Afence around a yard (50 yards by 40 yards) is 1.5 yards wide. what is the perimeter and area of the yard, including the fence? by what percent did the area increase?

The lenght of the yard plus a fence around this yard will be: 50 yd+2(1.5 yd)=53 yd
The width of the yard plus a fence a fence around this yard will be: 40 yd + 2(1.5 yd)=43 yd.

Data:
New lenght (yard +fence)=53 yd
New width (yard+fence)=43 yd

Perimeter of a rectangle=2(lenght) + 2(width)
Perimeter of this yard, including the fence: 2(53 yd)+2(43 yd)=192 yd

Area of a rectangle=lenght x width
Area of this yard, including the fence:(53 yd)(43 yd)=2279 yd².

Now, we calculate the increase area percent:
1)we calculate the old area:
Data:
lenght=50 yd
width=40 yd
Old area=(50 yd)(40 yd)=2000 yd²

2)we calculate: How many yards²  has the area of the yard increased?
Yards gained=2279 yd² - 2000 yd²=279 yd².

3) Now, at last,  What percent did the area increase:

2000 yd²100%
279 yd²  x
x=(279 yd² * 100%) / 2000 yd²=13.95%

The perimeter of the yard, including the fence would be 192 yd, the area of the yard, including the fence would be 2279 yd², and the percent of the area increase would be: 13.95%.

Hello!

the answer is:

the answer is:

why?

to solve the problem, we need to remember the quotient of power property, it's defined by the following relation:

if we have a quotienf of powers that have the same base, we need to keep the same base and subtract the exponent of the denominator power to the exponent of the numerator power.

so, we are given the expression:

then, calculating we have:

hence, the answer is:

have a nice day!

A: 18/15b : 7 - -8 = 15 , 12 - -6 = 18 the y variable always goes over the x variables, c : same thing a