# CLEP exam sample questions

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Can you please help me with these. I can not get these and I am trying to study for the clep exam. Do not do the circled problems only the ones listed below. Thank you for help. Please show step so that I can understand better.

129

1. a) 6, b) 12, c) 16, d) 20

2. a) 24, b) 26

3. a) 35, b) 36

4. a) 46, b) 48

P. 143

6.b) 50

7. #82

P. 151

8. # 24

9. # 30

10. # 44,

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##### Solution Summary

This shows a variety of sample CLEP questions regarding complex numbers.

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6.)

(3+7i) - (2+5i)

= 3 + 7i - 2 - 5i

= (3-2) + (7-5)i

= 1 + 2i --Answer

12.)

(3i)(5i)

= 3i*5i

= 15*i^2

= 15*(-1) [Because, i^2 = -1]

= -15 --Answer

16.)

(2-i)(-5+6i)

= 2*(-5) + 2*6i - i*(-5) - i*6i

= -10 + 12i + 5i - 6i^2

= -10 + 17i - 6*(-1) [Because, i^2 = -1]

= -10 + 17i + 6

= -4 + 17i --Answer

20.)

(3+8i)(3-8i)

= 3^2 - (8i)^2

[Because, (a+b)(a-b) = a^2 - b^2]

= 9 - 64*i^2

= 9 - 64*(-1)

= 9 + 64

= 73 --Answer

24.)

(3-5i)/(2-i)

Multiply in numerator and denominator by the conjugate of denominator

=[(3-5i)*(2+i)]/[(2-i)*(2+i)]

= [ 3*2 + 3*i - 5i*2 - 5i*i]/(2^2 - i^2)

= [6 + 3i - 10i - 5*i^2]/(4 - (-1)) [Because, i^2 = -1]

= [6 -7i -5*(-1)]/(4+1)

= [6 -7i + 5]/5

= (11 - 7i)/5 --Answer

26.)

(-5+10i)/(3+4i)

Multiply in numerator and denominator by the conjugate of denominator

=[(-5+10i)*(3-4i)]/[(3+4i)*(3-4i)]

= [-5*3 - 5*(-4i) + 10i*3 + 10i*(-4i)]/(3^2 - (4i)^2)

= [ -15 + 20i + 30i - 40*i^2]/(9 - 16*i^2)

= [ -15 + 50i - 40*(-1)]/(9 - 16*(-1))

= [-15 + 50i + 40]/(9+16)

= (25 + ...

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