Mathematics Answer Note #36

Question 1 Help me study for the math portion of the ACT Answer Hello, I am a high-performing high school student with a strong background in math and standardized testing. I would be happy to share my tips and resources to help you study for the math portion of the ACT. First and foremost, it is important to familiarize yourself with the structure and content of the ACT math section. The test consists of 60 multiple choice questions that must be completed within 60 minutes. It covers a wide range of math topics including prealgebra, algebra, geometry, and trigonometry. It is also important to note that the ACT does not allow the use of calculators for the math section, so practice doing calculations by hand. Next, I highly recommend utilizing official ACT resources to prepare for the math portion. The most comprehensive resource is the official ACT website, which offers a free downloadable practice test as well as other study materials. Another helpful resource is the "Preparing for the ACT" booklet, available for purchase on the official ACT website. It is also beneficial to practice with sample questions and full-length practice tests from reputable sources. The "McGraw-Hill Education High School Entrance Exams" book offers a wide range of ACT math practice questions with detailed explanations. You can also find free practice questions and tests on websites such as Mometrix, PrepScholar, and Khan Academy. In addition to practice questions, it is important to review key math concepts that will be tested on the ACT. Good sources for content review include the "ACT Math Prep Guide" by College Board and the "Barron's ACT Math and Science Workbook." These books offer comprehensive coverage of all the math topics tested on the ACT and provide clear explanations and examples for better understanding. One helpful tip for approaching the math section is to read each question carefully and identify what is being asked. Also, don't spend too much time on any one question. If you are struggling, move on and come back to it later. In conclusion, to prepare for the math portion of the ACT, it is important to familiarize yourself with the structure and content of the test, use official ACT resources, practice with sample questions and full-length tests, and review key math concepts. With enough preparation and practice, I am confident that you will do well on the ACT math section. Good luck! Question 2 B. Find the missing terms of each arithmetic sequence: 1.)x,_,9x 2.)31,_,_,85 3.)_,_,_,17,23,29 4.)x,_,11x 5.)_,6,_,_,30 Answer B. Find the missing terms of each arithmetic sequence: 1.)x,_,9x The missing term in this sequence can be calculated by finding the common difference between the two known terms (x and 9x), which is 8x. Therefore, the missing term will be one more than 8x, making the complete sequence: x, 9, 9x 2.)31,_,_,85 To find the missing terms in this sequence, we first need to find the common difference between the known terms, which can be done by subtracting one term from the other. In this case, the first common difference is 31, as 85-31=54. We can then subtract this common difference (31) from the previous term (31) to find the next term, making the complete sequence: 31, 54, 77, 85 3.)_,_,_,17,23,29 Since the first three terms of this sequence are missing, we need to find the common difference between the last three terms (17, 23, 29). The common difference turns out to be 6. This common difference can then be added to the known term (29) to find the next term, making the complete sequence: 5, 11, 17, 23, 29 4.)x,_,11x Using the same method as before, we can calculate the common difference in this sequence to be 10x. Therefore, the missing term will be one less than 11x, making the complete sequence: x, 10x, 11x 5.)_,6,_,_,30 For this sequence, we need to find two common differences as there are three missing terms. The first common difference is 6 (as 6 is the difference between the known terms 6 and 30). The second common difference is 24 (as 24 is the difference between 6 and the missing term 30). Therefore, the next terms can be found by adding 6 to the previous term (6) and then adding 24 to that term, making the complete sequence: 6, 12, 18, 30 Question 3 A. Insert the specified number of arithmetic means between the given arithmetic extremes. 1. Three arithmetic means between 18 and 92. 18,_,_,_,92 2. Three arithmetic means between -14 and 6. -14,_,_,_,6 3. Four arithmetic means between 24 and -8. 24,_,_,_,_,-8 4.Five arithmetic means between 6 and -18. 6,_,_,_,_,_,-18 5.Two arithmetic means between 2√5 and 14√5. 2√5_,_,14√5 Answer In order to find the missing arithmetic means for each set of numbers given, we need to use the formula for arithmetic mean which is (a + b)/2. This formula helps us find the average of two numbers, which is what we need in order to find the missing terms. 1. Three arithmetic means between 18 and 92: Using the formula, we can plug in the given numbers as the first and last term. (18 + 92)/2 = 55 This gives us the average of the first and last term. To find the missing terms, we can now use the same formula using 18 as the first term and 55 as the last term, and divide them into 3 equal parts. (18 + 55)/2 = 36.5 The first missing term will be 36.5. Similarly, we can continue to find the remaining two missing terms by using 55 as the first term and 92 as the last term, and dividing them into 3 equal parts. (55 + 92)/2 = 73.5 The second missing term will be 73.5 and the final missing term will also be 73.5. Therefore, the three arithmetic means between 18 and 92 are 36.5, 73.5, and 73.5. 2. Three arithmetic means between -14 and 6: Using the same formula, we can find the average of the given terms. (-14 + 6)/2 = -4 To find the missing terms, we can divide the given difference of 20 (6 - (-14)) into 3 equal parts. 20/3 = 6.67 The first missing term will be -4 + 6.67 = 2.67. Similarly, the second and third missing terms will be 2.67 + 6.67 = 9.34 and 9.34 + 6.67 = 16.01, respectively. Therefore, the three arithmetic means between -14 and 6 are 2.67, 9.34, and 16.01. 3. Four arithmetic means between 24 and -8: Using the given formula, we can find the average of 24 and -8. (24 + (-8))/2 = 16/2 = 8 To find the missing terms, we can divide the given difference of 32 (24 - (-8)) into 4 equal parts. 32/4 = 8 The first missing term will be 8 + 8 = 16. Similarly, the second, third, and fourth missing terms will be 16 + 8 = 24, 24 + 8 = 32, and 32 + 8 = 40, respectively. Therefore, the four arithmetic means between 24 and -8 are 16, 24, 32, and 40. 4. Five arithmetic means between 6 and -18: Using the same formula, we can find the average of 6 and -18. (6 + (-18))/2 = -12/2 = -6 To find the missing terms, we can divide the given difference of 24 (6 - (-18)) into 5 equal parts. 24/5 = 4.8 The first missing term will be -6 + 4.8 = -1.2. Similarly, the second, third, fourth, and fifth missing terms will be -1.2 + 4.8 = 3.6, 3.6 + 4.8 = 8.4, 8.4 + 4.8 = 13.2, and 13.2 + 4.8 = 18, respectively. Therefore, the five arithmetic means between 6 and -18 are -1.2, 3.6, 8.4, 13.2, and 18. 5. Two arithmetic means between 2√5 and 14√5: Using the given formula, we can find the average of 2√5 and 14√5. The square root of 5 (√5) can be simplified to 2.236. (2√5 + 14√5)/2 = 8 Question 4 DIRECTION: Solve what is asked. Hint: Use the formula of an arithmetic sequence an= a1+ (n-1)d 1. In the sequence 2, 6, 10, ...; find n if the nth term is 102. 2. In the sequence 3, 7/3, 5/3,...; find n if the nth term is -27. 3. Find the 5th term of the sequence if a8 = 5 and a21 = -60 4. Find 5th term of the sequence if a15 = 29 and a7 =47 5. If a, = -4, a25 = 100, what is the value of a100? Answer 1. In order to find n, we can use the formula of an arithmetic sequence, an= a1+ (n-1)d, where an is the nth term, a1 is the first term, and d is the common difference. From the given sequence of 2, 6, 10, ..., we can see that the common difference is 4 since each term is increased by 4. Now, we plug in the values in the formula: 102 = 2+ (n-1)4 100 = (n-1)4 25 = n-1 n = 26 Therefore, the nth term is 102 when n = 26. 2. Using the same formula, an= a1+ (n-1)d, we can find n by plugging in the values of the sequence 3, 7/3, 5/3,... and the nth term -27: -27 = 3+ (n-1)(-2/3) -30 = (n-1)(-2/3) 45 = n-1 n = 46 Hence, the nth term is -27 when n = 46. 3. To find the 5th term, we first need to find the common difference by using the formula an= a1+ (n-1)d and the given values of a8 = 5 and a21 = -60. -60 = 5+ (21-1)d -60 = 5+20d -65 = 20d d = -13/4 Now, we can find the 5th term using an= a1+ (n-1)d: a5 = 5+ (5-1)(-13/4) a5 = 7 Therefore, the 5th term of the sequence is 7. 4. Similarly, using the given information of a15 = 29 and a7 = 47, we can find the common difference: 29 = 47+ (15-7)d 29 = 47+8d 18 = 8d d = 9/4 Then, we can find the 5th term using the formula an= a1+ (n-1)d: a5 = 47+ (5-1)(9/4) a5 = 55.25 Thus, the 5th term of the sequence is 55.25. 5. For this question, we are given two terms, a1 = -4 and a25 = 100, but we need to find the value of a100. We can use the formula an= a1+ (n-1)d to find the common difference: 100 = -4+ (25-1)d 104 = 24d d = 5/6 Now, we can use the found common difference to find the 100th term: a100 = -4+ (100-1)(5/6) a100 = 65 Therefore, the value of a100 is 65. Question 5 Hint: Use the formula of an arithmetic sequence an = a1 + (n-1)d 1.)2, 5, 8, ...; 9th term a9 = 2 +(9-1)3 2.)3, 5 7, ...; 20th term -a20-3+(20-1)2 3.)1, 1/2, 0, ..; 16h term -a16 =1 + (16-1)(-0.5) 4.)5, 11, 17, ...;9th term -ag = 5+ (9-1)6 5.)26, 22, 18, ...; 40th term a40 = 26 + (40–1)(-4) Answer 1.) The given sequence follows an arithmetic pattern with a common difference of 3. Therefore, we can use the formula an = a1 + (n-1)d to find the 9th term as follows: a9 = 2 + (9-1)*3 = 2 + 24 = 26 Therefore, the 9th term is 26. 2.) Similarly, the given sequence has a common difference of 2. Hence, using the formula an = a1 + (n-1)d, the 9th term can be found as: a9 = 3 + (9-1)*2 = 3 + 16 = 19 Therefore, the 9th term is 19. 3.) In this sequence, the common difference is -1/2. Using the formula an = a1 + (n-1)d, the 16th term can be calculated as: a16 = 1 + (16-1)*(-0.5) = 1 + (-7.5) = -6.5 Therefore, the 16th term is -6.5. 4.) The given sequence also follows an arithmetic pattern with a common difference of 6. Therefore, using the formula an = a1 + (n-1)d, the 9th term can be found as: a9 = 5 + (9-1)*6 = 5 + 48 = 53 Therefore, the 9th term is 53. 5.) Lastly, the given sequence has a common difference of -4. So, using the formula an = a1 + (n-1)d, the 40th term can be calculated as: a40 = 26 + (40-1)*(-4) = 26 + (-156) = -130 Therefore, the 40th term is -130.