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____ [38]
3 years ago
8

What is the value of x in the equation below? 1/3 (12x- 24 -24) = 16 A 2 B 6 C 8 D 10​

Mathematics
1 answer:
Alexandra [31]3 years ago
5 0

Answer: C. 8

Step-by-step explanation:

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5.2 more than the product of seven and a number
vlabodo [156]
It would be 12.2 because 7+5.2 is 12.2
7 0
3 years ago
Read 2 more answers
The square of the sum of two consecutive positive even integers is greater than the sum of their squares by 48. Find the two int
koban [17]

Answer:

  • 4 and 6 or -6 and -4

Step-by-step explanation:

Let the numbers are 2x and 2x + 2.

<u>We have:</u>

  • (2x + 2x + 2)² = (2x)² + (2x + 2)² + 48
  • 4(2x + 1)² = 4x² + 4(x + 1)² + 48
  • (2x + 1)² = x² + (x + 1)² + 12
  • 4x² + 4x + 1 = x² + x² + 2x + 1 + 12
  • 2x² + 2x - 12 = 0
  • x² + x - 6 = 0
  • (x - 2)(x + 3) = 0
  • x = 2, x = -3

<u>The numbers are:</u>

  • 4 and 6 or - 6 and - 4

3 0
2 years ago
Need help answering this
makkiz [27]

Answer:

add the 2 equations together

Step-by-step explanation:

i was in class when I heard this so it should be correct

5 0
3 years ago
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:

  1. Switch any two rows
  2. Multiply a row by a nonzero constant
  3. 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}

7 0
3 years ago
How do I do this problem? Evaluate each expression <br> (-4)-(-6)-(2-1)
natita [175]

Answer:

1

Step-by-step explanation:

(-4) - (-6) - (2-1)

-4 + 6 -2 +1

1

6 0
3 years ago
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