Working simultaneously at their respective constant rates, machines a and b produce 800 nails in x hours. working alone at its constant rate, machine a produces 800 nails in y hours. in terms of x and y, how many hours does it take machine b, working alone at its constant rate, to produce 800 nails?

(a) x/(x+y)
(b) y/(x+y)
(c) xy/(x+y)
(d) xy/(x-y)
(e) xy/(y-x)

  (E)  xy/(y-x)

Step-by-step explanation:

The sum of the individual rates is the total rate. Each machine's rate is nails per hour is the inverse of its rate in hours per nail.

  total rate = (800 nails)/(x hours) = 800/x nails/hour

  A rate = (800 nails)/(y hours) = 800/y nails/hour

We want to find the time "b" such that ...

  B rate = (800 nails)/(b hours) = 800/b nails/hour

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As we said, the total rate is the sum of the individual rates:

  800/x = 800/y + 800/b

Multiplying by xyb/800, we get

  yb = xb + xy

Solving for b, we have ...

  yb -xb = xy

  b(y -x) = xy

  b = xy/(y-x) . . . . . matches choice E

It takes Machine B xy/(y-x) hours to produce 800 nails.

three-fourths of an hour

36÷48=0.75

0.75=

answer: since the question is talking about growth you have to use the growth formula which is inital amount (1 + growth)^time

step-by-step explanation:

the intial amount is 1200, the growth is 4.8, and the time is 7. plug those numbers into the equation, also i think you have to change 4.8 into a percent. but, i have forgotton how to do that. if this is a online class go on to the lesson.