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MaRussiya [10]
4 years ago
6

Melvin is making a large poster for the school carnival. he skeched his design on a 9 inch by 12 inch sheet of paper then he exp

anded his design by a scale of 3 what are the dimensions of the poster. What is the area of the poster
Mathematics
1 answer:
Solnce55 [7]4 years ago
8 0
Because the design is scaled by 3, the new dimensions are 9*3 X 12*3 = 27 X 36

Since the dimensions are 27 X 36, the area is equal to 27*36 = 972 square inches.
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#24, i am getting <img src="https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%260%260%26%5Cfrac%7B18%7D%7B7%7D%20
sveticcg [70]

It looks like you're talking about row-reducing an augmented matrix to solve the system of equations. Your answer is almost correct. The last row should read 0, 0, 1, 2/7.

The given system translates to

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

Eliminate x from the last two rows by combining -2 (row 2) and row 1, and -2 (row 3) and row 1; that is,

(2x - 3y + z) - 2 (x - y + 2z) = 2 - 2 (2)

2x - 3y + z - 2x + 2y - 4z = 2 - 4

-y - 3z = -2

and

(2x - 3y + z) - 2 (x + 2y - 3z) = 2 - 2 (4)

2x - 3y + z - 2x - 4y + 6z = 2 - 8

-7y + 7z = -6

In augmented matrix form, this step yields

\left[ \begin{array}{ccc|c} 2 & -3 & 1 & 2 \\ 0 & -1 & -3 & -2 \\ 0 & -7 & 7 & -6 \end{array} \right]

I'll omit these details in the remaining steps.

Eliminate y from the last row by combining -7 (row 2) and row 3 :

\left[ \begin{array}{ccc|c} 2 & -3 & 1 & 2 \\ 0 & -1 & -3 & -2 \\ 0 & 0 & 28 & 8 \end{array} \right]

Multiply the last row by 1/28 :

\left[ \begin{array}{ccc|c} 2 & -3 & 1 & 2 \\ 0 & -1 & -3 & -2 \\ 0 & 0 & 1 & 2/7 \end{array} \right]

Eliminate z from the second row by combining 3 (row 3) and row 2 :

\left[ \begin{array}{ccc|c} 2 & -3 & 1 & 2 \\ 0 & -1 & 0 & -8/7 \\ 0 & 0 & 1 & 2/7 \end{array} \right]

Multiply the second row by -1 :

\left[ \begin{array}{ccc|c} 2 & -3 & 1 & 2 \\ 0 & 1 & 0 & 8/7 \\ 0 & 0 & 1 & 2/7 \end{array} \right]

Eliminate y and z from the first row by combining 3 (row 2), -1 (row 3), and row 1 :

\left[ \begin{array}{ccc|c} 2 & 0 & 0 & 36/7 \\ 0 & 1 & 0 & 8/7 \\ 0 & 0 & 1 & 2/7 \end{array} \right]

Multiply the first row by 1/2 :

\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 18/7 \\ 0 & 1 & 0 & 8/7 \\ 0 & 0 & 1 & 2/7 \end{array} \right]

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