Arectangular garden measures 16 feet by 12 feet. a path of uniform width is to be added so as to surround the entire garden. the landscape artist doing the work wants the garden and path to cover an area of 320 square feet. how wide should the path be?

For this problem, please see the attached picture to illustrate the problem. The garden is represented by the inner rectangle having dimensions of 16×12 feet. A path is added that surrounds the garden. The path creates a uniform margin around the garden with equal spacing denoted as x feet. To determine x, we use the total area of the outer rectangle which is equal to 320 square feet.

We know that the area of a rectangle is equal to length times width. The length to be used here is 16 feet subtracted with two x feet of path accounting for the two corners. The same applies for the width. Therefore, the equation for the area is

A = (16 - 2x)(12 - 2x)

Finally, we equate area to 320 to solve for x:

320 = (16 - 2x)(12 - 2x)
320 = 192 - 124x - 32x + 4x²
320 = 4x² - 56x + 192

Dividing the whole equation by 4,
x² - 14x + 48 = 80
x² - 14x + 48 - 80 = 0
x² - 14x - 32 = 0

Factoring the polynomial,
(x-16)(x+2) = 0
x = 16, -2

From the two roots, we choose the positive root as our solution. Therefore, the path should be 16 ft.

The answer would he -4,-2 because the points are just on the opposite side. since the x axis is +4 it's going to he -4 and since the y axis is -2 it stays there because whaen rotated it stays in the third quadrant