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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 #31 : Distributing Exponents (Power Rule)

Simplify:

$(24^{12})^{\frac{1}{18}}$ $\displaystyle (24^{12})^{\frac{1}{18}}$

Possible Answers:

$9\sqrt[3]{4}$ $\displaystyle 9\sqrt[3]{4}$

$4\sqrt[3]{9}$ $\displaystyle 4\sqrt[3]{9}$

$\displaystyle 12$

$24\sqrt[3]{2}$ $\displaystyle 24\sqrt[3]{2}$

$\displaystyle 16$

Correct answer:

$4\sqrt[3]{9}$ $\displaystyle 4\sqrt[3]{9}$

Explanation:

When exponents are being raised by another exponent, we just multiply the powers.

$(24^{12})^{\frac{1}{18}}=24^{12*\frac{1}{18}}=24^{\frac{2}{3}}$ $\displaystyle (24^{12})^{\frac{1}{18}}=24^{12*\frac{1}{18}}=24^{\frac{2}{3}}$

When we have fractional exponents, we convert like this:

$x^{\frac{a}b{}}=\sqrt[b]{x^a}$ $\displaystyle x^{\frac{a}b{}}=\sqrt[b]{x^a}$ . $\displaystyle b$ is the index of the radical which is the denominator of the fractional exponent, $\displaystyle a$ is the power that will raise the base which is the numerator of the fractional exponent and $\displaystyle x$ is the base.

$24^{\frac{2}{3}}=\sqrt[3]{24^2}=\sqrt[3]{576}$ $\displaystyle 24^{\frac{2}{3}}=\sqrt[3]{24^2}=\sqrt[3]{576}$ The perfect cube we can get is $\displaystyle 64$ .

$\sqrt[3]{576}=\sqrt[3]{64*9}=4\sqrt[3]{9}$ $\displaystyle \sqrt[3]{576}=\sqrt[3]{64*9}=4\sqrt[3]{9}$

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Example Question #3681 : Algebra Ii

Evaluate:

(2^9)^4 $\displaystyle (2^9)^4$

Possible Answers:

$2^{36}$ $\displaystyle 2^{36}$

$4^{10}$ $\displaystyle 4^{10}$

4^8 $\displaystyle 4^8$

8^9 $\displaystyle 8^9$

$2^{13}$ $\displaystyle 2^{13}$

Correct answer:

$2^{36}$ $\displaystyle 2^{36}$

Explanation:

When dealing exponents being raised by a power, we multiply the exponents and keep the base.

$(2^9)^4=2^{9*4}=2^{36}$ $\displaystyle (2^9)^4=2^{9*4}=2^{36}$

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Example Question #3682 : Algebra Ii

Evaluate:

$(5^{6})^7$ $\displaystyle (5^{6})^7$

Possible Answers:

$5^{13}$ $\displaystyle 5^{13}$

$5^{42}$ $\displaystyle 5^{42}$

25^5 $\displaystyle 25^5$

$5^{28}$ $\displaystyle 5^{28}$

$5^{20}$ $\displaystyle 5^{20}$

Correct answer:

$5^{42}$ $\displaystyle 5^{42}$

Explanation:

When dealing exponents being raised by a power, we multiply the exponents and keep the base.

$(5^{6})^7=5^{6*7}=5^{42}$ $\displaystyle (5^{6})^7=5^{6*7}=5^{42}$

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Example Question #3683 : Algebra Ii

Evaluate:

(9^9)^9 $\displaystyle (9^9)^9$

Possible Answers:

$81^{9}$ $\displaystyle 81^{9}$

$9^{99}$ $\displaystyle 9^{99}$

$81^{18}$ $\displaystyle 81^{18}$

$9^{81}$ $\displaystyle 9^{81}$

$9^{18}$ $\displaystyle 9^{18}$

Correct answer:

$9^{81}$ $\displaystyle 9^{81}$

Explanation:

When dealing with exponents being raised by another exponent, we multiply the powers and keep the base the same.

$(9^9)^9=9^{9*9}=9^{81}$ $\displaystyle (9^9)^9=9^{9*9}=9^{81}$

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Example Question #3684 : Algebra Ii

Evaluate:

$(3^{15})^7$ $\displaystyle (3^{15})^7$

Possible Answers:

$3^{22}$ $\displaystyle 3^{22}$

$3^{42}$ $\displaystyle 3^{42}$

$9^{15}$ $\displaystyle 9^{15}$

$9^{12}$ $\displaystyle 9^{12}$

$3^{105}$ $\displaystyle 3^{105}$

Correct answer:

$3^{105}$ $\displaystyle 3^{105}$

Explanation:

When dealing with exponents being raised by another exponent, we multiply the powers and keep the base the same.

$(3^{15})^7=3^{15*7}=3^{105}$ $\displaystyle (3^{15})^7=3^{15*7}=3^{105}$

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Example Question #3685 : Algebra Ii

Evaluate:

$\displaystyle (18^5)^5$

Possible Answers:

$18^{25}$ $\displaystyle 18^{25}$

$90^{5}$ $\displaystyle 90^{5}$

$18^{10}$ $\displaystyle 18^{10}$

$90^{25}$ $\displaystyle 90^{25}$

$18^{55}$ $\displaystyle 18^{55}$

Correct answer:

$18^{25}$ $\displaystyle 18^{25}$

Explanation:

When an exponent is raised by another power, we will multiply the exponents and keep the base the same.

Therefore:

$(18^5)^5=18^{5*5}=18^{25}$ $\displaystyle (18^5)^5=18^{5*5}=18^{25}$

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Example Question #3686 : Algebra Ii

Evaluate:

$\left(\left(\frac{1}{2}\right)^6\right)^{\frac{1}{2}}$ $\displaystyle \left(\left(\frac{1}{2}\right)^6\right)^{\frac{1}{2}}$

Possible Answers:

$\displaystyle 8$

$\displaystyle 16$

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

$\frac{1}{4}$ $\displaystyle \frac{1}{4}$

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

Correct answer:

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

Explanation:

When an exponent is raised by another power, we will multiply the exponents and keep the base the same.

Therefore:

$\left(\left(\frac{1}{2}\right)^6\right)^{\frac{1}{2}}=\left(\frac{1}{2}\right)^{6*\frac{1}{2}}=\frac{1}{2^{3}}=\frac{1}{8}$ $\displaystyle \left(\left(\frac{1}{2}\right)^6\right)^{\frac{1}{2}}=\left(\frac{1}{2}\right)^{6*\frac{1}{2}}=\frac{1}{2^{3}}=\frac{1}{8}$

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Example Question #37 : Distributing Exponents (Power Rule)

Simplify:

$\displaystyle (x+y)^2+(xy)^2$

Possible Answers:

$\displaystyle x^2+4xy+y^2$

$\displaystyle x^2+y^2+x^2y^2$

$\displaystyle 2x^2+2y^2$

$\displaystyle x^2+2xy+y^2+x^2y^2$

Correct answer:

$\displaystyle x^2+2xy+y^2+x^2y^2$

Explanation:

To start, we must examine the first term. Note that we are squaring a sum, so we cannot simply distribute the power of 2 to each term. We must take the sum times the sum:

$\displaystyle (x+y)(x+y)=x^2+2xy+y^2$

The second term, however, is a product, in which case we can distribute the power:

$\displaystyle (xy)^2=x^2y^2$

Adding the two together, we get

$\displaystyle x^2+2xy+y^2+x^2y^2$

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Example Question #3687 : Algebra Ii

Simplify: (2^5)^9 $\displaystyle (2^5)^9$

Possible Answers:

$2^{16}$ $\displaystyle 2^{16}$

$2^{88}$ $\displaystyle 2^{88}$

$2^{45}$ $\displaystyle 2^{45}$

$2^{24}$ $\displaystyle 2^{24}$

$2^{14}$ $\displaystyle 2^{14}$

Correct answer:

$2^{45}$ $\displaystyle 2^{45}$

Explanation:

When an exponent is raised by another exponent, we just multiply the exponents and keep the base the same.

$(2^5)^9=2^{5*9}=2^{45}$ $\displaystyle (2^5)^9=2^{5*9}=2^{45}$

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Example Question #3688 : Algebra Ii

Simplify: $(17^8)^{-4}$ $\displaystyle (17^8)^{-4}$

Possible Answers:

$17^{-4}$ $\displaystyle 17^{-4}$

$17^{-32}$ $\displaystyle 17^{-32}$

$17^{8}$ $\displaystyle 17^{8}$

$17^{-12}$ $\displaystyle 17^{-12}$

17^4 $\displaystyle 17^4$

Correct answer:

$17^{-32}$ $\displaystyle 17^{-32}$

Explanation:

When an exponent is raised by another exponent, we just multiply the exponents and keep the base the same.

$(17^8)^{-4}=17^{8*-4}=17^{-32}$ $\displaystyle (17^8)^{-4}=17^{8*-4}=17^{-32}$

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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #31 : Distributing Exponents (Power Rule)

Example Question #3681 : Algebra Ii

Example Question #3682 : Algebra Ii

Example Question #3683 : Algebra Ii

Example Question #3684 : Algebra Ii

Example Question #3685 : Algebra Ii

Example Question #3686 : Algebra Ii

Example Question #37 : Distributing Exponents (Power Rule)

Example Question #3687 : Algebra Ii

Example Question #3688 : Algebra Ii

All Algebra II Resources

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