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MATH 043
STUDY GUIDE
NOTES
WORKSHEETS
Notes contain some copyrighted material from Lial, Hornsby, McGinnis Introductory and Intermediate algebra Published by Pearson Education
Math 043 Survival GuideNeed a 70% for Math 043 – Must attend Final – Time and Attitude are the Keys
Commit to Checking off everything on this list and you will succeedHOMEWORK – 100%
All problems copied from the book Show all work needed to complete the problem and clearly indicate your answer Correct problems in class during homework time at the start of each class
Copying from the solutions manual or another student IS NOT completing the homework-no credit. No credit for answers without required work!
TUTORIALS – 100% Every problem must be done correctly at least once (Attempted is not the same as done!) ON TIME on due date. Problems must be 100% before exams!
The tutorials have guided solutions, videos, help me solve this, you can e-mail your instructor, and get help from the LRC tutors-There is no excuse to not finish tutorials!
ATTENDANCE – 100% Attend EVERY class. DO NOT MISS! Contact Instructor as soon as possible when emergencies arrive You may be able to make-up missed days by watching pencasts
TEST/QUIZ/FINAL – how to study/prepare Attend class that day Tutorials must be completed before exams Complete chapter review and study it Study and understand the directions for every type of problem, know when it is finished Study/Rework quizzes, homework, and tutorial problems “Practice Test” – complete and grade yourself Attend review sessions Ask tutors any questions you are not sure about
Make-ups & Retests Retest for scores below 85% - below 70% is failing and should be retested
Regular tests can be retaken to improve scores and/or for skill mastery (except the last test)Retests must be approved by an LRC tutor – a blue sheet is required
Allowed 1 make-up test to be done in the assessment centerMake-up test should be finished within one week of the original test day
USE ALL AVAILABLE RESOURCESLRC tutors – Tutorials – Study Guides – Review Sessions - Teachers
Syllabus and Survival Guide Worksheet NAME:_______________________
1. What is your instructor’s name (first and last)?
2. Where can you find free tutoring for this course?
3. What supplies are needed for this course?
4. What is the lowest grade that you can receive in this course and still progress to the next course?
5. What procedures must you follow when doing homework?
6. Why is attendance important?
7. When must you complete the tutorials for each chapter?
What happens if the tutorials are not done before the test?
8. What does it mean for the tutorials to be complete before each test?
9. What is the procedure for making up a test?
10. What is the process for taking a Retest?
11. What is the difference between a make-up test and a Retest?
12. List three things that may be considered cheating in math class.
13. Summarize you teacher’s cell phone policy.
14. Where do you take make-up tests and Retest? What will you be required to show there?
15. How many late homeworks can you have?
16. What is the final exam policy?
MA 043 8.1: Review of Linear EquationsGoals: Review how to solve linear equations and inequalities in one variable.
Recall that to solve a linear equation we follow these steps:
1. Do all distributive property2. Combine like terms without crossing the equal sign3. Combine like terms by adding the opposite to both sides of the equation
a. Move the variables first, by moving the smallest one to the largestb. Move the numbers to the ‘other’ side of the equal sign
4. Multiply or divide to get a coefficient of 1 on the variable5. Check your answer!
Practice: Solve and check1) 3p + 2p + 1 = -24
2) 4x + 8x = 17x – 9 – 1
3) 5p + 4(3 – 2p) = 2 + p – 10
4) -2 + 3(x + 4) = 8x
Clearing Fractions and Decimals: For equations involving fractions, we are allowed to multiply both sides of the equation by the Least Common Denominator (LCD) of the fractions to obtain integer coefficients. Similarly, we can multiply equations involving decimals by a power of 10 in order to obtain integer coefficients.
5)
x4−2 x
3=5
6)
x+76
+ 2 x−82
=−4
7) 0.04x + 0.06(20 – x) = 0.05(50)
8) 0.09k + 0.13(k + 300) = 61
Special Cases. What kind of solutions to the following have?
9) 5(x + 2) – 2(x + 1) = 3x + 1
10) 6(2y + 4) = 4(3y + 5) + 4
*Review MA 043 Ch 5: PolynomialsThese notes represent a very brief review of polynomials. Students in 043 should already be familiar with these operations. If you need more instruction, see the book’s chapters 5-6, pencast lectures and/or visit the LRC.
Goals: Add and Multiply polynomials. Apply Rules of Exponents.
To Add polynomials, just combine like terms, by adding their coefficients.1) (2x2 + 5x – 2) + (3x2 – x + 5)
2) Add 5z2 – 10z – 12 3z 2 + 7z – 9
To Subtract, you must distribute the negative and then add.3) (14y3 – 6y2 + 2y – 5) – (2y3 – 7y2 – 4y + 6)
4) Subtract 3mn + 2m – 4n -mn + 4m + n
When multiplying all monomials, multiply coefficients and add exponents.1 ) (8k3 y )(9 ky3 )(−k2 y2)
When multiplying a monomial by a polynomial distribute.
2) −7 rx (5r−2rx+7 x 2)
When multiplying two binomials, distribute both terms, the acronym FOIL may be used.3) (4 a−5 )(3 a+7 )
Another Method for multiplying is to work vertically.
4)
(3 k−4 m)¿(k+2m)
Multiplying a sum and a difference of two binomials:
( x+ y )( x− y )=x2− y2
1) (4−5u )(4+5u)
2) (8 x2+ y )(8 x 2− y )
To Square a Binomial: 1.) Square the first term
( x+ y )2=x2+2 xy+ y2
( x− y )2=x2−2 xy+ y2 2.) Double the Product of 1st term and 2nd term
3.) Square the second term
3) (3 p−4 )2
4)( y+6 m)2
Caution: ( x+ y )n≠xn+ y n
Using order of operations on polynomials.
5) (4−x )2+3 x ( x−4 )
6) (4 x+1)( x−2)( x+1)
7) (2 x+5 y )(2 x−5 y )−( x2+3 xy+4 y2)+4 x (−2 y )
Exponent Rules A Zero Exponent Rule
a0=1
B Negative Exponent Rules
a−n= 1an 1
a−n =an ( ab )
−n
=( ba )
n
C Product Rule
aman=am+n (bases must be the same)
D Quotient Rule
am
an =am−n
(bases must be the same)E Power Rules
(am )n=amnPower to a Power: Multiply exponents
(ab )n=an bn or
( ab )
n= an
bn
Every factor is raised to exponent outside parentheses.
These rules only involve multiplication, division, and powers; not addition or subtraction.am+an≠am+n am−an≠am−n
Simplify each exponential expression using the rules of exponents.
1)(−2 x0 y6 z
y 4 )3
2)( 3 x−4 y
x−3 y2 )−2
3) ( x−2 y−3 )−2(3 x−1 y−2 )34)
(−5 ab3 )−2
(−2 a8 b−3 )3
5)
a−12 b−10 c3
a3 b−5 c36) (6 x−3 y−5 x0 )(−2 x3 y−6 y 0)2
Math 043 Ch 6: Factoring ReviewFactoring out GCF’sExample: Factor 5y2 + 10y The GCF is 5y, to write the polynomial as factors we are really dividing out the common factor and rewriting the polynomial using the
distributive property. 5 y ( 5 y2+10 y
5 y )=5 y ( y+2 )
Practice: Always look for ‘Taking out the Greatest Common Factor (GCF)’ FIRST!
1) 32 m2+24 m=
2) 16 t3+24 t 2−40 t
3) 12a2b2+15 ab+6 a3−24 ab2=
Sometimes the GCF may be a ‘grouping’.
4) r ( a+3)+4( a+3) (a + 3) is common to both, this will be one factor
= (a+3 )(r+4 ) What was in front of the (a + 3) will be the other.
5) x2( x+7 )+9( x+7)
Now we will add a step. You will need to group these 4 term polynomials into 2 pairs. Factor the GCF out of each pair and repeat the steps from 7-9.Example: pq + 5q + 2p + 10
pq + 5q + 2p + 10 q(p + 5) + 2(p + 5)
(p + 5)(q + 2)
Grouping (by pairs) method:
10) ax−ay+2 bx−2by=
11) 10 p2+15 p−12 p−18=
12) Sometimes a rearrangement may be necessary4 y−xz+xy−4 z=
13) 6 m2n+36 m+12 mn+18 m2=
TRINOMIALS SHORT PRODUCT/SUM METHOD: x2+bx+c Try to find a pair of numbers that have a product of c and a sum of b .
Example:
x2+9 x+20 , the numbers 4 and 5 have a product of 20 and sum of 9, use them to split 9xx2+4x +5x+20 , now use the grouping method from 6 . 1x ( x+4 )+5( x+4 )=( x+4 )( x+5)
Note that there is an easy shortcut here, but this only works if the coefficient of x2 is 1.
Practice:
p2+6 p+5
b2−7 b+10
m2−m−6
Using this method will also cause any ‘extra’ letters to fall into place
r2+2rs−8 s2
x2 y2+9 xy+20
y2−8 y+6
5 x4−5 x3−100 x2
Trinomials Long Product/Sum Method Example: 3 y2+7 y+2 , We are looking for two numbers with a product of 3*2 = 6 and sum of 7 . 1 and 6 work,.3 y2+ y+6 y+2 , Use 1 and 6 to split 7y into two parts, then use grouping to factory (3 y+1 )+2(3 y+1 )(3 y+1)( y+2)
Practice:
3 y2−11 y−4
6 k 2−19 k+10
12 y2−5 yz−2 z2
When the leading coefficient is negative, it is helpful but not necessary to factor out a -1 first.
−2 x2−5 x+3
Remember to always look for GCF first!
30 y5−55 y4−50 y3
Special Cases of Factoring
Difference of Squares:
1) 9 a2−16 b2=
2) 242 y3−128 y=
3) 81 m4−16=
4) x16−1=
Difference/Sum of Cubes:
7) x3−1000=
8) 27 u3−64=
9) 8 k3+125 y6=
10) 2 x8+54 x2=
Perfect Square Trinomial:
11) z2+12 z+36=
12) 9 a2+48 ab+64 b2=
13) 25 y4−70 y2 z3+49 z6=
Solving Equations by Factoring
Goals: Use zero-factor property and factoring to solve polynomials.
Zero-Factor Property: If ab=0 , then a=0 or b=0 .
Solve each equation.
1) x ( x−1)=0 2) ( x−2 )(3 x+1)=0
Quadratic Equation Form: ax 2+bx+c=0
1. Write the equation in standard form, ax 2+bx+c=0 .2. Factor the polynomial.3. Apply the zero-factor property by setting each factor equal to zero.
5) 4 k2−49=0 6) p2=12 p
7) 15 m2+7 m=2 8) −x3+x2=−6 x
Quick Check For Factoring Polynomials
Number Of TermsCHECK FOR Two Three Four or
MoreGreatest Common Factor X X X
Difference of Squares
a2−b2=(a+b)( a−b )X
Sum of Cubes
a3+b3=(a+b )(a2−ab+b2)X
Difference of Cubes
a3−b3=( a−b )(a2+ab+b2 )X
Perfect Square Trinomial
a2+2 ab+b2=(a+b)2
a2−2ab+b2=( a−b )2
X
Product Sum Method(‘Grouping Number’ Method) or Reverse
FOIL
X
Grouping by Pairs XGrouping by Cubes or Squares X
Also check your factoring by multiplying back out, and check that you have factored completely.
Factoring Decision Tree
Math 043 Notes 7.1: Rational ExpressionsRational ExpressionsGoals: Find domain, reduce, multiply and divide rational expressions.
A rational expression is a quotient of two numbers, variables, or polynomials. Below are examples of rational expressions. You will notice a rational expression is commonly called a fraction.
34
3 x2
4 x+2 6 x2−4 x+1
x 2 x+ y
3 x−4 y 2 x2−3 x
4 x2+6 x+11
Recall that the set of x values is called the domain. A zero denominator makes an undefined rational expression. Any value of x that makes the denominator be zero is NOT in the domain. A) What values of x would have to be restricted from the domains of these functions? B) Write the Domain in set notation
1)2 x+1x−5 4)
82 x2−4 x
2)3+2x
5 5)4 x
x2+2
Check GCF
How Many Terms are there?
2 terms
Check Difference of Squares
Check Sum/Difference
of Cubes
3 terms
Check Perfect Square Trinomial
Use Product/Sum Method
4 terms
Check Factor by grouping pairs
Check Factor by grouping 3 and 1
3)x+6
x2−x−6
Writing Rational Expressions in Lowest Terms
The fraction 2428 is not in lowest terms (reduced). To get it in lowest terms, divide
the numerator and denominator by the greatest common factor (GCF). The key word here is factor. Factors are multiplied together; terms are added/subtracted together. FACTORS CAN BE CANCELLED; TERMS CANNOT BE CANCELLED!
2428
= 4∗64∗7
=67
2428
≠48
or 12
by canceling the 2's
Never, Never Cancel Terms! Only Cancel Factors!
So to reduce rational expressions, we must first factor.3 x−125 x−20
=3 (x−4 )5 (x−4 )
=35 You CAN NOT CANCEL TERMS!
3 x−125x−20
≠3−35−5
To write a rational expression in lowest terms:1. Factor both numerator and denominator.2. Cancel common (or opposite) factors.3. Multiply across and leave in factored form.
Write the following rational expressions in lowest terms.
1)
2 y+4y2−4
=2)
2 y+4y2+4
=
3)
x2+2 x−3x2−3 x+2
=4)
8 m2n6 mn3 =
5)
a2−93 a+9
=
CAUTION:
a−33
≠a or a−1
6)
ac−ad+bc−bdc2−d2
=
7)
3 z2+z18 z+6
=
Opposites in a numerator and denominator equals -1.−33
=−1
2 x−2 x
=−1
−(a−b )(a−b )
=−1
Rational expressions of the form a−bb−a
=−1
7)
2 (n−4 )(1− p)(4−n)(1+ p)( p−1)
=
8) x2−11−x
9)
2r 2−r−64−r2 =
Multiplication or Division of Rational Expressions
To perform multiplication or division with rational expression:1. Factor each numerator and denominator.2. For division only, change to multiply by the reciprocal of the divisor.3. Cancel common (or opposite) factors.4. Multiply across and leave in factored form.
Multiply or Divide.
1)
c2+2 cc2−4
⋅c2−4 c+4c2−c
=
2)
r2−16r+2
⋅ r3
r 2+4 r=
3)
4 x+26 x
÷15 x+65 x
=
4)
y2+2 y5+ y
÷ 4− y2
3 y−6=
5)
4 a2−492 a2+a−21
⋅3 a2+14 a+156 a2−11a−35
=
6*) ( 2 x3+3 x2−2 x
3 x−15÷ 2 x3−x2
x2−3 x−10 )⋅ 5 x2−10 x3 x2+12x+12
Math 043 Notes 7.2: Adding and Subtracting Rational ExpressionsGoals: Add and subtract rational expressions with the same denominator. Find Least common denominator. Add and subtract rational expressions with different denominators.
Addition/Subtraction of Rational ExpressionsIn order to add or subtract rational expressions (fractions), there must be a common denominator.
310
+ 510
= 810
or 45
4x+ y
x=4+ y
xSteps to add or subtract rational expressions with common denominator:
1. Combine the numerators terms together and keep the denominator.2. Factor the numerator, if possible.3. Write in lowest terms, if possible, and leave in factored form.
Examples:
1)
2 ab
+ 3ab
+ 5b=
2)
2 y−1y2+ y−2
− yy2+ y−2
=
3)
3 x2
x2−4−12
x2−4=
4)
3m2 n
− 2 nm2 n
=
If there are not common denominators, a least common denominator must be found and the numerators ‘rewritten’ for that LCD. Essentially, both numerator and denominators are multiplied by the same quantity.
310
+ 415
=3(3)+4(2 )30
=1730
45 x
+ m2 xy
=4 (2 y )+m(5 )10 xy
=8 y+5 m10 xy
Steps for find the lowest common denominator:1. Factor each denominator.2. Multiply each factor together the GREATEST NUMBER OF TIMES A FACTOR
APPEARS IN A DENOMINATOR.3. Leave in factored form (except for numbers).
1) 72,54
72=23⋅32
54=2⋅33
Numbers can be factored using a ‘factor tree’ or repeated division.
LCD = 23⋅33=8(27 )=216
2) 75 x,60 x2 75=3⋅52
60=22⋅3⋅5
LCD = 22 (3 )(52 )x2=300 x2
3) 2m2 n2,6 m3 n
4) r,
r+6
5) 6 x−18,4 x2−12 x
6) 2 x2−6 x,2 x2−18
,x2+6 x+9
7) 2 x2+5 x−3,6 x2+x−2
8) x3−x2−x+1,
x2−x−2
Steps to add or subtract rational expressions without common denominators:1. Use the previous steps to find the LCD.2. Rewrite the numerators by multiplying by the ‘missing’ factor necessary to make the
LCD.3. Multiply and combine these terms and place over the LCD.4. Factor the numerator, if possible, and write in lowest terms, if possible.5. Leave in factored form.
Examples:1)
3 x2 x−6
+ 4x+2
− −20x2−x−6
Factor each denominator first, if necessary .
2( x−3 ) ( x−3 )( x+2 )
LCD=2( x+2)( x−3 ) Rewrite each numerator by multiplying by the ‘missing’ factor.
¿3 x (x+2 )+4 (2)( x−3 )−(−20 )(2)2( x+2)( x−3 )
=3x2+6 x+8x−24+402( x+2 )( x−3 )
=3 x2+14 x+162( x+2 )( x−3)
¿(3 x+8 )( x+2 )2( x+2)( x−3 )
=3 x+82( x−3 )
2)
6y+ 1
4 y=
3)
5x+3
+ x+2x
+ −6x2+3 x
=
4)
2m
− 4m−2
=
5)
2r−2
− r+3r−1
=
6)
yy−2
+ 82− y
=These denominators are opposites. Whenever there are
opposite factors in denominators, multiply one of the rational expressions by−1−1 .
This makes the factors be the same.y
y−2+ 8
2− y⋅−1−1
= yy−2
+ −8y−2
= y−8y−2
7)
x+3x2−6 x+9
− x+2x2−9
− 53−x
=
8)
−aa2+3 a−4
− 4 aa2+7 a+12
=
9)4+ 2
u− 3u
u+5=
10) *
2 x+6x2+6 x+9
+ 5xx2−9
+ 7x−3
=
*Note: The first rational expression can be in lower terms before addition.
11)
3 yy2+xy−2 x2
+ 4 y−xy2−x2
=
Math 043 Notes 7.3: Complex Fractions
Goals: Simplify complex fractions. Simplify rational expressions with negative exponents
Complex Fractions: expression having a fraction in the numerator, denominator or both.
Examples:
1+ 12
34−2
3
,
4y
6− 3y
,
m2−9m+1m+3m2−1
It is not acceptable to leave fractions in their complex forms. There are two methods for simplifying them.
Method 1 Simplify numerator and denominator. Then use reciprocal to divide
4+1
x
3+2x2
=
4 xx
+1x
3 x2
x2 +2x2
=
4 x+1x3 x2+2x2
=4 x+1x ÷
3 x2+2x2 =
4 x+1x ⋅
x2
3 x2+2=
( 4 x+1) x3 x2+2
=4 x2+x3 x2+2
Method 2 Multiply numerator and denominator by LCD. Simplify if possible.
4+1
x
3+2x2
=(4+1
x) x2
(3+2x2 )x2
=4 x2+1
xx2
3 x2+2x2 x2
=4 x2+x3x2+2
Practice: 1) 2)
a+25aa−37 a
2x−35x2−9
3) 4)
3+1
z
3−1z
5y
+7
83 y −1
5) 6)
1w
+1w+1
1w −
2w+1
2 p+5
p−1
3 p−2p
Simplify a rational expression with negative exponents.Start by rewriting the expression with positive exponents then use methods above.7) 8)
m−1+ p−2
2m−2−p−1 =
1m
+1p2
2m2 −1
p
b−4
b−5+2
Math 043 Notes 7.4: Solving Rational Equations
Goals: Determine domain of rational expression. Solve rational equations. Graph rational functions.
Any possible solution for a rational equation that ‘makes’ a denominator be zero, must be disregarded. It can not be a solution.
What x values would have to be excluded from the Domain (or disregarded as possible solutions)?
1)
35 x
+ 2x−1
=52)
12 x( x−2 )(3 x+1)
= x+1x (3 x+1 )(x+5 )
To solve a rational equation:1. Factor the denominators and find the LCD.2. Find all values that would make that LCD be zero. These values cannot be solutions.3. Multiply each term of the equation by the LCD.4. Solve the resulting equation.5. Disregard any possible solutions that are values found in step 2.
Solve the following equations.
1)
−320
+ 2x= 5
4 x LCD = x≠¿ ¿
2)
3x+1
= 1x−1
− 2x2−1 LCD =
x≠¿ ¿
3)
4x2+x−6
− 1x2−4
= 2x2+5 x+6 LCD =
x≠¿ ¿
4)
2x+3
− 1x−1
= −x2−3 xx2+2 x−3 LCD =
x≠¿ ¿
5)
4 x−12 x+3
=12 x−256 x−2 LCD =
x≠¿ ¿
6)
xx−3
+ 2x+3
=129−x2
LCD =x≠¿ ¿
7)
10m2
− 3m
=1LCD =m≠¿ ¿
8)
5 x+14x2−9
=−2 x2−5 x+2x2−9
+ 2 x+4x−3 LCD=
x≠¿ ¿
Graph of f(x) = 1/x Graph g(x)=
−2x−3
Asymptote:
Math 043 Notes 7.5 (Part 1)
Solve for the given variable. Start by clearing the denominator using the LCD.
1)
1f= 1
p+ 1
q for p
2)
32 a
+ 5b=1
c for b
3)L= nE
R+nr for r
4)S=nd
2(a+L )
for n
Name:_________________________
GROUP WORK OVER RATIONALEXPRESSIONS AND EQUATIONS
Find all values that would make each rational expression be undefined; in other words, the values that x could not equal.
1) r+7r2−25
2) a+7a3−3 a2+2 a
Write each rational expression in lowest terms.
3)−8 x6 y5
24 x 4 y7 4)11r2−22 r3
6−12r
5)r3−m3
r2−m2 6)8 x2+6 x−916 x2−9
Perform the indicated operations.
7)9k−186k+12
⋅ 3k+615 k−30 8)
4 z+122 z−10
÷ z2−9z2−z−20
9)p+8
p2−16 p+64− 1
p−8 10)m
m−1− 9
1−m
11)
1x
−1x+1
1x +
3x+1
12)b
a2−b2− a
a2−b2
13)4 x−94 x2−9
÷4 x2−13 x+94 x2−12 x+9 14)
5+1
y
5−1y
15)2 x−1
x2+x−2− x
x2+ x−2 16)4− y2
8 y⋅16 y2+8 y
2 y2−3 y−2
17)16−r2
r2+2 r−8÷ r2−2r−8
4−r2 18)4 p−3
3 p− p−5
2 p+ 2
p
Solve each rational equation.
19)9x=5−1
x
20)2
x−1+ 1
x+1= 5
4
21)3
r+1− 1
r−1= 2
r2−1
22)5−a
a+ 3
4=7
a
23)x+3
x− x+4
x+5=15
x2+5 x
24)x
x+4=2− 4
x+4
25)5
x−2+ 2
2−x= 4
x+1
Math 043 Notes 7.5: Applications of Rational Equations
Proportion Problems1) Biologist tagged 600 fish in a lake on the first day of May. One month later they returned and collected a random sample of 100 fish, 3 of which had been previously tagged. Approximately how many fish does the lake support based on this experiment?
Variable: x = total fish in lake
Plan: taggedfish1totalfish1
=taggedfish 2totalfish2
Equation: 600
x= 3
100
Solution: x = 20000 fish
2) The distance between Lafayette and Vincennes is 160 miles, it is represented by 6 inches on my map. If the distance between Lafayette and Fort Wayne is 120 miles, how far apart are they on the map?
3) If the two triangles are similar, find the value for x and all missing sides of the triangles.
Motion Problems
3) A plane travels 100 miles against the wind in the same time that it takes to travel 120 miles with the wind. The wind speed averages 20 miles per hour. Find the cruising speed of the plane (sometimes called ground speed).
rate time distanceagainst wind x - 20 100
with wind x + 20 120
A
B C
D
E F
2x + 44x - 57
10.5
x
4) A canal has a current of 2 miles per hour. Find the speed of Casey’s boat in still water, if the boat travels 11 miles down the canal in the same time that it goes 8 miles up the canal.
rate time distance
down
up
6) While on vacation, Jim and Annie decided to drive to Houston. During the first part of their trip, which was on the interstate, they averaged 60 miles per hour. When they got to Houston, traffic caused them to slow down considerably and they only averaged 30 miles per hour. The distance in Houston was 100 miles less than the distance on the interstate. What was their TOTAL distance, if they spend 50 minutes more on the interstate than they did in Houston?
rate time distanceinterstate 60
Houston 30
7) Lucy drove 300 miles north from San Antonio, mostly on the interstate. She usually averaged 55 miles per hour, but an accident slowed her speed through Dallas to 15 miles per hour. If her trip took 6 hours total, how many miles did she drive at the reduced speed?
rate time distanceinterstate 55
Dallas 15
Total 6
Job or Work Problems
8) Stan needs 45 minutes to do the dishes, while Bobby can do them in 30 minutes. How long will it take them, if they work together?
rate time (together) fractional partStan 1
45x
Bobby 130
x
9) Ben and Patty must clean up often after their grandson has made a mess in his playroom. Ben could always do the job in 15 minutes alone while Patty, working alone, could always clean up in 12 minutes. How long will it take them, if they work together?
rate time (together) fractional partBen
Patty
10) Ron can paint his den in 6 hours working alone. If Dee helps him, the job takes 4 hours. Estimate how long it would take Dee alone to paint the den.
rate time partRon
Dee
11) Ann can clean the bathroom alone in 20 minutes. One day her twin sister, Amy, helps Ann and they complete the cleaning in 8 minutes. How long would it take Amy alone to clean the bathroom?
rate time distance
Ann
Amy
12) A vat of acid can be filled by an inlet pipe in 10 hours and emptied by an outlet pipe in 25 hours. How long will it take to fill the vat, if both pipes are left open while acid is filling the vat?
rate time partinlet pipe 1
10outlet pipe − 1
25
ASA Math Test Self Assessment
Name:
You should complete this form (front and back) after you have received your graded test from your instructor. Your teacher may add up to 2 points to your test score, if you complete this form to their satisfaction. Form must be completed within one week of the test being returned to your class.
First check your exam preparation. Put a checkmark on what you did to study for this test, and what you plan on doing for the next test.
Method of Preparation This Test Next TestAttended every class session.
Completed all book homework, by copying, working and checking problems.
Worked tutorials ahead of time to study.
Took notes.
Read the textbook.
Asked questions about confusing parts of the assignment.
Visited tutors in the LRC.
Visited instructor’s office hours.
Attended help session, or supplemental instruction.
Watched pencasts, or MyMathLab videos.
Worked practice test.
Worked in a study group.
Used online tutor, or e-mailed instructor.
Estimate the hours, outside of class, that you spent studying in the week before this test:______
Do you feel that you adequately prepared for this exam? If no what changes could you make to your study habits to improve in the future? __________________________________________________
Were there any study methods that you found helpful on this exam that you will try again?__________
___________________________________________________________________________________
over →
37
What type of errors did you make on this exam? Review the different types of common errors below, and then categorize your errors by listing problem numbers from your test in the right column.
Problem # from your exam
1.) Misread directions Skipped instructions or Misread directions
2.) Careless errors Arithmetic errors or sign errors
3.) Concept errors Did not know how to do the problem, or did not understand the underlying principle, or did not know a formula
4.) Application errors Understood the arithmetic/algebra, but could not apply it to this problem. (common on word problems and graph problems)
5.) Test taking errors Ran out of time, missed problems early on the test due to anxiety or rushing, didn’t finish problem(s), changed answer to incorrect answer, turned test in early without checking answers
6.) Study errors Never learned materials missed during an absence, waited until the last minute to finish tutorials, did not complete homeworks or reviews
For more on test taking skills see: “Winning at Math” by Paul Nolting.
What test taking error cost you the most points for this exam? __________________________________
What steps can you take to fix some of your errors on future tests? ______________________________
____________________________________________________________________________________
What is the most challenging topic for you from this test? ______________________________________
What is the easiest topic for you on this exam? ______________________________________________
Additional comments/questions.__________________________________________________________
MA 043 Notes 9.1: Radical ExpressionsGoals: Find square roots, higher roots. Use calculator to estimate roots. Graph.
38
We are already familiar with square numbers:12 = 1 22 = 4 32 = 9 42 = 1652 = 25 62 = 36 72 = 49 82 = 6492 = 81 102 = 100 112 = 121 122 = 144
In this chapter we will be interested in the opposite process.
If a2 = 64, then a = If a2 = 81, then a = If a2 = ¼, then a = In each case, to find a we ask ourselves “what times itself will equal the
number?” Because we are finding the number that is at the root of the square, we call
this finding square roots.
Find all square roots of the following numbers:a) 49
b) 169
c) 25/196
d) .49
e) 18 ?
Principal Root:
Negative Root:
Find the following Roots:
Irrational Numbers:
39
√225 −√169
√3625
−√. 81
√−16 √17≈4 . 123 √37≈¿ ¿
Use a calculator to approximate:
Higher Roots:Similar to square roots, there are cubed roots.
We know that 27 = 33 , and we can find the cubed or third root is 3, by asking what times itself 3 times is 27?
What are the following roots?
Definition of a Root: if and only if
Radical Sign: n: the index ‘a’: the radicand
Signs for higher roots:For roots with an even index, we will follow the same convention as for square roots: is the principal, is the negative root, and the nth root of a negative number is not a real number
Find the roots below:
For roots with an odd index, there will be 1 unique root, and we can take the root of negative numbers.
Special Properties of Roots:If the index = the exponent, the following is generally true.
40
3√1254√163√216
bn=an√a=b√
n√ - n√
4√16−√814√−625
3√−27 3√85√−32−3√125195√−1
n√a n=a
The exception is if a is negative and n is even.
Be very careful when there is a negative radicand.
If the exponent is a multiple of the index, it is easy to simplify using division.
Approximate these examples using your calculator.
Example: The time for one complete swing or rotation of a pendulum is given by the following formula, where t = time in seconds, L = the length of the pendulum in feet, and g = 32 ft/sec2.
Find the time to the nearest tenth of a second for a pendulum 2 feet long.
Graphing: Graph and give the domain and range.
41
√−72 is not real .√(−7 )2 is 7, because √49 is 7Therefore, √ x2=|x| in general for even n, n√an=|a|
a ) 5√510
b ) 3√r9
c ) √m8
√153√104√2
t=2 π √ Lg
f ( x )=3√ xf ( x )=√ x
Graph and give the domain and range.
MA 043 Lesson 9.2: Rational ExponentsGoals: Convert between radical and rational exponent notation. Evaluate expressions of the form am/n Apply rules of exponents to rational exponents.
42
f ( x )=3√ x−1f ( x )=√ x+2
What would be the value of
Definition of rational exponent:
Evaluate Each by first changing to the radical expression:
Expanded Definition of rational exponents: The numerator becomes the powerAnd the denominator becomes the index of the root.
Convert the following to radical notation:
Convert the following to rational exponents:
Evaluate using definition of rational exponents:
Simplify using rules of exponents. Leave in rational exponent form:
43
91 /2?
a1/n=n√a if the root is defined .
321/5=−811/4=(−81 )1/4=
641/2=
(127 )1 /3
=
(0 .16 )1/2=
am/n=n√am or ( n√a )m
m3/4=53 /8=
5√ g2=(√7 )3=
a ) 253/2
b ) (−27 )2/3
c ) −163/4
d ) (−81)5/2
(21/2 ) ( 21/3 )=
Simplify by using rational exponents, write answer as radical:
Simplify by first converting to rational exponents, then using rules of exponents, write answers with rational exponents.
Math 043 Notes 9.3: Simplifying Radical ExpressionsGoals: Use product and quotient rules for radicals. Simplify radicals. Use Pythagorean theorem and distance formulas.
44
(−64125 )−
23=
a5/3
a7 /3 =
(m1/2 n2 /3 )4=
( x2 y )1/2
x3 /4 y−1/4 =
n3 /4 (2n5 /4−3n1/4 )=
( x−2/3
y−3/4 )4
( x−2/3 y1/4 )−2=
√212
3√49
√√81
3√ x⋅√x3√ t4
5√ t4
6√ y5⋅3√ y2
3√5√√ y
Product Rule for radicals: √a⋅√b=√ab
Use product rule:
1) √5√13 =
2) √7√x =
3)3√2 3√7=
4)6√8 r2 6√2 r3=
5)3√7√5=
Quotient Rule for radicals:
√a√b
=√ ab
Use quotient rule:
1) √10081
=
2)
3√18125
=
3)
4)
3√ x2
r 12=
5)
√18√2
=
45
Use the product rule to SIMPLIFY each radical.
1) √32=√16⋅2=√16⋅¿√2=4√2 or ¿
√32=√2⋅2⋅2⋅2⋅2=2⋅2⋅√2=4 √2
2) √300=
3) √18=
4) √48=
5) Is 3√8=6 √2 ? Which is considered simplified?
6)3√54=
7) 3√48=
8)4√243=
9) √35=
10) √25 p7=
11) √72 x3 y5=
12) √32 a5 b11=
13)3√ x12 y7 z5=
46
14)
4√32 x12 y7 z5=
Simplify by getting a smaller index, if possible.
1)12√23=
2)4√49=
3)9√a3 b6c15=
Don’t forget to simplify radical too!
Multiplying radicals with different indexes...a) Convert to exponentsb) Get a common denominator for exponentsc) Convert back to radical form
1) √5 3√4=
2)3√2 4√3=
Formulas with radicals.... Pythagorean Theorem: a2+b2=c2
1) 2)
Distance between 2 points: d=√( x2−x1 )2+( y2− y1 )
2
3) (2, -1)(5, 3)
47
8
14 20
5
4) (-3, 2)(0, -4)
MA 043 Notes 9.4: Adding and Subtracting Radicals
Goals: Simplify radical expressions using addition and subtraction.
Combining radicals with addition and subtraction follows the same rules as combining like terms. The radicals must have exactly the same radicand and index in order to add them.
Combine where possible:
1) 3√5−6√5+12√5−√5=
2) 5 3√12 y+ 3√12 y−14 3√12 y=
You must simplify these radicals first.
3) 2√11−√99+3√44=
4) 5 3√54−6 3√24+ 3√16−8 3√375=
5) 7 3√81+3 3√24−11 3√3=
48
6) 3√20 x−6√45 x+2√80 x=
7) −2 4√32−7 4√162+9 4√2=
8) 4 √20+5√18=
9) 2 xy3√ x2 y+
3√8x5 y 4=
10) √12
16+√48
64
11)
3√ 5x6 + 3√ 4
x9
49
12) Find the perimeter of this rectangle. √18
√98
MA 043 Notes 9.5: Multiplying and Rationalizing RadicalsGoals: Multiply radical expressions. Rationalize denominators. Write quotients in lowest terms.
Review: 1) √2∙√3=¿ 2) 4 ∙√5=¿ 3) √2∙√2=¿
4) (a+b )2=¿ 5) (a+b ) (a−b )=¿
Multiplication of radicals:Multiply ‘outside’ coefficients together and ‘inside' radicands together using product rule. Don’t forget to keep simplifying radicals too!
1) (4 √5 )(2√5 )=
2) (2 3√6 )(−3 3√4 )(6 3√5 )=
3) (4 √2 )(8√8)(−2√5 )=
Distributing with radicals:4) √3(2+√2)
5) √2¿)
FOIL with radical binomials:1) (3+√5)¿
50
2) (2+3√3)(4−2√5)=
3) (√6−2√5)2=
4) (2√5−3√3 )(4 √5+2√3 )=
5) (√5+√2)(√5−√2)=
6) (4+3√3)( 4−3√3 )=
7) (3+ 3√5)(3−3√5)
Division of radicals:Reduce ‘outside’ coefficients, then Reduce ‘inside’ radicands using quotient rule.
1)
14√82√2
=2)
15√610
=
3)
3√1625 3√54
=
Rationalizing the denominator: So that all answers are in standard form, we can not leave radicals in the denominator. If there is a single radical in the denominator, multiply the top and bottom by the offending radical.
51
4)
8√7
5) √136
Always simplify AND rationalize. If you simplify radicals first, numbers that reduce at the end of the problem will be smaller.
6)
3√48
7)−√ 8
45
8) √72y
9) √200 y6
k 7
Binomial denominators: Must use conjugates to rationalize.
52
This text will not distribute if the numerator is a monomial, but will distribute if the numerator is a binomial.
10)
−42+√5
11)
15√x+√2
12)
√2+√5√2−√5
Simplifying radical expressions with binomials: If the numerator and denominator have a common factor, then the quotient can be reduced by that factor. But You Must Factor the numerator first!
1)
6+2√54
=2(3+√5 )
4=3+√5
2
2)
24−36√716
=
Still simplify radicals!
3)
2√27+48
=
53
4)
4 x−6√x3
2 x2=
5)
189+√18
Simplifying Radical Expressions
There are three rules that you must ALWAYS follow when Radicals are involved:
1.) Simplify all radicals Examples √18=√9 ∙ 2=3√2 or √45 x3=√9 ∙5 ∙ x2 ∙ x=3 x√5 x
2.) Rationalize Denominators
Examples 8√7
= 8√7
∙ √7√7
=8√77 or 5
1+√3= 5
1+√3∙ 1−√31−√3
=5 (1−√3 )1−3
=−5 (1−√3)2
3.) Reduce all Fractions
Examples 2√34
=√32
or 4+2√56
=2¿¿
54
MA 043 Notes 9.6: Solving Radical EquationsGoals: Solve radical equations using power rule. Solve radical equations with indexes greater than 2.
Radical equation: An equation that contains one or more radical expressions.
Examples: √ x−4=8 , √5 x+12=3√2 x−1 ,3√6+x=27
Power Rule: If both sides of an equation are raised to the same power, all solutions of the original equation are also solutions of the new equation.Note: This does not say that all solutions of the new equation are solutions of the original, i.e some solutions you get may be extraneous so you must check answers!
Solving an Equation with radicals1.) Isolate the radical2.) Apply the power rule3.) Solve the equation, if there is still a radical repeat steps 1 and 24.) Check the solution to determine if they are extraneous or not
1.) √5 x+1=4
2.) √6 k−1=1
55
3.) √5 q−1+3=0
4.) √9 a−4=√8 a+1
5.) 3√ z−1=2√2 z+2
6.) √5−x−1=x
7.) √1−2 p−p2=p+1
Using the power rule, squaring twice
8.) √2x+3+√x+1=1
56
Higher powers:
9.) 3√2 m−1=3√m+13
10.) 3√r+1+1=0
11.) 4√8 z−3+2=0
12.) √1+√24−10 x=√3 x+5
57
13.) Do not actually solve. What would the first step be?
√ x+3=3√5+4 x
Solving Formulas:14.) Solve for k
V=√ 2km
MA 043 Notes 9.7: Complex NumbersGoals: Simplify numbers of the form √−b . Perform operations on complex numbers.
Imaginary Numbers: i=√−1and i2=−1
For any number b, √−b=i√b
Practice: Simplify using imaginary numbers.
1) √−100=10 i
2) √−37
3) √−49
4) −√−36
5) √−32
Note: It is customary to write the i before the radical so that i √2 is not confused
with √2i where i is under the radical.
When finding products like √−4⋅√−9 we must convert to i form first because √−1⋅√−1=i≠√+1=1 Therefore √−4⋅√−9=2i⋅3 i=6i2=−6
58
Practice:
1) √−16⋅√−25
2) √−2⋅√−10
3) √−15⋅√−2
4)
√−75√−3
5)
√−50√2
Complex Numbers: If a and b are real numbers, then any number of the form: a +bi is called a complex number. a is the real part and b is the imaginary part.Since i is a square root, all operations follow the same rules used for simplifying radical expressions.
Add:1) (4 + 5i) + (7 + 3i)
2) (-2 + i) + (5 – 7i)
Subtract:3) (6 + 5i) – (8 + 11i)
4) [(4 – 2i) – (3 + 5i)] – (2 + i)
Multiply:4) 4i(2 + 3i)
5) (2 + i)(3 + 2i)
6) (4 + 6i)(3 – 2i)
7) (3 + 2i)(3 – 2i)
59
Divide: For quotients use the ‘complex conjugate’ to clear i from the denominator and write your answer in the form a + bi
8)
2+ i3+2i
=(2+i)(3+2 i )
⋅(3−2i )(3−2i )
=6−4 i+3 i−2 i2
9−4 i2 = 6−i+29+4
=8−i13
= 813
− 113
i
9)
8−4 i1−i
10)
53−2 i
11)
5−ii
Simplifying powers of i: Powers of i are cyclic. Any power of i can be simplified to i,-1, -i, 1.
If you know these first 4 powers of i, you can figure out any power of i.
i1 = ii2 = -1i3 = i2i1=-ii4 = i2i2= 1
Watch what happens as we continue:
i5 = i4i = 1i = ii6 =i4i2 =1(-1) = -1i7 = i4i3=-ii8 = i4i4= 1
The powers of i cycle every fourth power. We can use this to find any power of i.
i 14= i12i2 = (i4)3i2=1(-1) = -1
i33 = i32i = 1i = i
60
+1
+i
-1
-i
i12
i31
i-2
i50
61
ASA Math Test Self Assessment
Name:
You should complete this form (front and back) after you have received your graded test from your instructor. Your teacher may add up to 2 points to your test score, if you complete this form to their satisfaction. Form must be completed within one week of the test being returned to your class.
First check your exam preparation. Put a checkmark on what you did to study for this test, and what you plan on doing for the next test.
Method of Preparation This Test Next TestAttended every class session.
Completed all book homework, by copying, working and checking problems.
Worked tutorials ahead of time to study.
Took notes.
Read the textbook.
Asked questions about confusing parts of the assignment.
Visited tutors in the LRC.
Visited instructor’s office hours.
Attended help session, or supplemental instruction.
Watched pencasts, or MyMathLab videos.
Worked practice test.
Worked in a study group.
Used online tutor, or e-mailed instructor.
Estimate the hours, outside of class, that you spent studying in the week before this test:______
Do you feel that you adequately prepared for this exam? If no what changes could you make to your study habits to improve in the future? __________________________________________________
Were there any study methods that you found helpful on this exam that you will try again?__________
___________________________________________________________________________________
over →
62
What type of errors did you make on this exam? Review the different types of common errors below, and then categorize your errors by listing problem numbers from your test in the right column.
Problem # from your exam
1.) Misread directions Skipped instructions or Misread directions
2.) Careless errors Arithmetic errors or sign errors
3.) Concept errors Did not know how to do the problem, or did not understand the underlying principle, or did not know a formula
4.) Application errors Understood the arithmetic/algebra, but could not apply it to this problem. (common on word problems and graph problems)
5.) Test taking errors Ran out of time, missed problems early on the test due to anxiety or rushing, didn’t finish problem(s), changed answer to incorrect answer, turned test in early without checking answers
6.) Study errors Never learned materials missed during an absence, waited until the last minute to finish tutorials, did not complete homeworks or reviews
For more on test taking skills see: “Winning at Math” by Paul Nolting.
What test taking error cost you the most points for this exam? __________________________________
What steps can you take to fix some of your errors on future tests? ______________________________
____________________________________________________________________________________
What is the most challenging topic for you from this test? ______________________________________
What is the easiest topic for you on this exam? ______________________________________________
Additional comments/questions.__________________________________________________________
63
MA 043 Notes 10.1: Square root PropertyGoals: Use Square root property. Solve quadratics with non-real roots.
Quadratic Equations: Equation of the form ax 2+bx+c=0where a, b, c are real numbers and a≠0.
We have already solved some quadratic equations by factoring, but as we know not all such equations can be factored.
If the quadratic equation is of the form x2=k then it can be solved by taking the square root of
both sides. x=±√k This can be justified using our knowledge of solving by factoring:
x2=kx2−k=0( x−√k )( x+√k )=0x=√k , x=−√k
1.) x2=5 2a.) x
2=64 2b.) x2=12
3.) ( x−5 )2=36 4.) 3 x2−54=0
5.) ( x−4 )2=3 6.)(3 k+1 )2=2
Solve: These equations have imaginary roots.1.) x2+1=0 2.) x
2=−4
3.) (4 m−7 )2=−27 4.) ( t+2)2=−16
64
MA 043 Notes 10.2: Completing the Square Goals: Use the completing the square method to solve quadratics.
Recall from section 10.6, if you have( x+k )2=m
where k and m are numbers, it will be easy to solve. Most problems do not come in this form, but we can ‘complete the square’ to put any quadratic into this form.
Examples of perfect squares: x2 + 4x + 4 = (x + 2)2 y2 – 6y + 9 = (y – 3)2
Now you try to fill in the blanks: 1.) x2 + 2x __ = (x ___)2 2.) u2 – 32u ___ = (u __)2
3.) n2 + 3n __ = (n ___)2 4.) p2 + ⅓p ___ = (p ___)2
1.) Now we will try this on an equation to solve:x2+8x+10=0 This can't be factored, so move the 10 to the other sidex2+8 x+ = -10 What number can we put in the blank spot to make the left side a perfect square?x2+8 x+16=−10+16 since 8 is the middle number if we take 8/2=4 and square it . So add 16 to both( x+4 )2=6 So we completed the square, now use square root property to solve itx+4=±√6x=−4±√6
2.) x2−2 x−10=0 3.) k
2+5 k−1=0
4.) 2 x2−4 x=5 5.) 8 x2−4 x=2
65
Now we can generalize this idea for any quadratic equation ax 2+bx+c=0
ax2+bx+c=0 First divide by a coefficient
x2+ba
x+ca
=0 Now get the number part(c/a ) on the other side
x2+ba
x =−ca
to complete the square take b/a divide by 2 and square
x2+ba
x+b2
4 a2 =−ca
+b2
4 a2 Add the fractions on the right, and rewrite left as a square
( x+b2a
)2=b2−4 ac4 a2 Now apply the square root property and solve
x+b2a
=±√b2−4 ac4 a2
x=-b±√b2−4 ac2 a
Now you have the quadratic formula
MA 043 Notes 10.3: Quadratic FormulaGoals: Solve quadratic equations by using quadratic formula. Use the discriminant to determine the number and type of solutions.
Quadratic formula x=−b±√b2−4 ac
2 a Where ax 2+bx+c=0
1) x2+20 x+100=0
2) 15 x2+11 x=14
3) 6 x2−5 x−4=0
66
4) 3q2−9q+4=0
5) 2 k2+19=14 k
6) (n+1 )(n−7 )=1
7) m2=4 m−5
The discriminant is the radical part of the quadratic formula, b2−4ac . It will determine how many and what types of solutions a quadratic equation will have.
Value of Discriminant Number of Solutions Types of Solutions0 1 Rational
Positive perfect square 2 RationalPositive non-perfect square 2 Irrational
Negative 2 ImaginaryUse the discriminant to describe how many and what types of solutions (roots) there will be?
8) 4 x2−28 x+49=0
9) 4 x2=4 x+3
67
10) 18 y2+60 y+82=0
MA 043 Notes 10.4: Quadratic Formula and ApplicationsGoals: Solve equations that must be put in quadratic form by using quadratic formula. Solve application problems.
Clear the denominators by using the LCD, then apply quadratic formula:
1) 4−7
r− 2
r2=0
2)
32 x
− 12 x+4
=1
3)
1x2
+1=−1x
Remember that if you apply the power rule, you must also check your solutions.
4) z=√5 z−4
5) 2 x=√ x+168
SubstitutionsThe following may be solved using substitutions and factoring, or substitutions and the quadratic formula. Let u = the middle variable(s) in the quadratic form, solve the new quadratic for u, then ‘undo’ your substitution and solve for the original variable.
1.) 9 y4−37 y2+4=0
2.) 5(r+3 )2+9(r+3 )=2
3.) 4 m2
3−3 m1
3−1=0
69
ApplicationsJob Problems1) Two pipes can fill a tank together in 5 hours. It takes one of the pipes two hours longer than the other to fill the tank alone. How long does it take each pipe to fill the tank alone?
= time alone for the fast pipe = time alone for the slow pipe
Rate Time PartFast pipe 5
Slow pipe 5
2) Carl can complete a certain lab test in two hours less time than Jamie can. If they work together, they can finish the job in two hours. How long would it take each of them working alone to complete the job?
x = time alone for Jamiex – 2 = time alone for Carl
Rate Time PartJamie
Carl
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Motion (d = rt) Problems3) In Indiana, Lafayette and Evansville are 200 miles apart. If Lennie drives his Lexus 25 miles per hour faster than Minnie drives her Mercedes, What is Lennie’s average speed if he travels from Lafayette to Evansville in 1 ⅓ hours less time than Minnie?
Rate Time DistanceLennie 200
Minnie 200
4) Lisa flew her plane for 6 hours at a constant cruising speed. She traveled 810 miles with the wind, then turned around and traveled 720 miles against the wind. The wind speed was a constant 15 miles per hour. Find the constant cruising speed of the plane.
x = cruising speed of the planeRate Time Distance
With wind x + 15 810
Against wind x – 15 720
Time with wind + Time against wind = Total Time
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5) In a total of 1 ¾ hours, Ken rows his boat 5 miles upriver and returns back to his starting point. He rows at a constant rate and the current is flowing at 3 miles per hour. What is his ‘rowing speed’ (as if in still water)?
x = rowing speed
MA 043 Notes 10.5: Applications of QuadraticsGoals: Solve formulas. Solve application problems involving quadratic equations.
Formulas: Solve the formulas involving squares.
1) A=πr 2 for r 2) D=√kh for h
3) Use the quadratic formula to solve: s=2t 2+kt for t
Projectile Problems1) A toy rocket is launched upward from ground level. Its distance from the ground (s) in feet after t seconds is given by the equations=−16 t2+208t . After how many seconds will the rocket be 550 feet above the ground?
550=−16 t2+208 t16 t2−208 t +550=0
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2) The Toronto Dominion Center is 407 feet high. A ball is projected upward from the top of the Center and its position in feet (s) above the ground after t seconds is given by the equation s=−16 t2+75t +407 . How many seconds have elapsed when the ball reaches a height of 450 feet above the ground?
2a) (Refer to the previous problem.) After how many seconds will the ball hit the ground?
2b) When will the ball reach 500 feet? How can you explain this answer?
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Pythagorean Theorem3) A ladder is leaning against a house. The distance from the bottom of the ladder to the house is 5 feet. The distance from the top of the ladder to the ground is 14 feet less than double the length of the ladder. How long is the ladder? x = ladder
4) Two cars left an intersection at the same time, one heading due north, and the other due west. Some time later, they were exactly 100 miles apart. The car heading north has gone 20 miles farther than the car heading west. How far has each car traveled?
Area Problems5) A piece of cardboard has a length that is 4 in less than twice the width. A square piece 2 in on each side is cut from every corner so that the piece of cardboard’s sides can be turned up to form a box of volume 256 in3 Find the length and width of the original piece of cardboard.
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MA 043 Notes 10.6: Graphing Quadratic Functions
Goals: Graph quadratic functions. Use standard form to predict graphs of quadratic functions.
A Quadratic Function has the form f ( x )=ax 2+bx+c or y=ax2+bx+c .
1) Graph f ( x )=x2 x -3 -2 -1 0 1 2 3
y or x2
The shape above is called a parabola. All graphs of quadratic functions will be parabolas. The point at the bottom is called the vertex. A parabola has symmetry about a vertical axis.
2) Graph F (x )=x2−1 x -3 -2 -1 0 1 2 3x2
x2−1
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Notice: The graph has been 'shifted' down 1 unit from the graph of f(x) in example 1.
A graph of the form F (x )=ax 2±k has the same shape as the graph of f ( x )=ax 2, but has
been shifted k units down (with the -) or k units up (with the +).
Predict what x2 + 2 would look like.
3) Graph F (x )=( x−1 )2
x -3 -2 -1 0 1 2 3 4
y or ( x−1 )2
This parabola has been 'shifted' one unit to the right.
A graph of the form F (x )=a (x±h)2 has the same shape as the graph of f ( x )=ax 2
, but has been shifted h units right (with the -) or h units left (with the +). Notice: Hop the opposite direction of the sign. HOPPOSITE Rules!
A graph could be shifted both up or down and left or right.
4) Graph F (x )=( x+3 )2+2 x -3 -2 -1 0 1 2 3
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ySubtract 3 from the x and add 2 to the y value.
5) Graph f ( x )=− 1
2x2
x -3 -2 -1 0 1 2 3y
The graph now opens downward instead of upward. The negative sign in the front caused a vertical reflection over the x axis.
A graph of the form F (x )=−ax 2 is a vertical reflection of the graph of f ( x )=ax 2
.
x -3 -2 -1 0 1 2 3
6.) Graph f ( x )=2 x2y
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Notice that the graph on Example 5 became wider with a coefficient of -1/2 and narrower in Example 6 with a coefficient of 2. Can you generalize this rule?
In general the standard form of a graph of the quadratic function is given by F (x )=a (x−h)2+k (h,k) will be the vertex, and a will determine the direction the parabola opens and how wide it is.
These transformations will work for all functions. Prove that is works for lines by considering the equations: y = x, y = x + 1, y = 2x, y = ½ x,
Ma 043 Notes 10.7: More on Graphing ParabolasGoals: Find vertex of parabola. Graph quadratic functions. Find x-intercepts. Solve max/min.
In the previous section the general the standard form of a graph of the quadratic function was given by F (x )=a (x−h)2+k Where (h,k) was the vertex, and a will determine the direction the parabola opens and how wide it is.
But the standard quadratic equation is given as y=ax2+bx+c
So to make the general quadratic equation easier to graph, we will use completing
the square to rewrite it as F (x )=a (x−h)2+k
Example: Use completing the square to find the vertex of f ( x )=x2−4 x+5x2−4 x+5 This is not an equation, so when completing the square we must x2−4 x +5 keep everything on one side. But we still take the x coefficient x2−4 x+4 -4+5 and divide by 2 then square it to complete the square, but now ( x−2 )2+1 we have to add and subtract the number to maintain equality. In this new format, we can easily see that the vertex is (2, 1)
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Find the vertex of the following parabolas
1.) g( x )=x2+4 x−9 2.) f ( x )=x2+6 x+7
3.) p( x )=2x2−4 x+1
4.)h( x )=−3 x2+6 x−1
Alt Method for Finding the Vertex, use a formula.Similar to the way we generalized completing the square to find the quadratic formula, we can generalize completing the square to find a formula for the vertex:
f ( x )=ax 2+bx+c→ f ( x )=a( x2+ba
x )+c→ f ( x )=a( x2+ba
x+b2
4 a2 −b2
4 a2 )+c→
f ( x )=a ( x2+ba
x+b2
4 a2 )+c−b2
4 a2 → f (x )=a (x+b2 a
)2+4 ac−b2
4 a
The most useful way to use this information is to see that the x-part of the vertex will be –b/2a . We can then plug this value back into the given quadratic formula to find the y part of the vertex.
Example:
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f ( x )=x2+6 x+7
Has x-part of vertex =−62(1)
=−3
now find f (−3) to find the y-partf (−3 )=(−3)2+6(−3)+7=−2So vertex ( -3, -2)
Practice: Find the vertex
1) g( x )=x2−4 x+1
2) f ( x )=4 x2−x+5
Finding the x-intercepts: Another useful piece of information about the parabola would be where it intersects the x-axis. This would be the x-intercepts, to find them set y = 0 and solve for x.
Example:f ( x )=x2−x−6
From the above examples this has vertex (½, -6¼)
We can also see that if
0=x2−x−60=( x−3 )( x+2)x=3 , x=−2 So we have points (3, 0 ), (-2, 0 )
If you can’t solve by factoring, just use the quadratic equation. With just those 3 points and symmetry we can get a fairly good graph.
Practice: Find the x-intercepts.
1) f ( x )=x2+6 x+8
2)g( x )=x2−3 x−3
3) h( x )=x2+2 x+3
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Recall that the discriminant is the b2 – 4ac part of the quadratic formula. How could the discriminant help us predict the number of x-intercepts for an equation?
How would you find the y intercepts?
Practice graphing some of the Parabolas from this section.
Max and MinThe vertex of an upward facing parabola is the parabola’s lowest points, and the vertex of a downward facing parabola is the highest points. Using this fact we can answer application problems involving maximums and minimums when the relationship can be modeled with a quadratic equation.
Example: A rocket is launched from a 60 foot platform with an initial velocity of 112 feet. It’s height is as a function of time is given by: h( t )=−16 t2+112 t+60
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What is the maximum height it will reach?
We will actually find the time at which it reaches the max by using the vertex
formula for the first coordinate: x=t=−b
2 a= −112
2(−16 )=3 . 5
So at time 3.5 seconds the rocket is at its max height, which we find by:
h(3 . 5)=−16(3 .5 )2+112 (3. 5 )+60=256 feet
Practice: A farmer has 120 feet of fence to enclose a rectangular area next to a building, using the building as one side of the rectangle. Find the maximum area he can enclose and the width he will use.
Practice: Find a pair of numbers that adds to 80 and whose product is a maximum.
Practice: A charter boat company charges $50 per passenger plus a $5 additional dollars per person fee for any unfilled seats. The boat has a capacity of 100 people, let x represent the number of no-shows.
a) Find the cost function f(x) = cost*number of people = (50 + 5x)(100 – x)
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b) What number of no-shows will cost you the most?
c) What would this cost be?
Practice: A company finds that their employee’s productivity over a quarter can be modeled with the quadratic equation: P ( x )=3 x2−6 x+1345 Where x is days and P is units produced.Find the vertex, what do your values represent?
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ASA Math Test Self Assessment
Name:
You should complete this form (front and back) after you have received your graded test from your instructor. Your teacher may add up to 2 points to your test score, if you complete this form to their satisfaction. Form must be completed within one week of the test being returned to your class.
First check your exam preparation. Put a checkmark on what you did to study for this test, and what you plan on doing for the next test.
Method of Preparation This Test Next TestAttended every class session.
Completed all book homework, by copying, working and checking problems.
Worked tutorials ahead of time to study.
Took notes.
Read the textbook.
Asked questions about confusing parts of the assignment.
Visited tutors in the LRC.
Visited instructor’s office hours.
Attended help session, or supplemental instruction.
Watched pencasts, or MyMathLab videos.
Worked practice test.
Worked in a study group.
Used online tutor, or e-mailed instructor.
Estimate the hours, outside of class, that you spent studying in the week before this test:______
Do you feel that you adequately prepared for this exam? If no what changes could you make to your study habits to improve in the future? __________________________________________________
Were there any study methods that you found helpful on this exam that you will try again?__________
___________________________________________________________________________________
over →
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What type of errors did you make on this exam? Review the different types of common errors below, and then categorize your errors by listing problem numbers from your test in the right column.
Problem # from your exam
1.) Misread directions Skipped instructions or Misread directions
2.) Careless errors Arithmetic errors or sign errors
3.) Concept errors Did not know how to do the problem, or did not understand the underlying principle, or did not know a formula
4.) Application errors Understood the arithmetic/algebra, but could not apply it to this problem. (common on word problems and graph problems)
5.) Test taking errors Ran out of time, missed problems early on the test due to anxiety or rushing, didn’t finish problem(s), changed answer to incorrect answer, turned test in early without checking answers
6.) Study errors Never learned materials missed during an absence, waited until the last minute to finish tutorials, did not complete homeworks or reviews
For more on test taking skills see: “Winning at Math” by Paul Nolting.
What test taking error cost you the most points for this exam? __________________________________
What steps can you take to fix some of your errors on future tests? ______________________________
____________________________________________________________________________________
What is the most challenging topic for you from this test? ______________________________________
What is the easiest topic for you on this exam? ______________________________________________
Additional comments/questions.__________________________________________________________
MA 043 8.1.2: Review of Linear Inequalities
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Solving inequalities follows the same rules with one major exception. If you multiply or divide both sides by a negative number, you must reverse the inequality symbol.
Practice: Solve the following inequalities, graph and give answer in interval notation.
graphs intervals1.) x – 5< -10
2.) 3x + 3 ≥ 2x – 1
3.) -m + 6 < 2m
4.) 8k ≤ -40
5.) -7s > -63
6.) 6(x – 1) + 3x ≥ -x – 3(x + 2)
7.)
14(m+3 )+2≤3
4(m+8 )
Three part inequalities follow the same rules but you must do each operation to all three parts8.) 3 < x + 10 ≤ 8
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9.) 5 ≤ 3x – 4 < 8
10.) 4 ≤ 5 – 9x < 8
11.) 2(2x+1) ≤ 4x+5
MA 043 Notes 8.2: Compound InequalitiesGoals: Distinguish between AND and OR. Solve and graph compound inequalities.
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The word AND in math means both criteria must be true at the same time, this will be the intersections of the graphs. The word OR means either criteria is true, this will be the union of the graphs.
Use the Venn diagram to answer the questions:
Who is in Math 043 AND Eng 111?Which students are in Math 043 OR Eng 111?
To Solve compound inequalities, First solve each inequality, then determine if the solution is an intersection or a union.Examples:x + 2 < 5 AND x – 3 ≥ -4
Copyright © 2010 Pearson Education, Inc. All rights reserved.
3x + 7 ≤ 1 OR -2x < -10 x ≤ -2 OR x > 5
Practice: Solve, Graph and give interval solution.1.) x + 3 < 1 and x – 4 ≥ -12
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John Joe Jesse
JamesJenny
JeffJuan
Math 043 Eng 111
My Advisees
2.) x + 2 > 3 or 2x + 1 < -3
3.) x – 1 > 2 or 3x + 5 ≤ 2x + 6
4.) 3x – 2 ≤ 13 and x + 5 ≤ 7
5.) x < 3 and x ≥ 6
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6.) 3x ≤ x + 12 or 3x – 8 ≥ 10
MA 043 Notes 8.3: Absolute Value
Goals: Solve equations and inequalities involving absolute value.
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Absolute Value: |x| is the distance from x to 0.
If |x| = 5, then x = 5 or x = -5. Copyright © 2010 Pearson Education, Inc. All rights reserved.
Because absolute value represents the distance from zero, |x| > 5 will be all numbers that are more than 5 units from 0.
Therefore, |x| > 5 has solution: x > 5 OR x < -5. Which is (-∞, -5) U (5, ∞)
For |x| < 5, the solution will be all numbers that are less than 5 units from zero, which will be all numbers between -5 and 5, which is -5 < x < 5 x > -5 AND x < 5
Note: We can check the solutions by plugging any value in the solution set into the original to see if it does make the equation / inequality true.
Based on the above examples, all equations / inequalities that involve absolute value will require us to solve 2 separate equations / inequalities.
Solving Absolute Value Equations and InequalitiesLet k be a positive real number, and p and q be real numbers.
Case 1: To solve |ax + b| = k, solve ax + b = k OR ax + b = -kThis case will usually result in an answer of the form {p, q}.
Case 2: To solve |ax + b| > k, solve ax + b > k OR ax + b < -kThis case will result in a solution set of the form (-∞, p) U (q, ∞)
Case 3: To solve |ax + b| < k, solve -k < ax + b < k This case will result in an answer of the form (p, q)Practice: Solve the following equations/inequalities, graph and give the set/interval solution.1) |x| = 7 2) |x + 3| = 5
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3) |3x – 4| = 11 4) |1 - ¾x| = 7
5) |x| > 7 6) |x + 3| > 5
7) |3x – 4| ≥ 11 8) |5 – x| > 8
9) |x| < 7 10) |x + 3| < 5
11) |3x – 4 | ≤ 11 12) |2s – 6| < 6
Note: The absolute value must be isolated in order to solve13) |x + 3| + 5 = 12 14) |5x + 2| - 9 = -7
93
15) |x + 2| - 3 ≥ 2 16) |3x + 2| + 4 < 15
Solving an equation with two absolute valuesTo solve an equation of the form |ax + b|= |cx + d| solve the compound equation :
ax + b = cx + d OR ax + b = -(cx + d)
Example: Solve |3x + 1| = |2x + 4|3x + 1 = 2x + 4 OR 3x + 1 = -(2x + 4) x + 1 =4 3x + 1 = -2x – 4 x = 3 5x = -5 → x = -1
Solution: {-1, 3}Check: |3(-1) + 1| = |2(-1) + 4| →|-2| = |2| or |3(3) + 1| = |2(3) + 4| → |10| = |10| Yes!
Practice:1) |x – 1| = |5x + 7| 2) |z + 6| = |2z – 3|
3) |x + 2| = |x – 5|
Special Cases with Absolute ValueNote: Absolute value can never be negative, and is equal to zero only when the expression in the absolute value is equal to zero.
Examples: |2x + 7| = -3 Impossible! Therefore, No Solution94
|4z – 10| < -13 Impossible! Therefore, No Solution
|5w – 3| = 0 Only If 5w – 3 = 0. Therefore w = 3/5
|3d – 2 | > -1 This will always be true! Therefore, All Real Numbers
|x + 2| ≤ 0 Need only solve for = 0. Therefore x = -2
Practice:1) |5x – 2| ≥ 0 2) |8n + 4| ≤ -4
3) |3r + 6| = 0 4) |4 + 9y| + 5 ≥ 4
5) |5n – 2| + 2 < 1 6) |3w + 9| + 2 = 2
Review of Linear EquationsStandard Form:
Slope:
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Point-Slope Form:
Slope Intercept Form:
Example A: Find an equation in standard form for a line with Slope -3 and through (10, 8)
Example B: Find an equation in slope-intercept form with Slope 2/5 and through (3, -4)
Example C: Find an equation in standard form for a line through (-2, 6) and (1, 4)
Example D:What is the slope and y-intercept of the line with the equation: 2x – 5y = 1
Example E: Find an equation in slope-intercept form through (-8, 3) and perpendicular to 2x – 3y = 10
Example F: Find an equation in Standard form through (2, 1) and parallel to 4x – 3y = 16
Horizontal lines:
Vertical lines:
Example G: Find the equation of a horizontal line through the point (-2, 6)
Example H: Find the equation of a vertical line though the point (-2, 6)
Review of Graphing y=mx+bTo quickly graph a line with this form, we first plot the y-intercept which is the point (0,b) Then we use that point and the slope m to find another point. Example: Graph
96
y=35
x−2
First graph the y-intercept which is the point (0, -2)
Now use the y-intercept as a starting point and the rise/run definition of slope to find more points.
Practice: Use slope-intercept to graph
y=−2x+6 3x – 5y = 25
y = 5 Graph a line with = ½ through ( 2, -3)
MA 043 Notes 4.1: Solving Systems of Linear Equations(Graphing)
Goals: Determine it a point is a solution of a set of equations. Solve systems by graphing.
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System of Linear Equations: A set of two or more linear equations containing the same variables.
Solution of a system of linear equations: An ordered pair that makes BOTH equations true at the same time.
Is the ordered pair (3, -12) a solution of the system:
Is the ordered pair (2, 5) a solution of the system:
Solve the system by graphing:
How Many points might a set of lines have in Common?
Solve the system by graphing:
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2 x+ y=−6x+3 y=2
x+ y=52 x− y=4
3 x−2 y=−45 x+ y=15
Solve the system by graphing:
Solve the system by graphing:
If two lines have the same slope, but a different intercept, how many solutions will there be?
If two lines have the same slope and intercept, how many solutions are there?
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2 x− y=46 x−3 y=18
x+ y=84 x+4 y=32
x−3 y=−24 x+ y=5
If two lines have different slopes, how many solutions are there?
Practice: Without actually graphing, determine how many solutions the set has:
See figure 1: The blue line represents the Sales in millions of digital cameras since 2000. The red line represents the sales of conventional cameras in millions since 2000. Where do these lines intersect and what does this ordered pair represent?
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
MA 043 Notes 4.2: Solving Systems of Linear Equations (Substitution)Goals: Solve systems of equations by substitution. Solving by graphing may not be accurate, and can be time consuming. Another method to solve systems of equations is substitution.
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2 x−3 y=53 y=2 x−7
2 x+ y=4x+2 y=9
Solving by SubstitutionExample
This system is a good candidate for substitution, we can substitute the y = 5x from the second equation into the first equation, in orderTo get a linear equation in one variable to solve.
Now substitute the x-value into any equation to get the corresponding y-value
Practice Problems: Solve the systems by substitution and check.1)
2)
3)
4)
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2 x− y=6y=5 x
2 x−5 x=6→−3 x=6→ x=−2
y=5 (−2 )=−10Solution: (-2, -10)
5 x−3 y=−6x=2− y
4 x+ y=52 x−3 y=13
3 x+2 y=13 x−4 y=−11
x=5−2 y2 x+4 y=6
5)
6)
7)
MA 043 Notes 4.3: Solving Systems of Linear Equations (Elimination)Goals: Solve systems of equations by elimination.
Substitution is a good method, because it always works, but it can also become cumbersome if one of the variables is not easy to solve for. A more sophisticated method for solving systems of equations is the Elimination method.
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2 x=6−2 yx+ y=3
0 .08 x−0 .01 y=1 .30 .22 x+0 .15 y=8 . 9
12
x+14
y=14
15
x−110
y=310
Elimination Method:Consider the system
Since the left side equals the right side of each equation, the left sides can be ‘added’ and the right sides can be ‘added’ and a new equation is formed. When this is done, the ‘y’ terms are eliminated. ‘Plug’ this value for x in either of the original equations and the matching y value is found.
The solution is (2,1).
Practice:1)
This method is called elimination because when we added the equations, one of the variables was eliminated. The elimination happened because the coefficients were opposite. If the coefficients are not opposite, we must make them opposite by multiplying the equations to form equivalent equations.Example
It does not matter which variable you eliminate. I chose x.2 and 3 have an LCM of 6, so I multiply the first equation by 3And the second equation by -2, to eliminate x.
Practice: Solve by elimination method1)
*2)
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2 x+3 y=73 x−3 y=3
5 x=10x=2
2(2)+3 y=74+3 y=73 y=3y=1
−2 x+3 y=−102 x+2 y=5
2 x+3 y=193 x−7 y=−6
6 x+9 y=57−6 x+14 y=12
3(2 x+3 y=19)→−2(3 x−7 y=−6)→
23 y=69y=3→ 2 x+3 (3)=19
2 x=10x=5
(5,3)12 x+10 y=78 x−15 y=10
12 x+10 y=78 x−15 y=10
3)Hint: you are going to multiply anyway, why not clear the fractions?
4) Hint: you need to have like terms lined up!
5)
MA 043 Notes 4.4: Applications of Systems of Equations
Goals: Solve application problems using two equations and two unknowns.
For these application problems: (1) Define two variables.(2) Write two equations to relate the information and variables.(3) Use the elimination or substitution methods.
104
56
x−13
y=23
12
x+34
y=134
2 x+ y=6−8 x=4 y−24
x=4−3 y2 x+6 y=−3
Example: A rectangular soccer field has a perimeter of 360 yards. Its length is 20 yards more than its width. What are the dimensions of the field?
Variables: W = width, L = length
Plan: P = 2L + 2W and Length is 20 yards more than width
Equations: 2L + 2W = 360 L = W + 20Solution: by substitution method: 2(W + 20) + 2W = 3602W + 40 + 2W = 360 4W = 320 Solution: Width 80 yards, Length 100
yards W = 80L = 80 + 20 = 100
Practice: 1) Tennis courts are 42 feet longer than they are wide, and have perimeters of 228 ft. What are the dimensions of a standard Tennis court?
2) During the summer my family visited a waterpark. For our family 3 kids and 2 adult tickets cost $105.05, for our friends 2 kids tickets and 1 adult ticket cost $58.12. What were we charged for each kind of ticket?K = price per kids ticketA = price per adult ticket
3) A grocer has some $4 per pound coffee and some $8 per pound coffee, which she will mix to make 40 pounds of coffee selling for $5.60 per pound. How many pounds of each kind of coffee should she use?
Number of Pounds
unit price/pound final worth
cheaper coffee x 4expensive coffee y 8final mix 40 5.6
105
65
x
y2x+10
4) During the 2009 MLB season, the Padres played 162 games. They lost 2 more games than they won. What was their record that year?
5) See the drawing. Find the measures of angles x and y.
6) How many liters each of 15% acid and 33% acid should be mixed to get 40 liters of 21% acid?
+ =
106
7) My brother’s lives 200 miles away. If we leave our houses at the same time and I drive 20 mph faster than he does, it takes us 2 hours to meet at a State Park located in between our cities. How fast were we driving?
Rate Time DistanceMe
Brother
8) In his motorboat, Nigel travels upstream at top speed to his favorite spot, a distance of 36 miles, in two hours. Returning downstream, still at top speed, the trip only takes 1.5 hours. Find the top speed of the boat (as if it was in still water) and the speed of the current.
Rate Time Distanceupstream
downstream
Math 043 Notes 3.6: Introduction to FunctionsGoals: Define relations and Functions, find Domain and Range, use vertical line test, use function notation.
Relations and FunctionsDefinition: A relation is a set of ordered pairs. Relations can be represented by lists (tables, sets, mappings), graphs, or equations of two variables.
All of these are relations.
x y {(0 ,−4 ) ,(9,9 ), ( 23
,−8 ) ,(3,9)}2 -3
0 5
107
3-24
-2
8
½ -2
y=2 x3+3 x2−5 x
Definition: A function is a relation where each x value is paired with one y value. This means that for each x input, there can be only one y output. The x value is called the independent variable and the y value is called the dependent variable.
A) Determine if these relations are functions.
1) {(2,3) ,(4,8 ) ,(−1,6 ) ,(0 ,−2)}
2)
3) x 4 2 5 6 y 3 1 1 7
B) To determine if an equation is a function, First write it in the form y = a quantity with the variable x. Consider what happens for each input x, is there only 1 output y ?
Are these equations functions?
1) 4 x+2 y=12 2) x2− y=0
3) y2=x 4) y=√x−3
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-4-39
12310
5) y=±3x 6)y=3 x−5
2 x
C If a graph passes the ‘vertical line test’, it is a function.Vertical Line Test: If every vertical line intersects the graph of a relation only once, then the relation is a function.
Are these graphs functions?
1) 2) 3)
4) 5)
D Definition: The domain of a function is the set of x values. The range of a function is the set of y values.
Find the domain and range of each function.1) {(−3,0 ) ,(−2,5 ) ,(6,0 ),(2,3 )}
2)
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(-3,-3)(8,-3)
(0,10)
Find Domain only
3)y=3
x 3b.) y= x+2
x2−4
4) y=√x 4b) y=√x+2
E Function Notation replaces the variable y with f(x) We read this as f of x, it does not mean f times x, rather it gives our function a name and acts as a template for the function. We can use almost any letter besides f, but f, g, h are the most common.
Given the following function, find the values.
1)f ( x )=5−3 x
2 f (7) means use the given template and replace the x's with a 7 and evaluate,
therefore f (7 )=5−3(7 )2
=5−212
=−162
=−8 , so f (7 )=−8
f (−3 )=f (0)=
2)g( x )=5 x2−1
g(−1)=g(2 )=g(u )g( x+1)
3) Suppose f ( x )is a function defined by this set: {(2,3 ),(5 ,−1) ,(6,0 ),(−2,9 )}
Find:
f (5)=f (6 )=
4) Suppose g( x )is a function defined by this mapping:
Find:
g(3 )=g( 4 )=
Write these equations using function notation, then evaluate f(2).
1) 3 x−4 y=18 2) 2 y=3 x2−2110
234
16-1
Many real life applications can be put into function form.
If r represents the length of the femur, the height of a man is determined by the function h(r )=69 . 09+2 .24 r . (Both r and h(r) are in centimeters.) Find the height of the man, if his femur bone measures 57 cm.
Johnny gets paid $10 per hour, plus a $1 safety bonus. So his pay is a function of how many hours he works p ( x )=10 x+1 What is p(15) and what does it represent?
ASA Math Test Self Assessment
Name:
You should complete this form (front and back) after you have received your graded test from your instructor. Your teacher may add up to 2 points to your test score, if you complete this form to their satisfaction. Form must be completed within one week of the test being returned to your class.
First check your exam preparation. Put a checkmark on what you did to study for this test, and what you plan on doing for the next test.
Method of Preparation This Test Next TestAttended every class session.
111
Completed all book homework, by copying, working and checking problems.
Worked tutorials ahead of time to study.
Took notes.
Read the textbook.
Asked questions about confusing parts of the assignment.
Visited tutors in the LRC.
Visited instructor’s office hours.
Attended help session, or supplemental instruction.
Watched pencasts, or MyMathLab videos.
Worked practice test.
Worked in a study group.
Used online tutor, or e-mailed instructor.
Estimate the hours, outside of class, that you spent studying in the week before this test:______
Do you feel that you adequately prepared for this exam? If no what changes could you make to your study habits to improve in the future? __________________________________________________
Were there any study methods that you found helpful on this exam that you will try again?__________
___________________________________________________________________________________
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What type of errors did you make on this exam? Review the different types of common errors below, and then categorize your errors by listing problem numbers from your test in the right column.
Problem # from your exam
1.) Misread directions Skipped instructions or Misread directions
2.) Careless errors Arithmetic errors or sign errors
3.) Concept errors Did not know how to do the problem, or did not understand the underlying principle, or did not know a formula
4.) Application errors Understood the arithmetic/algebra, but could not apply it to this problem. (common on word problems and graph problems)
5.) Test taking errors Ran out of time, missed problems early on the test due to anxiety or rushing, didn’t finish problem(s), changed answer to incorrect answer,
112
turned test in early without checking answers6.) Study errors Never learned materials missed during an
absence, waited until the last minute to finish tutorials, did not complete homeworks or reviews
For more on test taking skills see: “Winning at Math” by Paul Nolting.
What test taking error cost you the most points for this exam? __________________________________
What steps can you take to fix some of your errors on future tests? ______________________________
____________________________________________________________________________________
What is the most challenging topic for you from this test? ______________________________________
What is the easiest topic for you on this exam? ______________________________________________
Additional comments/questions.__________________________________________________________
MA 043 Notes Section 5.7: Dividing a Polynomial by a PolynomialGoals: Divide a polynomial using long division. Apply division to geometry problems.
First let us review how to do long division with numbers:
23 will not go into 5, but it will go into 52, 2 times, subtract 46 2×23 = 46 from 52, 61 23 will only go into 61 2 times, now subtract 46 from 61.
46 61 46
15 23 will not divide into 15, so we will write it as a remainder, over the divisor of 23
Answer:
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23 2|521
23 22|521
22 1523
Now let’s try the same thing with polynomials:
Start by looking at p, it will go into p3 , p2 times, giving:
Now we will subtract.-4p2 – 5p + 9 And continue the division process.
p goes into -4p2 , -4p times
-4p2 – 5p + 9 Now subtract, remember to change the signs. -4p 2 -8p 3p + 9 continues --->
p goes into 3p +9 3 times
-4p2 – 5p + 9 -4p 2 -8p Subtracting leaves 3, which p does not divide into
3p + 9 So we will write 3 as a remainder 3p + 6
3Answer:
Practice: 1)
114
p3−2 p2−5 p+9p+2
=p+2|p3−2 p2−5 p+9
p+2 p2 |p3−2 p2−5 p+9
p3+2 p2
p+2 p2−4 p |p3−2 p2−5 p+9
p3+2 p2
p+2 p2−4 p+3 |p3−2 p2−5 p+9
p3+2 p2
p2−4 p+3+ 3p+2
x2−x−6x+2
2)
3)
4) If a rectangle has area x3 + 4x2 +8x + 8, and Width x +2, What is its Length?
115
x3−1x−1
3 k3+9k−14k−2