Find three consecutive positive integers such that the square of the first increased by twice the second is three less than four times the third.

  3, 4, 5

Step-by-step explanation:

I like to let x represent the middle integer. Then the first is x-1 and the third is x+1. The given relation is ...

  (x-1)^2 +2x = 4(x+1) -3

We can put this equation into standard form and solve.

  x^2 -2x +1 +2x = 4x +4 -3

  x^2 -4x = 0 . . . . . subtract 4x+1

  x(x -4) = 0

So, x=0 is a solution to the equation, but an extraneous solution with respect to the problem.

x = 4 is also a solution to the equation, indicating the integers are 3, 4, 5.

Check

3^2 +2·4 = 4·5 -3

17 = 17 . . . . answer checks OK

d

step-by-step explanation:

hello and welcome!

so, when we see the words "proportional" along with a table, we need to know that we're looking numbers that divide by the same thing.

looking at this table, we simply have to do a trial and error, and divide to see which numbers have the same quotient. if you look at choice a & c, they both have equal quotients for three of the numbers, with one exception, so those can't be the answer. looking at b, there is no similarity, so that can't be it. then, we see d; if you divide all the numbers in the table, you'll see that they all divide by two, which is a proportional relationship, and our answer!

also, always remember that the x will always be the number that goes into y, and somewhere behind this table, is a formula. for this specific table, the formula would be:

since our x-value is being divided by two to get the y-value. and if they were to ask for more numbers in the series, this is the formula you would use to find them.