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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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    https://brainmass.com/math/complex-analysis/clep-exam-sample-questions-15272

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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 + ...

    Solution Summary

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

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