5, 16, 6, 16, 7, 16, 9
d
NTFS — New Technology File System
Fat — file allocation table
NTFS having a quota, commpress system with securites base as
administrator giving, multipal user, groups, to set permission
11
Perimeter of semi circular = πr+2r
Where r is radius , π = 22/7
πr+2r= 144 (given)
(22/7)r +2r = 144
22r+14r = 144*7 (multiply both sides by 7)
36r = 144*7
r= 144*7/36 = 28
radius =28 cm
Area =( 1/2)πr^2
Area=( 1/2 )*(22/7)*28*28
= 1232cm^2
Ans : area is 1232 cm^2
Profit ratio will be : 4:5:6
5x=150
X=300
X share is 1200
Y share is 1800
Difference will be = x-y =600
16, 25, 36, 72, 144, 196, 225
72
Because it is not a square number
4 days
The work done by A in 8 days is = 8/ 12 = 2/3
Means A alone completes 2/3 part of work.
Remaining work which is (1–2/3) = 1/3 is completed by B in 8–2 = 6 days
So the complete work done by B in 6/(1/3) = 18 days.
B alone can complete the work in 18 days.
45 km/hr
B. 70%
400%
e.g:-
l = 5 b = 2
Area= l*b =10
New after 100% increament
l=10 b = 4
Area = 10*4
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
12, 10