Answer: yes, it is
Step-by-step explanation:
A number is divisible by 6 if it is divisible by 2 and 3 simultaniously.
n = k(k+1)(k-1)
If k-1 is a multiple of 3, n is divisible by 3, so one of the requirements is ok.
Now, if k-1 is a multiple of 3, it can be even or odd.
if k-1 is even, then it is divisible by 2 and as it is divisible by 3 as well, n is divisible by 6
if k-1 is odd, then k and k+1 is even, hence, divisible by 2.
As n = k(k+1)(k-1), n is also divisible by 6.
Answer:
A graph shows zeros to be ±3. Factoring those out leaves the quadratic
(x-2)² +1
which has complex roots 2±i.
The function has roots -3, 3, 2-i, 2+i.
Step-by-step explanation:
<span> 1.04 x 10^2 in standard form = 104
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