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How do you make 55 with numbers 1 2 3 4?
2*(4! + 3) + 1 = 2*(24 + 3) + 1 = 2*27+1 = 54+1 = 55
Find the rational zeros of the equation 2x4 - x3 - 21x2 plus 9x plus 27 equals 0?
2x4 - x3 - 21x2 + 9x + 27 = 0 Let f(x) = 2x4 - x3 - 21x2 + 9x + 27. All possible rational zeros = (factors of the constant term, 27)/(factors of the leading coefficient, 2) = ±1, ±3, ±9, ±27)/(±1, ±2) = ±1, ±3, ±9, ±27, ±1/2, ±3/2, ±9/2, ±27/2. Using synthetic division, test all possible rational zeros: -1] 2 -1 -21 9 27 (2)(-1) = -2; -1 + -2 = -3 (-3)(-1) = 3; -21 + 3 = -18 (-18)(-1) = 18; 9 + 18 = 27 (27)(-1) = -27; 27 + -27 = 0 (the zero remainder shows that -1 is a zero) -3] 2 -1 -21 9 27 (2)(-3) = -6; -1 + -6 = -7 (-7)(-3) = 21; -21 + 21 = 0 (0)(-3) = 0; 9 + 0 = 9 (9)(-3) = -27; 27 + -27 = 0 (the zero remainder shows that -3 is a zero) 3/2] 2 -1 -21 9 27 (2)(3/2) = 3; -1 + 3 = 2 (2)(3/2) = 3; -21 + 3 = -18 (-18)(3/2) = -27; 9 + -27 = -18 (-18)(3/2) = -27; 27 + -27 = 0 (the zero remainder shows that 3/2 is a zero) 3] 2 -1 -21 9 27 (2)(3) = 6; -1 + 6 = 5 (5)(3) = 15; -21 + 15 = -6 (-6)(3) = -18; 9 + -18 = -9 (-9)(3) = -27; 27 + -27 = 0 (the zero remainder shows that 3 is a zero) Thus, the rational zeros are -3, -1, 3/2, and 3.
How do you write .27 as a fraction?
27 = 27/1 = (2*27)/2 = (3*27)/3 = ...
3 thirds divided by 27?
0.1111
What sum do you do to make 27 out of 1234?
Assuming the order of the numbers is not important, then ((4 x 2) + 1) x 3 = 27.
