What are three consecutive integers whose product is 480 more than their sum?

Mathematics

1 answer:

Natali [406]3 years ago

4 0

Answer:

The three numbers are 7 8 and 9

Step-by-step explanation:

Givens

Let the first number be n - 1
Let the second number be n
Let the third number = n + 1

Equation

(n - 1)(n)(n + 1) - (n-1 + n + n+1) = 480

Solution

Multiply (n - 1) and (n + 1) = (n - 1)*(n + 1) = n^2 - 1

Multiply the second integer by the result of the first and third: n (n^2 - 1)

Add the three integers together: (x - 1) + (n - 1) + n = 3n Combine these 2 steps

n(n^2 - 1) - 3n = 480 Remove the brackets

n^3 - n - 3n = 480

n^3 - 4n = 480

n^3 - 4n - 480 = 0

Graph

The graph shows that the intercept point is n =8. This is the only way I can see to solve this cubic. There are no other real roots.

Answer

n - 1 = 7

n = 8

n + 1 = 9

Check

Product 7*8*9 = 504

Sum = 7 + 8 + 9 = 24

504 - 24 = 480 Which checks.

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Which graph shows the solution to the system of linear inequalities?<br> y< 2x-5<br> y>-3x + 1

trapecia [35]

Answer:

See explanation

Step-by-step explanation:

Plot the solution sets to both inequalities.

1. For the inequality $y$ First, plot the dotted line $y=2x+5$ (dotted because sign is without notion "or equal to"), then choose correct part by substitution coordinates of the origin.

$2\cdot 0-5=-5$

so the origin does not belong to the needed part. Shade the part, which does not include origin.

2. For the inequality $y>-3x+1.$ First, plot the dotted line $y=-3x+1$ (dotted because sign is without notion "or equal to"), then choose correct part by substitution coordinates of the origin.