If (x - 2) and (x + 1) are factors of
x + px? + qx + 1, what is the sum of p and q?

p + q = -3

Step-by-step explanation:

First we need to take the original equation, and factor it to a form that's easier to get two binomial factors from (i.e., let's get a quadratic):

x^3 + px^2 + qx + 1

= x (x^2 + px + q) + 1

Now that we have factored out the x, we have a quadratic trinomial which we know can be broken down into two linear binomials.  The problem gives us two linear binomials, so let's take a look.

(x - 2) (x + 1) = (x^2 + px + q)

x^2 - 2x + x -2 = x^2 + px + q

Now let's solve.

x^2 - x - 2 = x^2 + px + q

-x - 2 = px + q

From here, we can easily see that p = -1 (the coefficient of x) and q = -2.

Hence, p + q = -1 + -2 = -3.

Cheers.

D1 & d2 are the diagonal lengths 
-side of a rhombus, if diagonals are given, can be calculated by 
s = sqrt(d1^2 + d2^2) 
  this is just pythagorean theorem 

side  = sqrt(24^2 + 18^2) 
side = sqrt(900) 
side = 30 cm 

perimeter = 4*s 
perimeter = 4*30 
perimeter = 120 cm
thats all  i can with, idk about the distance part