Respuesta :
To solve this problem, let's set up a proportion, letting x represent the unknown number of gallons the faucet loses in 48 hours.
5 gallons / 20 hours = x gallons / 48 hours
To solve this proportion, we are going to use cross-products. This is the products of numerator of one fraction and the denominator of the other fraction set equal to one another. It results in the equation:
5 * 48 = 20 * x
Next we simplify by performing the operations.
240 = 20x
Finally, we divide both sides by 20 to get x alone on the right side of the equation.
12=x
Therefore, 12 gallons of water will be lost in 48 hours.
5 gallons / 20 hours = x gallons / 48 hours
To solve this proportion, we are going to use cross-products. This is the products of numerator of one fraction and the denominator of the other fraction set equal to one another. It results in the equation:
5 * 48 = 20 * x
Next we simplify by performing the operations.
240 = 20x
Finally, we divide both sides by 20 to get x alone on the right side of the equation.
12=x
Therefore, 12 gallons of water will be lost in 48 hours.
5gallons leakout every 20 hours
5g:20h
fractionify
5g/20h=g/4h
g:4h
how many gallons in 48 hours
xgallons:48
assume constant rate of flow so
g:4h=x:48
fractionfy
g/4h=x/48h
multiply both sides by 48h
12g=x
answer is 12 gallons
5g:20h
fractionify
5g/20h=g/4h
g:4h
how many gallons in 48 hours
xgallons:48
assume constant rate of flow so
g:4h=x:48
fractionfy
g/4h=x/48h
multiply both sides by 48h
12g=x
answer is 12 gallons