A train normally travels 240 km at a certain speed. One day, due
1 answer:
Normal-travel DATA:
distance = 240 km
rate = x km/h
time = d/r
= 240/x hrs
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Bad weather DATA:
distance = 240 km
rate = (x-20) km/h
time = 240/(x-20) hrs
______________________
Bad time - normal time = 2 hrs
240/(x-20) - 240/x = 2
120/(x-20) - 120/x = 1
120x -120(x-20) = x(x-20)
20 = x^2 -20x
x^2 - 20x - 20 = 0
x = [20 +- sqrt(20^2 - 4*1*-20)]/2
x = [20 +- sqrt(480)]/2
x = [20 +- 4sqrt(30)]/2
x = 10 +- 2sqrt(30)
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Answer:
Area of the given triangle = 6.51 ft²
Step-by-step explanation:
If we redraw the triangle as shown in the figure attached we get the height of the triangle as x feet and base = 4.3 feet
Therefore area of ΔABC = (1/2) × Height (AD) × Base(BC)
From ΔABD
sin 53 = AD/AB = x/3.8
x = 3.8 × sin 53
= 3.8 × 0.7986 = 3.03 feet
Now area of ΔABC = (1/2)×3.03×4.3 = 6.51 feet²
