H(t)=(t+3)^2+5 over which interval does h have a negative average rate of change?

( -∞, -3 ) or any subinterval of  ( -∞, -3 )

Step-by-step explanation:

h(t)=(t+3)^2+5 has the form

h(t) =(t-h)^2 + 5, indicating that h(t)=(t+3)^2+5 represents a parabola with vertex at (-3,5) that opens up.  This (-3,5) is the minimum point of the graph.

To the left of t = -3, h(t) is decreasing; to the right of t = -3, h(t) is increasing.

Thus, it is on any interval to the left of t = -3 that the rate of change of this function is negative (the function is decreasing  on ( -∞, -3 ).

93.8

step-by-step explanation:

lets say, first number = x

second number = 3x+7

total = 49

(x)+(3x+7)= 49

4x+7 = 49

4x = 42

x = 42/4

x= 10.5

x= 10.5

3x+7 = 3(10.5)+7 = 93.8

hope this