Let one positive integer be x, so the other is x+3. The sum of the squares of the two integers is 89, so we have:x2 + (x+3)2 = 89x2 + (x2+6x+9) = 892x2 + 6x + 9 = 892x2 + 6x +9 - 89 = 02x2 + 6x - 80 = 0x2 + 3x - 40 = 0(X+8)(x-5)=0x = 5 (the only positive integer)The 2 positive integers are 5 & 8. Squaring both would be 25 & 64, which the sum is 89.
The numbers are 81, 82 and 83. Also, 80, 82 and 84
80... since -65 is a negative number, adding positive 80 will produce a value of positive 15.
40 * * * * * The correct answer is 6400.
200
17 , 19 , 21 and 23 are the odd integers whose sum is 80 and the least integer is 17
The product of the two integers is -80.
The integers are 17, 19, 21 and 23.
They are 79+80 = 159
Ten of them.
They are positive integers.
Let one positive integer be x, so the other is x+3. The sum of the squares of the two integers is 89, so we have:x2 + (x+3)2 = 89x2 + (x2+6x+9) = 892x2 + 6x + 9 = 892x2 + 6x +9 - 89 = 02x2 + 6x - 80 = 0x2 + 3x - 40 = 0(X+8)(x-5)=0x = 5 (the only positive integer)The 2 positive integers are 5 & 8. Squaring both would be 25 & 64, which the sum is 89.
At most -80 (negative 80).
There are 80 such integers.
The answer is -10 and -8. -8 + -10= -18 -8x-10 =80.
The set is 81, 82 and 83. There is also a set of consecutive even integers whose sum is 246. That set is 80, 82 and 84.
The numbers are 81, 82 and 83. Also, 80, 82 and 84