One thousand seventy two point thirty nine thousandths.
So the given series is "16, 06, 68, 88, __"
Count all the cyclical opening in each of these numbers. For example in 16, there is a one cyclical loop present in it(the one in 6), similarly in 06 it is two(one in zero and one in 6), going ahead, in 68 it is 3(one in 6 and two in 8).
From here on things become simple: hence, the cyclical figures in these equations written down becomes 1,2,3,4,_,3.
Let's now try solving the above sequence, going by the logical reasoning the only number that can fill in the gap should be 4.
Answer:
x²
Step-by-step explanation:
Given the values x² and x^9
The greatest common factor is the factors common to two or more compared values :
Factors of :
x² = x * x
x^9 = x * x * x * x * x * x * x * x * x
Multiplying the Factors in both are : x * x = x²
Similarly :
___|x² | x^9
_ x | x | x^8
_ x | 1 | x^7
There is no factor which can reduce both further simultaneously, Hence the G. C. F = (x * x) = x²
Solution:
We have to find the remainder when f(x) is divided by 
x²-1=0
x=
So, remainder is 13 and -13.
