Mat 221
Argenia L. McCray
Instructor: Farhad Abrishamkar
August 18, 2014
Compounded Semiannual Interest However in this paper I will be giving three problems to be figure out. Although two of the problems we are using compound interest, and with the others problem we will be dividing two polynomials. Although in this paper I will discuss the steps I am going to use to solving these problems. Although with the following formula I will use P (1 + r/2) ^2. Even though with these fact, they will have an exponent, and we must remove this exponent in order to get an accurate answer. Now we will set the formula up as (1 + r/2) * (1 + r/2). Because when we use the FOIL method we will be using the multiplication. Now we are going to use (A + B)*(C + D) = (1 + r/2)*(1 + r/2) where A = 1, B = r/2, C = 1, and D = r/2. Now we are now using A is multiplied by C, this actually comes down to me using 1 and then multiplied by 1 and the results which is 1. Consequently there we will multiply A by D 1 time’s r/2, and that equals to r/2. Then last we can multiply B by D this would be the result r/2 which we multiply this by r/2. Then the result for this two fractions and the result of the product and the term of the numerator will be R and R will become r squared there are two R’s. Now the product for these two denominators are 2 times 2 will be 2 squared, now the squared becomes 4. Now we put everything where it is needed to be and we create this fraction r squared over 4, or r^2/4. Because when we uses the FOIL method, this will then creates AC + AD + BC + BD the same as 1 + r/2 + r/2 + r^2/4. Now you can see, that r/2 and r/2 are the same, like terms we then add them together. Nevertheless as we think of the R being a possible of 1 one can make one think r/2 as ½ would make both as equal to 1. It is easy to place R back into the equation and create 1 + r + r^2/4 = 1 + r + 0.25r^2 where 1/4 can equal to 0.25. On the other hand we should then multiply the equation by P to get P + P r + 0.25Pr^2. This word problem we are using we must find that the amount of $200 after the first year for the annual interest of 10% are the compounded semiannually interest. Now we are using P as our principal $200 and then the r as the rate of 10% then we place the .10 since 10% dividend by the 100% which is .10. Now I will plug the values of the equation into the P + P r + 0.25Pr^2. After saying this we having the equation of 200 + 200*0.10 + 0.25*200*0.01. Although going from the left and then to right we first multiply the 200 by .10 and then get 20. Now we have the 200 + 20 + 0.25*200*0.01. Next we can then multiply .25 by 200 and by .01 this will then create .50. Now we have 200 + 20 + 0.50. Now we have the equation in descending order. After this we add the terms all together and get $220.5 which will then become the answer for after 1 year $200 I have invested at 10% and the interest becoming as compounded semiannually. Now I can move to the next word problem. This problem states that we find the amount after one year of $5,670 dollars at the annual interest rate of 3.5%, and our compounded semiannually. Now we use the P for our principal in this equation we are using $5,670 as our principal. Now the R we use as our rate, this rate of 3.5% will be divided by 100% and we will get .035. Now from here we can place the new information which we have find in the equation P +Pr+0.25Pr^2 this will then create 5670 + 5670*0.035 + 0.25*5670*0.035^2. Not like our previous problem we go from the left to the right and we then will get rid of the exponent first. Now looking at the .0035^2 as .0035 in times the .0035 we use to help create .001225. Now we have 5670 + 5670*0.035 + 0.25*5670*0.001225. Now we are going from the left to the right then we will use
Multiplication, and we will