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50 has no zeros. It's equal to 50 under all conditions.
x3 + x2 - 17x + 15 = (x - 1)(x - 3)(x + 5). Thus, the zeros are 1, 3, and -5. All three zeros are rational.
If leading zeros are allowed, then there are ten thousand possible: 0000 through 9999. If leading zeros are not allowed, then it is 1000 through 9999, eliminating 0000-0999, and now there are nine thousand possibilities.
No. All whole numbers are integers.
Not necessarily. The denominator need not have any real zeros, for example x2+1. Not necessarily. The denominator need not have any real zeros, for example x2+1. Not necessarily. The denominator need not have any real zeros, for example x2+1. Not necessarily. The denominator need not have any real zeros, for example x2+1.