maria created a graph of ( b(t) ), the temperature over time. for the interval between ( t = 3 ) and ( t = 7…

maria created a graph of ( b(t) ), the temperature over time. for the interval between ( t = 3 ) and ( t = 7 ), the average rate of change in her graph of ( b(t) ) is 8. which statement must be true?\nthe temperature was 8 degrees higher when ( t = 7 ) than when ( t = 3 ).\nthe temperature was 8 times higher when ( t = 7 ) than when ( t = 3 ).\nthe temperature was 32 degrees higher when ( t = 7 ) than when ( t = 3 ).\nthe temperature was 2 degrees higher when ( t = 7 ) than when ( t = 3 ).

maria created a graph of ( b(t) ), the temperature over time. for the interval between ( t = 3 ) and ( t = 7 ), the average rate of change in her graph of ( b(t) ) is 8. which statement must be true?\nthe temperature was 8 degrees higher when ( t = 7 ) than when ( t = 3 ).\nthe temperature was 8 times higher when ( t = 7 ) than when ( t = 3 ).\nthe temperature was 32 degrees higher when ( t = 7 ) than when ( t = 3 ).\nthe temperature was 2 degrees higher when ( t = 7 ) than when ( t = 3 ).

Answer

Explanation:

Step1: Recall the formula for average rate of change

The formula for the average rate of change of a function (y = B(t)) over the interval ([a,b]) is (\frac{B(b)-B(a)}{b - a}). Here, (a = 3), (b=7), and the average rate of change is (8).

Step2: Substitute values into the formula

Substitute into (\frac{B(7)-B(3)}{7 - 3}=8).

Step3: Simplify the denominator

Since (7-3 = 4), the equation becomes (\frac{B(7)-B(3)}{4}=8).

Step4: Solve for (B(7)-B(3))

Multiply both sides of the equation (\frac{B(7)-B(3)}{4}=8) by (4). We get (B(7)-B(3)=8\times4).

Step5: Calculate the product

(8\times4=32), so (B(7)-B(3) = 32). This means the temperature was (32) degrees higher when (t = 7) than when (t=3).

Answer:

The temperature was 32 degrees higher when (t = 7) than when (t = 3).