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Dmitriy789 [7]
3 years ago
6

A chemical manufacturer wants to lease a fleet of 28 railroad tank cars with a combined carrying capacity of 564,000 gallons. Ta

nk cars with three different carrying capacities are​ available: 6,000 ​gallons,12,000 gallons, and 24,000 gallons. How many of each type of tank car should be​ leased?
Let x1 be the number of cars with a 6,000 gallon​ capacity, x2 be the number of cars with a 12,000 gallon​ capacity, and x3 be the number of cars with a 24,000 gallon capacity. Select the correct choice below and fill in the answer boxes within your choice.
Mathematics
1 answer:
yan [13]3 years ago
3 0

You forgot to give the choices so I will solve the problem in detail step-by-step explanation and then you can select the correct choice and fill in the answer boxes within your choice correctly.

Given the criteria stated in the problem, we can obtain the  following linear system:

\left \{ {{x_{1} + x_{2} + x_{3}  = 28} \atop {6000x_{1} + 12000x_{2}  + 24000x_{3} =564000}} \right.

The corresponding augmented matrix is

\left[\begin{array}{ccc|c}1&1&1&28\\6000&12000&24000&564000\end{array}\right]

\left[\begin{array}{ccc|c}1&1&1&28\\6000&12000&24000&564000\end{array}\right]  \frac{1}{6000} R_{1}→ R_{2}\left[\begin{array}{ccc|c}1&1&1&28\\1&2&4&94\end{array}\right] R_{2} +(-1)R_{1}

→R_{2} \left[\begin{array}{ccc|c}1&1&1&28\\0&1&3&66\end{array}\right]  R_{1} +(-1)R_{2}→R_{1} \left[\begin{array}{ccc|c}1&0&-2&-38\\0&1&3&66\end{array}\right]

The linear system corresponding to the reduced form is

\left \{ {{x_{1} -2x_{3} =-38} \atop {x_{2}+3x_{3}  =66}} \right.

Since x_{3} is the free variable, we let x_{3} =t, where t is any real number. Then the general solution can be written as follows:

\left \{ \begin{array}{ccc}x_{1} =&-38&+2t\\x_{2} =&66&-3t\\x_{3} =&t\end{array}\right

where t is any real number.

However, the definitions of x_{1},x_{2} and x_{3} imply that the solutions must be non-negative integers.  Therefore, we need to derive the possible range of values of t such that the general solution make sense for this problem:

\left \{ \begin{array}{ccc}-38&+2t&\geq 0\\66&-3t&\geq 0\\t&&\geq 0\end{array}\right

Solving the inequalities, we finally obtain all the relevant solutions  to the linear system:

\left \{ \begin{array}{ccc}x_{1} =&-38&+2t\\x_{2} =&66&-3t\\x_{3} =&t\end{array}\right

where t is any integer such that 19 \leq t \leq 22.

Thus, you just need to apply the above solution to the choices given to you and fill in the answer boxes correctly.

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Which ordered pairs are solutions to the inequality 2x+y>−4?
hichkok12 [17]

we will proceed to resolve each case to determine the solution

we have

2x+y>-4

y>-2x-4

we know that

If an ordered pair is the solution of the inequality, then it must satisfy the inequality.

<u>case a)</u> (5,-12)

Substitute the value of x and y in the inequality

-12>-2*5-4

-12>-14 ------> is True

therefore

the ordered pair (5,-12) is a solution of the inequality

<u>case b)</u> (-3,0)

Substitute the value of x and y in the inequality

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0>2 ------> is False

therefore

the ordered pair (-3,0) is not a solution of the inequality

<u>case c)</u> (-1,-1)

Substitute the value of x and y in the inequality

-1>-2*-1-4

-1>-2 ------> is True

therefore

the ordered pair(-1,-1) is a solution of the inequality

<u>case d)</u> (0,1)

Substitute the value of x and y in the inequality

1>-2*0-4

1>-4 ------> is True

therefore

the ordered pair (0,1) is a solution of the inequality

<u>case e)</u> (4,-12)

Substitute the value of x and y in the inequality

-12>-2*4-4

-12>-12 ------> is False

therefore

the ordered pair (4,-12) is not a solution of the inequality

<u>Verify</u>

using a graphing tool

see the attached figure

the solution is the shaded  area above the line

The points A,C, and D lies on the shaded area, therefore the ordered pairs A,C, and D are solution of the inequality


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mash [69]

ANSWER

r = 10

EXPLANATION

To solve for r first we have to put r in the numerator. To do that, we have to multiply both sides of the proportion by r:

\begin{gathered} \frac{28}{35}\cdot r=\frac{8}{r}\cdot r \\ \frac{28}{35}r=8 \end{gathered}

Now, we have to multiply both sides by 35:

\begin{gathered} \frac{28}{35}r\cdot35=8\cdot35 \\ 28r=280 \end{gathered}

And finally divide both sides by 28:

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