MONKEY=YEKNOM then each letter is moved backward =XDJMNL so for TIGER=REGIT the each of the reversed letters are moved backward=QDFHS
Ans-44
4x+8=6Y
5X=5Y
X=Y
FROM EQ 1
2X=8
X=4
6Y+5Y=11*4=44
3/7
(x+2)^2 -x^2 = 84
X=20
So (20,22)
Sum= 42
BBMEHE
length of the base=3 times height
b=3h;
h=b/3;
so, 1/2bh=24
1/2b(b/3)=24
(b^2)/6=24
b^2=144
b=12
16
1+1=2
2+2=4
3+4-7
4+7=11
5+11=16
Simple interest Formula:
A=P(1+rt)
Therefore,
815=P(1+3r)
P+3r=815
854=P(1+4r)
P+4r=854
Solve the sums using Elimination method.
P+4r=854
P+3r=815
r=39.
Toget Principal
P+4r=854
P+4*39=854
P=854-117
P=698
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
I agree with @TUMWINE PETER
F+V-E =2;
F= faces;V= vertices;E = number of edges