Answer:
Step-by-step explanation:
Answer:
15) K'(t) = 5[5^(t)•In 5] - 2[3^(t)•In 3]
19) P'(w) = 2e^(w) - (1/5)[2^(w)•In 2]
20) Q'(w) = -6w^(-3) - (2/5)w^(-7/5) - ¼w^(-¾)
Step-by-step explanation:
We are to find the derivative of the questions pointed out.
15) K(t) = 5(5^(t)) - 2(3^(t))
Using implicit differentiation, we have;
K'(t) = 5[5^(t)•In 5] - 2[3^(t)•In 3]
19) P(w) = 2e^(w) - (2^(w))/5
P'(w) = 2e^(w) - (1/5)[2^(w)•In 2]
20) Q(W) = 3w^(-2) + w^(-2/5) - w^(¼)
Q'(w) = -6w^(-2 - 1) + (-2/5)w^(-2/5 - 1) - ¼w^(¼ - 1)
Q'(w) = -6w^(-3) - (2/5)w^(-7/5) - ¼w^(-¾)
In normal line integration, from what I understand, you are measuring the area underneath (,)
f
(
x
,
y
)
along a curve in the -
x
-
y
plane from point
a
to point
b
.
V = pi r^2 h
But it says 1/3 full so
1/3 pi r^2 h
1/3 * 3.14 * 4^2 * 6
100.48ft cubed
Whichrounds off to 100.5
