A tank with a square base contained water to a height of 10cm. Jason poured another 392 cm3 of water and the height of the water level increased to 12cm. How much water was in the tank at first?
10x^2 + 392 = 12x^2
find x, and the volume was 10x^2
thank you
To find the initial amount of water in the tank, we need to determine the volume of the tank.
The tank has a square base, and we know the height and the increase in height after Jason poured water. Let's denote the side length of the square base as "x."
We can calculate the initial volume of the tank by finding the volume of the water added. The increase in height from 10 cm to 12 cm is equal to 12 cm - 10 cm = 2 cm.
The volume of water added is given as 392 cm^3. Therefore, the volume of the tank can be calculated as (2 cm) * (x cm) * (x cm) = 392 cm^3.
Simplifying the equation, we have:
2x^2 = 392
To find the value of x, we can solve this quadratic equation. First, divide both sides of the equation by 2:
x^2 = 392/2
x^2 = 196
Next, take the square root of both sides of the equation to solve for x:
x = sqrt(196)
x = 14
Now that we know the side length of the square base (x = 14 cm), we can calculate the initial volume of the tank by multiplying the base area by the initial height:
Initial volume = (14 cm) * (14 cm) * (10 cm)
Initial volume = 1960 cm^3
Therefore, the initial amount of water in the tank was 1960 cm^3.