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3 years ago

What is the sum of 2.5 × 104 and 3.4 × 104?

Mathematics

2 answers:

Bond [772]3 years ago

7 0

260 + 353.6 = 613.6 that's your answer

steposvetlana [31]3 years ago

7 0

613.6 would be the answer. I got it by multiplying 2.5 and 104 together to get 260, then I multiplied 3.4 and 104 together getting 353.6. I then added 260 and 353.6 together to get 613.6.

2.5 * 104 = 260
3.4 * 104 = 353.6
353.6 + 260 = 613.6

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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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3 years ago

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yarga [219]

Answer:$3207

Step-by-step explanation

A=p(1+r/100)^n

A=1000×(1+6/100)^20

A=1000×(1.06)^20

A=1000×3.207

A=$3207

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shusha [124]

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Yuri [45]

Answer:

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$y - 6$

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