Will give and 50 pts
1. x+x^2y+x^3y^2 2. 3p^2r^2+6pr+p

1. x(x^2y^2 + xy + 1)

2. p(3pr^2 + 6r + 1)

Step-by-step explanation:

1.

x + x^2y + x^3y^2 =

First, try factoring out a common factor. x is a common factor.

= x(1 + xy + x^2y^2)

Now, write the trinomial in descending order of degree.

= x(x^2y^2 + xy + 1)

x^2y^2 is the square of xy.

1 is the square of 1.

2 * xy * 1 = 2xy. The middle term is just xy, so the trinomial is not factorable.

x(x^2y^2 + xy + 1)

2.

3p^2r^2 + 6pr + p =

Factor out the common factor, p.

= p(3pr^2 + 6r + 1)

c. 27 units.

step-by-step explanation:

we are told that in circle a, ∠bae ≅ ∠dae.

we can see from our given diagram that in and ;

, as these are radii of our given circle.

therefore, by sas congruence postulate.

hence, side be will be equal to side de as corresponding parts of congruent triangles are congruent. so we can set an equation to find the value of x as:

now let us substitute x=17 in the expression for the length of be.

therefore, length of be will be 27 units and option c is the correct choice.