Please answer in order

For this first problem, for the first part, you would take the derivative of the given equation s = 7t + 4t^2, because v = ds/dt. So, this would give you. . .v = 7+8tSo, after 3 seconds, the velocity would be. . .v = 7+8*3 = 31 m/s.For the second part, you would plug in 47 for the left side of our derivative, for v, then find t. So, we would have. . .47 = 7+8tSolving for t, we get t = 5 sec.For the second problem, you have to take the derivative of each function first. So, we need to find P'(x) and Q'(x). For P, that's going to be a derivative of a product. That follows a formula that reads something like. . .[F(x)G(x)]' = F(x)G'(x) + F'(x)G(x)
Now, not given the formulas, we have to read the values off the graph. For the derivatives, those are the slopes of the lines at those points.
So, F(2) = 3 and G(x) = 3. But, for the derivatives, we need the slope. For F'(x), the slope is horizontal, so F'(x) = 0. And, for G'(x), that is a constant 1/2 there. So, plugging all of these numbers in. . .P'(2) = [F(2)G(2)]' = 3*1/2 + 0*3 = 3/2We do similarly for Q'(x), except this would be a derivative of a quotient. So, that would be. . .[F(x)/G(x)]' = [G(x)*F'(x) - F(x)*G'(x)]/[G(x)^2]Plugging in these numbers. . .Q'(7) = [F(7)/G(7)]' = [G(7)*F'(7) - F(7)*G'(7)]/[G(7)^2]Reading the values off the graphs. . .Q'(7) = [F(7)/G(7)]' = [1*1/4 - 5*(-2/3)]/[1^2]= [(1/4) + (10/3)]/1 = 43/12