X = 0 ; y = 10 ; 10 = a(0) + b(0) + c
c = 10
x=2 ; y = 15 ; 15 = a(4) + b(2) + 10 ; 5 = 4a+2b
x=4 ; y=18 ; 18 = a(16) + b(4) + 10 ; 8 = 16a + 4b
2(5) = (4a+2b)2
-
<u> 8 = 16a + 4b
</u> 2 = -8a
a = -0.25
b = 2
y = (1/4)x^2 + 2x + 10 ; 4y = x^2 + 8x + 40
Alrighty, so 280.05 rounded to the nearest integer is 280. 280.05 rounded to tenths is 280.1. add those two together and you get 560.1.
When you have something like this, all you need to do is substitute the values, the last is for what value of x
For the first one;
((x^2+1)+(x-2))(2)
(x^2+x-1)(2)
(2)^2+(2)-1
4+2-1
5
For the second one;
((x^2+1)-(x-2))(3)
(x^2-x+3)(3)
(3)^2-(3)+3
9-3+3
9
For the last one;
3(x^2+1)(7)+2(x-2)(3)
3((7)^2+7)+2((3)-2)
3(49+7)+2(3-2)
3(56)+2(1)
168+2
170
Answer:i think there was a pic to go with this to
B(answer)
