2006-09-15 19:43:04 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
No integer base satisfies the question.
384 base 10 = 3014 base 5
384 base 10 = 1440 base 6
384 base 10 = 1056 base 7
So it would have to be a base between 6 and 7.
x^3 + 2x^2 + 3x + 4 = 384
x^3 + 2x^2 + 3x - 380 = 0
Solving this 3rd order polynomial we get a value of
x = 6.50882
So 384 is equal to 1234 in base 6.50882
2006-09-15 19:54:00 · answer #1 · answered by Puzzling 7 · 1⤊ 0⤋
I234 in base 5 = 194 in base10, base 6 = 310, and base 7 = 466. X should not be a non-integer between 6 and seven. Oh, i'm getting it! Who says it should be an integer? you may remedy the polynomial equation a million(x^3) + 2(x^2) + 3(x) + 4 = 384. it rather is between 6.5 and six.6 all and dissimilar know a thank you to remedy it different than successive approximation?
2016-12-12 09:17:44 · answer #2 · answered by ? 4 · 0⤊ 0⤋
1xN^3 + 2xN^2 + 3xN^1 + 4xN^0 = 384
N^3 + 2N^2 + 3N - 380 = 0
Thus N = 6.5088
Whats the point of this question, anyway?
2006-09-15 19:58:13 · answer #3 · answered by ctyuang 1 · 1⤊ 0⤋
binary system
2006-09-15 19:45:14 · answer #4 · answered by Anonymous · 0⤊ 1⤋