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umka21 [38]
3 years ago
15

Find the value of x

Mathematics
1 answer:
balu736 [363]3 years ago
6 0

Equals 180 because the exterior angle in question is equal to the sum of the other two angles in the triangle. In other words, the other two angles in the triangle (the ones that add up to form the exterior angle) must combine with the angle in the bottom right corner to make a 180 degree angle.

-----------------------------------

Add whole numbers.

2x + 1 + 37 + 118 = 180

Add whole number (if there is any).

2x + 38 +118 = 180

Subtract 156 from both sides to have one side have a variable.

2x + 156 = 180

     -156     -156

Divide both sides by 2 because you don't need the whole number with the variable.

2x = 24

---    ----

2       2

You get the answer as 12.

x = 12

Check:

2(12) + 1 + 37 + 118 = 180

24 + 1 + 37 + 118 = 180

25 + 37 + 118 = 180

62 + 118 = 180

180 = 180

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Step-by-step explanation:

The student made a mistake by identifying the maximum point by the x coordinate value of the vertex. Minimum or maximum points are determine using the y coordinate value.The student can determine the difference between a maximum and minimum by identifying the y coordinate of the vertex.

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Which strategy best explains how to solve this problem? Rebecca bought a meal and two snacks for lunch each day. The meal cost $
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Given:
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C = 5.5 + 2(0.75) = 5.5 + 1.5 = 7

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Based on the choices, the best strategy would be:

<span> A. Make a table. Write the numbers 1 to 12 in the top row of the table (the number of days). In the first box on the second row, write $7. This is how much Rebecca spends in 1 day. In each of the next boxes in the second row, write the amount Rebecca spends by adding $7 to the previous amount. The answer in box 12 is the total amount Rebecca spent after 12 days.</span>

6 0
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Please help if you can pic attatched
MA_775_DIABLO [31]

Answer:

Step-by-step explanation:

CI=\left[\begin{array}{ccc}1&6&0\\0&1&2\\1&-1&3\end{array}\right] \left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]

Subtract row 3 from row 1:

\left[\begin{array}{ccc}1&6&0\\0&1&2\\0&7&-3\end{array}\right] \left[\begin{array}{ccc}1&0&0\\0&1&0\\1&0&-1\end{array}\right]

Subtract row 3 from 7 times row 2:

\left[\begin{array}{ccc}1&6&0\\0&1&2\\0&0&17\end{array}\right] \left[\begin{array}{ccc}1&0&0\\0&1&0\\-1&7&1\end{array}\right]

Divide row 3 by 17:

\left[\begin{array}{ccc}1&6&0\\0&1&2\\0&0&1\end{array}\right] \left[\begin{array}{ccc}1&0&0\\0&1&0\\\frac{-1}{17} &\frac{7}{17} &\frac{1}{17} \end{array}\right]

Subtract 2 of row 3 from row 2:

\left[\begin{array}{ccc}1&6&0\\0&1&0\\0&0&1\end{array}\right] \left[\begin{array}{ccc}1&0&0\\\frac{2}{17} &\frac{3}{17} &\frac{-2}{17} \\\frac{-1}{17} &\frac{7}{17} &\frac{1}{17} \end{array}\right]

Subtract 6 of row 2 from row 1:

\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right] \left[\begin{array}{ccc}\frac{5}{17}&\frac{-18}{17}&\frac{12}{17}\\\frac{2}{17} &\frac{3}{17} &\frac{-2}{17} \\\frac{-1}{17} &\frac{7}{17} &\frac{1}{17} \end{array}\right]

C^{-1}=\frac{1}{17} \left[\begin{array}{ccc}5&-18&12\\2&3&-2\\-1&7&1\end{array}\right]

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