Pand q are the solutions to the quadratic equation x^2 + 4x + 6 = 0.
p^2 and q^2 are the solutions to the quadratic equation x^2 + bx + c = 0.
find b + c.

p+q= -4

Step-by-step explanation:

If 'p' and 'q' are solutions to x^2 + 4x +6=0 then that same eq. can be written as (x-p)*(x-q).

Hence we have the following relationship

(x-p)*(x-q)=x^2 + 4x +6

Distributing terms x^2 -xq-xp+pq=x^2 + 4x +6

We have obtained 2 equations if we look at the coefficients of the quadratic equation:

From x coefficients we have the equation: -p-q=4

and from independent terms we have: p*q= 6

From the first eq. we have -(p+q)=4, so p+q= -4

r(0,90)(x,y) . t(-5,0)

last option

step-by-step explanation:

∆def to ∆d"e"f"

- rotate 90 counterclockwise degrees

- then translated 5 to the left (x --> x - 5)

answer is the last option

51

step-by-step explanation:

73-22=51 the   mass number minus the amount of protons gives you the amount of neutrons