1 : 2
speed of the train respect to man
= (63 – 3) km/hr
= 60 km/hr
= 60 * 1000 / 3600 m/sec
= 50/3 m/sec
time
= distance/speed
= 500 * 3/ 50
= 30 sec
A Roman was born the first day of the 35th year before Christ means before the start of year 0, 35 years has been past and died the first day of the 35th year after Christ means 35th year is about to start at his death so only 34 years has been completed so total years = 69
C.1175
1km is equal to 1000 meter
Therefore 225 meter is how many km
Then we have to cross multiply
1km – 1000 meter
? – 225 meter
0.225 km
Then we have to apply formula
Speed = Distance/Time
Speed = 0.225/12 sec
= 0.01875×3600sec
= 67.5
Therefore speed of train is = 67.5
2:1
Error is at 48. Correct once is 58
To determine how many consecutive zeros the product of S will end with, we need to find the highest power of 10 that divides the product. This is equivalent to finding the highest power of 5 that divides the product, since the number of factors of 2 will always be greater than the number of factors of 5.
The primes in S are {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}.
There are 24 primes in S, so the product of S is:
2 x 3 x 5 x 7 x 11 x 13 x 17 x 19 x 23 x 29 x 31 x 37 x 41 x 43 x 47 x 53 x 59 x 61 x 67 x 71 x 73 x 79 x 83 x 89 x 97
We need to find the highest power of 5 that divides this product. To do this, we count the number of factors of 5 in the prime factorization of each number in S.
5 appears once: 5
5 appears once: 25
5 appears once: 35
5 appears once: 55
5 appears once: 65
5 appears once: 85
So, there are six factors of 5 in the product of S. However, we also need to consider the powers of 5 that arise from the factors 25, 35, 55, and 65.
25 = 5 x 5 appears once: 25
35 = 5 x 7 appears once: 35
55 = 5 x 11 appears once: 55
65 = 5 x 13 appears once: 65
Each of these numbers contributes an additional factor of 5 to the product of S. Therefore, there are 6 + 4 = 10 factors of 5 in the product of S.
Since each factor of 5 corresponds to a factor of 10, we know that the product of S will end with 10 zeros. Therefore, the product of S will end with 10 consecutive zeros
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