# Week One Written Assignment

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Week One Written Assignment Shereka Pierce Mat 126 Elizabeth Stepp December 6, 2011 We have been learning how to develop our skills, in speaking, reading, and writing the English language. Did you know that when we were in math class, we were also learning how to speak, read, and write the language of mathematics? Mathematics uses numbers and number systems instead of the alphabet, but it’s also a language: a language of patterns and symbols. Mathematics can help you recognize, understand, describe and identify changes in patterns. Problem35.

A person hired a firm to build a CB radio tower. The firm charges \$100 for labor for the first 10 feet. After that, the cost of the labor for each succeeding 10 feet is \$25 more than the preceding 10 feet. That is, the next 10 feet will cost \$125, the next 10 feet will cost \$150, etc. How much will it cost to build a 90-foot tower? Here is how we think about it: We see that there is a new price every ten feet as they build the tower. After that the cost of the labor for each succeeding 10 feet is \$25 more than the preceding 10 feet.

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That is, the next 10 feet will cost \$125, the next 10 feet will cost \$150, etc. How much will it cost to build a 90-foot tower? n= the number of terms altogether n=9 d= the common difference d=25 a1= the first term a1=100 aN= the last term aN=a9 (yet to be computed) Next, we need to compute what a9 is. The next step is to find the nth term of the sequence, or the 9th term in this case. aN=a1+(n-1)d a9=100+(9-1)25 a9=100+(8)25 a9=100+200 9=300 Now that we know what a9 is, we need to know what the sum of the sequence is from a1 to a9. The next step is to find the answer to this question. Sn= n(a1+a9)/2 S9= 9(100+300)/2 S9= 9(400)/2 S9=3600/2 S9=1800 Problem 37 A person deposited \$500 in a savings account that pays 5% annual interest that is compounded yearly. At the end of 10 years, how much money will be in the savings account? Here is how we think about it: Each year 5% of the balance is added to the balance, that would look like: B+ (. 03)B B(1+. 05) B(1. 05)

In other words each year the existing balance is multiplied by 1. 05. First, we need to identify the following numbers: n= the number of terms n=10 r= the common ratio r=1. 05 a1= the first term a1=500(1. 05)=525, the balance at the end of the first year, thus a1. In a savings account, the total balances at the end of each year form the sequence, so we don’t need to add up all the terms in the sequence. We just need to find out what the balance is at the end of 10 years, so we are looking for the value of a10.

The next step is to find the balance in the savings account at the end of 10 years. aN=a1(rN-1) a10=525(1. 05) 9 a10=525(4961. 25) a10=2,604,656. 25 Thus, the balance in the savings account at the end of 10 years is \$2,604,656. 25. In this conclusion I will tell you how I got my answer. For example, the first thing that I did was figure out which numbers go where n=9 that is the total of terms altogether d=25 that is the common difference a1= the first term and aN= the last term. The next thing that I did was igure out which numbers go where as stated earlier I took the one hundred added it with eight multiplied twenty-five then, the next step is to add one hundred plus two hundred and I got three hundred. The next step is to figure out the sequence is. The next step is to Sn=9 is n plus S9= nine multiplied by one hundred plus three hundred divided by two. The answer is nine multiplied by four hundred divided by two. The next step is to divide three thousand into two and you get 1800. The next problem stated earlier is I have to find out how much money will be in the savings account in ten years.

The first thing that I do is breakdown the percentage into decimal which I do by taking the percent and dropping the percent sign and changing it to a decimal like I did earlier then I multiply \$1. 05 by \$500 and I get \$525. Then I figure out what the sequence of the next step by figuring out what a10 is by multiplying five twenty five times one point five to the ninth power to get \$2,604,656. 25. References Bluman, A. G. (2005). Mathematics in our world. (Ashford University Custom Edition). United States: McGraw-Hill.

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