Thank you so much! :) It's funny how it all comes back to you once someone's explained one thing.

You're welcome! :)

I'm really not sure if this counts, but I really need some one to explain this to me.

((x²)/(a²))-((y²)/(b²))=1

How am I supposed to know which way the long way goes? Also, how so I know if the lines are horizontal or vertical? If any of this doesn't make sense please just tell me, and I'll try to explain this better.

What do you mean by, "the long way"? The graph you're talking about right now is a hyperbola, maybe you were asking for an ellipse, x²/a²+y²/b²=1?

If you're stuck you can go to khanacademy.com. The person that runs the site gives math tutorials on everything. It saved me so much during school.

In which case a and b give the magnitude of the "radius" on the corresponding axis. So whichever is the larger is the "long way".

No. I'm talking about hyperbolas (hyperboli?).These are my notes on the subject. I'm pretty good at math, and when he puts notes on the board to copy, I can pretty much figure out what I'm doing at the end of problem A, but I think the next to last day of school just made my brain go on automatic shutdown.

I would but my computer doesn't support video files.

Hyperbolas goes left-right when the y² term is negative, and they go up-down when the x² term is negative.

The asymptotes of an up-down hyperbola are y=bx/a,-bx/a, and y=ax/b,-ax/b for left-right.

Hyperbolas only have vertical and horizontal asymptotes when they are a reciprocal function, y=a/f(x).

Amazing what a 10 minute session of fiddling around with hyperbolas on WolframAlpha can teach you.

Really, if you need to know more, the best website that I know of is Wikipedia.

I totally think I get this now.. Now I must go to see if I can actually graph this..

Conics question.

P(a cosÓ¨, b sinÓ¨) and Q[a cos(-Ó¨), b sin(-Ó¨)] are the extremeties of the latus rectum x=ae of the ellipse x²/a² + y²/b² = 1

(a) Show that cosÓ¨=e

This one was pretty easy. You just let a cosÓ¨ (the x value of P) = to ae (the x value of the latus rectum) so a cosÓ¨ = ae. Therefore cosÓ¨ = e

(b) Show that PQ has length 2b²/a

I used point gradient formula to find PQ and am stuck with 2b sinÓ¨. I have no idea how I'm meant to get sinÓ¨ = b/a which is obviously what I'm required to find. Any ideas? :/

Edit: I would normally ask the teacher the next day, but the exam is tomorrow. Urgh.

Never mind. Got it.

Just going to throw this one out there... I Love math lol.

Alright so I have like three problems I keep getting wrong and I need some help, so here I go

Concrete can be bought by the cubic yard. How much will it cost to pour a slab 17ft by 17ft by 2 inches for a patio if concrete costs 40.00$ per cubic yard? (I keep getting an answer that is way to high.)

Another one is a triangular prism with a regular square base that has all sides of the base 24ft and a slant height of 20 ft. I need the volume, so I need to find the actual height.


Last one for now: The volume of a sphere is 3000Ï€m³. What is the surface area to the nearest square. (I keep getting way off from the options, which are 2158m squared, 37699m squared, 165m squared and 1079m squared)

(17 ft)(1 yd/3 ft)(17 ft)(1 yd/3 ft)(2 in)(1 yd/36 in)($40/yd³) = $71.358

V = 4/3 Ï€r³ = 3000Ï€ m³
r = ((3/4)(3000Ï€ m³)/Ï€)^1/3 = 13.1037 m
SA = 4Ï€r² = 2157.72 m²

For the second problem, you have two sides of the triangle, 24 ft and 20 ft. Do you know that it's an equilateral triangle?

Edit: Never mind, of course it can't be equilateral... >_>

$40 per 27 cubic feet.
17 * 17 * 1/6 = 289/6 = 48.16... cubic feet
48.16... / 27 = 1.78 batches of concrete
Since you can only purchase in increments of cubic yards, you round up to 2
2 * $40 = $80
EDIT: I must have typoed when I initially did 17 * 17 * 1/6
Wouldn't a triangular prism have a triangular base? I am confused.


Find the radius
Volume = 4/3 * Ï€ * r³
3000Ï€ = 4/3 Ï€ * r³
2250 = r³
13.1 = r

Surface Area = 4 * Ï€ * r²
A = 4 * 3.14159 * 171.7
A = 2157.7 m²

Lol, we got different answers for the first one, Blake. :P

Edit: Okay, now our answers are the same, except that we made different assumptions about whether fractions of a cubic yard could be purchased.

Double-edit: Oops, you're right. I missed that condition in the problem.

I posted this in the ask section as well, and got two different answers, none of which were what I got.

What is the product of this series?
(x – a) (x – b) (x – c) … (x – z)

The answers I received were "that is the product, in factored form" and "X^26"

But the answer I got was 0, because I'm assuming that the series would eventually get to (x - x) which would be 0, and everything multiplied by that would then also be 0. But I'm wondering if I'm misunderstanding the problem in the first place...

Whoops, yeah it's not a triangluar prism, it's a pyramid. And thanks :D

Edit: I kept converting 17ft into inches instead of the other way around and was getting a huge number. But wouldn't I be able to do it that way also?

@k, the answer is certainly not x²⁶. Assuming that the series of factors is all the letters of the alphabet, including "x", and that the "x" towards the end of the series is the same as the "x" in each of the terms, the answer would indeed be 0. More of a math riddle than a practical problem I suppose.

SuperJesus, look at the left half of the pyramid from the side. It forms a right triangle, and you're given the hypotenuse (20 ft) and can find the lower leg length by dividing the base side length by 2 to get 12 ft. By the Pythagorean theorem, (20 ft)² = (12 ft)² + height², giving you a convenient height of 16 ft. Can you find the volume from there?

Dammit, I was looking at the thing wrong, was doing the other thing as the hyp

k, if the (x - x) term cancels to 0, it is a simple math riddle, as described above. If, however, the (x - x) term is not zero, you have a scary expression that is best left in factored form; anything else is too unwieldy for mere mortals.

Under what algebra is (x - x) not 0? o_O

Find all real and complex solutions to x^5-1=0.

Isolate the x component.
x⁵=1
Take the fifth root of each side.

And then apply de Moivre generously.

(cos x + i sin x)ⁿ = cos(nx) + i sin(nx)

x can be 0, 2pi/5, 4pi/5, 6pi/5, or 8pi/5
cos(0) + i sin(0) = 1
cos(2pi/5) + i sin(2pi/5) = .309 + .951i
etc.

When they're different x's.