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MA 15300 Final Exam, Form 01 Fall 2014
1
1. If π₯ > 0 and π¦ < 0, which of the following inequalities is/are true? Lesson 1
2. Rationalizing the denominators and simplify. Lesson 2
β1
18π₯3π¦4
3. Perform the indicated operations and simplify: Lesson 3
π₯
π₯ + 1β
1
π₯ β 1
A. β2π₯
12π₯3π¦2
B. 1
3π₯π¦2
C. 1
6π₯2π¦2
D. β2π₯
6π₯2π¦2
E. 1
9π₯32π¦2
I. π₯2π¦ < 0
II. π¦βπ₯
π₯π¦> 0
III. π¦(π₯ β π¦) > 0 A. I and II only
B. I and III only
C. II and III only
D. I, II, and III are all true
E. I, II, and III are all false
A. π₯β1
2
B. 1
π₯+1
C. 1
D. π₯β1
π₯+1
E. π₯2β2π₯β1
π₯2β1
MA 15300 Final Exam, Form 01 Fall 2014
2
4. Simplify; do not include negative exponents in your final answer. Lesson 6
π₯π¦β1
(π₯ + π¦)β1
5. Solve π + π =πΆ+2
πΆ for πΆ. Lesson 7
6. Solve for π₯: Lessons 10 and 11
(π₯ β 2)(π₯ + 1) = 3
A. π₯+π¦
π₯π¦
B. π₯π¦
π₯+π¦
C. π₯(π₯+π¦)
π¦
D. π₯2
π₯+π¦
E. None of the above
A. πΆ =π+π
2
B. πΆ =2
π+πβ1
C. πΆ =2
π+π
D. πΆ =ππ
2
E. πΆ =2
ππβ1
A. π₯ = β1, 2
B. π₯ = 2, 5
C. π₯ =1
2Β±
β5
2
D. π₯ = β1
4Β±
β21
4
E. None of the above
MA 15300 Final Exam, Form 01 Fall 2014
3
7. Solve for π₯. Choose the answer that best describes the solution(s). Lesson 14
π₯ + β5π₯ + 19 = β1
8. Solve the inequality: Lesson 15
2|β11 β 7π₯| β 2 β₯ 10
9. Describe the set of all points (π₯, π¦) in the coordinate plane, such that π¦
π₯> 0. Lesson 16
A. There are two solutions. One is positive and one is negative.
B. There are two solutions. Both are positive.
C. There are two solutions. Both are negative.
D. There is one solution. It is positive.
E. There is one solution. It is negative.
A. [β17
7, β
5
7]
B. (ββ, β17
7] βͺ [β
5
7, β)
C. [5
7,
17
7]
D. (ββ,5
7] βͺ [
17
7, β)
E. None of the above
A. The set of all points in quadrants
II and IV only.
B. The set of all points in quadrants
I and III only.
C. The set of all points in quadrants
I and II only.
D. The set of all points in quadrants
III and IV only.
E. None of the above
MA 15300 Final Exam, Form 01 Fall 2014
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10. Give the equation of the line in slope-intercept form which is perpendicular to the line 2π₯ β 3π¦ = 7 and
contains the point (4, β1). Lesson 18
11. If π(π₯) = βπ₯2 + π₯ + 2, find π(π₯+β)βπ(π₯)
β. Lessons 20 and 21
12. The point π(5, β3) is on the graph of a basic function, π¦ = π(π₯). Find the corresponding point on the
graph of π¦ = 4π (β1
3π₯) β 2. Lesson 22
A. β2π₯ β β
B. ββ2
C. β2π₯ β β2 + β
D. ββ + 1
E. β2π₯ β β + 1
A. π¦ =3
2π₯ β 7
B. π¦ = β2
3π₯ +
5
2
C. π¦ =2
3π₯ β
11
3
D. π¦ = β3
2π₯ + 5
E. None of the above
A. (β15, β14)
B. (β5
3, β14)
C. (β15, β20)
D. (β5
3, β20)
E. None of the above
MA 15300 Final Exam, Form 01 Fall 2014
5
13. Which of the following is the correct graph of the piecewise defined function π? Lesson 24
π(π₯) = {π₯2 + 1; π₯ β€ β1
βπ₯ + 1; π₯ > β1
A. B.
C.
D.
E.
MA 15300 Final Exam, Form 01 Fall 2014
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14. Which of the following is the graph of the function π(π₯) = π₯2(π₯ β 1)(π₯ + 1)2? (each tick mark represents one unit on the graph) Lesson 29
B. A.
C. D. E.
MA 15300 Final Exam, Form 01 Fall 2014
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15. Solve the system of equations, then indicate the number of times the graphs intersect. Lessons 31 and 32
{π₯2 + π¦2 = 25
π¦ = π₯2 β 5
16. Suppose π and π are one-to-one functions such that π(2) = 7, π(4) = 2, π(2) = 5, and π(3) = 2.
Which of the following compositions is/are true: Lessons 27 and 34
17. Find the inverse of the function π(π₯) = ln(π₯ β 2) Lessons 34, 35, and 36
A. 0
B. 1
C. 2
D. 3
E. 4
A. (π β πβ1)(7) = 5
B. (π β πβ1)(5) = 4
C. (πβ1 β πβ1)(2) = 7
D. More than one of the above
E. None of the above
A. πβ1(π₯) = ππ₯+2
B. πβ1(π₯) = 10π₯+2
C. πβ1(π₯) = ππ₯ + 2
D. πβ1(π₯) = 10π₯ + 2
E. None of the above
MA 15300 Final Exam, Form 01 Fall 2014
8
18. If π(π₯) = 2π₯, which of the following graphs represents π(π₯ β 3) + 1? (each tick mark represents one unit on the graph) Lessons 22, 23, and 35
-5 -4 -3 -2 -1 0 1 2 3 4 5
-5 -4 -3 -2 -1 0 1 2 3 4 5
B. A.
-5 -4 -3 -2 -1 0 1 2 3 4 5
-5 -4 -3 -2 -1 0 1 2 3 4 5
-5 -4 -3 -2 -1 0 1 2 3 4 5
C. D. E.
MA 15300 Final Exam, Form 01 Fall 2014
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19. Which of the following statements is true about the function π(π₯) = log2 π₯, and its graph? Lessons 36 and 37
20. If π₯ < 0, which of the following functions is/are undefined using real numbers only?
Lessons 20, 36, and 37
π(π₯) = 1
π₯
π(π₯) = βπ₯
β(π₯) = log π₯
π(π₯) =π₯
2
A. π and π only
B. π and β only
C. π and β only
D. π and π only
E. β and π only
A. The domain of π is (ββ, β)
B. There is no π¦-intercept
C. The function is positive on the
interval (0, β)
D. The graph of π is decreasing
throughout its domain
E. There no zeros
MA 15300 Final Exam, Form 01 Fall 2014
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21. Given the function π(π‘) = 100,000π0.022π‘, find the value of π‘ when π(π‘) = 140,000. Lesson 37
22. Express as one logarithm: Lesson 38
log (π₯2
π¦3) β log(π₯π¦) β 4 log βπ¦
23. Solve for π₯: Lesson 39
log3 β2π₯ + 3 = 2
A. π‘ < 5
B. 5 < π‘ < 10
C. 10 < π‘ < 15
D. 15 < π‘ < 20
E. π‘ > 20
A. log (π₯
π¦6)
B. log (π₯
π¦2)
C. log π₯3
D. log (π₯
π¦8)
E. β4 log (π₯2
π¦3β π₯π¦ β βπ¦)
A. π₯ < 1
B. 1 < π₯ < 5
C. 5 < π₯ < 10
D. 10 < π₯ < 20
E. π₯ > 20
MA 15300 Final Exam, Form 01 Fall 2014
11
24. Solve for π₯: Lesson 40
25βπ₯ = 6
25. The base of a triangle is three inches more than its height. If both the base and the height are increased
by 3 inches the area is 14 square inches. Find the length of the original base (π) in inches. Lesson 9
26. Temperature readings on the Fahrenheit and Celsius scales are related by the formula πΉ =9
5πΆ + 32.
Determine when the temperature reading on the Fahrenheit scale is twice the temperature reading on the
Celsius scale. Lesson 19
A. π = 1
B. π =7
2
C. π = 4
D. π = 8
E. π = 9
A. π₯ = log2(6) β 5
B. π₯ = 5 β log6(2)
C. π₯ =log(2)
log(6)β 5
D. π₯ = 5 βlog(6)
log(2)
E. None of the above
A. When the Fahrenheit reading is
less than 0Β°
B. When the Fahrenheit reading is
between 0Β° and 100Β°
C. When the Fahrenheit reading is
between 100Β° and 200Β°
D. When the Fahrenheit reading is
between 200Β° and 300Β°
E. When the Fahrenheit reading is
greater than 300Β°
MA 15300 Final Exam, Form 01 Fall 2014
12
27. Eight hundred feet of chain-link fence is to be used to construct six animal cages, as shown in the figure.
Find the value of π₯ that maximizes the enclosed area. (hint: express π¦ in terms of π₯ first) Lesson 26
28. A woman has $216,000 to invest and wants to generate $12,000 per year in interest income. She can
invest in two tax-free funds. The first is stable, but pays only an average 4.5% interest per year. The
second pays an average of 9.25% interest per year, but has greater risk. If π₯ represents the amount of
money invested in the fund that averages 4.5% interest per year, which of the following best describes
the value of π₯? Lesson 8 or Lesson 33
A. π₯ =3
800
B. π₯ =400
3
C. π₯ =800
3
D. π₯ = 100
E. π₯ = 200
π₯
π¦
A. π₯ is less than $50,000
B. π₯ is between $50,000 and $60,000
C. π₯ is between $60,000 and $70,000
D. π₯ is between $70,000 and $80,000
E. π₯ is more than $80,000
F.
MA 15300 Final Exam, Form 01 Fall 2014
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29. Parents of a newborn baby are given a gift of $10,000 and will choose between two options to invest for
their childβs college fund. Option 1 is to invest the gift in a fund that pays an average annual interest
rate of 11% compounded quarterly; option 2 is to invest the gift in a fund that pays an average annual
interest rate of 10.75% compounded continuously. Calculate the value of each investment using the
formulas π΄ = ππππ‘ and π΄ = π (1 +π
π)
ππ‘
. Assume the investments have terms of 18 years and round
your answers to the nearest dollar. Lesson 35
30. A drug is eliminated from the body through urine. Suppose that for a dose of 10 milligrams, the amount
π΄(π‘) remaining in the body π‘ hours later is given by π΄(π‘) = 10(0.7)π‘. What is the half-life of the drug?
Lesson 40
A. Option 1 = $69,240
Option 2 = $70,517
B. Option 1 = $72,427
Option 2 = $69,240
C. Option 1 = $70,517
Option 2 = $69,240
D. Option 1 = $67,494
Option 2 = $69,240
E. Option 1 = $67,494
Option 2 = $72,427
A. Between 0 and 0.5 hours
B. Between 0.5 and 1 hour
C. Between 1 and 1.5 hours
D. Between 1.5 and 2 hours
E. Between 2 and 2.5 hours