...Pythagorean Quadratic MAT221 Introduction to Algebra Pythagorean Quadratic Week five of this class has been a complete challenge for me, from start to finish. Trying to master everything that we have been taught over the five weeks has truly been a test. I know there are benefits to knowing these principals, however, it stresses me to think about having to use it in real life circumstances. This problem involves using the Pythagorean Theorem to find distance between several points in our textbook (Dugopolski, 2012). Ahmed has half of a treasure map,which indicates that the treasure is buried in the desert 2x + 6 paces from Castle Rock. Vanessa has the other half of the map. Her half indicates that to find the treasure, one must get to Castle Rock, walk x paces to the north, and then walk 2x + 4 paces to the east. If they share their information, then they can find x and save a lot of digging. What is x? We need to look at the equation so we can know how far Ahmed will have to walk, which is 2x+6 paces from Castle Rock. Even though Vanessa’s half of the map does not indicate in which direction the 2x + 4 paces should go, it can be assumed that her’s and Ahmed’s paces should end up in the same place. When sketched on scratch paper, a right triangle is formed with 2x + 6 being the length of the hypotenuse, and x and 2x + 4 being the legs of the triangle. When a right triangle is involved, the Pythagorean Theorem helps solve for x. The Pyhagorean Theorem states that...
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...Pythagorean Quadratic Treasure Hunters Pythagorean Quadratic Treasure Hunters Introduction to Algebra Treasure Hunters Ahmed and Vanessa both have possession of one half of a complete treasure map. Ahmed’s map shows the treasure is buried in the desert 2x + 6 paces from Castle Rock. Vanessa’s map shows the treasure buried at x paces to the north and 2x + 4 paces to the east. When the two combine information, the location of the buried treasure is going to be a lot easier to find and they can share in the booty loot that they discover. Castle Rock is the lowest left point of the hypotenuse and at the bottom of the left leg and the treasure is at the furthest right point of the right leg. To factor the equation we start with the following, X2+(2x + 4)2 = (2x+6)2 Using the Pythagorean Theorem, a2+2ab+b2 i get a compound X2 +(4x216x+16)=4x2+24x+36 equation. It is then necessary to simplify using the quadratic 5x2+16x+16=4x2+24x+36 equation ax2-bx+c=0 so that I can factor. (x2+2)(x-10)=0 everything is set to zero for the zero factor X = 10 solve for x Plugging the x value for a, b, and c to the legs or the hypotenuse and what this does is it gives me the equation of how many paces it is to the treasure A= 10 B=2(10)+4 = 20+4 = 24 C=2x+6 = 2(10)+6 = 26 In conclusion, castle rock is located at the bottom left of a right hand triangle, and the treasure is 26 paces northeast of Castle...
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...ground instruction, first-time jumpers can be ready to make a solo jump. Without the assistance of oxygen, skydivers can jump from as high as 14,000 feet and reach speeds of more than 100 miles per hour as they fall toward the earth. Jumpers usually open their parachutes between 2000 and 3000 feet and then gradually glide down to their landing area. If the jump and the parachute are handled correctly, the landing can be as gentle as jumping off two steps. Making a jump and floating to earth are only part of the sport of skydiving. For 5 5.1 Factoring Out Common Factors Special Products and Grouping Factoring the Trinomial ax2 bx c with a 1 Factoring the Trinomial ax2 bx c with a 1 Difference and Sum of Cubes and a Strategy Solving Quadratic Equations by Factoring 5.2 Chapter 5.3 5.4 5.5 5.6 example, in an activity called “relative work skydiving,” a team of as many as 920 free-falling skydivers join together to make geometrically shaped formations. In a related exercise called “canopy relative work,” the team members form geometric patterns after their parachutes or canopies have opened. This kind of skydiving takes skill and practice, and teams are not always successful in their attempts. The amount of time a skydiver has for a free fall depends on the height of the jump and how much the skydiver uses the air to slow the fall. In Exercises 85 and 86 of Section 5.6 we find the amount of time that it takes a skydiver to fall from a given height. 322 Chapter...
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...A Treasure Hunt at Castle Rock using Pythagorean Quadratic June Tye-Patterson Math 221: Introduction to Algebra Instructor: Shenita Talton 07-13-2014 A Treasure Hunt at Castle Rock using Pythagorean Quadratic For this week assignment we are given a word problem and the use of the Pythagorean Theorem to solve it. We will be helping Ahmed and Vanessa, who both have a half of a map, find buried treasure in the desert somewhere around a place named Castle Rock. Ahmed map says the treasure is 2x+6 paces from Castle rock, whereas, Vanessa map says in order to find the treasure, go to Castle Rock, walk x paces to the north and then walk 2x+4 paces to the east. In order to discover the location of the treasure, we need to factor down the three quadratic expressions by putting the measurements into the Pythagorean Theorem. The first thing we need to do is to write an equation by inserting the binomials into the Pythagorean Theorem, which also states that every right triangle with legs of length have the relationship of a^2+b^2=c^2 x^2+ (2x+4)^2=(2x+6)^2 The binomials into the Pythagorean Theorem. x^2(2x+4) (2x+4)=(2x+6) (2x+6) The equation squared. x^2 4x+8x+8x+16=4x^2+12x+12x+36 Equation FOILED or distributed. x^2+4x^2=5x^2 First two terms added...
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...Running head: Pythagorean Quadratic Pythagorean Quadratic Sharlee M. Walker MAT 221 Instructor Xiaolong Yao December 2, 2013 Pythagorean Quadratic Ahmed’s half of the map doesn’t indicate which direction the 2x + 6 paces should go, we can assume that his and Vanessa’s paces should end up in the same place. I did this out on scratch piece of paper and I saw that it forms a right triangle with 2x + 6 being the length of the hypotenuse, and x and 2x + 4 being the legs of the triangle. Now I know how I can use the Pythagorean Theorem to solve for x. The Pythagorean Theorem states that in every right triangle with legs of length a and b and hypotenuse c, these lengths have the formula of a2 + b2 = c2. Let a = x, and b = 2x + 4, so that c = 2x + 6. Then, by putting these measurements into the Theorem equation we have x2 + (2x + 4)2 = (2x + 6)2. The binomials into the Pythagorean Thermo x2 + 4x2 + 16x + 16 = 4x2 + 24x + 36 are the binomials squared. Then 4x2 on both sides of the equation which can be (-4x2 -4x2) subtracted out first leaving the equation to be x2 + 16x + 16 = 24x + 36. Next we should subtract 16x from both sides of equation, which then leaves us with: x2 +16 = 8x + 36. The next step would then be to subtract 36 from both sides to get a result of. x2 -20= 8x. Finally we need to subtract 8x from both sides to get x2 – 8x...
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...Buried Treasure Allen Raikes MAT 221 DR. Steven Flanders Ahmed and Vanessa has a treasure that needs to be located. It’s up to me and to help find it, I will do that by using the Pythagorean quadratic. On page 371 we learned that the Ahmed has a half of the map and Vanessa has the other half. Ahmed half in say the treasure is buried in the desert 2x+6 paces from Castle Rock and Vanessa half says that when she gets to Castle Rock to walk x paces to the north, and then walk 2x+4 paces to the east. So with all the information I have I need to find x. the Pythagorean Theorem states that in every right triangle with legs of length a and b and hypotenuse c, which have of a relationship of a2+b2=c2. In this problem I will let a=x, and b= 2x+4, and c=2x+6. So know it time to put the measurements into the Theorem equation; 1) X2+ (2x+4)2=(2x+6)2 this is the Pythagorean Theorem 2) X2+4x2+16x+16 = 4x2+ 24x+36 are the binomials squared 3) 4x2 & 4x2 on both sides can be subtracted out. 4) X2+16x+16 = 24x +36 subtract 16x from both sides 5) X2+16 = 8x+36 now subtract 36 from both sides 6) X2-20 = 8x 7) X2-8x-20=0 this is the quadratic equation to solve by factoring using the zero factor. 8) (x-)(x+) Since the coefficient of x2 is 1 we have to start with pair of () is the 20 in negative there will be one + and one – in the binomials. 9) -2, 10: -10,2: -5,4; -4, -5 10) Looks I’m going to use -10 and 2 is...
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...Pythagorean Quadratic Melissa Hernandez MAT221: Introduction to Algebra Instructor Srabasti Dutta August 4, 2014 Pythagorean Quadratic Ever since I can remember when I was a little girl full of curiosity, I enjoyed the thought of finding a buried treasure and thus set out on treasure hunts with my sisters. Depending on how big your imagination is, you can take yourself to exotic locations, around town, or in your very own backyard. Finding buried treasures is a fun activity to do on your own or in a group. In this activity, we will be finding a buried treasure near Castle Rock with Ahmed and Vanessa. The assignment reads; Buried treasure. Ahmed has half of a treasure map, which indicates that the treasure is buried in the desert 2x + 6 paces from Castle Rock. Vanessa has the other half of the map. Her half indicates that to find the treasure, one must get to Castle Rock, walk x paces to the north, and then walk 2x + 4 paces to the east. If they share their information, then they can find x and save a lot of digging. What is x? (Dugopolski, 2012, p. 371) In this problem, we will use the Pythagorean Theorem which says that when a triangle has a right angle of 90 degrees, and squares are made on each of the three sides, then the biggest square has the exact same area as the other two squares put together. The Pythagorean Theorem can be written as: a^2 + b^2 = c^2 with “c” being the longest side or otherwise called the hypotenuse of the triangle, and “a” and “b” are the...
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...Pythagorean Quadratic Diane Todd MAT 221 Introduction to Algebra Instructor Alicia Davis September 29, 2013 Treasure hunts have always been a big deal in our home. Having raised five boys, anything to do with an adventure was exciting. Actually, this past June I planned one of my grandsons birthday parties around the theme of pirates and treasure hunting. I had never considered the math that went behind the maps in which I made up. Needless to say, when I saw the question entitled “buried treasure” in our math book, it brought back numerous memories. Ahmed has half of a treasure map, which indicates that the treasure is buried in the desert 2x + 6 paces from Castle Rock. Vanessa has the other half of the map. Her half indicates that to find the treasure, one must get to Castle Rock, walk x paces to the north, and then walk 2x + 4 paces to the east. If they share their information, then they can find x and save a lot of digging. What is x? Even though Ahmed’s half of the map does not tell him which direction the 2x + 6 paces should go, Ahmed can assume that his and Vanessa’s paces should end up in the same place. If I sketch out this scenario on paper, I see that I have a right triangle with 2x + 6 being the length of the hypotenuse, and 2x + 4 being the legs of the triangle. I now can use the Pythagorean Theorem to solve for x. The Pythagorean Theorem states that in every right triangle with legs of length a and b and hypotenuse of c, these...
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...Pythagorean Theorem: Finding Treasure Patricia Diggs MAT 221 Introduction to Algebra Instructor Bridget Simmons May 12, 2013 Pythagorean Theorem: Finding Treasure In this paper I will attempt to use the Pythagorean Theorem to solve the problem which reads Ahmed has half of a treasure map which indicates that the treasure is buried in the desert 2x+6 paces from Castle Rock. Vanessa has the other half of the map. Her half indicates that to find the treasure, one must get to Castle Rock, walk x paces to the north, and then walk 2x+4 paces to the east. If they share their information they can find x and save a lot of digging. What is x? The Pythagorean Theorem states that in every right triangle with legs the length a and b and hypotenuse c, these lengths have the relationship of a2 + b2=c2. a=x b=(2x+4)2 c=(2x+6)2 this is the binomials we will insert into our equation x2+(2x+4)2=(2x+6)2 the binomials into the Pythagorean Theorem x2+4x2+16x+16=24x36 the binomial squared. The 4x2can be subtracted out first x2+16x+16=24x+36 now subtract 24x from both sides x2+-8x+16=36 now subtract 36 from both sides x2-8x-20=0 this is a quadratic equation to solve by factoring and using the zero factor. (x- )(x+ ) the coefficient of x2 is one (1). We can start with a pair of parenthesis with an x each. We have to find...
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...the treasure, one must get to Castle Rock, walk x paces to the north, and then walk 2x + 4 paces to the east. If they share their information then they can find x and save a lot of digging. What is x? The key to solving this equation is to use the Pythagorean Theorem which has a right angle in the equation with a & b equaling side c. Based on the example we know that we will have to use the formula A^2 + B^2 = C^2. What we know is that Ahmed’s half of the map is 2x + 6 = treasure, which would be how far he would need to walk, (A^2). Vanessa’s map needs to show her how to get to the North of Castle Rock, which is 2x + 4, (C^2). We are trying to solve for X (side B). Vanessa is forming a 90 degree angle from point B and walk (2x + 4) until she made it to C. The formula we would use to solve for X is the following: (2x + 6)^2 = x^2 + (2x + 4)^2 4x^2 + 24x + 36 = x^2 +4x^2 + 16x + 16 The next step would be to combine like terms, multiplying, adding, and subtracting. 24x + 36 = x^2 + 16x + 16 -24x -24x 36 = x^2 – 8x + 16 -36 -36 Now we have a quadratic equation. Treasure = x^2 – 8x – 20. We have simplified the equation, now we can solve for X. Inside the quadratic equation, x^2 – 8x – 20 = 0, we can solve by factoring and using the zero factor. This is called the compound equation, because we know that the two factors we have are 10 and -2, so we can solve for each binomial. The equation is (x-a)(x+b)=0 or (x...
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...Term Paper On Role of the Pythagoras in the field of mathematics Business Mathematics code Submitted By Team Harmony 1. Faisal Enayet (B1506003) 2. HafijulHasan (B1506007) 3. Plato Khisa (B1506035) 4. FarhanajAnchal (B1506075) 5. K.HusFariha (B1506120) 6. SumaiyaMeher(B1506155) Submitted To Lecturer AKTER KAMAL Business Mathematics Bangladesh University of Professionals Submission on Date: 02/05/2016 BBA 2015; SEC- C LETTER OF TRANSMITTAL 02 may 2016 Akter Kamal Lecturer Faculty of Business Studies Bangladesh University of Professionals Subject: Submission of term paper on “The role of Pythagoras in the field of mathematics” Respected Sir, We the students of BBA, section C, we are very glad to submit you the term paper on the topic of “The role of Pythagoras in the field of mathematics” that you asked us to submit, which is a part of our course requirement. For the purpose of completing the term paper we did a simple research on the provided topic. We have completed our research and assessment on our term paper topic according to your specification and regulation. We have tried our best to gather information according to the requirements and our ability. There may be a few mistakes, because we are still beginner in this line of work but we hope that in future this term paper will remind us not to make the same mistakes again and so this will become a great learning in experience. At last, we would like to thank to you...
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...[pic] COURSE SYLLABUS Semester: Fall, Year: 2012 College Mission Statement Richard J. Daley College provides high-quality education which leads to academic success, career development, and personal enrichment that fulfill diverse community needs. Mathematics Department Mission Statement Our mission is to deliver excellent service and to provide learning opportunities by offering a wide range of mathematics courses, which will help our diverse student population to reach their goals in their path of preference such as baccalaureate transfer, workforce development, adult or continuing education. Our dedicated faculty will guide our students in constructing the necessary elements that will help them succeed in their math classes, and also encourage and motivate them to participate in college wide activities. We are committed to preparing our students to be productive, contributing members of their community with problem solving and critical thinking skills. We provide our students with the motivation to use the power, beauty, and utility of mathematics to successfully prepare themselves for global citizenship. |Math 99 RW | |[pic] | |“Intermediate Algebra with Geometry” ...
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... | | |Axia College | | |MAT/117 Version 7 | | |Algebra 1B | Copyright © 2010, 2009, 2007 by University of Phoenix. All rights reserved. Course Description This course explores advanced algebra concepts and assists in building the algebraic and problem-solving skills developed in Algebra 1A. Students solve polynomials, quadratic equations, rational equations, and radical equations. These concepts and skills serve as a foundation for subsequent business coursework. Applications to real-world problems are also explored throughout the course. This course is the second half of the college algebra sequence, which began with MAT/116, Algebra 1A. Policies Faculty and students/learners will be held responsible for understanding and adhering to all policies contained within the following two documents: • University policies: You must be logged into the student website to view this document. • Instructor policies: This document is posted in the Course Materials forum. University policies are subject to change. Be sure to read the policies at the beginning...
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...TLFeBOOK WHAT READERS ARE SAYIN6 "I wish I had had this book when I needed it most, which was during my pre-med classes. I t could have also been a great tool for me in a few medical school courses." Or. Kellie Aosley8 Recent Hedical school &a&ate "CALCULUS FOR THE UTTERLY CONFUSED has proven to be a wonderful review enabling me t o move forward in application of calculus and advanced topics in mathematics. I found it easy t o use and great as a reference for those darker aspects of calculus. I' Aaron Ladeville, Ekyiheeriky Student 'I1am so thankful for CALCULUS FOR THE UTTERLY CONFUSED! I started out Clueless but ended with an All' Erika Dickstein8 0usihess school Student "As a non-traditional student one thing I have learned is the Especially in importance of material supplementary t o texts. calculus it helps to have a second source, especially one as lucid and fun t o read as CALCULUS FOR THE UTTERtY CONFUSED. Anyone, whether you are a math weenie or not, will get something out of this book. With this book, your chances of survival in the calculus jungle are greatly increased.'I Brad &3~ker, Physics Student Other books i the Utterly Conhrsed Series include: n Financial Planning for the Utterly Confrcsed, Fifth Edition Job Hunting for the Utterly Confrcred Physics for the Utterly Confrred CALCULUS FOR THE UTTERLY CONFUSED Robert M. Oman Daniel M. Oman McGraw-Hill New York San Francisco Washington, D.C. Auckland Bogoth Caracas Lisbon...
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...Algebra 1 Pacing Plan 2012-2013 1st Quarter Tools of Algebra 1.1 Using Variables 1.2 Exponents and Order of Operations 1.3 Exploring Real Numbers 1.4 Adding Real Numbers 1.5 Subtracting Real Numbers 1.6 Multiplying and Dividing Real Numbers (1.3 – 1.6 mini lessons based on need) 1.7 The Distributive Property 1.8 Properties of Numbers Solving Equations and Inequalities 2.1 Solving One-Step and Two-Step Equations 3.1 Inequalities and Their Graphs 3.2 Solving Inequalities using Addition and Subtraction 3.3 Solving Inequalities using Multiplication and Division 2.2 Solving Multi-Step Equations 3.4 Solving Multi-Step Inequalities 2.4 Ratios & Proportions (mini lessons) 3.5 Compound Inequalities 3.6 Absolute Value Equations and Inequalities 2.5 Equations and Problem Solving 2.6 Mixture Problems and Work Problems California Content Standards 1.0, 1.1, 2.0, 4.0, 10.0, 24.1, 24.3, 25.0, 25.1, 25.2 2.0, 3.0, 4.0, 5.0, 24.2, 24.3, 25.0, 25.2, 25.3 2nd Quarter Linear Equations and Their Graphs 4.1 Graphing on the Coordinate Plane 5.1 Rate of Change and Slope 5.1.1 Graphing Using Input-Output Table (Supplemented) 5.1.2 Graphing Using Slope-Intercept Form (Supplemented) 5.1.3 Graphing Using...
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