At noon, ship A is 50 nautical miles due west of ship B. Ship A is sailing west at 25 knots and ship B is sailing north at 24 knots. How fast (in knots) is the distance between the ships changing at 6 PM? (Note: 1 knot is a speed of 1 nautical mile per hour.)

The rate of change in distance between Ship A and Ship B is 0.16 Knots per hour.

Step-by-step explanation:

Change in time, t = 18.00hrs - 12.00hrs = 6.00hours

As given in the question,

i) Ship A, sA is 50nm due West of Ship B, sB;

ii) sA sails further West at 25nm/hr (t=6.0hrs) = 150nm West of sB in (i) above.

iii) sB sails North at 24nm/hr (t=6.0hrs) = 144nm North of sB in (i) above.

iv) rate of change in distance between sA and sB is the slope of the graph of the trips embarked upon by both Ship, given by;

(Change in position of sB ÷ Change in position of sA)/time, t

⇒ (144nm - 0nm) ÷ (200nm - 50nm) / 6

⇒( 144nm/150nm) / 6 = 0.96/6

= 0.16nm/hr

no

step-by-step explanation:

2-7=-5

7-2=5

when you combing like terms for ech expression, 5 and -5. these numbers are not equal to each other.

3. (5x + 1) - (-10x + 6)

simplify the answer choice. combine like terms. remember to distribute the negative to the terms inside the second parenthesis. also remember that two negatives = one positive, and one of each sign = negative.

- (-10x + 6) = + 10x - 6

simplify. combine like terms:

5x + 1 + 10x - 6

5x + 10x + 1 - 6

15x - 5

∴ it is your answer

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