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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 #3211 : Algebra Ii

Evaluate: $9^{-1}+3^{-3}$

Possible Answers:

$\textup{The answer is not given.}$

$\frac{244}{27}$

$\frac{82}{27}$

$\frac{4}{27}$

$\frac{2}{9}$

Correct answer:

$\frac{4}{27}$

Explanation:

The negative exponents can be simplified as follows:

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

Rewrite the expression.

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

Convert the one-ninth to $\frac{3}{27}$ to match denominators, and add the fractions.

$\frac{3}{27}+\frac{1}{27}= \frac{4}{27}$

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

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Example Question #3212 : Algebra Ii

Simplify: $-6-3^{-2}$

Possible Answers:

$-\frac{1}{18}$

$-\frac{55}{9}$

$-\frac{35}{6}$

$-\frac{37}{6}$

$-\frac{53}{9}$

Correct answer:

$-\frac{55}{9}$

Explanation:

The negative exponent can be rewritten into a fraction.

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

Convert the expression.

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

Ignoring the negative sign, multiply the denominator with the whole number and add the numerator will give an improper fraction. The denominator stays the same.

$-\frac{(9\times 6)+1}{9}$

The answer is: $-\frac{55}{9}$

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Example Question #81 : Understanding Exponents

Solve the following: $8^{-2}-2^{-5}$

Possible Answers:

$-\frac{1}{32}$

$-\frac{3}{32}$

$-\frac{1}{16}$

$-\frac{3}{64}$

$-\frac{1}{64}$

Correct answer:

$-\frac{1}{64}$

Explanation:

In order to simplify this expression, we will need to rewrite the negative exponents.

The negative exponents can be rewritten into fractional form.

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

$8^{-2}-2^{-5} = \frac{1}{8^2}-\frac{1}{2^5} = \frac{1}{64}- \frac{1}{32}$

Convert the second fraction to match the denominator of the first fraction.

$\frac{1}{64}- \frac{1(2)}{32(2)} = \frac{1}{64}-\frac{2}{64}= -\frac{1}{64}$

The answer is: $-\frac{1}{64}$

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Example Question #82 : Understanding Exponents

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

Possible Answers:

$\frac{27} {4}$

$\frac{1}{2}$

$\frac{27}{8}$

$\frac{9}{2}$

$\frac{9}{8}$

Correct answer:

$\frac{27}{8}$

Explanation:

Rewrite the negative exponent as the reciprocal of the positive power.

The negative exponent can be rewritten into a fraction as follows:

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

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

Simplify the complex fraction.

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

The answer is: $\frac{27}{8}$

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Example Question #83 : Understanding Exponents

Solve: $-4^{-3}-2^{-3}$

Possible Answers:

$-\frac{9}{64}$

$-\frac{9}{513}$

$-\frac{513}{64}$

$-\frac{513}{8}$

$\frac{11}{64}$

Correct answer:

$-\frac{9}{64}$

Explanation:

The negative exponents can be rewritten into a fraction.

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

Rewrite both terms given in the problem.

$-4^{-3}-2^{-3}=-\frac{1}{4^3}-\frac{1}{2^3} = -\frac{1}{64}-\frac{1}{8}$

Find the least common denominator of both fractions.

$-\frac{1}{64}-\frac{1}{8} = -\frac{1}{64}-\frac{1(8)}{8(8)}$

Simplify the fractions.

$-\frac{1}{64}-\frac{8}{64}=-\frac{9}{64}$

The answer is: $-\frac{9}{64}$

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Example Question #84 : Understanding Exponents

Solve: $-(-8^{-2})+8$

Possible Answers:

$7\frac{1}{8}$

$8\frac{1}{16}$

$4\frac{1}{64}$

$8\frac{1}{64}$

$7\frac{15}{16}$

Correct answer:

$8\frac{1}{64}$

Explanation:

Simplify the negative exponent as follows:

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

$8^{-2} = \frac{1}{8^2} = \frac{1}{64}$

Rewrite the expression.

$-(-8^{-2})+8 = -(-\frac{1}{64})+8 = \frac{1}{64}+8$

The answer is: $8\frac{1}{64}$

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Example Question #85 : Understanding Exponents

Solve: $9^{-2}\times\frac{3}{ 2^{-4}}$

Possible Answers:

$\frac{1}{9}$

$\frac{1}{432}$

$\frac{8}{27}$

$\frac{16}{27}$

$\frac{3}{16}$

Correct answer:

$\frac{16}{27}$

Explanation:

Change all negative exponents into a fractional form by the following property.

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

$9^{-2}\times\frac{3}{ 2^{-4}} = \frac{1}{9^2} \times \frac{3}{\frac{1}{2^4}} = \frac{1}{81} \times \frac{3}{\frac{1}{16}}$

Simplify the fractions.

$\frac{1}{81} \times \frac{3}{\frac{1}{16}} = \frac{1}{81} \times 48 = \frac{48}{81}$

$\frac{48}{81} =\frac{16 \times 3}{27 \times 3}$

The answer is: $\frac{16}{27}$

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Example Question #551 : Mathematical Relationships And Basic Graphs

Evaluate: $7^{-3}$

Possible Answers:

$-\frac{1}{7^3}$

7^3

-7^3

$\frac{1}{7^3}$

Correct answer:

$\frac{1}{7^3}$

Explanation:

When dealing with negative exponents, we write $x^{-a}=\frac{1}{x^a}$ . Therefore $7^{-3}=\frac{1}{7^3}$ .

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Example Question #552 : Mathematical Relationships And Basic Graphs

Evaluate: $(x+5)^{-3}$

Possible Answers:

$\frac{1}{x^3+5^3}$

$\frac{x}{5^3}$

$\frac{5^3}{x}$

$\frac{1}{(x+5)^3}$

$\frac{5^3}{(x+5)^3}$

Correct answer:

$\frac{1}{(x+5)^3}$

Explanation:

When dealing with negative exponents, we write $x^{-a}=\frac{1}{x^a}$ . Therefore $(x+5)^{-3}=\frac{1}{(x+5)^3}$

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Example Question #553 : Mathematical Relationships And Basic Graphs

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

Possible Answers:

$-\frac{4}{9}$

$-\frac{9}{4}$

$\frac{4}{9}$

$\frac{9}{4}$

Correct answer:

$\frac{9}{4}$

Explanation:

When dealing with negative exponents, we write $x^{-a}=\frac{1}{x^a}$ . Therefore $(\frac{2}{3})^{-2}=\frac{1}{(\frac{2}{3})^2}=\frac{1}{\frac{4}{9}}=\frac{9}{4}$ .

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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #3211 : Algebra Ii

Example Question #3212 : Algebra Ii

Example Question #81 : Understanding Exponents

Example Question #82 : Understanding Exponents

Example Question #83 : Understanding Exponents

Example Question #84 : Understanding Exponents

Example Question #85 : Understanding Exponents

Example Question #551 : Mathematical Relationships And Basic Graphs

Example Question #552 : Mathematical Relationships And Basic Graphs

Example Question #553 : Mathematical Relationships And Basic Graphs

All Algebra II Resources

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