Question

From a full 50-liter container of a 40% concentration of acid, x liters are removed and replaced with 100% acid. (A) Write the amount of acid in the final mixture as a function of x (B) Determine the domain and range of the function (C) Determine if the final mixture is 50% acid PLEASE EXPLAIN...WILL MARK BRAINLIEST FOR BEST ANSWER

Answer 1

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