Andrew is about to leave for school. if he walks at the speed of 50 meters per minute he will arrive 3 minutes after the bell rings. if he runs at the speed of 80 meters per minute he will arrive 3 minutes before the bell rings. in how many minutes will the bell ring?

Explanation:

x = time it takes for the bell to ring, time is in minutes

d = distance from his house to the school, in meters

distance = rate*time

If Andrew walks at a speed of 50 meters per minute, then he'll be 3 minutes late. This means he took x+3 minutes to get to class. The x represents the amount of time it takes if he arrived on time, then the +3 is those 3 minutes he is late. We then write

d = 50(x+3)

d = 50x+150

If he runs to school, then he's 3 minutes early. So he took (x-3) minutes to get to class. His rate is 80 meters per minute

d = 80(x-3)

d = 80x-240

We're referring to the same distance (d) value, which allows us to perform substitution to solve the two equations we formed

80x-240 = 50x+150

80x-50x = 150+240

30x = 390

x = 390/30

x = 13

Andrew has 13 minutes to get to class.

Checking the

If he walks at a speed of 50 m per minute, then he'll be 3 minutes late. So he really took 13+3 = 16 minutes to arrive. His distance traveled is d = r*t = 50*16 = 800 meters.

If he ran 80 m per minute, and is 3 minutes early, then he took 13-3 = 10 minutes to get to class. His distance is the same as before d = r*t = 80*10 = 800 meters. The fact we get 800 both times helps confirm we have the right answer.

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation: