The sum of the first ( n ) positive integers can be calculated using the formula ( \frac{n(n + 1)}{2} ). For the first 80 positive integers, this becomes ( \frac{80(80 + 1)}{2} = \frac{80 \times 81}{2} = 3240 ). Therefore, the sum of the first 80 positive integers is 3240.
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.
To find the sum of 54, 26, and 19, you can group the addends strategically. For example, you can first add 54 and 26 to get 80, and then add 19 to that sum: 80 + 19 equals 99. Therefore, the total sum is 99.
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
They are positive integers.
Ten of them.
At most -80 (negative 80).
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.
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