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4 years ago

7

Vanessa earns a base salary of $400.00 dollar every week with an additional %5 percent commission on everything she sells. Vanes

sa sold $1650.00 signs worth of items last week.
What was Vanessa's total pay last week?

1 answer:

GalinKa [24]4 years ago

5 0

5% is $82.5. So total pay last week is 482.5.

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Cassandra earns $12.98 an hour working at Key Food. How much money will she

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12.98x36.7=476.366

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3 years ago

Please help me it dues in 51 minutes

AfilCa [17]

Answer:

y=3x-4

Step-by-step explanation:

Slope intercept form is :

y=mx+b

where m is the slope, and b is the y intercept.

We are given the slope, it is 3. we are also given the y intercept, it is (0,-4). For this form, the 0 in (0,-4) is ignored, and we consider the y intercept to be -4.

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y=3x-4

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A 7/16 inch long bolt is used in a machine. what is the length of the bolt written as a decimal?

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3 years ago

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3 years ago

[10] In the following given system, determine a matrix A and vector b so that the system can be represented as a matrix equation

irina1246 [14]

Answer:

$y=-\frac{158}{579}$

Step-by-step explanation:

To find the matrix A, took all the numeric coefficient of the variables, the first column is for x, the second column for y, the third column for z and the last column for w:

$A=\left[\begin{array}{cccc}1&1&2&2\\-7&-3&5&-8\\4&1&1&1\\3&7&-1&1\end{array}\right]$

And the vector B is formed with the solution of each equation of the system: $b=\left[\begin{array}{c}3\\-3\\6\\1\end{array}\right]$

To apply the Cramer's rule, take the matrix A and replace the column assigned to the variable that you need to solve with the vector b, in this case, that would be the second column. This new matrix is going to be called $A_{2}$ .

$A_{2}=\left[\begin{array}{cccc}1&3&2&2\\-7&-3&5&-8\\4&6&1&1\\3&1&-1&1\end{array}\right]$

The value of y using Cramer's rule is:

$y=\frac{det(A_{2}) }{det(A)}$

Find the value of the determinant of each matrix, and divide:

$y==\frac{\left|\begin{array}{cccc}1&3&2&2\\-7&-3&5&-8\\4&6&1&1\\3&1&-1&1\end{array}\right|}{\left|\begin{array}{cccc}1&1&2&2\\-7&-3&5&-8\\4&1&1&1\\3&7&-1&1\end{array}\right|} =\frac{158}{-579}$

$y=-\frac{158}{579}$

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4 years ago