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Finite State Machine with binary numbers having even parity?
Checkstate 0 1s0 S1 S2S1 S1 S2S2 S2 S1 final
Justin has 3 pairs of trouses 2 shirts and 2 pairs of boots How many different outfits can he wear?
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
Write a program to print a right angle triangle containing digit in java?
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.....");}}
2n plus 3 is an arithmetic sequence how many terms should be taken starting from the first term so that the ratio of the sum of the first third of these terms to the rest of the terms is 7 to 44?
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.
What patterns can you see of the multiples of 5?
It's going by 5's1
