# SOLUTION: please show the working. if 0.25 inches^3 of paint is applied to a circular disk (top only) with a diameter of 20 inches what is the thickness of the paint? At noon car A is

Algebra ->  Calculators  -> Pythagorean Theorem Solver, Cartoon Lesson, and practice -> SOLUTION: please show the working. if 0.25 inches^3 of paint is applied to a circular disk (top only) with a diameter of 20 inches what is the thickness of the paint? At noon car A is       Log On

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 Click here to see ALL problems on Geometry Word Problems Question 419104: please show the working. if 0.25 inches^3 of paint is applied to a circular disk (top only) with a diameter of 20 inches what is the thickness of the paint? At noon car A is traveling north at 30 mph and is located 20 miles north of car B. Car B is traveling west at 50 mph. what is the distance between the cars at 12:45 pm, rounded to the nearest mile?Answer by lwsshak3(6494)   (Show Source): You can put this solution on YOUR website!if 0.25 inches^3 of paint is applied to a circular disk (top only) with a diameter of 20 inches what is the thickness of the paint? At noon car A is traveling north at 30 mph and is located 20 miles north of car B. Car B is traveling west at 50 mph. what is the distance between the cars at 12:45 pm, rounded to the nearest mile? .. First problem: Volume = Area*height or Area*thickness thickness=volume/area Area of a circle= pir^2=3.14*10^2=314 in^2 thickness=.25 in^3/314 in^2=.000796178 inches ans: Thickness of the paint=.000796178 inches .. second problem: Distance = speed*time of travel At 12:45, car A will have traveled 30*(45/60)=90/4=45/2 miles which is added to the 20 miles it has already traveled for a total of 20+45/2 =42.5miles north. At 12:45, car B will have traveled 50*(45/60)=75/2=37.5 miles west. We now have a right triangle with legs of 42.5 and 37.5 miles. Using the pathagorean theorem to solve: x=sqrt(42.5^2+37.5^)=56.68 miles ans: Distance between the cars at 12:45 PM = 57 miles