Answer:
There are a total of 2011 integer divisors.
Step-by-step explanation:
The only primes p such that 1/p has finite spaces after the coma are 2 and 5. If we divide a number with last digit odd we will obtain 1 extra digit after the decimal point and if we divide a number by 5 we will obtain 1 more digit if that number has a last digit in the decimal which is not a multiple of 5.
If we take powers of those primes we will obtian one more digit each time. In order to obtain more digits it is convinient to divide by a power of 2 instead of a power of 5, because the resulting number will be smaller.
If we want 2010 digits after the decimal point, we need to divide 1 by 2 a total of 2010 times, hence f(2010) = 2²⁰¹⁰, which has as positive integer divisors every power of 2 between 0 and 2010, hence there are a total of 2011 integer divisors of f(2010).
The answer is the last one
15 + (3/2)t ≤ 12
Y = 5x + 2
3x = y + 10
3x = y + 10
3x = 5x + 2 + 10
3x = 5x + 12
3x - 5x = 12
-2x = 12
x = -6
substitute x = -6 into y = 5x + 2
y = 5(-6) + 2
y = -30 + 2
y = -28
x = -6, y = -28
hope this helped, God bless!
