9. Find (f.g)(a) if f(a) = . and g(a) = 2². Fully simplify your answer.

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 a separable differential equation. 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

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

4 0

3 years ago

Solve the proportion below. x/29=2/5.8

Bond [772]

X/29 = 2/5.8

(X) x (5.8). (2) x (29)

5.8x = 58

58/5.8

X = 10

5 0

3 years ago

I just need help with this table

Mashutka [201]

Answer:

from the looks of it, all you have to do is, for f(x), is plug it in as an exponent. in order (top to bottom), it should be: 64, 2048, 4096, 8192.

g(x) is being squared and then multiplied, so it should be (from top to bottom): 720, 2420, 2880, 3380

Step-by-step explanation:

6 0

3 years ago

The perimeter of a college athletic field is 100 meters and the length is 16 m more than the width. Find the length and the widt

Inga [223]

P=100
L=16+w
W=w
so we know that Perimeter is all the sides added so 2l+2w
so 100=(2(16+w))+(2(w))
100=32+2w+2w
100=32+4w
68=4w
w=17
so L=16+w therefore L=16+17 L=33
so L=33 and W=17

4 0

3 years ago

The decimal -2.97 is the solution to which subtraction problem?

Natalija [7]

Answer:

Option 4: -6.1 - (-3.13)

Step-by-step explanation:

Let's solve each option individually to see if -2.97 fits as the solution:

-6.1 - 3.13 = -9.23

-6.1 - (-31.3) = 25.2

-0.61 - (-3.13) = 2.52

-6.1 - (-3.13) = -2.97

As you can see, -2.97 is the answer to Option 4.

So, the decimal -2.97 is the solution to the subtraction problem in Option 4.

5 0

2 years ago

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