Um, is there angles to look at, or an actual problem?
False. It’s should be 3/4
For this problem, all you need to do is find the three #'s that add up to 156.
So, lets look at the answers and add them up.
A. 50, 52, 54
50 + 52 + 54 = 156
B. 51,52,53
51 + 52 + 53 = 156
C. 49,50,51
49 + 50 + 51 = 150
D. 49,51,53
49 + 51 + 53 = 153
We get the answers (50,52,54) and (51,52,53)
Now, consecutive numbers are numbers that in order, like 1,2,3.
Therefore, the answer is (51,52,53)
If one of the numbers we multiply (factors) has zeros at the end, and the other isn't a fraction: all those zeros will stay in the product.
But there might be additional zeros if the other numbers in the factors (the numbers which aren't 0) mupliply to "end" in zero and this is the case here:
8*5=40.
so the product will be 40 and the zeros of the 5000:
40 000
the number of zeros in the product will be bigger than the number of zeros in the factors if the non-zero parts of the fractions multiply to a number with 0 at the end.
Answer:
The zeros are x=0,3,-2
There is a multiplicity of 1 for all of them.
Step-by-step explanation:
