Select one of the factors of x^2y^3 - 11x^2y + 6y^2 - 66.

a.) (x^2y + 6)
b.) (x^2y - 6)
c.) (y^2 + 11)
d.) (y - 3)

The correct answer is:

A.) (x^2y + 6)  

Explanation:

In order to factor this, we will use factoring by grouping.  To do this, we first break the polynomial into two groups by placing parentheses around the first two terms and around the second two terms:

(x²y³-11x²y)+(6y²-66)

Next, we factor out the GCF of each group.  For the first group, both terms have an x² and a y; there are no numbers in common (other than 1).  Factoring this out,

x²y(y²-11)+(6y²-66)

The GCF of the second group is 6, as both terms are divisible by 6:

x²y(y²-11)+6(y²-11)

We look to the remaining terms in parentheses.  If we have worked correctly, they should be the same.  They are.  This is now the GCF of the polynomial, since it is common to both terms:

(y²-11)(x²y+6)

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation: