$\left[\begin{array}{ccc}9&7\\-4&2\end{array}\right] -4M=\left[\begin{array}{ccc}-3&3\\4&-18\end{array}\right]\qquad(*) \\\\M=\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]\to4M=\left[\begin{array}{ccc}4a&4b\\4c&4d\end{array}\right]$

$(*)\qquad\left[\begin{array}{ccc}9&7\\-4&2\end{array}\right] -\left[\begin{array}{ccc}4a&4b\\4c&4d\end{array}\right]=\left[\begin{array}{ccc}-3&3\\4&-18\end{array}\right]\\\\\left[\begin{array}{ccc}9-4a&7-4b\\-4-4c&2-4d\end{array}\right]=\left[\begin{array}{ccc}-3&3\\4&-18\end{array}\right]\\.\qquad\qquad\qquad\qquad\ \ \ \Downarrow\\(_{11})\qquad9-4a=-3\ \ \ |-9\\-4a=-12\ \ \ \ |:(-4)\\a=3\\\\(_{12})\qquad7-4b=3\ \ \ \ |-7\\-4b=-4\ \ \ \ |:(-4)\\b=1$

$(_{21})\qquad-4-4c=4\ \ \ \ |+4\\-4c=8\ \ \ \ |:(-4)\\c=-2\\\\(_{22})\qquad2-4d=-18\ \ \ \ |-2\\-4d=-20\ \ \ \ \ |:(-4)\\d=5\\\\Answer:\ A.\ M=\left[\begin{array}{ccc}3&1\\-2&5\end{array}\right]$

3 0

3 years ago

An above-ground swimming pool in the shape of a cylinder has a diameter of 18 feet and a height of 4.5 feet. If the pool is fill

laila [671]

The volume of a cylinder is

   (pi) (radius²) (height) .

Radius = 1/2 diameter.
Radius of this pool = (1/2) (18 ft) = 9 ft

The pool is a cylinder with height of 4.5 feet.
The water in it is also a cylinder, but only 4 ft high.

Volume of the water =

   (pi) x (radius²) x (height)

   = (pi) x (9 ft)² x (4 ft)

   = (pi) x (81 ft²) x (4 ft)

   = (pi) x (324 ft³) = 1,017.9 ft³ .

7 0

2 years ago

Read 2 more answers

YEAR 9 MATHEMATUCS ALGEBRA

WINSTONCH [101]

Answer:

Your answer is :

B. 6a^2 + 3ab

Markn my answer as brainlist . Follow me for more answer.

Step-by-step explanation:

7 0

2 years ago