The Quadratic Function has the domain as the set of all real numbers.
For the range, start from minimum value to maximum value.
But because the parabola is downward as a < 0. Thus, there are no minimum value but the maximum value instead.
Therefore the range is y <= -4
Answer:
-50 ml of 3% alcohol solution and 150 ml of 8% alcohol solution
Step-by-step explanation:
For us to solve this type of mixture problem, we must represent the problem in equations. This will be possible by interpreting the question.
Let the original volume of the first alcohol solution be represented with x.
The quantity of the first alcohol solution needed for the mixture is 3% of x
⇒
= 0.03x
Let the original volume of the second alcohol solution be represented with y.
The quantity of the second alcohol solution needed for the mixture is 5% of y
⇒
= 0.05y
The final mixture of alcohol solution is 6% of 100 ml
⇒
= 6 ml
Sum of values of two alcohol solutions = Value of the final mixture
0.03x + 0.05y = 6 ml ..........(1)
Sum of original quantity of each alcohol solution = Original volume of the of mixture
x + y = 100 ml ..........(2)
For easy interpretation, I will be setting up a table to capture all information given in the question.
Component Unit Value Quantity(ml) Value
3% of Alcohol solution 0.03 x 0.03x
8% of Alcohol solution 0.08 y 0.08y
Mixture of 100ml of 6% 0.06 100 6
x + y = 100 0.03x + 0.08y =6
Looking at the equations we derived, we have two unknowns in two equations which is a simultaneous equation.
0.03x + 0.05y = 6 ml ..........(1)
x + y = 100 ml ..........(2)
Using substitution method to solve the simultaneous equation.
Making x the subject of formula from equation (2), we have,
x = 100 - y ..........(3)
Substituting x = 100 - y from equation (3) into equation (1)
0.03(100 - y) + 0.05y = 6
3 - 0.03y + 0.05y = 6
Rearranging the equation,
0.05y - 0.03y = 6 - 3
0.02y = 3
y = 150 ml
Substituting y = 150 ml into equation (3) to get x
x = 100 - 150 ml
x = - 50 ml
The quantity of the first alcohol solution needed for the mixture for 3% is - 50 ml
The quantity of the second alcohol solution needed for the mixture for 5% is 150 ml
This solution means 50 ml of the first alcohol solution must be removed from the mixture with 150 ml of the second alcohol solution to get a final mixture of 100 ml of 6% alcohol solution.