Answer:
See explanation
Step-by-step explanation:
Given:
50 liter container
40% concentration of acid
x liters are removed and replaced with 100% acid
(A)
Acid in 50-liter container = 40% of 50 liters
= (40/100) * 50
= 0.4 * 50
= 20
So there are 20 liters amount of acid in 50 liter container.
Now x liters are removed and replaced with 100% acid
The acid in x-liter
40% of x liters = (40/100) * x = 0.4 * x
When x liters are removed then amount of acid that remained in container:
20 - 40% of x liters = 20 - (40/100) * x = 20 - 0.4 * x
This can also be written as:
(50-x) (40/100) = (40*50/100) - (40*x/100) = 2000/100 - 40x/100 = 20 - 2x/5
Since x liters replaced with 100% acid so this becomes:
20 - 40% of x liters + 100% of x liters
= 20 - (40/100) * x + 100/100 * x
= 20 - (40/100) x + (100/100)x
= 20 + (100-40/100)x
= 20 + (60/100)x
= 20 + 0.6 x
This can also be written as:
20 - 2x/5 + x = 20 + 3x/5 = 20 + 0.6 x
Hence the amount of acid in the final mixture as a function of x:
f(x) = 20 + 0.6 x
(B)
The value of x can not be greater than 50 liters and can not be lesser than 0 liters.
so, range is [0, 50] where both 0 and 50 are inclusive.
(C)
Since final mixture is 50% acid So
Acid = 50% of 50 litres =
= (50/100) * 50
= 1/2 * 50
= 50/2
= 25
Put 25 in the computed function of x
20 + 0.6 x = 25
0.6x = 25 - 20
0.6x = 5
x = 5/0.6
x = 8.333333
x = 8.3 liters
r^2 with 'r' being your radius to get the area of the base