1 and 20.
the 2 numbers are 51 and 19
72 plus 19 = 91
Name the two numbers x & y , you can form these equations:x + y = 21x - y = 19Solve the system by substitution:x - y = 19 --> x = 19 + y19 + y + y = 2119 + 2y = 212y = 2y = 1But, X + y = 21X + 1 = 21x = 20
The numbers are: (19+square root of 131) and (19-square root of 131)
1 and 20.
the 2 numbers are 51 and 19
16 and 35
Find 3 consecutive numbers where the product of the smaller two numbers is 19 less than the square of the largest number.
19 and -7
Surely the question is wasn't meant to say what it literally said, but a solution to its literal statementis still possible.The two consecutive numbers are: 18 and 19.The smaller two numbers are: 18 and 19.Their product is 342 .The largest number is 19.Its square is 361.342 is 19 less than 361.
20 and 1. 20 + 1 = 21 20 - 1 = 19
72 plus 19 = 91
4x4=16, and 5x5=25, so 4 and 5 are the whole numbers closest to the square root of 19. [And incidentally of all the square roots of all the whole numbers between 16 and 25.]
The sum of two numbers witha difference of two equals forty can be set up and solved this way: x + (x - 2) = 40 2x - 2 = 40 2x = 42 x = 21 x - 2 = 19 The two numbers are 21 and 19.
How about: 14 and 11
20 and 1 Seems to be a common problem. Add the sum to the difference and divide by 2 gives one number. Then start over and subtract the smaller of the two givens from the larger and then divide by 2 to get the other number. So here you have (21+19)/2 and (21-19)/2