It looks as if each number is equal to the previous number, multiplied by -1/2. Therefore, one way to write the formula for the n'th term is:-24 x (-1/2)^n
Per the formula in the related link, substitution into the formula yields (3*12)1/2, which is (36)1/2 or 6.
The quadratic formula is.... -b +/- sqrt(b^2 -4ac)/2a in this problem.... ( 3X^2 + 6X + 2 = 0 ) a = 3 b = 6 c = 2 so..... -6 +/- sqrt(6^2 - 4(3)(2))/2(3) -6 +/- sqrt(36 -24)/6 -6 +/- sqrt(12)/6 since square root 12 is irrational -1 +/- sqrt(12) is the answer
The sequence 0, 3, 6, 9, 12 is an arithmetic sequence where the first term is 0 and the common difference is 3. The formula for the nth term can be expressed as ( a_n = 3(n - 1) ) or simply ( a_n = 3n - 3 ). This formula generates the nth term by multiplying the term's position (n) by 3 and adjusting for the starting point of the sequence.
12 + (2 x 3) - 7 = 12 + 6 - 7 = 11
x=(6±sqrt(36-4*6))/2 x=3±(sqrt(12)/2) x=3±sqrt(3)
The answer depends on what the explicit rule is!
Per the formula in the related link, substitution into the formula yields (3*12)1/2, which is (36)1/2 or 6.
Each term is 3 times greater than the previous term and so the next term will be 486
Good Question! After 6 years of math classes in college, and 30+ years of teaching (during which I took many summer classes) I've never seen an explicit formula for the nth term of the Fibonacci sequence. Study more math and maybe you can discover the explicit formula that you want.
12 is the LCM of any of the following 44 sets:{12},{1, 12}, {2, 12}, {3, 4}, {3, 12}, {4, 6} , {4, 12}, {6, 12},{1, 2, 12}, {1, 3, 4}, {1, 3, 12}, {1, 4, 6}, {1, 4, 12},{1, 6, 12}, {2, 3, 4}, {2, 3, 12}, {2, 4, 6}, {2, 4, 12},{2, 6, 12}, {3, 4, 6}, {3, 4, 12}, {3, 6, 12}, {4, 6, 12},{1, 2, 3, 12}, {1, 2, 4, 12}, {1, 2, 6, 12}, {1, 3, 4, 12},{1, 3, 6, 12}, {1, 4, 6, 12}, {1, 2, 4, 6}, {1, 3, 4, 6},{1, 2, 3, 4}, {2, 3, 4, 6}, {2, 3, 4, 12}, {2, 3, 6, 12},{2, 4, 6, 12}, {3, 4, 6, 12},{1, 2, 3, 4, 6}, {1, 2, 3, 4, 12}, {1, 2, 3, 6, 12},{1, 2, 4, 6, 12}, {1, 3, 4, 6, 12}, {2, 3, 4, 6, 12},{1, 2, 3, 4, 6, 12}.
The quadratic formula is.... -b +/- sqrt(b^2 -4ac)/2a in this problem.... ( 3X^2 + 6X + 2 = 0 ) a = 3 b = 6 c = 2 so..... -6 +/- sqrt(6^2 - 4(3)(2))/2(3) -6 +/- sqrt(36 -24)/6 -6 +/- sqrt(12)/6 since square root 12 is irrational -1 +/- sqrt(12) is the answer
There are 12 hydrogen atoms is C6H12O6 (glucose).
The sequence 0, 3, 6, 9, 12 is an arithmetic sequence where the first term is 0 and the common difference is 3. The formula for the nth term can be expressed as ( a_n = 3(n - 1) ) or simply ( a_n = 3n - 3 ). This formula generates the nth term by multiplying the term's position (n) by 3 and adjusting for the starting point of the sequence.
12 + (2 x 3) - 7 = 12 + 6 - 7 = 11
(6/3)/12 = 2/12 = 1/6 or 6/(3/12) = 6/(1/4) = 24
The LCM of 3, 4, 6, and 12 is 12
x=(6±sqrt(36-4*6))/2 x=3±(sqrt(12)/2) x=3±sqrt(3)