Going against the current, a boat takes 6 hours to make a 120-mile trip. when the boat travels with the current on the return trip, it takes 5 hours. if x = the rate of the boat in still water and y = the rate of the current, which of the following systems could be used to solve the problem? a) 6(x - y) = 120 b) 5(x + y) = 120 c) 6(x + y) = 120 d) 5(x - y) = 120 e) 6x - 5y = 120 f) x + y = 120

You can use A and B systems

Step-by-step explanation:

Lets call z the total speed of the boat. If the boat goes against the current, then, the current will drop the boat natural speed, and therefore z is obtained from substracting y from the rate of the boat on still water, x. Thus, z = x-y. If The boat goes in favor of the current, then the current will raise the speed, and we obtain z by adding y to x. If we want to calculate x and y, we know that:

On the first trip, z = x-y, and it took 6 hours to finish the 120 mile trip, therefore 6z = 120, or equivalently, 6(x-y)=120

On the return trip, z = x+y, and it took 5 hours to finish the trip, so we have 5z = 5(x+y) = 120.

Thus, in order so solve the problem, we can use system A and B.

Note that system A is equivalent to the equation x-y = 20, obtained by dididing everything by 6. If we divide by 5 the second equation we obtain that x+y = 24. We have

By summing this equations it follows that 2x = 44, therefore x = 22. Since x-y = 20, we obtain that y = 2.

The correct answer is 6(x - y) = 120 and 5(x + y) = 120

24

step-by-step explanation:

c.

step-by-step explanation:

she should of added 9^2 and 12^2. that is because it is a^2+b^2=c^2. c^2 is the hypotenuse and should always be bigger than the other two numbers. which is why she should of added 9 squared and 12 squared together instead of 15 squared.