This problem deals with eye color. an inherited trait. for purpose of this problem, assume that only two eye colors are possible, brown and blue. we use b to represent a blue eye gene and B a brown eye gene. if any B genes are present, the person will have brown eyes. the table shows the four possibilities for the children of two Bb( brown-eyed) parents, where each parent has one of each eye color gene. Second parent: B b
First parent: B BB Bb
b Bb bb

step-by-step explanation:

always plug in x. in this case, x equals 4.

i'm going to first solve 4(2+3x)

1) 4(2+3(4)) use pemdas to solve.

i'm going to multiply whats in the parenthesis first, so 3 times 4=12.

2)4 (2+12) i'm then going to multiply 4 by both numbers in the parenthesis (because multiplying comes before adding)

4 times 2= 8.   4 times 12= 48

4) (8+48)= 56 (answer)

i'm then going to solve 8+12x and plug in 4 for x.

1) 8 +12(4)

multiplying comes before adding, so i'm gonna multiply 12 and 4.

12 times 4= 48.

2) 8+48=56

both equal the same!

b

step-by-step explanation: