For this case we have the following polynomial:
Applying the formula of the resolvent we have:
Substituting values we have:
Rewriting we have:
Therefore, the roots of the polynomial are given by:
Answer:
The values of x that are roots of the polynomial are:
Answer:
(a) The amount of pure acid in 60 oz of 20% acid solution is 12 oz.
(b) The amount of pure acid in 60 oz of 25% acid solution is 15 oz.
(c) The amount of pure acid in 60 oz of 30% acid solution is 18 oz.
(d) The amount of pure acid in 60 oz of 50% acid solution is 30 oz.
Step-by-step explanation:
* <em>Lets explain how to solve the problem</em>
- The solution is a mixture of two or more substances
- A container of liquid contains 60 oz of solution
- The solution contains pure acid by different concentrations
- (a) 20% (b) 25% (c) 30% (d) 50%
- We want to find the number of ounces of the pure acid in each
concentration
- Assume that the solution is 100%
- That means 100% represents 60 oz of solution
a)
- The percentage of pure acid is 20%
∵ solution ⇒ pure acid
100% 20%
60 The number of ounces
- By using cross multiplication
∴ The number of ounces =
(a) The amount of pure acid in 60 oz of 20% acid solution is 12 oz.
b)
- The percentage of pure acid is 25%
∵ solution ⇒ pure acid
100% 25%
60 The number of ounces
- By using cross multiplication
∴ The number of ounces =
(b) The amount of pure acid in 60 oz of 25% acid solution is 15oz.
c)
- The percentage of pure acid is 30%
∵ solution ⇒ pure acid
100% 30%
60 The number of ounces
- By using cross multiplication
∴ The number of ounces =
(c) The amount of pure acid in 60 oz of 30% acid solution is 18 oz.
d)
- The percentage of pure acid is 30%
∵ solution ⇒ pure acid
100% 50%
60 The number of ounces
- By using cross multiplication
∴ The number of ounces =
(d) The amount of pure acid in 60 oz of 50% acid solution is 30 oz.