factor

[frail patrol]

Trying to factor x^3-x^2+8x+60 But I'm not sure what to do I put it in a factoring calculator and tried to work backwards but I'm not sure how they did it. I'll put my work and problem/ solution below

[upper kelp]

Did they give you any clue in the question as to what may be the root of the function? (if you know that x = -3 is a root, then you can use polynomial division to factor the polynomial.)

Also, perhaps you're meant to guess small numbers and check whether they are roots of the function.

[frail patrol]

yes they told me a zero was = -3

[upper kelp]

Oh, I see. Are you familiar with polynomial division?

[frail patrol]

yes

I can do it easily

[upper kelp]

In that case:
If you know that x = -3 is the root of a polynomial, then you know that it can be written as f(x) = (x + 3)·g(x), where g is a polynomial of a degree 1 lower than that of f's.
You can find g using polynomial division.

Generally speaking, if x = α is a root of a polynomial f, then you can write that polynomial f as:
f(x) = (x - α) · g(x)
where deg (g) = deg (f) - 1
and you can find g through polynomial division.

[frail patrol]

so did you just solve if x= -3 you just added 3 and made x+3=0 then just used polynomial remainder theorem

k thx

[frail patrol]

I ran into a problem there was a question similar to that but I dont think you can you polynomial remainder theorem the question gave me a zero of 2+3i and f(x)=x^3-3x^2+9x+13 what do I divide f(x) by to get a solution because I don't think you can use imaginary numbers in poly rem theorem All I really need is the first step

[graceful ledge]

I ran into a problem there was a question similar to that but I dont think** you can you **polynomial remainder theorem the question gave me a zero of 2+3i and f(x)=x^3-3x^2+9x+13 what do I divide f(x) by to get a solution because I don't think you can use imaginary numbers in poly rem theorem All I really need is the first step
@frail patrol what

[uncut mulch]

what exactly is the question asking for?

[frail patrol]

How do I factor f(x)

[uncut mulch]

apply conjugate root theorem

[frail patrol]

k

[graceful ledge]

Genius

[frail patrol]

thanks I was able to solve it

[burnt osprey]

ssdsaw

[vale basin]

What is the best way to find the angle between 2 tangents

[echo wagon]

Although no one will agree with me, I always use the dot product. Find a vector on each line, and use the two forms of the dot product to get the angle between the vectors (which is the same as the angle between the lines)

Since this is precalculus, if you don't know what a dot product is, ignore my answer

[vale basin]

I know what it is 😄

But i need to find 2 vectors and then say cos(alpha) = |u•v|/|u|•|v|

Or am i wrong 😅

[echo wagon]

Exactly

That's what I would do. But I know other people prefer other nonsense

[vale basin]

Thanks man

[dark sky]

how to do this

[still meadow]

What have you tried?

Also what are the options?

[dark sky]

oh i got it

is it 3?

in my first attempt i tried completing the square but i did it wrong lol

[still meadow]

This question is very poorly written lmao

I think it’s -3.

[echo wagon]

3 is right

[still meadow]

Hmm?

[hallow thunder]

yeah 3 is right

[echo wagon]

Expand (x-2)^2 and you get x^2 - 4x + 4, which is three more than what you were given

[still meadow]

Oh yeah smh my head.

[scarlet juniper]

Can someone help me factorize (sp?) 8+ 10x -3x[squared]

[proven marten]

nani?

$-3x^2 +10x +8$?

[obsidian monolith]

JC Denton:

[scarlet juniper]

yes

[proven marten]

First take out the negative

I would advise

What would you get?

[scarlet juniper]

$3x^2 +10x +8$ ?

[obsidian monolith]

Venx:

[proven marten]

No, sadly not

Ok wait

So, $-1(x+1)=(-x-1) right$?

[obsidian monolith]

JC Denton:

[scarlet juniper]

right

[proven marten]

Ok so if we were to reverse that

for $-3x^2 +10x +8$

[obsidian monolith]

JC Denton:

[proven marten]

What do you think it would be

[scarlet juniper]

$-1(3x^2 -10x -8$

[obsidian monolith]

Venx:

[proven marten]

Yes nicely done

Dont forget a bracket at the right

[scarlet juniper]

yep

[proven marten]

Ok now you think of the divisors of 3, it would only be 3 and 1 so:
$-1(3x ± a)(x ± b)$

[obsidian monolith]

JC Denton:

[proven marten]

Do you get that step?

Notice that $3x(x) = 3x^2$

[obsidian monolith]

JC Denton:

[proven marten]

Which is the first part of the expansion

[scarlet juniper]

Yes I follow

[proven marten]

Ok now you think of the divisors of 8

[scarlet juniper]

8 and 1 or 2 and 4

[proven marten]

Correct so now you also think of where you would substitute the 8 and 1 or 2 and 4 in $-1(3x ± a)(x ± b)$

[obsidian monolith]

JC Denton:

[scarlet juniper]

so theres a bit of trial and error to that process

[proven marten]

So that when adding 3xb and ax add up it gives -10

[lime bolt]

Just use the grid method

It is very quick

[proven marten]

and when multiplied it gives -8

I think this is quicker when you get used to it tbh

You could also use the 'down under' method

[lime bolt]

Wait I think this is essentially the same method but less systematic

[proven marten]

Wait let me show you

[lime bolt]

It is Looking for a pair which adds to 10 and times to 8, from the factors of 3 and 8

[echo wagon]

Subtract to 10

[proven marten]

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

This is kind of what I meant

except you would keep it all over 18

Just use the grid method
@lime bolt Is the grid method the one with the X?

[scarlet juniper]

$-1(3x -4)(x -2)$

[obsidian monolith]

Venx:

[lime bolt]

Tbh it isn’t really a grid it literally just writing the down factors of 3 and 8 but like in a grid

[proven marten]

Yes

[lime bolt]

The grid doesn’t have a purpose

[proven marten]

So the X method?

Wait can I see what you mean kane

[lime bolt]

I guess, I don’t know the names of these things

[echo wagon]

@scarlet juniper No, you need to get -8 not +8

[proven marten]

@scarlet juniper You seem to be familiar with this

[echo wagon]

So one bracket has a + the other a -

[proven marten]

^

When 2 negatives multiply it produces a positive

keep that in mind

[scarlet juniper]

yup

[proven marten]

kane wdym by grid method

[lime bolt]

Ehh I don’t really know how to explain it that well

[proven marten]

$-4*-2=8$

[obsidian monolith]

JC Denton:

[lime bolt]

I could send a pic or something

[proven marten]

not -8

Yeah pic would be fine

[echo wagon]

I'm sure he means list the pairs of factors of a and c in a grid and then systematically multiply and add them until you get b somehow.

[lime bolt]

Yea

[proven marten]

Ive never heard of that method before though

[echo wagon]

That is more systematic, but can be excessive when there are obvious factors that can be left out or when intuition can tell you what the right factors are

[proven marten]

^

[scarlet juniper]

$-1(3x +2)(x -4)$

[obsidian monolith]

Venx:

[proven marten]

Yes

Gj

[echo wagon]

Hooray

[scarlet juniper]

Thank you

[fleet yew]

amazing

[scarlet juniper]

so if y = -1(3x +2)(x -4), y = 0 when...

[fleet yew]

i personally would have factored out the LC but your way works too ig!!

[proven marten]

Whats LC

[echo wagon]

LC??

[proven marten]

y=0 would mean the roots/solutions/x-intercepts of the graph

[scarlet juniper]

if the -1 wasnt there, y = 0 when x = 4

[fleet yew]

Leading

Coefficient

[proven marten]

huh

how would you do that then

[echo wagon]

The -1 doesn't change anything

[proven marten]

Yes it does?

The parabola concaves down

[echo wagon]

It doesn't change the roots I mean

[proven marten]

Oh yes

My bad

Leading
@fleet yew So, factor out -3?

[fleet yew]

yes

but nvm keep doing your thing

you're doing great

[echo wagon]

AMD be factorizing with fractions

[fleet yew]

sorry to interrupt

[proven marten]

You didnt interrupt

AMD be factorizing with fractions
@echo wagon I dont think my mental capacity can factor fractions

[echo wagon]

Haha

Maybe he meant factor out the 3 afterwards

[scarlet juniper]

imma guess with the above, y = 0 when x is -4

[proven marten]

Yes

[echo wagon]

No

[proven marten]

I mean he's not wrong

He just missed another solution

[echo wagon]

He is, lol

Look at the equation again

[proven marten]

Oh no do I have to relearn quadratics

LMAO

I messed up on quadratics

Yeah so

You want the bracket to equal to 0

[scarlet juniper]

so the -1 doesnt actually change it

[proven marten]

so basically you're solving for

$x-4 = 0$

[obsidian monolith]

JC Denton:

[proven marten]

My bad

Thanks Luna

[echo wagon]

Venx: If you have any product that equals 0, it is true only when one of the factors is zero. So if abc = 0 then a = 0 or b = 0 or c = 0

$-1(3x +2)(x -4) = 0$

[obsidian monolith]

Lunasong:

[echo wagon]

You have three factors, but obviously you cannot have -1 = 0

So you set the other two factors equal to 0

[scarlet juniper]

my instinct with that -1 is just to oppositize everything

[echo wagon]

No

Follow the logic. A product is 0 if one of the factors is zero

So set the different factors equal to 0 and solve for x

Don't just guess what you should do

[scarlet juniper]

well when x = 4 then, and

when x = -2/3

[proven marten]

not 3x

Yes

Those are your roots

Are you graphing a parabola?

[scarlet juniper]

Im learning about inequalities

[echo wagon]

Where is the inequality lol?

[scarlet juniper]

Im just linking formulas like that to the correct notation

[proven marten]

Ah quadratic inequalities

[scarlet juniper]

oh plot twist it was >0

[proven marten]

Yes

[scarlet juniper]

but works the same

[proven marten]

Yes

[echo wagon]

Well, finding the roots are the same, but the solution is not the same

[scarlet juniper]

thanks guys

what is a 'root'

the point at which it =0?

[echo wagon]

A root is a value of x for which a function is 0

Yes

[proven marten]

no problem

[scarlet juniper]

you are so precise in your language

im still tongue tied over all the terms

[proven marten]

As mathematician he has to be

[echo wagon]

So if it was > 0, your answer should be - 2/3 < x < 4

Do you understand why?

[scarlet juniper]

I think I grok it

[echo wagon]

Okay

[scarlet juniper]

im doing exercises from an EDx course

[proven marten]

grok

[scarlet juniper]

know the term?

the lecturer glossed over how he factorized a similar problem

I couldnt figure out how he knew what to do

let me try and dig it up

$y = 2x^2 +5x -3$ ?

[obsidian monolith]

Venx:

[scarlet juniper]

so the first step he did was to turn that into this:

$y = 2x^2 - x +6x -3$

[obsidian monolith]

Venx:

[scarlet juniper]

which factorises to

$(2x-1)(x+3)$

[obsidian monolith]

Venx:

[scarlet juniper]

so anyway I don't understand why 5x was turned into 6x - x

I can see how x and 6x works nicely with the bracket multiplication, I guess I can't see how you'd know to do that before you factorized

unless he was just being illustrative

[proven marten]

Oh that method

I dont understand how people do that method

its so tedious sometimes

[scarlet juniper]

is it just a case of splitting x's until you find a pair that multiply nicely?

like if -x + 6x didnt work, next you'd try it with -2x + 7x etc?

[proven marten]

Idk how to do that method but I have a textbook that does

so here you go

[viscid thistle]

denton

you posted examples not the explanation

lmao

@scarlet juniper google factor by grouping

[scarlet juniper]

the example helped actually, thanks both

[fleet yew]

nice

great job guys

good effort

[viscid thistle]

thanks @fleet yew

[scarlet juniper]

my first time posting here, great community by the looks of things

[proven marten]

great job guys
@fleet yew tyty

the example helped actually, thanks both
@scarlet juniper I felt like you grasped math easier, so I just decided to give you examples

[obsidian monolith]

Yes:

[harsh smelt]

for the first part it is unclear what they want from you

[viscid thistle]

it wants you to show work

[harsh smelt]

work

here it is

[viscid thistle]

is there something about question 2. that explains it more zeek?

as it stands that question is just 2 sentences

not sure what work it's asking for

[harsh smelt]

(would not earthquake they want be 490 richters lol)

[fleet yew]

`0.......+3

3

3

3

3

3

[blissful ridge]

The first part is just a sentence

[fleet yew]

3

3

3

3

[harsh smelt]

4

[fleet yew]

3

3

3

3

3

[fleet yew]

Wtf

Sorry guys i was accidentally mashing my keyboard

Just happened to be on the 3 key

[lethal zephyr]

3

[zealous slate]

3

[stuck lark]

三

[willow bear]

III

[stuck lark]

州

[smoky pagoda]

Tres

[proven marten]

dri

[blissful ridge]

Teen

[wide lynx]

idk where to put this question but idk where to start

[blissful ridge]

What have you done so far?

[wide lynx]

well i dont really knopw where to start

sorry i went to go eat

[blissful ridge]

Start by making a diagram

[blissful ridge]

Diameter is 33

Not the length

[wide lynx]

but i thought thats

the point

well i guess the orientation matters

i have no idea how to approach that then lol

[blissful ridge]

Join the diagonal

[lime bolt]

Yes draw a right triangle with radius, half width and length

[blissful ridge]

The diagonal would pass through centre