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4 years ago

How to find the unit rate 15:3

Mathematics

1 answer:

Feliz [49]4 years ago

5 0

The answer is 5. In other words just do 15 divided by 3

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How much is the jacket without sales tax

ExtremeBDS [4]

105%= 1.05
Let p= the price of the jacket without the tax.
1.05p= 31.50
p= 30
Therefore the price of the jacket without the sales tax is $30.

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3 years ago

On Monday Sarah had homework in 7⁄10 of her classes, Tuesday 3⁄5 , Wednesday 9⁄11 , and Thursday 1⁄2 . Which day did she have th

slava [35]

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4 years ago

Read 2 more answers

Solve the following systems of equations using the matrix method: a. 3x1 + 2x2 + 4x3 = 5 2x1 + 5x2 + 3x3 = 17 7x1 + 2x2 + 2x3 =

lara [203]

Answer:

a. The solutions are

$\left[\begin{array}{c}x_1&x_2&x_3\\\end{array}\right]=\begin{pmatrix}\frac{11}{13}\\ \frac{50}{13}\\ -\frac{17}{13}\end{pmatrix}$

b. The solutions are

$\left[\begin{array}{c}x&y&z\\\end{array}\right]=\begin{pmatrix}\frac{54}{235}\\ \frac{6}{47}\\ \frac{24}{235}\end{pmatrix}$

c. The solutions are

$\left[\begin{array}{c}x_1&x_2&x_3&x_4\\\end{array}\right]=\begin{pmatrix}\frac{22}{9}\\ \frac{164}{9}\\ \frac{139}{9}\\ -\frac{37}{3}\end{pmatrix}$

Step-by-step explanation:

Solving a system of linear equations using matrix method, we may define a system of equations with the same number of equations as variables as:

$A\cdot X=B$

where X is the matrix representing the variables of the system, B is the matrix representing the constants, and A is the coefficient matrix.

Then the solution is this:

$X=A^{-1}B$

a. Given the system:

$3x_1 + 2x_2 + 4x_3 = 5 \\2x_1 + 5x_2 + 3x_3 = 17 \\7x_1 + 2x_2 + 2x_3 = 11$

The coefficient matrix is:

$A=\left[\begin{array}{ccc}3&2&4\\2&5&3\\7&2&2\end{array}\right]$

The variable matrix is:

$X=\left[\begin{array}{c}x_1&x_2&x_3\\\end{array}\right]$

The constant matrix is:

$B=\left[\begin{array}{c}5&17&11\\\end{array}\right]$

First, we need to find the inverse of the A matrix. To find the inverse matrix, augment it with the identity matrix and perform row operations trying to make the identity matrix to the left. Then to the right will be inverse matrix.

So, augment the matrix with identity matrix:

$\left[ \begin{array}{ccc|ccc}3&2&4&1&0&0 \\\\ 2&5&3&0&1&0 \\\\ 7&2&2&0&0&1\end{array}\right]$

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

$\left[ \begin{array}{ccc|ccc}1&0&0&- \frac{2}{39}&- \frac{2}{39}&\frac{7}{39} \\\\ 0&1&0&- \frac{17}{78}&\frac{11}{39}&\frac{1}{78} \\\\ 0&0&1&\frac{31}{78}&- \frac{4}{39}&- \frac{11}{78}\end{array}\right]$

And the inverse of the A matrix is

$A^{-1}=\left[ \begin{array}{ccc} - \frac{2}{39} & - \frac{2}{39} & \frac{7}{39} \\\\ - \frac{17}{78} & \frac{11}{39} & \frac{1}{78} \\\\ \frac{31}{78} & - \frac{4}{39} & - \frac{11}{78} \end{array} \right]$

Next, multiply $A^ {-1}$ by $B$

$X=A^{-1}\cdot B$

$\left[\begin{array}{c}x_1&x_2&x_3\\\end{array}\right]=\left[ \begin{array}{ccc} - \frac{2}{39} & - \frac{2}{39} & \frac{7}{39} \\\\ - \frac{17}{78} & \frac{11}{39} & \frac{1}{78} \\\\ \frac{31}{78} & - \frac{4}{39} & - \frac{11}{78} \end{array} \right] \cdot \left[\begin{array}{c}5&17&11\end{array}\right]$

$\left[\begin{array}{c}x_1&x_2&x_3\\\end{array}\right]=\begin{pmatrix}-\frac{2}{39}&-\frac{2}{39}&\frac{7}{39}\\ -\frac{17}{78}&\frac{11}{39}&\frac{1}{78}\\ \frac{31}{78}&-\frac{4}{39}&-\frac{11}{78}\end{pmatrix}\begin{pmatrix}5\\ 17\\ 11\end{pmatrix}=\begin{pmatrix}\frac{11}{13}\\ \frac{50}{13}\\ -\frac{17}{13}\end{pmatrix}$

The solutions are

$\left[\begin{array}{c}x_1&x_2&x_3\\\end{array}\right]=\begin{pmatrix}\frac{11}{13}\\ \frac{50}{13}\\ -\frac{17}{13}\end{pmatrix}$

b. To solve this system of equations

$x -y - z = 0 \\30x + 40y = 12 \\30x + 50z = 12$

The coefficient matrix is:

$A=\left[\begin{array}{ccc}1&-1&-1\\30&40&0\\30&0&50\end{array}\right]$

The variable matrix is:

$X=\left[\begin{array}{c}x&y&z\\\end{array}\right]$

The constant matrix is:

$B=\left[\begin{array}{c}0&12&12\\\end{array}\right]$

The inverse of the A matrix is

$A^{-1}=\left[ \begin{array}{ccc} \frac{20}{47} & \frac{1}{94} & \frac{2}{235} \\\\ - \frac{15}{47} & \frac{4}{235} & - \frac{3}{470} \\\\ - \frac{12}{47} & - \frac{3}{470} & \frac{7}{470} \end{array} \right]$

The solutions are

$\left[\begin{array}{c}x&y&z\\\end{array}\right]=\begin{pmatrix}\frac{54}{235}\\ \frac{6}{47}\\ \frac{24}{235}\end{pmatrix}$

c. To solve this system of equations

$4x_1 + 2x_2 + x_3 + 5x_4 = 0 \\3x_1 + x_2 + 4x_3 + 7x_4 = 1\\ 2x_1 + 3x_2 + x_3 + 6x_4 = 1 \\3x_1 + x_2 + x_3 + 3x_4 = 4\\$

The coefficient matrix is:

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

The variable matrix is:

$X=\left[\begin{array}{c}x_1&x_2&x_3&x_4\\\end{array}\right]$

The constant matrix is:

$B=\left[\begin{array}{c}0&1&1&4\\\end{array}\right]$

The inverse of the A matrix is

$A^{-1}=\left[ \begin{array}{cccc} - \frac{1}{9} & - \frac{1}{9} & - \frac{1}{9} & \frac{2}{3} \\\\ - \frac{32}{9} & - \frac{5}{9} & \frac{13}{9} & \frac{13}{3} \\\\ - \frac{28}{9} & - \frac{1}{9} & \frac{8}{9} & \frac{11}{3} \\\\ \frac{7}{3} & \frac{1}{3} & - \frac{2}{3} & -3 \end{array} \right]$