Answer:
they are all multiples of 5
Step-by-step explanation:
when you count by five you will meet each number =D
Yes it can. It doesn't <em>have to be</em>, but it can be without too much trouble.
In fact, ALL rhombuses are quadrilaterals.
Answer:
f(x) = 8x⁴-8x²+1
Step-by-step explanation:
I will assume that f(cos θ) = cos(4θ). Otherwise, f would not be a polynomial. lets divide cos(4θ) in an expression depending on cos(θ). We use this properties
- cos(2a) = cos²(a) - sin²(b)
- sin(2a) = 2sin(a)cos(a)
- sin²(a) = 1-cos²(a)
cos(4θ) = cos(2 * (2θ) ) = cos²(2θ) - sin²(2θ) = [ cos²(θ)-sin²(θ) ]² - [2cos(θ)sin(θ)]² = [cos²(θ) - ( 1 - cos²(θ) ) ]² - 4cos²(θ)sin²(θ) = [2cos²(θ)-1]² - 4cos²(θ) (1 - cos²(θ) ) = 4 cos⁴(θ) - 4 cos²(θ) + 1 - 4 cos²(θ) + 4 cos⁴(θ) = 8cos⁴(θ) - 8 cos²(θ) + 1
Thus f(cos(θ)) = 8 cos⁴(θ) - 8 cos²(θ) + 1, and, as a result
f(x) = 8x⁴-8x²+1.
352.16 this should indefinite like this should be the ace in the game.
