Agravel path is to be built around an 18‘ x 15‘ rectangle garden if the width remains constant how why can the path be if there is enough gravel for 328 ft.²

P = 2*(18 + 13) = 62ft

Now to figure out the width of the path we have to do some geometry. There are 3 basic shapes we will use for the path, they are the 13 ft long rectangle, the 18 ft long rectangle, and the squares formed at the corners. These squares' areas are equal to the width squared, whereas the rectangles are only equal to the length times the width. Our total area then becomes:

A = 2*(13*w) + 2*(18*w) + 4*(w*w)

A = 4w^2 + 62w

And since we know the available area is 516 ft^2, putting this in for A and solving the quadratic equation we can get the width:

516 = 4w^2 + 62w

0 = 4w^2 + 62w - 516

Using the quadratic formula: (w = -b2+/-SQRT(b^2 - 4ac)/2a, where a = 4, b = 62, and c = -516)

w = 6, -21.5

Since we obviously cannot have a negative width, 6 must be our answer. Therefore the path can be 6 feet wide!

We can check this answer by doing the following. Please note that by having a 6ft wide path you would actually be adding 12 ft to both dimensions, since you add 6 feet of length to both sides. Your total area would then become:

18+(2*6) x 13+(2*6) = 30x25 = 750 ft^2

Subtracting the area of the garden gives you the are of the gravel required:

750ft^2 - (18*13)ft^2 = 516ft^2

Therefore a width of 6 feet is correct.

Step-by-step explanation:

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation: