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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

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10 Diagnostic Tests 630 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #3241 : Algebra Ii

Evaluate: $(\frac{3^{-2}}{6^{-1}})^{-2}$

Possible Answers:

$\frac{9}{4}$

2916

$\frac{1}{2916}$

$\frac{9}{16}$

$\frac{16}{9}$

Correct answer:

$\frac{9}{4}$

Explanation:

To evaluate a negative exponent, it is the reciprocal of the positive power.

$x^{-y} = \frac{1}{x^y}$

Evaluate each term.

$3^{-2} = \frac{1}{3^2} =\frac{1}{9}$

$6^{-1} = \frac{1}{6}$

Divide both fractions.

$(\frac{3^{-2}}{6^{-1}}) = \frac{1}{9}\div \frac{1}{6} = \frac{1}{9}\times \frac{6}{1} = \frac{2}{3}$

The original expression was: $(\frac{3^{-2}}{6^{-1}})^{-2}$

Replace the term and repeat the process for solving negative exponents.

$(\frac{3^{-2}}{6^{-1}})^{-2} = (\frac{2}{3})^{-2} = \frac{1}{(\frac{2}{3})^2} = \frac{1}{\frac{4}{9}}$

The answer is: $\frac{9}{4}$

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Example Question #101 : Exponents

Solve: $3^{-1}+3^{-2}-3^{-3}$

Possible Answers:

$\frac{13}{27}$

$\frac{11}{27}$

$\frac{7}{27}$

$-\frac{11}{27}$

$\frac{4}{27}$

Correct answer:

$\frac{11}{27}$

Explanation:

Convert all terms with negative exponents to fractional form.

$x^{-Y} = \frac{1}{x^Y}$

$3^{-1}+3^{-2}-3^{-3} =\frac{1}{3^{1}}+\frac{1}{3^2}-\frac{1}{3^3}$

$\frac{1}{3}+\frac{1}{9}-\frac{1}{27}$

Convert all fractions to a common denominator. Visually, the LCD can be seen as 27, since this value is divisible by all three terms.

$\frac{1(9)}{3(9)}+\frac{1(3)}{9(3)}-\frac{1}{27} = \frac{9}{27}+ \frac{3}{27}- \frac{1}{27}$

The answer is: $\frac{11}{27}$

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Example Question #111 : Negative Exponents

Simplify: $\frac{2^{-3}}{3^{-2}}+\frac{3^{-1}}{2}$

Possible Answers:

$\frac{43}{24}$

$\frac{19}{18}$

$\frac{1}{6}$

$\frac{31}{24}$

$\frac{13}{72}$

Correct answer:

$\frac{31}{24}$

Explanation:

Simplify each term. The negative exponents can be rewritten as a fraction.

$x^{-A} = \frac{1}{x^A}$

$\frac{2^{-3}}{3^{-2}} = \frac{\frac{1}{2^3}}{\frac{1}{3^2}}= \frac{\frac{1}{8}}{\frac{1}{9}}= \frac{1}{8}\div \frac{1}{9} = \frac{1}{8}\times \frac{9}{1} =\frac{9}{8}$

$\frac{3^{-1}}{2} = \frac{\frac{1}{3}}{2} = \frac{1}{3}\div 2 = \frac{1}{3}\times \frac{1}{2} =\frac{1}{6}$

Add the fractions.

$\frac{9}{8}+\frac{1}{6} = \frac{9(3)}{8(3)}+\frac{1(4)}{6(4)} = \frac{31}{24}$

The answer is: $\frac{31}{24}$

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Example Question #112 : Negative Exponents

Solve: $\frac{1}{12}+6^{-1}-6^{-2}$

Possible Answers:

$\frac{7}{26}$

$\frac{5}{36}$

$\frac{2}{9}$

$\frac{8}{9}$

$\frac{4}{9}$

Correct answer:

$\frac{2}{9}$

Explanation:

Convert each negative exponent to a fraction. Use the rule below to rewrite the terms.

$x^{-A} = \frac{1}{x^A}$

$\frac{1}{12}+6^{-1}-6^{-2}=\frac{1}{12}+\frac{1}{6^{1}}-\frac{1}{6^{2}}$

The expression becomes:

$\frac{1}{12}+\frac{1}{6}-\frac{1}{36}$

Convert the denominators to a common LCD.

$\frac{1(3)}{12(3)}+\frac{1(6)}{6(6)}-\frac{1}{36} = \frac{3}{36} +\frac{6}{36} -\frac{1}{36}=\frac{8}{36}$

Reduce this fraction.

The answer is: $\frac{2}{9}$

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Example Question #113 : Negative Exponents

Simplify the exponents: $\frac{3^{-3}}{\frac{1}{9}}$

Possible Answers:

$\frac{1}{9}$

243

$\frac{1}{27}$

$\frac{1}{3}$

Correct answer:

$\frac{1}{3}$

Explanation:

The negative exponent can be rewritten as a fraction.

$x^{-A} = \frac{1}{x^A}$

$3^{-3} = \frac{1}{3^3} = \frac{1}{27}$

Divide this fraction with one-ninth.

$\frac{\frac{1}{27}}{\frac{1}{9}}= \frac{1}{27}\div \frac{1}{9} = \frac{1}{27}\times \frac{9}{1}$

Reduce the fraction.

The answer is: $\frac{1}{3}$

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Example Question #114 : Negative Exponents

Solve: $\frac{7^{-1}}{5^{-2}}$

Possible Answers:

$\frac{7}{25}$

$\frac{25}{7}$

$\frac{25}{49}$

175

$\frac{1}{175}$

Correct answer:

$\frac{25}{7}$

Explanation:

The negative exponents must be rewritten in fractional form by the following rule.

$a^{-b} = \frac{1}{a^b}$

This expression becomes:

$\frac{7^{-1}}{5^{-2}} = \frac{\frac{1}{7^{1}}}{\frac{1}{5^{2}}} = \frac{\frac{1}{7}}{\frac{1}{25}}$

Rewrite this complex fraction using a division sign. Take the reciprocal of the second term, and change the division sign to multiplication.

$\frac{1}{7}\div \frac{1}{25}=\frac{1}{7}\times \frac{25}{1}$

The answer is: $\frac{25}{7}$

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Example Question #115 : Negative Exponents

Solve: $\frac{(\frac{2}{3})^{-3}}{2^{-1}}$

Possible Answers:

$\frac{4}{9}$

$\frac{16}{27}$

$\frac{27}{4}$

$\frac{4}{3}$

Correct answer:

$\frac{27}{4}$

Explanation:

Evaluate both terms on the numerator and denominator separately.

We can rewrite the negative exponents as the reciprocal of the positive exponent.

Use the multiplicative rule of division to simplify the complex fraction.

$(\frac{2}{3})^{-3} = \frac{1}{(\frac{2}{3})^3} = 1\div (\frac{8}{27}) = 1\times \frac{27}{8}= \frac{27}{8}$

$2^{-1} = \frac{1}{2}$

Divide the two fractions.

$\frac{27}{8}\div \frac{1}{2} = \frac{27}{8}\times \frac{2}{1} =\frac{27}{4}$

The answer is: $\frac{27}{4}$

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Example Question #116 : Negative Exponents

Simplify: $[(2^{-3})\cdot (5^{-2})]^{-1}$

Possible Answers:

200

$\frac{1}{200}$

$\frac{1}{100}$

$\frac{8}{25}$

$\frac{25}{8}$

Correct answer:

200

Explanation:

The negative exponents can be rewritten as fractions based on the following rule:

$x^{-Y}=-\frac{1}{x^Y}$

Rewrite all the terms in the parentheses.

$[(2^{-3})\cdot (5^{-2})]^{-1} =[(\frac{1}{2^3})\cdot (\frac{1}{5^{2}})]^{-1} =[(\frac{1}{8})\cdot (\frac{1}{25})]^{-1}$

Simplify the expression.

$[\frac{1}{200}]^{-1} = 1\div \frac{1}{200} = 1\times \frac{200}{1} = 200$

The answer is: 200

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Example Question #117 : Negative Exponents

Simplify: $\frac{3^{-2}}{8}$

Possible Answers:

$\frac{1}{36}$

$\frac{1}{72}$

$\frac{1}{17}$

$\frac{1}{48}$

$\frac{1}{24}$

Correct answer:

$\frac{1}{72}$

Explanation:

The negative exponent can be rewritten as follows:

$x^{-Y} = \frac{1}{x^Y}$

$3^{-2} = \frac{1}{3^2} = \frac{1}{9}$

Rewrite the expression.

$\frac{3^{-2}}{8} =\frac{ \frac{1}{9}}{8} = \frac{1}{9}\times \frac{1}{8} = \frac{1}{72}$

The answer is: $\frac{1}{72}$

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Example Question #3242 : Algebra Ii

Simplify the following expression:

$\frac{4x^{-2}y^{3}}{18x^{-4}y^{2}}$

Possible Answers:

$\frac{2y}{9x^2}$

$\frac{2x^{2}y}{9}$

$\frac{y}{4x^{2}}$

$\frac{x^{6}y^{5}}{4}$

Correct answer:

$\frac{2x^{2}y}{9}$

Explanation:

$\frac{4x^{-2}y^3}{18x^{-4}y^{2}}=\frac{4x^{4}y^{3}}{18x^{2}y^{2}}=\left (\frac{4}{18} \right )x^{4-2}y^{3-2}=\left (\frac{4}{18} \right )x^{2}y=\frac{2x^{2}y}{9}$

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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #3241 : Algebra Ii

Example Question #101 : Exponents

Example Question #111 : Negative Exponents

Example Question #112 : Negative Exponents

Example Question #113 : Negative Exponents

Example Question #114 : Negative Exponents

Example Question #115 : Negative Exponents

Example Question #116 : Negative Exponents

Example Question #117 : Negative Exponents

Example Question #3242 : Algebra Ii

All Algebra II Resources

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