SOLUTION: I am in dire need or assitance with the following questions and if anyone could assist me i WOULD BE SO GREATFUL!!! (1)you are traveling North for 35km the you travel East 65km.

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Question 72776: I am in dire need or assitance with the following questions and if anyone could assist me i WOULD BE SO GREATFUL!!!
(1)you are traveling North for 35km the you travel East 65km. how far are you from your starting point? N and E ca be considered the direstions of the y and x axis. Round to the tenths place.
Answer:
Show work:

(2)the volume of a cube is given by V=S^3. find the length of a side of a cube if the volume is 729cm^3.
Answer:
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(3)suppose you deposit $10,000 for two years at a rate of 10%.calculate the return (A) if the bank compounds annually(n=1). round answer to the hundreths place.
Answer:
Show work:(use ^ to indicate the power).
(A)calculate the return(A) if the bank compounds quarterly(n=4). round answer to the hundreths place.
Answer:
Show work:
Bank compounds monthly(n=12). Round answer to hundreths place.
Answer:
Show work:
Bank compound daily(n=365).round answer to hundreths.
Answer:
Show work:
what observation can you make about the size of the increase in your return as compounding increases more frequently?
Answer:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
1) Use Pythagoreans theorem
%2835%29%5E2%2B%2865%29%5E2=d%5E2
1225%2B4225=d%5E2
5450=d%5E2
d=sqrt%285450%29
d=73.8Approximately


2)To undo a exponent of 3, take a cube root
3sqrt%28729%29=3sqrt%28s%5E3%29
s=9So the side's length is 9


3)
A)
For annual compounding let n=1 and t=2 (two years)
use the equation
A=P%281%2Br%2Fn%29%5E%28nt%29Where A is return
A=10000%281%2B.1%29%5E2
A=12100So the annual return is $12,100
B)
For interest compounded quarterly let n=4 and t=2 (two years)
A=P%281%2Br%2Fn%29%5E%28nt%29
A=10000%281%2B.1%2F4%29%5E%288%29
A=10000%281.025%29%5E8
A=12184.03
C)
For interest compounded monthly let n=12 and t=2 (two years)
A=P%281%2Br%2Fn%29%5E%28nt%29
A=10000%281%2B.1%2F12%29%5E%282%2A12%29
A=10000%281.008333%29%5E24
A=12203.81
D)
For interest compounded daily let n=365 and t=2 (two years)
A=P%281%2Br%2Fn%29%5E%28nt%29
A=10000%281%2B.1%2F365%29%5E%282%2A365%29
A=10000%281.0002739%29%5E730
A=12213.05
So as the frequency of compounding increases it approaches a finite number. Notice how the increase gets smaller as we increase the frequency of compounding. So as the compounding frequency increases, it approaches a finite number. It actually approaches the continuous value of the continuous compound formula. Continuous compounding has the form
A=Pe%5E%28rt%29 Where e is a constant e=2.71828...
So if it was contiuously compounded for 2 years at 10% then
A=10000%282.718%5E%28.1%2A2%29%29
A=10000%281.221403%29
A=10000%281.221403%29
A=12214.03
So over 2 years it continuously compounds to $12,214.03
Hope this helps. Feel free to ask about any step.