How many ml of a 78% acid solution must be mixed with 12ml of a 10% acid solution to make 30% solution?

Approximately 0.6 ml of 78% acid solution has to be mixed with 12ml of a 10% acid solution to make 30% solution

First, set up a table. Fill in the unknowns with variable "x"

"x" ml of 78% acid solution must be mixed with 12 ml of 10 % acid solution to make 30% acid solution

Let us solve for "x"

The table is attached below

From the table below, we can set up two equations

Sum of values of two salts = Value of mixture

Sum of values of two salts = 0.78x + 1.2

Value of mixture = 0.3(x + 12)

Sum of values of two salts = Value of mixture  

Then, 0.78x + 1.2 = 0.3(x + 12)  

0.78x + 1.2 = 0.3x + 3.6  

0.78x – 0.36x = 3.6 – 1.2  

4.2x = 2.4  

x = 0.5714  ≈ 0.6

Hence, approximately 0.6 ml of 78% acid solution has to be mixed

problem:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

f(5/2) = 10/3

i hope this !