The distance between the ships changing at 92.29 Knots
- Using the position of ship A as the reference point, at time t measured in hours past noon, ship A is 18 t miles west of this point and ship B is 40 + 17t north of this point.
- The distance between ships is then

The rate of change of distance is -

after putting t = 5 into this rate of change ,
we get, answer = 92.29
To learn more about differentiation from the given link
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Answer:
-3
Step-by-step explanation:
got it right on edg
Answer:
D. N(a) = (a + 20)/3
Step-by-step explanation:
A(n) = 3n - 20
A is a function of n. The inverse will be: N is a function of a.
Change A(n) to y and n to x.
y = 3x - 20
Switch x and y.
x = 3y - 20
Solve for y.
x + 20 = 3y
3y = x + 20
y = (x + 20)/3
Now switch y to N(a) and x to a.
Answer: N(a) = (a + 20)/3
3 times as much as the sum of 3/4 and 2/6
A, If I'm not mistaken Celsius is the same thing as Fahrenheit