(n + 6)(n - 4) = 0
n = -6 or n = 4
Solve by factoring. n2 + 2n – 24 = 0 (1 point)
–12, 2 –2, 12 –6, 4 –4, 6
3 answers
One more rectangular-shaped piece of metal siding needs to be cut to cover the exterior of a pole barn. The area of the piece is 30 ft2. The length is 1 less than 3 times the width. How wide should the metal piece be? Round to the nearest hundredth of a foot.
(1 point)
3.33 ft 4.3 ft 1 ft
30 ft
(1 point)
3.33 ft 4.3 ft 1 ft
30 ft
Let's define the width of the metal piece as x ft.
Given that the length is 1 less than 3 times the width, we can represent the length as 3x - 1 ft.
The area of a rectangular piece of metal siding is given by the formula: Area = Length * Width
Therefore, in this case, 30 = (3x - 1) * x
Expanding the equation:
30 = 3x^2 - x
Rearranging the equation:
3x^2 - x - 30 = 0
Let's solve this quadratic equation by factoring:
3x^2 - 10x + 3x - 30 = 0
x(3x - 10) + 3(3x - 10) = 0
(x + 3)(3x - 10) = 0
Setting each factor to zero:
x + 3 = 0 or 3x - 10 = 0
Solving for x:
x = -3 or 3x = 10 => x = 10/3 = 3.33 ft (rounding to the nearest hundredth)
Therefore, the width should be approximately 3.33 ft.
Given that the length is 1 less than 3 times the width, we can represent the length as 3x - 1 ft.
The area of a rectangular piece of metal siding is given by the formula: Area = Length * Width
Therefore, in this case, 30 = (3x - 1) * x
Expanding the equation:
30 = 3x^2 - x
Rearranging the equation:
3x^2 - x - 30 = 0
Let's solve this quadratic equation by factoring:
3x^2 - 10x + 3x - 30 = 0
x(3x - 10) + 3(3x - 10) = 0
(x + 3)(3x - 10) = 0
Setting each factor to zero:
x + 3 = 0 or 3x - 10 = 0
Solving for x:
x = -3 or 3x = 10 => x = 10/3 = 3.33 ft (rounding to the nearest hundredth)
Therefore, the width should be approximately 3.33 ft.