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3 years ago

Find the distance between the points (0, 6) and (2, 3).

Mathematics

2 answers:

bogdanovich [222]3 years ago

7 0

Distance= √((x2-x1)²+(y2-y1)²) = √((6-3)²+(0-2)²)=√(9+4)=√13 ≈ 3.61

lisov135 [29]3 years ago

5 0

Sqrt13=3.60555127...........

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Which relation is not a function?

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Answer: UPC

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Find the set of solutions for the linear system.

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Answer:

The system has infinitely many solutions.

$\left\begin{array}{ccc}x_1&=&-\frac{1}{3}x_2-\frac{16}{9}x_4-\frac{2}{9} \\x_2&=&s_1\\x_3&=&-\frac{4}{3}x_4+\frac{1}{3} \\x_4&=&s_2\end{array}\right$

Step-by-step explanation:

To find the solution for this system of linear equations $-3x_1-x_2+4x_3=2\\-3x_3 - 4x_4 = -1$ you must:

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

A matrix is a rectangular arrangement of numbers into rows and columns.

A system of equations can be represented by an augmented matrix.

In an augmented matrix, each row represents one equation in the system and each column represents a variable or the constant terms.

This is matrix that represents the system

$\left[ \begin{array}{ccccc} -3 & -1 & 4 & 0 & 2 \\\\ 0 & 0 & -3 & 4 & -1 \end{array} \right]$

The augmented matrix can be transformed by a sequence of elementary row operations to the matrix.

There are three kinds of elementary matrix operations.

Interchange two rows (or columns).
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Multiply a row (or column) by a non-zero number and add the result to another row (or column).

Using elementary matrix operations, we get that

Row Operation 1: Multiply the 1st row by -1/3

Row Operation 2: Multiply the 2nd row by -1/3

Row Operation 3: Add 4/3 times the 2nd row to the 1st row

$\left[ \begin{array}{ccccc} 1 & \frac{1}{3} & 0 & \frac{16}{9} & - \frac{2}{9} \\\\ 0 & 0 & 1 & \frac{4}{3} & \frac{1}{3} \end{array} \right]$

Step 2: Interpret the reduced row echelon form

The reduced row echelon form of the augmented matrix is

$\left[ \begin{array}{ccccc} 1 & \frac{1}{3} & 0 & \frac{16}{9} & - \frac{2}{9} \\\\ 0 & 0 & 1 & \frac{4}{3} & \frac{1}{3} \end{array} \right]$

which corresponds to the system

$x_1+\frac{1}{3}x_2+ \frac{16}{9}x_4=-\frac{2}{9} \\x_3+ \frac{4}{3}x_4=\frac{1}{3}$

We see that the variables $x_2, x_4$ can take arbitrary numbers; they are called free variables. Let $x_2=s_1$ , $x_4=s_2$ . All solutions of the system are given by

$\left\begin{array}{ccc}x_1&=&-\frac{1}{3}x_2-\frac{16}{9}x_4-\frac{2}{9} \\x_2&=&s_1\\x_3&=&-\frac{4}{3}x_4+\frac{1}{3} \\x_4&=&s_2\end{array}\right$

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Answer:

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