√(-10 - 5n2 - 330) = i√(5n2 + 340) = i√[(5(n2 + 68)]
5n2 + 17n + 6 is a quadratic expression. It can be factorised as 5n2 + 17n + 6 = (5n + 2)(n + 3). The expression can be expressed as equal to to a fixed or variable amount when it becomes a function of n. Example : 5n2 + 17n + 6 = 7 or y = 5n2 + 17n + 6
5 * * * * * No. It should be 5n2
(5n + 1)(n + 7)
-4n3 + 8n2 - 4n + 7
√(-10 - 5n2 - 330) = i√(5n2 + 340) = i√[(5(n2 + 68)]
5n2 + 17n + 6 is a quadratic expression. It can be factorised as 5n2 + 17n + 6 = (5n + 2)(n + 3). The expression can be expressed as equal to to a fixed or variable amount when it becomes a function of n. Example : 5n2 + 17n + 6 = 7 or y = 5n2 + 17n + 6
5 * * * * * No. It should be 5n2
(5n + 1)(n + 7)
5(n2 + 2n + 4)
-4n3 + 8n2 - 4n + 7
86. Generated by the cubic t(n) = n3 - 5n2 + 9n - 4 for n = 1, 2, 3, ...
Take 5 out. If the missing signs are pluses, it becomes 5(n2 + 2n + 4) If the missing signs are minuses, it becomes 5(n2 - 2n - 4)
The reaction equation for the oxidation of acetylene (C2H2) with nitrous oxide (N2O) is: 2C2H2 + N2O -> 2CO2 + H2O + N2
There are infinitely many rules that can generate this sequence. As imple one is Un = 5n2 - 21n +72 for n = 1, 2, 3, ... And then n = 4 gives U4 = 68
There are infinitely many possible options. It could, for example, be n to -0.1n5 + n4 - 3.5n3 + 5n2 + 0.6n + 3 for n = 0, 1, 2, ... The simplest, though is n to 3(n + 1)
5n2 + 2n + 6 = 0 is a quadratic equation in the variable n.The equation does not have any real roots. The roots are the complex conjugate pair -0.2 ± 1.077i where i is the imaginary square root of -1.