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3 years ago

Answer Please! I hate khan academy lol

Mathematics

1 answer:

Zanzabum3 years ago

6 0

Answer:

C; None of the above.

Step-by-step explanation:

Pretend x=1, y=0, and z=2.

You have to match an answer to 1+0-2 (equals to -1).

The first equation would be 2+0+1. When you add them all together you get 3, and that isn't equivalent to -1, so therefore this one is not the answer.

The second equation becomes 1+2+0, and when you solve for the answer, you see that the answer is also 3, and so we can conclude that neither A or B matches with the equation above, so the answer is C, None of the above.

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Make a conjecture about the next item in the sequence. 6, 9, 7, 10, 8?

inessss [21]

Answer:

"The next term should be 11"

Step-by-step explanation:

If we look closely at the sequence we can see:

second term is 3 more than previous

third term is 2 less than previous

fourth term is 3 more than previous

fifth term is 2 less than previous

We can see a pattern. To get next terms, we add 3, to get next to that, we subtract 2. It goes on like this.

Since 4th term to 5th term is "subtracting 2", logically 5th to 6th term (the term we are wanting) should be "add 3". So the term after 8 would be 11

The next term would be 11

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3 years ago

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Answer:

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3 years ago

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3 years ago

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3 years ago

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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

Answer Please! I hate khan academy lol​

Answer Please! I hate khan academy lol