Answer:
R = ![\left[\begin{array}{ccc}-3&-2\\1&-3\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-3%26-2%5C%5C1%26-3%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
P - Q + R = I ( I is the identity matrix )
![\left[\begin{array}{ccc}2&4\\3&5\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%264%5C%5C3%265%5C%5C%5Cend%7Barray%7D%5Cright%5D)
- ![\left[\begin{array}{ccc}-2&2\\4&1\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%262%5C%5C4%261%5C%5C%5Cend%7Barray%7D%5Cright%5D)
+ R = ![\left[\begin{array}{ccc}1&0\\0&1\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%5C%5C0%261%5C%5C%5Cend%7Barray%7D%5Cright%5D)
( subtract corresponding elements )
![\left[\begin{array}{ccc}2-(-2)&4-2\\3-4&5-1\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2-%28-2%29%264-2%5C%5C3-4%265-1%5C%5C%5Cend%7Barray%7D%5Cright%5D)
+ R =
![\left[\begin{array}{ccc}4&2\\-1&4\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%262%5C%5C-1%264%5C%5C%5Cend%7Barray%7D%5Cright%5D)
+ R =
R = - =
We know that
if <span>the probability of hitting the blue circle is the same as the probability of hitting the green region
then
the area of the blue circle is equal to the area of the green region
Let
x----> diameter of the blue circle
area of the blue circle=pi*(x/2)</span>²----> (pi/4)*x² m²-----> equation 1
area of the green region=area of the larger circle-area of the blue circle
area of the green region=pi*(1/2)²-(pi/4)*x²
=(pi/4)-(pi/4)*x² m²----> equation 2
equate equation 1 and equation 2
(pi/4)*x²=(pi/4)-(pi/4)*x² -----> divide by (pi/4)---> x²=1-x²
2x²=1-----> x²=1/2----> x=1/√2-----> x=√2/2 m
the diameter of the blue circle is √2/2 m
The zeros of the function f(x) = x^3 + 3x^2 + 2x are x = 0, x = -1 and x = -2
<h3>How to determine the zeros of the function?</h3>
The function is given as:
f(x) = x^3 + 3x^2 + 2x
Factor out x in the above function
f(x) = x(x^2 + 3x + 2)
Set the function to 0
x(x^2 + 3x + 2) = 0
Factorize the expression in the bracket
x(x + 1)(x + 2) = 0
Split the expression
x = 0, x + 1 = 0 and x + 2 = 0
Solve for x
x = 0, x = -1 and x = -2
Hence, the zeros of the function f(x) = x^3 + 3x^2 + 2x are x = 0, x = -1 and x = -2
Read more about zeros of function at
brainly.com/question/20896994
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