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The only that has the shape of the one below.
y = -4(x^3 + 7x^2+8x - 16)
Let x = 1
y = - 4(1 + 7 + 8 - 16)
y = 0
y = -4(x -1)(x^2 + 8x + 16)
y = -4(x - 1)(x + 4)^2
y = -4(x^3 + 7x^2+8x - 16)
Let x = 1
y = - 4(1 + 7 + 8 - 16)
y = 0
y = -4(x -1)(x^2 + 8x + 16)
y = -4(x - 1)(x + 4)^2
Answer:
I) |xz| ≈ 28.6 km
II) |yz| ≈ 34.8 km
Step-by-step explanation:
Let's assume that the position of ship due south of x is z (aà pictor representation of the question is attached)
|xy| = 20 km, |xz| = ?, |yz| = ?, θ(y) = 55°
Using Trigonometric ratio - SOHCAHTOA
I) Tan θ = |xz| ÷ |xy| ⇒ Tan 55° = |xz| ÷ 20
|xz| = 20 * Tan 55 = 20 * 1.428
|xz| = 28.56 km
|xz| ≈ <u>28.6 km</u>
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II) Cos θ = |xy| ÷ |yz| ⇒ Cos 55° = 20 ÷ |yz|
|yz| * Cos 55° = 20 ⇒ |yz| = 20 ÷ Cos 55°
|yz| = 20 ÷ 0.574 = 34.84 km
|yz| ≈ <u>34.8 km</u>
