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First Pass: (5 1 4 2 8 ) –> (1 5 4 2 8 ), Here, algorithm compares the first two elements, and swaps since 5 > 1. ( 15 4 2 8 ) –> ( 14 5 2 8 ), Swap since 5 > 4 ( 1 45 2 8 ) –> ( 1 42 5 8 ), Swap since 5 > 2 ( 1 4 25 8 ) –> ( 1 4 25 8 ), Now, since these elements are already in order (8 > 5), algorithm does not swap them. Second Pass: (1 4 2 5 8 ) –> (1 4 2 5 8 ) ( 14 2 5 8 ) –> ( 12 4 5 8 ), Swap since 4 > 2 ( 1 24 5 8 ) –> ( 1 24 5 8 ) ( 1 2 45 8 ) –> ( 1 2 45 8 ) Now, the array is already sorted, but our algorithm does not know if it is completed. The algorithm needs onewhole pass withoutany swap to know it is sorted. Third Pass: (1 2 4 5 8 ) –> (1 2 4 5 8 ) ( 12 4 5 8 ) –> ( 12 4 5 8 ) ( 1 24 5 8 ) –> ( 1 24 5 8 ) ( 1 2 45 8 ) –> ( 1 2 45 8 )
I. √(74x-250 )– x=15
II. √(3y²-37y+18)+ 2y=18
I. 2y2 + 11y + 15 = 0
II. 3x2 + 4x - 4= 0
I. 8x² + 2x – 3 = 0
II. 6y² + 11y + 4 = 0
I. 8y2- 2y - 21 = 0
II. 2x2+ x - 6 = 0
LCM of 'x' and 'y' is 30 and their HCF is 1 such that {10 > x > y > 1}.
I. 2p²- (x + y) p + 3y = 0
II. 2q² + (9x + 2) = (3x + y) q
I. 40 x² - 93 x + 54 = 0
II. 30 y² - 61 y + 30 = 0
I. 2x² - 7x + 3 = 0
II. 8y² - 14y + 5 = 0
I. 10x2 + 33x + 9 = 0
II. 2y2 + 13y + 21 = 0
I. 3x2 – 16x + 21 = 0
II. y2 – 13y + 42 = 0