Checkstate 0 1s0 S1 S2S1 S1 S2S2 S2 S1 final
t1 s1 b1 t1 s1 b2 t1 s2 b1 t1 s2 b2 t2 s1 b1 t2 s1 b2 t2 s2 b1 t2 s2 b2 t3 s1 b1 t3 s1 b2 t3 s2 b1 t3 s2 b2 TOTAL 12 combinations OR 2(3 x 2) = 12
let s1,s2,s3 be three sides of a triangle.import java.lang.*;import java.io.*;import java.util.*;class Triangle{public static void main(String args[]){boolean test=false;int s1,s2,s3;Scanner input = new Scanner(System.in);System.out.println("enter the side1 of triangle");s1=input.nextInt();System.out.println("enter the side2 of triangle");s2=input.nextInt();System.out.println("enter the side3 of triangle");s3=input.nextInt();if((s1*s1)==(s2*s2)+(s3*s3)){test=true;}else if((s2*s2)==(s1*s1)+(s3*s3)){test=true;}else if((s3*s3)==(s1*s1)+(s2*s2)){test=true;}if(test==true)System.out.println("Entered sides form a right angle triangle.....");elseSystem.out.println("Entered sides dosn't form a right angle triangle.....");}}
The sum of an arithmetic series of n terms is: S(n) = n(2a1 + (n - 1)d) / 2 where a1 is the first term and d is the difference between terms. The key to solving this is to realise that two sums are required, S1 and S2 such that: S1 = S(r) S2 = S(3r) - S(r). But that this second sum can also be stated as: S2 = sum of 2r terms starting at term ar+1 So that the problem then becomes: Find two sums S1 and S2 of the sequence an = 2n + 3 such that the first (S1) has r terms starting at a1 and the second (S2) has 2r terms starting at ar+1, with the ratio of S1 : S2 = 7 : 44. This gives: S1 = r (2a1 + (r - 1)d) / 2 = r(2(2x1 + 3) + (r - 1)2) / 2 = r(2r + 8) / 2 = r(r + 4) S2 = 2r(2ar+1 + (2r - 1)d) = 2r(2(2(r + 1) + 3) + (2r - 1)2) / 2 = 2r(2(2r + 5) + (2r - 1)2) / 2 = 2r(2r + 5 +2r - 1) = 2r(4r + 4) = 8r(r + 1) As S1 : S2 = 7 : 44, 44 S1 = 7 S2 => 44 r(r + 4) = 7 x 8r(r + 1))0 => 11(r + 4) = 14(r + 1) => 11r + 44 = 14r + 14 => 3r = 30 => r = 10 As r is the number of terms in S1 it is a third of the number of terms required, so 3 x 10 = 30 terms are needed. To check: a1 = 2 x 1 + 3 = 5 S(30) = 30(2 x 5 +(30-1) x 2) / 2 = 30 x 34 = 1020 S(10) = 10(2 x 5 + (10-1) x 2) / 2 = 10 x 14 = 140 Sum of 1st third S1 = S(10) = 140, sum of rest S2 = S(30) - S(10) = 1020 - 140 = 880 Ratio of S1 : S2 = 140 : 880 = 14 : 88 = 7 : 44 Note that the second sum S2 is also 20 terms starting at a11 = 2 x 11 + 3 = 25: S2 = 20(2 x 25 + 19 x 2)/2 = 20 x 44 = 880.
It's going by 5's1
Control and support are the two responsibilities of the S1.
brief soldiers on BN/Unit level SRP and mobilization process
The responsibilities of the S1, according to AR 600-6-101, are support and control. Support involves increasing morale involves awarding honors and helping families. Control is in place to check who goes in and out of the unit, along with tracking casualties.
5th lumber vertebrae and first sacral vertebra.
strlen(s1) to find the length of the string s1 strcpy(s1,s2) copy source string to destination string(i.e copies s2 to s1,s2 remain unchanged) strcmp(s1,s2) compares s1 and s2 and prints 0 if s1 and s2 are equal,-1 if s2 is greater, 1 if s1 is greater strcat(s1,s2) combines string s1 and s2 to a single word and stores it in s1 strupr() converts lower case string to upper case strlwr() converts upper case string to lower case
Statement S2 is anti-dependent on statement S1 if S2 follows S1 in program order and if the output of S2 overlaps the input of S1. The anti-dependence S1 to S2 define as cross arrow such as S1 |-> S2.
S1 and S2, representing the closure of the valves of the atria, and then the ventricles, respectively.
Hard drive and monitor.
s1 : Continiuos running
S1
what is the treatment for a desiccated L5-S1
Definition: A set S1 is a superset of another set S2 if every element in S2 is in S1. S1 may have elements which are not in S2.