The answer is D if you combine like terms
4 quarts per gallon
<span>10 gallons: 4x10 = 40 quarts. </span>
<span>Add the 11 quarts to get 51 quarts. </span>
<span>51 divided by 3 = 17 quarts, or 4 gallons + 1 quart.</span>
So y=mx+b
m=slope
y=-3x+b
subsitutue 1 for x and 4 for y
4=-3(1)+b
4=-3+b
add 3 to both sides
7=b
the equaiton is y=-3x+7
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 =

It’s already in ascending order.
3/5= .6
5/8= .625
5/6= .833
7/4= 1.75