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Inga [223]
3 years ago
13

A strip of wood 78 inches long is to be cut into pieces 3 3/4 inches long. How many pieces can be cut?. . A. 12. B. 26. C. 20. D

. 21.
Mathematics
2 answers:
mafiozo [28]3 years ago
6 0
20 A strip of wood 78 inches long can be cut into 20 3 3/4 inch pieces. (78/3.75 = 20.8 Since you can't have a .8 piece of wood, the answer is 20)
nika2105 [10]3 years ago
5 0
The whole length of wood is measured to be 78 inches. It is to be cut into smaller pieces measuring 3 (3/4) or 3.75 inches long. To calculate how many pieces the strip of wood can be cut, we divide 78 into the smaller measurement.

78 / 3.75 = 20.8 or 21

Therefore, the correct answer is option D, 21.
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Answer:

Each piece is 4/5 m long

Step-by-step explanation:

Kofi has 4 meters of fabric

he cuts them into 5 pieces

4meters/ 5 pieces = length of each piece

4/5 meters long

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What are the coefficients for the binomial expansion of (p+q)^6
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Answer:

1 6 15 20 15 6 1

Step-by-step explanation:

To figure this out, we need to look at Pascal's Triangle, which is a tricky little way to find the coefficients for any binomial expression like this! Check the attached photo.

Because this is to the sixth, we need the 6th row, which is <u>1 6 15 20 15 6 1.</u> From this, we know that those numbers are the coefficients!

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Step-by-step explanation:

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2 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)

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