MTH 100 College Algebra Essex County College Division of Mathematics Sample Review Questions 1 Created June 6, 2011



Similar documents
MTH 092 College Algebra Essex County College Division of Mathematics Sample Review Questions 1 Created January 17, 2006

MTH 086 College Algebra Essex County College Division of Mathematics Sample Review Questions 1 Created January 20, 2006

1.1 Practice Worksheet

Higher Education Math Placement

Answer Key for California State Standards: Algebra I

Math 0980 Chapter Objectives. Chapter 1: Introduction to Algebra: The Integers.

Algebra I Vocabulary Cards

Math Placement Test Study Guide. 2. The test consists entirely of multiple choice questions, each with five choices.

Review of Intermediate Algebra Content

MATH 60 NOTEBOOK CERTIFICATIONS

1) (-3) + (-6) = 2) (2) + (-5) = 3) (-7) + (-1) = 4) (-3) - (-6) = 5) (+2) - (+5) = 6) (-7) - (-4) = 7) (5)(-4) = 8) (-3)(-6) = 9) (-1)(2) =

MATD Intermediate Algebra Review for Pretest

MATH 095, College Prep Mathematics: Unit Coverage Pre-algebra topics (arithmetic skills) offered through BSE (Basic Skills Education)

Florida Math Correlation of the ALEKS course Florida Math 0028 to the Florida Mathematics Competencies - Upper

What are the place values to the left of the decimal point and their associated powers of ten?

Algebra 1 Course Title

Vocabulary Words and Definitions for Algebra

Big Bend Community College. Beginning Algebra MPC 095. Lab Notebook

Algebra and Geometry Review (61 topics, no due date)

LAKE ELSINORE UNIFIED SCHOOL DISTRICT

Factoring Polynomials

MA107 Precalculus Algebra Exam 2 Review Solutions

Thnkwell s Homeschool Precalculus Course Lesson Plan: 36 weeks

of surface, , , of triangle, 548 Associative Property of addition, 12, 331 of multiplication, 18, 433

6.1 Add & Subtract Polynomial Expression & Functions

Mathematics Placement

1.3 Algebraic Expressions

HIBBING COMMUNITY COLLEGE COURSE OUTLINE

COMPETENCY TEST SAMPLE TEST. A scientific, non-graphing calculator is required for this test. C = pd or. A = pr 2. A = 1 2 bh

TSI College Level Math Practice Test

Florida Algebra 1 End-of-Course Assessment Item Bank, Polk County School District

Definition 8.1 Two inequalities are equivalent if they have the same solution set. Add or Subtract the same value on both sides of the inequality.

ALGEBRA 2: 4.1 Graph Quadratic Functions in Standard Form

Geometry and Measurement

Algebra 1 If you are okay with that placement then you have no further action to take Algebra 1 Portion of the Math Placement Test

Algebra 1. Curriculum Map

Mathematics Online Instructional Materials Correlation to the 2009 Algebra I Standards of Learning and Curriculum Framework

Extra Credit Assignment Lesson plan. The following assignment is optional and can be completed to receive up to 5 points on a previously taken exam.

Indiana State Core Curriculum Standards updated 2009 Algebra I

Quick Reference ebook

MATH 100 PRACTICE FINAL EXAM

MATH 21. College Algebra 1 Lecture Notes

Students will be able to simplify and evaluate numerical and variable expressions using appropriate properties and order of operations.

COWLEY COUNTY COMMUNITY COLLEGE REVIEW GUIDE Compass Algebra Level 2

Midterm 2 Review Problems (the first 7 pages) Math Intermediate Algebra Online Spring 2013

This unit has primarily been about quadratics, and parabolas. Answer the following questions to aid yourselves in creating your own study guide.

PERT Mathematics Test Review

Summer Math Exercises. For students who are entering. Pre-Calculus

MATH 0110 Developmental Math Skills Review, 1 Credit, 3 hours lab

Mathematics Placement Examination (MPE)

In this section, you will develop a method to change a quadratic equation written as a sum into its product form (also called its factored form).

Blue Pelican Alg II First Semester

Algebra Cheat Sheets

Brunswick High School has reinstated a summer math curriculum for students Algebra 1, Geometry, and Algebra 2 for the school year.

POLYNOMIAL FUNCTIONS

a. all of the above b. none of the above c. B, C, D, and F d. C, D, F e. C only f. C and F

SPECIAL PRODUCTS AND FACTORS

BookTOC.txt. 1. Functions, Graphs, and Models. Algebra Toolbox. Sets. The Real Numbers. Inequalities and Intervals on the Real Number Line

How do you compare numbers? On a number line, larger numbers are to the right and smaller numbers are to the left.

Definitions 1. A factor of integer is an integer that will divide the given integer evenly (with no remainder).

SUNY ECC. ACCUPLACER Preparation Workshop. Algebra Skills

Administrative - Master Syllabus COVER SHEET

Free Pre-Algebra Lesson 55! page 1

The program also provides supplemental modules on topics in geometry and probability and statistics.

CRLS Mathematics Department Algebra I Curriculum Map/Pacing Guide

ALGEBRA I (Created 2014) Amherst County Public Schools

Expression. Variable Equation Polynomial Monomial Add. Area. Volume Surface Space Length Width. Probability. Chance Random Likely Possibility Odds

Algebra 1 End-of-Course Exam Practice Test with Solutions

Algebra Academic Content Standards Grade Eight and Grade Nine Ohio. Grade Eight. Number, Number Sense and Operations Standard

Mathematics Common Core Sample Questions

This is a square root. The number under the radical is 9. (An asterisk * means multiply.)

How To Understand And Solve Algebraic Equations

FACTORING QUADRATICS and 8.1.2

Math Review. for the Quantitative Reasoning Measure of the GRE revised General Test

Virginia Placement Test Practice Questions and Answers

Prentice Hall Mathematics: Algebra Correlated to: Utah Core Curriculum for Math, Intermediate Algebra (Secondary)

SUFFOLK COMMUNITY COLLEGE MATHEMATICS AND COMPUTER SCIENCE DEPARTMENT STUDENT COURSE OUTLINE Fall 2011

Algebra 2: Themes for the Big Final Exam

Veterans Upward Bound Algebra I Concepts - Honors

Pre-Algebra Academic Content Standards Grade Eight Ohio. Number, Number Sense and Operations Standard. Number and Number Systems

Virginia Placement Test Practice Questions and Answers

Algebra 2 Chapter 1 Vocabulary. identity - A statement that equates two equivalent expressions.

MATH Fundamental Mathematics IV

12) 13) 14) (5x)2/3. 16) x5/8 x3/8. 19) (r1/7 s1/7) 2

IV. ALGEBRAIC CONCEPTS

Algebra I Credit Recovery

Anchorage School District/Alaska Sr. High Math Performance Standards Algebra

Algebra I. In this technological age, mathematics is more important than ever. When students

Polynomial Operations and Factoring

CORRELATED TO THE SOUTH CAROLINA COLLEGE AND CAREER-READY FOUNDATIONS IN ALGEBRA

Factoring and Applications

Number Sense and Operations

APPLICATIONS AND MODELING WITH QUADRATIC EQUATIONS

PRACTICE PROBLEMS IN ALGEBRA, TRIGONOMETRY, AND ANALYTIC GEOMETRY

Lyman Memorial High School. Pre-Calculus Prerequisite Packet. Name:

Math 120 Final Exam Practice Problems, Form: A

Scope and Sequence KA KB 1A 1B 2A 2B 3A 3B 4A 4B 5A 5B 6A 6B

Chesapeake College. Syllabus. MAT 031 Elementary Algebra (silcox87517) John H. Silcox

Algebra 1 Course Information

Understanding Basic Calculus

Transcription:

MTH 00 College Algebra Essex County College Division of Mathematics Sample Review Questions Created June 6, 0 Math 00, Introductory College Mathematics, covers the mathematical content listed below. In order to place out of Math 00 or to prepare for the final examination in this course, one should be extremely comfortable with all of these items. NOTE: Calculators are permitted in this course, but should only be used when appropriate. The content listed below is covered in chapters through 8 and 0 in Algebra for College Students, Custom Version for Essex County College, Second Edition, by Bittinger & Ellenbogen (published by Addison Wesley, Inc.), the current textbook used in Math 00. Simplifying expressions involving negative integer and rational exponents. Simplifying, adding, subtracting, multiplying, and dividing rational expressions, radical expressions, and complex numbers. Multiplying and dividing polynomials. Factoring polynomials including greatest common factors (GCF), differences of squares, differences and sums of cubes, and quadratic trinomials. Evaluating expressions involving function notation. Finding the domains of functions. Solving equations, including ones containing rational or radical terms. Solving literal equations or formulas for a specified variable in terms of other variables. Solving inequalities and systems of equations. Solving quadratic equations by using the zero-product rule, the quadratic formula, and the square-root method in conjunction with completing the square. Solving application problems (including, but not limited to, total value, mixture, motion, work, area, and determine-the-number problems). Finding equations of lines through given points, with given slopes, perpendicular or parallel to given lines, and/or with given intercepts. Finding slopes and x and y-intercepts of lines. Finding vertices and lines of symmetry of parabolas and centers and radii of circles. Finding equations of circles with given centers and radii. Graphing lines, parabolas, and circles. Finding the distance between two points. This document was prepared by Ron Bannon using L A TEX ε and is a slight modification of Susan Gaulden s MTH 00 Content worksheet.

Review Problem Set At Essex County College you should be prepared to show all work clearly and in order, ending your work by boxing the answer. Furthermore, justify your answers algebraically whenever possible. These questions are for review only, and placement tests are not limited to these problems alone. Solutions and work are provided for each question. Please feel free to email rbannon@mac.com with questions or comments pertaining to this document.. Simplify. (a) ( x y z 0) ( x y z 0) ( ) y ( x = = x y ) = x4 9y 6 (b) 4a b 5 c 6a c 4a b 5 c 6a c = a5 b 5 c (c) ( m 4 ) n 6mn ( m 4 ) ( n = ) = ( m 5 n ) = 8m 5 n 6mn m 5 n. Simplify. (a) ( 5x y 4) ( 5x y 4) = 5 x y (b) ( 8a 6 b 9 c ) ( 8a 6 b 9 c ) = 4a 4 b 6 c These are sample problems and you should not limit your study to just these problems.

. Simplify (a) (b) x 5x + 6 x + x 8 6x 7 4x 5 + x 4x 5 x 5x + 6 (x ) (x ) x = + x 8 (x + 4) (x ) = x x + 4, x 6x 7 4x 5 + x 6x 7 + x 8x 0 (4x 5) = = = 4x 5 4x 5 4x 5 (4x 5) =, x 5 4 (c) x + + 5x x + x + x + x + + 5x x + x + x + = = (x + ) + 5x (x + ) (x + ) (x + ) (x + ) 5x (x + ) + (x + ) (x + ) (x + ) (x + ) (x + ) (x + ) = x + 6 + 5x + 5x (x + ) (x + ) = 5x + 8x + 5 (x + ) (x + ) (d) x 4 x + 6x + x 7x + 0 x 4 x + 6x + x 7x + 0 = = (x + ) (x ) (x + ) (x + ) (x ) (x 5) (x + ) (x 5), x, x = x + 4 x 5, x, x

(e) x + 4x + x x 0 x x x 9x + 0 x + 4x + x x 0 x x x 9x + 0 = x + 4x + x x 0 x 9x + 0 x x (x + ) (x + ) (x 5) (x 4) = (x + ) (x 5) (x + ) (x 4) = x +, x, x 4, x 5 x + 4. Simplify. (a) 9x + x 4 9x 9x + x 4 9x = 9x + x 4 9x 9x 9x = + x 9x 4, x 0 ( + x) = (x + ) (x ), x 0 = x, x 0, x (b) x x + x + + 5 x x x + x + + 5 x = = = x x + (x ) (x + ) x + + 5 (x ) (x + ) x (x + ) (x ) (x ) + 5 (x + ), x, x x + x +, x, x 4

5. Simplify. (a) 4 m 7 n 8 p 4 m 7 n 8 p = n mp 4 m (b) 80x y 5xy 5 Note, the restriction here is that x 0 and y 0. 80x y 5xy 5 = 6x y 4 = 4 x y (c) 4 7 5 + 4 7 + 4 7 5 + 4 7 + = 4 4 7 (d) 48 + 7 4 48 + 7 4 = 4 + 9 8 = 5 (e) 40 0 40 0 = 00 = 4 00 = 0 (f) 5 4a 4 bc 4 5 8a b c 5 4a 4 bc 4 5 8a b c = 5 a 7 b c 5 = ac 5 a b (g) 0xy 5 5x y The restriction here is that x > 0 and y > 0. 0xy 5 5x y = 0xy 5 4y 5x y = 4 x = y x 5

(h) 5 + 5 + = 5 + 5 5 = 7 8 5 = 7 8 = 7 8 6. Rationalize the denominator 9 0x y. The restriction here is that x 0 and y > 0. 9 9 0x y = 0x y = x 5y = 5y x 5y = 5y 5y 0 x y 7. Simplify. (a) 4 + 5 6 4 + 5 6 = i + 5 4i = 6i + 0i = 6i (b) ( + i) + (5 6i) ( + 4i) ( + i) + (5 6i) ( + 4i) = 7 8i (c) 5i (7 4i) 5i (7 4i) = 5i 0i = 5i + 0 = 0 + 5i (d) (4 + i) ( i) (4 + i) ( i) = 8i + i i = 5i + = 4 5i (e) 4 + i i 4 + i i = 4 + i i + i + i 0 + i 0 + i = = = 0 9 4i + i 6

8. Simplify. ( (a) x x + 4 ) ( x + 5x ) ( x x + 4 ) ( x + 5x ) = x 4 + 5x x 6x 0x + 4x + x + 0x 8 = x 4 x + 4x 8 (b) x 5x x 4 x 5x x 4 = (x + ) (x 4) (x 4) = x +, x 4 (c) ( x + 7x 4 ) (x + ) Found by long division. x + 6 x + 9. Factor. (a) 5x y 5x y 4 + 0x y 5x y 5x y 4 + 0x y = 5x y ( x y + 4xy ) (b) m 4 6 m 4 6 = ( m 4 ) ( m + 4 ) = (m ) (m + ) ( m + 4 ) (c) 7x 7x = (x ) ( 9x + x + ) (d) 5x + x + 5x + x + = (5x + ) (x + ) 7

(e) x + 64 x + 64 = (x + 4) ( x 4x + 6 ) 0. Evaluate. (a) f ( ) given that f (x) = x x + 5. f ( ) = ( ) ( ) + 5 = 9 9 + 5 = 4 (b) g () given that g (x) = x x + 0. g () = + 0 = 8 + 0 = 6 = 4 (c) h (0) given that h (x) = 4x + x + 8. h (0) = 4 0 + 0 + 8 = 8 =. Find the domain of each function. (a) f (x) = 5 x R, x 0 (b) g (x) = x + 5 x R, x (c) h (x) = x x + 7 R 8

. Solve. (a) x 8 = 5 (x + 4) x x 8 = 5 (x + 4) x x 8 = 5x + 0 x x 8 = 4x + 0 8 0 = 4x x 8 = x 4 = x (b) So, x = 4. 6 x + 4 x = 5 So, x =. 6 x + 4 x = 5 x 6 x + x 4 x = x 5 6 + 4 = 5x 0 = 5x = x (c) x + 4 = x = x + 4 x (x + 4) (x ) = (x + 4) (x ) (x + 4) (x ) (x ) = (x + 4) x 9 = x + 8 x x = 9 + 8 x = 7 So, x = 7. 9

(d) 50 x 6 x = 0 x So, x =. 50 x 6 x = 0 x 50 x (x ) x x (x ) 6 x = x (x ) 0 x x 50 (x ) 6 = (x ) 0 50x 6x + = 0x 60 4x + = 0x 60 4x 0x = 60 4x = 9 x = (e) x + 7 + 5 = x x + 7 + 5 = x x + 7 = x 5 ( x + 7 ) = (x 5) x + 7 = x 0x + 5 0 = x x + 8 0 = (x ) (x 9) It appears that x = and x = 9 are solutions, however, x = is extraneous (i.e. doesn t work), so x = 9. (f) x = x = ( x ) = ( ) x = 7 x = 5 So, x = 5. 0

. Solve. (a) A = P ( + rt) for r. A = P ( + rt) A = P + P rt A P = P rt A P P t = r So, r = A P P t (b) v = d + d t or r = A P t t. for d v = d + d t vt = d + d vt d = d So, d = vt d. (c) K = rt r t for t K = rt r t K (r t) = rt Kr Kt = rt Kr = rt + Kt Kr = t (r + K) Kr r + K = t So, y = Kr r + K.

4. Solve. (a) 4x 7 > x + 4x 7 > x + 4x x > 7 + x > 8 x > 6 So, x > 6, and in interval notation, (6, ). (b) x (x 4) x + 5 (x + ) x (x 4) x + 5 (x + ) x x + 8 x + 5x + 5 x + 8 7x + 5 8 5 7x x 6x x So, x [ ), and in interval notation,,. 5. Solve. (a) { x + y = y = x + 7 Using the method of substitution. x + y = x + (x + 7) = x + 4x + 4 = 5x + 4 = 5x = 5 x = So, x = and since y = x + 7 = ( ) + 7 =, y =. (b) { x y = 7 x + 5y = 8 Using the method of elimination, by multiplying the first row by and the second row by, and then adding the two equations together. { { (x y) = (7) 6x + 4y = 4 9y = 8 y =. (x + 5y) = ( 8) 6x + 5y = 4

Using y = to back-substitute into any of these equations to find x. x y = 7 x + 4 = 7 x = x = So, finally, the solution, in ordered-pair notation, is (, ) 6. Solve. (a) x 9 = 0 Using the square-root rule. x 9 = 0 x = 9 x = ± 9 x = ± (b) x = x x = x x + x = 0 x (x + ) = 0 So, by the zero product rule, x = 0 or x =. (c) x x 0 = 0 x x 0 = 0 (x + 5) (x ) = 0 So, by the zero product rule, x = 5 or x =. (d) x 8x + 9 = 0 Since x 8x + 9 is not factorable over the rational numbers, you ll need to either complete the square or use the quadratic formula. Here I am using the technique of completing the square. x 8x + 9 = 0 x 8x = 9 x 8x + 6 = 9 + 6 (x 4) = 7 x 4 = ± 7 x = 4 ± 7 So, x = 4 + 7 or x = 4 7.

(e) x 6x = 5 Since x 6x 5 is not factorable over the rational numbers, you ll need to either complete the square or use the quadratic formula. Here I am using the quadratic formula. x 6x = 5 x 6x 5 = 0 x = 6 ± 6 4 () ( 5) 4 = 6 ± 76 4 = 6 ± 9 4 So, x = + 9 or x = 9. 7. Find the equation of the line through the point (0, 5) with slope /. Graph the line. Using the point slope form. y 5 = (x 0) Or, the slope intercept form. y = x + 5 Or, standard form. Here s the graph. x + y = 5 0 5-5 -0-5 0 5 0 5-5 -0 Figure : Graph of x + y = 5. 4

8. Find the equation of the the line through the point (, 5) line with slope. Graph the line. Using the point slope form. y 5 = (x + ) Or, the slope intercept form. Or, standard form. Here s the graph. y = x + x + y = 0 5-5 -0-5 0 5 0 5-5 -0 Figure : Graph of x + y =. 9. Find the equation of the line through the points (4, ) and (, ). Graph the line. First you ll need to find the slope. m = y y x x = 4 = = Using the point slope form with the point (4, ). y = (x 4) Using the point slope form with the point (, ). y + = (x ) Or, the slope intercept form. Or, standard form. y = x x y = 5

Here s the graph. 0 5-5 -0-5 0 5 0 5-5 -0 Figure : Graph of x y =. 0. Find the equation of the line through the point (0, ) that is perpendicular to the line x y =. Graph the line. The slope of the line whose equation is x y = can be found easily by solving for y. x y = y = x So the slope of the line perpendicular to this line is /. Using the point slope form. y = (x 0) Or, the slope intercept form. y = x + Or, standard form. Here s the graph. x + y = 6 0 5-5 -0-5 0 5 0 5-5 -0 Figure 4: Graph of x + y = 6. 6

. Find the equation of the line through the point (, 4) that is parallel to the line x y = 5. Graph the line. The slope of the line whose equation is x y = 5 can be found easily by solving for y. x y = 5 y = x 5 So the slope of the line parallel to this line is /. Using the point slope form. y 4 = (x ) Or, the slope intercept form. y = x + 7 Or, standard form. Here s the graph. x + y = 7 0 5-5 -0-5 0 5 0 5-5 -0 Figure 5: Graph of x + y = 7.. Find the equation of the line through the point (0, 4) and (, 0). Graph the line. First you ll need to find the slope. m = y y x x = 0 4 0 = 4 = 4 Using the point slope form with the point (0, 4). y 4 = 4 (x 0) Using the point slope form with the point (, 0). y 0 = 4 (x + ) 7

Or, the slope intercept form. y = 4 x + 4 Or, standard form. Here s the graph. 4x + y = 0 5-5 -0-5 0 5 0 5-5 -0 Figure 6: Graph of 4x + y =.. Find the slope of the line given by x 4y = 6. The slope of the line whose equation is x 4y = 6 can be found easily by solving for y. x 4y = 6 y = 4 x So the slope of this line is 4. 4. Find the slope of the line through the points (4, ) and (, 5). Using the formula. m = y y x x = 5 4 = 4 6 = 5. Find the x-intercept and the y-intercept of the line given by x 4y =. To find the x-intercept, set y = 0 and solve for x. x = x = 4 So the x-intercept is (4, 0). To find the y-intercept, set x = 0 and solve for y. 4y = y = So the y-intercept is (0, ). 8

6. Determine the vertex and the line of symmetry of the parabola given by y = (x ) 4. Graph the parabola. By inspection the vertex is (, 4) and the axis-of-symmetry is x =. - - - 0 4 - - - -4 Figure 7: Graph of y = (x ) 4. 7. Determine the vertex and the line of symmetry of the parabola given by y = x 4x +. Graph the parabola. Rewrite into standard form. y = x 4x + y = ( x 4x + 4 ) 4 + y = ( x ) By inspection the vertex is (, ) and the axis-of-symmetry is x =. 4-0 4 5 6 - Figure 8: Graph of y = x 4x +. 9

8. Determine the equation of the circle with center (, ) and radius. Graph the circle. (x ) + (y + ) = 9 - - - 0 4 - - - -4-5 Figure 9: Graph of (x ) + (y + ) = 9. 9. Determine the center and radius of the circle given by (x + ) + (y ) = 4. Graph the circle. The center is (, ) and radius 5 4-4 - - - 0 Figure 0: Graph of (x + ) + (y ) = 4. 0

0. Determine the center and radius of the circle given by x + y + 6x + 4y + = 0. Graph the circle. Rewrite into standard form. x + y + 6x + 4y + = 0 x + 6x + y + 4y = x + 6x + 9 + y + 4y + 4 = + 9 + 4 (x + ) + (x + ) = The center is (, ) and radius -5-4 - - - 0 - - - -4 Figure : Graph of x + y + 6x + 4y + = 0.. Determine the distance between the points (, 5) and (5, ). Using the distance formula. (5 ) + ( + 5) = 6 + 9 = 5 = 5. Determine the distance between the points (0, 6) and (4, ). Using the distance formula. (4 0) + ( 6) = 6 + 49 = 65

. The sum of two numbers is 8. The difference between four times the smaller number and seven is equal to the sum of two times the larger number and five. Find the numbers. Let s be the smaller number and l be the larger number. Resulting in this system of equations: s + l = 8 4s 7 = l + 5. Solving this system, s = 8 and l = 0. So the two numbers are 8 and 0. 4. The square of a negative number is fifteen more than twice the negative number. Find the number. Let x be the negative number. Resulting equation is: x = x + 5. Solving this equation gives two solutions, x = 5 or x =. So the number is. 5. The degree measures of the angles in a triangle are three consecutive integers. Find the measure of the angles. Let x, x + and x + be three consecutive integers. Resulting in this equation: x + x + + x + = 80. Solving this equation, x = 59. So the measures of the angles are 59, 60, and 6. 6. The perimeter of a rectangle is 50 meters. The width of the rectangle is five meters less than the length. Find the length and width of the rectangle. Let l be the measure of the length of the rectangle, the width of the rectangle, in terms of l, is l 5. Resulting in this equation: l + (l 5) = 50. Solving this equation, l = 50. So the length is 5 meters, and the width is 0 meters. 7. The width of a rectangle is five meters less than the length. The area of a rectangle is 76 square meters. Find the length and width of the rectangle. Let l be the measure of the length of the rectangle, the width of the rectangle, in terms of l, is l 5. Resulting in this equation: l (l 5) = 76. Solving this equation, l = 6 or l =, the later being extraneous because we cannot have a negative length. So the length is 6 meters, and the width is meters.

8. A steel rod 90 centimeters long is to be cut into two pieces, each to be bent to make an equilateral triangle. The length of a side of one triangle is to be twice the length of a side of the other. How should the rod be cut? Let x be the measure of the length, in centimeters, where the rod is cut, resulting in two pieces x and 90 x centimeters. Resulting in this equation: x = 90 x Solving this equation, x = 0. So the 90 centimeter rod should be cut at 0 centimeters. 9. A 6-foot ladder is leaning against a building. How high on the building will the ladder reach when the bottom of the ladder is 5 feet from the base of the building? Let x be the measure, in feet, how high on the building the ladder will reach when the bottom of the ladder is 5 feet from the base of the building. Resulting in this equation: 6 = 5 + x Solving this equation, x = ±, but again, due to the physical nature of the problem, we can only accept the positive value. So the 5.99 feet is how high the ladder will reach. 40. Cheap Cars rents cars for $8.00 a day and $0.0 for every mile driven. Save-On Cars rents cars for $0.00 a day and $0.08 per mile driven. You want to rent a car for one week. For how many miles will it cost you less to rent from Cheap Cars than from Save-On Cars? Let x be the number of miles driven. The resulting inequality is: 7 8 + 0.0x < 7 0 + 0.08x. Solving this inequality, x < 700. So, you have to drive less than 700 miles for Cheap Cars to be less than Save-On Cars. 4. An investment of $,500 is divided between two simple interest accounts. On one account, the annual simple interest rate is 5%, and on the second account the annual simple interest rate is 7.5%. How much should be invested in each account so that the total interest earned from the two accounts is $5? Let x be the amount of money invested at 5%, and 500 x be what remains to invest at 7.5%. The resulting equation is: 0.05x + 0.075 (500 x) = 5. Solving this equation, x = 900. So, of the $,500 you would need to invest $,900 at 5% and $,600 at 7.5% This is approximately 5 feet, 8 inches.

4. A motorboat leaves a harbor and travels at an average speed of 9 miles per hour (mph) toward a small island. Two hours later, a cabin cruiser leaves the same harbor and travels at an average speed of 8 mph toward the same island. In how many hours after the cabin cruiser leaves will the cabin cruiser be alongside the motorboat? Let x be the number of hours after the cabin cruiser leaves. equation is: 9 (x + ) = 8x. The resulting Solving this equation, x =. So, it will take hours. 4. A goldsmith combined an alloy that cost $4.0 per ounce with an alloy that cost $.80 per ounce. How many ounces of each alloy were used to make a mixture of 00 ounces costing $.50 per ounce? Let x be the number of ounces of the more expensive alloy, and 00 x be the number of ounces of the cheaper alloy. The resulting equation is: 4.0x +.8 (00 x) =.50 00. Solving this equation, x = 56. So you will need to use 56 ounces of the expensive alloy and 44 ounces of the cheaper alloy. 44. A chemist needs to make liters of an 8% acid solution by mixing a 0% acid solution and a 5% acid solution. How many liters of each solution should the chemist use? Let x be the number of liters of the 0% acid solution, and x be the number of liters of the 5% acid solution. The resulting equation is: 0.0x + 0.05 ( x) = 0.08 Solving this equation, x =.. So you will need to use. liters of the 0% acid solution and 0.8 liters of the 5% acid solution. 45. One grocery clerk can stock a shelf in 0 minutes, whereas a second clerk requires 0 minutes to stock the same shelf. How long would it take to stock the shelf if the two clerks worked together? Let t be the number of minutes that the two clerks work together on the given task. The resulting equation is: 0 t + 0 t =. Solving this equation, t =. So, it will take minutes for the two clerks, working together, to stock the shelf. 4