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DochEvi [55]
3 years ago
12

3) 5 fl. oz. of a 2% alcohol solution was mixed with 11 fl.

Mathematics
1 answer:
dangina [55]3 years ago
3 0

Answer:

<em>The concentration of the new mixture is 46%</em>

Step-by-step explanation:

The first solution of 5 fl. oz. is concentrated at 2% alcohol. The amount of alcohol in the solution is:

5*2/100=0.1 fl. oz.

The second solution of 11 fl. oz. is 66% alcohol. The amount of alcohol in the solution is:

11*66/100=7.26 fl. oz.

The total alcohol in the mixture is

0.1 + 6.6 = 7.36 fl. oz.

The total amount of the mixture is:

5 + 11 = 16 fl.oz.

The concentration of the new mixture is:

7.36 / 16 * 100 = 46%

The concentration of the new mixture is 46%

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Answer:

1) Option A) is correct

The given rational exponent expression is not simplified correctly as a radical expression is

x^{\frac{7}{4}}=\sqrt[7]{x^4}

2)Option A) is correct

That is (729x^3y^{-18})^{\frac{1}{6}}=\frac{3\sqrt{x}}{y^3}

Step-by-step explanation:

1) Given that x^{\frac{7}{4}}=(\sqrt{x^7})^\frac{1}{4}

x^{\frac{7}{4}}=\sqrt[4]{x^7} is the correct answer but in the given problem they gave the RHS as wrong.

Therefore the given rational exponent expression is not simplified correctly as a radical expression is

x^{\frac{7}{4}}=\sqrt[7]{x^4}

2)Given that the rational exponent expression is (729x^3y^{-18})^{\frac{1}{6}}

To find it as a radical expression:

(729x^3y^{-18})^{\frac{1}{6}}=(729)^{\frac{1}{6}}(x^3)^{\frac{1}{6}}((y^{-18})^{\frac{1}{6}})

=3(x^{\frac{3}{6}})(y^{\frac{-18}{6}})

=3(x^{\frac{1}{2}})(y^{-3})

=\frac{3\sqrt{x}}{y^3}

Therefore (729x^3y^{-18})^{\frac{1}{6}}=\frac{3\sqrt{x}}{y^3}

Therefore Option A) is correct

That is (729x^3y^{-18})^{\frac{1}{6}}=\frac{3\sqrt{x}}{y^3}

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