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raketka [301]
3 years ago
9

What is the volume in cubic feet of a greenhouse that is 20 feet long 15 feet wide and 10 feet high

Mathematics
1 answer:
9966 [12]3 years ago
8 0
<h2>Greetings!</h2>

Answer:

3000 ft³

Step-by-step explanation:

The area of a rectangle is:

Height x length x width

So simply plug the given values in:

20 x 15 x 10 = 3000

<h3>So the volume is 3000 ft³</h3>
<h2>Hope this helps!</h2>
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Determine whether the given system has a unique solution, no solution, or infinitely many solutions.
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The system has no solution.

Step-by-step explanation:

To find the solution to this system of linear equations

\left\begin{array}{ccccc}-3x_1&+x_2&-2x_3&=&8&\\x_1&+5x_2&-x_3&=&4&\\-x_1&+11x_2&-4x_3&=&1&\end{array}\right

First, state the problem in matrix form, this means, extracting only the numbers, and putting them in a box.

\left[ \begin{array}{ccc|c} -3 & 1 & -2 & 8 \\\\ 1 & 5 & -3 & 4 \\\\ -1 & 11 & -4 & 1 \end{array} \right]

This is called an augmented matrix. The word “augmented” refers to the vertical line, which we draw to remind ourselves where the equals sign belong

Next, transform the augmented matrix to the reduced row echelon form with the help of Row Operations.

Row Operation 1: multiply the 1st row by -1/3

\left[ \begin{array}{ccc|c} 1 & - \frac{1}{3} & \frac{2}{3} & - \frac{8}{3} \\\\ 1 & 5 & -1 & 4 \\\\ -1 & 11 & -4 & 1 \end{array} \right]

Row Operation 2: add -1 times the 1st row to the 2nd row

\left[ \begin{array}{ccc|c} 1 & - \frac{1}{3} & \frac{2}{3} & - \frac{8}{3} \\\\ 0 & \frac{16}{3} & - \frac{5}{3} & \frac{20}{3} \\\\ -1 & 11 & -4 & 1 \end{array} \right]

Row Operation 3: add 1 times the 1st row to the 3rd row

\left[ \begin{array}{ccc|c} 1 & - \frac{1}{3} & \frac{2}{3} & - \frac{8}{3} \\\\ 0 & \frac{16}{3} & - \frac{5}{3} & \frac{20}{3} \\\\ 0 & \frac{32}{3} & - \frac{10}{3} & - \frac{5}{3} \end{array} \right]

Row Operation 4: multiply the 2nd row by 3/16

\left[ \begin{array}{ccc|c} 1 & - \frac{1}{3} & \frac{2}{3} & - \frac{8}{3} \\\\ 0 & 1 & - \frac{5}{16} & \frac{5}{4} \\\\ 0 & 0 & 0 & -15 \end{array} \right]

Row Operation 5: add -32/3 times the 2nd row to the 3rd row

\left[ \begin{array}{ccc|c} 1 &- \frac{1}{3}  & \frac{2}{3} & - \frac{8}{3} \\\\ 0 & 1 & - \frac{5}{16} & \frac{5}{4} \\\\ 0 & 0 & 0 & -15 \end{array} \right]

Row Operation 6: multiply the 3rd row by -1/15

\left[ \begin{array}{ccc|c} 1 &- \frac{1}{3}  & \frac{2}{3} & - \frac{8}{3} \\\\ 0 & 1 & - \frac{5}{16} & \frac{5}{4} \\\\ 0 & 0 & 0 & 1 \end{array} \right]

Row Operation 7: add -5/4 times the 3rd row to the 2nd row

\left[ \begin{array}{ccc|c} 1 &- \frac{1}{3}  & \frac{2}{3} & - \frac{8}{3} \\\\ 0 & 1 & - \frac{5}{16} & 0 \\\\ 0 & 0 & 0 & 1 \end{array} \right]

Row Operation 8: add 8/3 times the 3rd row to the 1st row

\left[ \begin{array}{ccc|c} 1 &- \frac{1}{3}  & \frac{2}{3} & 0 \\\\ 0 & 1 & - \frac{5}{16} & 0 \\\\ 0 & 0 & 0 & 1 \end{array} \right]

Row Operation 9: add 1/3 times the 2nd row to the 1st row

\left[ \begin{array}{cccc} 1 & 0 & \frac{9}{16} & 0 \\\\ 0 & 1 & - \frac{5}{16} & 0 \\\\ 0 & 0 & 0 & 1 \end{array} \right]

The reduced row echelon form of the augmented matrix is

\left[ \begin{array}{cccc} 1 & 0 & \frac{9}{16} & 0 \\\\ 0 & 1 & - \frac{5}{16} & 0 \\\\ 0 & 0 & 0 & 1 \end{array} \right]

which corresponds to the system

\left\begin{array}{ccccc}x_1&&+\frac{9}{16} x_3&=&0&\\&1x_2&-\frac{5}{16}x_3&=&0&\\&&0&=&1&\end{array}\right

Equation 3 cannot be solved, therefore, the system has no solution.

7 0
3 years ago
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