“Projects”Mathematics in Our World. You should be concise in your reasoning. For Project #1, work only equations (a) and (c), but complete all 6 steps

“Projects”Mathematics in Our World. You should be concise in your reasoning. For Project #1, work only equations (a) and (c), but complete all 6 steps (a-f) as shown in the example. For Project #2, select at least five numbers; 0 (zero), two even, and two odd. Make sure you organize your paper into separate projects. The assignment must include (a) all math work required to answer the problems as well as (b) introduction and conclusion paragraphs. Your introduction should include three to five sentences of general information about the topic at hand. The body must contain a restatement of the problems and all math work, including the steps and formulas used to solve the problems. Your conclusion must comprise a summary of the problems and the reason you selected a particular method to solve them. It would also be appropriate to include a statement as to what you learned and how you will apply the knowledge gained in this exercise to real-world situations. The assignment must be formatted according the APA (6th edition) style, which includes a title page and reference page. ————————————————— (I try to solve the math problems myself to save some time, not sure if they are all correct) Project 1# Equation (C) : x² + 12x – 64 = 0 (a) Move the constant term to the right side of the equation. x² + 12x – 64 = 0 x² + 12x – 64 + 64 = 0 + 64 x² + 12x + 0 = 0 + 64 x² + 12x = 0 + 64 x² + 12x = 64 (b) Multiply each term in the equation by four times the coefficient of the x squared term. The coefficient of the x² term is 1. x² + 12x = 64 (4 * 1) * (x² + 12x = 64) (4) * (x² + 12x = 64) (4)*x² + (4)*(12x) = (4)*(64) 4x² + 48x = 256 (c) Square the coefficient of the original x term and add it to both sides of the equation. The coefficient of the original x term is 12. (12)² = 144 4x² + 48x = 256 4x² + 48x + 144 = 256 + 144 4x² + 48x + 144 = 400 (d) Take the square root of both sides. 4x² + 48x + 144 = 400 Sqrt(4x² + 48x + 144) = Sqrt(400) Sqrt(2x + 12)² = Sqrt(20²) Sqrt(2x + 12)² = ±20 2x + 12 = ±20 (to simplify both sides of the equation, divide each side of the equation by 2) (2x + 12)/2 = ±20/2 x + 6 = ±10 (e) Set the left side of the equation equal to the positive square root of the number on the right side and solve for x. x + 6 = 10 x + 6 – 6 = 10 – 6 x + 0 = 4 x = 4 (f) Set the left side of equation equal to the negative square root of the number on the right side of the equation and solve for x. x + 6 = -10 x + 6 – 6 = -10 – 6 x + 0 = -16 x = -16 the final solution is: x ∈ {-16, 4} check the solution by substituting the two numerical values of x into the original equation for x = -16 x² + 12x – 64 = 0 (-16)² + 12(-16) – 64 = 0 256 – 192 – 64 = 0 256 – 256 = 0, OK for x = 4 x² + 12x – 64 = 0 (4)² + 12(4) – 64 = 0 16 + 48 – 64 = 0 64 – 64 = 0 0 = 0, OK Project 2: Formula that yields prime numbers. x² – x + 41 select at least five numbers; 0 (zero), two even, and two odd. x values selected: 0, 6, 8, 11, 13 if x = 0 x² – x + 41 = 0² – 0 + 41 = 41 41 is a prime number if x = 6 x² – x + 41 = 6² – 6 + 41 = 71 41 is a prime number if x = 8 x² – x + 41 = 8² – 8 + 41 = 97 97 is a prime number if x = 11 x² – x + 41 = 11² – 11 + 41 = 151 151 is a prime number if x = 13 x² – x + 41 = 13² – 13 + 41 = 197 197 is a prime number However the expression x² – x + 41 does yield composite numbers as well: If x = 45 x² – x + 41 = 45² – 45 + 41 = 2021 2021 is a COMPOSIT number. Its factors are: 43 and 47 2021 can be factored into at least two smaller positive integers which not equal to 1.

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