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Algebra II : Mathematical Relationships and Basic Graphs

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Example Questions

Example Question #45 : Imaginary Numbers

Simplify: $\frac{i^3+1}{i^{9}-2}$

Possible Answers:

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

$\frac{3}{5}i-\frac{3}{5}$

$\frac{1}{5}i-\frac{4}{5}$

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

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

Correct answer:

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

Explanation:

Write out the power terms of .

$i=\sqrt{-1}$

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

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

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

Then,

$i^9 = (i^4)^2\cdot i =(1)^2 \cdot i =i$

Replace the terms.

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

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

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

Simplify both terms by the FOIL method.

(1-i)(i+2) = (1)(i)+(1)(2)+(-i)(i)+(-i)(2)

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

(i-2)(i+2) = i^2-4 = (-1)-4 = -5

Divide the numerator by the denominator.

$\frac{3-i}{-5}$

The answer is: $\frac{1}{5}i-\frac{3}{5}$

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Example Question #46 : Imaginary Numbers

Evaluate: $\frac{i^2-i-3}{i}$

Possible Answers:

1-4i

4i-1

2i-1

-3i

2i-2

Correct answer:

4i-1

Explanation:

Simplify the numerator. Recall that $i=\sqrt{-1}$ and $i^2= i\cdot i = -1$ .

Replace the value of i^2 .

i^2-i-3 = -1-i-3 = -i-4

Rewrite the fraction and split it into two terms.

$\frac{-i-4}{i} = -\frac{i}{i}-\frac{4}{i}$

Simplify both terms.

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

To simplify $\frac{4}{i}$ , multiply on the top and bottom.

$\frac{4}{i}\cdot \frac{i}{i} = \frac{4i}{i^2} = \frac{4i}{-1} = -4i$

This means that: $-1-\frac{4}{i}=-1-(-4i) = 4i-1$

The answer is: 4i-1

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Example Question #47 : Imaginary Numbers

Solve: $\frac{6}{4i-3}$

Possible Answers:

$-\frac{4}{5}i-\frac{9}{5}$

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

$-\frac{18}{25}i-\frac{24}{25}$

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

$-\frac{24}{25}i-\frac{18}{25}$

Correct answer:

$-\frac{24}{25}i-\frac{18}{25}$

Explanation:

Multiply the numerator and denominator by the conjugate of the denominator.

$\frac{6}{4i-3} \cdot \frac{4i+3}{4i+3}$

Simplify the top and bottom.

6(4i+3) = 24i+18

(4i-3)(4i+3) = 16i^2-9

The fraction becomes: $\frac{24i+18}{16i^2-9}$

Recall that i^2 =-1 since $i=\sqrt{-1}$ . Replace the value.

$\frac{24i+18}{16i^2-9}=\frac{24i+18}{16(-1)-9} = \frac{24i+18}{-25}$

The answer is: $-\frac{24}{25}i-\frac{18}{25}$

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Example Question #41 : Complex Imaginary Numbers

(2+i)(4+3i)

Possible Answers:

-5-10i

None of these.

5-10i

5+10i

-5+10i

Correct answer:

5+10i

Explanation:

(2+i)(4+3i)

FOIL:

8+6i+4i+3i^2

Simplify like terms:

8+10i+3i^2

Simplify i^2 :

8+10i+3(-1)

$\mathbf{5+10i}$

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Example Question #42 : Imaginary Numbers

Simplify: $\frac{2}{(2i+1)(2i-1)}$

Possible Answers:

$\frac{8}{25}i-\frac{6}{25}$

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

$-\frac{8}{25}i-\frac{6}{25}$

$-\frac{2}{5}$

$-\frac{2}{3}$

Correct answer:

$-\frac{2}{5}$

Explanation:

Use the FOIL method to simplify the denominator.

(2i+1)(2i-1) = (2i)(2i)+(2i)(-1)+(1)(2i)+(1)(-1)

= 4i^2-1

The imaginary term is defined as: $i=\sqrt{-1}$

This means that: $i^2= i\cdot i = -1$

Replace the term.

4i^2-1 = 4(-1)-1=-5

The answer is: $-\frac{2}{5}$

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Example Question #4661 : Algebra Ii

Simplify: 7i^3*6i^2

Possible Answers:

42i

-42i

42i^5

Correct answer:

42i

Explanation:

We know that $i=\sqrt{-1}$ and i^2=-1 . Therefore 7i^3*6i^2=42i^5 . We know i^5=i^2*i^2*i . Therefore we will have just . Our final answer is 42*i=42i .

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Example Question #4662 : Algebra Ii

Simplify: $\frac{i^5}{4}*\frac{16}{i^7}$

Possible Answers:

-4i

16i

Correct answer:

Explanation:

We know that $i=\sqrt{-1}$ and i^2=-1 . $\frac{i^5}{4}*\frac{16}{i^7}$ simplifies to $\frac{4}{i^2}$ . Since i^2=-1 our final answer is $\frac{4}{-1}=-4$ .

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Example Question #2001 : Mathematical Relationships And Basic Graphs

Simplify: $\frac{1}{4+i}$

Possible Answers:

$\frac{4-i}{17}$

$\frac{4+i}{4-i}$

$\frac{16}{4+i}$

$\frac{4-i}{15}$

$\frac{16}{4-i}$

Correct answer:

$\frac{4-i}{17}$

Explanation:

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

$\frac{1}{4+i}=\frac{1}{4+i}*\frac{4-i}{4-i}=\frac{4-i}{16-i^2}=\frac{4-i}{16-(-1)}=\frac{4-i}{17}$

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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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Algebra II : Mathematical Relationships and Basic Graphs

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #45 : Imaginary Numbers

Example Question #46 : Imaginary Numbers

Example Question #47 : Imaginary Numbers

Example Question #41 : Complex Imaginary Numbers

Example Question #42 : Imaginary Numbers

Example Question #4661 : Algebra Ii

Example Question #4662 : Algebra Ii

Example Question #2001 : Mathematical Relationships And Basic Graphs

Example Question #2002 : Mathematical Relationships And Basic Graphs

Example Question #4672 : Algebra Ii

All Algebra II Resources

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