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GeometryCPESummerPacket

RAMAPO-INDIANHILLSSCHOOLDISTRICT

DearRamapo-IndianHillsStudent:

Pleasefindattachedthesummerpacketforyourupcomingmathcourse.Thepurposeofthesummerpacketistoprovideyouwithanopportunitytoreviewprerequisiteskillsandconceptsinpreparationforyournextyear’smathematicscourse.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‘Ž = 2, 𝑏 = βˆ’3, 𝑐 = βˆ’1, and 𝑑 = 4.

1.) 2π‘Ž + 𝑐 2.) !"!!

3.) !!!!!

4.) !!!!!!

5.) (𝑐 + 𝑏)! 6.) 𝑐 + 𝑏!

PEMDAS – the order in which you evaluate expressions P – Parenthesis E – Exponents M – Multiplication (from left D – Division to right) A – Addition (from left S – Substraction to right)

GeometryCPESummerPacketSection 0.5: Linear Equations

Solve each equation.

7.) π‘Ÿ + 11 = 3 8.) !!π‘Ž = βˆ’6 9.) !

!"+ 15 = 21

10.) 9𝑛 + 4 = 5𝑛 + 18 11.) βˆ’2𝑦 + 17 = βˆ’13 12.) βˆ’2(𝑛 + 7) = 15

Example: Solve the equation. !!𝑛 + 1 = 11

!!𝑛 + 1 = 11 Subtract 1 from each side.

βˆ’1 βˆ’ 1 (3) !

!𝑛 = 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:

𝐴 (3,5) - Quadrant I

𝐡(βˆ’3, 6) - Quadrant II

𝐢(βˆ’3,βˆ’5) - Quadrant III

𝐷(3,βˆ’5) - Quadrant IV

GeometryCPESummerPacketSection 0.8: Systems of Linear Equations

Solve by graphing.

19.) 𝑦 = βˆ’π‘₯ + 2

𝑦 = βˆ’ !!π‘₯ + 1

20.) 𝑦 – 2π‘₯ = 1

2𝑦 – 4π‘₯ = 1

Solving Systems by Graphing Example: Solve the system of equations by graphing.

𝑦 = 2π‘₯ βˆ’ 1 π‘₯ + 𝑦 = 5

Step 1: Graph each line. The second equation here needs to be changed to 𝑦 = π‘šπ‘₯ + 𝑏. π‘₯ + 𝑦 = 5 βˆ’π‘₯ βˆ’ π‘₯ 𝑦 = βˆ’π‘₯ + 5 Step 2: Find the point of intersection. If the lines overlap – infinitely many solutions If lines are parallel – no solution Solution: (2,3)

GeometryCPESummerPacket

Solve by substitution.21.) βˆ’5π‘₯ + 3𝑦 = 12 22.) π‘₯ – 4𝑦 = 22 π‘₯ + 2𝑦 = 8 2π‘₯ + 5𝑦 = βˆ’21

Solving Systems by Substitution

Example: Solve the system of equations by substitution.

𝑦 βˆ’ 3π‘₯ = βˆ’3 βˆ’2π‘₯ βˆ’ 4𝑦 = 26 Step 1: Solve for a variable for either equation. (It is ideal to pick the variable with a coefficient of 1) 𝑦 βˆ’ 3π‘₯ = βˆ’3 +3π‘₯ + 3π‘₯ 𝑦 = 3π‘₯ βˆ’ 3 Step 2: Plug the expression 3π‘₯ βˆ’ 3 in for 𝑦 of the OTHER equation. βˆ’2π‘₯ βˆ’ 4𝑦 = 26 βˆ’2π‘₯ βˆ’ 4(3π‘₯ βˆ’ 3) = 26 Step 3: Solve for π‘₯. βˆ’2π‘₯ βˆ’ 4(3π‘₯ βˆ’ 3) = 26 βˆ’2π‘₯ βˆ’ 12π‘₯ + 12 = 26 βˆ’14π‘₯ + 12 = 26 βˆ’14π‘₯ = 14 π‘₯ = βˆ’1 Step 4: Plug in x for either equation to solve for y. 𝑦 = 3π‘₯ βˆ’ 3 𝑦 = 3(βˆ’1) βˆ’ 3 𝑦 = βˆ’6 Final Solution: (βˆ’1,βˆ’6)

GeometryCPESummerPacket

Solve by elimination.23.) βˆ’3π‘₯ + 𝑦 = 7 24.) βˆ’4π‘₯ + 5𝑦 = βˆ’11 3π‘₯ + 2𝑦 = 2 2π‘₯ + 3𝑦 = 11

Solving Systems by Elimination

Example: Solve the system of equations by elimination.

4π‘₯ βˆ’ 3𝑦 = 25 βˆ’3π‘₯ + 8𝑦 = 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‘₯ βˆ’ 3𝑦 = 25) Γ  12π‘₯ βˆ’ 9𝑦 = 75 4(βˆ’3π‘₯ + 8𝑦 = 10) βˆ’12π‘₯ + 32𝑦 = 40 Step 3: Add the columns of like terms. 12π‘₯ βˆ’ 9𝑦 = 75 βˆ’12π‘₯ + 32𝑦 = 40 23𝑦 = 115 𝑦 = 5 Step 4: Plug in 𝑦 for either equation to solve for π‘₯. 4π‘₯ βˆ’ 3𝑦 = 25 4π‘₯ βˆ’ 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

● Product Property for two numbers, ,

● Quotient Property for any numbers a and b, where β‰₯ 0,

Ex. 1 Simplify √45 = √9 β‹… 5 = √3 β‹… 3 β‹… 5= 3√5

Ex.2 Simplify !2516= √25

√16= 54

GeometryCPESummerPacketRatios and Proportions

Solve each proportion.

31.) !!"= !

!" 32.) !!

!= !

!

33.) !.!!= !.!

! 34.) !

!!!= !"

!"

35.) !!!!"

= !"!

36.) !!!!= !!

!

GeometryCPESummerPacketMultiplying Polynomials

Find each product.

37.) (𝑛 + 8)(𝑛 + 2) 38.) (π‘₯ βˆ’ 3)(π‘₯ + 3)

39.) (4β„Ž + 5)(β„Ž + 7) 40.) (5π‘š βˆ’ 6)(5π‘š βˆ’ 6)

FOIL:Tomultiplytwobinomials,findthesumoftheproductsofthe:

F–firsttermsO–outertermsI–innertermsL–lastterms

(π‘₯ βˆ’ 2)(π‘₯ + 4)

(π‘₯)(π‘₯)+ (π‘₯)(4) + (2)(π‘₯) + (βˆ’2)(4)

F O I L

π‘₯! + 4π‘₯ βˆ’ 2π‘₯ βˆ’ 8

π‘₯! + 2π‘₯ βˆ’ 8

Double-Distribution:Distributebothofthetermsinthefirstparenthesistothetermsinthesecondparenthesis.

(π‘₯ βˆ’ 2)(π‘₯ + 4)

π‘₯(π‘₯ + 4) βˆ’ 2(π‘₯ + 4)

π‘₯! + 4π‘₯ βˆ’ 2π‘₯ βˆ’ 8

π‘₯! + 2π‘₯ βˆ’ 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€’ Theslopeofalineistheratioofthechangeinthey–coordinatestothecorrespondingchangeinthex–coordinates.

β€’ Slope=m=!!!!!!!!!!

Ex1:Giventhepoints(-4,3)and(2,5)

π‘š =5 βˆ’ 3

2βˆ’ (βˆ’4) =5βˆ’ 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‘₯! βˆ’ 16π‘₯ + 64 = 0 46.)π‘₯! βˆ’ 11π‘₯ + 30 = 0

47.)π‘₯! βˆ’ 4π‘₯ βˆ’ 21 = 0 48.)π‘₯! βˆ’ 6π‘₯ βˆ’ 16 = 0

Factoringisusedtorepresentquadraticequationsinthefactoredformofa(x–p)(x–q)=0,andsolvethisequation.

FactoringGCF

InaquadraticequationyoumayfactorouttheGreatestCommonFactor.

Ex1:16π‘₯! + 8π‘₯ = 0. GCF=8x

8x(2x+1)=0 ZeroProductRule

8x=0or2x+1=0

x=0orx=-1/2

π‘Žπ‘₯! + 𝑏π‘₯ + 𝑐

Factoringwherea=1

𝐄𝐱 𝟐: π‘₯! + 9π‘₯ + 20 = 0 ToFactorwewanttofindtwonumbersthatmultiplyto20andaddto9.

(x+ 5)(π‘₯ + 4) = 0 5+4=9and5*4=20

(π‘₯ + 5) = 0 π‘œπ‘Ÿ (π‘₯ + 4) = 0 ZeroProductRule

π‘₯ = βˆ’5 π‘œπ‘Ÿ π‘₯ = βˆ’4

GeometryCPESummerPacket

Solve the following by factoring.

49.)15π‘₯! βˆ’ 8π‘₯ + 1=0 50.)3π‘₯! + 2π‘₯ βˆ’ 5=0

Factoringfunctionswherea>1involvesadifferentprocess.Somewilluseguessandcheck.Belowisamethodthatwillalwayswork.

π‘Žπ‘₯! + 𝑏π‘₯ + 𝑐

Ex2:6π‘₯! + 13π‘₯ βˆ’ 5=0 Multiplya*c.6(βˆ’5) = βˆ’30

-2,15 Findtwonumbersthatmultiplythe–30andsumto13

6π‘₯! βˆ’ 2π‘₯ + 15π‘₯ βˆ’ 5 = 0 GrouptheTerms

(6π‘₯! βˆ’ 2π‘₯) + (15π‘₯ βˆ’ 5) = 0FactorGCF

2π‘₯(3π‘₯ βˆ’ 1) + 5(3π‘₯ βˆ’ 1) = 0Usethe(3x–1)andfactoredterms

(2π‘₯ + 5)(3π‘₯ βˆ’ 1) = 0 ZeroProductRule

2π‘₯ + 5 = 0 π‘œπ‘Ÿ 3π‘₯ βˆ’ 1 = 0 Solve

x=-5/2orx=1/3