hello from mrbilldoesmath!
didn't get an answer from you so i'll assume the cheetah runs at 70 miles/hour. since 1 mile = 5280 feet,
70 miles/hour * 5280 feet/mile = (note how the "mile" units cancel)
70 * 5280 feet/hour =
there are 15 - 3000lb cars and 3 - 5000lb cars.
we can solve a problem with two variables by setting up a system of equations based on the information given in the problem.
since we know that there are 18 cars total, our first equation is:
a + b = 15, where a = the amount of 3000lb cars and b = the amount of 5000lb cars.
we also know that the total weight of the cars is 30 tons, or 60,000lbs. so, the second equation is:
3000a + 5000b = 60,000
using our first equation to solve for 'b', we get b = 15 - a. we can then substitute this expression in for 'b' in the second equation:
3000a + 5000(18 - a) = 60,000
distribute and combine like terms: 3000a + 90,000 - 5000a = 60,000 or 90,000 - 2000a = 60,000
subtract 90,000 from both sides and divide by -2000 gives us a = 15.
15 + b = 18, so b = 3.