Suppose h is a function such that h(2) = −3, h'(2) = 3, h''(2) = 4, h(5) = 5, h'(5) = 6, h''(5) = 12, and h'' is continuous everywhere. evaluate 5 h''(u) du.

3

Step-by-step explanation:

It seems like your question is not complete.

So I will asume that it is like the one in the image.

But probably the answer is not 3 if the range is different from (1-2), eventhough, you just have to follow the same steps.

logv3(x+2)=1

rewrite in exponential form:

logb(x) = y is equal to b^y =x

b = 3, y = 1

now solve for x:

3^1 = x+2)

x = 3-2

x = 1

see the picture for the graph.

51

step-by-step explanation:

73-22=51 the   mass number minus the amount of protons gives you the amount of neutrons