precalculus

[valid violet]

I have to go now

I claim that your data are very appropriate to be approximated by a line

think about way, bye

think about why*

[vapid torrent]

i dont think we are answering th eright hting here

[valid violet]

You found the answer,im asking you why the question writer is justified in using just one number to describe the slope of the whole histogram

In your first question you asked for help with

The slope wouldn't be a meaningful number

[vapid torrent]

i dont think thats the answer though

im not sure thats what the question is asking @valid violet

[valid violet]

The question is asking you to analyze the ratio of the frequencies on x compared to x+1

That is the slope

[viscid thistle]

What effects tension of a cord between 2 masses on an inclined plane? Coefficient, the angle, or the masses?

[vagrant stirrup]

how do i know if its a local min or local max by my 2nd derivative test? my value is postive and what does it being local min/max tell us?

[heady field]

Can you attempt to learn pre algebra alone?

Next year I am taking pre algebra with trig and I want to learn a good amount to get a good grade to get into the AP class next year

Get a good grasp on it I guess I mean

[glossy raven]

Hello! I am studying average rate of change with functions and variables
I understand the steps until variables are thrown in
How would you approach this problem to start?

[valid violet]

@heady field try Khan academy for self learning

You definitely can with all the sources online there are

@glossy raven how would you do it from x=1 to 2?

Do you know how?

[heady field]

Ok thank you

[valid violet]

That's all you needed? Great lol

[glossy raven]

Not sure, I figured I substituted 1 and 1+h into the x value..

With 1 to 2, I would use the normal formula f(x)² - f(x)¹ over x² - x¹, and simplify, but with the variables I'm confused..

[viscid thistle]

help pls TT TT

[coarse storm]

@viscid thistle

What do you need help with?

@glossy raven

It is the exact same as $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

[obsidian monolith]

Lachlan:

[glossy raven]

I'm still confused on how to write the problem out...am I solving for the variables at all?

[coarse storm]

No, all you are wanting to do is to determine the average rate of change of f(x) from x=1 to x=1+h. h is any real value.

This is the same as determining the slope between those points

so using the slope formula above, we get that the average rate of change from x=1 to x=1+h $\frac{f\left(1+h\right)-f\left(1\right)}{\left(1+h\right)-\left(1\right)}$

[obsidian monolith]

Lachlan:

[coarse storm]

Now what is f(1)?

[glossy raven]

Wouldn't that be f?
I'm a bit confused

[coarse storm]

Ok

It seems like you are stuck on notation

f(x)=... is the same as y=..., so f(x)=2x-4 is the same as y=2x-4

But we tend to use f(x) because if we write f(a) that means the we substitute in x=a in f(x).

So f(2)=2(2)-4

or f(-5)=2(-5)-4

f(a)=2a-4

f(x) is just a way of notation a function called f.

We don't have to use f

We could called it g

so it would g(x)=2x-4

But in this case, it is called f

[glossy raven]

I get that when we use f(x)
X is the value that's being substituted for the most part

[coarse storm]

yes

But f(1) means we substitute in x=1

[glossy raven]

That makes sense

[coarse storm]

So now,

What is f(1)?

[glossy raven]

I must be overthinking it...
I'm sure that in f(1) 1 is being substituted for x

[coarse storm]

Yep

Ok

Let us go through it together

f(x)=2x-4

So f(1) all we do is substitute x=1 into 2x-4

so f(1)=2(1)-4

f(1)=2-4

f(1)=-2

[glossy raven]

Yeah, I think I got that far when solving it
But when substituting 1+h into X, how do I go about doing that?

[coarse storm]

Exact same

f(1+h)=2(1+h)-4

and so simplifying it we get

f(1+h)=2+2h-4

f(1+h)=2h-2

[glossy raven]

So first, is it 2(1h)?

[coarse storm]

What?

Where did you get that from?

[glossy raven]

I thought we distribute 2 into the 1+h

[coarse storm]

this is for f(1+h)?

[glossy raven]

Yes, I'm a bit lost of the transition from 2 + 2h -4 to 2h - 2

[coarse storm]

Yeah, we distribute the 2 into (h+1)

because we had 2(h+1) -4

So we get 2h+2 - 4

[glossy raven]

Oh nevermind

[coarse storm]

Still there

Haha

I have to go rn

But I'll be on in 15 or so.

Maybe

[glossy raven]

Ok that part makes sense now
Thanks for all of your help thus far! I appreciate your patience with me🙏

[coarse storm]

@glossy raven

Were you able to get there?

[azure junco]

I guess this goes here

But can someone explain this question

Oh wait wrong one

Yeah this

Q14

And Q13

[azure igloo]

Can someone help me with finding logarithmic equations from tables?

[azure junco]

Uh what is that lol

[azure igloo]

can you find an equation from this?

[azure junco]

No sorry I'm a beginner I was just wondering what you were talking about thanks for classifying

Lol

[azure igloo]

xD okokok

help please

[viscid thistle]

hi, i just started trying to learn advanced functions, do even functions have this property f(x)= -f(x)?

or is it odd functions?

[azure junco]

Ok so for differentiation, do you write the answer as Dy/DX=

Or

Just whatever they gave u =

For example this

It says the answer is y'

=

6x+7

[sonic jay]

If it’s given as y = blah blah

You’d write it as y’

[azure junco]

Yea

[sonic jay]

If it’s f(x)

[azure junco]

Andywah ik

F'(X)

No caps

I'm on phone

[sonic jay]

You can also write it as dy/dx

[azure junco]

Same thing?

[sonic jay]

Same thing

[azure junco]

So if its written as j=blah blah

It would be j'?

[sonic jay]

Or dj/dx

[azure junco]

And is it the same with integration

[sonic jay]

I mean

[azure junco]

Like if it's given as y' you write it as y

[sonic jay]

If youre given dy/dx as blah blah blah

Yah

Or dy/dx probably still y

It’s all the same thing and probably would never lose you marks or anything

Unless you have a strict teacher/marker

[azure junco]

Ah ok

So safer to write as y

Also differentiation is always making it without the apostrophe right

And vice versa with integration

[sonic jay]

Okay so y’ is gradient function right

Differ Mia y to get y’

Damn autocorrect

And intergrate y’ to get y

[azure junco]

Ah ok

Why would you need to differentiate and integrate though

Like in calculus what can you do with them

Just wondering

[lucid hearth]

Because a lot of things has relation to graphs, and calculus helps us to manipulate graphs

One application is to derive formulas, for example the volume of a cone that you’ve known since years ago is actually by revolving a line through the y or x axis using integrals

But it is also used to derive formulas and relations on more complex things

[gilded mirage]

@viscid thistle not sure if anyone answered your question, even functions will have that f(x)=-f(x) property (y-axis reflection)

and you test it out through f(-x)

[stuck lark]

even functions will have that f(x)=-f(x)
you mean f(-x)=f(x) @gilded mirage

[gilded mirage]

yes

[rigid beacon]

@azure junco the main application of integrals is finding net change

if you integrate f'(x) from a to b

you find the net change of f(x) (not f'(x)) from a to b

so like

velocity is the derivative of position

I can integrate the velocity of an object to find it's net change in position

[dawn light]

I am really confused about this equation

f(x) + 3x * f(1/x) = 2(1 + x)

solve for f(2003)

i tried to solve for f(x), but I ended up getting some recursive function

[patent beacon]

With just that equation, you won't be able to solve for f(x). But you can get another equation with a clever trick

Replace x with 1/x

@dawn light

[dawn light]

ok

im still a bit confused

what option does flipping every x give us

[patent beacon]

Basically we can get the equation:
f(1/x) + 3f(x)/x = 2 + 2/x

There's a f(1/x) in each, so you can equate the equations

[dawn light]

im sorry im still confused

[patent beacon]

In the second equation:
f(1/x) = 2 + 2/x - 3f(x)/x

Subbing that into the first:
f(x) + 3[2 + 2/x - 3f(x)/x] = 2 + 2x

f(x)(1 - 9/x) + 6 + 6/x = 2 + 2x

f(x) = (-4 + 2x - 6/x) / (1 - 9/x)

I can even clean it up a bit by multiplying by x/x
f(x) = (2x² - 4x - 6) / (x - 9)

[dawn light]

thank you :>

[harsh cipher]

Qestion

Question

we are adding rational expressions with binomial denominators

Why are we multiplying the numerator by cox/1?

and not 1/cosx

[jagged wind]

@harsh cipher cancer fraction over fraction over fraction

Cot is 1/tan which is cos/sin

Sec is 1/cos

So u get fraction divided by fraction and shit

Do the algebra and u done

[leaden stratus]

How can I verify this identity?

[tawdry current]

by using the fact that $$\cos x =\sin \left(x+\dfrac{\pi}{2}\right)$$

[obsidian monolith]

Probably Still Ele:

[leaden stratus]

🤔 why? @tawdry current
Associated angles?

[tawdry current]

Yeah complementary angles

If you draw a right triangle with angles π/2, α and β then you can see that sinα=cosβ and sinβ=cosα

[leaden stratus]

Ok 👍

[full garden]

hey guys a very stupid question but when we find the phase shift do we like expand the formula or factor because right here they expanded the formula in order to find the phase shift

but in this video they factored the formula in order to find the phase shift

here this one too

aren't phase shift and horizontal shift the same

[pale bison]

i think phase is used almost exclusive to trig functions

i might be wrong

and generally speaking, phase shifts is bounded by how big the period is

whereas, horz. shifts in f(x)=x^2 can be unbounded

[soft night]

No

They are the same

In math

you should talk to your instructor

about which definition

In introductory math courses like pre cal, they mean the same thing but in physics they are different

[pale bison]

oh ok, my bad

[plucky fractal]

hey guys how do i do this

i dont know how to start it

[lost mesa]

i think it wants you to use trig identities? not sure

[soft night]

Use the double angle formula

[soft night]

^

Now do it

[lost mesa]

but just replace the u with x

[soft night]

Yup

[plucky fractal]

for 1a is that answer 2 radical 63/64

[daring yarrow]

i solved it but didnt bother simplifying much cuz im too tired rn

,rotate 180

[harsh cipher]

question

i think im being silly but i must know

couldn't get the answer on the right

[willow bear]

multiply num and denom of the big fraction by cos(x)

[odd helm]

How would I approach this problem

[uncut mulch]

doing the substitution: let u = log_5(x)
might make things clearer

note that there is more than one solution; only one of which are listed in the options

[odd helm]

Ah wait I got it

I got that the logarithm could equal 2 or 1.5

Why can’t I just cancel the log(p) for this problem?

[willow bear]

what do you mean, "cancel the log(p)"

how are you getting from $\log(p) = \frac{2 - \log(p)}{\log(p)}$ to $1 = 2 - \log(p)$

[obsidian monolith]

Ann:

[willow bear]

@odd helm

[odd helm]

I did this

[heady jewel]

hMmM

[odd helm]

Wait wth

Ohh that’s stupid

Mk got it now

That was a stupid mistake lmao

Got logp = -2 or 1

So 1/100 is the answer

[daring yarrow]

@odd helm so did you end up solving it?

[viscid thistle]

http://web.iitd.ac.in/~pmvs/courses/mel705/curvefitting.pdf

Yo so on pg 9 they got a table and are looking to create a second order polynomial. What I’m confused on is how they end up filling in the values for the following matrix

Sum x sub i sum x^2 sub i and so on

[viscid thistle]

If anyone knows anything about polynomial regression I need your power

[sharp marsh]

Can someone tell me why this is wrong?

[hardy abyss]

what you wrote is equal to 18
not to x^2 + 10x + 18

[sharp marsh]

I don't get it

[hardy abyss]

so you found the roots, call them a and b
how can you write down the quadratic if you know the roots?

[sharp marsh]

er

(x - square root of 7) (x + square root of 7?)

but where does the 5 go

.

[hardy abyss]

close, but "roots" here means the roots of g(x), not the square roots of 7

you already found the roots a = -5 + sqrt(7) and b = -5 - sqrt(7)

then the polynomial is (x - a)*(x - b), right?

[sharp marsh]

oh

so is it uh

(x + 5 sqrt(7)) (x -5 (sqrt(7))?

[hardy abyss]

no

if a = -5 + sqrt(7) then x - a = x - (-5 + sqrt(7))

you should be able to finish it from here

[sharp marsh]

er

x -a = (x + 5 - sqrt (7))?

[hardy abyss]

yes

[sharp marsh]

x -b = x + 5 + sqrt 7?

[hardy abyss]

no

[sharp marsh]

I meant plus

[hardy abyss]

yes, the edit is correct

now g(x) = (x - a)*(x - b), and that is....?

[sharp marsh]

I got it

[sour inlet]

So I found the inverse of the function which is sqrt 3(x+3) / 3. When I plug in 24 to the inverse function I get the value of 3. But in the original function it says x <= 0, does that mean I would get a value of -3 ?

[lost mesa]

i think so

shouldn't the inverse function be -sqrt3(x+3)/3

bc the original function hits (-1, 0)

and the inverse i gave hits (0, -1)

[sour inlet]

ah you're right

I was using the positive version

thank you

[lost mesa]

np

[sharp marsh]

x^3 − 6x^2 + 11x − 6

How do I factor this?

[uncut mulch]

start with rational root theorem

[serene heath]

u can see that 1 is a root

so try work with that

[uncut mulch]

^

[sharp marsh]

my brain hurty on radical root therom

thereom

Also how do u know it's 1

[pale kettle]

just plug in 1

[sharp marsh]

so it's always 1

er

[uncut mulch]

not always 1.

[daring yarrow]

not always 1 no

its nice to try plugging simple numbers like 1 and 2 in tho

sometimes they give you a solution

[sharp marsh]

oh

How do I uh synthetic division

[sharp marsh]

so it's fancy adding?

also what happened to the -2?

[full garden]

guys I have a question I saw this formula for cos(-x)=cos(x) so how come here f(-t)=cost(-t+pi/2)

[uncut mulch]

looks logarithmic. try determine the base yourself

[viscid thistle]

No idea

[patent beacon]

I have no idea what y = 3 is supposed to represent

The useful equations are:
x1 = 10(t - 0.1)
x2 = 100 - 9t
@viscid thistle

[viscid thistle]

Right

[patent beacon]

Now, what time does x1 capture the flag? That is, for what t is x1 = 100?

[viscid thistle]

Algebra time

[willow bear]

50

not 100

[viscid thistle]

Yeah because it’s between the

[patent beacon]

Oh haha. Thought they had to get to the opposite side

[vernal spindle]

I need help finding the period and function

[woeful sierra]

the period is the time it takes for one cycle to go. so in this case it's 2

and b is 2pi/period. so the func would be
f(x)=2sin(pi * x)+1

@vernal spindle

[vernal spindle]

the function was wrong

It gave me a new problem to attempt

[woeful sierra]

oh mb it's cosine

[vernal spindle]

Is this one cosine also

[woeful sierra]

yes

and it has a reflection over the x-axis

the prompt asks for a sine function tho, so maybe u have to add a horizontal translation if it doesn't allow you to do cosine

[vernal spindle]

ahh

so 3sin(x)-1

but i need something with the x

[woeful sierra]

so to find the factor of the horizontal strech/shrink. you have to find the period. and then do (2pi)/period to find the factor.

[vernal spindle]

pi?

[woeful sierra]

right. the factor of the shrink is pi

make sure u add the horizontal translation

also when you're adding the translation. make sure it goes before the horizontal stretch, so:
sin(pi(x+1/2)) but not sin(pix + 1/2)

[vernal spindle]

this was wrong

[woeful sierra]

you forgot the reflection

notice the point on the y-axis is the min

-3sin(pi(x+1/2))-1

[vernal spindle]

Got it!

[hollow crane]

So I have a teacher that makes us write a bunch of Never Actually explains the stuff

[daring yarrow]

idk if htis goes here but

how to say if (n^2)^k has a higher rate of growth than 2^(2*n)?

n is a variable and k is a constant (not given)

$$how to say if (n^2)^k has a higher rate of growth than 2^(2*n)$$

[obsidian monolith]

piece:

[daring yarrow]

what

[sonic jay]

How to say if $$ (n^2)^k $$ has a higher rate of growth than $$ 2^(2n) $$

[obsidian monolith]

DuskyMountain:

[sonic jay]

Hmm

[final rover]

You mean $n^{2n}$ ?

[obsidian monolith]

WWM resident:

[sonic jay]

I assume he does

[final rover]

If k is constant, then wouldn't it be the opposite ?

[novel dirge]

Hello

Can I get help with rhis problem?

I know how to do usual integrals but I have no idea how to do it with e and ln

[stuck lark]

what'd you try

[novel dirge]

I was thinking abot trying tobsolve the root

But idk what to do tih the e

[stuck lark]

wdym solve the root

[novel dirge]

Then I thought about subbing it, but I dont know if thatvwill work either

By solving I meant trying to see if I can do anything with or extract anything

[stuck lark]

the best way to see if a sub works is to just do it

[novel dirge]

Doing it now

I get dx/t=dt/e^x

The integral of that

[stuck lark]

show work?

[novel dirge]

That was a * instead of =

Messed up while writing on dis

[stuck lark]

,rccw

[novel dirge]

Any idea what else i can do?

[stuck lark]

you didn't do anything after picking your sub

rewrite the integral using your sub then see what else you can do

[novel dirge]

,rccw

[stuck lark]

you swapped the bounds

[novel dirge]

I forgot the root

[stuck lark]

also you want to rewrite the integral entirely in terms of t

that too

[novel dirge]

What do you mean I swaed the bounds?

Am I supposed to change them?

[stuck lark]

you had ln(8) to ln(3). for some reason you now have ln(3) to ln(8)

[novel dirge]

You meam from the fkrst pic?

I mesed up here

The last 2 are correct

[stuck lark]

the real q is $\int_{\ln(3)}^{\ln(8)}\frac{\dd x}{\sqrt{e^x+1}}$?

[obsidian monolith]

RokettoJanpu:

[novel dirge]

yes

[stuck lark]

ok what's the integral after rewriting in terms of t?

[novel dirge]

I have to redo it since I forgot the root

,rccw

[novel dirge]

not really sure if I even did this correctly

[stuck lark]

rewrite entirely in terms of t, ie rewrite e^x in terms of t

[novel dirge]

not sure how to do that

do I extract the t from the first equation?

where I am setting the sub

e^x = t-1 ?

[stuck lark]

yep

[novel dirge]

I get t^x - t now

instead of e^x * t

[stuck lark]

t^x-t how

[novel dirge]

e^x = t-1

so it will be (t-1) * t

so t^2 -t

there is still the rest of the problem

[stuck lark]

you said t^x-t before

[novel dirge]

oh

my bad

I copy pasted e^x and forgot to change the x

cause I cant find the ^ on my keayboard

,rccw

[novel dirge]

I tried extracting t^1/2

but i finally got 1/((t^3/2) - (t^1/2))

but then i would need to do a sub again

and I am afraid that I will overcomplicate things for myself

[stuck lark]

best way to see if a sub works is to just do it

[novel dirge]

it will probably work but I am not sure if there simpler ways to do it

I complicate the problem unnecessarily way too many times

[stuck lark]

thinking if a sub will work probably will take more time than testing the sub

[novel dirge]

testing it rn

if I sub t^1/2 = g

I will get dt = 2*sqrt(t) * dg

and whn I use that in the integral

what do I do with the t^3/2

?

[stuck lark]

t^(1/2)=g implies t^(3/2)=g^3

[novel dirge]

so it will be g^3 - g?

[stuck lark]

just show what you've worked out on paper related to this sub

[novel dirge]

sending the pic

taking a while to load

,rccw

[novel dirge]

and I have a feeling that I overcomplicated

since I had dtsqrt(t) / t^2 -t

but now I have this abomination

with both t and g

[stuck lark]

rewrite sqrt(t) wrt g

[novel dirge]

?

not sure I understand

[stuck lark]

rewrite sqrt(t) in terms of g

whenever you do a sub, you must rewrite everything in terms of the new variable

[novel dirge]

oh

ok

so since g = sqrt*(t) I just write 2g*dg

should I write g^3 - g as g*(g-1)*(g+1) ?

[stuck lark]

if it helps

[novel dirge]

yes

I think it does

now got an integral of 1/g^2 - 1 which I can sub

damn there is a lot of subbing here

idk how I will return it all later

[stuck lark]

sub what?

[novel dirge]

g^2 -1

[stuck lark]

ok try

[novel dirge]

or should I do something else?

from your question I have a feeling that there is a better option

[stuck lark]

i'll repeat. best way to see if a sub works is to try. thinking whether it works usually takes longer than trying the sub

[novel dirge]

hmm

found this

and I think I can use 11.

[stuck lark]

i advise you stick to the g integral

so yeah. smth other than a sub will work

[novel dirge]

,rccw

[novel dirge]

this is what I got when I used that formula

and returned the sub from g to t

so should I continue like this or instead try to do another sub in the previous step

[stuck lark]

i would've suggested partial frac decomp but that formula covers this particular case

[novel dirge]

so I can continue like this?

[stuck lark]

sure

[novel dirge]

Or should I wait until the end after I found the limits

To return the sub from t

@stuck lark

[obsidian monolith]

Mr.Pancake:

[stuck lark]

you can either unsub all the way back to x then plug in the x bounds or swap to other bounds

[novel dirge]

I understand the first part

that is what I did here

but what do you mean by swap to other bounds?

[stuck lark]

if for some integral you have x bounds 1 to 5 and you sub u=2x, then your u bounds are 2(1)=2 to 2(5)=10

[novel dirge]

ok

so if I want to continue like I did and first return to x

I get ln*(sqrt*((e^x + 1)-1)

so when I apply the limits, do I go like e^ln(3) ?

[stuck lark]

just do it on paper

[novel dirge]

I want to

but I am not sure how to

[stuck lark]

If you want to stick to x bounds, unsub g to t then t to x

[novel dirge]

that is what I did already

[stuck lark]

what’s next?

[novel dirge]

now I have to include the limits

idk how to say it in eng

but I sub the x with the limits

did I do it correctly?

[stuck lark]

you didn’t finish

[novel dirge]

I know

but did I start correctly?

[stuck lark]

surely you’re familiar with this step in definite integrals

[novel dirge]

yes

when they don't have e in them

but idk what to do with the e

[stuck lark]

ok so an algebra problem

[novel dirge]

yes

[stuck lark]

exp and ln are inverse functions, meaning $e^{\ln(x)}=x$ for $x>0$

[obsidian monolith]

RokettoJanpu:

[novel dirge]

so this e^ln8 is just 8?

[stuck lark]

yes

[novel dirge]

ohh

this makes it much much easier

thank you

[stuck lark]

Very good. You’re welcome

[novel dirge]

finished

thanks

and just one more question, not related to math

do you get a notification when someone @ you?

[blazing parrot]

yes usually

[novel dirge]

makes sense

the laTex formula thing is still cool though

[stuck lark]

very cool

[rose trellis]

Anyone have a precalc textbook

PD

F

that i can use to self study ezez

[viscid thistle]

What is $\sum_{n=1}^{\infty}\frac{k}{n}=?$

[obsidian monolith]

εpsilon:

[viscid thistle]

Where k is any constant

[pale kettle]

What do you think?

For example, what if k = 1?

[patent beacon]

You can also write this as
k Σ (1/n)
Because that's how sums do.

[fading token]

What if k = 0

[daring yarrow]

can someone check for me if this proof is done correctly? its the hardest one in the series according to my teacher but it went pretty well so i feel like i messed up somewhere to make it easier (have to make the big fraction = 1/(1+tan²(x) which is equal to cos²(2x) by my calculations)

[flat grove]

how to calculate the perimeter of a parallelogram based on its segments without substituting a value?

[distant grotto]

https://cdn.discordapp.com/attachments/581478405422448650/684505321347219482/unknown.png guys is the suggested answer for this example wrong? (idk if this is the right place to post cos idk US curriculum)

[flat grove]

@ornate cairn would you help me?

😉

[neon garden]

Can anyone help me with 35?

[flat grove]

@neon garden I would help you but mathematics in English and in other units of measurement is impossible for me 😄

[neon garden]

The unit of measurement doesn’t effect the actual problem though?

[flat grove]

No

it doesn't

[valid violet]

@neon garden do you understand the x³ term of V?

[neon garden]

The volume of the cube?

@valid violet

[valid violet]

@neon garden i don't know, is it?

[rose trellis]

Anyone happen to have pre-calc textbook thats PDF I can use or sm?

[viscid thistle]

c(x)=0.6x^2-420x+87919

i need some help

[valid violet]

That is indeed a function

[remote veldt]

@viscid thistle what do you need help with. we aren't telepathic - what is the question?

since you're trying to find the min

put that parabola into vertex form

the vertex is the min

[viscid thistle]

A supply company manufactures copy machines. The unit cost (the cost in dollars to make each copy machine) depends on the number of machines made. If machines are made, then the unit cost is given by the functionc(x)=0.6x^2-420x+87919 . How many machines must be made to minimize the unit cost?
Do not round your answer.

so would the answer just be the minimum or is there more to it

bc i have the minimum

wouldnt the minimum be 919?

[remote veldt]

Read the question, identify what it is asking for, and then you'll know

[valid violet]

The wording is ambiguous whether c(x) is the total cost incurred or the marginal cost of one more

[viscid thistle]

so 350

?

[valid violet]

Probably means the former

[viscid thistle]

Could anyone help out and explain how did he get those nth terms?

I know how to find the a1 and d or r, but I don't understand how he found the nth terms

for both of those problems

[valid violet]

I dont know what a1 and d are but

[viscid thistle]

A1 is just the first term

and d is the arithmetic sequence number (common diff.)

while r is for the geometric ratio

[valid violet]

You can factor out 1/3 first

[viscid thistle]

What?

[valid violet]

To get (1/3)(6+8+10+...+208+210)

For me that would be easier to handle

Then factor out a 2

(2/3)(3+4+5+...+104+105)

Which is 2/3 times (105)(106)/2 - (2+1)

That's how i would do it

[viscid thistle]

How does that give n=103?

[valid violet]

I dont know what n is, im telling you that the sum is (2/3)((105)(106)/2-3)

What's n supposed to be

You just add the missing numbers 1+2 to make it n(n+1)/2 and then subtract the extra 1+2, do you see

[viscid thistle]

Not really..

[valid violet]

Do you agree that

1+2+3+...+104+105=(105)(106)/2

?

[viscid thistle]

To be frank, I have no clue how you got to that equation. All I needed was to find the first term, the common difference, and the n in order to plug it in the arithmetic sequence eqation to evaluate it

That's what I'm trying to do

Basically.

[valid violet]

That doesn't look very useful

[viscid thistle]

I'm trying to find how many terms are in the sequence

[valid violet]

The main formula is

[viscid thistle]

That's all that I don't know how to find

[valid violet]

1+2+3+4+...+(n-1)+n=n(n+1)/2

Sorry i dont know how to use your formula, it looks useless

¯\_(ツ)_/¯

[viscid thistle]

:/

[remote veldt]

@viscid thistle you're using the formula for the nth term, not the sum of the first n terms

[viscid thistle]

Hm..

[remote veldt]

in an arithmetic sequence, the nth term is a1 + (n - 1)d

[viscid thistle]

right

[remote veldt]

which makes sense

you start at (a1) then add the common difference (n- 1) times

[viscid thistle]

yeah, but i just dont understand how she got n=103

[remote veldt]

the sum of the first n terms is a totally different formula

[viscid thistle]

oh.

[remote veldt]

you'll want to work with that

[viscid thistle]

so what formula is it

[valid violet]

1+2+3+4+...+(n-1)+n=n(n+1)/2

Except your guy is multiplied by 2/3

Forget the missing terms

2/3(1+2+3+4+...+n)

[viscid thistle]

i understood the first one on my own

70 = 2 + (n - 1) (2/3)

i solve for n

which gives me 103

now i gotta figure out the second one

with the geometric sequence

[willow bear]

what's giving you trouble

[gilded mirage]

I don't know where to start

[willow bear]

if b was 0, i.e. if you were simply asked to find the limit of xe^-x at positive infinity, would you be able to do it

[gilded mirage]

well it'd be xe^-x which is 0?

[willow bear]

wording

[gilded mirage]

limit of xe^-x as x approaches infinity is 0

[willow bear]

uh huh

okay great

and what is the limit of e^-x itself as x approaches infinity?

[gilded mirage]

well it's a exponential with a reflection so limit of e^-x as x approaches infinity is 0

so it would just be 0?

0+b(0)

[willow bear]

i don't know, would it

do the limit laws you ought to be familiar with still hold

[gilded mirage]

sum theory holds i think

[willow bear]

"sum theory"

[viscid thistle]

sequence? -2 4 -8 16 -32

f(n) = ?

[agile gyro]

Theres a common factor for these

I see it being to the power of n

[viscid thistle]

n^n-1 * 2

?

[agile gyro]

The common factor?

[willow bear]

the blind leading the blind

[agile gyro]

Okay

[thorny bluff]

Any tips on how to actually get trigo identities right constantly? Everytime I just go a different "route" instead of the answer

[slate vessel]

hai, what will be
lim x ___> 1 (Sinx/x)³

[fluid shore]

[willow bear]

$\lim_{x \to 1} \paren{\frac{\sin(x)}{x}}^3$

[obsidian monolith]

Ann:

[willow bear]

this?

@slate vessel

[viscid thistle]

direct sub

[slate vessel]

Yes sir, already got the answer thanks

[viscid thistle]

just plug in 1

[willow bear]

don't call me sir

[slate vessel]

Ok chad

[willow bear]

........somehow that's even worse

[viscid thistle]

Lana Del Rey

[slate vessel]

your excellency

[willow bear]

why's it so hard to use my name???

[novel dirge]

what do I do with sin3x ?

the 3rd problem

[serene heath]

wdym what do you do with it

also this goes in #calculus

[novel dirge]

oh

ok

just not sure how to solve this integral

I know how to do with normal trig functions

like sinX

but not sure how sin3X is supposed to behave

moved it to calculus

[candid lance]

wtf.

(x-2)(x+3i)(x-3i)

?

[serene heath]

whats your doubt?

[molten forge]

I'm leaning about matrices

How do I do these?

We have a sub and we're kinda just learning on our own

Just these 2 pages

[patent beacon]

@molten forge
Two matricies are equal if they are the same size, and have all the same entries

So 17-20 are very easy, don't over think them

[molten forge]

@patent beacon ok thank you :)

And I think I can get the other pages

I googled how to do inverse

:)

[patent beacon]

@molten forge
2 does ask you to find an inverse, you'll need it for that.

However 3 and 4 just wants you to show that the matricies are inverses of eachother. All you have to do is multiply them together and show you get I

[novel dirge]

Need help with this one

I dont really understand the process behind it

,rccw

[novel dirge]

I understand that I can divide the sin^2x into sinx*sinx

And then use partial integration

But I am confused by the end of the 2nd row, in the integral part

How do I get the +X in the 3rd row?

[uncut mulch]

cos^2(x) = 1 - sin^2(x)

also missing +C

[novel dirge]

Yep

Figured it out

But thanks

[uncut mulch]

IBP is a bit overkill for this

[novel dirge]

What would be a better approach?

[uncut mulch]

double angle identities

[novel dirge]

Didnt do that in school

[uncut mulch]

really? trig is before calculus

cos(2x) = 1 - 2sin^2(x)
look familiar?

[novel dirge]

Yeah, but we barely did anything

And I am terrible ith trig

[uncut mulch]

calculus involves quite a lot of trig so you should probably review that

[odd dagger]

$\lim_{x\to 0^{+}} \left( \frac{1}{x} + \frac{1}{3} \tanh{x} - \coth{x} \right) $

[obsidian monolith]

tadders:

[odd dagger]

how do i do this so i don't get a infinity - infinity indeterminate form

i tried factoring out 1/(6xsinhxcoshx)

i know cothx = 1/x for small x because of laurent series?

there's gotta be a better way tho

[serene heath]

uhh

I'd change it exponential

might make it easier

[odd dagger]

right yh

$\coth{x} = 1 + \frac{1}{e^x -1} - \frac{1}{e^x + 1}$

[obsidian monolith]

tadders:

[odd dagger]

@serene heath

[harsh cipher]

Question

why is there a square root on y= +- x^2 - a^2

equation on the right

[candid lance]

Write the polynomial as the product of linear factors.
f(x) = x3 − 1x2 + 3x + 5

.

[heady jewel]

youre taking the square root of both sides @harsh cipher

@candid lance you can use hit and trial with trivial values

x=-1 works

so you now know that x+1 is a factor

now you can get a quotient with this factor

[fluid shore]

@candid lance

$x^3-x^2+3x+5 = x^3-x^2-2x+5(x+1) = x(x^2-x-2)+5(x+1) = x(x-2)(x+1)+5(x+1) = (x+1)(x^2-2x+5)$

[obsidian monolith]

Abhijeet Vats:

[reef jasper]

how would u do tan(pi/12)

[fleet yew]

@reef jasper sin 15 / cos 15

[reef jasper]

how’d u do that

[fluid shore]

Consider the following identity:

$\tan(2x) = \frac{2\tan(x)}{1-\tan^2(x)}$

[fleet yew]

Magic

[obsidian monolith]

Abhijeet Vats:

[reef jasper]

tf

[fluid shore]

Have you not learned trigonometry?

[reef jasper]

ok i forgot...

[fleet yew]

Honestly pi/12 is such a small angle relatively speaking

Sin theta=theta if theta is about 0, cos theta=1 if theta is about 0

[harsh cipher]

I asked a silly question above lol

double identity is really messing with my head...

is csc theta

hypotenuse/opposite

or opposite/hypotenuse

this guy in the video makes frequent mistakes.

omg

I think i've gone crazy

cant do 1/1/sinx/1

sinx =oop/hyp

hyp/opp

I didn't get why that comes out as x^2

how did he get x^2....

nm im so dumb

[hollow edge]

Probably should have put this in the prealg but whatever

[willow bear]

what's giving you trouble

[hollow edge]

so like I understand I am suppose to use pythagorean theorem but I don't really understand what I am suppose to do

[willow bear]

you're ASKED to find y in terms of x

and there is only one right triangle here

so perhaps you might want to write down the Pythagorean theorem for that one triangle, especially given that you said you knew it'd come into play

do you know the theorem itself

@EmmaL20#1525

[hollow edge]

It's a^2 + b^2 = c^2 right? So the 100m would be one of the sides

[willow bear]

what are a, b and c

in words

what sides are they, in the context of the right triangle

[hollow edge]

a and b are the sides and c is the hypotenuse

[willow bear]

the LEGS.

they're all sides

but a and b are the LEGS

so

what are the legs of your triangle

this question should not be taking that long to answer.

[stray ibex]

Cant you solve this as a vector?
officer(0,-100) and car(x,0)
OC = <x-0,0-(-100)>

[willow bear]

@stray ibex please do not interrupt.

[hollow edge]

sorry for the dumb question..

[willow bear]

do you see the triangle. is it in front of you right now.

yes or no.

[hollow edge]

in this one a or b would be 100m right?

but that's all we are given

[willow bear]

do you see the triangle. is it in front of you right now.

[hollow edge]

yes

[willow bear]

do you see its right angle.

yes or no.

[hollow edge]

it looks like a 45-45-90 but one side is as the cars are driving towards the sign

yes

[willow bear]

do you see the two sides touching the right angle.

yes or no.

[hollow edge]

yes

[willow bear]

tell me their lengths.

[hollow edge]

should I assume both are 100m^2 or only one of them? Isn't that only for one side

[willow bear]

the lengths are 100 meters and x meters.

you should not assume anything. all i asked you to do was, in essence, to READ these off for me.

that's it. but you overthought it to hell and back.

[hollow edge]

sorry about that and thanks for being patient with me

[willow bear]

so

now that we've gotten past the near insurmountable hurdle of reading off the lengths of the legs

what's the length of the hypotenuse

[hollow edge]

y meters?

[willow bear]

there we go

so with the legs as 100 and x, and the hypotenuse as y, what will the pythagorean theorem look like?

[hollow edge]

100m^2+x^2=y^2

[willow bear]

no

y^2 = 100^2 + x^2, yes.

so can you now turn that into a formula for y?

please don't say y=100+x please don't say y=100+x please don't say y=100+x please don't say y=100+x please don't say y=100+x

[hollow edge]

umm y = square root of x^2 + 100^2?

[willow bear]

parentheses

how am i meant to know if you meant $y = \sqrt{x^2+100^2}$ or $y = \sqrt{x^2} + 100^2$ or god forbid $y = (\sqrt{x})^2 + 100^2$

[obsidian monolith]

Ann:

[hollow edge]

sorry I didn't know how to type in the square root and forgot the parentheses but I meant the first one

[willow bear]

y = sqrt(x^2 + 100^2), then

well there you have it, don't you

[hollow edge]

is that all it's asking for? thanks for explaining it step by step

for me

[willow bear]

yes that is all it's asking for

now was that really so hard once you stop overthinking it

[hollow edge]

yea I really did overthink it

well I really need to practice trigonometry

[viscid thistle]

I need help

[willow bear]

@viscid thistle you should post your question rather than just saying "i need help" and waiting for someone to respond

[viscid thistle]

I’ll send later

[willow bear]

...

why not now

[viscid thistle]

.

Hello!

I have this!

My question is, to get the y- intercept. Should I plug every x intercept of the polynomial?

[fluid shore]

No. That doesn't make sense

Let p(x) be your polynomial. The y-intercepts are the points where x = 0. So, compute p(0).

If you want to find the x-intercepts, you need to solve p(x) = 0.

[viscid thistle]

But I already have my x intercepts

I wasn't paying attention and now I can't get help

Brilliant move...

[fluid shore]

Lul

[viscid thistle]

I'm here!

You see? I already have x-intercepts

Wait a minute

Na na, yeah, x intercepts

But idk how to get the y intercepts

[fluid shore]

Did you understand what i said above?

There's a recipe for getting the y-intercepts

[viscid thistle]

So plug 0 into the polynomials?

[fluid shore]

Let x = 0. Then, find p(0). Do you understand why that is what needs to be done to get the y-intercepts?

[viscid thistle]

This?

[fluid shore]

I mean yea, just substitute x = 0 into your polynomial

[fleet yew]

Yea

[atomic talon]

can anyone help me set this up? im so confused

[lilac pier]

@atomic talon Use the formula $a^{3} + b^{3} = (a+b)(a^{2} - ab + b^{2})$

[obsidian monolith]

Sup?:

[lilac pier]

Then factorize $a^{2} - ab + b^{2}$ so that you have two more linear terms. The previous is (a+b).

[obsidian monolith]

Sup?:

[willow bear]

that's unfactorable over the reals

[uncut mulch]

still considered linear right?

[willow bear]

[viscid thistle]

help @atomic talon

[odd dagger]

use a^3 + b^3 = (a+b)(a^2 - ab + b^2) @viscid thistle

then complete the square on a^2 - ab + b^2 and it's a difference of squares which will give you two complex linear factors

[proud gate]

I have a question about a khanacademy video if i could ask? Its in the precalculus section

[serene heath]

go ahead

[proud gate]

https://www.khanacademy.org/math/precalculus/x9e81a4f98389efdf:complex/x9e81a4f98389efdf:complex/x9e81a4f98389efdf:complex-mul-div-polar/v/exponential-form-to-find-complex-roots?modal=1

Well i cant explain the question very well since i have no idea what to take from the video

1 = 120 and 240 and 360 degrees??

I only ever heard of e this section and dont know anything much about it

[dim jungle]

Learn polar coordinates

And like the different ways to express it

[proud gate]

Thats what im trying.
I just have no idea how 1 which became e^(i2pi) kept changeing angles

[viscid thistle]

e^ix = cos(x) + i*sin(x)

So ix = cos(x) and iy = sin(x).

FOr example:

In [51]: np.angle(1+1j)/pi*180
Out[51]: 45.0```

[stuck lark]

if you're gonna use x for the real part, you might wanna call the angle smth else

[viscid thistle]

oh, you're right

that was silly

In [56]: np.angle(exp(1j*pi/4))/pi*180
Out[56]: 44.99999999999999```

Basically, adding k*2π will give you the same angle, because of this sin/cos expansion.

[proud gate]

@viscid thistle but then why in that video did the angle on the complex plane change from 0 to 8pi/12 to 16pi/12 24pi/12

It just kept getting 120 added every 2pi added

From 0

[viscid thistle]

Sorry, I don't know.

I don't know which values they used in the video.

But you should be able to verify all of them by interpreting the real portion as an x-value and the imaginary as the y-value, like a cartesian coordinate system.

Similar to how you would do on a Mandelbrot fractal, domain coloring, etc.

And 1+1i will have the same angle as 2+2i. They are all 45 degrees.

Or, imagine they are vectors.

In [61]: j=[(1+0.1j)**k for k in range(1000)]
In [62]: plot(real(j),imag(j));show()```

You see that they are basically rotations.

[viscid thistle]

I’m riding my bike at a constant speed of 20 miles per hour on flat land when I ride over a dot of fresh yellow paint, thereby getting a yellow spot on my front wheel, which is 2 feet in diameter. I continue riding my bike in a straight line at 20 miles per hour. When t seconds have passed since running over the dot, how many feet is the yellow spot on my tire from the yellow dot I ran over?

can someone explain this to me

i don't know what is happening at all

[viscid thistle]

<@&286206848099549185>

[viscid thistle]

any helpers here?

AA

[viscid thistle]

yo

[willow bear]

okay so

this may be best visualized by setting up a coordinate system

for convenience let's put the dot at the origin and have the x axis run parallel to the road and the y axis vertically

[viscid thistle]

seems reasonable

[willow bear]

so let R denote the radius of the wheel in feet and v denote its speed in feet per second

i want to not involve their values for a moment

[viscid thistle]

ok

[willow bear]

alright so we're gonna assume the wheel never slides and always just rolls

obv

oh and we're also gonna define t=0 to be the point in time when you rode over the yellow dot

so the center of the wheel starts out at (0, R) and simply moves horizontally forward at v

so after t seconds it's gonna be at (vt, R)

that make sense?

[viscid thistle]

what's v

[willow bear]

so let R denote the radius of the wheel in feet and v denote its speed in feet per second

[viscid thistle]

oh

yeah

ok

[willow bear]

notice that vt is also the length that the contact point of the wheel with the ground has traveled along its rim, if we look at the wheel's point of view

this may be not the best wording

hold on

[viscid thistle]

ok

[willow bear]

here's what i mean

the red straight line is the distance that the wheel's covered so far

and the red arc is where the point of contact has been on the wheel

[viscid thistle]

i think you got it the other way around

[willow bear]

with its topmost point being the yellow dot

[viscid thistle]

unless you mean the wheel is going backward

[willow bear]

no the wheel is going forward

[viscid thistle]

clockwise right

[willow bear]

the wheel is rotating clockwise in this scenario yes

my point is that the red line and the red arc have the same length

namely, vt

[viscid thistle]

ok i imagined it wrong at first

ok

got it

[willow bear]

yeah so

the wheel has rotated through an angle of θ := vt/R

does that make sense

[viscid thistle]

well

not too much

[willow bear]

the arc length is vt and the radius is R

[viscid thistle]

so

[willow bear]

obviously i'm measuring θ in radians

[viscid thistle]

wouldn't theta just be vt then

[willow bear]

no it wouldn't

that's only if the radius is 1

well

okay i guess R = 1 foot here but i didn't really pay much attention to that

[viscid thistle]

what if it wasn't 1 foot

[willow bear]

then θ = vt/R

the radian measure of an angle is by definition the ratio of its subtended arc to the radius

[viscid thistle]

okay

i get it

it's because i've only been dealing with radii of 1

i didn't get this at first

[willow bear]

yeah so the position of the yellow spot relative to the wheel's center is (-Rsin(θ), Rcos(θ))

of which you can convince yourself by looking at my pic again

and so its position relative to our origin (the yellow dot on the ground) would be
(vt - Rsin(vt/R), R - Rcos(vt/R))

and the distance from that to (0,0) is the answer to your problem

[viscid thistle]

why (-Rsin(θ), Rcos(θ)) now?

[willow bear]

ah my bad

it's (-R sin(θ), -R cos(θ))

[viscid thistle]

why??

[willow bear]

take a look at the picture

i'm on my phone rn but ig if you insist i can recreate it in more detail