For #10
x=2 w(x)= 5
x=0 w(x)=-9
x=1 w(x)=-1
x=4 w(x)=7
To get these answers, you must use the p(x) table. When x=2, p(x)=6. So you would substitute p(x) for 6 and solve. That would be what you would do for the rest of teh problems.
For #12, the answer would be 5x^2+15x+5
Just mutiply the areas of the out side and inside and subtract them
Answer:
241
Step-by-step explanation:
solve using pemdas
2^(8)-10-15/3
256-10-15/3
256-10-5
246-5
=241
Let’s solve for “a” and “s”.
a = 927 - 45s
a = 792 - 36s
927 - 45s = 792 - 36s
927 - 792 = 45s - 36s
135 = 9s
s = 15
a = 927 - (45 x 15) = 252
Option C: center (1,2) and radius sqrt 5
263.27 m^2
rectangle : 15 x 10 = 150 m^2
the half circle at the top has a diameter of 6m. therefore, it’s radius is 3m. if it is cut in half, the part removed from the triangle height is 3m. therefore, the height of the triangle is 35 -15 = 20; 20 - 3 = 17. the height of the triangle is 17. the triangle area formula is a = 1/2(bh).
a = 1/2 (10)(17)
a = 1/2 (170)
a = 85 m^2
the circle would be a = πr^2 but divided by 2 bc it’s a half circle
a = π(3)^2
a = π9
a = 28.27 m^2
150 + 85 + 28.27 = 263.27 m^2
