2013 amc10b. 2013 AMC 10B (Problems • Answer Key • Resources) Preceded by Pro...

2012 AMC10B Solutions 2 1. Answer (C): There are 18−

The test was held on Wednesday, February 5, 2020. 2020 AMC 10B Problems. 2020 AMC 10B Answer Key. Problem 1. Problem 2. Problem 3. Problem 4.17 Kas 2022 ... AMC10B 1st = Leo Ahtaridis 2nd = Junwon Park 3rd = Jack Berrien. AMC12B ... Copyright ©2013-2015. All rights reserved. Contact | Directions ...2019 AMC 10B Problems and Answers. The 2019 AMC 10B was held on Feb. 13, 2019. Over 490,000 students from over 4,600 U.S. and international schools attended the contest and found it very fun and rewarding. Top 20, well-known U.S. universities and colleges, including internationally recognized U.S. technical institutions, ask for AMC scores on ...Are you looking for the problems and solutions of the 2019 AMC 10B, a prestigious math contest for students in grades 10 and below? Visit the Art of Problem Solving wiki page to find them, along with other useful resources and tips.2012 AMC10B Problems 2 1. Each third-grade classroom at Pearl Creek Elementary has 18 students and 2 pet rabbits. How many more students than rabbits are there in all 4 of the third-grade classrooms? (A) 48 (B) 56 (C) 64 (D) 72 (E) 80 2. A circle of radius 5 is inscribed in a rectangle as shown. The ratio of the length of the rectangle to its ...Jan 1, 2021 · 7. 2013 AMC 10B Problem 24: A positive integer n is nice if there is a positive integer m with exactly four positive divisors (including 1 and m) such that the sum of the four divisors is equal to n. How many numbers in the set {2010, 2011, 2012, ..., 2019} are nice? Resources Aops Wiki 2015 AMC 10B Problems Page. Article Discussion View source History. Toolbox. Recent changes Random page Help What links here Special pages. Search. GET READY FOR THE AMC 10 WITH AoPS Learn with outstanding instructors and top-scoring students from around the world in our AMC 10 Problem Series online course.Amc 10b 2013 Art Of Problem Solving. 1905 women's cooking class at the University of Arizona. Source: University of Arizona, "UA History & Traditions: 1905. The Cooking Class at Work," arizona.edu (accessed Sep. 26, 2013) The College of Education Manages programs that will help shape the future of FAMU! The Art Institute campuses are ...2006 AMC 12B. 2006 AMC 12B problems and solutions. The first link contains the full set of test problems. The rest contain each individual problem and its solution. 2006 AMC 12B Problems. Answer Key. Problem 1.Solution. Let the population of the town in 1991 be p^2. Let the population in 2001 be q^2+9. Let the population in 2011 be r^2. 141=q^2-p^2= (q-p) (q+p). Since q and p are both positive integers with q>p, (q-p) and (q+p) also must be positive integers. Thus, q-p and q+p are both factors of 141.AMC 10B DO NOT OPEN UNTIL Thursday, February 15, 2018 **Administration On An Earlier Date Will Disqualify Your School’s Results** 1. All the information needed to administer this exam is contained in the AMC 10/12 Teacher’s Manual. PLEASE READ THE MANUAL BEFORE February 15, 2018. 2.Solving problem #18 from the 2013 AMC 10B test.Solution 3. Since there are an equal number of juniors and seniors on the debate team, suppose there are juniors and seniors. This number represents of the juniors and of the seniors, which tells us that there are juniors and seniors. There are juniors and seniors in the program altogether, so we get Which means there are juniors on the debate ...The 2021 AMC 10C will be administered in November 2021. Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10C&oldid=152970"A. Use the AMC 10/12 Rescoring Request Form to request a rescore. There is a $35 charge for each participant's answer form that is rescored. The official answers will be the ones blackened on the answer form. All participant answer forms returned for grading will be recycled 80 days after the AMC 10/12 competition date.Radius of new jar = 1 + 1/4. Area of new base = pi * (1 + 1/4) ^ 2. Suppose new height = x * old height. Old Volume = New Volume = area of base * height. h = (1 + 1/4) ^ 2 * x * h. x = 1 / (1 + 1/4) ^ 2 = 16/25. Comparing x*h with h, we see the difference is 9/25, or 36%. The key to not get confused is to understand that if a value x has ...As the unique mode is 8, there are at least two 8s. Suppose the largest integer is 15, then the smallest is 15-8=7. Since mean is 8, sum is 8*8=64. 64-15-8-8-7 = 26, which should be the sum of missing 4 numbers.The first link contains the full set of test problems. The rest contain each individual problem and its solution. 2002 AMC 10B Problems. 2002 AMC 10B Answer Key. 2002 AMC 10B Problems/Problem 1. 2002 AMC 10B Problems/Problem 2. 2002 AMC 10B Problems/Problem 3. 2002 AMC 10B Problems/Problem 4.Resources Aops Wiki 2013 AMC 10A Page. Article Discussion View source History. Toolbox. Recent changes Random page Help What links here Special pages. Search. 2013 AMC 10A. 2013 AMC 10A problems and solutions. The test was held on February 5, 2013. ... 2013 AMC 10B: 1 ...Timestamps for questions0:06 212:32 224:09 239:20 2412:58 25Problems and Answers with detailed solutions, 美国数学竞赛AMC10,历年真题,视频完整讲解,真题解析,视频讲解 ...(AMC 10A 2013) Central High School is competing against Northern High School in a backgammon match. Each school has three players, and the contest rules ... (2006 AMC10B) In a triangle with integer side lengths, one side is three times as long as a second side, and the length of the third side is 15. ...The United States of America Mathematical Olympiad (USAMO) is a highly selective high school mathematics competition held annually in the United States.Since its debut in 1972, it has served as the final round of the American Mathematics Competitions.In 2010, it split into the USAMO and the United States of America Junior Mathematical Olympiad (USAJMO).2004 AMC 10A, AMC 10B, AMC 12A & AMC 12B by Gender AMC 10A & AMC 10B AMC 12A & AMC 12B by Grade AMC 10A ...Resources Aops Wiki 2015 AMC 10B Problems Page. Article Discussion View source History. Toolbox. Recent changes Random page Help What links here Special pages. Search. GET READY FOR THE AMC 10 WITH AoPS Learn with outstanding instructors and top-scoring students from around the world in our AMC 10 Problem Series online course.The test was held on Tuesday, November , . 2021 Fall AMC 10B Problems. 2021 Fall AMC 10B Answer Key. Problem 1. Problem 2. Problem 3. Problem 4. AMC10 strategy and problem solving skills. Specifically overcoming the fear of the unfamiliar problem type in Math Competitions.EDIT: so somehow I forgot I h...Created Date: 2/8/2020 5:27:00 PMThe first link contains the full set of test problems. The rest contain each individual problem and its solution. 2006 AMC 10A Problems. 2006 AMC 10A Answer Key. 2006 AMC 10A Problems/Problem 1. 2006 AMC 10A Problems/Problem 2. 2006 AMC 10A Problems/Problem 3. 2006 AMC 10A Problems/Problem 4.If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100? Solution. The nth item for the sequence is: An=An-1+4n. We add increasing multiples of 4 each time we go up a figure. So, to go from Figure 0 to 100, we add. 4 *1+4*2+...+4*99+4*100=4*5050=20200.(2013 AMC10B Question 21) Two non-decreasing sequences of non-negative integers have. different first terms. Each sequence has the property that each term beginning with the third is the. sum of the previous two terms, and the seventh term of each sequence is N . What is the smallest.2021 AMC 10A The problems in the AMC-Series Contests are copyrighted by American Mathematics Competitions at Mathematical Association of America (www.maa.org).2007 AMC 10B - Art of Problem Solving. This page contains the problems and solutions of the 2007 AMC 10B, a 25-question, 75-minute multiple choice test for students in grades 10 and below. The test covers topics such as algebra, geometry, number theory, and combinatorics. If you are looking for a challenge and want to improve your problem-solving skills, check out this page and see how you ... 2016 AMC 12A problems and solutions. The test was held on February 2, 2016. 2016 AMC 12A Problems. 2016 AMC 12A Answer Key. Problem 1. Problem 2. Problem 3. Problem 4. Problem 5.2007 AMC 10B - Art of Problem Solving. This page contains the problems and solutions of the 2007 AMC 10B, a 25-question, 75-minute multiple choice test for students in grades 10 and below. The test covers topics such as algebra, geometry, number theory, and combinatorics. If you are looking for a challenge and want to improve your problem …Resources Aops Wiki 2012 AMC 10B Problems Page. Article Discussion View source History. Toolbox. Recent changes Random page Help What links here Special pages. Search. GET READY FOR THE AMC 10 WITH AoPS ... 2013 AMC 10A Problems: 1 ...LeRoy and Bernardo went on a week-long trip together and agreed to share the costs equally. Over the week, each of them paid for various joint expenses such as gasoline and car rental. At the end of the trip, it turned out that LeRoy had paid dollars and Bernardo had paid dollars, where . How many dollars must LeRoy give to Bernardo so that ...Amc 10b 2013 Art Of Problem Solving, Curriculum Vitae Da Compilare E Scaricare, Essay Just Lather Thats All, Professional Blog Post Ghostwriting Sites Ca, Write My Best Creative Essay On Hacking, How To Write A Memorandum University, Top School Thesis Proposal SampleThe test was held on February 20, 2013. 2013 AMC 12B Problems. 2013 AMC 12B Answer Key. Problem 1. Problem 2. Problem 3. 2013 AMC 12A (Problems • Answer Key • Resources) Preceded by 2012 AMC 12A, B: Followed by 2013 AMC 12B,2014 AMC 12A, B: 1 ... 2013 AMC 10B Problems/Problem 25. The following problem is from both the 2013 AMC 12B #23 and 2013 AMC 10B #25, so both problems redirect to this page. Contents.Problem. What is the sum of all the solutions of ?. Solution. We evaluate this in cases: Case 1. When we are going to have .When we are going to have and when we are going to have .Therefore we have .. Subcase 1 . When we are going to have .When this happens, we can express as .Therefore we get .... AMC10B,https://artofprobl . ... 2013AMC12A、AMC10B、AMC12B、AIME,https://math.pro/db/thread-1532-1 ...The test was held on February 17, 2016. 2016 AMC 12B Problems. 2016 AMC 12B Answer Key. Problem 1. Problem 2. Problem 3. Problem 4.2012 AMC 10A. 2012 AMC 10A problems and solutions. The test was held on February 7, 2012. 2012 AMC 10A Problems. 2012 AMC 10A Answer Key. Problem 1. Problem 2. Problem 3.Since after B's trip, the 2 circles have the points of tangency, that means A's circumference is an integer multiple of B's, ie, 2*100*pi/2*r*pi = 100/r is an integer, or r is a factor of 100. 100=2^2*5^2, which means 100 has (2+1) (2+1) = 9 factors. 100 itself is one of the 9 factors, which should be excluded otherwise B = A. So the answer is 8.Jan 10, 2021 · Solving problem #19 from the 2013 AMC 10B test. 2011 AMC 10A. 2011 AMC 10A problems and solutions. The test was held on February 8, 2011. The first link contains the full set of test problems. The rest contain each individual problem and its solution. 2011 AMC 10A Problems. 2013 AMC10B Solutions 6 Note that the quadratic equation x2 +(4¡2 p 3)x+7¡4 p 3 satisfles the given conditions. 20. Answer (B): The prime factorization of 2013 is 3 ¢ 11 ¢ 61. There must be a factor of 61 in the numerator, so a1 ‚ 61. Since a1! will have a factor of 59 and 2013 does not, there must be a factor of 59 in the denominator ... The straight lines will be joined together to form a single line on the surface of the cone, so 10 will be the slant height of the cone. The curve line will form the circumference of the base. We can compute its length and use it to determine the radius. The length of the curve line is 252/360 * 2 * pi *10 = 14 * pi.Solution 2. Since A-B and A+B must have the same parity (both odd or both even), and since there is only one even prime number (number 2), it follows that A-B and A+B are both odd. Since A+B is odd, one of A, B is odd and the other is even, ie prime even 2.The first link contains the full set of test problems. The rest contain each individual problem and its solution. 2006 AMC 10A Problems. 2006 AMC 10A Answer Key. 2006 AMC 10A Problems/Problem 1. 2006 AMC 10A Problems/Problem 2. 2006 AMC 10A Problems/Problem 3. 2006 AMC 10A Problems/Problem 4.2013 AMC 10B Printable versions: Wiki • AoPS Resources • PDF: Instructions. This is a 25-question, multiple choice test. Each question is followed by answers ...欢迎使用瀚海学院为大家整理的历年AMC10真题模考(附2020年AMC10真题). 2020 AMC10A ... 2013 AMC10B 真题 下载 · 2012 AMC10A 真题 下载 · 2012 AMC10B 真题 下载 · 2011 ...Solution 2. Note that we can divide the polynomial by to make the leading coefficient 1 since dividing does not change the roots or the fact that the coefficients are in an arithmetic sequence. Also, we know that there is exactly one root so this equation must be of the form where . We now use the fact that the coefficients are in an arithmetic ...The test was held on February 7, 2018. 2018 AMC 10A Problems. 2018 AMC 10A Answer Key. Problem 1. Problem 2. Problem 3. Problem 4.AMC B The proles i the AMC -Series Cotests are opyrighted y Aeri a Matheais Copeiios at Matheaial Assoiaio of Aeri a Á Á.aa.org. For ore praie ad resour es, isit zil.aretee.orgAMC 10 Problems and Solutions. AMC 10 problems and solutions. Year. Test A. Test B. 2022. AMC 10A. AMC 10B. 2021 Fall. 2011 AMC 10A. 2011 AMC 10A problems and solutions. The test was held on February 8, 2011. The first link contains the full set of test problems. The rest contain each individual problem and its solution. 2011 AMC 10A Problems.The first link contains the full set of test problems. The rest contain each individual problem and its solution. 2007 AMC 10B Problems. Answer Key. 2007 AMC 10B Problems/Problem 1. 2007 AMC 10B Problems/Problem 2. 2007 AMC 10B Problems/Problem 3. 2007 AMC 10B Problems/Problem 4. 2007 AMC 10B Problems/Problem 5.View 2013 AMC 10B.pdf from MATH BC at Seven Lakes High School. 10/22/2017 Art of Problem Solving 2013 AMC 10B Problems Contents 1 Problem 1 2 Problem 2 3 Problem 3 4 Problem 4 5 Problem 5 6 Problem(2013-amc10b-23) let n be a positive integer greater than 4 such that the decimal representation of n! ends in k zeros and the decimal representation of (2n)! ends in 3k zeros. let s denote the sum of the four least possible values of n. what is the sum of digits of s?2003 AMC 10B Answer Key 1. C 2. D 3. B 4. A 5. C 6. D 7. B 8. B 9. B 10. C 11. A 12. C 13. E 14. D 15. E 16. E 17. B 18. D 19. E 20. D 21. C 22. B 23. D 24. E 25. B . THE *Education Center AMC 10 2003 A clock chimes once at 30 minutes past each hour and chimes on the hour according to the hour. For example, at 1 PM there is one chime and at ...THE *Education Center AMC 10 2012 A bug travels from A to B along the segments in the hexagonal lattice pictured below. The segments marked with an arrow can be traveled only in the direction of the arrow, and the2016 AMC 10 9 All three vertices of 4 ABC lie on the parabola de ned by y = x 2, with A at the origin and BC parallel to the x -axis. The area of the triangle is 64.Resources Aops Wiki 2012 AMC 10B Page. Article Discussion View source History. Toolbox. Recent changes Random page Help What links here Special pages. Search. 2012 AMC 10B. 2012 AMC 10B problems and solutions. The test was held on February 22, 2012. ... 2013 AMC 10A, B: 1 ...Amc 10b 2013 Art Of Problem Solving. 1 Customer reviews. ID 19300. Essay (Any Type), Biology (and other Life Sciences), 7 pages by Mitrofan Yudin. Emery Evans. #28 in Global Rating. 630. Finished Papers.2013 AMC 12A (Problems • Answer Key • Resources) Preceded by 2012 AMC 12A, B: Followed by 2013 AMC 12B,2014 AMC 12A, B: 1 ...School Search for AMC Contest Proctoring. This search page can assist in finding a school where you could take one of the AMC contests, displaying those who registered for the past or upcoming contest. You are responsible for contacting the school to see if they are willing to accept you to take the contest at their location. All contest ...Resources Aops Wiki 2022 AMC 10B Problems Page. Article Discussion View source History. Toolbox. Recent changes Random page Help What links here Special pages. Search. GET READY FOR THE AMC 10 WITH AoPS Learn with outstanding instructors and top-scoring students from around the world in our AMC 10 Problem Series online course.Problem 1. A taxi ride costs $1.50 plus $0.25 per mile traveled. How much does a 5-mile taxi ride cost? Solution. Problem 2. Alice is making a batch of cookies and needs cups of sugar. Unfortunately, her measuring cup holds only cup of sugar. How many times must she fill that cup to get the correct amount of sugar?The number 2013 has the property that its units digit is the sum of its other digits, that is 2 + 0 -l- 1 = 3. How many integers less than 2013 but greater than 1000 share this property? (A) 33 (B) 34 (C) 45 (D) 46 (E) 58 The real numbers c, b, a form an arithmetic sequence with a > b > c > 0. The quadratic a:r2 + + c has exactly one root. https://ivyleaguecenter.org/ Tel: 301-922-9508 Email: [email protected] Page 1 201 9 AMC 10 B Problem 1 Alicia had two containers. The first was2010. 188.5. 188.5. 208.5 (204.5 for non juniors and seniors) 208.5 (204.5 for non juniors and seniors) Historical AMC USAJMO USAMO AIME Qualification Scores.Solution. Let be the number of two point shots attempted and the number of three point shots attempted. Because each two point shot is worth two points and the team made 50% and each three point shot is worth 3 points and the team made 40%, or . Because the team attempted 50% more two point shots then threes, .AMC10 2014,MATH,CONTEST. We note that the 6 triangular sections in triangle ABC can be put together to form a hexagon congruent to each of the seven other hexagons (In the diagram I draw, the area of yellow triangle is same as 3 side triangles combined).2015 AMC 10B Problems/Problem 25; See also. 2015 AMC 10B (Problems • Answer Key • Resources) Preceded by 2014 AMC 10A, B: Followed by 2016 AMC 10A, B: 1 ... 2004 AMC 12A. 2004 AMC 12A problems and solutions. The test was held on Tuesday, February 10, 2004. The first link contains the full set of test problems. The rest contain each individual problem and its solution. 2004 AMC 12A Problems.A Mock AMC is a contest intended to mimic an actual AMC (American Mathematics Competitions 8, 10, or 12) exam. A number of Mock AMC competitions have been hosted on the Art of Problem Solving message boards. They are generally made by one community member and then administered for any of the other community members to take. Sometimes, the administrator may ask other people to sign up to write ...The number 2013 has the property that its units digit is the sum of its other digits, that is 2 + 0 -l- 1 = 3. How many integers less than 2013 but greater than 1000 share this property? (A) 33 (B) 34 (C) 45 (D) 46 (E) 58 The real numbers c, b, a form an arithmetic sequence with a > b > c > 0. The quadratic a:r2 + + c has exactly one root. View AMC 10B 2012 .pdf from MATHEMATIC 252 at Rutgers University. 12/4/2017 Art of Problem Solving 2012 AMC 10B Problems Problem 1 Each third-grade classroom at Pearl Creek Elementary has 18 students. Upload to Study. Expert Help. Study Resources. ... Preceded by 2012 AMC 10A Problems Followed by 2013 AMC 10A Problems 1 ...The first link contains the full set of test problems. The rest contain each individual problem and its solution. 2001 AMC 10 Problems. 2001 AMC 10 Answer Key. 2001 AMC 10 Problems/Problem 1. 2001 AMC 10 Problems/Problem 2. 2001 AMC 10 Problems/Problem 3. 2001 AMC 10 Problems/Problem 4. 2001 AMC 10 Problems/Problem 5.The test was held on Tuesday, November , . 2021 Fall AMC 10B Problems. 2021 Fall AMC 10B Answer Key. Problem 1. Problem 2. Problem 3. Problem 4.Resources Aops Wiki 2010 AMC 10B Page. Article Discussion View source History. Toolbox. Recent changes Random page Help What links here Special pages. Search. 2010 AMC 10B. 2010 AMC 10B problems and solutions. The test was held on February 24 th, 2010. The first link contains the full set of test problems. The rest contain each individual ...AMC10 strategy and problem solving skills. Specifically overcoming the fear of the unfamiliar problem type in Math Competitions.EDIT: so somehow I forgot I h...Solution 4. Assume WLOG that Elmer's old car's range is miles. So, Elmer's new car's range is miles. Also, assume that the gas Elmer's old car uses is , which means that diesel will cost . Now we can deduce that Elmer's old car uses per mile, and Elmer's new car uses per mile. Therefore, Elmer's new car saves more money than his old car.The first link contains the full set of test problems. The rest contain each individual problem and its solution. 2007 AMC 12B Problems. Answer Key. 2007 AMC 12B Problems/Problem 1. 2007 AMC 12B Problems/Problem 2. 2007 AMC 12B Problems/Problem 3. 2007 AMC 12B Problems/Problem 4. 2007 AMC 12B Problems/Problem 5.2015 AMC 10B 难题讲解 #20-25. 美国数学竞赛AMC10,历年真题,视频完整讲解。. 真题解析,视频讲解,不断更新中. 你的数学竞赛辅导老师。. YouTube 频道 Kevin's Math Class. 上线第一天,800%爆率_神装坐骑全靠打!. 破洞容器装水,答案居然不是过三个洞截面?. 泉州质检 ...2014 AMC10B Solutions 4 Notice that AE = 3 since AE is composed of a hexagon side (length 1) and the longest diagonal of a hexagon (length 2). Triangle ABE is 30-60-90 , so BE = √3 3 = √ 3. The area of ˚ABC is AE ·BE = 3 √ 3. 14. Answer (D): Let m be the total mileage of the trip. Then m must be a multiple of 55.Solution (s): Consider the following diagram where the lighter colored area makes up region S: The circles can be in only two locations. We first place the largest circle and then the second largest circle in the opposite location. After this, the circle of radius \ (3\) must be placed on one of the two sides.Amc 10b 2013 Art Of Problem Solving, Singer Speciesism Essay, How To Indent Paragraph In An Essay, Past And Present Life Essay, Argumentative Essay On Hamlet, Sea Cucumber Farming Business Plan, A compelling thesis: the thesis is the main premise of your argument. You want to tie together the entire essay with a common thread and then create a ...Amc 12b 2016 cutoff AMC10/12 Cutoff scores for AIME Qualification AMC 10A AMC 10B AMC 12A AMC 12B 2022 93 94.5 85.5 81 2021 Nov 96 96 91.5 84 2021 Feb 103.5 102 93 91.5 2020 103.5 102 87 87 2019 103.5 108 84 94.5 2018 111 108 93 99 2017 112.5 120 96 100.5 2016 110 110 92 100 2015 106.5 120 99 100 2014 120 120 93 100 2013 108 120 88.5 93 20122013 AMC 10B Printable versions: Wiki • AoPS Resources • PDF: Instructions. This is a 25-question, multiple choice test. Each question is followed by answers ... The 2022 AMC 10B neither featured exceedingly simple questions nor contained extremely difficult problems that were nearly unsolvable. In past exams, the first ten questions were generally straightforward, allowing most students to score points easily. However, in the 2022 exam, the difficulty of the initial ten questions increased, challenging ...If we can find this N, then the next number, N+1, will make P (N)<321/400. You can do a few tries as above (N=5, 10, 15, etc.), and you will see that the ball "works" in places. from 1 to 2/5 * N + 1, and places 3/5 * N +1 to N+1. This is a total of 4/5 * N + 2 spaces, over a total of N+1 spaces: (4/5 * N + 2)/ (N + 1) Let the above = 321/400 .... View AMC 10B 2012 .pdf from MATHEMATIC 252 at Rutgers UniversityResources Aops Wiki 2022 AMC 10B Problems Pa View 2013 AMC 10B.pdf from MATH 0277 at Obra D. Tompkins High School. AMC For B ore pra ti e a d resour es, isit zi l.aretee .org The pro le s i the AMC-Series Co tests are opyrighted y A eri a Mathe. Upload to Study. ... 2011 AMC 10B.pdf. Obra D. Tompkins High School. MATH 0277. Blue booth: give 3 blue, get 1 silver, 1 red. Supp 9. The knights in a certain kingdom come in two colors: 2 7 of them are red, and the rest are blue. Furthermore, 1 6 of the knights are magical, and the fraction of red knights who are magical is 2 times the fraction of blue knights 5 Eki 2020 ... 2012: Contest A - Problems | Solution...

Continue Reading