precalculus

[leaden glade]

consider
[1 1 1]
[0 1 1]
[0 0 1]

you have to clean the previous ones up

[fringe fossil]

how would I go about solving for x for 6^{3x+1} = 2^{2x-3} ?

Its just mixing and matching log & exponent rules till its replaced with something where i can sub in x to equal 0 right?

[faint acorn]

Yes

If you take the log of both sides, the power drops to the front

And then you can rearrange it easily

[fringe fossil]

think I just figured it out, answer is -log48/(log54)

this feels ass backwards

[faint acorn]

=pup 6^(3x+1) = 2^(2x-3)

Rip mathbot

You're correct though

[granite stirrup]

Do more with Wolfram|Alpha pro

Query made by

@faint acorn

[fringe fossil]

can i represent this word problem with this 2800 = 1500 [ 1+ (r/12) ]^(12x6) ?

im getting r= 7/1620 but that seems bad

log(28/15) = log(1+(r/12))^72

log(28/15) = 72log(1+(r/12))

log(28/15) = log72 + log(72r/12)

log(28/15) = log72 + log(6r)

log(28/15) - log72 = log(6r)]

log(28/15 x 1/72) =log6r

log(28/1080) =log6r

log(7/270) = log6r

10^x = 7/270 , 10^x = 6r

so 7/270 = 6r which means 7/1620 = r , the annual rate of interest

[granite stirrup]

Wolfram|Alpha didn't send a result back.
Maybe your query was malformed?

Wolfram|Alpha didn't send a result back.
Maybe your query was malformed?

Wolfram|Alpha didn't send a result back.
Maybe your query was malformed?

Wolfram|Alpha didn't send a result back.
Maybe your query was malformed?

[hybrid wave]

someone is having a stroke

[hexed ermine]

Your writeup is correct

==2800/1500

[granite stirrup]

28/15 = 1.86666666666667

[hexed ermine]

==(28/15)^(1/72)

[granite stirrup]

15^(71/72)×2^(1/36)×7^(1/72)/15 = 1.00870649279038

[hexed ermine]

==.00870649279038*12

[granite stirrup]

1305973918557/12500000000000 = 0.10447791348456

[hexed ermine]

Roughly 10.4%

And yeah, your 4th line is where you messed up with logs

Idk what you did but it is not correct

I would've simply evaluated the LHS, divide by 72, then get into exponential form then some algebra

[fringe fossil]

thank you

yeah I was trying not to get a decimal while solving for some reason

[brave void]

Okay

I need some help

How on earth do I do this

[wind igloo]

Here's one method: Take the difference of two of the points to a third, then dot them together to find the angle between them.

[brave void]

to a third

?

[wind igloo]

If that angle is 0 (ie if a . b = |a||b|) then the points are collinear.

You have three points.

Pick two. Calculate the vector difference between each of them to the third point.

[brave void]

But what about that 3rd axis

[wind igloo]

What about it?

The dot product works fine in 3 dimensions.

[brave void]

I was never taught the dot product

That's the problem

[wind igloo]

Ah.

Ok. So start off the same way.

But then see if there is some scalar c such that c (P-R) = (Q-R)

If there is, then the points are collinear. If there is not, then they are not.

[brave void]

I don't mean to offend you

But I got no clue

do i just ignore Z @wind igloo

[wind igloo]

No.

What is it you think I'm suggesting you do? Because I'm a little confused why you keep mentioning a third dimension.

I mean, how would you calculate P-R ?

[brave void]

I don't know what to do with z

I don't know what P or R are

[wind igloo]

P is the point (-3, 2, 4) and R is (-3, 2, 0)

[brave void]

Okay

[wind igloo]

So what is P - R ?

[brave void]

(0,0,-4)

4*

(0,0,4)

[wind igloo]

Right.

And what is Q - R ?

[brave void]

(15,1,-1)

[wind igloo]

Is there any number that you can multiply P-R with to get Q-R ?

[brave void]

Uh

No?

[wind igloo]

Right. Therefore, the points are not collinear.

[brave void]

All of them?

[wind igloo]

You can always draw a line through any two points.

So it takes at least three for the set to be non-collinear.

[brave void]

what would be a case where you could multiply p - r to get q- r, cus a vector of <15,1,-5>

[wind igloo]

Then they would be collinear

[brave void]

like

I need an example

Because I don't know what multiplying coordinates

is

[wind igloo]

One second.

P(7, 13, 3): Q(31, 61,11) and R(13, 25, 5) should be collinear.

If I didn't mess something up.

[brave void]

okay

why

P - R would be

-6, -12, -2

q - r would be

18. 36, 6

You could multiply by -3

So they would be collinear?

[wind igloo]

Yes.

They are.

[faint peak]

Hey is anyone here i need some help

[eternal folio]

Just ask.

[faint peak]

im currently doing graphing with horizontal and vertical shifts on functions

and im starting to get lost on what a horizontal shift looks like on a y=sin and cosx graph

[fast crystal]

do this

when you have something like sin (3x +2) or whatever

solve 3x+2 = pi/2 and 3x+2 = 5pi/2

and those will be the points of minima and maxima

i mean both will be points of maxima

for cosine you have to use 0 and 2pi

@faint peak

if you come back i can explain a bit more if you dont im expliaining for nothing

[faint peak]

sorry

i was helping my parents with someone

something

[brave void]

I need some help

How to calculate?

Am I to assume this starts from 0?

[forest canopy]

Is 2*5^0.5 a possible answer ?

[brave void]

How what that

How

What

How

That is an answer

wtf

[forest canopy]

the magnitude is its length

so use length formula

[brave void]

okay

[forest canopy]

(2^2 + (-4)^2)^0.5

It makes 2*5^0.5

[brave void]

wpah

woah

I need a minute

@forest canopy What about

9?

[forest canopy]

3

[brave void]

oh yteah

forgot root

[eternal folio]

=pup sqrt((-3)^2 + 0^2)

[granite stirrup]

Do more with Wolfram|Alpha pro

Query made by

@eternal folio

[brave void]

An answer

But it says its wrong

I got the top to be 16

The bottom to be the two magnitudes

root 35 and root 22

Then I took the cosign

got .83

and it was wrong

[brave void]

<@&286206848099549185>

[forest canopy]

Is the real cos near 0.51 ?

[brave void]

what do you mean

[forest canopy]

you were wrong, so I want to know if what I found is the right answer

[wind igloo]

I think you want the arccos

[brave void]

where'd I go wrong

[forest canopy]

"Then I took the cosign
got .83
and it was wrong"

[brave void]

Those are my selections

[forest canopy]

I found 59.4

[brave void]

please tell me how

[wind igloo]

You did cos( a . b / (|a| * |b|) )?

Or arccos?

[brave void]

Yes

I got 16

For a . b

[forest canopy]

a . b = 16
|a| = 45^0.5
|b| = 22^0.5

[brave void]

aww

fck

I wrote 35 accidentally

thx man

So I multiplyed these

I got -1 as a . b

bottom values I got root 30 times root 29

.88 ain't the right answer tho

[wind igloo]

It's is only asking for the dot product

[brave void]

I'm not getting any of these questions right

And I don't know wtf I'm doing

It's always an error in some radical

or some other thing

a.b is -2

for that one

For the radicals I have

root 113

and root 94

but I get some weirdass answer

[wind igloo]

Those all look right to me

[brave void]

Then how do I evaluate

I took cosign of it

And it's wrong

[eternal folio]

$$ \cos(\theta) = \frac{10 \cdot 3 + 2 \cdot (-7) + 3 \cdot (-6)}{\sqrt{10^2 + 2^2 + 3^2} \cdot \sqrt{3^2 + (-7)^2 + (6)^2}} $$

[wind igloo]

== arccos(-2/(sqrt(113)*sqrt(94)))

[granite stirrup]

Parse error
On line 1 at position 20

arccos(-2/(sqrt(113)sqrt(94)))
                   ^

[wind igloo]

== arccos(-2/(sqrt(113)*sqrt(94)))

[granite stirrup]

Runtime error in iterm_2
On line 1 at position 1

arccos(-2/(sqrt(113)*sqrt(94)))
^
Failed to access variable arccos

[brave void]

== cos(-2/(sqrt(113)*sqrt(94)))

[granite stirrup]

cos(sqrt(10622)/5311) = 0.999811717450766

[brave void]

== arcos(-2/(sqrt(113)*sqrt(94)))

[granite stirrup]

Runtime error in iterm_4
On line 1 at position 1

arcos(-2/(sqrt(113)*sqrt(94)))
^
Failed to access variable arcos

[brave void]

I don't know

This is what I choose from

I've been on this pre quiz for over 2 hours, and I'm still remotely confused

[eternal folio]

91.1

arccos(-2 / (sqrt(113)*sqrt(94))) = 91.1119... degrees

[brave void]

Why do I get 1.5 then

@eternal folio Where did you plug that in

[wind igloo]

Is your calculator in Radians?

=pup convert 1.5 radians to degrees

[granite stirrup]

Do more with Wolfram|Alpha pro

Query made by

@wind igloo

[brave void]

nope

god dammit

[wind igloo]

Maybe a mistake in entry? Missing parentheses or something?

[eternal folio]

=pup convert arccos(-2 / (sqrt(113)*sqrt(94))) to degrees

[granite stirrup]

Do more with Wolfram|Alpha pro

Query made by

@eternal folio

[faint peak]

Does anyone here use a ti 84 plus calcullator

[hexed ermine]

I have a ti 84 plus ce

[faint peak]

im trying to figure out how to do graphs on trigometric functions on mine

because i keep running into the issue off figuring out how to graph tangents and its causing me to lose my mind

[late pond]

Anyone here good with oscillating waves that exponentially decay?

I'm supposed to do word problem -> equation but the textbook is shit and the professor ignores my messages

Correction: The textbook doesn't align with the homework

[fast crystal]

whats the problem

[late pond]

A spring is attached to the ceiling and pulled 10 cm down from equilibrium and released. The amplitude decreases by 11% each second. The spring oscillates 14 times each second. Find an equation for the distance, D the end of the spring is below equilibrium in terms of seconds, t.

also does this go here or in #help-1

[fast crystal]

nevermind idont know

[late pond]

oof

[fast crystal]

what does the answer say

[wind igloo]

It'll be something like e^{-kt}cos(wt)

Where w is related to the oscillation frequency and k is the decay constant.

So all you should need to do is calculate the appropriate values for k and w, based on the information in the question.

[late pond]

Don't have the answer

So w is 14/sec and decay would be like 0.89^t?

[wind igloo]

Something like that.

[late pond]

How would I express 14/s

14t?

Where t=seconds obv

[wind igloo]

What is the usual periodicity of cos?

[late pond]

2pi/b

[wind igloo]

So cos(t) oscillates every 2 pi seconds, it t is in seconds.

How would you make oscillate every 1/14 second?

[late pond]

2pi/b=1/14?

[late pond]

Okay my book wanted it in a really strange long-form that it didn't specify in the textbook in question

I think I got it now, thanks

[viscid thistle]

hi guys

[eternal folio]

Hi.

[viscid thistle]

damn carpla

I need better stance when I am spamming LoL

or I get this silly ache

[brave void]

Is there an easy way to do this

[eternal folio]

x1 = -4 - 2s
y1 = -19 - 4s
z1 = -16.5 - 3s

x2 = -13 - 4t
y2 = 1.5 + 3t
z2 = -12.5 - t

and you want

x1 = x2
y1 = y2
z1 = z2

@brave void

If you want the easy solution, just type it in Wolframalpha.

[brave void]

How can I solve this

By setting each one equal to each other?

I don't have wolfram alpha on my tests

[eternal folio]

-4 - 2s = -13 - 4t
-19 - 4s = 1.5 + 3t
-16.5 - 3s = -12.5 - t

and solve for s and t

[brave void]

then?

Or is that the intersection

[eternal folio]

If you have s and t that work in all three equations, then you have the intersection.

[brave void]

the lecture that this program gave me was ass, so I'm trying to find how to solve these equations easier

u clarified

ty

I can't find a single Youtube video on this

[eternal folio]

Three dimensional analytic geometry?

[brave void]

Yes

I rather just put it into wolf

Considering I have to do 10 of these in a row

And my time is limited

But there is no wolf calc on this

I don't know what to do when I find t and s

Like for this

I found t is 1

and s is 4

But what do I do

[eternal folio]

Check if they satisfy all three equations. And then to get the intersection point, you just substitute s or t into one of the two groups of three equations to get x1, y1 and z1.

Those are the coordinates of the intersection.

[brave void]

I still don't get how you're getting numbers

I'm just getting 1=1

for the first equation

Where are you generating coordinates

[eternal folio]

You know that x1 = 0 + 2s, if you know s, just plug that in.

[brave void]

Where did you get

0 + 2s

[eternal folio]

r1 = 0i + 3j - 3k + s(2i + j + 4k) => r1 = i(0 + 2s) + j(3 + s) + k(-3 + 4s), and because the i component is basically the x-coordinate, x1 = 0 + 2s.

And similarly for all the others.

[brave void]

well shit

Because x1 doesnt reflect on any of my answers

I got x1 =8

But my only possible x1s would be 0 or 4

[eternal folio]

You could also do the following:

when they intersect, then

r1 = r2
0i + 3j - 3k + s(2i + j + 4k) = i - 2j + 4k + t(i - 3j - k)
i(0 + 2s) + j(3 + s) + k(-3 + 4s) = i(1 + t) + j(-2 + 3t) + k(4 - t)

and then you match the coefficients of 'i', 'j' and 'k'

0 + 2s = 1 + t
3 + s = -2 - 3t
-3 + 4s = 4 - t

and then you solve that, check if 's' and 't' are actually solutions, then plug 's' in r1, or 't' in r2 to get the intersection

[brave void]

Do you have the like to wolf alpha?

I can't do it.

[eternal folio]

Nah, what can't you do?

[brave void]

Everything

I can't get any things they want me to get

I can't get their fucking coordinates

I don't know

And I'm genuinely confused

[eternal folio]

Take a break, breathe some fresh air, drink a glass of lemon juice, watch a funny video, and come back after 15 minutes.

[brave void]

k

okay

Are we trying to find the values of i j and k

[eternal folio]

No, they are our basis vectors.

We are trying to find s and t such that r1 = r2.

[brave void]

okay

okay

So I want to plug in a t value

I would do it

In r2

r2 would be

2i - j + 3k

I plug s in r1

I get r1= 8i+7j+13k

what do I do from here

I have r1 coordinates (8,7,13) and r2 as (2,-1,3)

Uno

@eternal folio

Are they skew?

[eternal folio]

What value did you plug into r2?

[brave void]

1

[eternal folio]

Hmmm... something must have gone wrong while solving the equations. At least two of the three coordinates have to match.

[brave void]

:/

Can I please have the link to wolf

[eternal folio]

=pup solve 0 + 2s = 1 + t, 3 + s = -2 - 3t, -3 + 4s = 4 - t

[granite stirrup]

Do more with Wolfram|Alpha pro

Query made by

@eternal folio

[eternal folio]

It seems there is no intersection

[brave void]

😫

=pup solve 3+2s=7-2t, -2+5s= 8+t, 1-s=-1+2t

[granite stirrup]

Do more with Wolfram|Alpha pro

Query made by

@brave void

[brave void]

This is a god machine

With s and t, I just plug those in

And the coordinates I get

If they both match

I have an intersect?

[eternal folio]

Yes.

[brave void]

YESSSS

why does math have to be like this 😭

When they add letters Im out

=pup solve 2+8s=3+8t, 6-8s=5-3t, 10s= 6+6t

[granite stirrup]

Do more with Wolfram|Alpha pro

Query made by

@brave void

[brave void]

@eternal folio Ty for help out this math weeb

[eternal folio]

No problem.

[brave void]

=pup solve 2+8s=3+8t, 6-8s=5-3t, 10s= 6+6t

[granite stirrup]

Do more with Wolfram|Alpha pro

Query made by

@brave void

[brave void]

=pup solve -4-2s= -13-4t, -19-4s=1.5+3t, -16.5-3s=-12.5-t

[granite stirrup]

Do more with Wolfram|Alpha pro

Query made by

@brave void

[brave void]

=pup solve (-5/2)(-2i-4j-3k)

[granite stirrup]

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@brave void

[eternal folio]

There is nothing to solve for in that equation.

=pup (-5/2)(-2i-4j-3k)

[granite stirrup]

Do more with Wolfram|Alpha pro

Query made by

@eternal folio

[eternal folio]

And there you have the coordinates.

[brave void]

aye

[eternal folio]

WolframAlpha is a very useful tool, but be aware that you need to practice these problems if you don't want to fail.

[brave void]

There are usually 1 of these

On a huge test I take

doing 10 of them in a row is unnecessary

But as long as I can replicate the concept onto the test

I think I'm gucci

=pup solve 2+4s= 2+2t, 3=3+2t, 1-s=1+t

[granite stirrup]

Do more with Wolfram|Alpha pro

Query made by

@brave void

[brave void]

=pup solve 2+8s= 3+8t, 6-8s=5-3t, 10s=6+t

[granite stirrup]

Do more with Wolfram|Alpha pro

Query made by

@brave void

[viscid thistle]

.

[brave void]

3?

[hexed ermine]

Yep

[brave void]

I'm reviewing pretest answers

Would this be D

<@&286206848099549185>

Ultra rip

halp

[hexed ermine]

Not vertical

[brave void]

Horizontal

By 1/2

[hexed ermine]

Consider 2x^2 and x^2

yes

[brave void]

ty m8

How do it ell

If horizontal

Or vertical

[hexed ermine]

wait I think I got it backwards

a*f(x) = Vertical Shrink or Stretch

f(a*x) = Horizontal Shrink or Stretch

[slender river]

yes

if it's a function on x it affects the domain

if it's on the y (otherwise outside of the function at least) then its vertical

[glass harbor]

Not sure if this is the right section but how would I go about this, I got x^2 6 to the power root of 2yz

I checked Photomath and it’s a whole different answer

[true vigil]

it's the same

$$\sqrt[6]{x^{15}} = x^2 \sqrt[6]{x^3} $$

[glass harbor]

Would this be correct as well?

I’ve checked different sites to see if my answer was right, but there all different

[true vigil]

sure

[terse prism]

can someone tell me how to find whether the vectors v=3i-j and w=6i-2j are parallel, orthogonal, or neither?

[hexed ermine]

Vector dot product

Or just multiply v by 2 and notice the similarity

[fringe fossil]

im stuck on this trig identiy question

"determine exact value of tan2x when you know cosx= -12/13 and pi =< 0 =< 3pi/2 without calculating the value of x"

i narrowed it down to tan2x = 2SinxCosx / cos^2x - sin^2x, but im not sure what to do from this point on.

cant seem to find a way to get rid of sinx to just have the cosx's

any help or hints would be appreciated, thanks

[fringe fossil]

im retarded 120/119

[limpid plover]

$$tan2A= \pm \sqrt{\frac{1-cosA}{1+cosA}}$$

[granite stirrup]

Support the bot on Patreon: https://www.patreon.com/dxsmiley

[limpid plover]

Lemme confirm though

My bad

It's the other way round

$$tanA= \pm \sqrt{\frac{1-cos2A}{1+cos2A}}$$

[limpid plover]

You can find the perpendicular using Pythagoras theorem

Let $$b=12$$ and $$h=13$$ where b is the base and h is the hypotenuse

[limpid plover]

$$\sqrt{13^2-12^2}=5$$

[limpid plover]

Hence, $$sinA=\pm\frac{5}{13}$$

[limpid plover]

Just find if it is + or - depending on the angle

@fringe fossil

[fringe fossil]

what do you mean?

[lost cipher]

i think he means to check the quadrant

which quadrant the angle is in

0~180 is positive sine and 180~360 is negative sine

(look at the y-coordinate of a unit circle, it tells you the sign of sine, x-coordinate for the sign of cosine)

[fringe fossil]

but doesn't it have to be in the last quadrant cause the question says its between 3pi/2 and 2pi (270 degrees and 360) ?

[lost cipher]

so that means it's negative

[fringe fossil]

i did the the thing and i got positive 120/119

since its cos and not sin

i think

[lost cipher]

yeah if it's cosine it's positive

although it shouldn't go above 1?

oh oops didn't see the original q

[fringe fossil]

im not sure why @limpid plover brought up sinx, i thought it was just tanx= 5/12 (cause of the pythagorean thing), and then input that into tan2x= 2tanx/(1-tan^2x)

where i got 120/119

since tanx is just opp/adjacent and i know the other values from the given equation cosx = - 12/13

[lost cipher]

is that not the right answer?

(dunno just asking)

[fringe fossil]

i am not sure

[lost cipher]

seems right to me

we can check by finding the actual angle

angle is approx. 3.54 radians

tan (2x) = 1.02309775859

(using x = 3.54 radians)

[fringe fossil]

oh

[lost cipher]

120/119 = 1.0084033613

so it's probably right?

more accurately tan(2x) = 1.00700594315 (a bit off since i couldn't find exact values for arccos)

so im pretty sure your answer is right

[viscid thistle]

Is there a way to algebraically solve this?
Find the coordinates of all points whose distance from (1, 0) is sqrt(10) and whose
distance from (5, 4) is sqrt(10).

I solved it by graphing but I want to be able to do it without graphing.

[lost cipher]

im thinking to set up two equations with two variables and solve

using pythagorean theorem

[viscid thistle]

or maybe using the general form?

[lost cipher]

oh wait

[viscid thistle]

x2 + y2 + Dx + Ey + F = 0

[lost cipher]

are these two different questions?

[viscid thistle]

No, it's just one

[lost cipher]

i think you would be using the circle formula

[viscid thistle]

(x - h)^2 + (y - k)^2 = r^2?

[lost cipher]

yeah

[viscid thistle]

I'll give it a go

Cause I hate graphing lol

[lost cipher]

https://math.stackexchange.com/questions/256100/how-can-i-find-the-points-at-which-two-circles-intersect

[fringe fossil]

😌

[limpid plover]

@fringe fossil The only reason I brought up sinx was i narrowed it down to tan2x = 2SinxCosx / cos^2x - sin^2x, but im not sure what to do from this point on. and cant seem to find a way to get rid of sinx to just have the cosx's

[hexed ermine]

What's the original equation?

Because those both are sin and cosine double angle formulas, idk if youve simplified it to that point or what

[fringe fossil]

oh ok duckinator i understand

@hexed ermine original question is "determine exact value of tan2x when you know cosx= -12/13 and pi =< 0 =< 3pi/2 without calculating the value of x"

[hexed ermine]

Hmmm okay

So its located in quadrant III, so tan is positive

Use the Pythagorean identity to find your sin value and then use the double angle formula for tangent

I believe its 2tan(a) / 1-tan^2a

And then just sub tan for sin/cos and use some algebra to find the exact value of tan(2a)

[faint peak]

Do ln cause a logirithmic graph to flip if not what does

ive been trying to do domain and had no hope

[lean ocean]

aye can i have help

with that

[patent beacon]

They will collide if their x's and y's match for any t

[fringe fossil]

i have question

for f(f^-1(5))

how would i evaluate it, if im given f(x) = 9x+3

am i putting in the y value or the x value from the inverse function back into f(x)

f(y)=3(5)+1 or 5 = 3y+1 ?

is it just 5

[unreal relic]

The function and it's inverse cancel each other out when made into a compound function

yea

f • f(^-1) = 1

[fringe fossil]

i see

thank u

[raven basin]

Hello. i need help in this. Find a mathematical model that represents the statement. (Determine the constant of proportionality.)
v varies jointly as p and q and inversely as the square of s.
(v = 3.5 when p = 3.7, q = 6.3 and s = 1.2.) my answer is v=.216pq/s^2 and it wrong

[unreal relic]

its asking you for the constant of proportionality, no? not the equation with the constant put back in

[silent wraith]

I has question:
Solve the following equation for the exact value of x:
(pi^2)/6=pi*tan^-1(x/pi)
I really don't know where to start...

[wind igloo]

=tex \frac{\pi^2}{6} = \pi\cdot\arctan{\frac{x}{\pi}}

[wind igloo]

I recommend starting by dividing both side by pi.

[unreal relic]

divide both sides by pi, use tan() on each side, multiply both sides by pi

[silent wraith]

ok thanks!

[winged abyss]

Not sure if this is the right category, but been out of math a few years and trying to work on something. The derivative of x^3 is 3x^2, right? And derivatives are the slope of a function, right? So, why does x^3 not increase incrementally according to 3x^2?

[lost cipher]

it should

[winged abyss]

1,8,27,64,125...

[lost cipher]

so at x=1, it has a slope of 3

oh i see what you mean

it's because derivative talks about the instantaneous rate of change

[winged abyss]

that's 7, 19,37,61,91

[lost cipher]

which is the rate of change at a specific point

you're talking about the change between two points like 1 and 8 or 8 and 27

[winged abyss]

yeah

I don't remember much about the slope of non-linear graphs

[lost cipher]

derivatives talk about change between 1 and 1.000000001

(the idea is like that)

more specifically it talks about the change at a point, not between points

[winged abyss]

how could I figure out the growth between two points? I just bruteforced 3x^2+3x+1 but don't know how to get there

[lost cipher]

where did you get the 3x+1?

[winged abyss]

just fit.

actually, thinking about this, I might be able to do it with y(x+1)-y(x) or something... not sure if that makes sense

[lost cipher]

what do you mean by growth between two points?

[winged abyss]

but the difference between sequential points of y=x^3 matches 3x^2+3x+1

like I said, 1,8,27,64...

the difference is 7, 19, 37, 61...

matches 3x^2+3x+1

[lost cipher]

true

you're right

however, that's different from derivatives

[past jay]

What do you mean by "sequential"?

[lost cipher]

you're talking about the growth from 1 to 8

he's talking about a sequence that follows n^3

[winged abyss]

by sequential I mean increasing by 1

[lost cipher]

where n is an integer

[winged abyss]

yeah

[past jay]

ah.

[lost cipher]

but derivatives usually deal with domain R, meaning x can be any real number, like a decimal

so you're talking about what happens to y when x increases by 1

[winged abyss]

yeah

I kept telling myself, "This is slope, right?" but I guess it's more complicated with non-linear equations

[lost cipher]

but derivatives talks about what happens to y when x increases by a really small number

(like 0.00000000000001, if you want to imagine it in your head)

it is a slope

but not between two points

it's at a single point

(but you can imagine that it's between two really close points)

(idea is to get two points so close that it becomes one point)

[grizzled hull]

Just popped in; I have no context. Why is talk of derivatives in the precalculus channel?

[slender river]

dipping feet in a pool

[grizzled hull]

Ah, well, nothing wrong with exploration. As you were!

[lost cipher]

dunno lol

[slender river]

yeah lowkey that's a really good question

[winged abyss]

I don't know what "pre-calc" means I guess

[slender river]

i mean it's okay

the concept of the difference quotient is pre cal

[grizzled hull]

Usually, it's a grouping of high school algebra concepts, and trigonometry.

[past jay]

The way you get to it,

$$\Delta y = y(x + \Delta x) - y(x) = (x + \Delta x)^3 - x^3$$

[past jay]

In this case, since you're using integers, Delta x is 1.

[winged abyss]

where should I have put this, for future reference? Calc-analysis?

[past jay]

You can replace it with h, if you want.

I mean with you not knowing it was calculus I can't see how you could be faulted

and technically it's more like juuuust before calculus.

[winged abyss]

that was my thinking

[grizzled hull]

They can't be, really. It's at the point where derivatives were brought in that I raised an eyebrow.

[winged abyss]

I think I had the right idea, then, thanks

[past jay]

You basically would just expand and simplify

$$(x+\Delta x)^3 - x^3 = x^3 + 3x^2\Delta x + 3x\Delta x^2 + \Delta x^3 - x^3$$

[past jay]

$$\Delta y = \Delta x^3 + 3x\Delta x^2 + 3x^2\Delta x$$

[past jay]

And since you were using only integer x, that would mean Delta x, the difference between x values, is 1.

So it simplifies to just 3x^2 + 3x + 1

which is what you got

[winged abyss]

gonna try x^4 on my own. Thanks a lot

[grizzled hull]

I looked more closely at what all is being done. This discussion is usually handled rigorously for the first time in Calculus II in the American system. This looks more or less like an application of the finite method of differences being mapped to polynomials. If that isn't what's going on, see also the finite method of differences for identifying polynomial representations of sequences.

[winged abyss]

Looking up finite method of differences, some of it makes sense, some of it looks like Calc III (Taylor Polynomials) and Diff Eq which is where I hit a brick wall. Actually one of the first videos that comes up is one of my old professors. But, the method SpiderString showed made sense to me

[past jay]

Wait, you know calc?

Then, why did you act like you didn't know what a derivative is...?

[lost cipher]

he's going back over the content that he did years ago

[past jay]

uh... ok. It's just he acted like he didn't know a derivative was based on the slope formula

[fringe fossil]

hey, quick question, is (f(g(x)) the same as f(g(x)) ?

Cant tell if this is some kinda of notation im unfamiliar with, or just inconsistent formatting

[lost cipher]

yeah

outer brackets can be removed

[fringe fossil]

sweet

[lost cipher]

unless it's h(f(g(x)))

[fringe fossil]

Thanks! also, another quick question lol, are brackets around pi/6 x t implied here or is this cos(pi/6) x t ?

cause if it was implied why wouldnt the t be with the pi yaknow? instead of outside

[lost cipher]

hmm

i would guess it's cos(pi/6 * t)

but not sure

[faint acorn]

pretty sure it is

[lost cipher]

might know if we know what the formula is for

[faint acorn]

i assume t is for time

and its like some oscillation or whatever

[lost cipher]

oh true, if t is outside, they can just simplify it

[fringe fossil]

the function models the population of deer in a park, and "scientists recently discovered that the deer pop. is decreasing by 25 deer each month" so im supposed to make a new function

i guess it would be weird if a population was a trig function

[faint acorn]

wtf

If its decreasing why is it a trig function

[lost cipher]

indeed, something's up

[faint acorn]

Should be an exponential if anything

[fringe fossil]

this is the complete question

[lost cipher]

makes sense

[faint acorn]

yeah makes sense now

[fringe fossil]

it doesnt say!

the previous unrelated question (thats in the same section) goes by hours

[faint acorn]

probably month since the question says 25 deer per month

so your new function would be in terms of months

[fringe fossil]

not sure if i should email my prof over this or accept the 50/50 chance of losing marks here lol

[lost cipher]

email

nothing to lose

[fringe fossil]

assignment due in 6 hours so 🤞 he wakes up early lol

[lost cipher]

...

rough times

[fringe fossil]

if im asked for the average rate of change over the first 10 days, does that mean i should use an x value of 1 or 0 along with the 10?

like f(10) - f(0)/10-0 or f(10) - f(1)/10-1 ? are 0 days part of "the first 10 days"? is 0 a first day?

[lost cipher]

i would say 10-0

since f(0) refers to the beginning of day 1

and f(10) is the end of day 10

[past jay]

Yeah that's... Normally they specify "with t=0 being day 1" or something...

[raven basin]

A small theater has a seating capacity of 2000. When the ticket price is $15, attendance is 1500. For each $1 decrease in price, attendance increases by 100.
(a) Write the revenue R of the theater as a function of ticket price x.

I wrote -100x^2+22500 and its wrong plz help

[gritty blaze]

Maybe: $$y=100(15 - x)+1500$$.

[patent beacon]

Let the price of the tickets be x.

Then the number of people in attendance is
100(15 - x) + 1500
= 1500 - 100x + 1500
= 3000 - 100x

The revenue is the cost of the tickets × the number of tickets sold:
R = x(3000 - 100x)
R = 3000x - 100x²

[raven basin]

ohh I understand now thanks

[patent beacon]

Np, feel free to ask if you have anything else

[waxen sedge]

need help plz

[slender river]

owo

[patent beacon]

2sinθ - 1 = 0
sinθ = 1/2

θ = π/6 + 2πk, 5π/6 + 2πk

[limpid plover]

$$\theta= n\pi+(-1)^n \frac{\pi}{6}$$

[rocky nexus]

[limpid plover]

[fringe fossil]

For a rational function like f(x)= x^2 / (x-4)(x+3)(x-1), why does it have x and y intercepts if 0/0 is undefined?

[eternal folio]

It has a y-intercept because f(0) is defined, and it has an x-intercept because f(x) = 0 for some x values.

[fringe fossil]

but f(0) is 0/0, and if 0/0 is undefined, then f(0) is undefined?

[viral lynx]

f(0) = 0^2 / (0-4)(0+3)(0-1) ≠ 0/0

but yeah if f(0) were something/0 then it wouldnt have a y intercept; f(0) would be undefined

in this case it's 0/12 though which is just 0

[fringe fossil]

oh whoops missed that, so f(x) = x^2/x would be undefined?

[eternal folio]

Then f(0) would be undefined.

[viral lynx]

yea it'd look like a y=x graph but have a hole where it's undefined at (0, 0) so it wouldnt have an x or y intercept

[fringe fossil]

oh i see, thank you

[hexed ermine]

Yeah x^2/x you can cancel out a factor of x, given you a f(x) = x but you must still account for discontinuity at x=0, since that's where its undefined

It'll create a hole rather than a vertical asymptote

[fringe fossil]

another question if its ok

apparently the correct answer is c, but i dont get why it says x=/=4 instead of x>4

how can the graph be decreasing at an interval of x=/=4, what does that mean?

[eternal folio]

There are two intervals, over which this function is decreasing. 1<x<4 and 4<x.

Combinig that gives 1<x where x<>4, because the function is not defined at x=4.

@fringe fossil

[fringe fossil]

ohhhh when x is greater than 1 at every rational number except 4

thank u

[eternal folio]

Yes. Exactly.

[proud raven]

im probably stupid but is writing y=x²+1 in polar/in terms of r easy in some way

[patent beacon]

@proud raven
rsinθ = r²cos²θ + 1
r²cos²θ - rsinθ + 1 = 0

By quadratic formula:
r = [sinθ ± √(1 - 5cos²θ)] / 2cos²θ

[proud raven]

@patent beacon thanks!

[proud raven]

Weird, in the answers my teacher had it as not a function

but then it is?

not that it matters

[patent beacon]

@proud raven
It isn't a function. That ± means there's two r for certain θ, which is a no-no

[rare flame]

"Quadratic Functions" You have a wire that is 56 cm long. You wish to cut it into two pieces. One piece will be bent into the shape of a square. The other piece will be bent into the shape of a circle. Let A represent the total area enclosed by the square and the circle. What is the circumference of the circle when A is minimum.

[stray wave]

So let L be the length of the wire

[reef dirge]

don't

[stray wave]

You’re splitting it into two pieces so their lengths would be x and L-x

[reef dirge]

answer

duplicate post

#help-3

[stray wave]

O

Sorry

[reef dirge]

don't post in multiple channels @rare flame

[raven basin]

i need help i dont know what it wants me to put inside

[patent beacon]

I'm guessing the linear asymptote doesn't count?

It wants the vertical and horizontal ones

[raven basin]

but this a slant asymptote

[patent beacon]

If you just put x = -2, what does it do?

[raven basin]

ohh wow it worked

thanks man

[patent beacon]

You're right, that slant asymptote exists. I guess it just didn't want it

[raven basin]

thanks for the clarifcation

[ebon leaf]

anyone wanna help with some quick maffs? its easy but i just forgot how to do it since the last time i did so

if you're willing to help please pm me

[tall granite]

[ebon leaf]

sorry for the late reply, people were talking to me

are you ok with pm's?

[tall granite]

No

Just ask here

[glass venture]

nothing bad will happen if you just ask here, more people may be able to help lol

[tall granite]

Worst thing that can happen is someone makes fun of you and your questions

So no worries dude :)

[dusk kettle]

try changing basis to x and y

then A is just

[severe verge]

it's like Fibonacci sequence ~

but changed slightly

[dusk kettle]

01
12``` I think

[severe verge]

yep

[dusk kettle]

then A^2 is 12 25 and A^4 is 5 12 12 29 so A^5 is 12 29 29 70

So A^5x = [12 29]

[ebon leaf]

How would i be able to graph a sine function from the equation? Ex: y=-2sin(2/3x)

[slender river]

note the amplitude and the period

since there is no term added to the end there is no vertical translation

i like to imagine the 2/3 inside the parens of the sin function as another kind of function of x that affects the "speed" of x

this might be a little confusing at first but you can imagine that the sin function of yours is 2/3rds as fast

[ebon leaf]

ok

i got that part

[slender river]

so the period or total x length of 1 cycle (?) would be 3/2 times one normal period (2pi)

anyway

ye

the amplitude

there's this 2 in front of the entire sin bit

so that means the max and min are 2 and -2 from the centre (which is 0)

but it's negative so the function is flipped around the x axis

so instead of initially going up it starts by going down

[ebon leaf]

yeah

thank you!

[slender river]

np

[slate niche]

yay

im totally using this server for advanced pre calc

if i need

[rocky bison]

ok

thank you!

[olive star]

help pls

[violet bane]

the answer is that the top-right entry and the top-left entry have to equal 0

[olive star]

wutt

i have to explain how i got to the answer lol

[violet bane]

the way i did it was to just create two general matrices (matrices with unknown entries) and found a condition that ensured that it was the same regardless of which matrix is first

[olive star]

so a b c d for A and e f g h for B

[violet bane]

yeah

from there it's somewhat tedious, and you kind of just need to make an educated guess what the answer is, but it's not especially hard, i would say

[viral lynx]

SO(2) is abelian though

if your matrices are of the form
[cos(θ) -sin(θ)]
[sin(θ) cos(θ)]
they'll also commute

and diagonal matrices also commute with each other (there your condition of "the top-right entry and the top-left entry have to equal 0" doesnt hold)

[vast crater]

Please help meeee!

Im having a problem in conic sectionssss

[tall granite]

Just

Post

Your

Question

🙂

[limpid plover]

Just posting your question is really hard

It's impossible without asking if you can ask something

Almost impossible

[reef dirge]

hey can I ask if I can ask if I can ask something?

[limpid plover]

Yes, you can ask

Probably

Lemme check

Rip

Sorry bro

The answer is No.

No, you cannot ask if you can ask if you can ask something

That results in a ban

[silent wraith]

(23/144)(x^2)+(13√3/72)(xy)−(1/48)(y^2)=1
Solve for Vertices:

I rotated and then solved for the vertices but I don't see my answers among any of the correct choices, so I am wondering if I should solve for the x-prime,y-prime coordinates of the vertices.

Pls help

[frosty thicket]

Hi guys would you please help me here?

[worthy estuary]

What are you stuck with?

Plug in the x values into your function

[frosty thicket]

is this what you mean?

what could be the the limit? - infinity or 0?

[worthy estuary]

What do you notice about the numbers?

[granite folio]

Taylor expand to get answer
guesses can be misleading

also you function values do not seen to be correct

[viscid thistle]

l hooooopital

[reef dirge]

I hope it all too

[limpid plover]

I hope it all three

[torn nebula]

where did the n come from

I don't understand the 2n+1/2 (pi)

[spring thunder]

do you agree that the solutions of cosx=0 on [0,2pi] are

$$x=\frac{\pi}{2}$$ or $$x=\frac{3\pi}{2}$$

[torn nebula]

yes

cosine is 0 when it is at pi/2 and 3pi/2

[spring thunder]

yeah ok

so now with the trig circle

when you do one turn around it (so 2pi rad)

you get to the same angle

[torn nebula]

yes

[spring thunder]

so $$cos(x+2\pi)=cos(x)$$ (=0 in this case)

[torn nebula]

I'm confused aout cos(x+2pi)

[spring thunder]

you have the same angle (in principal measure)

so you have the same cosine right?

[torn nebula]

ohhh is it because the value repeats every 2pi

[spring thunder]

yeah

so now you can rewrite the solutions as

$$x=\frac{\pi}{2}+2k\pi$$ or $$x=\frac{3\pi}{2}+2k\pi; k \in \mathbb{Z}$$

[spring thunder]

this just means we can add any number of times 2pi to each solution

and it will still be a solution (cause we're just doing whole turns around the trig circle)

[torn nebula]

ohh ok I get it tysm

[spring thunder]

^^

[torn nebula]

in the picture I sent how did it become 2n+1/2 pi

[spring thunder]

i thought you got it lol

let me continue then

[torn nebula]

I get the pi/2+2kpi

but I just confused in picture haha

[spring thunder]

yeah ok

let's just fix k for a second

[torn nebula]

ok

[spring thunder]

to pass from pi/2 + 2k*pi to 3pi/2+2kpi you add pi right?

[torn nebula]

yes

[spring thunder]

yeah and 3pi/2+2kpi you add pi again

you get 5pi/2+2kpi

=pi/2 +2pi + 2kpi

so it's still cyclic

but of period pi this time (and not 2pi)

because you can just add (or remove) pi to get to the next solution each time

[torn nebula]

I understand the cyclic part

I just don't know how those 2 equations turned into this

[spring thunder]

yeah 2mins

[torn nebula]

or maybe I'm just dumb ahaha

[spring thunder]

i'm arriving to it

[torn nebula]

okay ty

[spring thunder]

so in fact

$$x=\frac{\pi}{2}+2k\pi$$ or $$x=\frac{3\pi}{2}+2k\pi; k \in \mathbb{Z}$$

[spring thunder]

becomes

$$x=\frac{\pi}{2}+k\pi; k \in \mathbb{Z}$$

[spring thunder]

(that's the new cycle we get from combining the 2 previous solutions)

[torn nebula]

because cosine is 0. every pi

[spring thunder]

yeah

now we factor by pi

$$x=\pi(\frac{1}{2}+k) k \in \mathbb{Z}$$

[torn nebula]

ohhhhhhh I get it

thank u

its just a different form

😄

[spring thunder]

yeah exactly

^^

[torn nebula]

if sinx=squaroot of 1/2

shouldn't it be pi /6 and 5pi/6?

[spring thunder]

that's if sin x = 1/2

sin x = sqrt(2)/2 <=> x= pi/3 or x=2*pi/3 (on [0,2pi])

[torn nebula]

ohhh my bad

ty

ohhh its equals to sqt 2/ 2

ah tysm I should really think more

also when we do these kind of problems like finding x do we only use the top half of the circle

only quad 1 and 2?

or we use the whole circle

for tanx=0 it only shows 1 solution when 2pi could also be a solution?

wait never mind its 0 every pi

[spring thunder]

the sols are combined again^^

it's like x=2npi and x=2npi+pi, n in Z

so ...,-pi,0, pi, 2pi,3pi,...

[dark crane]

a matrix * its transpose is the same as multiplying all its columns together, right?

because I have a function that checks if the matrix * its transpose = identity. so I can just check if every column * itself = 1, and every column * every OTHER column = 0, right?

[tall granite]

yeah

in other words you need to check that all the column vectors are orthonormal

[bold hatch]

is it column vectors or row vectors?

[tall granite]

oh

[dark crane]

column

no, the opposite

[tall granite]

transpose * matrix would be column vectors

[dark crane]

but in any case, is that correct?

[tall granite]

matrix * transpose would be row vectors

[dark crane]

Transpose(M)*M = I

is teh requirement

so it's column vectors right?

[tall granite]

then it's column vectors, yeah

[dark crane]

so I can just check if every column * self is 1, and * every other column is 0, right?

[tall granite]

bc in matrix multiplication youre multiplying rows of the 1st matrix with columns of the 2nd matrix

yes

that is correct

[dark crane]

no need to do any crazy transpose stuff

    bool IsUnitary() const {
        for (int col1Idx = 0; col1Idx < 4; ++col1Idx) {
            // 1. If any column is the zero vector, it cannot produce the identity, so short circuit.
            if (isZeroVector(_matrix[col1Idx]))
                return false;

            // 2. The diagonal must consist of 1s, so check dot product with self
            if (dotProduct(_matrix[col1Idx], _matrix[col1Idx]) != 1.f)
                return false;

            // 3. Check if dotproduct of all NON-diagonal row/columns is equal to 0
            for (int col2Idx = 0; col2Idx < 4; col2Idx++) {

                // We already checked the diagonals.
                if (col2Idx == col1Idx)
                    continue;

                if (dotProduct(_matrix[col1Idx], _matrix[col2Idx]) != 0.f)
                    return false;
            }
        }
        return true;
    }```

does this look right?

notice I save some work by first checking if it is zero vector, bc then it would automatically fail

[gritty blaze]

Is the matrix the identity matrix ?

[dark crane]

what do you mean?

thats what this function is meant to check

so theoretically this should work, right?

[gritty blaze]

Is _matrix the identity matrix ?

[dark crane]

no, _matrix is just the local data of the class

it could be anything

so this function is meant to check whether or not _matrix * TRANSPOSE(_matrix) = identity

[dark crane]

btw, is a matrix of all 0s orthonormal?

"A matrix with orthonormal bases is a matrix in which result of the dot
product between any two columns is either one or zero. "

[wind igloo]

That definition should -probably- include the word "nonzero" in it.

[dark crane]

rhx

[wide yoke]

Hey could I get help on onne of my questions?

I forget how to get the secant and cosine etc of trig

[eternal folio]

Just ask.

[wide yoke]

Okay sorry

I know cos=x and sin=y but how do I get csc cot sec and tan from cos and sin?

[slender river]

inverse

[eternal folio]

tan = sin/cos
cot = 1/tan = cos/sin
sec = 1/cos
csc = 1/sin

[wide yoke]

Okay thank you I just didn't have my regular trig sheet with me. Thank you so much

[eternal folio]

You'll find this page really helpful: https://en.wikipedia.org/wiki/List_of_trigonometric_identities . @wide yoke

[wide yoke]

thank you once again @eternal folio

[cedar chasm]

hey im learning a bit of pre calc before school starts

and i have been reading on the unit circle

but I just dont get whats the point of the unit circle

it has a radius of 1 and stuff

and if u move from it, y cordinate is sine and x is cosine and stuff

tangent is sin/cos

but I just dont get whats the point

is there something big im missing?

[eternal folio]

The unit circle is used to define the sine and cosine for all real values.

If you take a point on the unit circle, then its coordinates can be expressed as cos(t) and sin(t).

So the coordinates of a point on the unit circle are the cosine and sine values of a particular angle.

[past jay]

I mean... I'd argue that triangles are used to define trig functions and they just have a relationship to the unit circle but.... >.>

But yeah, without going into too much detail,

You can right the formula for a circle as either
x^2 + y^2 = r^2
or
sin^2(theta) + cos^2(theta) = 1

Using that fact, you can solve for the coordinates of any point on the unit circle and find the trig value at that point because a circle has that relationship. You could also just view it as drawing a triangle inside the unit circle.

As for why the radius is 1? Just so it's simpler and you don't have to change denominators.

[olive meteor]

Guys I need help on this word problem

A real estate office handles an apartment complex with 50 units. when the rent per unit is $580 per month. All 50 units are occupied. However when the rent is $625 per month. The average number of occupied units drops to 47. Assume that the relationship between the monthly rent p and the demand x is linear

Write the equation of the line giving the demand x in terms of the rent P.

[patent beacon]

@olive meteor
There exists an equation:
x = aP + b
For some a, and b.

You also know that (580,50) and (625,47) are points on your line. You can write,
50 = 580a + b
47 = 625a + b

[olive meteor]

@patent beacon Thank you for saving my sanity

[patent beacon]

Can you go from there?

[main geyser]

Yo

So I have a question

For all American students here

Can you immediately go from precalc to AP Cal BC

Or are you limited to AP calc AB?

Or regular calc

[viral lynx]

i went from precal to AP BC

doing that was pretty common at my school. i cant speak for other schools tho

[slender river]

wow

my school would never allow that

not even to calc ab

jesus

impressive

or did you take precal and then go to bc because that sounds kind of o k a y ?

idk

[viral lynx]

yeah i did precal but tbh we didnt do anything new; it was a waste of a year

it was algebra 2 (which we took the year before precal) plus like... partial fraction decomposition and some imaginary number stuff

[slender river]

yep

i call precal "algebra 3"

very disappointing class imo

[main geyser]

So

Does that mean it is a class worth skipping if you test out of it?

In your opinion

Because I’m enrolled in algebra II this year

And I want to advance through my high school math requirements

And I didn’t know whether algebra II or precalc was worth skipping

Because from what you guys have been saying, it basically seems like precalc is a refresher course for algebra II

[viral lynx]

that's how it was at my school; id ask some upperclassmen at your school or maybe even the teachers to see what they recommend

[main geyser]

Ok

[dense zealot]

So

@main geyser

Like

All alg 2 does is

Teach u conic sections which u will never use again

Untill u get to like

Elliptical integration and hyperbolas

circles u use too

U basically forget them until calculus tho

Precalc just teaches u basic linear algebra and some unit circle stuff

u memorize formulas and do parametrizations

tbh u need both for calc

U need the formulas for trig in calc and alg 2 for conics

Y not just take a summer class instead of skipping if u don't wanna learn it all ahead

[pure flax]

What is this called? Is there a good khan academy vid on it?

I have no clue what to do.

[viscid thistle]

Wait what are you asking about

what is the csc/cot called?

[pure flax]

No

How to solve

[viscid thistle]

oh

You know cotan x = 1/tan x right?