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Here is my attempt:

1024^2 ==1,048,576 

There are: 1^6,  2^6,  3^6,  4^6,  5^6,  6^6,  7^6,  8^6,  9^6,  10^6 =10 perfect squares that can be expressed as 6th powers of some integers.

Since there are 1024 perfect squares, then the probability is: [10 / 1024] * [9 / 1023] =90 / 1,047,552= 15 / 174592

(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32)

(1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625, 676, 729, 784, 841, 900, 961, 1024) =perfect squares of above 32 numbers

There are only 2 perfect squares that can be expressed as 6th powers of some integers. They are:

64 =2^6   and   729 =3^6

Therefore, the probability is: 2 / 32 =1 / 16

20130118_{12}  mod 18 (base 10) =2

20130118_{12}  mod 18 (base 12) =0

8 C 3 ==56 combinations

(2, 3, 4) , (2, 4, 5) , (2, 5, 6) , (2, 6, 7) , (2, 7, 8) , (3, 4, 5) , (3, 4, 6) , (3, 5, 6) , (3, 5, 7) , (3, 6, 7) , (3, 6, 8) , (3, 7, 8) , (4, 5, 6) , (4, 5, 7) , (4, 5, 8) , (4, 6, 7) , (4, 6, 8) , (4, 7, 8) , (5, 6, 7) , (5, 6, 8) , (5, 7, 8) , (6, 7, 8) , Total =  22 combinations meet the condition for 3 sides of a triangle.

Therefore, the probability is: 22 / 56 == 11 / 28

Find all real numbers a that satisfy.

Hello Guest!

\(1/(a^3 + 7) - 7 = -a^3/(a^3 + 7) + 5\\ \dfrac{1}{a^3+7}-7=-\dfrac{a^3}{a^3+7}+5\\ \dfrac{1+a^3}{a^3+7}=12\\ 1+a^3=12a^3+84\\ 11a^3=-83\\ \color{blue}a=\sqrt[3]{-\frac{83}{11}}=-1.961\)

  !

the sum of all possible values of x

Hello Guest!

\(2x(x-10)=-50+x^2-30x\\ 2x^2-20x+50-x^2+30x=0\\ x^2+10x+50=0\\ \color{blue}x\in \{-5-5i,-5+5i\}\)

The sum of all possible values of x is  -10 .

  !

That is true. There are: 1215 even products + 81 odd products ==1,296 total products.

\({1 \over (a-1)} + {1 \over (b-1)} = {(b-1) \over (a-1)(b-1)} + {(a-1) \over (a-1)(b-1)} \)

\({(b-1) \over (a-1)(b-1)} + {(a-1) \over (a-1)(b-1)} = {a + b -2 \over {ab−a−b+1}}\)

\({a + b -2 \over {ab−a−b+1}} = {a + b -2 \over {ab−(a+b)+1}}\)

\(2x^2-7x+2 = x^2-11x+1\)

\(x^2 + 4x + 1 = 0\)

\(a + b = -{b \over a} = -4\)

\(ab = {c \over a} = 1\)

\({(-4) - 2 \over (1) - (-4) + 1} = {-6 \over 6} = \color{brown}\boxed{-1}\)

The product is different (i think) because to get an even product, you need even x even or even x odd; odd only happens with odd x odd. 

[BCE] = 14.

1/(a - 1) + 1/(b - 1) = -13/2.

a = 7/2.

The radius is 3*sqrt(17)

The answer is 19/2.

(a,b,c) = (-1,5,12)

The probability is 2/45.

Yes my first answer is wrong because of a careless error   4!/(2!2!) = 6   [not 3]

so     1+6+1=8   the answer is 8/16 = 1/2

Angle HAC = 155 degrees.

Angle RAC = 38 degrees.

I agree with BuilderBoi:  648 / 1296 = 1/2

Builderboi? What about the product

There are just as many ways to roll even as odd, so the answer is 1/2