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3 years ago

10

Joe is building a fort for his little brother out of boxes. He is wondering how much room there will be inside the

2 answers:

vladimir2022 [97]3 years ago

8 0

The answer is 1600 square inches

katen-ka-za [31]3 years ago

3 0

The answer is 1600....

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Which point is exterior to APD

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Point E is exterior to APD

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3 years ago

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Determine the slope and y-intercept of the line: y=-4x

lapo4ka [179]

Equation y=-4x
m=-4. Slope is -4 y-intercept is 0

y-12=2(x-1)
y-12=2x-2
y=2x-2+12
y=2x+10

7 0

3 years ago

Find the sine of both angle A and angle B.

olchik [2.2K]

Sin = opposite/hyp

Sin A = 10/26 = 5/13

Sin b = 24/26 = 12/13

The answer is option A

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3 years ago

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How can I calculate a square root without guess and check?

Darina [25.2K]

By multiplying the same numbers.

2*2=4 *4 square root is 2

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3 0

3 years ago

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

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

4 0

3 years ago