MATHEMATICS HIGH SCHOOL

The quotient of x^2+x-6/x^2-6x+5*x^2+2x-3/x^2-7x+10 has ___ in the numerator and ______ in the denominator.

Answers

Answer 1
Answer:

Answer:

So the quotient of (x^(2) + x - 6)/(x^(2) -6x + 5) X (x^(2) + 2x - 3)/(x^(2) -7x + 10) has (x + 3)² in the numerator and (x + 5)² in the denominator.

Step-by-step explanation:

(x^(2) + x - 6)/(x^(2) -6x + 5) X (x^(2) + 2x - 3)/(x^(2) -7x + 10)

Factorizing the expressions we have

(x^(2) + 3x -2x - 6)/(x^(2) -x - 5x + 5) X (x^(2) + 3x - x - 3)/(x^(2) -2x -5x + 10)

(x(x + 3)- 2(x + 3))/(x(x -1) - 5(x - 1)) X (x(x + 3) - 1(x + 3))/(x(x - 2) - 5(x - 2))

((x + 3)(x - 2))/((x - 5)(x - 1))X((x + 3)(x - 1))/((x - 2)(x - 5))

Cancelling out the like factors, (x -1) and (x - 2), we have

((x + 3)(x + 3))/((x - 5)(x - 5))

= ((x + 3)^(2) )/((x + 5)^(2) )

So the quotient of (x^(2) + x - 6)/(x^(2) -6x + 5) X (x^(2) + 2x - 3)/(x^(2) -7x + 10) has (x + 3)² in the numerator and (x + 5)² in the denominator.

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Answers

Let's write an equation:
6x-7=41
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6      6
x= 8

Answer:

Step-by-step explanation:

X=8

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Answers

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