prove that the square of any whole number is 1 greater than the product of the preceding and following whole numbers.

Let x be the whole number. x^2 is the square of the number. x-1 is the preceding number. And x+1 is the following number.

(x+1)(x-1)=

x^2 + x - x - 1

x^2 -1

Your answer is x^2-1 which is one less that x^2. This also means that the square of any whole number (x^2) is one greater than the product of the preceding (x-1) and the following (x+1).

=====================================

explanation

the graph has a slope of 1/3 because each time you move up 1, you go to the right 3. so, slope = rise/run = 1/3. we can cross choice a off the list since we want a slope of 3, rather than 1/3.

the first table shown has a slope of -3. this is because the rise is -6 and the run is 2, so slope = rise/run = -6/2 = -3. note how x increases by 2 each time (run) and y drops by 6 each time (rise). choice b can be eliminated.

the second table has a slope of 3. each time x goes up by 1, the y values increase by 3. slope = rise/run = 3/1 = 3. alternatively you can use the slope formula to prove that the slope is 3 here. use any two rows of values to generate the two points. choice c is one of the answers

the third table also has a slope of 3. let's use the slope formula to prove as such. i'm going to use the first two rows of the table. so i'll use the two points (2,1) and (5,10)

m = slope

m = (y2-y1)/(x2-x1)

m = (10-1)/(5-2)

m = 9/3

m = 3 so we get a slope of 3

choice d is the other answer