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Ganezh [65]
3 years ago
14

A box contains 4 red pencils and 6 black pencils. What fraction of he pencils are red? What fraction of pencils are black? Draw

a model.
Mathematics
1 answer:
Scorpion4ik [409]3 years ago
3 0

Answer:

The fraction of the pencils that are red is 4/10. The fraction of the pencils that are black is 6/10.

Step-by-step explanation:

[I can't draw you a model but I reccomend that when you do, you seperate it into 10 pieces and mark 4 as red pencils and 6 as black pencils. You could also simplify 4/10 to 2/5 and 6/10 to 3/5, but instead you'd seperate the model into 5 pieces and mark 2 as red pencils and 3 as black pencils.]

First, let's add this up so we have our denominator.

4 + 6 = 10.

Now that we know that the denominator is 10, we apply it to the question.

<em>The box has 4/10 red pencils and 6/10 black pencils.</em>

<em>The box has 4/10 red pencils and 6/10 black pencils.If you were to simplify, it'd be 2/5 red pencils and 3/5 black pencils.</em>

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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:

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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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Step-by-step explanation:

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