Factor the gcf
3x(x^2-8x+12)
Then factor the inner trinomial
3x(x-6)(x-2)
Final answer: 3x(x-6)(x-2)
Complete question is;
How many 10 - digit numbers are there, such that the sum of the digits is divisible by 2?
Answer:
There are 4500000000 ten digit numbers whose sum is divisible by 2.
Step-by-step explanation:
Since we are dealing with 10 digit numbers, the first 10 digit number whose sum is divisible by 2 is; 1000000001.
While the last 10 digit number whose sum is divisible by 2 is 9999999999.
Now,to find how many 10 digit numbers whose sum are divisible by 2, we will simply divide the difference of both numbers by 2 and then add 1 to the answer. This is because subtracting the minimum from the maximum gives us the number of 10 digit numbers we have. Then dividing by 2 gives the number of ten digit numbers with sum divisible by 2 in between. Then adding 1 to the result to cover the number not included gives;
((9999999999 - 1000000001)/2) + 1 = 4500000000
Answer:
13/6 or 2 1/6
Step-by-step explanation:
6 1/2 = 13/2 (Improper Fraction)
1/3 = 2/6 (1/3 is just simplified)
1/3 x 1 + 1 + 4 + 6 + 1 = Answer
(13/2 x 1/3 = 13/6)
Answer:
6
Step-by-step explanation:
your answer is 6 because she did 12 in 1hr and the other 12 in 3hr so you take 3=1+4 and divide 24 by 4 which is 6
