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alexira [117]
3 years ago
13

At noon, ship a is 180 km west of ship

Mathematics
1 answer:
lawyer [7]3 years ago
5 0
If we let the coordinates of each ship be (x, y) with the positive directions of these coordinates corresponding to East and North, then the positions of the ships t hours after noon are

Ship A
  (-180+35t, 0)

Ship B
  (0, 30t)

The distance between them is the "Pythagorean sum" of the difference in their coordinates:
  d = √((-180 +35t)² +(-30t)²)
    = √(32400 -12600t +2125t²)
The rate of change of this distance is
  dd/dt = (2125t -6300)/√(32400 -12600t +2125t²)

At 4 pm, the value of this rate of change is
  (2125*4 -6300)/√(32400 -12600*4 +2125*4²)
  = 2200/√16000
  ≈ 17.39 km/h

The distance between the ships is increasing at about 17.39 km/h at 4 pm.
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Kamila [148]

Answer:

Option A - 7.3.

Step-by-step explanation:

Given : Number 7.323.

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Solution :

Rounded to the nearest tenth rule :

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4 0
3 years ago
Can someone please help me?<br> If g(x) = x2 - 5x + 2, find g(0).
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Now you need to solve according to the rules of PEMDAS:

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0 - 0 + 2

0 + 2

2

g(0) = 2

Hope this helped!

~ Just a girl in love with Shawn Mendes

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