Let x = length of the park
Let y = width of the park
Because the area is 392392 ft², therefore
xy = 392392 (1)
Because three sides of fencing measure 5656 ft, therefore
2x + y = 5656 (2)
That is
y = 5656 - 2x (3)
Substitute (3) into (1).
x(5656 - 2x) = 392392
5656x - 2x² = 392392
2x² -5656x + 392392 = 0
x² - 2828x + 196196 = 0
Solve with the quadratic formula.
x = (1/2)*[2828 +/- √(2828² - 4*196196)]
= 2756.83 or 71.17
Answer:
The possible dimensions are 2756.8 ft and 71.2 ft (nearest tenth)
Answer:
length = 6 cm
width = 1 cm
Step-by-step explanation:
l = length
w = width
l = 4 + 2w
2(4 + 2w) + 2w = 14
8 + 4w + 2w = 14
8 + 6w = 14
6w = 6
w = 1
find length:
l = 4 + 2(1)
l = 6
2km=2000m
25m=25m
2500cm=25m
3000mm=3m
Hence the longest is 2km=2000m
I think it would be 45-x x= minutes spent getting dressed
