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3 years ago

Evaluate the following expression. 10−9×5+6×(−2)

1 answer:

7 0

According to the order of operations, evaluate all multiplication first.
10-45+-12

Then add and subtract from left to right
-35-12
-47

Final answer: -47

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If $17,800 is invested at an interest rate of 8% per year, find the value of the investment at the end of 5 years for the follow

marta [7]

Answer:

7120

Step-by-step explanation:

6 0

2 years ago

What is the equation of the following line? Be sure to scroll down first to see

guajiro [1.7K]

Answer:

$y = \frac{2}{7} x$

Step-by-step explanation:

The slope is

$slope = \frac{rise}{run} = \frac{2}{7}$

The y-intercept is 0

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2 years ago

25 points!!!!!!!!

podryga [215]

Answer:

A is your answer, my guy

Step-by-step explanation:

4x1=4

3x-1=-3

5x1=5

4+(-3)+5=6

6x1=6

8x-1=-8

6x1=6

6+(-8)+6=4

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6x1=6

4+(-2)+6=8

8 0

3 years ago

PLEASE HELP ME I DONT UNDERSTAND THIS AT ALL

Delicious77 [7]

What do you not understand

4 0

3 years ago

Find the inverse of the given matrix, if it exists.Aequals=left bracket Start 3 By 3 Matrix 1st Row 1st Column 1 2nd Column 0 3

BabaBlast [244]

Answer:

$A^{-1}=\left[ \begin{array}{ccc} \frac{1}{9} & \frac{4}{27} & - \frac{2}{27} \\\\ \frac{8}{9} & \frac{5}{27} & \frac{11}{27} \\\\ - \frac{4}{9} & \frac{2}{27} & - \frac{1}{27} \end{array} \right]$

Step-by-step explanation:

We want to find the inverse of $A=\left[ \begin{array}{ccc} 1 & 0 & -2 \\\\ 4 & 1 & 3 \\\\ -4 & 2 & 3 \end{array} \right]$

To find the inverse matrix, augment it with the identity matrix and perform row operations trying to make the identity matrix to the left. Then to the right will be inverse matrix.

So, augment the matrix with identity matrix:

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

Subtract row 1 multiplied by 4 from row 2

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ -4&2&3&0&0&1\end{array}\right]$

Add row 1 multiplied by 4 to row 3

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&2&-5&4&0&1\end{array}\right]$

Subtract row 2 multiplied by 2 from row 3

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&0&-27&12&-2&1\end{array}\right]$

Divide row 3 by −27

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]$

Add row 3 multiplied by 2 to row 1

$\left[ \begin{array}{ccc|ccc}1&0&0&\frac{1}{9}&\frac{4}{27}&- \frac{2}{27} \\\\ 0&1&11&-4&1&0 \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]$

Subtract row 3 multiplied by 11 from row 2

$\left[ \begin{array}{ccc|ccc}1&0&0&\frac{1}{9}&\frac{4}{27}&- \frac{2}{27} \\\\ 0&1&0&\frac{8}{9}&\frac{5}{27}&\frac{11}{27} \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]$

As can be seen, we have obtained the identity matrix to the left. So, we are done.

6 0

3 years ago