Burning one gallon of gasoline releases 20 pounds of co2 into the air. harry’s compact car can travel 25 miles on one gallon of gas. if harry drives 15,000 miles in a year, his vehicle will add pounds of co2 to the atmosphere. nextreset

If Harry drives 15,000 miles in a year, and his car can travel 25 miles on one gallon of gas, then you first need to find out how many gallons of gas he would use in 15,000 miles.

15,000 / 25 = 600

600 gallons of gas in 15,000 miles.

If one gallon of gas releases 20 pounds of Co2, then you need to find out how much Co2 is released by 600 gallons.

600 x 20 = 12,000

His vehicle will add 12,000 pounds of Co2 to the atmosphere.

the direct variation says that:

then the equation is of the form is given by:

            where k is the constant variation.

from the given table:

consider any value of x and m(x)=y

x = 2 and m(x)=y = 3

substitute these value in [1]; we have;

divide both sides by 2 we get;

⇒

check:

put x = -1 and m(x) = -1.5

substitute in the equation:

-1.5 = -1.5                   true

therefore, the function m(x) represents a direct variation

graph of this function as shown below: