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Algebra II : Algebra II

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

Example Questions

Example Question #41 : Negative Exponents

Evaluate $21^{-2}$ $\displaystyle 21^{-2}$

Possible Answers:

441 $\displaystyle 441$

$\displaystyle -42$

$\displaystyle -441$

$\frac{1}{441}$ $\displaystyle \frac{1}{441}$

$-\frac{1}{441}$ $\displaystyle -\frac{1}{441}$

Correct answer:

$\frac{1}{441}$ $\displaystyle \frac{1}{441}$

Explanation:

When expressing negative exponents, we rewrite as such:

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

in which $\displaystyle a$ is the positive exponent raising base $\displaystyle x$.

$21^{-2}=\frac{1}{21^2}=\frac{1}{441}$ $\displaystyle 21^{-2}=\frac{1}{21^2}=\frac{1}{441}$

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Example Question #50 : Understanding Exponents

Evaluate $-4^{-4}$ $\displaystyle -4^{-4}$

Possible Answers:

$\frac{1}{16}$ $\displaystyle \frac{1}{16}$

$-\frac{1}{256}$ $\displaystyle -\frac{1}{256}$

$\displaystyle 16$

$\frac{1}{256}$ $\displaystyle \frac{1}{256}$

$-\frac{1}{16}$ $\displaystyle -\frac{1}{16}$

Correct answer:

$-\frac{1}{256}$ $\displaystyle -\frac{1}{256}$

Explanation:

When expressing negative exponents, we rewrite as such:

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

in which $\displaystyle a$ is the positive exponent raising base $\displaystyle x$.

$-4^{-4}=-\frac{1}{4^4}=-\frac{1}{256}$ $\displaystyle -4^{-4}=-\frac{1}{4^4}=-\frac{1}{256}$

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Example Question #51 : Negative Exponents

Evaluate $(-5)^{-4}$ $\displaystyle (-5)^{-4}$

Possible Answers:

$-\frac{1}{625}$ $\displaystyle -\frac{1}{625}$

$\frac{1}{20}$ $\displaystyle \frac{1}{20}$

$\displaystyle -20$

$\displaystyle 20$

$\frac{1}{625}$ $\displaystyle \frac{1}{625}$

Correct answer:

$\frac{1}{625}$ $\displaystyle \frac{1}{625}$

Explanation:

When expressing negative exponents, we rewrite as such:

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

in which $\displaystyle a$ is the positive exponent raising base $\displaystyle x$.

$(-5)^{-4}=\frac{1}{(-5)^4}=\frac{1}{625}$ $\displaystyle (-5)^{-4}=\frac{1}{(-5)^4}=\frac{1}{625}$

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Example Question #51 : Negative Exponents

Evaluate $\frac{1}{2}^{-3}$ $\displaystyle \frac{1}{2}^{-3}$

Possible Answers:

$\frac{1}{8}$ $\displaystyle \frac{1}{8}$

$-\frac{1}{6}$ $\displaystyle -\frac{1}{6}$

$\displaystyle -8$

$-\frac{1}{8}$ $\displaystyle -\frac{1}{8}$

$\displaystyle 8$

Correct answer:

$\displaystyle 8$

Explanation:

When expressing negative exponents, we rewrite as such:

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

in which $\displaystyle a$ is the positive exponent raising base $\displaystyle x$.

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

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Example Question #53 : Understanding Exponents

Evaluate $\frac{1}{6}^{-\frac{1}{2}}$ $\displaystyle \frac{1}{6}^{-\frac{1}{2}}$

Possible Answers:

$\frac{\sqrt{6}}{6}$ $\displaystyle \frac{\sqrt{6}}{6}$

$\sqrt{6}$ $\displaystyle \sqrt{6}$

$\displaystyle 6$

$-\sqrt{6}$ $\displaystyle -\sqrt{6}$

$\frac{1}{36}$ $\displaystyle \frac{1}{36}$

Correct answer:

$\sqrt{6}$ $\displaystyle \sqrt{6}$

Explanation:

When dealing with fractional exponents, we rewrite as such:

$x^\frac{a}{b}=\sqrt[b]{x^a}$ $\displaystyle x^\frac{a}{b}=\sqrt[b]{x^a}$

in which $\displaystyle b$ is the index of the radical and $\displaystyle a$ is the exponent raising base $\displaystyle a$.

When expressing negative exponents, we rewrite as such:

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

in which $\displaystyle a$ is the positive exponent raising base $\displaystyle x$.

$\frac{1}{6}^{-\frac{1}{2}}=\frac{1}{(\sqrt\frac{1}{6})}=\frac{1}{\frac{1}{\sqrt{6}}}=\sqrt{6}$ $\displaystyle \frac{1}{6}^{-\frac{1}{2}}=\frac{1}{(\sqrt\frac{1}{6})}=\frac{1}{\frac{1}{\sqrt{6}}}=\sqrt{6}$

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Example Question #51 : Exponents

Evaluate $12^{-\frac{1}{2}}$ $\displaystyle 12^{-\frac{1}{2}}$

Possible Answers:

$\frac{\sqrt{3}}{6}$ $\displaystyle \frac{\sqrt{3}}{6}$

$\frac{1}{12}$ $\displaystyle \frac{1}{12}$

$2\sqrt{3}$ $\displaystyle 2\sqrt{3}$

$6\sqrt{3}$ $\displaystyle 6\sqrt{3}$

$\frac{\sqrt{3}}{2}$ $\displaystyle \frac{\sqrt{3}}{2}$

Correct answer:

$\frac{\sqrt{3}}{6}$ $\displaystyle \frac{\sqrt{3}}{6}$

Explanation:

When dealing with fractional exponents, we rewrite as such:

$x^\frac{a}{b}=\sqrt[b]{x^a}$ $\displaystyle x^\frac{a}{b}=\sqrt[b]{x^a}$

in which $\displaystyle b$ is the index of the radical and $\displaystyle a$ is the exponent raising base $\displaystyle a$.

When expressing negative exponents, we rewrite as such:

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

in which $\displaystyle a$ is the positive exponent raising base $\displaystyle x$.

$12^{-\frac{1}{2}}=\frac{1}{(\sqrt{12})}=\frac{1}{2{\sqrt{3}}}=\frac{\sqrt{3}}{6}$ $\displaystyle 12^{-\frac{1}{2}}=\frac{1}{(\sqrt{12})}=\frac{1}{2{\sqrt{3}}}=\frac{\sqrt{3}}{6}$

Remember when getting rid of radicals, just multiply top and bottom by that radical.

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Example Question #55 : Understanding Exponents

Evaluate $48^{-\frac{2}{3}}$ $\displaystyle 48^{-\frac{2}{3}}$

Possible Answers:

$\frac{\sqrt[3]{6}}{24}$ $\displaystyle \frac{\sqrt[3]{6}}{24}$

$\frac{\sqrt{3}}{24}$ $\displaystyle \frac{\sqrt{3}}{24}$

$\frac{\sqrt{3}}{6}$ $\displaystyle \frac{\sqrt{3}}{6}$

$\frac{\sqrt[3]{36}}{16}$ $\displaystyle \frac{\sqrt[3]{36}}{16}$

$\frac{\sqrt{6}}{16}$ $\displaystyle \frac{\sqrt{6}}{16}$

Correct answer:

$\frac{\sqrt[3]{6}}{24}$ $\displaystyle \frac{\sqrt[3]{6}}{24}$

Explanation:

When dealing with fractional exponents, we rewrite as such:

$x^\frac{a}{b}=\sqrt[b]{x^a}$ $\displaystyle x^\frac{a}{b}=\sqrt[b]{x^a}$

in which $\displaystyle b$ is the index of the radical and $\displaystyle a$ is the exponent raising base $\displaystyle a$.

When expressing negative exponents, we rewrite as such:

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

in which $\displaystyle a$ is the positive exponent raising base $\displaystyle x$.

$48^{-\frac{2}{3}}=\frac{1}{\sqrt[3]{48^2}}=\frac{1}{\sqrt[3]{2304}}$ $\displaystyle 48^{-\frac{2}{3}}=\frac{1}{\sqrt[3]{48^2}}=\frac{1}{\sqrt[3]{2304}}$

Let's find a perfect cube which is

$\displaystyle 64 =(4^3)$.

$\frac{1}{\sqrt[3]{2304}}=\frac{1}{\sqrt[3]{64}*\sqrt[3]{36}}=\frac{1}{4\sqrt[3]{36}}$ $\displaystyle \frac{1}{\sqrt[3]{2304}}=\frac{1}{\sqrt[3]{64}*\sqrt[3]{36}}=\frac{1}{4\sqrt[3]{36}}$

To simplify, we need to multiply top and bottom by an appropriate cubic root. We know 36=6^2 $\displaystyle 36=6^2$ so if we multiply top and bottom by $\sqrt[3]{6}$ $\displaystyle \sqrt[3]{6}$ we will get an integer in the denominator.

$\frac{1}{4\sqrt[3]{36}}*\frac{\sqrt[3]{6}}{\sqrt[3]{6}}=\frac{\sqrt[3]{6}}{4*6}=\frac{\sqrt[3]{6}}{24}$ $\displaystyle \frac{1}{4\sqrt[3]{36}}*\frac{\sqrt[3]{6}}{\sqrt[3]{6}}=\frac{\sqrt[3]{6}}{4*6}=\frac{\sqrt[3]{6}}{24}$

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Example Question #56 : Understanding Exponents

Evaluate $12^{-\frac{3}{4}}$ $\displaystyle 12^{-\frac{3}{4}}$

Possible Answers:

$\frac{\sqrt[4]{12}}{8}$ $\displaystyle \frac{\sqrt[4]{12}}{8}$

$\frac{\sqrt[4]{12}}{12}$ $\displaystyle \frac{\sqrt[4]{12}}{12}$

$\frac{\sqrt[3]{36}}{16}$ $\displaystyle \frac{\sqrt[3]{36}}{16}$

$\frac{\sqrt{6}}{12}$ $\displaystyle \frac{\sqrt{6}}{12}$

$\frac{\sqrt[3]{12}}{12}$ $\displaystyle \frac{\sqrt[3]{12}}{12}$

Correct answer:

$\frac{\sqrt[4]{12}}{12}$ $\displaystyle \frac{\sqrt[4]{12}}{12}$

Explanation:

When dealing with fractional exponents, we rewrite as such:

$x^\frac{a}{b}=\sqrt[b]{x^a}$ $\displaystyle x^\frac{a}{b}=\sqrt[b]{x^a}$

in which $\displaystyle b$ is the index of the radical and $\displaystyle a$ is the exponent raising base $\displaystyle a$.

When expressing negative exponents, we rewrite as such:

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

in which $\displaystyle a$ is the positive exponent raising base $\displaystyle x$.

$12^{-\frac{3}{4}}=\frac{1}{\sqrt[4]{12^3}}=\frac{1}{\sqrt[4]{1728}}$ $\displaystyle 12^{-\frac{3}{4}}=\frac{1}{\sqrt[4]{12^3}}=\frac{1}{\sqrt[4]{1728}}$

Let's find a perfect fourth power which is

$\displaystyle 16 = (2^4)$.

$\frac{1}{\sqrt[4]{1728}}=\frac{1}{\sqrt[4]{16}*\sqrt[4]{108}}=\frac{1}{2\sqrt[4]{108}}$ $\displaystyle \frac{1}{\sqrt[4]{1728}}=\frac{1}{\sqrt[4]{16}*\sqrt[4]{108}}=\frac{1}{2\sqrt[4]{108}}$

To simplify, we need to multiply top and bottom by an appropriate fourth root. We know $\displaystyle 108=3^3*2^2$.

We need to complete the numbers to the fourth power. It we multiply top and bottom by $\sqrt[4]{3*2^2}=\sqrt[4]{12}$ $\displaystyle \sqrt[4]{3*2^2}=\sqrt[4]{12}$ we will get an integer in the denominator.

$\frac{1}{2\sqrt[4]{108}}*\frac{\sqrt[4]{12}}{\sqrt[4]{12}}=\frac{\sqrt[4]{12}}{2*6}=\frac{\sqrt[4]{12}}{12}$ $\displaystyle \frac{1}{2\sqrt[4]{108}}*\frac{\sqrt[4]{12}}{\sqrt[4]{12}}=\frac{\sqrt[4]{12}}{2*6}=\frac{\sqrt[4]{12}}{12}$

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Example Question #51 : Exponents

Simplify $(4x^{2}y^{-6})(2x^{-7}y^{^3})$ $\displaystyle (4x^{2}y^{-6})(2x^{-7}y^{^3})$.

Possible Answers:

$8x^{5}y^{3}$ $\displaystyle 8x^{5}y^{3}$

$\frac{1}{8x^{14}y^{18}}$ $\displaystyle \frac{1}{8x^{14}y^{18}}$

$\frac{8}{x^{5}y^{3}}$ $\displaystyle \frac{8}{x^{5}y^{3}}$

$\frac{1}{8x^{5}y^{3}}$ $\displaystyle \frac{1}{8x^{5}y^{3}}$

$\frac{8}{x^{14}y^{18}}$ $\displaystyle \frac{8}{x^{14}y^{18}}$

Correct answer:

$\frac{8}{x^{5}y^{3}}$ $\displaystyle \frac{8}{x^{5}y^{3}}$

Explanation:

First multiply the like terms, remembering that when multiplying terms that have exponents, you add the exponents.

$(4x^{2}y^{-6})(2x^{-7}y^{3})=(4\cdot 2)(x^{2}\cdot x^{-7})(y^{-6}\cdot y^{3})$ $\displaystyle (4x^{2}y^{-6})(2x^{-7}y^{3})=(4\cdot 2)(x^{2}\cdot x^{-7})(y^{-6}\cdot y^{3})$

$=8x^{2-7}y^{-6+3}$ $\displaystyle =8x^{2-7}y^{-6+3}$

$=8x^{-5}y^{-3}$ $\displaystyle =8x^{-5}y^{-3}$

Negative exponents indicate that the term should be in the denominator, so the final answer is:

$\frac{8}{x^{5}y^{3}}$ $\displaystyle \frac{8}{x^{5}y^{3}}$

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Example Question #51 : Understanding Exponents

Simplify: $6^{-2}+3^{-2}$ $\displaystyle 6^{-2}+3^{-2}$

Possible Answers:

$\frac{1}{81}$ $\displaystyle \frac{1}{81}$

$\frac{5}{36}$ $\displaystyle \frac{5}{36}$

$\sqrt{6}+\sqrt{3}$ $\displaystyle \sqrt{6}+\sqrt{3}$

$\frac{5}{18}$ $\displaystyle \frac{5}{18}$

$\displaystyle -3$

Correct answer:

$\frac{5}{36}$ $\displaystyle \frac{5}{36}$

Explanation:

Convert each negative exponent into fractional form.

$a^{-n} = \frac{1}{a^n}$ $\displaystyle a^{-n} = \frac{1}{a^n}$

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

Simplify the denominators.

$\frac{1}{36}+\frac{1}{9}$ $\displaystyle \frac{1}{36}+\frac{1}{9}$

Convert the second fraction with a common denominator of 36.

$\frac{1}{36}+\frac{4}{36}= \frac{5}{36}$ $\displaystyle \frac{1}{36}+\frac{4}{36}= \frac{5}{36}$

The answer is: $\frac{5}{36}$ $\displaystyle \frac{5}{36}$

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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #41 : Negative Exponents

Example Question #50 : Understanding Exponents

Example Question #51 : Negative Exponents

Example Question #51 : Negative Exponents

Example Question #53 : Understanding Exponents

Example Question #51 : Exponents

Example Question #55 : Understanding Exponents

Example Question #56 : Understanding Exponents

Example Question #51 : Exponents

Example Question #51 : Understanding Exponents

All Algebra II Resources

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