Atrain ticket has a perimeter of 72 centimeters and an area of 315 square centimeters. what are the dimensions of the ticket?

You can work out the perimeter by working out the sum of twice the length (l) and twice the width (w).

2l + 2w = 72

The area is length (l) * width (w)

l * w = 315.

Now we just need to do some rearrangement and isolate one of the terms so we can substitute it back into the area equation.

2l + 2w = 72

2l = 72 - 2w

l = 36 - w

l * w = 315

(36 - w) * w = 315

36w - w² = 315

w² - 36w + 315 = 0

(w - 21)(w + 15) = 0

w = -15 cannot be right, because it can't be negative, so w = 21.

Now that we know the width is 21, we can substitute it into the perimeter equation to work out the length.

2l + 2w = 72

2l + 2(21) = 72

2l + 42 = 72

2l = 72 - 42

2l = 30

l = 15.

Therefore, the width is 21 cm and the length is 15 cm.

step-by-step explanation:

try remembering the steps

1. when "a" is a root of a polynomial, (x -a) is a factor of it. for the three roots given, the factors of the desired polynomial are (x +2)(x -1/2)(x -3).

in order to make the leading coefficient be negative, we need to multiply this product by a negative number. any negative number will do, but we choose a small (magnitude) value that will eliminate the fraction: -2.

then

f(x) = -2(x +2)(x -1/2)(x -3) = -(x +2)(2x -1)(x -3)

= -(2x² +3x -2)(x -3)

= -(2x³ -3x² -11x +6)

f(x) = -2x³ +3x² +11x -6

2. a graph created by the desmos on-line graphing calculator is shown, and the zeros are highlighted.

3. as indicated in part 1, the multiplier of this equation can be anything and the zeros will remain the same. you want a negative leading coefficient, so the "anything" is restricted to any of the infinite number of numbers that will make that be the case.