Find the dimensions of a rectangle whose width is 6 miles less than its length and whose area is 55 square miles.

Let
x-----------> the length of a rectangle
y-----------> the width of a rectangle

we know that
area of a rectangle =length*width
area=55 mi²
55=x*y--------> equation 1
y=x-6-------> equation 2
substitute equation 2 in equation 1
55=x*[x-6]------> 55=x²-6x----------> x²-6x-55=0

using a graph tool-----> to resolve the second order equation
see the attached figure

the solution is
x=11 mi
y=x-6-----> y=11-6-----> y=5 mi

the answer is
 the length of a rectangle is 11 miles
the width of a rectangle is 5 miles

Which of the following values are in the range is the function graphed below? check all that applya.3b.1c.0d.-2e.5f.2a