The battery power available to run a satellite is given by the formula P=50e^(-t/250), where P is the power in watts and t is time in days. for how many days can the satellite run if it requires 15 watts of power?

The power left after 378 days is 14.18 watts.

Step-by-step explanation:

50x5=$250

step-by-step explanation:

- "runner b is moving at a slower rate."

- "the graphs shown are graphs of distance(km) against time(min).

the slopes of this graphs are are given by:

slope = (change in y)/(change in x)

= (change in distance)/(change in time)

this is the formula for finding the speed. so, the slope of these graphs represent the speed of the runner.

taking each division in the graph to represent 1 on both axis:

runner a ⇒ speed = (2-0)/(1-0)

= 2/1

= 2 km/min

runner b ⇒ speed = (3-0)/(3-0)

= 3/3

= 1 km/min

this shows that   runner b is moving at a slower rate."