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3 years ago

The area of the triangle is 28.7 square centimeters. What is the base length of the triangle in centimeters?

Mathematics

1 answer:

Montano1993 [528]3 years ago

3 0

In order to derive the base of a triangle from its area, you need its height as well.

In fact, if we solve the area formula for the base, we have

$A=\dfrac{bh}{2} \iff 2A = bh \iff b=\dfrac{2a}{h}$

So, the base length would be

$b=\dfrac{2\cdot 28.7}{h} = \dfrac{57.4}{h}$

where h is the height relative to the base you're interested in.

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Solve the given system of equations using either Gaussian or Gauss-Jordan elimination. (If there is no solution, enter NO SOLUTI

cricket20 [7]

Answer:

The system has infinitely many solutions

$\begin{array}{ccc}x_1&=&-x_3\\x_2&=&-x_3\\x_3&=&arbitrary\end{array}$

Step-by-step explanation:

Gauss–Jordan elimination is a method of solving a linear system of equations. This is done by transforming the system's augmented matrix into reduced row-echelon form by means of row operations.

An Augmented matrix, each row represents one equation in the system and each column represents a variable or the constant terms.

There are three elementary matrix row operations:

Switch any two rows
Multiply a row by a nonzero constant
Add one row to another

To solve the following system

$\begin{array}{ccccc}x_1&-3x_2&-2x_3&=&0\\-x_1&2x_2&x_3&=&0\\2x_1&+3x_2&+5x_3&=&0\end{array}$

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

$\left[ \begin{array}{cccc} 1 & -3 & -2 & 0 \\\\ -1 & 2 & 1 & 0 \\\\ 2 & 3 & 5 & 0 \end{array} \right]$

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

Row Operation 1: add 1 times the 1st row to the 2nd row

Row Operation 2: add -2 times the 1st row to the 3rd row

Row Operation 3: multiply the 2nd row by -1

Row Operation 4: add -9 times the 2nd row to the 3rd row

Row Operation 5: add 3 times the 2nd row to the 1st row

to the matrix

$\left[ \begin{array}{cccc} 1 & 0 & 1 & 0 \\\\ 0 & 1 & 1 & 0 \\\\ 0 & 0 & 0 & 0 \end{array} \right]$

The reduced row echelon form of the augmented matrix is

$\left[ \begin{array}{cccc} 1 & 0 & 1 & 0 \\\\ 0 & 1 & 1 & 0 \\\\ 0 & 0 & 0 & 0 \end{array} \right]$

which corresponds to the system

$\begin{array}{ccccc}x_1&&-x_3&=&0\\&x_2&+x_3&=&0\\&&0&=&0\end{array}$

The system has infinitely many solutions.

$\begin{array}{ccc}x_1&=&-x_3\\x_2&=&-x_3\\x_3&=&arbitrary\end{array}$

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Answer:

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Answer:

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