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)
Answer:
.2 lbs
Step-by-step explanation:
4.5/22.5
Answer:
See Below.
Step-by-step explanation:
We want to verify the equation:

We can convert sec(α) to 1 / cos(α):

Multiply both layers of the first fraction by cos(α):

Create a common denominator. We can multiply the first fraction by (1 - cos(α)):

Simplify:

From the Pythagorean Identity, we know that cos²(α) + sin²(α) = 1 or equivalently, 1 - cos²(α) = sin²(α). Substitute:

Subtract:

Distribute:

Rewrite:

Split:

Factor the second fraction, and substitute sin²(α) for 1 - cos²(α):

Factor:

Cancel:

Divide the second fraction by cos(α):

Hence proven.