Question 72776
1) Use Pythagoreans theorem 
{{{(35)^2+(65)^2=d^2}}}
{{{1225+4225=d^2}}}
{{{5450=d^2}}}
{{{d=sqrt(5450)}}}
{{{d=73.8}}}Approximately
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2)To undo a exponent of 3, take a cube root
{{{3sqrt(729)=3sqrt(s^3)}}}
{{{s=9}}}So the side's length is 9
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3)
A)
For annual compounding let n=1 and t=2 (two years)
use the equation
{{{A=P(1+r/n)^(nt)}}}Where A is return
{{{A=10000(1+.1)^2}}}
{{{A=12100}}}So the annual return is $12,100
B)
For interest compounded quarterly let n=4 and t=2 (two years)
{{{A=P(1+r/n)^(nt)}}}
{{{A=10000(1+.1/4)^(8)}}}
{{{A=10000(1.025)^8}}}
{{{A=12184.03}}}

C)
For interest compounded monthly let n=12 and t=2 (two years)
{{{A=P(1+r/n)^(nt)}}}
{{{A=10000(1+.1/12)^(2*12)}}}
{{{A=10000(1.008333)^24}}}
{{{A=12203.81}}}
D)
For interest compounded daily let n=365 and t=2 (two years)
{{{A=P(1+r/n)^(nt)}}}
{{{A=10000(1+.1/365)^(2*365)}}}
{{{A=10000(1.0002739)^730}}}
{{{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^(rt)}}} Where e is a constant e=2.71828...
So if it was contiuously compounded for 2 years at 10% then
{{{A=10000(2.718^(.1*2))}}}
{{{A=10000(1.221403)}}}
{{{A=10000(1.221403)}}}
{{{A=12214.03}}}
So over 2 years it continuously compounds to $12,214.03

Hope this helps. Feel free to ask about any step.