Ratio And Proportion If 8 engines consume 12000 kgs. of coal when each running 10
hours per day, how much coal is required for 7 engines out
which 5 are running 12 hours a day and the rest 6 hours a day?​

11200 for 7 engines for a total of 80 hours

Step-by-step explanation:

You have 8 engines burning 12000 .for 10 hours.

Divide 10 hours into 12000=1200 PER hour.

1200 ÷8 engines = 140 per hour per engine.

140 × # of hours (12)=1680 for 12 hours per engine.

# of engines × 1680 for 12 hour/per engine

5×1680 = 8400 of fuel for 12 hours/5 engine

Then you have 2 that runs 10 hours

2 × 140 per hour per engine =280 × 10 =2800 for 2 engines for 10 hours

8400. + 2800= 11200 of fuel for 6 engines

f(x) = y

f(-3) = -5

step-by-step explanation:

answer: all of the odd numbers were opened

the first person was the one that was opening all of the lockers. since one is odd, this causes all of the odd number lockers to be opened in the end.