All of them but A because hexagon as no acute angel
Let the boy be B and the girl be G
there are six possible outcomes
BGG, GBB, GBG, BGB, BBB, and GGG
therefore the probability of getting one boy and two girls is

also the probability of getting 3 girls is
<h3>
Answer: f( h(x) ) = 2x - 4</h3>
Work Shown:
f(x) = x - 7
f( h(x) ) = h(x) - 7
f( h(x) ) = 2x+3 - 7
f( h(x) ) = 2x - 4
Explanation:
In the second step, I replaced every x with h(x). In the next step, I replaced the h(x) on the right hand side with 2x+3. From there I combined like terms.
Solve for x:
x^2 + 10 x + 12 = 36
36 = 36:
x^2 + 10 x + 12 = 36
Subtract 12 from both sides:
x^2 + 10 x = 24
Add 25 to both sides:
x^2 + 10 x + 25 = 49
Write the left hand side as a square:
(x + 5)^2 = 49
Take the square root of both sides:
x + 5 = 7 or x + 5 = -7
Subtract 5 from both sides:
x = 2 or x + 5 = -7
Subtract 5 from both sides:
Answer: x = 2 or x = -12 Thus the Answer is A.