Answer:
14.4 acres for corn (if you need to round, it's 14 acres.)
Step-by-step explanation:
40% = 19.2 acres
20% = 9.6 acres
10% = 4.8 acres
40 + 20 + 10 = 70
100-70 = 30
30% = 14.4 acres
Answer:

Step-by-step explanation:
The function that could model this periodic phenomenon will be of the form

The tide varies between 3ft and 9ft, which means its amplitude
is

and its midline
is
.
Furthermore, since at
the tide is at its lowest ( 3 feet ), we know that the trigonometric function we must use is
.
The period of the full cycle is 14 hours, which means


giving us

With all of the values of the variables in place, the function modeling the situation now becomes

Answer:4.5
Step-by-step explanation: Check the file for my explanation
Find the LCD = 18
5/6 = 15/18
3/9 = 6/18
15/18 + 6/18 = 21/18 = 7/5 or 1 2/5
Since there are more parakeets than canaries, it is not possible to have only 1 of each bird in each cage <u>and</u> have the same number of birds in each cage.
He could use 42 cages, putting a canary in with the parakeet in 18 of them. Then he would have 18 cages with 2 birds each, and 24 cages with 1 bird each.
The only way to have the same number of birds (1) in all cages is to have 60 cages, 42 of which have 1 parakeet, and 18 of which have 1 canary.
_____
If more than 1 of each kind of bird can be put in the cage, the collection of birds could be put into 6 cages, each of which would be home to 7 parakeets and 3 canaries.