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2 years ago

The question and answers are in this photo Thank you for the help!

Mathematics

2 answers:

arlik [135]2 years ago

5 0

Answer:

Step-by-step explanation

Just pick a

julsineya [31]2 years ago

3 0

I would assume B

but im not 100% sure

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I have to solve for X?

VashaNatasha [74]

yes

95+97 =192

192+x=180

x= -12

4 0

3 years ago

How do you make 1 1-2/1-6-7/8 a mixed number

AveGali [126]

Answer:

17/8

Step-by-step explanation:

2/1=2

11-2-6-7/8

9-6-7/8

3-7/8

24/8-7/8

17/8

8 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

What is the nth term for the sequence 5 , 14 , 29 , 50 , 77

Marina86 [1]

The answer would be 3n^2 + 2.

This can be found/proven by replacing "n" with term number (1,2,3,4...), then solving to get the final number. For example 3 * 1^2 + 2. You would first do 1^2, which is 1. Next, you would multiply 1 by 3, to get 3. Finally, you'd and the 2 to get 5. 5 is the 1st term, and you can use this same equation to get the rest of the terms you need.

I hope this helps!

8 0

2 years ago

Solve for y. <br> 2x+6y=12

Veronika [31]

2x + 6y = 12

2x - 2x + 6y = 12 - 2x

6y = 12 - 2x

6y/6 = 12/6 - 2x/6

Y = 2 - 2/6 X

Y = 2 - 1/3 X or

Y = -1/3 X + 2.

7 0

3 years ago