Your answer's 25
You have to divide 125 times 5 to get your answer of 25
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Hope this helps</u></h2>
Answer:
Multiply the length times width:
30 1/2 x 9 1/3= 284.6
Answer:
32
Step-by-step explanation:
8/100 x 4
=
32/400
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
![L - Ak = e^{kt + C'} \\L - Ak = e^{kt}e^{C'}\\L - Ak = C"e^{kt} (C" = e^{C'} )\\Ak = L - C"e^{kt}\\A = \frac{L}{k} - \frac{C"}{k} e^{kt}](https://tex.z-dn.net/?f=L%20-%20Ak%20%3D%20e%5E%7Bkt%20%2B%20C%27%7D%20%5C%5CL%20-%20Ak%20%3D%20e%5E%7Bkt%7De%5E%7BC%27%7D%5C%5CL%20-%20Ak%20%3D%20C%22e%5E%7Bkt%7D%20%20%20%20%20%20%28C%22%20%3D%20e%5E%7BC%27%7D%20%29%5C%5CAk%20%3D%20L%20-%20C%22e%5E%7Bkt%7D%5C%5CA%20%3D%20%5Cfrac%7BL%7D%7Bk%7D%20%20-%20%5Cfrac%7BC%22%7D%7Bk%7D%20e%5E%7Bkt%7D)
When t = 0, A(0) = 0 (since the forest floor is initially clear)
![A = \frac{L}{k} - \frac{C"}{k} e^{kt}\\0 = \frac{L}{k} - \frac{C"}{k} e^{k0}\\0 = \frac{L}{k} - \frac{C"}{k} e^{0}\\\frac{L}{k} = \frac{C"}{k} \\C" = L](https://tex.z-dn.net/?f=A%20%3D%20%5Cfrac%7BL%7D%7Bk%7D%20%20-%20%5Cfrac%7BC%22%7D%7Bk%7D%20e%5E%7Bkt%7D%5C%5C0%20%3D%20%5Cfrac%7BL%7D%7Bk%7D%20%20-%20%5Cfrac%7BC%22%7D%7Bk%7D%20e%5E%7Bk0%7D%5C%5C0%20%3D%20%5Cfrac%7BL%7D%7Bk%7D%20%20-%20%5Cfrac%7BC%22%7D%7Bk%7D%20e%5E%7B0%7D%5C%5C%5Cfrac%7BL%7D%7Bk%7D%20%20%3D%20%5Cfrac%7BC%22%7D%7Bk%7D%20%5C%5CC%22%20%3D%20L)
![A = \frac{L}{k} - \frac{L}{k} e^{kt}](https://tex.z-dn.net/?f=A%20%3D%20%5Cfrac%7BL%7D%7Bk%7D%20%20-%20%5Cfrac%7BL%7D%7Bk%7D%20e%5E%7Bkt%7D)
So, D = R - A =
![D = \frac{L}{k} - \frac{L}{k} - \frac{L}{k} e^{kt}\\D = \frac{L}{k} e^{kt}](https://tex.z-dn.net/?f=D%20%3D%20%5Cfrac%7BL%7D%7Bk%7D%20-%20%5Cfrac%7BL%7D%7Bk%7D%20%20-%20%5Cfrac%7BL%7D%7Bk%7D%20e%5E%7Bkt%7D%5C%5CD%20%3D%20%5Cfrac%7BL%7D%7Bk%7D%20e%5E%7Bkt%7D)
when t = 0(at initial time), the initial value of D =
![D = \frac{L}{k} e^{kt}\\D = \frac{L}{k} e^{k0}\\D = \frac{L}{k} e^{0}\\D = \frac{L}{k}](https://tex.z-dn.net/?f=D%20%3D%20%5Cfrac%7BL%7D%7Bk%7D%20e%5E%7Bkt%7D%5C%5CD%20%3D%20%5Cfrac%7BL%7D%7Bk%7D%20e%5E%7Bk0%7D%5C%5CD%20%3D%20%5Cfrac%7BL%7D%7Bk%7D%20e%5E%7B0%7D%5C%5CD%20%3D%20%5Cfrac%7BL%7D%7Bk%7D)