Recent content by happyg1

With ONE coin, If you get a T first, you are looking for another T. If that doesn't happen and you still get a T first, then get a H, you are looking for another H for the 2 in a row. Here ORDER MATTERS. That's a permutation. YOu have to count all of them. In the poker problem, you have 52...

4/12 does not equal 6/2. Reread the problem.

\frac{8}{3}=\int_{0}^{\sqrt b} b-x^2 dx =bx-\frac{1}{3}x^3 from 0 to \sqrt b

set it equal to 8/3! It works!

Yes, My tex skills are a little rusty. i edited it and fixed it..

I found the desired area to be 8/3 (I just chopped it in half and did 0 to 2). Then integrated \frac{8}{3}=\int_{0}^{\sqrt b} b-x^2 dx Solved for b...

You have to account for the possibility that 2 or 3 or 4 are all dealt the same suit...that's where the exactly one, one particular comes in.

expand your parenthesis and put the -2 that results in with the constant. You have the same answer, just in a different form.

I did this and it turns out kinda ugly. I got 0 for the coefficient of the squared term (well, 1.5x 10^{-14} )I got 1 for the coefficient of the x term; and I got 3x10^{-14} for the constant. Looks pretty linear there...although not EXACTLY linear...but I may have made a mistake. EDIT: I...

I thought that i=\sqrt{-1} I'm confused...:)

The identity doesn't have 3^2, it has 1 \sin^2 T + \cos ^2 T=1 , not 9. You are mixing up the the identity with the triangle. So far you have \frac{2x}{3}\sqrt{1-\frac{x^2}{9} Then you find the common denominator under the radical...

You said that you used sin^2+cos^2=1 to solve for the cos T. It should be \cos T= \sqrt{1-sin^2 T} \cos T=\sqrt {1-\frac{x^2}{3^2}}. Find a common denominator under the radical and a 3 pops out in the denominator.

for the first one, see if this helps... By Fermat's Theorem (assuming that p and p are relatively prime...) we have q^p=q mod p . We can cancel q from both sides since gcd(q,p) = 1, so q^{(p-1)} = 1 (mod p). Also p^{(q-1)} = 0 (mod p) we get p^{(q-1)} + q^{(p-1)} = 1 (mod p) can...

That's the method that I learned for calculating the determinate in HS. You copy the first column over on the Right side so the diagonals are all nice and in line...then you calculate the number as shown above...forgot the + signs on the last line there. (3)+(2)+(1) - (3)(2)(1) should have + in...

Hey, You have to drop your absolute value bars on the second line of your solution up there. The equation of the line from -1 to 0 is -x and from 0 to 1 it's just x...that changes the sign on your 1/2 aand gets you what you know is correct. CC