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ANTONII [103]
3 years ago
12

HELP ILL MARK YOU BRAINLIEST (40PTS)!! Which number line best shows how to solve −6 − (−8)? A number line from negative 10 to 10

is shown, with numbers labeled at intervals of 2. An arrow is shown from point 0 to negative 6. Another arrow points from negative 6 to 2. A number line from negative 10 to 10 is shown, with numbers labeled at intervals of 2. An arrow is shown from point 0 to negative 6. Another arrow points from negative 6 to 8. A number line from negative 10 to 10 is shown, with numbers labeled at intervals of 2. An arrow is shown from point 0 to 6. Another arrow points from 6 to negative 8. A number line from negative 10 to 10 is shown, with numbers labeled at intervals of 2. An arrow is shown from point 0 to negative 6. Another arrow points from negative 6 to negative 2.
Mathematics
1 answer:
GalinKa [24]3 years ago
8 0

Answer:

i think it is 6 but not sure

Step-by-step explanation:

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siniylev [52]
Thé transformation is that the function moved 6 units down
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3 years ago
Solve for XX. Assume XX is a 2×22×2 matrix and II denotes the 2×22×2 identity matrix. Do not use decimal numbers in your answer.
sveticcg [70]

The question is incomplete. The complete question is as follows:

Solve for X. Assume X is a 2x2 matrix and I denotes the 2x2 identity matrix. Do not use decimal numbers in your answer. If there are fractions, leave them unevaluated.

\left[\begin{array}{cc}2&8\\-6&-9\end{array}\right]· X·\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right] =<em>I</em>.

First, we have to identify the matrix <em>I. </em>As it was said, the matrix is the identiy matrix, which means

<em>I</em> = \left[\begin{array}{ccc}1&0\\0&1\end{array}\right]

So, \left[\begin{array}{cc}2&8\\-6&-9\end{array}\right]· X·\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right] =  \left[\begin{array}{ccc}1&0\\0&1\end{array}\right]

Isolating the X, we have

X·\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]= \left[\begin{array}{cc}2&8\\-6&-9\end{array}\right] -  \left[\begin{array}{ccc}1&0\\0&1\end{array}\right]

Resolving:

X·\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]= \left[\begin{array}{ccc}2-1&8-0\\-6-0&-9-1\end{array}\right]

X·\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]=\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]

Now, we have a problem similar to A.X=B. To solve it and because we don't divide matrices, we do X=A⁻¹·B. In this case,

X=\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]⁻¹·\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]

Now, a matrix with index -1 is called Inverse Matrix and is calculated as: A . A⁻¹ = I.

So,

\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]·\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]=\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]

9a - 3b = 1

7a - 6b = 0

9c - 3d = 0

7c - 6d = 1

Resolving these equations, we have a=\frac{2}{11}; b=\frac{7}{33}; c=\frac{-1}{11} and d=\frac{-3}{11}. Substituting:

X= \left[\begin{array}{ccc}\frac{2}{11} &\frac{-1}{11} \\\frac{7}{33}&\frac{-3}{11}  \end{array}\right]·\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]

Multiplying the matrices, we have

X=\left[\begin{array}{ccc}\frac{8}{11} &\frac{26}{11} \\\frac{39}{11}&\frac{198}{11}  \end{array}\right]

6 0
3 years ago
911 x 65 what is 911 x 65 not just the answer the work to
alexandr402 [8]

Answer:

Step-by-step explanation:

911

<u>x65</u>

5x1=5

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5x900=4500

60x1=60

60x10=600

60x900=54000

5+50+4500+60+600+54000=59,215

6 0
3 years ago
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A line includes the points (10,-5) and (-10,-1). What is the equation in slope intercept form?
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3 years ago
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