Arectangular parking lot has a perimeter of 820 ft. the area of the parking lot measures 42,000 ft2. what is the width of the parking lot?

A rectangle, let length=a and width=b

perimeter: a+a+b+b=820

2a+2b=820  --> a+b = 410

Make a or b the subject: a=410-b

Area: ab=42 000 (area of a rectangle, length x width)

Sub a in.

b(410-b) = 42000

Expand: 410b - b^2 = 42000

Rearrange and use quadratic formula: b^2 -410b + 42000=0

b= 210 (can only be positive because length cannot be negative)

Sub this value back into perimeter or area equation: a + 210 = 410 --> a =200

Therefore, a= 200 and b=210.

Depending what you made the width is, it should be 200ft.

(I may have made some calculation errors as i'm not used to typing it out, but the process should be correct) I hope helps! :)

The answer is C. 210

answer: 2 and 1

step-by-step explanation: got this question

since your problem is x^4 = 81, what you have to undeo is the "to the fourth power" part.

the inverse (opposite operation) of "to the fourth power" is "to the fourth root".

  (x^4) ^ ¼   = 81 ^ ¼  

  81 = 3^4     so (3 ^ 4) ^ ¼   =   3 ^ 1   = 3

by hand: (3 ^ 4) ^ ¼   =   3 ^ (4 x ¼) = 3 ^ 1   = 3   (when you take a power to a power, multiply the exponents.)

or, do it by calculator. one way of finding the fourth root is to use ^ (1/4)

step-by-step explanation: