Answer:
Answer Counterclockwise rotation about the origin by 180 degrees followed by reflection about the y-axis explanation the given figure has vertices p(1,2),q(2,1),r(3,2),s(3,3)recall that for counterclockwise rotation about the origin, (x,y)\rightarrow (-x,-y)when we rotate figure pqrs counterclockwise through an angle of 180° about the origin, the coordinates will become p1( - 1, - 2),q1( - 2, - 1),r1( - 3, - 2),s1( - 3, - 3)next, we reflect in the y-axis by negating the x-coordinates of the resulting figure to obtain,p'(1, - 2),q'(2, - 1),r'(3, - 2),s'(3, - 3)
Step-by-step explanation:
Answer:
Step-by-step explanation:
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Answer:
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Step-by-step explanation:
1. 
Taking LCM of (x+3) and (x+5) which is: (x+3)(x+5)

Prove closure: The value of x≠-3 and x≠-5 because if there values are -3 and -5 then the denominator will be zero.
2. 
Factors of x^2-16 = (x)^2 -(4)^2 = (x-4)(x+4)
Factors of x^2+5x+6 = x^2+3x+2x+6 = x(x+3)+2(x+3) =(x+2)(x+3)
Putting factors

Prove closure: The value of x≠-2 and x≠4 because if there values are -2 and 4 then the denominator will be zero.
3. 
Factors of x^2-9 = (x)^2-(3)^2 = (x-3)(x+3)
Factors of x^2-5x+6 = x^2-2x-3x+6 = x(x-2)+3(x-2) =(x-2)(x+3)
Putting factors

Taking LCM of (x-3)(x+3) and (x-2)(x+3) we get (x-3)(x+3)(x-2)


Prove closure: The value of x≠3 and x≠-3 and x≠2 because if there values are -3,3 and 2 then the denominator will be zero.
4. 
Factors of x^2-5x+6 = x^2-3x-2x+6 = x(x-3)-2(x-3) = (x-2)(x-3)
Factors of x^2-16 = (x)^2 -(4)^2 = (x-4)(x+4)

Converting ÷ sign into multiplication we will take reciprocal of the second term

Prove Closure: The value of x≠2 and x≠4 because if there values are 2 and 4 then the denominator will be zero.
I think your answer would be D.0562