The ship is 88.8 miles far from the port at 4 pm.
Given,
The displacement from 1 to 3 in the afternoon.
D1 = 30 miles/hr × 2 hr
= 60 miles to the north
The displacement from 3 till 4 in the afternoon.The ship changes its course 20 degrees eastward at 3 o'clock.
Therefore, D2 = 30 miles/hr × 1 hr
= 30 miles to the 20° north eastward
By combining two vectors, the resulting displacement:
D = √((D1 ₊ D2 × cos(20°))² ₊ (D2 × sin(20°))²
D = √((60 ₊ 30 × cos (20°))² ₊ (30 × sin(20°))²
D = 88.8 miles
Hence the ship is 88.8 miles far away from the port at 4 pm.
Learn more about "Law of sines and cosines" here-
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Answer:
<h2>m∠2 = m∠1</h2><h2>m∠1 = 75°</h2>
Step-by-step explanation:
Look at the picture.
Vertical angles, corresponding angles, alternate exterior angles are congruent.
Supplementary angles add up to 180°.
Answer: g(x)-f(x)=8x-16
Explanation:
f(x)=-3x+4 (I am assuming "-3+4" is a typo and it is "-3x+4")
g(x)=5x-12
g(x)-f(x)=(5x-12)-(-3x+4)=5x+3x-12-4=8x-16
Surface area of a sphere = 4πr²
Surface area of the sphere = 4π(0.8)² = 8.0 ft²
Volume of a sphere = 4/3 πr³
Volume of the sphere = 4/3 π(0.8)³ = 2.1 ft³
Answer: Surface area = 8.0 ft² and volume - 2.1 ft³ (Answer C)
The answer would be 2 to the 2nd power
