Lisa started to drive from Boston to Washington, DC, which is 440 miles away, at 8:00 AM. For the first four hours of her trip, Lisa was driving at a speed of 50 mph. What was her average speed during the second part of the trip, if she reached Washington at 4:00 PM?

60 mph

Step-by-step explanation:

Given;

Total distance covered d = 440 miles

Average speed in the first 4 hours v1 = 50 mph

Total time taken for the whole trip t = 8:00 to 4:00pm

t = 16:00-8:00 = 8 hours

Firstly we need to calculate the distance covered during the first 4 hours;

d1 = average speed × time = v1 × t1

t1 = 4

d1 = 50 × 4 = 200 miles

Then we need to calculate the distance covered in the second period of the journey;

d = d1 + d2

d2 = d - d1

Substituting the values;

d2 = 440 - 200

d2 = 240 miles

The time taken for the second period of travel t2;

t2 = t -t1 = 8-4

t2 = 4 hours

Average speed = distance travelled ÷ time taken

The average speed for the second part of the trip v2 is;

v2 = d2 ÷ t2

Substituting t2 and d2;

v2 = 240 miles ÷ 4 hours

v2 = 60 mph

The average speed during the second part of the trip is 60 mph

1.5%

step-by-step explanation:

(sin 42)/31. = sinx)/37

sinx = 37 (sin42)/31

= 37× 0.66/31 = 0.787

x= 51.9

other angle opposite to y = 180-51.9-42

= 180- 93.9 =86.1

so

sin42)/31 = sin 86.1)/y

0.021 = 0.997 /y

y= 0.997/0.021 =47.47