Fluent Essays

To MATH 1810 students, Due to COVID-19, we are forced to switch to a remote delivery. To make this communication process smooth and clear, you are required to use the internal MSU D2L platform. Our entire communication will be held on D2L only. In what follows please find the changes that we are forced to make compared to the regular on-ground delivery process: ◼ Each Monday (starting May 25th) by 10:00am, I will e-mail everybody a weekly message with detailed Week Instructions on how to ease your approach to the course material that is assumed to be covered during that week; ◼ During the course, five Week Tests (one test per week) will be offered. The last of those is the Final Exam. Each test is required to be turned in on Friday (May 29 th , June 5th, June 12th, June 19th, and June 26th ) by 5:00pm; ◼ Make sure that you promptly submit all five tests. Your submissions must be written down with a black pen. This will make it easier to check them out; ◼ This point is the most important of all. While turning in the Week Tests and the Final Exam, you are absolutely required to show each-and-every step of your work!!! Just problem answers are meaningless, because your tests ARE NOT going to be multiple choice. Partial credits will be appropriately given!!! ◼ Maximum 100 points will be possible to get for each of the Week Tests (the Final Exam included). This makes 500 points as total maximum possible number of points accumulated for the course; ◼ The course grading policy will be announced in the Course Syllabus. Hope you will stay focused and try to find at least three-four hours a day for learning materials of our course. I believe that working together as a team, we will be able to overcome all the difficulties of the nowadays. Please keep in mind that your course instructor will be opened for contacts, once a question or concern arises, do not hesitate to get in touch on D2L. Good luck to everybody, Sincerely,-------------------------------------------------------------------------------------------------------------------------------------------------------------------------the questions are on the last page. please follow the instructions. __week_1_instruction.pdf Unformatted Attachment Preview Week 1 Instruction 05/26/2020 Well, let us get started with the first week of our remote process. The following lecture is supposed to prepare students to Test 1. It briefly covers all sections of Chapter 1 of our textbook. You are recommended to read this message in detail, and to review Chapter 1, paying special attention to its Examples. Note that we are going to write down in red answers of the problems that will be discussed in this lecture. We will focus, first, on a very important subject of our course. That is the simplification of algebraic expressions. As an illustrative example, let us simplify the following expression 𝑝2 − 9 𝑝2 −8𝑝+16 ∙ 𝑝−4 (1) 5𝑝 + 15 Recalling the standard algebraic identities (𝑎 ± 𝑏)2 = 𝑎2 ± 2𝑎𝑏 + 𝑏 2 and 𝑎2 − 𝑏 2 = (𝑎 + 𝑏)(𝑎 − 𝑏), we are able to ultimately simplify (1) as 𝑝2 − 9 𝑝2 −8𝑝+16 ∙ 𝑝−4 5𝑝 + 15 = (𝑝+3)(𝑝−3) (𝑝−4)2 ∙ 𝑝−4 5(𝑝 +3) = 𝑝−3 5(𝑝−4) (2) . Another important course issue that is related to Test 1 is the solution of algebraic equations. For illustration, let us solve the following algebraic equation 2(3𝑥−1) 4𝑥−1 = . (3) 5𝑥+1 3𝑥+2 To solve this equation means to determine value(s) of the variable 𝑥 that make it true. To proceed with the solution, we transform (3) into 2(3𝑥 − 1)(3𝑥 + 2) = (5𝑥 + 1)(4𝑥 − 1) which is mathematically equivalent to 18𝑥 2 + 6𝑥 − 4 = 20𝑥 2 − 𝑥 − 1 or 2𝑥 2 − 7𝑥 + 3 = 0 (4) The equality in (4) represents a quadratic equation. Therefore upon using the quadratic formula, we obtain two solutions for (4) as 𝑥1,2 = −𝑏±√𝑏2 −4𝑎𝑐 2𝑎 = −(−7)±√(−7)2 −4∙2∙3 2∙2 = 7±5 4 . Thus, the two distinct solutions of the equation in (3) are 𝑥1 = 3 and 𝑥2 = 1/2 . Another important type of equations (that Test 1 deals with) is radicals containing equations, as the one below √𝑥 + 15 − √𝑥 − 1 = 4 . (5) just for example. To solve this equation, we rewrite it in the following equivalent form (6) √𝑥 + 15 = 4 + √𝑥 − 1 . And then we square both sides of (6), keeping in mind that its left side represents the sum of two additive terms, whose square must be obtained with the aid of the first of the standard identities shown in (2). This results into 𝑥 + 15 = 16 + 8√𝑥 − 1 + (𝑥 − 1) or 8√𝑥 − 1 = 0 , from which it follows that the single solution of the equation in (5) is 𝑥 = 1 . Another important subject covered in Test 1 is the solution of inequalities. Before we go any further with inequalities, let us point out the formal difference between solutions of equations, on one hand, and inequalities, on the other hand. The point is that solution set of an equation is a limited set of numbers (𝑥1 , 𝑥2 , 𝑥3 , . . . , 𝑥𝑛 ), where n is a number of distinct solutions. Solution of an inequality represents, in contrast, an interval of the 𝑥 − axis . As an example, we consider the following inequality |2𝑥 + 17| ≤ 9 In order to solve it, we recall that for the inequality |𝑎| ≤ 𝑏 to be true, the following two inequalities must be true 𝑎 ≥ −𝑏 and 𝑎≤𝑏. With this in mind, for the inequality in (7) we have 2𝑥 + 17 ≥ −9 and 2𝑥 + 17 ≤ 9 . From (8), it follows that 𝑥 ≥ −13 whereas from (9), we have 𝑥 ≤ −4 . Thus, summarizing, the solution for (7) is −13 ≤ 𝑥 ≤ −4 Or, in other words, it is the closed interval [−13, −4] on the 𝑥 − 𝑎𝑥𝑖𝑠. (7) (8) (9) The last covered in Test 1 subject is the rationalization of radical-containing expressions. To illustrate this subject, we solve a problem. That is, let us rationalize the denominator of the following ordinary fraction type expression 3 − √7 (10) 3 + √7 To rationalize the denominator, we multiply both the numerator and denominator by the conjugate of the denominator. This does not, of course, change the quantity of (10), just transforming it to (3 − √7)(3 − √7) (11) (3 + √7)(3 − √7) In compliance with the second of the standard algebraic identities in (2), the denominator in (11) represents difference of squares 32 − (√7)2 = 9 − 7 = 2 while the numerator in (11) is square of difference 32 − 2 ∙ 3 ∙ √7 + (√7)2 = 16 − 6√7 Wherefore, the whole fraction in (10) reads as 8 − 3√7 . The following is Test 1 (the first of the five tests to be offered during this five week-long remote session). You have to turn it in by 5:00pm on May 29th. Please, send it to me as a scan or a picture of your work via the MTSU D2L email. And the following is just a friendly reminder. Make sure that you provide me with your stepby-step work on each-and-every offered problem in the test. I do not accept just answers. Test 1 1. Simplify the algebraic expression 3𝑚−12 𝑚2 +8𝑚+16 ∙ 𝑚+4 𝑚2 −16 . 2. Solve the given algebraic equation 2𝑥−3 𝑥+2 = 3𝑥−2 5𝑥−8 . 3. Solve the radical containing equation √𝑥 + 4 − √𝑥 − 5 = 3 . 4. Find solution of the given inequality |5𝑥 − 3| ≤ 8 . 5. Rationalize the denominator 5+√3 2−√3 . ... Purchase answer to see full attachment