Answer:
C
Step-by-step explanation:
3*-4
Answer:
D = L/k
Step-by-step explanation:
Since A represents the amount of litter present in grams per square meter as a function of time in years, the net rate of litter present is
dA/dt = in flow - out flow
Since litter falls at a constant rate of L grams per square meter per year, in flow = L
Since litter decays at a constant proportional rate of k per year, the total amount of litter decay per square meter per year is A × k = Ak = out flow
So,
dA/dt = in flow - out flow
dA/dt = L - Ak
Separating the variables, we have
dA/(L - Ak) = dt
Integrating, we have
∫-kdA/-k(L - Ak) = ∫dt
1/k∫-kdA/(L - Ak) = ∫dt
1/k㏑(L - Ak) = t + C
㏑(L - Ak) = kt + kC
㏑(L - Ak) = kt + C' (C' = kC)
taking exponents of both sides, we have
When t = 0, A(0) = 0 (since the forest floor is initially clear)
So, D = R - A =
when t = 0(at initial time), the initial value of D =
If your just looking for x then x=2
Alright, let's convert to decimals for these calculations: fractions could get messy.
1/5 = .2
1/2 = .5
Imagine a rectangular solid in your mind. How many sides does it have?
6. In the image shown you have 3 facing you. In addition to those, there are 3 not shown. They will correspond to the calculations you make on the front, though, so all you have to do is double the values you get.
Area is length * width, or length * height, or width * height. So:
A = 3.2 (length) * 4.5 (height) for the front face of the rectangular solid(as well as the back face.)
A = 5 (width) * 4.5 (height) for the right face of the rectangular solid (and the one on the left, away from you.)
A = 3.2 (length) * 5 (width) for the top of the solid(and the bottom).
Calculating these values, we get that
A=14.4
A=22.5
A= 16
So that's 3 out of 6 values for the full surface area.
Like I said though, these values can merely be doubled for the complete area.
Add these three together:
14.4+22.5+16=52.9
multiply by 2 to account for the other 3 sides
52.9 * 2 = 105.8
105.8 is the surface area.
Answer:
thousandths
Step-by-step explanation: