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Usimov [2.4K]
3 years ago
15

Which Matrix is equal to A-2B+C

Mathematics
2 answers:
anastassius [24]3 years ago
8 0
It is convenient to use the matrix capability of your calculator to do this. The result is ...
\left[ \begin{array}{cc}-1&4\\5&-7\end{array} \right]

lawyer [7]3 years ago
5 0
\bf A=
\begin{bmatrix}
5&2\\3&0
\end{bmatrix}\qquad B=
\begin{bmatrix}
4&3\\-1&6
\end{bmatrix}\qquad C=
\begin{bmatrix}
2&8\\0&5
\end{bmatrix}\\\\
-------------------------------\\\\
2B\implies 2\begin{bmatrix}
4&3\\-1&6
\end{bmatrix}\implies \begin{bmatrix}
2\cdot 4&2\cdot 3\\2\cdot -1&2\cdot 6
\end{bmatrix}\implies 
\begin{bmatrix}
8&6\\-2&12
\end{bmatrix}

\bf \stackrel{A-2B}{\begin{bmatrix}
5&2\\3&0
\end{bmatrix}-\begin{bmatrix}
8&6\\-2&12
\end{bmatrix}}\implies 
\begin{bmatrix}
-3&-4\\5&-12
\end{bmatrix}
\\\\\\
\stackrel{A-2B+C}{\begin{bmatrix}
-3&-4\\5&-12
\end{bmatrix}+\begin{bmatrix}
2&8\\0&5
\end{bmatrix}}\implies 
\begin{bmatrix}
-1&4\\5&-7
\end{bmatrix}
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Arrange the fractions from the largest to the smallest.
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Answer:

14/25, 9/20, 2/7

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7 0
3 years ago
Dale bought a map of his city.it uses a scale of 1 inch = 8miles. dales house and school are 1 1/2 inches apart on the map. how
Elza [17]
If you would like to know how far apart would Dale's house and school be on the map if the scale were 1 inch = 6 miles, you can calculate this using the following steps:

1 inch ... 8 miles
1 1/2 inches = 3/2 inches ... x miles = ?

1 * x = 8 * 3/2
x = 8 * 3/2
x = 12 miles

1 inch ... 6 miles
y inches = ? ... 12 miles

1 * 12 = 6 * y
12 = 6 * y     /6
y = 12 / 6
y = 2 inches

The correct result would be C. 2 inches.
6 0
3 years ago
Let r be the region bounded by y = x 2 and y = 9. find a and b so that the horizontal lines y = a and y = b split the region r i
Arturiano [62]

y = x² has a lower bound at the vertex (0, 0); <em> y = 0</em>

y = 9 has an upper bound at y = 9

x² = 9   ⇒  x = +/- 3  the region is bounded on the left at x = -3 and on the right at x = 3

Solve the integral from -3 to 3 to find the area under the curve, then multiply by 1/3 to calculate 1/3 of the area.

\int\limits^3_3 (9 - {x^{2}}) \, dx = (9x - x³/3) | from -3 to 3

= (9(3) - 3³/3) - (9(-3) - (-3)³/3)

= (27 - 9) - (- 27 - (-9))

= 16 - (-16)

= 32

*** 32/3 is 1/3 of the area ****

Now that we know what 1/3 of the area is, we can find the height <em>at y = a</em> which will satisfy 1/3 of the area. We do this by solving the integral from "0 to a" and setting it equal to 1/3 of the area (32/3).

\int\limits^a_0 (9 - {x^{2}}) \, dx - a(9 - a²)

=  (9x - x³/3) | from 0 to a   - (9a - a³)

= (9a - a³/3) - (9(0) - (0)³/3) - 9a + a³

= 9a - a³/3 - 9a + a³

= a³ - a³/3

= 2a³/3

32/3 = 2a³/3

32 = 2a³

16 = a³

∛16 = a  

2∛2 = a  (This is the distance from the upper bound (y = 9)

Answer: y = 9 - 2∛2

Follow the same steps using 2/3 of the area (64/3) to find "b".

64 = 2b³

32 = b³

∛32 = b   3.17

2∛4 = b  (This is the distance from the upper bound (y = 9)

Answer: y = 9 - 2∛4


8 0
3 years ago
Point V is on line segment UW. Given VW = x, UV = 3x – 1, and
creativ13 [48]

Answer:

4 units

Step-by-step explanation:

Since V lies on UW,

UV + VW = UW

3x - 1 + x = 5x - 5 \\ 5x - x - 3x = 5 - 1 \\ x = 4

Therefore, <u>VW = 4 units</u>

6 0
3 years ago
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