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tangare [24]
3 years ago
6

How do i solve 2k^2-5k-18=0

Mathematics
1 answer:
devlian [24]3 years ago
5 0

Answer:

factor left the side of the equation

(2k_9)(k+2)=0

set factors equal to 0

2k_9=0 or k+2=0

the answer is K= 9/2 or K =-2

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Answer:

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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

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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

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Does the commutative property of addition apply when you add to negative integers?
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