GeometryCPESummerPacket
RAMAPO-INDIANHILLSSCHOOLDISTRICT
DearRamapo-IndianHillsStudent:
Pleasefindattachedthesummerpacketforyourupcomingmathcourse.Thepurposeofthesummerpacketistoprovideyouwithanopportunitytoreviewprerequisiteskillsandconceptsinpreparationforyournextyearβsmathematicscourse.Whileyoumayfindsomeproblemsinthispackettobeeasy,youmayalsofindotherstobemoredifficult;therefore,youarenotnecessarilyexpectedtoanswereveryquestioncorrectly.Rather,theexpectationisforstudentstoputforththeirbesteffort,andworkdiligentlythrougheachproblem.
Tothatend,youmaywishtoreviewnotesfrompriorcoursesoron-linevideos(www.KhanAcademy.com,www.glencoe.com,www.youtube.com)torefreshyourmemoryonhowtocompletetheseproblems.Werecommendyoucircleanyproblemsthatcauseyoudifficulty,andaskyourteacherstoreviewtherespectivequestionswhenyoureturntoschoolinSeptember.Again,giventhatmathbuildsonpriorconcepts,thepurposeofthispacketistohelpprepareyouforyourupcomingmathcoursebyreviewingtheseprerequisiteskills;therefore,thegreatereffortyouputforthonthispacket,thegreateritwillbenefityouwhenyoureturntoschool.
PleasebringyourpacketandcompletedworktothefirstdayofclassinSeptember.Teacherswillplantoreviewconceptsfromthesummerpacketsinclassandwillalsobeavailabletoanswerquestionsduringtheirextrahelphoursafterschool.Teachersmayassessonthematerialinthesesummerpacketsafterreviewingwiththeclass.
Ifthereareanyquestions,pleasedonothesitatetocontacttheMathSupervisorsatthenumbersnotedbelow.
Enjoyyoursummer!
RamapoHighSchoolMichaelKaplanmkaplan@rih.org201-891-1500x2255IndianHillsHighSchoolAmandaZielenkieviczazielenkievicz@rih.org201-337-0100x3355
GeometryCPESummerPacket
Name: ___________________________________
Previous Math Course: _____________________
Ramapo- Indian Hills High School Summer Math Packet
Geometry CPE
To the students: The following set of review problems were designed to prepare you for your Geometry CP/CPE course. You can either print out the problems or complete them on a separate piece of paper. Please bring the packet and your completed work at the beginning of the new school year.
Thank you.
GeometryCPESummerPacketSection 0.4: Algebraic Expressions
Evaluate each expression if π = 2, π = β3, π = β1, and π = 4.
1.) 2π + π 2.) !"!!
3.) !!!!!
4.) !!!!!!
5.) (π + π)! 6.) π + π!
PEMDAS β the order in which you evaluate expressions P β Parenthesis E β Exponents M β Multiplication (from left D β Division to right) A β Addition (from left S β Substraction to right)
GeometryCPESummerPacketSection 0.5: Linear Equations
Solve each equation.
7.) π + 11 = 3 8.) !!π = β6 9.) !
!"+ 15 = 21
10.) 9π + 4 = 5π + 18 11.) β2π¦ + 17 = β13 12.) β2(π + 7) = 15
Example: Solve the equation. !!π + 1 = 11
!!π + 1 = 11 Subtract 1 from each side.
β1 β 1 (3) !
!π = 10(3) Multiply each side by 3.
!!!= !"
! Divide each side by 2.
GeometryCPESummerPacketSection 0.7: Ordered Pairs
Write the ordered pair for each point shown in the coordinate plane.
13.) B = ( , ) 14.) C = ( , )
15.) E = ( , ) 16.) A = ( , )
17.) D = ( , ) 18.) F = ( , )
Important Notes:
β’ Points in the coordinate plane are known as ordered pairs. β’ Ordered pairs are written in the form . β’ The x-axis and y-axis divide the coordinate plane into four quadrants. β’ The point of intersection of the axes is the origin. β’ The origin is located at .
Example:
π΄ (3,5) - Quadrant I
π΅(β3, 6) - Quadrant II
πΆ(β3,β5) - Quadrant III
π·(3,β5) - Quadrant IV
GeometryCPESummerPacketSection 0.8: Systems of Linear Equations
Solve by graphing.
19.) π¦ = βπ₯ + 2
π¦ = β !!π₯ + 1
20.) π¦ β 2π₯ = 1
2π¦ β 4π₯ = 1
Solving Systems by Graphing Example: Solve the system of equations by graphing.
π¦ = 2π₯ β 1 π₯ + π¦ = 5
Step 1: Graph each line. The second equation here needs to be changed to π¦ = ππ₯ + π. π₯ + π¦ = 5 βπ₯ β π₯ π¦ = βπ₯ + 5 Step 2: Find the point of intersection. If the lines overlap β infinitely many solutions If lines are parallel β no solution Solution: (2,3)
GeometryCPESummerPacket
Solve by substitution.21.) β5π₯ + 3π¦ = 12 22.) π₯ β 4π¦ = 22 π₯ + 2π¦ = 8 2π₯ + 5π¦ = β21
Solving Systems by Substitution
Example: Solve the system of equations by substitution.
π¦ β 3π₯ = β3 β2π₯ β 4π¦ = 26 Step 1: Solve for a variable for either equation. (It is ideal to pick the variable with a coefficient of 1) π¦ β 3π₯ = β3 +3π₯ + 3π₯ π¦ = 3π₯ β 3 Step 2: Plug the expression 3π₯ β 3 in for π¦ of the OTHER equation. β2π₯ β 4π¦ = 26 β2π₯ β 4(3π₯ β 3) = 26 Step 3: Solve for π₯. β2π₯ β 4(3π₯ β 3) = 26 β2π₯ β 12π₯ + 12 = 26 β14π₯ + 12 = 26 β14π₯ = 14 π₯ = β1 Step 4: Plug in x for either equation to solve for y. π¦ = 3π₯ β 3 π¦ = 3(β1) β 3 π¦ = β6 Final Solution: (β1,β6)
GeometryCPESummerPacket
Solve by elimination.23.) β3π₯ + π¦ = 7 24.) β4π₯ + 5π¦ = β11 3π₯ + 2π¦ = 2 2π₯ + 3π¦ = 11
Solving Systems by Elimination
Example: Solve the system of equations by elimination.
4π₯ β 3π¦ = 25 β3π₯ + 8π¦ = 10 Step 1: Decide which variable you want to eliminate and find the LCM of the two coefficients for that variable. Eliminate π₯ Γ 4 and -3 have an LCM of 12 Step 2: Multiply each equation by the number that will make the x-terms have a coefficient of 12. One must be negative and the other must be positive. 3(4π₯ β 3π¦ = 25) Γ 12π₯ β 9π¦ = 75 4(β3π₯ + 8π¦ = 10) β12π₯ + 32π¦ = 40 Step 3: Add the columns of like terms. 12π₯ β 9π¦ = 75 β12π₯ + 32π¦ = 40 23π¦ = 115 π¦ = 5 Step 4: Plug in π¦ for either equation to solve for π₯. 4π₯ β 3π¦ = 25 4π₯ β 3(5) = 25 π₯ = 10 Final Solution: (10, 5)
GeometryCPESummerPacket0.9: Square Roots and Simplifying Radicals Simplify the following radicals. Remember, no radicals can be left in the denominator.
25.) 23
26.) 32
27.) 50 β 10 28.) 16 β 25
29.) 8149
30.) 1027
β Product Property for two numbers, ,
β Quotient Property for any numbers a and b, where β₯ 0,
Ex. 1 Simplify β45 = β9 β 5 = β3 β 3 β 5= 3β5
Ex.2 Simplify !2516= β25
β16= 54
GeometryCPESummerPacketRatios and Proportions
Solve each proportion.
31.) !!"= !
!" 32.) !!
!= !
!
33.) !.!!= !.!
! 34.) !
!!!= !"
!"
35.) !!!!"
= !"!
36.) !!!!= !!
!
GeometryCPESummerPacketMultiplying Polynomials
Find each product.
37.) (π + 8)(π + 2) 38.) (π₯ β 3)(π₯ + 3)
39.) (4β + 5)(β + 7) 40.) (5π β 6)(5π β 6)
FOIL:Tomultiplytwobinomials,findthesumoftheproductsofthe:
FβfirsttermsOβoutertermsIβinnertermsLβlastterms
(π₯ β 2)(π₯ + 4)
(π₯)(π₯)+ (π₯)(4) + (2)(π₯) + (β2)(4)
F O I L
π₯! + 4π₯ β 2π₯ β 8
π₯! + 2π₯ β 8
Double-Distribution:Distributebothofthetermsinthefirstparenthesistothetermsinthesecondparenthesis.
(π₯ β 2)(π₯ + 4)
π₯(π₯ + 4) β 2(π₯ + 4)
π₯! + 4π₯ β 2π₯ β 8
π₯! + 2π₯ β 8
GeometryCPESummerPacketSlope
Find the slope of the line given the following points.
41.)(1,-3)and(3,5) 42.)(8,11)and(24,-9)
β’ Theslopeofalineistheratioofthechangeintheyβcoordinatestothecorrespondingchangeinthexβcoordinates.
β’ Slope=m=!!!!!!!!!!
Ex1:Giventhepoints(-4,3)and(2,5)
π =5 β 3
2β (β4) =5β 32+ 4 =
26 =
13
GeometryCPESummerPacketFactoring
- add multiplying polynomials (Foiling) (Elizabeth)
-add factoring (Matt)
Solve each equation.
43.)20π₯! + 15π₯ = 0 44.)6π₯! + 18π₯! = 0
45.)π₯! β 16π₯ + 64 = 0 46.)π₯! β 11π₯ + 30 = 0
47.)π₯! β 4π₯ β 21 = 0 48.)π₯! β 6π₯ β 16 = 0
Factoringisusedtorepresentquadraticequationsinthefactoredformofa(xβp)(xβq)=0,andsolvethisequation.
FactoringGCF
InaquadraticequationyoumayfactorouttheGreatestCommonFactor.
Ex1:16π₯! + 8π₯ = 0. GCF=8x
8x(2x+1)=0 ZeroProductRule
8x=0or2x+1=0
x=0orx=-1/2
ππ₯! + ππ₯ + π
Factoringwherea=1
ππ± π: π₯! + 9π₯ + 20 = 0 ToFactorwewanttofindtwonumbersthatmultiplyto20andaddto9.
(x+ 5)(π₯ + 4) = 0 5+4=9and5*4=20
(π₯ + 5) = 0 ππ (π₯ + 4) = 0 ZeroProductRule
π₯ = β5 ππ π₯ = β4
GeometryCPESummerPacket
Solve the following by factoring.
49.)15π₯! β 8π₯ + 1=0 50.)3π₯! + 2π₯ β 5=0
Factoringfunctionswherea>1involvesadifferentprocess.Somewilluseguessandcheck.Belowisamethodthatwillalwayswork.
ππ₯! + ππ₯ + π
Ex2:6π₯! + 13π₯ β 5=0 Multiplya*c.6(β5) = β30
-2,15 Findtwonumbersthatmultiplytheβ30andsumto13
6π₯! β 2π₯ + 15π₯ β 5 = 0 GrouptheTerms
(6π₯! β 2π₯) + (15π₯ β 5) = 0FactorGCF
2π₯(3π₯ β 1) + 5(3π₯ β 1) = 0Usethe(3xβ1)andfactoredterms
(2π₯ + 5)(3π₯ β 1) = 0 ZeroProductRule
2π₯ + 5 = 0 ππ 3π₯ β 1 = 0 Solve
x=-5/2orx=1/3
Top Related