Pangolin14

We must have that the two lines intersect at x=3. Thus, x+2 = 2x-3a, so 5 = 6-3a, and a = 1/3. Thus, 2x-1 and x+2 intersect at (3,5) which is the endpoint of y in x>3, so a = 1/3.

Just function bash; 1/59 for 1st, 59/57 for second -57/61 for third, 61/175 for fourth, 175/53 for fifth, and finally -53/297

Exponent rules, 8^x = 9^x*3, so x = x = -log(3)/log(9/8)

I have already answered #1

#2: Evaluate and use PEMDAS, so 3^3 = 27, 27 - 5 + 8 = 22 + 8 = 30, so 2 * 30 = 60

first, 2^3 = 8, so 8 * (8+4-10). Next, we do the things inside parenthesis, so 8+4-10 = 2. Next, 8x2 = 16.

ABC is obtuse with obtuse angle at A. Therefore, the orthocenter is outside the triangle.

The trick is that you have to find AB and EC in terms of DE. Let DE be x. Then BD is 3x and BE = 2x√2. 

As the midpoint of the hypotenuse is center of the circumcircle of a right triangle, AC is the diameter, so BD, AD,DC are all equal and radii of the same circumcircle. Thus, we have AC = 6x. Thus, AE = 4x and EC = 2x. So, AB = sqrt(16x^2 + 9x^2) = 5x, so, AB/EC = 5/2

-i is not real. Thus, we square it. -i^2 = i^2 =-1. -1 is real. Thus, we square 1. After a whole chain of squaring 1 and -1, we end up eventually with 1, so the answer is 1.

I seem to have overlooked the 4,0,0,0 case... 

I might be wrong, but for the 2,1,1,0 case the number of ways should be the number of arrangements of 2,1,1,0 * number of ways to pick, and the number of arrangements is 4!/2! = 12, and the number of ways to pick the bracelets: 4C2 for the first pair, 2C1 for the second, 1C1, multiply, 6 * 2 = 12, then * 24 = 288, so perhaps the pair of 1s was left out in your solution?

The polynomial is 2^x, so g(0) = 1.