Which second degree polynomial function has a leading coefficient of –1 and root 4 with multiplicity 2
2 answers:
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Answer:
θ = (60° )
Step-by-step explanation:
Using the identity
sin²x + cos²x = 1 ⇒ sin²x = 1 - cos²x
Given
cos²θ - sin²θ = 2 - 5cosθ
cos²θ - (1 - cos²θ) = 2 - 5cosθ
cos²θ - 1 + cos²θ = 2 - 5cosθ
2cos²θ - 1 = 2 - 5cosθ ( subtract 2 - 5cosθ from both sides )
2cos²θ + 5cosθ - 3 = 0 ← in standard form
(cosθ + 3)(2cosθ - 1) = 0 ← in factored form
Equate each factor to zero and solve for θ
cosθ + 3 = 0
cosθ = - 3 ← not possible as - 1 ≤ cosθ ≤ 1
2cosθ - 1 = 0
cosθ = , so
θ = (
) =
( or 60° )
Answer:
OPTION A: Divide both the sides by 5
Step-by-step explanation:
We are given: 5z - 3 = 12
When we add 9 to both sides we get:
5z - 3 + 9 = 12 + 9
⇒ 5z + 6 = 21
⇒ 5z = 21 - 6 = 15
⇒ 5z = 15
To find the value of z we have to divide both the sides by 5.
i.e.,
⇒ z = 3.