imaginary numbers. put into standard form?

(6-7i)(2+5i)

Please solve and put into standard form. Thanks,

4 Answers

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  • 1 decade ago
    Favorite Answer

    12+30i+14i+35i^2

    12+44i+35i^2

    Source(s): I'm smart. =]
  • 4 years ago

    The 'standard form' for an imaginary number is i*b where i is the imaginary unit (sqrt(-1) and b is the magnitude of the imaginary value expressed as a 'normal' (or real) value. The i is usually written first to 'alert' the reader that what follows is an imaginary value. Also, in electrical engineering, the j is usually used for the imaginary unit since a lower case i is used for instantaneous current. Remember that sqrt(a*b) = sqrt(a) * sqrt(b) so....... In your problem, factor out the -1 from each of the radicals and write the imaginary value as i*(2*sqrt(49) + 3*sqrt(64)) and simplify it to i*(2*7 + 3*8) = i*38 Doug

  • 1 decade ago

    12 +16i -35i2

    = -17 +16i

  • Anonymous
    1 decade ago

    erm how about you do it yourself

    that's what other people do

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