precalculus

[drowsy quail]

@sharp marsh why is this in pre-calc? If you can't do that get back to pre-alg

[lost mesa]

oh true this is precalc

[drowsy quail]

That's like 5 year old stuff lol

[lost mesa]

maybe they had that in a precalc question ¯_(ツ)_/¯

[drowsy quail]

True haha

[carmine elbow]

Soo, circumference?

[tardy ridge]

you don't know how to calculate the circumference of a circle?

2r*pi

[carmine elbow]

It's been awhile

[tardy ridge]

and 2r is 2* the radius

you are given the diameter which is 2* the radius

also there is google

[drowsy quail]

$C=πd$

[obsidian monolith]

MisterLJGibson:

[tardy ridge]

The real question is why is this in precalc

[lost mesa]

yeah this is more of an algebra question

or geometry

[drowsy quail]

Where $d=2r$

[carmine elbow]

because it's part of my school's precalc lesson

[drowsy quail]

I would say this is more pre-alg

[lost mesa]

you mean $d=2r$?

[obsidian monolith]

andrwxlin:

[drowsy quail]

Ofc, what an idiot I am, sorry lol

[lost mesa]

lol np

[obsidian monolith]

MisterLJGibson:

[drowsy quail]

@lost mesa I'm solving a fractional calculus problem atm where and I need to find the $πth$ derivative of a function, so I have π in my head rn haha

[obsidian monolith]

MisterLJGibson:

[lost mesa]

lol

[past meadow]

angular speed by radius probably

[drowsy quail]

Yes

[carmine elbow]

oh wait, I know it

nvm

[drowsy quail]

@lost mesa do you want to try the problem?

[lost mesa]

sure

i dont think i've gotten into fractional calc

but it might be fun to try

$\pi'th$ derivative tho

[obsidian monolith]

andrwxlin:

[drowsy quail]

Ok, I'm gonna try and use the Text bot

[carmine elbow]

This is the question

[lost mesa]

we just answered your question

[drowsy quail]

Yup

[lost mesa]

find the circumference

it's going 7200 circumferences per minute

convert that to a circumferences per hour unit

then you have your answer

[drowsy quail]

Evaluate ${d^π}/{dx^π} of x at x=π$

[obsidian monolith]

MisterLJGibson:
Compile Error! Click the reaction for details. (You may edit your message)

[lost mesa]

$\frac{a}{b}$

[obsidian monolith]

andrwxlin:

[drowsy quail]

Oh thanks

[carmine elbow]

I'm still confused, but I'll try it

[lost mesa]

here let's move to #calculus

[drowsy quail]

Ok

[lost mesa]

ok you're given the length of the diameter

actually let's back up for a second

which part is confusing you

[carmine elbow]

All of it. I'm a visual learner, so if the question is broken down visually, I'll be able to understand

it's also easier to see where I'm getting stuck

[tardy ridge]

People also ask
What is the circumference of a circle definition?
In geometry, the circumference (from Latin circumferens, meaning "carrying around") of a circle is the (linear) distance around it. That is, the circumference would be the length of the circle if it were opened up and straightened out to a line segment.

[carmine elbow]

ooooh

ok

[tardy ridge]

Speed is a measure of how quickly an object moves from one place to another. It is equal to the distance traveled divided by the time

[carmine elbow]

got it

[lost mesa]

diameter is the width of the circle

[tardy ridge]

a circular disk is a disk that is a circle

[carmine elbow]

ok

[lost mesa]

ok here's a diagram

if we were to take that red circle

and stretch it out into a straight line, that's the length of the circumference

the green line, which goes from one point in the circle to another and passes through the center, is the diameter

the radius is half of the diameter and a straight line drawn from the center to the edge

now let's say that the point they're asking for in your question is represented by the black dot

[carmine elbow]

ok

[lost mesa]

when that black dot goes all the way around the circle and back to its original position, that's 1 rotation

when i say "around the circle", i mean it follows that red line back to its original position

you follow?

[carmine elbow]

yes

[lost mesa]

so in other words, that black dot just travelled the circumference of the circle

[carmine elbow]

sooo, in my case, the black dot goes around 7200 times?

[lost mesa]

per minute, yes

it's asking for the linear speed of that black dot per hour

linear meaning it's a constant speed throughout

speed = distance/time

speed is the measure of rate of change of distance in a certain time interval

so like a car goes 60 miles/hour, its speed is 60 and then unit is miles/hour

it goes 60 miles in one hour

[carmine elbow]

ok

[lost mesa]

so in order to find the distance unit, we need to find the circumference of the circle

the equation for circumference is $C = \pi d$, where $C$ is circumference and $d$ is the diameter

[obsidian monolith]

andrwxlin:

[carmine elbow]

c= pi(2.5)?

[lost mesa]

yes

and that's how far the black dot travelles every 1 rotation

in this case, it states that the disk rotates at 7200 revolutions per minute

and it asks for the speed (how far it's gone in a time interval) in miles per hour

so we have two ways of approaching this

we can either find the inches/minute speed, or we can find the inches/hour speed

[carmine elbow]

wait, before we go on, do i need to round the answer for the circumference

[lost mesa]

right now, we're keeping the answer for circumference in terms of pi

so we're keeping it as 2.5pi

[carmine elbow]

ok

[lost mesa]

just because that makes life easier

so let's say we want to find the inches/minute

if we know that it goes 2.5pi inches per revolution, and the disk goes 7200 revolutions per minute, how many inches does it go per minute?

[carmine elbow]

give me a minute to calculate

[lost mesa]

treat pi as a variable for now

[carmine elbow]

9047.79?

[lost mesa]

ignore the pi for now

[carmine elbow]

I divided the 2.5pi by the 7200 revolutions

[lost mesa]

you should be multiplying 2.5pi by 7200

because if it goes 2.5pi inches per 1 revolution, and we're finding how far it goes in 7200 revolutions, we have to multiply 2.5pi in/revolution by 7200 total revolutions

do you understand why?

[carmine elbow]

ummm, no

[lost mesa]

if im walking at a constant pace (linear speed) of 1 mile per minute

and you want to know how far i will be in 10 minutes

how do you calculate that?

[carmine elbow]

that's tricky

would that be ten miles?

[lost mesa]

yes

so now let's apply this same logic to the question at hand

if this disk is going at a linear speed of 2.5pi inches per revolution, then how far will it go in 7200 revolutions

[carmine elbow]

18000

Why don't we divide?

[lost mesa]

correct, 18000pi inches

[carmine elbow]

I'm asking because i want to understand

[lost mesa]

we dont divide because the original unit is smaller than the one we're finding

like back to the walking example i gave you

how did you find 10 miles?

here let me show you in fractions

[carmine elbow]

it took some time, but then I thought, If you are walking 1 mile per minute, and I want to know how far you have gone in 10 minutes, then you must have walked 10 miles

[lost mesa]

correct

$\frac{1 mile}{1 minute} = \frac{x miles}{10 minutes}$

[obsidian monolith]

andrwxlin:

[lost mesa]

so the easy way to solve for x is by getting rid of that 10 in the denominator

so we multiply the left and the right side by 10

then you get $\frac{10}{1} = x$

[obsidian monolith]

andrwxlin:

[carmine elbow]

ok

[lost mesa]

therefore, x=10

which shows that 1 mile/minute, then in 10 minutes i must have gone 10 miles

so in this case, we multiply to find the answer

let me show you a case when we would divide

$\frac{x mile}{1 minute} = \frac{10 miles}{10 minutes}$

[obsidian monolith]

andrwxlin:

[lost mesa]

if i asked you how many miles i have gone in 1 minute, then how would you find that?

remember, our unit of speed in this question is miles/1 minute

[carmine elbow]

would that be one?

[lost mesa]

yes

if i told you that i had gone 20 miles in 10 minutes, how many miles do i go per one minute?

[carmine elbow]

hmm

5?

[native crag]

2?

anyways, I also need some help

[lost mesa]

hm haven't worked with velocity questions in a while

what i would do is set time to the x-axis, set velocity to the y-axis, and plot those two points

since they're asking for a linear model, what i would do is find the equation of a line that passes through both points

and then plug in 4.9 to find the predicted velocity

does that make sense?

@native crag

[native crag]

hmm

I'll give it a try

[carmine elbow]

5?

[lost mesa]

@carmine elbow how did you get 5

[carmine elbow]

you said 20 miles in 10 minutes, so I thought it was 5 @lost mesa

[sharp marsh]

What do I do for this?

[lost mesa]

@sharp marsh evaluate each function at 2pi and pi, then express that in whatever format it's asking

i think express it as a range interval?

because sin(2pi) = 0, and sin(pi) = 0

but the curve of sin(theta) from pi to 2pi goes down to (3pi/2, -1)

[sharp marsh]

Oh?

[lost mesa]

i think my explanation was kinda crappy, but do you see what im trying to get at

[sharp marsh]

Nope

[lost mesa]

ok lemme try using the unit circle

have you learned this yet?

[carmine elbow]

you said you traveled 20 miles in 10 minutes, so I thought it was 5

[lost mesa]

heard you the first time

[carmine elbow]

oh

[lost mesa]

sorry, but i've tried my best explaining it

[carmine elbow]

it's fine

maybe we could just move on from 18000 and I could just figure how I got it later

[lost mesa]

kk

so 18000pi inches travelled in 7200 revolutions

7200 revolutions per one minute

therefore, 18000pi inches travelled in one minute

does that make sense?

[carmine elbow]

yes

[lost mesa]

and it's asking for the speed miles per hour

60 minutes/hour

18000pi inches per minute * 60 minutes = 1,080,000pi inches per hour

does that make sense?

[carmine elbow]

kind of

[lost mesa]

it moved 18000pi inches in the first minute, then another 18000pi inches in the second minute, etc etc for 60 minutes

so that's basically 18000pi added to itself 60 times

which is the same thing as 18000pi * 60

so we have 1,080,000pi inches per hour

but it asks for miles per hour

there are 63360 inches in 1 mile

so to convert 1,080,000 inches to miles, we divide 1,080,000pi inches by 63360 inches per mile

[fleet yew]

cora do you know about dimensional analysis

[carmine elbow]

no

[fleet yew]

let me show you

[lost mesa]

https://cdn.discordapp.com/attachments/363224154469826562/698999676027797616/196732296202682368.png this was her question

[fleet yew]

it's way easier to keep track of unit conversions

[carmine elbow]

I'm open to learning it

[fleet yew]

$\frac{18000\pi \space inches}{1 minute} \times \frac{60 minutes}{1 hour} \times \frac{1 miles}{63360 inches}$

[lost mesa]

i dont think latex counts spaces

[fleet yew]

eh good enough

[lost mesa]

\space i think

$\space$

[obsidian monolith]

andrwxlin:

[lost mesa]

yeah

i guess not lol

[fleet yew]

lmao it's fine

anyway the whole principle is that you can cancel units diagonally

just like numbers

[carmine elbow]

1 minute = 18pi in?

[fleet yew]

idk was it?

my bad

[lost mesa]

18000pi inches/minute

[obsidian monolith]

AMD:

[carmine elbow]

ok

[fleet yew]

if you look at the units they cancel diagonally

so you still end up with miles/hour

[carmine elbow]

so, you cancel the inches in 18000 and 63360?

and the minutes in 60 and 1?

[fleet yew]

Yeah exactly. You can treat units just like numbers in that sense

[carmine elbow]

so all that's left is 18000pi * 60 minutes * 1 mile/63360

?

[fleet yew]

Units cancel

So you are left with

(18000pi * 60) miles /(63360 hours)

Then you just simplify it so you have the rate per hour

[carmine elbow]

53.55?

[fleet yew]

looks good

[carmine elbow]

ok good!

Thank you for the help! Sorry it took soo long for me to get it @fleet yew

[harsh cipher]

hi

quick question!

is cotagent negative in quad 2 and 4

[novel dirge]

also a question from me, if we have 2 black rabbits and 4 white ones in a cage and they start to randomly leave it, what is the probability of the last reabbit leaving to be a white one?

[past meadow]

@harsh cipher yes

wouldnt that just be the same as the probability of picking a white rabbit?

[novel dirge]

Idk

cause it says it has to exist the last

so I am not sure if there are any other conditions

[past meadow]

im pretty sure that it should just be the prob of picking a white rabbit, but i would wait for someone else to give an answer

[novel dirge]

hopefully someone will come soon

can you try helping me with another one?

if there is a colored cube

and it is divided into 1000 equal parts

how many new cubes have 2 sides colored

[proud sparrow]

Yeah, same

[novel dirge]

I tried doing this by thinking of the cube as 10x10x10

but when I added everything up, got 1005 cubes

which is 5 too many

[proud sparrow]

You can always rearrange the events to get an equally likely scenario

what 1005?

calculations please

[past meadow]

2 sides exactly or 2 sides or more?

[novel dirge]

of, forgot to write

colored cue is divided into 1000 parts

how many of them have 3 sides colored, how many 2, how many 1 and how many 0

when I added everything up I got 1005

[past meadow]

oh, 2 would just be the amount of edges that arent on vertices

[novel dirge]

I divided the cube like 10x10x10

yes

for 2 sides I looked at edges

3 sides are the 8 corners

1 side are 12 sides * (8*8) ?

this leaves me with much more than just 1000

my main problem here is not knowing how to divide the big cube

because it seems that dividing it into 10x10x10 doesn't work

and for the rabbit problem, can I write out of 4 w rabbits choose 1 and out of 2 black ones chose 0 ?

because they already left

[harsh cipher]

@past meadow ty

[novel dirge]

you still here?

[past meadow]

huh

i am, what do you mean by

1 side are 12 sides * (8*8) ?

by side do you mean face?

2 sides is edges minus vertices, 3 sides should just be vertices, 1 side should be faces - vertices-edges

[novel dirge]

yeah, I meant faces

[past meadow]

you sure there are 12 faces on a cube

[novel dirge]

wait

wtf did I do

there are 6

right?

[past meadow]

yes

lol

[novel dirge]

Idk how I came up with 12

so it will be 6x8x8?

[past meadow]

is this for one side painted?

[novel dirge]

yes

just realized I am stupid

I wrote it in my notebook correctly

but read the part for 2 sided

[past meadow]

sounds right to me i think

[novel dirge]

which was 12x8

[past meadow]

so theres 12 edges on a cube

each of which is 10 wide

[novel dirge]

yes

[past meadow]

and there are 8 vertices

so 120-8?

[novel dirge]

and on each edge 2 are taken by the corners

[past meadow]

yes, there are 8 corners

so take the corner cubes away from the vertices right?

[novel dirge]

yes

and every edge is 8 cubes long

and there are 12 of those

[past meadow]

oh wait that makes more sense than what i was thinking

so should be 96 then, no?

[novel dirge]

yes

and for the non colored ones

it will be 8x8x8

which is 512

wtf

it fits now

[past meadow]

ye

[novel dirge]

8+96+384+512 = 1000

[past meadow]

yup

[novel dirge]

but last night when I tried this, I got 1005

didn't even change anything

[past meadow]

how the fuck did you get an odd number

[novel dirge]

I have no clue

that is why I was so confused

because I was pretty sure that I can divide a cube correctly

so I thought that maybe it shouldn't be divided into 10x10x10

can you check out what I wrote for the rabbits?

out of 4 w rabbits choose 1 and out of 2 black ones chose 0

[past meadow]

im not sure what you mean, i would have just thought 2/3

since you're choosing 1 white rabbit, essentially, the change of doing that is 4/6=2/3

[novel dirge]

Im a thinking this because it is supposed to be the last

and there can't be any black rabbits left

[past meadow]

why cant there be any black rabbits left?

[novel dirge]

because this one is the last one to leave

[past meadow]

im not sure what your problem is, it seems pretty obviously like it would be 2/3 to me but idk, im pretty sure element agreed. repost and ill give one last look at the problem, and someone else can try answer i guess idk

i could be wrong too

[manic sky]

nobodys helping in the calc channel ;-; how do I figure out which quantities vary, are constant, and which are parameters.

[novel dirge]

@past meadow if we have 2 black rabbits and 4 white ones in a cage and they start to randomly leave it, what is the probability of the last reabbit leaving to be a white one?

but the more I think about it the more logical your solution sounds

they probably said last just to confuse us

[manic sky]

@viscid thistle

pls

[willow bear]

why are you pinging this one random person

[manic sky]

its his niche

[novel dirge]

I just realized, arent the chances of a white rabbit being last, much smaller than the chances of picking any whjte rabbit, not regarding the order

[novel dirge]

Can anyone here help me with this?

[novel dirge]

would a valid approach be to calculate the probability of the 5 others getting out first?

and then getting the probability for the last from that?

[novel dirge]

This doesnt really make sense

That the probability would be so low

I know that I am probably annoying with the rabbits, but I am really confused

[muted steeple]

Hi guys, i got a question about "special factoring formula's for polynomials" specifically the $A^3+B^3$ one.

So my book states that $A^3+B^3 = (A+B)(A^2-AB+B^2)$

Should it not be $A^3+B^3 = (A+B)(A^2+2AB+B^2)$ because $A^3+B^3 = (A+B)(A^2+B^2)$ and $A^2+B^2 = A^2+2AB+B^2$?

[obsidian monolith]

꧁༺Vocal༻꧂:

[willow bear]

no

[uncut mulch]

$A^2+B^2 \umwhat A^2+2AB+B^2$

[obsidian monolith]

ramonov:

[past meadow]

A^2+B^2/=A^2+2AB+B^2

[willow bear]

$A^2 + B^2 \wtfeq A^2 + 2AB + B^2$

[obsidian monolith]

Ann:

[muted steeple]

oh, i meant $(A+B)^2$

[obsidian monolith]

꧁༺Vocal༻꧂:

[uncut mulch]

$\fdream$

[obsidian monolith]

ramonov:

[willow bear]

$A^3 + B^3 \wtfeq (A+B)(A+B)^2 = (A+B)^3$

[obsidian monolith]

Ann:

[muted steeple]

oh yes, that rule is for products.

okay, i forgot, thank you.

[past meadow]

vocal try expanding the (A+B)(A^2+2AB+B^2) out if you want

[muted steeple]

I still need to figure out why the factoring formula is the way it is.

[past meadow]

it wont be what you want

[uncut mulch]

distributive property

and/or long division

[willow bear]

expand out $(A+B)(A^2 - AB + B^2)$ and see.

[obsidian monolith]

Ann:

[muted steeple]

$A^3 - B^3$?

[obsidian monolith]

꧁༺Vocal༻꧂:

[past meadow]

no

[muted steeple]

okay, how do they came up with that right trinomial?

[viscid thistle]

@manic sky hahahaha

😂 😂

[uncut mulch]

factor theorem with long division and/or alternative methods

also that expression above doesn't expand to A^3 - B^3

[muted steeple]

$(A+B)(A^2-AB+B^2) = A(A^2-AB+B^2)+B(A^2-AB+B^2) = A^3-A^2B+AB^2+A^2B-AB^2+B^3$ it expands like this right?

[obsidian monolith]

꧁༺Vocal༻꧂:

[muted steeple]

oh

[past meadow]

Looks right to me

[muted steeple]

A^3+B^3

ty.

alright, hopefully it will tell me how they came up with these factors.

[novel dirge]

I gave up on the rabbits, now I am stuck on another one

if x can range from (0, 1/2], and y(1/2,1]

what are the chances of picking x and y randomly and fulfill the condition |y-x| < 1/3 ?

[willow bear]

are their distributions uniform?

[novel dirge]

what do you mean?

on a scale from 0 to 1, the points x and y are randomly picked

x=(0, 1/2] and y=(1/2,1]

and I have to find the probability of them fulfilling |y-x| < 1/3

[forest crypt]

Hey can any1 help me with this quest?

[novel dirge]

@willow bear any idea how I can do it?

sorry for bothering you, but i really need help

[willow bear]

@forest crypt your thing is false as stated bc the left hand side is undefined for x = -1

[brittle fable]

is it defined for x = 0?

[willow bear]

no

[brittle fable]

so yeah this thing is weird

[willow bear]

there's an x inside a log

@novel dirge ok so uniform

graph the region {0 < x <= 1/2 < y <= 1} in the plane and take its intersection with the strip {|y-x| < 1/3}

the ratio of their areas will be your probability

[novel dirge]

sorry, but can you explain that a bit more?

neved done something like that

[brittle fable]

draw an xy plot

[novel dirge]

idk how to do it

this is the best attempt so far, but I am not sure if it is correct and how to fill/use it

,rccw

[novel dirge]

added 1/3 because I need it in the condition

@willow bear @brittle fable

[willow bear]

bottom and left edges on the square should be dotted but it's not really important

[novel dirge]

so what I started is good?

[brittle fable]

yes, you can also make sure that zero on x is dotted and 1/2 on y is dotted

then use |y-x| < 1/3

[willow bear]

since we're concerned with areas here it really doesn't matter what's dotted and what isn't

[brittle fable]

thats true

[novel dirge]

so how exactly do I make that "path"?

the blue one

I can see that it is probably going from 1/3

[brittle fable]

you can start by expanding the absolute value

[willow bear]

you can rewrite |y-x| < 1/3 as -1/3 < y-x < 1/3

[novel dirge]

but do I just take some random samples that fit the criteria or what_

[willow bear]

and in turn, as x - 1/3 < y < x + 1/3

that's how to make the strip

[novel dirge]

I understand the first 2 parts

but not the last one

how do I get x and y from that?

[brittle fable]

well think of x+1/3 as a line

[willow bear]

what do you mean "get x and y"

x - 1/3 < y < x + 1/3 describes a region

[novel dirge]

ohh

never did that

so I am unfamilliar

[willow bear]

idk how you're expected to do it then

[novel dirge]

I am supposed to find the probability

and my teacher probably didn't even look at what problems she gave us

because she doesn't care if we learn anything

[willow bear]

if you could plot that strip, you could take the intersection of it with the square

i.e. that triangle

and then find its area

and divide that by the area of the square

[novel dirge]

why the square?

I just found something similar

but idk how it is done

[willow bear]

bc you said x ∈ (0, 1/2] and y ∈ (1/2, 1]

the set of all points (x,y) which meet those conditions is that square up there

[novel dirge]

This one says 2 people are meeteing eachother bwtween 7 and 8PM, and they made a deal that the first one to arrive will wait 20 minutes and go home

and what are the chances of that encounter happening

below is the solution

looks like the same kind of graph

[brittle fable]

think of the square as the set of all points x and y that you currently have and you're going to pick 2 random points x and y from that square such that |y-x|<1/3

that triangle is the set of all points x and y such that the distance between x and y is less than 1/3

so it makes sense that this set of points when divided by the total set of points gives you the probability

do you understand?

[novel dirge]

oh

yes

I get the square now

but not the rest

[brittle fable]

the triangle is formed from the inequality x-1/3 < y < x+1/3

so..

look at x+1/3 for a moment

set x=0, then you get y<1/3

[stable pasture]

btw my limit problem is done

[brittle fable]

same for the negative side

[novel dirge]

oh, so y is x+1/3?

[stable pasture]

by using sin x and cos x mc lauren expansion

[brittle fable]

y is inscribed in the region

bounded by x-1/3 and x+1/3

you can think of these two as the lines

y=x+ 1/3 and y = x-1/3

then draw the 2 lines

idk if you're familiar with lines

if you're not, then try taking y=x+1/3 and set x=0 and see which y you get, then plot that point, then try setting y=0 and see which x you get, plot that x, then join the two points together

[brittle fable]

thats the line y=x+1/3 you tried it yourself?

now you can do the same for y=x-1/3 , it doesn't really matter here because its outside the region, but its good practice

[novel dirge]

Yes

I'll try now

Is it supposed to pass through 1/3 for x?

[brittle fable]

yes

[novel dirge]

and the second line is y= x + (-1/3) ?

[brittle fable]

exactly

[novel dirge]

What now?

[brittle fable]

okay so remember what was the goal

you wanted to know how likely it is to pick x and y from that square such that |y-x| <1/3

[novel dirge]

to find the probability of x and y being that

while being inside their constraints

[brittle fable]

good

so now you have the given |y-x|<1/3

which is the region between the 2 lines

and you have the square, which is the thing that you 'pick' from

so the intersection of |y-x| <1/3 and the square is what?

[novel dirge]

the little triangle

with our desired values that fit both conditions

[brittle fable]

you know a pie chart right

imagine a small slice in a pie chart

it represents the probability

its kinda the same situation here

the square is the whole pie

and the triangle is the slice

so use the area of the triangle and the area of the square now

[warped dagger]

https://cdn.discordapp.com/attachments/501175742504239106/699246707321602098/674711032127553540.png wondering how to start solving this, just not sure cause theres many variables

[brittle fable]

this looks like a quadratic equation

you know about those?

[warped dagger]

Yeah, i do

[brittle fable]

try using exponent properties on 5^(2x-3) and 5^(x-2)

[novel dirge]

how am I supposed to do that?

I am terrible at geometry

and is it possible to do something similar to what is done in the pic I sent?

[brittle fable]

@novel dirge well the most important thing is that you understand why we've been doing all that

[novel dirge]

Up untill now, I understand what is done

[brittle fable]

the area formulas dont matter

you can look up the area of the triangle and the area of the square

[novel dirge]

but just not sure what to do next

[brittle fable]

check the length of the square's sides

and the height and base of the triangle

[novel dirge]

1/2 on all sides

not sure about the triangle

[brittle fable]

ok, the height atleast?

[novel dirge]

I have an idea

the base of the triangle is from x(1/6,1/2) to x(1/2,5/6)

so it is just 1/6 to 1/2

which is 1/3

so the base is 1/3

and the height is from 1/2 to 5/6

which is 1/3 again

and since this triangle has an angle of 90

it will be (1/3)^2 +(1/3)^2 = c^2 ?

[brittle fable]

the base of the triangle is from x(1/6,1/2) to x(1/2,5/6)
@novel dirge notice that the base is at a constant y value which is 1/2

so 5/6 cannot possibly be right

[novel dirge]

damn

wait

I said those coordinates

because that is the y for the x I want

because the sqare ends at x = 1/2

and y for that is 5/6

[brittle fable]

you want the length of the base, so from (1/6 ,1/2) to (1/2 , 1/2)

right?

[novel dirge]

yes

[brittle fable]

now the height is from (1/2 , 1/2) to (1/2 , 5/6)

[novel dirge]

yes

[brittle fable]

so the length of the base is 1/2 - 1/6 , right?

[novel dirge]

yes

I think so

[brittle fable]

and do the same for height

[novel dirge]

which is 1/3

for both

[brittle fable]

then use the area formula of the rectangle which is A = (1/2) * (base) * (height)

sorry, triangle

what do you get

[novel dirge]

1/18?

seems kinda small to me

[brittle fable]

correct, now calculate the area of the square which is length * width

[novel dirge]

(1/2)^2?

[brittle fable]

which is equal to

[novel dirge]

1/4

[brittle fable]

so you got the area of the triangle and the square now, what should you do with them?

[novel dirge]

divide the triangle by the square

[brittle fable]

thats the probability i hope you understood now

[novel dirge]

yes

I did

thank you very much

for both the patience and explanation

[tardy ridge]

@warped dagger and make 5^x as U and solve for U

uh I have imaginary number - edit from the future - because I did it wrong

[past meadow]

fishraider

index laws

i think

[tardy ridge]

no

look he has this thing

https://cdn.discordapp.com/attachments/501175742504239106/699246707321602098/674711032127553540.png

[past meadow]

oh

i misread the problem

my bad

[tardy ridge]

then I covert it to $(5^x)^2/(5^3)=2*5^x/(5^2)+3$

[obsidian monolith]

Fishraider:

[tardy ridge]

and I covert 5^x to U

and I get u is imaginary

oh I am dumb

I got it

(u+15)(u−25)

very nice, now it is goo

[obsidian monolith]

PhysicsMonster:

[stable pasture]

how do you do this

[viscid thistle]

try combining the fractions

actually you could try squeeze theorem

@stable pasture

[stable pasture]

it isnt working

have you tried, does it reduce to a solvable form?

[viscid thistle]

i mean you could use lhopitals rule

do you know what that is?

[serene heath]

$\lim_{x \to 0} \frac{x-\sin(x)}{x^3}$

[obsidian monolith]

the one n only:

[serene heath]

if you can evaluate this then your limit is straightforward

@stable pasture

[drowsy quail]

@stable pasture just keep applying L'Hopital's Rule

[stable pasture]

yes but I dont want a solution use Lhopital

[viscid thistle]

you can also do a u sub

u sub for arcsin(x)

[stable pasture]

if you can evaluate this then your limit is straightforward
yes I can evaluate this, using sin x mc lauren series

but how do I go from @serene heath series to the one I posted?

[viscid thistle]

u sub for arcsin(x)

[stable pasture]

um I mean I dont want a solution using Lhopital, as I'm sure there's a slick solution without it

[viscid thistle]

$\left(\frac{\sin\left(x\right)^{2}-x^{2}}{x^{2}\sin\left(x\right)^{2}}\right)$

[obsidian monolith]

sanath1237:

[viscid thistle]

if you can do this limit

then you can do the other one

they're equal

[undone pawn]

obtained by substituting x=siny

[serene heath]

@stable pasture yea what sanath said

[stable pasture]

ah nice, got it. I think I can do this now using mc lauren on sin x

I already have, its quite boring honestly

[drowsy quail]

What did you get?

@stable pasture

[stable pasture]

-1/3

[viscid thistle]

thats right?

[stable pasture]

yes

[drowsy quail]

I haven't done it yet

[viscid thistle]

also don't you need derivatives to find the mclaurin series of sin(x)

i don't see the point

[drowsy quail]

Indeed

[stable pasture]

ofc you dont

[drowsy quail]

What do you mean by that? @stable pasture

[stable pasture]

I mean you can now just apply McLauren for f(x)=sin x into what @viscid thistle said

[undone pawn]

I mean using the power series is kinda the same as applying lhopital

[stable pasture]

how tho

oh wait

yeah lol

[drowsy quail]

😆

[solid gazelle]

https://discordapp.com/channels/268882317391429632/326138772477575180/699362079781552160

Can somebody tell me how to proceed with that problem please?

[serene heath]

,rotate 90

[serene heath]

what have u tried

also this isnt calculus

nvm i cant read channel names it seems

[fallen crane]

I have a question

https://www.youtube.com/watch?time_continue=10&v=8VzcFMRe2n4&feature=emb_title

I needed help identifying the mistakes my teacher made

there are multiple, but the only one I can seem to find is that he factored the difference of squares of x^4 - 9y^4 incorrectly

can anyone help?

[serene heath]

it says spot the error

as in oke singular error

[fallen crane]

he is not exactly the best with wording, sorry

[serene heath]

I see only 1 error

the one u mentioned

[fallen crane]

is that it?

[serene heath]

seems so

[hardy kiln]

min 1:20

you cant factor x^4+9y^4

into what he did

only when its a minus

[serene heath]

yes

[hardy kiln]

u can

[serene heath]

that's the error

[fallen crane]

Thx guys 🙂

I had another question

How would I do this?

[hardy kiln]

okay so a difference of squares is when a square of a number - another square of a number can be factor as its conjugate pairs so you want some the powers on the variables to be 1

so you just gotta find numbers A and B that you can factor 4 or 3 times to get (cx-dy)

so like uh

exept i think in ur case ud actually write the 2^8 to its full number to get A

and B

but that is just an example

[fallen crane]

wouldnt the 2 and the 3 also divide?

oh wait, you added exponents to the A and B?

[hardy kiln]

ya so its easier to factor out,

but A = 2^8 and B=3^8

[fallen crane]

okay, thank you!

[viscid thistle]

oof big factoring

[eager path]

https://gyazo.com/e0fe53e2ee0493c8c28d98c0fa4b6778

[vague crystal]

@eager path are you asking for help?

[eager path]

I figured it out

[vague crystal]

okee dokee

[eager path]

convert sqrt(3)+i to trig notation

and then de moivre

[vague crystal]

Good job

De Moivre's is fun

I like complex numbers

[narrow peak]

i didnt know complex numbers was precalc

[lost mesa]

i'd say demoivre's theorem counts as precalc

and in precalc (at least in my US public high school) complex numbers are used

[harsh cipher]

Hi

[drowsy karma]

how do i factor a^2+1a/4+1/2

I'm trying to factor by inspection by finding a factor that adds to 1/4 and multiplies to 1/2 but I don't know if I'm doing it wrong or?

[past meadow]

,w a^2+a/4+1/2

[obsidian monolith]

Display results online and refine query

Requested by Sneaky#0928

[past meadow]

Yeah, that cant be factored in the reals

[drowsy karma]

So not possible right?

It's that for clarity

[past meadow]

Not possible

[drowsy karma]

thankyou

,w a^2-10a-24

[obsidian monolith]

Display results online and refine query

Requested by Stewie#8321

[past meadow]

You can figure out yourself whether it can be factored by considering the discriminant

[drowsy karma]

my brain no work rn

[wild cobalt]

tbh for lack of a better channel I’m going to throw this in here.

let’s say I have $15k in the stock market, expecting an average annual return of 10%. I want to purchase a new car for $15k, so I finance it at 0 down, 5% interest, over 5 years, which is a payment of $283. every month I’m making money from stock market growth, but also making the payments on the car, how would I model this mathematically?

[summer monolith]

Why am I being asked to fill the table without the calculator? How is it possible to do without it? The "hepful hint" is making no sense.

Nah...
Doesn't really matter. Forget it

[muted steeple]

Is this channel taken?

[willow bear]

guess not

[muted steeple]

Hi, could someone explain to me how to factor a polynomial where the coefficient of the term with the highest power is not 1?

Let me send an example.

$8x^2-14x-15$

[obsidian monolith]

꧁༺Vocal༻꧂:

[willow bear]

a quadratic specifically?

[muted steeple]

Aa yes.

[willow bear]

one way to do this is to just use the quadratic formula and get the roots

another is to divide out by the leading coefficient, factor that, then multiply the leading coefficient back in

[muted steeple]

I've seen someone suggest this, but i don't quite understand what they say divide out by the leading coefficient.

so, in this case it's 8.

?

[willow bear]

yes

factor x^2 - (14/8)x - (15/8)

then multiply the factorization by 8

[muted steeple]

oh, so now i need to look for factors which will sum up to 14/8 and give a product of 15/8?

[willow bear]

yes, though you might consider simplifying 14/8 to 7/4 before doing so.

er

you need the product to be -15/8

not 15/8

[tardy ridge]

You should do decomposition

another way is prime factoring the 8 and 15, but that takes a lot of guesswork

I think it's called the cross method or something

[viscid thistle]

How would I do this problem ?

#2

[tardy ridge]

I got it

first you want to try this

log a=p*logx

so a=x^p

similarily b=x^q

c=x^r

b^2=x^2q

then you divide and get the answer

@viscid thistle

[viscid thistle]

What would I divide by?

@tardy ridge

Never mind I figured it out, thank you!

[viscid thistle]

Could I get help with 3?

[tardy ridge]

idk what it means by simplest form because I have never done this subject

but you can subsitute 2^-2k as U

and then end up with something that looks like a/b^k

@viscid thistle

[viscid thistle]

Still kinda confused

[tardy ridge]

have you tried substituting $2^{-2k} = U$

[obsidian monolith]

Fishraider:

[tardy ridge]

you don't have to do that, but it makes it easier

also am I dumb

I can't solve question 4

[viscid thistle]

We have our days lol

[tardy ridge]

did you solve 3 yet

[viscid thistle]

Yeah, I had to factor

I had trouble with that

[tardy ridge]

wdym

2^-(2k+1) = U/2

you can do that with the others can get something that is nice

[lapis onyx]

how do i show a graph is symmetrical in the line y=x?

[acoustic talon]

are there asymptotes in a rose curve?

[tardy ridge]

@viscid thistlecan explain to me how you solved 4

[past meadow]

fish you can do a substitution for 4

u=3^x

@tardy ridge

[tardy ridge]

BRUH

I was telling him how to do 3

Which is literally the same idea as 4

Ahhhh

[past meadow]

lmao

[pale kettle]

uh

I don't think this is the right channel

#groups-rings-fields would be better

[honest spear]

Oh-, but I picked this book for pre-calculus so asking it in Advanced Maths category might be, um, awkward ?

[pale kettle]

Uh, what book is it?

It didn't seem like anything that would be about precalc

[honest spear]

Well, I did heard that it's bit too technical but good to catch up with US curriculum

It's Basic Mathematics

Serge Lang

[pale kettle]

What page was your picture from?

[honest spear]

371

[pale kettle]

ah yeah

then feel free and ask here

I can just answer it

[honest spear]

I just wanted to know that we arrived at that conclusion because no. of orbits of this permutation is invariant, right?

Or there is something else too

Dunno why just that + an odd/even integer bother me a bit

[pale kettle]

I'm not sure which conclusion you're talking about

[honest spear]

Well nevermind

I was thinking on it for a while

I was just confused on that last part

No. of orbits - n is either odd or even

I thought of it backwards so far and confused myself

Lels Thanks though to consider it !

[viscid thistle]

how do you fractional negative exponents

[past meadow]

like 2^(-1/2)? just like you would with any normal fractional or negative exponent

if you're talking -2^(1/2). then thats not a real number

although there are situations where it might be real, like -2^(-2/3)

[willow bear]

if you're talking -2^(1/2). then thats not a real number

au contraire, $-2^{1/2}$ is just $-\sqrt{2}$

[obsidian monolith]

Ann:

[willow bear]

you meant $(-2)^{1/2}$.

[obsidian monolith]

Ann:

[past meadow]

you know what i mean

although you're right

i should use parentheses

[viscid thistle]

yeah

25^-1/2

[willow bear]

"you know what i mean" is no excuse for improper notation

[viscid thistle]

=1/5

[misty heath]

ok not sure why that doesn't work

[obsidian monolith]

Sneaky:
Compile Error! Click the reaction for details. (You may edit your message)

[misty heath]

I don't know what's happening anymore

[buoyant hound]

am I insane or is this unsolvable

this is what we were given as a practice problem

there are 4 other questions like it

there's no other info on the pdf

literally none

it's a bunch of stuff about orthogonal angles

and then that

[serene heath]

its 7

[buoyant hound]

it's a true or false on the pdf

[serene heath]

post the whole thing

[buoyant hound]

this is the most crackhead study sheet I've ever seen

let me just post the other problems specifically as there's some school and location info elsewhere on the page

what am I supposed to be learning from this

[serene heath]

post the entire problem

[buoyant hound]

that is the entire problem

that's literally all there is on it

[serene heath]

thats not a problem

[buoyant hound]

I'm not making this up

that's it

yeah

I know

[serene heath]

thats just an equation

[buoyant hound]

that's why I'm asking

it's saying "true or false, circle one" beneath it

with no values

[serene heath]

so its not the whole thing then

[buoyant hound]

it's listed as a whole problem

and there are 4 of them

I guess it's just a weird error?

idk

no

nope

anyway can someone explain this to me

I don't really get the dot product stuff

and I haven't been able to ask the teacher about it because of eschool and shit

[serene heath]

what does it mean for 2 vectors to be parallel

[buoyant hound]

they have an angle of either 90 or 270 between them

[serene heath]

no

what

[buoyant hound]

okay I don't know then

I'm hella tired and there's been no instruction on this

[serene heath]

how are they related

in terms of their components

ok well if theyre parallel then theyre scalar multiples of each other

that should help you

[buoyant hound]

honestly, I still don't really get it. I probably sound stupid, but explain it to me like I know nothing past the basics of what vectors are, because I don't. There hasn't really been any instruction on this.

sorry if I sound stupid, as I said

also I just realized I meant "0 and 180" not "90 and 270"

[tardy ridge]

You should search it up

[buoyant hound]

I did, and didn't find anything specifically on the question there

I mean I guess I sorta get dot product I just don't get that problem specifically

god all this weird school closure stuff stuis messing me up

[proud gate]

What if imaginary numbers were included with a hyperbola? Could it still be graphed on a complex plane?

[willow bear]

what

those words make sense individually but not together

[copper vigil]

hello friends

this is a physics question but it is confusing the hell out of me

basically these are momentum vectors

and at the top you can see that the initial object collides with the object at rest

i've ruled out all the ones at angles because this is linear momentum

and i know that 4 and 7 are wrong for some reason

but i have no fucking clue how or why

it's gotta be 2

but i don't know why

[patent beacon]

@copper vigil
Conservation of momentum. The sum of the momentums before collision = the sum of the momentums after

Why do you say it's 2? I'd say it's very clearly not

There's no reason why angles can't happen. Let's say one ball hits another on the edge

[copper vigil]

@patent beacon

what do you mean by the sum of the vectors?

sum of their magnitudes?

or the actual sum of the vectors

[patent beacon]

The literal vector sum

[copper vigil]

what about an elastic collision though?

direction is not preserved?

[patent beacon]

Did I studder?

Jk fair question. Conservation of momentum still holds

You also get conversation of energy but that's not related here

Again, there's no restriction on angles

[copper vigil]

ohh ok

i was assuming center of mass

but i guess it could go off at an angle

OHH

It's number 6

because the upwards and downwards cancel out

and the horizontal length is the same

[stable pasture]

i think both the objects are meant to be point sized

also since the initial velocities of the objects is in the x direction only (also the centre of mass velocity), velocity at angles arent possible

unless they arent point sized ofc

[copper vigil]

yes that's exactly what i was thinking

4 is wrong, 7 is wrong, 6 is wrong

wtf

2 must also be wrong because there is no leftwards motion

[patent beacon]

A planet is point sized, if you take an approximation

[copper vigil]

it's gotta be 3 then?

[patent beacon]

3 is one of them!

[copper vigil]

this question is unfair lol

there's no way it could go off at an angle if it's point sized

and there's only velocity in the x direction

3 is right

wtf

i ended up having to use an online ruler to actually measure the pixel lengths

[patent beacon]

We use points to approximate non-point stuff

Wait really? That's dumn

[copper vigil]

i have a quick question about proving trig identities

why isn't it ok to manipulate both sides?

[proven marten]

could you give an example

[copper vigil]

let's look at something trivial like

tan * cos = sin

the way my teacher expects me to do it

is to keep the right side the same

and then arrive at

sincos/cos = sin

so that sin = sin

[uncut mulch]

that's dodgy too

generally you shouldn't be using
= like that for stuff that hasn't been confirmed to be true

[copper vigil]

but isn't it equally valid to say that if tan*cos = sin, then tancos/cos = sin/cos so that sin = sin

[uncut mulch]

you would need to introduce extra symbols

$\iff$

[obsidian monolith]

ramonov:

[copper vigil]

but here's the thing though. you still end up with the identity that sin=sin which is always true. if you don't end up with an identity after manipulating both sides then you would instead only have a conditional eq

[uncut mulch]

there are ways to mix sides, but its frowned upon by most teachers as the required notation isn't included by students.
and most up them just want you to do
LHS = ...
= ...
=RHS proofs

[copper vigil]

yeah i see what you're saying

they just want me to do it a certain way

but it's not technically invalid to manipulate both sides as long as you still end up with an identity though right?

[uncut mulch]

as long as you include the extra symbols and/or comments

[copper vigil]

so it would look something like this

$\tan(x) \times \cos(x) =\sin(x) \iff \tan(x) = \sin(x)/\cos(x)$

is that how you use the iff symbol?

[obsidian monolith]

EndTimes:

[uncut mulch]

yeh

[copper vigil]

and then you just use the definition of tan(x) which is sin(x)/cos(x) so then the identity is proven

i see thanks

so it's a matter of notation really

you can't use equal signs like that

[uncut mulch]

in cases like these, you may also need to clearly describe how you're getting that. (i.e. dividing both sides by cos(x))

[patent beacon]

You run the possibility of something like
x = 2
x² = 4

But x = -2 satisfies that, so -2 = 2

Where instead showing that both sides are literally the same thing is error-free

[copper vigil]

not an identity though

[inner spear]

anyone know how to do this w law of cosines?

[uncut mulch]

do you know the law?

[inner spear]

i have no idea what this even is my teacher is just giving us work and not teaching the material

like we arent even doing zoom classes or anything

i kind of know how the law of sines works

but this is cosine so idk if that helps me at all

[uncut mulch]

$c^2 = a^2+b^2 - 2ab\cdot\cos(C)$

[obsidian monolith]

ramonov:

[uncut mulch]

have you ever seen that before?

[inner spear]

😭😭 nope

u already lost me dawg

is thst the rule

[uncut mulch]

yes

[inner spear]

like how the sin has the formula rule thijg

aii bet

so how can i use it to solve these

[uncut mulch]

if you know the sin law already, you should know the convention for labelling sides and angles of a triangle right?

[inner spear]

yes

[uncut mulch]

in your diagram, you are given 2 sides and the angle between them right?

[inner spear]

si

[uncut mulch]

which would be represented by a,b and C in the formula

conveniently this form matches the placement of variables of the diagram and you can just sub in the values and solve for c

[inner spear]

ok thank you

[proud jetty]

C) it’s sine graph how would I go finding the b and c value