x and x+1 are the numbers
the squares are x^2 and (x^2+2x+1)
We need these to be less and or equal to 25 so we write
2x^2+2x+1< or = to 25
2x^2+2x-24< or equal to 0
Solve 2(x^2+x-12)=0
2(x+4)(x-3)=0
so x=-4 or x=3
Now note that 2x^2+2x-24 is the graph of a parabola opening toward the positive y direction, that is to say, opening upward.
Now thinking of f(x)=2x^2+2x-24 we want the x values such that y is < and = to 0. In fact we want the smallest values.
If x is greater than -4 than y is less than 0, if x is less than 3, y is greater than 0.
So the answer is for the inequality in interval notation is [-4,3]. If we want the smallest integer that will work, it will be -4.
Let's try it
x=-4 and the next consecutive integer is -3.
16+9=25 so that works!
If x is 3 than the integers are 3 and 4 and once again it is 25. For any numbers between -4 and 3, say 0, it clearly works also, but if x is -5 and the other number is -4 then the sum is 25+16 which is greater.
So the smallest integers are -4 and -3.
The smallest is 15.
The average of 33 consecutive whole numbers is 58, what is the smallest of these whole numbers? The answer is 42
The smallest is 14.
1
2520
No, it is impossible.
Consecutive whole numbers are integer pairs of the form n and n+1. There can be no integer, such as 110, between such numbers.
The smallest is 15.
The smallest is 55.
231 / 3 = 77So 76 + 77 +78 = 23177
The average of 33 consecutive whole numbers is 58, what is the smallest of these whole numbers? The answer is 42
The smallest set of such prime numbers is 5, 7, 11, 13, making a sum of 36.
If n is one integer, then the consecutive integer to it is n+1, and the next is n+2 and so on.
The smallest is 14.
The smallest is 55.
The smallest consecutive three numbers that total 120 are... 39, 40 & 41.
The smallest of the three is 13.