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Montano1993 [528]
4 years ago
8

Solve x^2 - 6x + 63 = 0 in a + bi form

Mathematics
1 answer:
creativ13 [48]4 years ago
5 0
The square root of a negative number does not exist in the set of real numbers 
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Simplify (3x)-1. A. 81x^4. B. 4x^6. C. 1/3x. D. 1/x^6
elena55 [62]
If you mean: (3x)^-1

When you have a power to a negative number (1 in this case), place the power on the numerator, with the base in the denominator.

(3x)^-1 = 1/3x

1/3x, or C) is your answer


hope this helps
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