MAT221- Week 1 Assignments

  

Hello, To ensure that all formulas were formatted correctly I have copied and pasted the assigment verbatim in a Word Document & attached it to this Homework Post. If you are unable to open it and read it please let me know and I will work with you to make necessary changes.

 

This task will also include completing Week 1’s Alesk lab and homework assignment as well as responding to at least 2 classmates initial post.

 

The 3 attachments label classmate 1,2, and 3 responses have been added to this order. You only have to choose 2 of the 3 to respond to. Answering the following questions:  Do you agree with how your classmates used the vocabulary? Did the student handle the negatives in the formulas accurately? If not explain why.

Simplifying Expressions
Read the following instructions in order to complete this assignment, and review the 

example

 of how to complete the math required for this assignment:

· Use the properties of real numbers to simplify the following expressions:

· 2a(a – 5) + 4(a – 5)

· 2w – 3 + 3(w – 4) – 5(w – 6)

· 0.05(0.3m + 35n) – 0.8(-0.09n – 22m)
 

· Write a two to three page paper that is formatted in APA style and according to the

Math Writing Guide

. Format your math work as shown in the Instructor Guidance and be concise in your reasoning. In the body of your essay, please make sure to include:

· Your solution to the above problem, making sure to include all mathematical work.

· Plan the logic necessary to complete the problem before you begin writing. For examples of the math required for this assignment, please review Elementary and Intermediate Algebra and the 
example
 of how to complete the math required for this assignment.

· Show every step of the process of simplifying and identify which property of real numbers was used in each step of your work.  Please include your math work on the left; the properties used on the right.

· A discussion of why the properties of real numbers are important to know when working with algebra. In what ways are they useful for simplifying algebraic expressions?

· The incorporation of the following five math vocabulary words into the text of your paper. Use bold font to emphasize the words in your writing (Do not write definitions for the words; use them appropriately in sentences describing your math work.):

· Simplify

· Like terms

· Coefficient

· Distribution

· Removing parentheses

CLASSMATE #1

The first thing I will be using will be my birthday, December 12, 1956, my integers will be as follows:
a=12
b=1
c=56
I will then plug in12 for variable a,and negative -1 for vairable b
1 to the third and 12 to the third equal 1728-negtive one
I will then raise the i
ntegers to the given exponent which is three.
1728- (-1)
Two negatives make a positive

1728 plus one
Final answer 1729.
I have again plugged in 12 for variable a and -1 for variable
In the beginning the two negatives became a positive
In the second the integers were squared and simplified: Adding two negatives became a positiveand  mutiplying a negative by a negative became a positive.
Final answer 1729
(b-c)
———-
(2b-a)
-1-56
———-
2(-1) – 12
-3-12 and reduce to the lowest term would be 
57
—–
14
All varibles was use in the expression 12 plugged in for a negative 1 for b
The first part of the equation the two negative was subtract thus resulting in a positive number. The divsor was also solved likewise, because the numerator and denominator is negative then you will get positive.
Final answer was reduced to its lowest term which is:
57
—–
14
I assume the answers was not a coincidence I would say this is the way math works that they are set to come to the same answer when the variables are plugged.

CLASSMATE #2

I will be using my birth date which is November 17, 1970. And the vocabulary words to describe my math discussion questions.

1. A= 11, B= -17, C= 70             My variables are going to be, a and b also my exponents are
                                                 going to be 3 for both a and b.
2. A3 – b3
113 -173                                    I will now raise my integers to the given exponents of 3.
1331+ 4913                                I subtracted a negative and it is now changed to a positive. 
                                                My final answer is 6,244 
 
3. (a-b)(a2+ab+b2)                    I will plug in variable 11 and variable -17
(11–17)(112+11x-17+-172)        I will then raise all of the integers and multiply all of my factors.
(11–17)(121+-187+289)            Next I will do my adding and subtracting.
28(223)                                     I will then multiply to get my final answer which is 6,244.

4. (b-c)                                     I will then plug in my -17 for my variable b, and also 70 for my 
                                               variable c, and A for 11.
(2b-a) 
                                                I changed my double negative the divisor to positive.
(-17-70)
2x-17+17
-87
-17
87
17                                           Both my numerator and denominator were negative so my answer is positive. 
                                              My final answer is a fraction and is in the lowest terms.

In my first two algebraic expressions my answers were the same. I am assuming the answers were not a coincidence. I am thinking if the algebraic expressions are plugged in the right way the answers are going to be the same.

CLASSMATE #3

For this exercise I will use my birth date which is January 10, 1981 (01-10-81). Therefore, I will be working with the following integers:
Let a = 1
       b = -10 (negative ten)
       c = 81
A) a³ – b³ = variable a and b and exponent on each of them
    1³ – -10³ = for variable a I plugged in one and for variable b I plugged in a negative ten.
    1 – -1000 = I then raised the integers to the given exponents which is three
    1 + 1000 = subtracting a negative integer would change it into addition 
    1,001= bringing me to my final answer
B)
 (a-b)(a² + ab + b²)
 [1 – (-10)][1² + 1 (-10) + (-10)²]
 [1 + 10][1 + (-10) + 100]
 11 (1 – 10 + 100)
 11(91)
 1,001
For B) I used variables a and b. In each case for variable a I plugged in one and variable b I plugged in negative ten. After that, the two negatives turn into a positive. I added and simplified my signs, then I did the squaring and multiplying and came up with my final answer of 1,001. 
C) for this expression I will use variables a, b and c. When I plug it in a equals one, b equals negative ten, and c equals eighty-one. It is a rational expression which uses a divisor of 2b – a. I then put the integers in for the variables, and evaluated the numerator and denominator, and both were negative giving me a positive answer. Which gives me an answer of 91/19 which is an improper fraction and at its lowest terms the answer is 13/3.
b – c
2b -a 
-10 – 81
2 (-10) – 1
 -91     13
 -21 =   3   final answer   (because they are both divisible by 7)  
I found it interesting that the results of a³ – b³ and (a + b) (a² + ab +b²) had the same answer of 1,001, in my case. I think it is two math problems (one in addition and one in subtraction) set up differently but still produces the same answer. I think this was done to show us students that there is more than one way to set up a problem and get the same answer.