Karl is putting a frame around a rectangular photograph. The photograph is 14 inches long and 10 inches wide, and the frame is t
1 answer:
<h2>The area of the framed photograph = 1156 
</h2>
Step-by-step explanation:
Given,
The length of photograph (l) = 14 inches and
The breadth of photograph (b) = 10 inches
To find, the area of the framed photograph = ?
We know that,
The area of the framed photograph = (l + 2b)(l + 2b)
= (14 + 2 × 10) (14 + 2 × 10)
= (14 + 20) (14 + 20)
= (34) (34)
= 1156
∴ The area of the framed photograph = 1156
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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 =