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.
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.