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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 = ...
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.
3 thirds divided by 27?
0.1111
How do you make 27 using 1 2 3 and 4 once?
3*(2*4 + 1) = 3*(8 + 1) = 3*9 = 27
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 many 3-digits number can you make from the digits 1 2 and 3?
33 = 27 with repetition, 3! = 3*2*1 = 6 without repetition.
Can 3 3 2 2 1 to equal 26?
(3 cubed +1 squared) - 2 = 26 also written as (3^3 + 1^2) - 2 = 26 or (27+1)-2 = 26 so 3,3,2,2 and 1 can make 26
What is the LCM of 32 54 80?
32, 54, 80 / 2 = 16, 27, 40 16, 27, 40 / 2 = 8, 27, 20 8, 27, 20 / 2 = 4, 27, 10 4, 27, 10 / 2 = 2, 27, 5 2, 27, 5 / 2 = 1, 27, 5 1, 27, 5 / 3 = 1, 9, 5 1, 9, 5 / 3 = 1, 3, 5 1, 3, 5 / 3 = 1, 1, 5 1, 1, 5 / 5 = 1, 1, 1 SCM = 2^5 x 3^3 x 5 = 4,320
Factor y6 27?
(y^2 + 3)(y^2 - 3y + 3)(y^2 + 3y + 3)
How do you write .27 as a fraction?
27 = 27/1 = (2*27)/2 = (3*27)/3 = ...
What are the multiplication factors of 27?
The divisors of 27 are: 1, 3, 9, 27.
What is one ninth of 27?
1/9 x 27/1 = 27/9 = 3
What are the factors of 54?
The factors of 54 are 1, 2, 3, 6, 9, 18, 27, and 54. The factor pairs of 54 are 1 x 54, 2 x 27, 3 x 18, and 6 x 9The factors of 54 are: 1 2 3 6 9 18 27 54.1, 2, 3, 6, 9, 18, 27, 54The factors of 54 are: 1, 2, 3, 6, 9, 18, 27, & 541, 2, 3, 6, 54, 27, 18, 9
What are the factors of 18 27 and 63?
The factors of 18 are 1, 2, 3, 6, 9, and 18 The factors of 27 are 1, 3, 9, and 27
