A plane leaves airport A and travels 550 miles to airport B on a bearing of N35°E. The plane later leaves airport B and travels to airport C 380 miles away on a bearing of S72°E. Find the distance from airport A to airport to the nearest tenth of a mile

d₃ = 803.9 Miles

Step-by-step explanation:

d₁ = 560 Miles

R₁ = N35⁰E

d₂ = 410 Miles

R₂ = S76⁰E

d₃ = ?

α = 35⁰+76⁰ = 111⁰

We can apply the Law of Cosines as follows

d₃ = √(d₁² + d₂² - 2*d₁*d₂* Cos α)

⇒  d₃ = √(560² + 410² - 2*560*410* Cos 111⁰) = 803.9 Miles

correct answer is infinitely many solutions

step-by-step explanation:

2(2x + 3) = -4 + 2(2x + 5)

solving the above equation:

4x + 6 = -4 +4x +10

combining like terms

4x -4x = -4+10-6

0=0

since both sides are 0=0, we end up into true statement it means it has infinite many solutions.

so correct answer is infinitely many solutions

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step-by-step explanation: