P(1+r)2 This is the first expression.
P=$200 and r=10% = .10 P= amount invested and r = % rate.
P+2Pr+Pr2 Formula expanded
200+2(200)(.10)+200(.10)2 Substituted values into formula
200+40+200(.01) .102=.01 and 2(200)(.10)=40
200+40+2 200(.01)=2
242 Formula solved.
Starting with $200 and compounding 10% interest annually, yields $42 in interest at the end of one year for a total of $242.00.
P(1+r)2 This is the first expression.
P=$5,670 and r=3.5% = .035 P= amount invested and r = % rate.
5,670+2(5,670)(.035)+5,670(.035)2 Substituted values into formula.
5,670+396.90+5,670(.001225) .0352=.001225 and 2(5,670)(.035)=396.90
5,670+396.90+6.94575 5,670(.001225) = 6.94575
6,073.84575 Formula solved.
Starting with $5,670 and compounding 3.5% interest once a year, yields $403.85 in interest at the end of one year for a total of $6,073.85. The above problems are applicable to my everyday life, because they show me how to compound the current interest that I have on some of my accounts. Therefore, if I calculate everything correctly, I will know how much interest my money has yielded over a period of time. The following is a completion of problem 70 on page 311 of Elementary and Intermediate Algebra. I will show how I divide like terms and the dividend by the divisor.
(-9x3+3x2-15x)/(-3x) Given problem -9x(x)(x)(x) = -9x(x)(x)(x) = 3x2
-3x -3x I divided -9x3 by -3x and used the strikeout font to show the canceling factors.