Download Segment 2 Exam

Algebra 2 Segment 2 Exam Review (v8)
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Describe how to determine whether a sequence is arithmetic or not. Give an example and explain WHY is constitutes
an ARITHMETIC sequence.
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, …
Describe how to determine whether a sequence is geometric or not. Give an example and explain WHY is constitutes
an GEOMETRIC sequence.
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?
Determine the probability of being born in a month that starts with the letter “M.”
In how many different ways can you arrange the letters of the word “Yesterday” taken 4 at a time?
Describe the differences between the mean, mode and the median of a set of data. When are they equivalent if ever?
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?
Find the probability of choosing two vowels from the word “trigonometry”, one at a time without replacing. Let “y” be
considered a vowel.
Which central tendency below is the greatest in the data {1, 4, 4, 7, 8, 10, 22}?
What is the difference between an even and odd function?
8
What is the horizontal asymptote of f(x) 
3?
x7
The equation 6x2 + 11x - 10 = 0 has how many solutions and what type of solutions are they?
What are the possible solutions to the equation 5x3 – 4x2 + 7x – 15=0?
x 2  4x  4
Simplify the expression
.
x2  4
Solve 3 x  2  9 for x.
x5
x5
x5
Find the LCD of: 2
,
,
2
x  9 x  6x  9 x  3
Solve the equation x3  5x2  2x  24  0 and list all the solutions.
Find the remainder when dividing 2x3  3x  5 by x-2
4x  16
3x  6
Multiply
and
3x  12
3x  12
8x 2
24x 4
Simplify
divided by
10y
14y
22. Add
x  9 3x  7

.
7y
7y
23. List the possible rational zeros for the polynomial P(x)  5x 3  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)  2x 3  3x 2  6x  8
26. Describe how to find the equation of the vertical asymptote(s) of a rational expression similar to, but not limited to
x5
.
x 2  49
27. Solve for x: 2log5 (2x)  log5 (8)  log5 (x)
a
28. Every part of a fractional exponent tells you something about the radical form. Describe how to write y b in radical
form. Which part becomes the radical index, which becomes the radicand, which becomes the exponent on the
radicand?
29. Solve 3x  27
30. Rewrite 2x  3 in logarithmic form to solve for x. Round your answer to two decimal places.
31. Solve 52x  253x4 for x without using logarithms.
Algebra 2 Segment 2 Exam Review (v8)
32. Solve log7 (x  9)  log7 (x  3)  5 for x using the properties of logarithms.
33. Rewrite 24  16 in logarithmic form.
34. Solve log6 (x  2)  2 for x using the properties of logarithms.
35. Simplify log5 (a)  log5 (x  y) using the properties of logarithms.
3
36. Solve 3x  9 8 for x.
1
37. Simplify 2log8 (x)  log8 (b)  log8 (w) using the properties of logarithms.
2
38. Give three examples of exponential functions and three examples of NON-exponential functions.
39. Consider the exponential function g(x)  7 x . 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
43. Find the sum of a geometric sequence whose first term is -125, whose last (nth) term is 
1
, and whose common ratio
5
1
.
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?
is 
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 time. So, we need to
n!
calculate: P(9, 4) using the formula P(n,r) =
(n  r)!
P(9, 4) =
7.
8.
9!
9! 9  8  7  6  5  4  3  2  1
 
 9  8  7  6 = 3024
(9  4)! 5!
5  4  3  2 1
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.
This is a combination, there are 7 choices and we will pick 3. So we need to calculate C(7, 3) using the formula C(n, r)
n!
=
r!(n  r)!
C(7, 3) =
7!
7!
7  6  5  4  3  2  1 7  6  5 210




= 35
3!(7  3)! 3!(4)! 3  2  1(4  3  2  1) 3  2  1
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Algebra 2 Segment 2 Exam Review (v8)
9.
The probability of picking a vowel the first time is
5
. That leaves 11 letters left, 4 of which are still vowels, so the
12
4
. That makes the probability of choosing two vowels from the word
11
5 4
20 10 5
“trigonometry”, one at a time without replacing
 


12 11 132 66 33
Mean = 8; Median = 7; Mode = 4. So, the mean is the greatest central tendency.
Odd functions have exponenbts that are NOT divisible by 2. Even functions have exponents that ARE divisible by 2.
x
This function is in the form f(x) 
 a , which means that the horizontal asymptote is at y = a. So, the horizontal
xb
asymptote ius at y = -3
The discriminant is b2  4ac  112  4(6)(10)  121  240  361 . 361 is positive which means there are two real
probability of picking a vowel this time is
10.
11.
12.
13.
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.
factors of 15
14. Find the factors of 5: 1, 5 and the factors of 15: 1,3,5,15. Then find the fraction combinations of 
.
factors of 5
1 1 3 3 5 5 15 15
1 3
Those would be  , , , , , , , . Now you simplify each to get: 1, ,3, ,5,1,15,3 Lastly, you do not need
1 5 1 5 1 5 1 5
5 5
1 3
to list repeats, so your final list would be: 1, ,3, ,5,15
5 5
15. Name restrictions by setting denominator equal to zero and solving for x. x  2,x  2 Factor the top and
bottom
(x  2)(x  2)
(x  2)
. Cancel common factors
. This is your final answer.
(x  2)
(x  2)(x  2)
3 x2 9
16.
x  2  3 Square both sides
x29
x  11
x5
17. 2
,
x 9
x5
,
x  6x  9
2
x5
x3
x5
x5
,
,
First factor the denominators of each (x  3)(x  3) (x  3)(x  3)
x5
x3
Now
you can clearly see that the common denominator between each will be (x + 3)(x + 3)(x – 3).
18. x3  5x2  2x  24  0 . Possibilities are:  1,  2,  3,  4,  6,  8,  12,  24. Trial & error through synthetic
2 1 -5 -2 24
division in order gives the first zero as -2
-2 14 -24 . Then you are left with (x  2)(x 2  7x  12)  0 and you
1 -7 12 0
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.
2 2 0 -3 5
19.
20.
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





3x  12 3x  12 3(x  4) 3 ( x  4 ) 3(x  4) 3x  12
Algebra 2 Segment 2 Exam Review (v8)
21.
112 x 2 y
8x 2 24x 4 8x 2 14y
7





10y 14y 10y 24x 4 240 x 4 y 15x2
22.
x  9 3x  7 x  9  3x  7 4x  2



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.
There are 3 sign changes when you plug in a positive, so there are 3 or 1 positive zeros.
1 2 3
26. Set the denominator equal to zero and solve for the variable.
2log5 (2x)  log5 (8)  log5 (x)
log5 (2x)2  log5 (8)  log5 (x)
 (2x)2
log5 
 8
(2x)2
x
8
27.
(2x)2  8x

  log5 (x)

4x 2  8x
x 2  2x
x 2  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 base stay under as the
a
radicand. y b  b y a
3 x  27
29.
3 x  33
3
x
3
3
x3
2x  3
log2 (3)  x
30.
log(3)
x
log(2)
.477
x
.301
x  1.58
Now you have to change the base
Algebra 2 Segment 2 Exam Review (v8)
52 x  253 x4
52 x  52( 3 x4)
5
2x
5
2( 3 x 4)
31. 2x  2(3x  4)
2x  6x  8
4x  8
x2
log7 (x  9)  log7 (x  3)  5
 x9
log7 
5
 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
34.
62  x  2
36  x  2
38  x
log5 (a)  log5 (x  y)
35. log5 (a(x  y))
log5 (ax  ay)
3
3x  9 8
3
2 
36.
3x  3  8 
x
6
 
3  3 8
x
6 3

8 4
Algebra 2 Segment 2 Exam Review (v8)
1
2log8 (x)  log8 (b)  log8 (w)
2
1
log8 (x)2  log8 (b) 2  log8 (w)
 2
(x)
log8  1
 2
 (b)
37.
 2
(x)
log8  1
 2
 (b)

  log (w)
8



(w) 


 wx 2 
log8 

 b 
 wx 2 b
log8 
 b

Can't have radicals in the denominator.



y  3 x 7
38. Examples:
y  ( 4)
yx
x
Non Examples:
y  2x 2  6
3x
y  x3
 1
y  
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
an  a1  (n  1)d
a 24  1  (24  1)( 3)
the common difference, -3. So, we use the formula as follows: a 24  1  (23)( 3)
a 24  1  ( 69)
a 24  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
n
common difference, d = +4, so we will use the formula Sn  2a1  d  n  1 . Since we need to find the sum of the
2
100
 2( 13)  (4) 100  1 
S100 
2 
 50  26  (4)  99  
first 100 terms, n = 100 and we can calculate S11 as follows:
 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 7.” So, for i=1, we get
i1
(2-7) = -5. I = 2, we get (4-7)=-3, … After calculating all of these you get:
7
2i  7   5   3   1  1   3   5   7
i1
7
Algebra 2 Segment 2 Exam Review (v8)
43. The formula for geometric series when we know the first AND last term is Sn 
-125, the last term is an = 
a1  ran
. We know the first term is a1 =
1 r
1
1
, and whose common ratio is  . So plugging these values in and simplifying we will
5
5
 1  1 
125       
 5  5 
Sn 
 1
1   
 5
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 and p=3. Plug these in
17!
P(17,3) 
(17  3)!
17!

to the formula and you get:
14!
17  16  15  14!

14!
 4080