an astronaut drops a rock off the edge of a cliff on the moon. the distance, d(t), in meters, the rock travels after t seconds can be modeled by the function d(t) = 0.8t^2. what is the average speed, in meters per second, of the rock between 5 & 10 seconds after it was dropped?

The average speed between the times given is 12 m/s

Step-by-step explanation:

Here, we want to calculate rate of change between the times given

Mathematically , we can get the average rate of change using the formula

f(b)-f(a)/b-a

in this case, b = 10 s and a = 5 s

f(b) is d(10) = 0.8(10)^2 = 80

f(a) is d(5) = 0.8(5)^2 = 20

Rate of change is thus;

(80-20)/(10-5) = 60/5 = 12 m/s

she will sew have sewn five dresses   and be 1/3 done with the sixth one in an hour.

step-by-step explanation:

3/4 of an hour=45 mins

lucy makes 1 dress per 11.25 minutes

45 mins+11.25 mins=56.25 mins

60 mins- 56.25 mins=3.75 mins

3.75 mins is 1/3 of of 11.25 mins

therefore she the other 1/3 would be part of the hour and she will have sewn 5 dresses and begun the sixth.

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