Perform the following operations and prove closure. Show your work. 1. x/x+3 + x+2/x+5
2. x+4/x^2+5x+6 • x+3/x^2−16
3. 2/x^2−9 − 3x/x^2−5x+6
4 . x+4/x^2−5x+6 ÷ x^2−16/x+3

1.  x/x+3 + x+2/x+5

2.   x+4/x²+5x+6 • x+3/x²−16

3. 2/x²−9 − 3x/x²−5x+6

4. x+4/x²−5x+6 ÷ x²−16/x+3

Step-by-step explanation:

1.   x/x+3 + x+2/x+5

Taking LCM of (x+3) and (x+5) which is: (x+3)(x+5)

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Therefore, to prove closure, The value of x≠-3 and x≠-5 because if their values are -3 and -5 then the denominator will be zero.

2.  x+4/x²+5x+6 • x+3/x²−16

Factors of x²-16 = (x)² -(4)² = (x-4)(x+4)

Factors of x²+5x+6 = x²+3x+2x+6 = x(x+3)+2(x+3) =(x+2)(x+3)

Putting factors

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Therefore, to prove closure, the value of x≠-2 and x≠4 because if their values are -2 and 4 then the denominator will be zero.

3.  2/x²−9 − 3x/x²−5x+6

Factors of x²-9 = (x)²-(3)² = (x-3)(x+3)

Factors of x²-5x+6 = x²-2x-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)

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Therefore, to 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.   x+4/x²−5x+6 ÷ x²−16/x+3

Factors of x²-5x+6 = x²-3x-2x+6 = x(x-3)-2(x-3) = (x-2)(x-3)

Factors of x²-16 = (x)² -(4)² = (x-4)(x+4)

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

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Therefore, to prove closure, the value of x≠2 and x≠4 because if their values are 2 and 4 then the denominator will be zero.

(d)

step-by-step explanation:

given: the line in the graph passes through points (5,30) and (6,26).

then, the equation of line is:

now, put y=a(n) and x=n,

hence, option d is correct.

c: lkllmn.

hope it's you