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Kryger [21]
3 years ago
9

The product of 8, and a number increased by 7. is 104. What is the number?

Mathematics
1 answer:
Lynna [10]3 years ago
4 0
Answer:
12.125
Step-by-step
8x+7=104
-7 -7
8x=97
/8 /8
x=12.125
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Find the area of the figure.( sides meet at right angles.)picture below
Tema [17]

You can divide the figure as follows:

  • A vertical rectangle, 5 meters tall and 2 meters wide, on the top right.
  • A horizontal rectangle, 2 meters tall and 5 meters wide, on the bottom left
  • A 2x2 meters square on the bottom right, where the two rectangles meet.

The area of the rectangles is 5x2=10 meters squared, while the area of the square is 2x2=4 meters squared.

So, the total area is 10+4=14 meters squared.

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3 years ago
Litter such as leaves falls to the forest floor, where the action of insects and bacteria initiates the decay process. Let A be
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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

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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}

When t = 0, A(0) = 0 (since the forest floor is initially clear)

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A = \frac{L}{k}  - \frac{L}{k} e^{kt}

So, D = R - A =

D = \frac{L}{k} - \frac{L}{k}  - \frac{L}{k} e^{kt}\\D = \frac{L}{k} e^{kt}

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}

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