document.write( "Question 72776: I am in dire need or assitance with the following questions and if anyone could assist me i WOULD BE SO GREATFUL!!!\r
\n" ); document.write( "\n" ); document.write( "(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.
\n" ); document.write( "Answer:\r
\n" ); document.write( "\n" ); document.write( "Show work:\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "(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.
\n" ); document.write( "Answer:
\n" ); document.write( "Show work:\r
\n" ); document.write( "\n" ); document.write( "(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.
\n" ); document.write( "Answer:\r
\n" ); document.write( "\n" ); document.write( "Show work:(use ^ to indicate the power).\r
\n" ); document.write( "\n" ); document.write( "(A)calculate the return(A) if the bank compounds quarterly(n=4). round answer to the hundreths place.
\n" ); document.write( "Answer:
\n" ); document.write( "Show work:\r
\n" ); document.write( "\n" ); document.write( "Bank compounds monthly(n=12). Round answer to hundreths place.
\n" ); document.write( "Answer:
\n" ); document.write( "Show work:
\n" ); document.write( "Bank compound daily(n=365).round answer to hundreths.
\n" ); document.write( "Answer:
\n" ); document.write( "Show work:
\n" ); document.write( "what observation can you make about the size of the increase in your return as compounding increases more frequently?
\n" ); document.write( "Answer:
\n" ); document.write( "

Algebra.Com's Answer #52322 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
1) Use Pythagoreans theorem
\n" ); document.write( "\"%2835%29%5E2%2B%2865%29%5E2=d%5E2\"
\n" ); document.write( "\"1225%2B4225=d%5E2\"
\n" ); document.write( "\"5450=d%5E2\"
\n" ); document.write( "\"d=sqrt%285450%29\"
\n" ); document.write( "\"d=73.8\"Approximately
\n" ); document.write( "


\n" ); document.write( "2)To undo a exponent of 3, take a cube root
\n" ); document.write( "\"3sqrt%28729%29=3sqrt%28s%5E3%29\"
\n" ); document.write( "\"s=9\"So the side's length is 9
\n" ); document.write( "


\n" ); document.write( "3)
\n" ); document.write( "A)
\n" ); document.write( "For annual compounding let n=1 and t=2 (two years)
\n" ); document.write( "use the equation
\n" ); document.write( "\"A=P%281%2Br%2Fn%29%5E%28nt%29\"Where A is return
\n" ); document.write( "\"A=10000%281%2B.1%29%5E2\"
\n" ); document.write( "\"A=12100\"So the annual return is $12,100
\n" ); document.write( "B)
\n" ); document.write( "For interest compounded quarterly let n=4 and t=2 (two years)
\n" ); document.write( "\"A=P%281%2Br%2Fn%29%5E%28nt%29\"
\n" ); document.write( "\"A=10000%281%2B.1%2F4%29%5E%288%29\"
\n" ); document.write( "\"A=10000%281.025%29%5E8\"
\n" ); document.write( "\"A=12184.03\"\r
\n" ); document.write( "\n" ); document.write( "C)
\n" ); document.write( "For interest compounded monthly let n=12 and t=2 (two years)
\n" ); document.write( "\"A=P%281%2Br%2Fn%29%5E%28nt%29\"
\n" ); document.write( "\"A=10000%281%2B.1%2F12%29%5E%282%2A12%29\"
\n" ); document.write( "\"A=10000%281.008333%29%5E24\"
\n" ); document.write( "\"A=12203.81\"
\n" ); document.write( "D)
\n" ); document.write( "For interest compounded daily let n=365 and t=2 (two years)
\n" ); document.write( "\"A=P%281%2Br%2Fn%29%5E%28nt%29\"
\n" ); document.write( "\"A=10000%281%2B.1%2F365%29%5E%282%2A365%29\"
\n" ); document.write( "\"A=10000%281.0002739%29%5E730\"
\n" ); document.write( "\"A=12213.05\"
\n" ); document.write( "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
\n" ); document.write( "\"A=Pe%5E%28rt%29\" Where e is a constant e=2.71828...
\n" ); document.write( "So if it was contiuously compounded for 2 years at 10% then
\n" ); document.write( "\"A=10000%282.718%5E%28.1%2A2%29%29\"
\n" ); document.write( "\"A=10000%281.221403%29\"
\n" ); document.write( "\"A=10000%281.221403%29\"
\n" ); document.write( "\"A=12214.03\"
\n" ); document.write( "So over 2 years it continuously compounds to $12,214.03\r
\n" ); document.write( "\n" ); document.write( "Hope this helps. Feel free to ask about any step. \n" ); document.write( "

\n" );