Asking the Math Gods...
550*.082+550=$595.10
$45.10 in tax 
        
                    
             
        
        
        
What we know:
Football field is 100 yards in length 
End zones are 10 yards each in length 
Perimeter between pylons is 306 2/3 yards
What we need to find:
a. Perimeter and area of one end zone
b. Perimeter and area of without end zones
c. Perimeter and area of playing field with end zones
 First we need to find the measurements of the field between pylons using the perimeter of 306 2/3 yards. We already know the length is 100 yards so we need to find width (w).
P=2l + 2w
306 2/3=2 (100) + 2w
306 2/3= 200 + 2w
300 2/3-200=200-200+2w
106 2/3=2w
106 2/3/2=2/2w
53 1/3=w 
a. Perimeter=2 (10)+2 (53 1/3)=126 2/3 yards
 Area=10×53 1/3=533 1/3 yd²
b. Perimeter=2 (100)+2 (53 1/3)=306 2/3 yards
 Area=100×53 1/3=5333 1/3 yd²
c. Perimeter=2 (120) + 2 (53 1/3)=346 2/3 yards
 Area=120×53 1/3=6400 yd²
        
             
        
        
        
Answer:
1) D. (2, -6)
2) B. (3, 1)
Step-by-step explanation:
idk how to explain just trust me
if u need any other questions answered message me brother
 
        
             
        
        
        
Assuming that all tiles must stay whole it would be:
1×54
2×27
3×18
6×9