Find three positive consecutive integers such that the product of the smallest and largest is 17 more than three times the median integer

explain.

  Three consecutive positive integers following the given condition will be 5, 6 and 7.

  Let three consecutive integers are x, (x + 1), (x + 2).

Given in the question → "product of the smallest and the largest is 17 more than 3 times the median integer".

Median number amongst these integers → x

Therefore, expression for the statement will be,

x(x + 2) = 3(x + 1) + 17

x² + 2x = 3x + 3 + 17

x² + 2x - 3x - 20 = 0

x² - x - 20 = 0

x² - 5x + 4x - 20 = 0

x(x - 5) + 4(x - 5) = 0

(x + 4)(x - 5) = 0

x = -4, 5

Therefore, consecutive integers will be

x = 5

(x + 1) = 6

(x + 2) = 7

         Hence, three positive consecutive integers will be 5, 6 and 7.

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Explanation:

Let A,B,C be the three positive consecutive integers such that A is the smallest, B is in the middle (aka median), and C is the largest.

If A = x, then B = x+1 and C = x+2. The x is some positive whole number. Eg: x = 2 leads to A = 2, B = 3, C = 4

"The product of the smallest and largest is 17 more than three times the median integer" means

(smallest)*(largest) = 3*(median)+17

A*C = 3*B+17

x(x+2) = 3(x+1)+17

Let's solve for x.

x(x+2) = 3(x+1)+17

x^2+2x = 3x+3+17

x^2+2x = 3x+20

x^2+2x-3x-20 = 0

x^2-x-20 = 0

(x+4)(x-5) = 0

x+4 = 0 or x-5 = 0

x = -4 or x = 5

We made x be positive, so ignore x = -4

The only practical solution here is x = 5

A = x = 5

B = x+1 = 5+1 = 6

C = x+2 = 5+2 = 7

The three consecutive integers are: 5, 6, 7

Check:

A*C = 3*B + 17

5*7 = 3*6 + 17

35 = 18 + 17

35 = 35

Answer is confirmed.

1,589.63 in.^3

step-by-step explanation:

i hope this you : d

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step-by-step explanation: