Exactly like the previous question you posted.
-3x² - 9x = 0
Divide each side of the equation by -3 :
x² + 3x = 0
Factor the left side:
x (x + 3) = 0
This is true if x=0 or if x=-3 .
Answer:
![\left[\begin{array}{ccc}2&-3\\1&-4\end{array}\right]=-5](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%26-3%5C%5C1%26-4%5Cend%7Barray%7D%5Cright%5D%3D-5)
Step-by-step explanation:
Given:
The given matrix is.
![\left[\begin{array}{ccc}2&-3\\1&-4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%26-3%5C%5C1%26-4%5Cend%7Barray%7D%5Cright%5D)
the determinant of the above matrix is.
![\left[\begin{array}{ccc}2&-3\\1&-4\end{array}\right]=2\times (-4) - (-3)\times 1](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%26-3%5C%5C1%26-4%5Cend%7Barray%7D%5Cright%5D%3D2%5Ctimes%20%28-4%29%20-%20%28-3%29%5Ctimes%201)
![\left[\begin{array}{ccc}2&-3\\1&-4\end{array}\right]=-8 - (-3)](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%26-3%5C%5C1%26-4%5Cend%7Barray%7D%5Cright%5D%3D-8%20-%20%28-3%29)
![\left[\begin{array}{ccc}2&-3\\1&-4\end{array}\right]=-8 +3](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%26-3%5C%5C1%26-4%5Cend%7Barray%7D%5Cright%5D%3D-8%20%2B3)
![\left[\begin{array}{ccc}2&-3\\1&-4\end{array}\right]=-5](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%26-3%5C%5C1%26-4%5Cend%7Barray%7D%5Cright%5D%3D-5)
Therefore, the determinant of the matrix is -5
48 cups = 12 quarts therefor, 48 greater than 11qt
<h3>Given:</h3>
<h3>Note that:</h3>
<h3>
To find:</h3>
The circumference of the circular playing field.
<h3>Solution:</h3>

We'll have to multiply 2, π and radius.
Substitute the values according to the formula.



<u>Therefore</u><u>,</u><u> </u><u>the</u><u> </u><u>circumference</u><u> </u><u>of</u><u> </u><u>the</u><u> </u><u>circular</u><u> </u><u>playing</u><u> </u><u>field</u><u> </u><u>is</u><u> </u><u>47.12</u><u> </u><u>feets</u><u>.</u>
46 × 12 = 46 × (10 + <em>2</em>) =
= (<em>46</em> × 10 ) + (<em>46</em> × 2)
= <em>460</em> + 92
= <em>552</em>