Arectangle with are 315 square inches has a length that is one less than four times the width. find the length and the width of the rectangle

l=35 inches

w= 9 inches

Step-by-step explanation:

Area is 315 square inches

Area=l*w=315     (Equation 1)

l=4w-1    (Equation 2)

Putting value of l from equation 2 in equation 1

(4w-1)w=315

4w^2-w-315=0

Solving by using quadratic equation

a=4, b=-1, c=-315

w= (-b+sqrt(b^2-4ac))/2a    and   w= (-b-sqrt(b^2-4ac))/2a

w= (1+sqrt(1+5040))/8         and    w= (1-sqrt(1+5040))/8

w= (1+sqrt(5041))/8             and    w= (1-sqrt(5041))/8

sqrt(5041)=71

w= (1+71)/8             and    w= (1-71)/8

w=72/8                  and     w=-70/8

as width can't be a negative number so we'll solve for w=72/8=9

width =9 inches

putting w in equation 2

l=(4*9)-1

l=35 inches

:

step-by-step explanation:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

answer:

f(5/2) = 10/3

i hope this you!