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100 C 2 ==4,950 ways of picking 2 numbers.

Of those 4,950 combinations, there are:

25  x  49 == 1225 ways of getting multiples of 4.

Therefore, the probability is: 1225 / 4950 ==49 / 198

tan (alpha + beta) = 2/3

tan gamma = 3/5

tan (alpha + beta + gamma) = 7/4

It works when x >= 0 and y >= 0.

There is NONE! Unless you have a mistake in your question. Check it carefully.

Why are you including the black royals???

(-2,2)

7^2 = 49

7^3  = 343

7^4 = 2401

7^5 = 16807

7^6 = 117649

7^7 = 823543 

7^8  = 5764801

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7^(2n).....when n is odd, the tens  digit  =  4

7^(2n).....when n is even, the tens digit  = 0

7^(2008)  = 7^(2 * 1004)  →   the tens  digit   =   0

2 i think

So 52 cards in a deck.

Red is half.

So 26.

Then the black J Q Ks

6 black J Q Ks

Meaning (26+6)/52 = 0.615 approx.

See my answer to the first one under your original post

tan α = BD / AD  =  6/2  =  3

tan β = ED / AD  =   2/2 =  1

tan ( β + γ)  = CD / AD  =  4/2 =  2

f(g(x))  =  3 ( 2x + 3) + 4 =  6x + 13  = h(x)

Write

y = 6x + 13      get x by itself

(y - 13)   /  6  = x        "swap" x and  y

y  = (x -13)  /  6   =   h^-1(x)

a + b = 12.

\(x - \frac{1}{x} = 3\)

\(x^2 - 1 = 3x\)

\(x^2 - 3x -1 = 0\)

\(x = \frac{3\pm\sqrt{13}}{2}\)

Because Triangle ADB is a right triangle, DB = \(\sqrt{81-16} = \sqrt{65}\). 

\(sin^{-1}(\frac{1}{2}) = \angle{DCA} = 30\), \(DC = 4\sqrt{3}\).

\(\angle{B} \cong \angle{B}, \angle{D} = \angle{E}, \) Triangle BDA is congruent to Triangle BEC. 

\(\frac{x}{CB} = \frac{4}{9} = sin(\angle{B})\),

 \(CB = \frac{16\sqrt{3}+4\sqrt{65}}{9}\). Use the Pythagorean Theorem to solve for AE. 

Tried my best. Please correct me if I did anything wrong. 

N=1991*1993*1995*1997*1999==3160183 6203377055

Sum==0 + 5 + 5 ==10

7^2008 mod10^10==8326964801 - these are the last 10 digits.

This is what a path would look like: 

We have the equation: \(x - {1 \over x} = 3\)

Solve as follows: 

\({x^2 \over x} - {1 \over x} = 3\)

\({x^2 - 1 \over x} = 3\)

\(x^2 - 1 = 3x \) 

\(x^2 - 3x - 1 = 0\)

Now, just solve the quadratic...