Answer:
Below.
Step-by-step explanation:
f) (a + b)^3 - 4(a + b)^2
The (a+ b)^2 can be taken out to give:
= (a + b)^2(a + b - 4)
= (a + b)(a + b)(a + b - 4).
g) 3x(x - y) - 6(-x + y)
= 3x( x - y) + 6(x - y)
= (3x + 6)(x - y)
= 3(x + 2)(x - y).
h) (6a - 5b)(c - d) + (3a + 4b)(d - c)
= (6a - 5b)(c - d) + (-3a - 4b)(c - d)
= -(c - d)(6a - 5b)(3a + 4b).
i) -3d(-9a - 2b) + 2c (9a + 2b)
= 3d(9a + 2b) + 2c (9a + 2b)
= 3d(9a + 2b) + 2c (9a + 2b).
= (3d + 2c)(9a + 2b).
j) a^2b^3(2a + 1) - 6ab^2(-1 - 2a)
= a^2b^3(2a + 1) + 6ab^2(2a + 1)
= (2a + 1)( a^2b^3 + 6ab^2)
The GCF of a^2b^3 and 6ab^2 is ab^2, so we have:
(2a + 1)ab^2(ab + 6)
= ab^2(ab + 6)(2a + 1).
Answer:
The solution is given in the photo
Answer:
A
Step-by-step explanation:
Using Pythagoras' identity in the right triangle.
The square on the hypotenuse is equal to the sum of the squares on the other 2 sides, that is
BC² = 9² + 12² = 81 + 144 = 225 ( take the square root of both sides )
BC = 
= 15 → A
