Question

9. Water is added to two containers for 20 minutes. The equations below model the ounces of
water, y, in each container after x minutes. At the time when the containers hold the same
amount of water, how much water do they hold?
Container A: y=62x+48
Container B: y=-2x2 +80x+120
234 ounces
482 ounces
792 ounces
920 ounces

Answer 1

Answer:

792 ounces

Step-by-step explanation:

At the time when the containers hold the same amount of water the value of y in the equations is the same,  so

$\large 62x+48=-2x^2+80x+120$

Solve the quadratic equation:

$\large 62x+48=-2x^2+80x+120\Rightarrow 2x^2-80x-120+62x+48=0\\\\2x^2-18x-72=0\Rightarrow x=\frac{-(-18)\pm\sqrt{(-18)^2-4(2)(-72)}}{2*2}=\\\\=\frac{18\pm\sqrt{324+576}}{4}=\frac{18\pm\sqrt{900}}{4}=\frac{18\pm 30}{4}$

Take only the positive solution since x is positive (minutes)

$\large x=\frac{18+30}{4}=\frac{48}{4}=12$

Now, compute the amount of water y by replacing x = 12 in any of the two equations, for example in container A which is easier to calculate

y = 62(12) + 48 = 792

The correct answer is then 792 ounces