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makkiz [27]
3 years ago
14

The length of a rectangular wall is 16.5 feet. The height of the wall is 8.625 feet. Cameron determines the area of the wall to

be 1423.125 square feet. Which best explains the reasonableness of Cameron’s solution?
Cameron’s solution is reasonable because there are four decimal places in the factors and four decimal places in the product.
Cameron’s solution is reasonable because there are three decimal places in the factors and three decimal places in the product.
Cameron’s solution is unreasonable because 17 times 9 is 153, and 153 is not close to his product.
Cameron’s solution is unreasonable because 10 times 10 is 100, and 100 is not close to his product
Mathematics
1 answer:
Drupady [299]3 years ago
4 0

Answer:

Cameron’s solution is unreasonable because 17 times 9 is 153, and 153 is not close to his product.

Step-by-step explanation:

Area of a rectangle:

The area of a rectangle, of length l and height h, is given by:

A = lh

The length of a rectangular wall is 16.5 feet. The height of the wall is 8.625 feet.

This means that l = 16.5, h = 8.625

Estimate of the area:

To use an "easy" multiplication, lets consider l = 17, h = 9. So

A = lh = 17(9) = 153

Cameron determines the area of the wall to be 1423.125 square feet.

Very far from the area of 153, which means that he made a confusion with the decimal places. Thus, the correct answer is:

Cameron’s solution is unreasonable because 17 times 9 is 153, and 153 is not close to his product.

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Answer:

The system has infinitely many solutions

\begin{array}{ccc}x_1&=&-x_3\\x_2&=&-x_3\\x_3&=&arbitrary\end{array}

Step-by-step explanation:

Gauss–Jordan elimination is a method of solving a linear system of equations. This is done by transforming the system's augmented matrix into reduced row-echelon form by means of row operations.

An Augmented matrix, each row represents one equation in the system and each column represents a variable or the constant terms.

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To solve the following system

\begin{array}{ccccc}x_1&-3x_2&-2x_3&=&0\\-x_1&2x_2&x_3&=&0\\2x_1&+3x_2&+5x_3&=&0\end{array}

Step 1: Transform the augmented matrix to the reduced row echelon form

\left[ \begin{array}{cccc} 1 & -3 & -2 & 0 \\\\ -1 & 2 & 1 & 0 \\\\ 2 & 3 & 5 & 0 \end{array} \right]

This matrix can be transformed by a sequence of elementary row operations

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

Row Operation 2: add -2 times the 1st row to the 3rd row

Row Operation 3: multiply the 2nd row by -1

Row Operation 4: add -9 times the 2nd row to the 3rd row

Row Operation 5: add 3 times the 2nd row to the 1st row

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\left[ \begin{array}{cccc} 1 & 0 & 1 & 0 \\\\ 0 & 1 & 1 & 0 \\\\ 0 & 0 & 0 & 0 \end{array} \right]

The reduced row echelon form of the augmented matrix is

\left[ \begin{array}{cccc} 1 & 0 & 1 & 0 \\\\ 0 & 1 & 1 & 0 \\\\ 0 & 0 & 0 & 0 \end{array} \right]

which corresponds to the system

\begin{array}{ccccc}x_1&&-x_3&=&0\\&x_2&+x_3&=&0\\&&0&=&0\end{array}

The system has infinitely many solutions.

\begin{array}{ccc}x_1&=&-x_3\\x_2&=&-x_3\\x_3&=&arbitrary\end{array}

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