Can someone me with my algebra fraction homework pages 8,9 and 10. you!

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

the correct option is 2.

step-by-step explanation:

according to the given information triangle abc is a right angle triangle with right angle at a. segment ab is 5 and segment ac is 7. point d is on segment bc and angles adb and adc are right angles.

it is given that triangles abd, cad, and cba are similar.

two triangles are similar if their corresponding sides are proportional.

in triangle cba, using pythagoras theorem

                      )

triangle abd and cba are similar,

                            (2)

triangle cad and cba are similar,

                      )

using (1), (2) and (3), we get

hence proved.

therefore option 2 is correct.