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Mariulka [41]
2 years ago
15

Banana trees yield 32 pounds of fruit per acre and strawberry plants yield 84 pounds of fruit per acre. Suppose Erica's farm has

b acres of banana trees, s acres of strawberry plants, and yields 1,200 pounds of fruit. What is the equation that represents this scenario?
84b + 32s = 1,200

32b + 84s = 1,200

116f = 1,200

b + s = 1,200
Mathematics
2 answers:
Usimov [2.4K]2 years ago
6 0

32b + 84s = 1,200 hope this helps

Mila [183]2 years ago
3 0

Answer:  The correct option is (B) 32b+84s=1200.

Step-by-step explanation:  Given that banana trees yield 32 pounds of fruit per acre and strawberry plants yield 84 pounds of fruit per acre.

Suppose Erica's farm has b acres of banana trees, s acres of strawberry plants and yields 1,200 pounds of fruit.

We are to select the equation that represents the given scenario.

The total pounds of fruits yield by banana trees is

32 \times b=32b

and

the total pounds of fruits yield by strawberry plants is

84\times s=84s.

Since Erica's farm yield a total of 1200 pounds of fruits, so the required equation that represents the scenario is given by

32b+84s=1200.

Thus, the required equation is 32b+84s=1200.

Option (B) is the correct option.

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A large corporation starts at time t = 0 to invest part of its receipts continuously at a rate of P dollars per year in a fund f
Andrews [41]

Answer:

A = \frac{P}{r}\left( e^{rt} -1 \right)

Step-by-step explanation:

This is <em>a separable differential equation</em>. Rearranging terms in the equation gives

                                                \frac{dA}{rA+P} = dt

Integration on both sides gives

                                            \int \frac{dA}{rA+P} = \int  dt

where c is a constant of integration.

The steps for solving the integral on the right hand side are presented below.

                               \int \frac{dA}{rA+P} = \begin{vmatrix} rA+P = m \implies rdA = dm\end{vmatrix} \\\\\phantom{\int \frac{dA}{rA+P} } = \int \frac{1}{m} \frac{1}{r} \, dm \\\\\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \int \frac{1}{m} \, dm\\\\\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \ln |m| + c \\\\&\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \ln |rA+P| +c

Therefore,

                                        \frac{1}{r} \ln |rA+P| = t+c

Multiply both sides by r.

                               \ln |rA+P| = rt+c_1, \quad c_1 := rc

By taking exponents, we obtain

      e^{\ln |rA+P|} = e^{rt+c_1} \implies  |rA+P| = e^{rt} \cdot e^{c_1} rA+P = Ce^{rt}, \quad C:= \pm e^{c_1}

Isolate A.

                 rA+P = Ce^{rt} \implies rA = Ce^{rt} - P \implies A = \frac{C}{r}e^{rt} - \frac{P}{r}

Since A = 0  when t=0, we obtain an initial condition A(0) = 0.

We can use it to find the numeric value of the constant c.

Substituting 0 for A and t in the equation gives

                         0 = \frac{C}{r}e^{0} - \frac{P}{r} \implies \frac{P}{r} = \frac{C}{r} \implies C=P

Therefore, the solution of the given differential equation is

                                   A = \frac{P}{r}e^{rt} - \frac{P}{r} = \frac{P}{r}\left( e^{rt} -1 \right)

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