All categories

2 years ago

Evaluate 9 P 2 ................

Mathematics

1 answer:

Svetradugi [14.3K]2 years ago

8 0

21 because the 9 is equal to 2 and don’t forget to carry the 5

You might be interested in

Find the measures of the numbered angles.

Arte-miy333 [17]

1 = 90
2 = 65
3 = 65
4 = 25

90+25=115
180-115=65

have a merry christmas :)

3 0

2 years ago

What is the scientific notation of 68000

Dmitry [639]

OOHHH!! I like sci not! Okay, so you take the decimal ( which in this case is at the end: 68000. ) and move it over as many times as needed until it gets in between the, in this case, 6 and 8. ( 6.8000 ). You need to show exactly how many times it was actually moved so you multiply that number time 10^ whatever place. In this question, it'll be to the 4th place because you move the decimal over 4 times. The answer will be 6.80 x 10^4 OR -written in short hand- 6.8x10E4

8 0

3 years ago

Read 2 more answers

HELP with part (b)!! will mark brainlest!!!!

Natasha2012 [34]

7.) (x, y) —> (1/2*x, 1/2*y)

3 0

3 years ago

Read 2 more answers

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

Please need help on this

IrinaVladis [17]

The first statement is true.

The second statement is false.

The third statement is true.

4 0

3 years ago

Read 2 more answers