The perimeter of a rectangular skating rink is 530 feet. If the length is 5 feet more than 3 times the width, what are the dimensions?
Please show work, I will make you brainleist

Width and length of rectangle is 65 feet and 200 feet

Given:

Perimeter of rectangular skating rink = 530 feet

Length is 5 feet more than 3 times width

Find:

Dimensions of rectangle

Computation:

Assume;

Width of rectangle = a feet

So,

Length of rectangle = (3a + 5) feet

Perimeter of rectangle = 2(l + b)

2[(3a + 5) + a] = 530

2[4a + 5] = 530

8a + 10 = 530

8a = 520

a = 65

Width of rectangle = a feet

Width of rectangle = 65 feet

Length of rectangle = (3a + 5) feet

Length of rectangle = 3(65) + 5 feet

Length of rectangle = 200 feet

Learn more:

link

length:  200

width:  65

Step-by-step explanation:

length:  3w + 5

width:  w

perimeter = 2L + 2W

530 = 2(3w + 5) + 2w

530 = 6w + 10 + 2w

530 = 8w + 10

520 = 8w

w = 65

length:  3w + 5 = 3(65) + 5 = 200

width:  w = 65

omg what happend to your question? if you could edit this question i can

step-by-step explanation:

the one that has a slope of 1/2 and is positive.

step-by-step explanation: 2 buckets 1 trip ; 4 buckets 2 trips and so on.