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expeople1 [14]
3 years ago
12

Hevesh wants to make 2 lines of dominoes that are both 21 inches long. Each line has dominoes spaced 1 and 1/2 inches apart. How

many dominoes will she need for both lines?
Mathematics
1 answer:
aliya0001 [1]3 years ago
3 0

Answer: 30 dominoes in total.

Step-by-step explanation:

Each line is 21 inches long.

the space between dominoes is 1 + 1/2 inches.

(assuming that the distance includes the domino itself) Now, we need at least two dominoes to see this distance,

The we have N + 1 dominoes per line, and the distance is:

N*(1 + 1/2)inches = 21 inches

If we find N, we can find the number of domineoes in each line:

N = 21/(1 + 1/2) = 14

This means that we have 14  + 1 = 15 dominoes in each line, and we have two lines, so we have 30 dominoes in total.

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Determine if AB¯¯¯¯¯¯¯¯ and CD¯¯¯¯¯¯¯¯ are parallel, perpendicular, or neither. Given: A (−1, 3), B (0, 5), C (2, 1), D (6, −1)
dolphi86 [110]

Answer:

perpendicular

Step-by-step explanation:

To determine if AB and CD are  parallel, perpendicular, or neither, we need to get the slope of AB and CD first

Given A (−1, 3), B (0, 5),

Slope Mab = 5-3/0-(-1)

Mab = 2/1

Mab = 2

Slope of AB is 2

Given C (2, 1), D (6, −1)

Slope Mcd = -1-1/6-2

Mcd = -2/4

Mcd = -1/2

Slope of CD is -1/2

Take their product

Mab * Mcd = 2 * -1/2

Mab * Mcd = -1

Since the product of their slope is -1, hence AB and CD are perpendicular

5 0
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
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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

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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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Step-by-step explanation:

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3y/x=5

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