Alan

Consider g(1) = f(1)^16  This will consist of the sum of the even degree coefficients and the sum of odd degree coefficients..  Now consider g(-1) = f(-1)^16.  This will consist of the sum of the even degree coefficients and the negative of the sum of the odd degree coefficients.  Add g(1) and g(-1).  This will eliminate the odd degree coefficients, but will leave twice the sum of the even degree coefficients. Hence to get the sum of the even degree coefficients you need:

                                                (g(1) + g(-1))/2  or  (f(1)^16  +  f(-1)^16)/2

Two points A and B are chosen at random on a circle of radius 1.  What is the probability that AB is greater than 1?

See Melody's reply to your earlier post.

See https://web2.0calc.com/questions/math-a-bit-of-help-needed#r1

The terms of a particular sequence are determined according to the following rules: * If the value of a given term is an odd positive integer $s$, then the value of the following term is $3s - 9$ * If the value of a given term is an even positive integer $t$, then the value of the following term is $2t - 7$. Suppose that the terms of the sequence alternate between two positive integers $(a,b,a,b,\ldots)$. What is the sum of the two positive integers?

2. Area of circle A = pi.r²    Area of circle/Area of segment = A/250pi.  But this must equal 360/100.

I'll leave you to take it from here.

I see hugo... has already done it!
 

1.   The distance XZ = XY + YZ = 6, so the radius of the largest semicircle is 3. Hence its area is (1/2)pi3² or A1 = 9pi/2.

The area of the next largest semicircle is A2 = (1/2)pi2² or A2 = 2pi

The area of the smallest semicircle is A3 = (1/2)pi1²  or A3 = pi/2

Hence the area of the shaded region = A1 - A2 - A3 = 9pi/2 - 2pi - pi/2 = 2pi

See the answer here: https://web2.0calc.com/questions/help-please_92732#r4 

By definition of a prime, the only possible products that result in a prime require two of the dice to be 1 and the other a prime.  So the possibilities are :

1 1 2, 1 2 1, 2 1 1    

1 1 3, 1 3 1, 3 1 1

1 1 5, 1 5 1, 5 1 1

9 possibilities out of a total of 216 possible throws.  So probability = 9/216