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Colt1911 [192]
4 years ago
13

A moving firm charges a flat fee of $35 plus $30 per hour. let y be the cost in dollars of using the moving firm for x hours. fi

nd the slope-intercept form of the equation
Mathematics
2 answers:
storchak [24]4 years ago
8 0
Y = 30x + 35 is a answer
mezya [45]4 years ago
7 0

Answer: y=35+30x

Step-by-step explanation:

The slope intercept form of a linear equation is given by :-

y=mx+c, where m is the slope of the line and c is the initial value.

Given : A moving firm charges a flat fee of $35 plus $30 per hour.

i.e.Slope (rater of change) =$30 per hour

Initial value = $35

Let 'y' be the cost in dollars of using the moving firm for 'x' hours,then we the following slope-intercept form of the equation :-

y=35+30x

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To see the steps to the diagonal form see the step-by-step explanation. The solution to the system is x =  -\frac{1}{9}, y= -\frac{1}{9}, z= \frac{4}{9} and w = \frac{7}{9}

Step-by-step explanation:

Gauss elimination method consists in reducing the matrix to a upper triangular one by using three different types of row operations (this is why the method is also called row reduction method). The three elementary row operations are:

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To solve the system using the Gauss elimination method we need to write the augmented matrix of the system. For the given system, this matrix is:

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

For this matrix we need to perform the following row operations:

  • R_2 - 1 R_1 \rightarrow R_2 (multiply 1 row by 1 and subtract it from 2 row)
  • R_3 + 1 R_1 \rightarrow R_3 (multiply 1 row by 1 and add it to 3 row)
  • R_4 - 1 R_1 \rightarrow R_4 (multiply 1 row by 1 and subtract it from 4 row)
  • R_2 \leftrightarrow R_3 (interchange the 2 and 3 rows)
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  • R_1 - 1 R_2 \rightarrow R_1 (multiply 2 row by 1 and subtract it from 1 row)
  • R_4 - 1 R_2 \rightarrow R_4 (multiply 2 row by 1 and subtract it from 4 row)
  • R_3 \cdot ( -1) \rightarrow R_3 (multiply the 3 row by -1)
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  • R_4 + 3 R_3 \rightarrow R_4 (multiply 3 row by 3 and add it to 4 row)
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After this step, the system has an upper triangular form

The triangular matrix looks like:

\left[\begin{array}{cccc|c}1 & 0 & 0 & -0.5 & -0.5  \\0 & 1 & 0 & -0.5 & -0.5\\0 & 0 & 1 & 2 &  2 \\0 & 0 & 0 & 1 &  \frac{7}{9}\end{array}\right]

If you later perform the following operations you can find the solution to the system.

  • R_1 + 0.5 R_4 \rightarrow R_1 (multiply 4 row by 0.5 and add it to 1 row)
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After this operations, the matrix should look like:

\left[\begin{array}{cccc|c}1 & 0 & 0 & 0 & -\frac{1}{9}  \\0 & 1 & 0 & 0 &   -\frac{1}{9}\\0 & 0 & 1 & 0 &  \frac{4}{9} \\0 & 0 & 0 & 1 &  \frac{7}{9}\end{array}\right]

Thus, the solution is:

x =  -\frac{1}{9}, y= -\frac{1}{9}, z= \frac{4}{9} and w = \frac{7}{9}

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