Algebra 2 Segment 2 Exam Review (V8)

Algebra 2 Segment 2 Exam Review (v8)

1. Describe how to determine whether a sequence is arithmetic or not. Give an example and explain WHY is constitutes an ARITHMETIC sequence. 2. Give the formula used to find a specific number term in an arithmetic sequence. Tell what each variable represents and then use the formula to find the 8th term of the arithmetic sequence 1, 4, 7, 10, 13, … 3. Describe how to determine whether a sequence is geometric or not. Give an example and explain WHY is constitutes an GEOMETRIC sequence. 4. The tribe on the show “Survivor” is down to 7 members. For the next challenge, they will compete one at a time. How many different ways can they be arranged for the competition? 5. Determine the probability of being born in a month that starts with the letter “M.” 6. In how many different ways can you arrange the letters of the word “Yesterday” taken 4 at a time? 7. Describe the differences between the mean, mode and the median of a set of data. When are they equivalent if ever? 8. The tribe on the show “Survivor” is down to 7 members. For the next challenge, only 3 will compete. How many different combinations are possible? 9. Find the probability of choosing two vowels from the word “trigonometry”, one at a time without replacing. Let “y” be considered a vowel. 10. Which central tendency below is the greatest in the data {1, 4, 4, 7, 8, 10, 22}? 11. What is the difference between an even and odd function? 8 12. What is the horizontal asymptote of f(x)= - 3? x- 7 13. The equation 6x2 + 11x - 10 = 0 has how many solutions and what type of solutions are they? 14. What are the possible solutions to the equation 5x3 – 4x2 + 7x – 15=0? x2 - 4x + 4 15. Simplify the expression . x2 - 4 16. Solve 3 x- 2 = 9for x. x+ 5 x + 5 x + 5 17. Find the LCD of: , , x2- 9 x 2 + 6x + 9 x - 3 18. Solve the equation x3- 5x 2 - 2x + 24 = 0and list all the solutions. 19. Find the remainder when dividing 2x3 - 3x + 5by x-2 4x- 16 3x+ 6 20. Multiply and 3x+ 12 3x- 12 8x2 24x4 21. Simplify divided by 10y 14y x+ 9 3x - 7 22. Add + . 7y 7y 23. List the possible rational zeros for the polynomial P(x)= 5x3 + 10x 2 - 6x + 40 24. If you find that you have a complex number in the form a + bi as a solution to a polynomial equation, is the complex conjugate of that number also a solution? 25. How many positive real zeros are there for the polynomial P(x)= 2x3 - 3x 2 + 6x - 8 26. Describe how to find the equation of the vertical asymptote(s) of a rational expression similar x- 5 to, but not limited to . x2 - 49

27. Solve for x: 2log5 (2x)- log 5 (8) = log 5 (x) 28. Every part of a fractional exponent tells you something about the radical form. Describe how a to write yb in radical form. Which part becomes the radical index, which becomes the radicand, which becomes the exponent on the radicand? 29. Solve 3x = 27 Algebra 2 Segment 2 Exam Review (v8)

30. Rewrite 2x = 3in logarithmic form to solve for x. Round your answer to two decimal places. 31. Solve 52x= 25 3x- 4 for x without using logarithms.

32. Solve log7 (x+ 9) - log 7 (x - 3) = 5 for x using the properties of logarithms. 33. Rewrite 24 = 16in logarithmic form.

34. Solve log6 (x- 2) = 2 for x using the properties of logarithms.

35. Simplify log5 (a)+ log 5 (x + y) using the properties of logarithms. 3 36. Solve 3x = 98 for x. 1 37. Simplify 2log (x)- log (b) + log (w) using the properties of logarithms. 82 8 8 38. Give three examples of exponential functions and three examples of NON-exponential functions. 39. Consider the exponential function g(x)= 7x . If the domain consists of all real numbers, what is the range? 40. Find the twenty-fourth term in the sequence 1, -2, -5, -8, -11, … 41. Find the sum of the first 100 terms of the arithmetic sequence -13, -9, -5, -1, 3, … 7 42. Find 2i- 7 i= 1 1 43. Find the sum of a geometric sequence whose first term is -125, whose last (nth) term is - , 5 1 and whose common ratio is - . 5 44. There are 17 students in an essay contest. In how many different ways can the 1st, 2nd and 3rd place awards be arranged? Algebra 2 Segment 2 Exam Review (v8)

Answers: 1. Each term in an arithmetic sequence, after the first is found by adding (or subtracting since subtraction is just addition of a negative) the same fixed number to each of the previous terms. In the example 5, 2, -1, -4, … is an arithmetic sequence because to get from 5 to 2, you subtract 3; to get from 2 to -1, you subtract 3; to get from -1 to -4, you subtract 3, etc. So, the common difference is -3.

2. FORMULA: an = a1 + (n-1)d an = the term value you are looking for; a1 = the first term in the sequence; n = the number of the term you are looking for; d = the common difference between each term. CALCULATION: The 8th term of the arithmetic sequence 1, 4, 7, 10, 13, …

an = what we are looking for, or a8

a1 = 1 n = 8 d = +3

a8 = a1 + (n-1)d

a8 = 1 + (8-1)(3)

a8 = 1 + 21

a8 = 22 3. Each term in a geometric sequence, after the first is found by multiplying (or dividing since division is just multiplication of the reciprocal) the same fixed number to each of the previous terms. In the example -1, 4, -16, 64, … is a geometric sequence because to get from -1 to 4 by multiplying by -4; you get from 4 to -16 by multiplying by -4; you get from -16 to 64 by multiplying by -4, etc. So, the common ratio is -4. 4. Using the fundamental counting principal, because order does not matter would be: 7! 7! = 7 X 6 X 5 X 4 X 3 X 2 X 1 = 5040 5. There are 12 months, 2 of them start with the letter “M,” so the probability of being born in either March or May would be: 2/12 which must be simplified to 1/6. 6. This is a permutation, there are 9 letters in the word yesterday and we are taking them 4 at a n! time. So, we need to calculate: P(9, 4) using the formula P(n,r) = (n- r)! 9! 9! 9鬃 8 7 鬃 6 5 鬃 4 3 鬃 2 1 P(9, 4) = = = = 9鬃 8 7 6 = 3024 (9- 4)! 5! 5鬃 4 3 鬃 2 1 7. The mean is the average of all numbers. The median is the dead center of the data, or the average of the two middle terms. The mode is the value that occurs most often. It is possible for all three to be equivalent, but they are not always. 8. This is a combination, there are 7 choices and we will pick 3. So we need to calculate C(7, 3) n! using the formula C(n, r) = r!(n- r)! 7! 7! 7鬃 6 5 鬃 4 3 鬃 2 1 7 鬃 6 5 210 C(7, 3) ======35 3!(7- 3)! 3!(4)! 3鬃 2 1(4 鬃 3 2 鬃 1) 3 2 1 6 Algebra 2 Segment 2 Exam Review (v8)

5 9. The probability of picking a vowel the first time is . That leaves 11 letters left, 4 of which 12 4 are still vowels, so the probability of picking a vowel this time is . That makes the 11 probability of choosing two vowels from the word “trigonometry”, one at a time without 5 4 20 10 5 replacing � = = 12 11 132 66 33 10. Mean = 8; Median = 7; Mode = 4. So, the mean is the greatest central tendency. 11. Odd functions have exponenbts that are NOT divisible by 2. Even functions have exponents that ARE divisible by 2. x 12. This function is in the form f(x)= + a, which means that the horizontal asymptote is at y = x- b a. So, the horizontal asymptote ius at y = -3 13. The discriminant is b2- 4ac = 11 2 - 4(6)( - 10) = 121 + 240 = 361. 361 is positive which means there are two real solutions. It is also a perfect square which means that those two solutions are integer solutions. So, the best answer is there are two real integer solutions. 14. Find the factors of 5: 1, 5 and the factors of 15: 1,3,5,15. Then find the fraction combinations factors of 15 1 1 3 3 5 5 15 15 of . Those would be , , , , , , , . Now you simplify each to get: factors of 5 1 5 1 5 1 5 1 5 1 3 1 3 1, ,3, ,5,1,15,3 Lastly, you do not need to list repeats, so your final list would be: 1, ,3, ,5,15 5 5 5 5 15. Name restrictions by setting denominator equal to zero and solving for x. x构 2,x- 2Factor the (x- 2)(x - 2) (x- 2) top and bottom . Cancel common factors . This is your final answer. (x- 2)(x + 2) (x+ 2) 3 x- 2 = 9 x- 2 = 3 Square both sides 16. x- 2 = 9 x= 11 x+ 5 x + 5 x + 5 17. , , x2- 9 x 2 + 6x + 9 x - 3 First factor the denominators of each x+ 5 x + 5 x + 5 , , Now you can clearly see that the common (x- 3)(x + 3) (x + 3)(x + 3) x - 3 denominator between each will be (x + 3)(x + 3)(x – 3). 18. x3- 5x 2 - 2x + 24 = 0. Possibilities are: 1, 2, 3, 4, 6, 8, 12, 24. Trial & error -21 -5 -2 24 through synthetic division in order gives the first zero as -2 -2 14 -24. Then you are left 1 -7 12 0

2 with (x+ 2)(x - 7x + 12) = 0and you can factor the second set of parentheses to get: (x+ 2)(x - 3)(x - 4) = 0 . Set each of these equal to zero and solve and you get x=-2, 3, & 4. Algebra 2 Segment 2 Exam Review (v8)

2 2 0 -3 5 19. 4 8 10 So, the remainder is 15. 2 4 5 15 4x- 16 3x + 6 4(x - 4) 3(x+ 2) 4(x+ 2) 4x + 8 20. � = = 3x+ 12 3x - 12 3(x+ 4) 3( x- 4) 3(x+ 4) 3x + 12 8x2 24x 4 8x 2 14y 112 x2 y 7 21. � � = 10y 14y 10y 24x4 240 x4 y 15x2 x+ 9 3x - 7 x + 9 + 3x - 7 4x + 2 22. + = = 7y 7y 7y 7y 23. Factors of 40: 1, 2, 4, 5, 8,10, 20, 40 Factors of 5: 1, 5 1 2 4 8 Possible Zeros: 1, , 2, , 4, , 5, 8, , 10, 20, 40 5 5 5 5 24. Yes. P(+ ) = ( + ) ( - ) ( + ) ( - ) 25. 1 2 3 There are 3 sign changes when you plug in a positive, so there are 3 or 1

positive zeros. 26. Set the denominator equal to zero and solve for the variable.

2log5 (2x)- log 5 (8) = log 5 (x) 2 log5 (2x)- log 5 (8) = log 5 (x) 骣(2x)2 log5琪 = log 5 (x) 桫 8 (2x)2 = x 8 27. (2x)2 = 8x 4x2 = 8x x2 = 2x x2 - 2x = 0 x(x- 2) = 0 x= 0 and x = 2 28. The denominator of the fraction becomes the index of the radical and the numerator and the

a base stay under as the radicand. yb = b ya 3x = 27 3x= 3 3 29. 3 x = 3 3 x= 3 Algebra 2 Segment 2 Exam Review (v8)

2x = 3

log2 (3)= x Now you have to change the base log(3) = x 30. log(2) .477 x .301 x 1.58 52x= 25 3x- 4 52x= 5 2(3x- 4) 2x 2(3x- 4) 5 = 5 31. 2x= 2(3x - 4) 2x= 6x - 8 -4x = - 8 x= 2

log7 (x+ 9) - log 7 (x - 3) = 5 骣x+ 9 log7 琪 = 5 桫x- 3 x+ 9 75 = x- 3 32. x+ 9 16807= x- 3 16807(x- 3) = x + 9 16807x- 50421 = x + 9 16806x= 50430 x 3.0002 24 = 16 33. log2 (16)= 4

log6 (x- 2) = 2 62 = x - 2 34. 36= x - 2 38= x

log5 (a)+ log 5 (x + y)

35. log5 (a(x+ y))

log5 (ax+ ay) 3 3x = 98

骣3 2琪 3x = 3 桫8

36. 骣6 x 琪 3 = 3桫8 6 3 x= = 8 4 Algebra 2 Segment 2 Exam Review (v8)

1 2log (x)- log (b) + log (w) 82 8 8 1 2 2 log8 (x)- log 8 (b) + log 8 (w)

骣 2 (x) log琪 + log (w) 8琪 1 8 桫(b)2

37. 骣 2 (x) log琪 (w) 8 琪 1 桫(b)2 骣wx2 log8 琪 Can't have radicals in the denominator. 桫 b 骣wx2 b log 琪 8 琪桫 b y= 3x+ 7 y= x y= ( - 4)x y= 2x2 + 6 38. Examples: 3x Non Examples: 骣1 3 y= 琪 y= x 桫2 39. The range is all of the y-values. Because the base is a positive, no matter what value you put for x, the result will be positive. So, the range will be all real numbers greater than zero. 40. We are looking for the 24th term. We know the first number, 1, we know the term we are looking for, 24 and we know the common difference, -3. So, we use the formula as follows:

an= a 1 + (n - 1)d

a24 = 1 + (24 - 1)( - 3)

a24 = 1 + (23)( - 3)

a24 = 1 + ( - 69)

a24 = - 68

41. We are looking for the sum of the first 100 terms, so we are looking for S100. We know the first

term, an = -13 and the common difference, d = +4, so we will use the formula n Sn=轾 2a 1 + d( n - 1) . Since we need to find the sum of the first 100 terms, n = 100 and we can 2臌

100 S100 =轾 2( - 13) + (4)( 100 - 1) 2 臌 =50轾 - 26 + (4) 99 calculate S11 as follows: 臌 ( ) = 50[ 370] = 18500

7 42. 2i- 7 meas “Find the sim of the terms found when you let I equal every number from 1 to i= 1 7.” So, for i=1, we get (2-7) = -5. I = 2, we get (4-7)=-3, … After calculating all of these you

7 2i- 7 =( - 5) +( - 3) +( - 1) +( 1) +( 3) +( 5) + ( 7) get: i= 1 = 7 Algebra 2 Segment 2 Exam Review (v8)

a- ra 43. The formula for geometric series when we know the first AND last term is S = 1 n . We n 1- r 1 1 know the first term is a1 = -125, the last term is an = - , and whose common ratio is - . So 5 5

轾 1骣 1 -125 - -琪 - 犏 5桫 5 S = 臌 n 骣 1 1-琪 - 桫 5 plugging these values in and simplifying we will get the following: -125 -[ .04] = 1.2 -125.04 = 1.2 = -104.2

44. In this case, order matters, so we will use a permutation. We will calculate P(12,3). Here, n=17

17! P(17,3) = (17- 3)! 17! = and p=3. Plug these in to the formula and you get: 14! 17鬃 16 15 14! = 14! = 4080