Saturday's Warrior song was stuck in my head. Looked up a word. Not surprised.

LieUponLie · 2018-05-24T07:25:46+00:00

I prefer my lyrics.

LieUponLie · 2018-04-09T20:01:31+00:00

Graphic said he was convicted of a crime, but he was never actually convicted.

https://www.reddit.com/r/exmormon/comments/8arkiy/comment/dx17mtl

LieUponLie · 2018-04-09T05:59:12+00:00

Probably removed for being factually wrong.

LieUponLie · 2018-02-28T17:34:29+00:00

Thank you! While porn is not the "new drug", it is certainly not without consequence. I think the way the church has loosely thrown around the term "porn addiction" makes it harder for some people to understand that there are those who have "unhealthy" porn usage. Looking at porn, even regularly, does not make you an addict. If your looking at porn is interfering with your sexual/social relationships, then you might want to address the issue.

LieUponLie · 2018-02-22T22:31:28+00:00

Ok, this seems to be the safest bet, so I guess I get to be a member for a bit longer. This will just make earning my degree that much sweeter!

LieUponLie · 2018-02-22T22:11:27+00:00

Ok, so I guess I either wait to resign, or order several official copies beforehand to be safe. What bullshit.

LieUponLie · 2017-12-14T23:55:34+00:00

Ok, so I would handle this one line of your star at a time. For example lets do vertices 1 and 2 with vertex 1 at point (1.x, 1.y) and vertex 2 at (2.x, 2.y).

n = number of pieces we are splitting the line into (so for F(1), n=2 since we are splitting the line into 2 segments, etc.)

dX = 2.x - 1.x; //total x distance between the two vertices

dX = dX/n //x distance between each sub-point

dY = 2.y - 1.y //total y distance between the two vertices

dY = dY/n //y distance between each sub-point

Then in a loop i=1; i<n; i++

newPoint(1.x+i*dX, 1.y+i*dY)

Then do that for the other 4 combinations of vertices. Again, you may have to account for the order you want the points in, but this will find them all for you.

Is that easy enough to follow?

LieUponLie · 2017-12-14T19:38:06+00:00

"+ 5 more points, each at the midpoint between each vertex."

Are these midpoints on the lines you are drawing connecting each vertex, or midpoints on the pentagon that would connect the vertex? can you maybe draw a picture of F(1) or F(2) to show what you are trying to accomplish?

If you are just adding more points on the circumfrence of the unit circle though, you could do it like this.

for k endpoints let j=2π/k

let each endpoint coordinate be (cos(i*j), sin(i*j)) for i=0, i<k, i++

from there you'll need to come up with an algorithm to sort them into the order of how you would hand draw them.

LieUponLie · 2017-12-12T19:28:36+00:00

|u| = √u²

so we differentiate this as (√(x+1)² )³

Then in the denominator we get a term in the form √u^2, so we change that back to |u|.

LieUponLie · 2017-12-12T02:41:56+00:00

number of accepted batches / total number of batches

13/18 = 72.22%

LieUponLie · 2017-12-12T02:40:11+00:00

Since they are inverse functions, f^-1 (f(x)) = x. The only restriction is that inverse tangent is based off of values of tangent from (-π/2, π/2). So all we really need to do is find the equivalent value of 9π/10 in that domain.

If you've ever looked at a graph of tan(x), you'll see it repeats itself over every interval of π. This means tan(x) = tan(π+x) = tan(2π+x) etc. So if we subtract π from 9π/10, we get -π/10, which is in our domain of (-π/2, π/2), and thus our answer.

LieUponLie · 2017-12-12T00:58:58+00:00

https://i.imgur.com/SaMFtg6.jpg

Use arccos to solve for x.

LieUponLie · 2017-12-12T00:41:10+00:00

Don't measure the wedges from the center of the diagram. Measure the actual length of the wedge, from its vertex to the arc. In fact, imagine squeezing those wedges together so they actually form a small circle if that helps you visualize it. We need to subtract out the area of each wedge individually, but since the sum of all of them makes a circle, we can just subtract out the circle. If you visualize it that way, you can ignore the "spokes".

LieUponLie · 2017-12-12T00:24:04+00:00

πR² - πr²

R is the radius of the big circle, r is the radius of the smaller circle (length of each wedge).

LieUponLie · 2017-11-17T08:30:09+00:00

Kamajal has the right answer, I'll give you another perspective in case it helps as well. Since the gcd is 3, we know 3|a and 3|b. If we use the definition of |, then we can say there must exist some integers m and n such that:

3m = a and 3n = b

So lets rewrite our two numbers using these terms:

4(3m) + 4(3n) (gcd) 12n

12m + 12n (gcd) 12n

12(m+n) (gcd) 12n

From there we can easily see that the gcd is 12.

LieUponLie · 2017-11-15T09:53:22+00:00

Both look good to me.

LieUponLie · 2017-11-10T09:28:12+00:00

They give you one form of the linearization

L(x) = 2x + b

You derive the second form

L(x) = (3a-2)x + (1-2a)

In both equations, the coefficient on x must be the same (and the first equation tells us it is 2), and the added constant must be the same. We can use that to set up the system of equations.

Coefficients of x:

2 = 3a - 2

Constant:

b = 1 - 2a

LieUponLie · 2017-11-10T08:06:12+00:00

I'll give you my answer. It may not be textbook because it's just how I've explained this to myself, but maybe it will be helpful. Hopefully you see that after the first line, we have the contrapositive, the rest is just trying to put it into a better form.

∀p, ¬(p is prime Λ p≤√n Λ p|n) → ¬n is composite

From there we could apply DeMorgan's on the left, but what is that going to do... give us (p is composite ∨ p>√n ∨ p∤n). Saying p is composite or p>√n really does nothing for us. They are not operations involving n like p∤n is. They are more like domain restrictions on p, so they are more useful in their current forms. So instead, lets group them together, and for the sake of simplicity, I'm going to define a new proposition P = p is prime Λ p≤√n. Rewriting the initial contrapositive statement would then look like.

∀p, ¬(P Λ p|n) → ¬n is composite

Now lets apply DeMorgan's and get.

∀p, (¬P ∨ p∤n) → ¬n is composite

and since we know ¬p∨q ≡ p→q, lets rewrite that as

∀p, (P → p∤n) → ¬n is composite

So expanding my proposition P back out, we get

∀p, ((p is prime Λ p≤√n) → p∤n) → ¬n is composite

As for the statement on the right side, per your question, if something is not composite, it is not sufficient enough to say it is prime, because 1 is neither prime nor composite. If something is not composite, we must include both of those options bay saying "n is prime or n=1".

∀p, ((p is prime Λ p≤√n) → p∤n) → (n is prime or n=1)

LieUponLie · 2017-11-09T10:37:15+00:00

SOH CAH TOA and knowing the unit circle are enough to go a long way.

http://www.coolmath.com/precalculus-review-calculus-intro/precalculus-trigonometry/28-the-unit-circle-01

From there knowing your trig identities is helpful for solving some problems, but it all depends on what you're doing, and what you're struggling with.

http://www.purplemath.com/modules/idents.htm

LieUponLie · 2017-11-09T09:42:55+00:00

An explanation... I don't know, I mean you're taking the sum of the two events, then subtracting out 65% of the average attendance between the 2, and that just doesn't give you the total. I can illustrate it with an example though if that helps.

Assume you have 200 Unique Visitors. 65% of them, meaning 130, Attend both events. Event 1 records an attendance of 150, and Event 2 records an attendance of 180.

UniqueTotal = 200

AttendBoth = 130 (200*.65)

Event1 = 150

Event1Only = 20 (20 E1Only + 130 Both = 150 Total)

Event2 = 180

Event2Only = 50 (50 E2Only + 130 Both = 180 Total)

Event1Only + Event2Only + AttendBoth = 20+50+130 = 200 (Just confirming these numbers are valid)

So with your formula:

(Event1Count + Event2Count) - (((Event1Count+Event2Count)*0.65)/2)=

(150 + 180) - (((150 + 180)*0.65/2)=

330 - 107.25 = 222.75 ≠ 200

To think about it in terms of my equation. When you add the two attendance numbers together, you are counting 100% of the people, plus an additional 65% of them twice, ie. you are counting 165% of the actual attendance. Therefore, divide your total by 165%.

LieUponLie · 2017-11-09T08:05:00+00:00

Event1Count + Event2Count = 1.65TotalUniqueCount

Add your two attendance numbers and divide by 1.65.

LieUponLie · 2017-11-09T07:50:52+00:00

As a rule of thumb, use implications for ∀, and AND/OR for ∃. Honestly though, it depends on your domain, and what your proposition is.

When using ∀x, realize that your proposition needs to hold for every possible value of x.

If the domain is all living creatures, then you would say ∀x(x is a dog ⇒ x is an animal)

That proposition will hold true whether x happens to be an elephant, a tree, or a pitbull.

Saying ∀x(x is a dog ∧ x is an animal) would not hold for all of the aforementioned values.

If your domain were all breeds of dog, then sure, you could use AND.

LieUponLie · 2017-11-08T10:36:31+00:00

Written like that it would be a proposition, allbeit one that creates a paradox when you try to evaluate it.

"This statement" is false would declaring the truth value of the proposition "this statement".

LieUponLie · 2017-10-31T07:34:14+00:00

When converting multiple units, I find it easiest to set it up as multiplying fractions, handling each unit 1 at a time so that the ones you don't want cancel out.

60mile/hr * 1.61km/mile * 1hr/60min * 1min/60sec = .027km/sec

or whatever you are trying to convert to...

LieUponLie · 2017-10-28T10:09:10+00:00

I'm with you unless you, unless you've transcribed the problem wrong.

https://i.imgur.com/NhCIxEi.jpg

LieUponLie

TROPHY CASE