Respuesta :

Answer:

Step-by-step explanation:

#8 the answer is A

using special products, fact the denominator of both fractions

q^2+5q+6 is the same as (q+2)(q+3)

q^2+3q+2 is the same as (q+2)(q+1)

no you have to find the lease common denominator of the two fractions

multiply the first fraction by (q+1)/(q+1)

multiply the second fraction by (q+3)/(q+3)

you will get (q^2+2q+3)/(q+1)(q+2)(q+3) after you combine like terms and stuff like that.

#9 the answer is C

using special products factor the numerator of the first fraction, and the denominator of the second fraction

r^2+7r+10 is the same as (r+2)(r+5)

r^2-5r-50 is the same as (r-10)(r+5)

then factor the numerator if the second fraction

3r-30 is the same as 3(r-10)

is the second fraction there is (r-10) in the numerator and denominator, simplify that.

then simplify the 3, and the (r+5)

you will be left with r+2 from the first fraction

#10 the answer is A

when you divide same base exponents, you subtract their powers

so

x^0/x^2 = x^-2       and

y^-3/y^-1=y^=2

so you are left with x^-2*y^-2 but that is not an option

we also know that negative exponents is the same thing as the reciprocal with the exponent in the denominator

x^-1 = 1/x

so

x^-2*y^-2 = 1/x^2*y^2