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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 #2002 : Mathematical Relationships And Basic Graphs

Simplify: $\frac{2-i}{3+i}$

Possible Answers:

$\frac{7-i}{8}$

$\frac{6-i}{10}$

$\frac{7-i}{10}$

$\frac{7+i}{8}$

$\frac{7+i}{10}$

Correct answer:

$\frac{7-i}{10}$

Explanation:

To get rid of a complex number, we multiply by the conjugate which is opposite the sign.

$\frac{2-i}{3+i}=\frac{2-i}{3+i}*\frac{3-i}{3-i}=\frac{6+2i-3i-i^2}{9-i^2}=\frac{6-i-(-1)}{9-(-1)}=\frac{7-i}{10}$

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Example Question #4672 : Algebra Ii

Simplify: $\frac{i^{150}-2i^{8}}{3i^{100}-6}$

Possible Answers:

$\textup{The expression cannot be factored.}$

2i-2

$\frac{1}{9}$

Correct answer:

Explanation:

Write out some of the imaginary terms.

$i=\sqrt{-1}$

$i^2= i\cdot i = -1$

$i^4 = i^2\cdot i^2 = (-1)\cdot (-1) = 1$

The powers of the imaginary number can be rewritten using the product of exponents.

$i^{150} = (i^2)^{75} = (-1)^{75} = -1$

i^8= (i^4)^2 = (1)^2 =1

$i^{100} = (i^4)^{25} = (1)^{25} = 1$

Replace all the terms back into the expression.

$\frac{i^{150}-2i^{8}}{3i^{100}-6} = \frac{-1-2(1)}{3(1)-6} = \frac{-3}{-3} = 1$

The answer is:

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Example Question #55 : Complex Imaginary Numbers

Evaluate: $\frac{i^6+1}{i-1}$

Possible Answers:

$\frac{1}{2}i-\frac{1}{2}$

Correct answer:

Explanation:

The imaginary term is equivalent to $\sqrt{-1}$ .

This means that:

i^6 =( i^2)^3 = (-1)^3 = -1

Substitute this term back into the numerator.

$\frac{i^6+1}{i-1}=\frac{-1+1}{i-1} = \frac{0}{i-1} =0$

There is no need to use extra steps such as multiplying by the conjugate of the denominator to simplify.

The answer is:

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Example Question #2011 : Mathematical Relationships And Basic Graphs

Solve: $\frac{2(2i^2-3)}{i^4-i^{12}}$

Possible Answers:

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

$-\frac{2}{5}i-\frac{4}{5}$

$-\frac{1}{2}i-5$

-3i-4

$\frac{3}{5}i+\frac{1}{5}$

Correct answer:

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

Explanation:

Evaluate by first evaluating the imaginary terms.

$i=\sqrt{-1}$

$i^2 = i\cdot i = -1$

$i^4= i^2\cdot i^2 =1$

We can also use the product rule of exponents to simplify the higher powered imaginary numbers.

$i^{12} = (i^4)^3=(1)^3 =1$

$\frac{2(2i^2-3)}{i^4-i^{12}}=\frac{2(2(-1)-3)}{1-1} =\frac{ -10}{0}$

Note that evaluating the denominator will give us a zero denominator.

This will indicate that the expression will approach to infinity.

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

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Example Question #57 : Complex Imaginary Numbers

Simplify: $\frac{i^{60}-1}{i^3-1}$

Possible Answers:

i-1

$\textup{Undefined.}$

i+1

Correct answer:

Explanation:

Write the first few terms of the imaginary term.

$i=\sqrt{-1}$

$i^2 = i\cdot i =-1$

$i^3 = i^2\cdot i = -i$

$i^4 = i^2\cdot i^2=1$

i^5 = (i^4)(i)=i

Notice that these terms will be in a pattern for higher order imaginary terms.

Rewrite the numerator using the product of exponents.

$\frac{i^{60}-1}{i^3-1} =\frac{(i^4)^{15}-1}{i^3-1} = \frac{1-1}{-i-1} = 0$

The answer is:

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Example Question #2012 : Mathematical Relationships And Basic Graphs

Evaluate: $(7i^{-400})^{3}$

Possible Answers:

343i

343

$\frac{1}{343}i$

$\frac{1}{343i}$

-343i

Correct answer:

343

Explanation:

Using the exponential property, we can expand the term in parentheses by multiplying the inner and outer powers together.

$(7i^{-400})^{3} = (7^3\cdot i^{-400\times 3}) = 7^3 \cdot i^{-1200} = 343 \cdot i^{-1200}$

Evaluate $i^{-1200}$ by breaking this up as a product of imaginary powers.

$i=\sqrt{-1}$

$i^2 = i\cdot i =-1$

$i^4 =i^2\cdot i^2 =1$

$i^{-1200} = (i^4)^{-300} = (1)^{-300} = 1$

This means that:

$343 \cdot i^{-1200}= 343 \cdot 1$

The answer is: 343

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Example Question #2013 : Mathematical Relationships And Basic Graphs

Evaluate: $\frac{5i}{6-2i}$

Possible Answers:

$\frac{3}{4}i-\frac{1}{4}$

$\frac{1}{4}i+\frac{5}{2}$

$\frac{3}{10}i-\frac{1}{10}$

$\frac{7}{10}i-\frac{9}{10}$

$\frac{1}{4}i-\frac{1}{2}$

Correct answer:

$\frac{3}{4}i-\frac{1}{4}$

Explanation:

Multiply the top and bottom by the conjugate of the denominator.

$\frac{5i}{6-2i}\cdot \frac{6+2i}{6+2i}$

Simplify the bottom using the FOIL method.

(6)(6)+(6)(2i)+(-2i)(6)+(-2i)(2i)

Simplify the expression by distributing all terms. Recall that $i=\sqrt{-1}$ and i^2=-1 . Replace the term.

36-4i^2 = 36-4(-1) = 36+4 =40

Simplify the top.

5i(6+2i)=30i+10i^2 = 30i+(10)(-1) = 30i-10

Divide this by forty.

$\frac{30i-10}{40} = \frac{3}{4}i-\frac{1}{4}$

The answer is: $\frac{3}{4}i-\frac{1}{4}$

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Example Question #53 : Imaginary Numbers

Evaluate: $\frac{2i}{5-2i}$

Possible Answers:

$\frac{10i-4}{29}$

$\frac{10i+4}{29}$

$i-\frac{2}{5}$

$\frac{10i-4}{5}$

$\frac{10i-4}{21}$

Correct answer:

$\frac{10i-4}{29}$

Explanation:

Multiply the top and bottom by the conjugate of the denominator.

$\frac{2i}{5-2i}\cdot \frac{5+2i}{5+2i}$

Distribute the numerator and FOIL the denominator.

$\frac{2i}{5-2i}\cdot \frac{5+2i}{5+2i} = \frac{10i+4i^2}{25-4i^2}$

Recall that $i=\sqrt{-1}$ . This means that $i^2 = i\cdot i = -1$ .

Replace the terms.

$\frac{10i+4(-1)}{25-4(-1)} = \frac{10i-4}{29}$

The answer is: $\frac{10i-4}{29}$

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Example Question #242 : Gre Subject Test: Math

Evaluate:

$\left (2 + 3i \right )^{3}$

Possible Answers:

-46 + 63 i

-4 6+ 9i

8 - 27i

-62 + 63 i

8 + 9i

Correct answer:

-4 6+ 9i

Explanation:

We can set A = 2, B = 3i in the cube of a binomial pattern:

$\left ( A + B\right ) ^{3} = A ^{3} + 3 A^{2 }B + 3AB^{2} + B^{3}$

$= 2 ^{3} + 3 \cdot 2 ^{2 } \cdot 3i + 3 \cdot 2 \cdot (3i)^{2} + (3i)^{3}$

$= 2^{3} + 3 \cdot 2^{2} \cdot 3i + 3 \cdot 2 \cdot 3 ^{2}\cdot i^{2} + 3^{3} \cdot i^{3}$

$= 8 + 36i + 54 i^{2} + 27 i^{3}$

= 8 + 36i + 54 (-1) + 27 (-i)

= 8 + 36i - 54 - 27 i

= -46 + 9i

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Example Question #2011 : Mathematical Relationships And Basic Graphs

Which of the following is equal to $-\sqrt{-1}$ ?

Possible Answers:

i^4

i^2

i^3

Correct answer:

i^3

Explanation:

Each of the following are true:

$i = \sqrt{-1}$

$i^2 = (\sqrt{-1})^2 = -1$

$i^3 = (\sqrt{-1})^3 = -\sqrt{-1}$

$i^4 = (\sqrt{-1})^4 = 1$

Therefore, the correct answer is i^3 .

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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #2002 : Mathematical Relationships And Basic Graphs

Example Question #4672 : Algebra Ii

Example Question #55 : Complex Imaginary Numbers

Example Question #2011 : Mathematical Relationships And Basic Graphs

Example Question #57 : Complex Imaginary Numbers

Example Question #2012 : Mathematical Relationships And Basic Graphs

Example Question #2013 : Mathematical Relationships And Basic Graphs

Example Question #53 : Imaginary Numbers

Example Question #242 : Gre Subject Test: Math

Example Question #2011 : Mathematical Relationships And Basic Graphs

All Algebra II Resources

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