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 value of f(-3) is 5.
In order to find this answer, we are going to input -3 in for x (this is what the notation f(-3) means).
f(x)=| x-2 |
f(-3)=| -3 -2 |
f(-3)=| -5 |
Now we have to take the absolute value.
f(-3) = 5
Answer: 11/15
Step-by-step explanation:
Answer: pt 2 HOPE THIS HELPS!
