Perimiter=2(L+W)
P=2(100+120)
P=2(220)
P=440
0.1 per foot
440*0.10=44
$44
Answer:
9x
Step-by-step explanation:
Since you are subtracting a negative from a positive you would get 9
Answer:
(a) 100 fishes
(b) t = 10: 483 fishes
t = 20: 999 fishes
t = 30: 1168 fishes
(c)

Step-by-step explanation:
Given


Solving (a): Fishes at t = 0
This gives:






Solving (a): Fishes at t = 10, 20, 30






Solving (c): 
In (b) above.
Notice that as t increases from 10 to 20 to 30, the values of
decreases
This implies that:

So:
The value of P(t) for large values is:




Answer: C, 3 HOURS
Step-by-step explanation:
she canoed for 1 hour and windsurfed for 2 hours