Group and factor
undistribute then undistribute again
remember
ab+ac=a(b+c)
this is important
6d^4+4d^3-6d^2-4d
undistribute 2d
2d(3d^3+2d^2-3d-2)
group insides
2d[(3d^3+2d^2)+(-3d-2)]
undistribute
2d[(d^2)(3d+2)+(-1)(3d+2)]
undistribute the (3d+2) part
(2d)(d^2-1)(3d+2)
factor that difference of 2 perfect squares
(2d)(d-1)(d+1)(3d+2)
77.
group
(45z^3+20z^2)+(9z+4)
factor
(5z^2)(9z+4)+(1)(9z+4)
undistribuet (9z+4)
(5z^2+1)(9z+4)
In terms of diameter (d), the area (A) of a circle is given by the formula
A = (π/4)d²
You have d=20 yd, so the area of the circle is
A = (π/4)×(20 yd)² . . . . substitute the given information into the formula
A = 100π yd² . . . . . . . . do the arithmetic
A ≈ 314.16 yd²
The answer is C to be informed about a new topic.
