Match each interval with the corresponding rate of change for the function h(x) = 2–x on the interval.

1. –2 ≤ x ≤ 0
6
–0.5000
2. 0 ≤ x ≤ 2
5
–1.
3. –1 ≤ x ≤ 0
4
–0.3750
4. 0 ≤ x ≤ 1
3
–1.5000
5. –2 ≤ x ≤ –1
2
–2.
6. –2 ≤ x ≤ 2
1
–0.9375

Function: h(x) = 2 ^ (-x)

In the table below is the complete calculation of the rate of change.

The last column has the resulting rate of chages for each interval.

Interval          change of h(x)                   change of x      rate of change

-2 ≤ x ≤ 0        2^0 - 2^(-(-2)) = -3            0 - (-2) = 2        - 3 / 2 = - 1.5000

0 ≤ x ≤ 2        2^(-2) - 2^0 = - 0.75           2 - 0 = 2             -0.75 / 2 = - 0.3750

-1 ≤ x ≤ 0      2^0 - 2(-(-1)) = - 1              0 - (-1) = 1          -1/1 = -1.0000

0 ≤ x ≤ 1      2^(-1) - 2^0 = - 0.5              1 - 0 = 1               -0.5 / 1 = -0.5000                 

-2 ≤ x ≤ -1   2^(-(-1)) - 2^(-(-2)) = -2        -1 - (-2) = 1          -2 / 1 = -2.0000

-2 ≤ x ≤ 2    2^(-2) - 2^(-(-2)) = -3.75       2 - (-2) = 4           -3.75/4 = -0.9375

The answer would be b ! explanation: the formula for sa *surface area * is 2×l×w+2×l×h+2×w×h so in this case l=4w=12h=8so you use the formula to solve it and it give you the answer of 352^cm^2