The factors of 33 are <u>1</u>, 3, <u>11</u>, and 33 .
The factors of 55 are <u>1</u>, 5, <u>11</u>, and 55 .
The <u>common</u> factors of 33 and 55 are 1 and 11 .
The<u> greatest</u> one is <em>11</em> .
$62.35 because 40*12=480 480 divided by 7.65=62.35 or just 62
I hope this helps:)
What you need to do is find the greatest amount (1) and the lowest amount (1/4) of sap collected and the find how many times the lowest can go into the greatest. (So how many times can 1/4 go into 1). I believe that this is how you would do that.
Answer:
P = $300
r = 0.15
n = 12
$544.61 (to the nearest cent)

$524.70 (to the nearest cent)
Step-by-step explanation:
P = principal amount = $300
r = annual interest rate in decimal form = 15% = 15/100 = 0.15
n = number of times interest is compounded per unit t = 12
<u>How much she'll owe in 4 years</u>
P = 300
r = 0.15
n = 12
t = 4

= $544.61 (to the nearest cent)
<u>Yearly compounding interest rate</u>

<u>How much she'll owe in 4 years at yearly compounding interest</u>

= $524.70 (to the nearest cent)