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Leap 2025 algebra i

Leap 2025 geometry Better Scores in ONE Day

Better Scores in ONE Day

Leap 2025 algebra i Better Scores in ONE Day

Our Louisiana LEAP 2025 Algebra 1 Boot Camp is designed specifically to increase the number of students scoring Basic and Mastery on the LEAP 2025 Algebra 1 assessment.

In just one day, students will learn: • Core skills for success in algebra • Pacing and time management • Test-taking and guessing strategies that really work • How to overcome test anxiety and put their best foot forward on test day

Why schedule a LEAP Algebra I Boot Camp? • Authentic, up-to-date practice questions • Students review exactly what they need in the “final hours” before the test. • Improves student confidence • Easy to schedule, during the school day or on the weekend • Makes test prep fun and less overwhelming for students

Implementation Models • Full-day workshop during school hours • After-school or Saturday programming • Virtual and in-person programs available

Table of Contents

Table of Contents Chapter 1: Leap 2025 aLgebra I OvervIew .................................................... 7 Chapter 2: SOLvIng aLgebraICaLLy ......................................................................... 13 Solving AlgebrAicAlly overview ......................................................................... 14 Mini-TeST one ...............................................................................................................16 Plug iT in ..........................................................................................................................18 DiSTribuTive ProPerTy: Show your work .......................................................... 19 Mini-TeST Two ............................................................................................................. 21 creATe A viSuAl ...........................................................................................................23 SubSTiTuTion ..................................................................................................................24 negATive PArAnoiA ..................................................................................................... 25 Mini-TeST Three .......................................................................................................... 26 Mini-TeST Four ........................................................................................................... 28 Try nuMberS ................................................................................................................. 30 Mini-TeST Five .............................................................................................................. 31 worD ProbleM TrAnSlATion .................................................................................. 33 Mini-TeST exPlAnATionS .......................................................................................... 34 Chapter 3: InterpretIng FunCtIOnS .....................................................................41 inTerPreTing FuncTionS overview ....................................................................... 42 Mini-TeST one ............................................................................................................. 43 geT reAl ........................................................................................................................ 45 Don’T overThink iT .................................................................................................... 46 Mini-TeST Two .............................................................................................................47 reAD The QueSTion .....................................................................................................50 DoMAin ............................................................................................................................. 51 Mini-TeST Three .......................................................................................................... 52

Chapter One: LEAP 2025 Algebra I Overview

uSe The AnSwer choiceS......................................................................................... 54 Mini-TeST Four ............................................................................................................ 55 Mini-TeST exPlAnATionS ........................................................................................... 57

Chapter 4: SOLvIng graphICaLLy and rate OF Change ...............................61 Solving grAPhicAlly AnD rATe oF chAnge overview ................................62 Mini-TeST one ............................................................................................................. 63 FinDing PercenTAgeS ................................................................................................. 66 eQuATionS oF lineS .....................................................................................................68 ProceSS oF eliMinATion ............................................................................................ 70 Mini-TeST Two .............................................................................................................. 71 Plug in PoinTS on A grAPh ..................................................................................... 73 MulTi-STeP PAnic ........................................................................................................75 Mini-TeST Three .......................................................................................................... 77 TrAnSlATionS & reFlecTionS ..................................................................................80 Mini-TeST exPlAnATionS .......................................................................................... 82 Chapter 5: wrap-up ...................................................................................................87 Chapter 6: Further praCtICe ..................................................................................91 PrAcTice SeT one .......................................................................................................92 PrAcTice SeT Two .......................................................................................................95 PrAcTice SeT Three ....................................................................................................99 PrAcTice SeT exPlAnATionS .................................................................................. 103

Chapter One: LEAP 2025 Algebra I Overview

Chapter 1 LEAP 2025 Algebra I Overview

LEAP 2025 ALGEBRA I OVERVIEW

Chapter One: LEAP 2025 Algebra I Overview LEAP 2025 ALgEbrA I OvErvIEw

What Is End-of-Course Testing? End-of-Course (EOC) testing measures your aptitude in a given subject after you have finished a course. Consider it a subject understanding checkup. Teachers use it to identify both your strengths and areas where improvement is needed. This helps ensure you are on track in developing the knowledge and skills needed for the next grade and, eventually, college and a career. In this Boot Camp, we’ll focus on the three most common concepts in Algebra I: Solving Algebraically, Interpreting Functions, and Solving Graphically and Rate of Change. Your understanding of each of these concepts will help you pass the LEAP 2025 Algebra I test.

Why Should You Care? •

Many schools require the EOC as part of your final grade in the course.

•

If your school uses the EOC as a final exam for the course, then doing well on this test can boost your GPA.

•

A good EOC score is a positive indicator that you are on track for college.

•

Mastering the foundational skills taught in this Boot Camp will help you succeed in more difficult math courses in the future.

•

Put in the effort now and save yourself from repeating a course or taking summer school.

NOTES:

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LEAP 2025 ALgEbrA I OvErvIEw

Orientation The three most common conceptual categories tested on LEAP 2025 Algebra I are Solving Algebraically, Interpreting Functions, and Solving Graphically and Rate of Change. The test is timed. All sections except the first one allow you to use a calculator. To do well on this test, it is important to be comfortable performing mathematical operations with and without a calculator. Here is a breakdown of possible ways the LEAP 2025 Algebra I assessment will test you in each of the three main categories. Each section of the test has questions from each category.

Solving Algebraically tests equations, expressions, and inequalities in the following ways: •

Evaluate problems with one or two variables

•

Create models to describe real-life situations and relationships

•

Understand and apply basic mathematical principles

Interpreting Functions measures your ability to interpret, understand, and build functions. Solving Graphically and Rate of Change assesses how well you can create graphs and solve problems involving lines, slope, intercepts, and solution sets.

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LEAP 2025 ALGEBRA OVERVIEW

Chapter One: LEAP 2025 Algebra I Overview

LEAP 2025 ALGEBRA I OVERVIEW

Chapter One: LEAP 2025 Algebra I Overview LEAP 2025 ALgEbrA I OvErvIEw

Orientation In Louisiana, your LEAP 2025 Algebra I score can count as a percentage of your final grade for the Algebra I course. The percentage is always between 15 and 30 percent of your grade, depending on your school district. If you know what areas you struggle in, compare them to the most important skills needed for the test. The table below gives the approximate contribution of each conceptual category to your score. This is just an estimate and can vary greatly from one test to another. Conceptual Category

Total Points

Percentage of Points

Solving Algebraically

40%

Interpreting Functions

25%

Solving Graphically and Rate of Change

35%

Total

100%

The test is administered in four sessions. The sessions are timed, so it is important to manage your minutes and avoid spending too much time on any one question. Usually, each single-part question is worth one point, and each multi-part question is worth one point per part. Questions that require you to show work or justify your answers are typically worth 2, 3, or even 4 points each. Each test section may include multiple choice, multiple select, constructed response, fill in the blank, or a variety of technology-assisted answering methods. For example, you might have to actually draw a graph on your computer. We provide practice for all of these different question types during the Boot Camp.

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Test Session

Calculator?

Number of Points

Time Limit

Session 1a

25 minutes

Session 1b

Yes

55 minutes

Session 2

Yes

80 minutes

Session 3

Yes

80 minutes

Total

–

240 minutes

leAP 2025 AlgebrA i booT cAMP

Chapter Two: Introductory Algebra

Chapter 2 Solving Algebraically

Chapter Two: Introductory Algebra SOLvIng aLgebraICaLLy:

overview

Solving Algebraically Overview The Solving Algebraically conceptual category tests your proficiency over a broad range of algebra skills. The

SOLVING ALGEBRAICALLY

skills that will be tested on your exam include but are not limited to the following:

Seeing Structure in Expressions •

Interpret the structure of expressions

•

Write expressions in equivalent forms to solve problems

Arithmetic with Polynomials and Rational Expressions •

Perform arithmetic operations on polynomials

•

Understand the relationship between zeros and factors of polynomials

Creating Equations •

Create equations that describe numbers or relationships

Reasoning with Equations and Inequalities •

Understand solving equations as a process of reasoning and explain the reasoning

•

Solve equations and inequalities in one variable

•

Solve systems of equations and inequalities

•

Represent and solve equations and inequalities graphically

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Chapter Two: Introductory Algebra SOLvIng aLgebraICaLLy:

overview

What Are Boot Camp Mini-Tests? During this Boot Camp you will take several mini-tests, which are small segments of an Algebra I test. While taking these mini-tests, it’s important to imagine that you are in an actual testing environment. The time limits assigned as you complete the mini-tests. In the mini-tests, we are focusing on only one category of questions at a time, but on the real assessment each test section will include questions from each major conceptual category. For these mini-tests, you have 10 minutes to answer several questions. Your instructor will signal when you are out of time. Try to get through all the questions within the time limit. Unless your instructor has provided you with an answer sheet, circle your answers directly in this book. The real test does not allow the use of cell phones, watches, or computers, so you shouldn’t use them on the mini-tests, either.

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SOLVING ALGEBRAICALLY

match the pace that you should try to keep during the actual test. Practice all of the skills that you have learned

Chapter Two: Introductory Algebra

Solving Algebraically - Mini-Test One 1.

Consider the function f, where f(t) = 2t2 + 8t – 10.

PART A What is the vertex form of f(t) ? A.

2(t – 2)2 – 18

2(t + 2)2 – 18

2(t – 4)2 – 14

2(t + 4)2 – 14

PART B What is a factored form of f(t) ?

(2t – 1)(t + 10)

(2t + 1)(t – 10)

2(t + 5)(t – 1)

2(t – 5)(t + 1)

Which expression is equivalent to (x – 2)(3x 2 – 5x + 9)? A.

3x 2 – 4x + 7

3x 2 – 4x + 9

3x 3 – 5x 2 + 9x – 2

3x 3 – 11x 2 + 19x – 18

Solving Algebraically - Mini-Test One

GO ON TO THE NEXT PAGE.

Chapter Two: Introductory Algebra

The boiling point of ethyl alcohol, T (measured in degrees), at an altitude above sea level, a (measured in feet), can be determined by the expression –0.0013a + 173. What is the meaning of the 173 in the expression? A.

The boiling point is 173 degrees at sea level.

The boiling point decreases by 173 degrees as the altitude increases by 1,000 feet.

The minimum altitude is 173 feet.

The maximum altitude is 173 feet.

Kedrick used the method of completing the square to solve a quadratic equation. His first two steps are shown below. Given: 3x2 + 30x + 21 = 0 Step 1: x2 + 10x + 7 = 0 Step 2: x2 + 10x = –7 Write numbers in each box to correctly complete the square in Step 3 (selecting from the options appearing below). Step 3: x2 + 10x +

Options: –32 –25 –18 –5 5 18

Solving Algebraically - Mini-Test One

STOP! END OF TEST. YOU MAY GO BACK AND CHECK YOUR WORK.

Chapter Two: Introductory Algebra SOLvIng aLgebraICaLLy:

Plug iT in

Plug It In If you don’t know exactly how to answer a question, try plugging in assumed values for the variables given and see if you can draw some conclusions.

SOLVING ALGEBRAICALLY

The boiling point is 173 degrees at sea level.

The boiling point decreases by 173 degrees as the altitude increases by 1,000 feet.

The minimum altitude is 173 feet.

The maximum altitude is 173 feet.

In this question, plug in experimental values for a to prove answer choices incorrect. Choice B says that an increase of 1000 in a will increase the value of the expression by 173. Test to see if this is true. Use 1000 and 2000 as values for a and see what happens. –0.0013(1000) + 173 = 171.7 and –0.0013(2000) + 173 = 170.4. This is not a difference of 173, so we can eliminate choice B. Choices C and D can be eliminated because you can plug in a value of 150 (below the “minimum altitude” described in C) and a value of 200 (above the “maximum altitude” described in D) and the expression still works. Therefore, choice A is correct. You can verify this by plugging in the value 0 (sea level) for a. This gives you a boiling point of –0.0013(0) + 173 = 173, which is exactly what choice A claims.

NOTES:

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Chapter Two: Introductory Algebra

Solving Algebraically - Mini-Test Two 1.

b1 + b2 h, where A is the area, b 1 and b 2 are the lengths of the 2

The formula for the area of a trapezoid is A = bases, and h is the height. Which is the formula for b 2?

2A h

b2= b1 −

= b2

2A − b1 h

b2 =

b1 + A h 2

b = 2b1 + 2 Ah 2

What is the solution to –3(6t + 4) – 6t ≥ –11t – (17t + 4)? A.

t≤2

t≥2

t≤4

t≥4

Solving Algebraically - Mini-Test Two

GO ON TO THE NEXT PAGE.

Chapter Two: Introductory Algebra

Rayan and Megan are playing a game. •

Rayan and Megan each started with 50 points.

•

At the end of each turn, Rayan’s points increased by 250.

•

At the end of each turn, Megan’s points doubled.

PART A Create a model that can be used to determine the total number of points between Rayan and Megan based on the number of turns that have passed. Write your model in the space below.

PART B At the end of the game, Megan has 400 points. How many points did Rayan and Megan score in total? Provide your answer in the space below. Show your work.

PART C At the end of which turn does Megan’s score first exceed Rayan’s score?

Solving Algebraically - Mini-Test Two

STOP! END OF TEST. YOU MAY GO BACK AND CHECK YOUR WORK.

Chapter Two: Introductory Algebra SOLvIng aLgebraICaLLy:

SubSTiTuTion

Substitution Freezing up on a challenging question is a common problem among students. When you come across a scary-looking question and don’t know where to begin, look for a formula. If you find a formula, see if there are any values to substitute into it. Use this as a starting point for figuring out the question. If the problem doesn’t give any values, you can assume (make up) values!

SOLVING ALGEBRAICALLY

For example, consider how this problem becomes manageable if you substitute made-up values into the formula provided.

The formula for the area of a trapezoid is A = of the bases, and h is the height.

b1 + b2 h, where A is the area, b 1 and b 2 are the lengths 2

Which is the formula for b 2? b1

b2 2A h

b2= b1 −

= b2

2A − b1 h

b2 =

b1 + A h 2

b = 2b1 + 2 Ah 2

8+ 4 · 2 = 12. You now have a full set of values for the variables. 2 If an answer choice doesn't work for those values, it's wrong. Plug these values into each answer choice and Assume that b1 is 8, b2 is 4, and h is 2. Then A =

eliminate the ones that don't work. Choice A can be eliminated because it gives the value for b2 as –4 instead of 4. b2 = 8 – Choice C can be eliminated because with the assumed values it gives b2 =

2(12) = –4 2

8 + 12 · 2 = 20. 2

Choice D can be eliminated because it gives b2 = 2(8) + 2(12)(2) = 64 with the assumed values. Only choice B checks out with the assumed values, so it is correct. b2 = 2(12) – 8 = 4 2 Many math questions can be answered using this strategy. Assume values, and then see what choices work.

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Chapter Two: Introductory Algebra SOLvIng aLgebraICaLLy:

negATive PArAnoiA

Negative Paranoia When solving problems loaded with negative signs, let the paranoia sink in. This test uses a lot of negative signs to trip you up. When you face a problem like this, you need to pay more attention to every calculation that involves a negative sign. It is likely that the trap answers are meant to catch you if you mess up a calculation with a negative sign.

SOLVING ALGEBRAICALLY

Let’s look at an example of when you should feel the negative paranoia:

What is the solution to –3(6t + 4) – 6t ≥ –11t – (17t + 4)? A.

t≤2

t≥2

t≤4

t≥4

Simplify both sides of the inequality and solve for t. Don’t forget to be paranoid about the negative signs! –3(6t + 4) – 6t ≥ –11t – (17t + 4) –18t – 12 – 6t ≥ –11t – 17t – 4 –24t – 12 ≥ –28t – 4 28t – 24t ≥ 12 – 4 4t ≥ 8 t≥2 Notice that if you forget to distribute the negative sign to the 4 at the end of the inequality, you will ultimately arrive at choice D, t ≥ 4. Don’t fall into these traps! Keep a close eye on all negative signs.

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Chapter Two: Introductory Algebra SOLvIng aLgebraICaLLy:

Mini-TeST exPlAnATionS

Mini-Test Explanations MINI-TEST ONE 1. Part A: The correct answer is B. The vertex form of a quadratic is given by f(x) = a(x – h)² + k, where (h, k) is the vertex.

SOLVING ALGEBRAICALLY

Given a quadratic ax² + bx + c, its vertex is found by computing h = – h =

−8 = 2(2)

b and then evaluating f(x) at h to find k. 2a

−2

2t2 + 8t – 10 = 2(t + 2)2 + k 2t2 + 8t – 10 = 2(t2 + 4t + 4) + k 2t2 + 8t – 10 = 2t2 + 8t + 8 + k k = –18 Thus the vertex form of this function is 2(t + 2)2 – 18. It’s t + 2 instead of t – 2 because h = –2 and the vertex form is given with x – h, not x + h. Part B: The correct answer is C. Since all three terms are multiples of 2, first factor out 2. Then factor the remaining quadratic. 2(t2 + 4t – 5) The binomial constants have a product of –5 (the last term in the quadratic) and a sum of 4 (the coefficient of the middle term in the quadratic). Their values must therefore be 5 and –1. 2(t + 5)(t – 1) 2. The correct answer is D. Distribute both terms in the first parentheses to all terms in the second parentheses and combine like terms. (x – 2)(3x2 – 5x + 9) 3x3 – 5x2 + 9x – 6x2 + 10x – 18 3x3 – 5x2 – 6x2 + 9x + 10x – 18 3x3 – 11x2 + 19x – 18 3. The correct answer is A. Sea level is an altitude of 0. When a = 0, then the expression for the boiling point is –0.0013(0) + 173 = 173. Therefore, 173 represents the boiling point at sea level. 2

 −b  4. The correct answers are 25 and 18. To complete Step 3, the term   is added to both sides of the  2a  2 equation, where a is the coefficient of the x term and b is the coefficient of the x term. In the quadratic equation provided, a = 1 and b = 10.

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Chapter Two: Introductory Algebra SOLvIng aLgebraICaLLy:

Mini-TeST exPlAnATionS

 −b    =  2a  2

 −10    =  2 (1) 

(–5)2 = 25 To complete Step 3, add 25 to both sides of the equation. Step 3: x2 + 10x + 25 = 25 – 7

SOLVING ALGEBRAICALLY

x2 + 10x + 25 = 18

MINI-TEST TWO 1. The correct answer is B. Solve the equation for b2. b +b A= 1 2 h 2 A b1 + b2 = h 2 2A = b1 + b2 h 2A b2 = − b1 h

2. The correct answer is B. Simplify both sides of the inequality and solve for t. –3(6t + 4) – 6t ≥ –11t – (17t + 4) –18t – 12 – 6t ≥ –11t – 17t – 4 –24t – 12 ≥ –28t – 4 28t – 24t ≥ 12 – 4 4t ≥ 8 t≥2 3. Part A Sample Correct Response: If p represents the number of points and t represents then number of turns passed, then: p = 50 · 2t + 250t + 50 Megan’s 50 points double every turn, which can be represented by 50 • 2t. Rayan’s points start at 50 and increase by 250 each turn, which can be represented by 50 + 250t. Combine these two expressions to model the total number of points. leAP 2025 AlgebrA i booT cAMP

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Chapter Three: Functions

Chapter 3 Interpreting Functions

Chapter Three: Functions InterpretIng FunCtIOnS:

overview

Interpreting Functions Overview The Interpreting Functions conceptual category tests your proficiency over a broad range of algebra skills. The skills that will be tested on your exam include but are not limited to the following:

Interpreting Functions •

Understand the concept of a function and its domain and range, and use function notation

•

Interpret functions that arise in applications in terms of the context

•

Analyze functions using different representations, including graphs and tables

INTERPRETING FUNCTIONS

Building Functions •

Build a function that models a relationship between two quantities

•

Build new functions from existing functions

Linear, Quadratic, and Exponential Models •

Construct and compare linear, quadratic, and exponential models and solve problems

•

Interpret expressions for functions in terms of the situation they model

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Chapter Three: Functions

Interpreting Functions - Mini-Test One 1.

ISP (Internet Service Provider) A charges a $15 installation fee and $0.10 per GB of data, x. ISP B charges $0.15 per GB of data and no installation fee. Which function below represents the difference in cost between ISP A and ISP B? A.

f(x) = –0.05x – 15

f(x) = –0.05x + 15

f(x) = 15x + 0.05

f(x) = 15x – 0.05

Ronaldo compared the y-intercept of the graph of the function f(x) = 2x + 7 to the y-intercept of the graph of the linear function containing the points in the table below. x

g(x)

–5

–3

–1

What is the difference when the y-intercept of f(x) is subtracted from the y-intercept of g(x)? A.

1.5

5.2

7.0

17.5

Interpreting Functions - Mini-Test One

GO ON TO THE NEXT PAGE.

Chapter Three: Functions

The graph below models a quadratic function f(x) that gives a projectile’s distance in meters from the ground x seconds from the moment it struck a target. PART A Which of these statements is true? A.

The projectile reaches a maximum distance of 20 meters.

The projectile reaches its maximum distance 6 seconds before impact.

The distance is decreasing on the interval –8 < x < –2.

The projectile is launched from the ground.

f(x)

–12

–10

–8

–6

–4

–2

-10

PART B The table above models the flight of the projectile from Part A. Which statements are true for the function represented by the table? Select all that apply.

The function has a zero at (–12, 0).

The function has a zero at (0, –12).

The values of f(x) are increasing on the interval –4 < x < 0.

The values of f(x) are decreasing on the interval –4 < x < 0.

The function has a minimum value between x = –6 and x = –4.

The function has a maximum value between x = –6 and x = –4.

The cost to generate z megawatt hours of electricity using solar panels can be represented by a function, S(z). Fill in the blanks using the options above and below each box: 0 1 5 198 396 If S(5) = 990, then

5.00

990.00

megawatt hours cost $

. 4950.00

Interpreting Functions - Mini-Test One

STOP! END OF TEST. YOU MAY GO BACK AND CHECK YOUR WORK.

Chapter Three: Functions InterpretIng FunCtIOnS:

geT reAl

Get Real If you aren’t sure about what equation to select as your answer choice, making up a real situation and a set of values for it can really simplify things. Only one equation will work with all of the circumstances you can dream up. 1.

f(x) = –0.05x – 15

f(x) = –0.05x + 15

f(x) = 15x + 0.05

f(x) = 15x – 0.05

For this question, let’s imagine that I used 20 GB of data. We will compare the cost between the two providers for 20 GB. Then we’ll see which equation predicts that same difference in cost. ISP A costs 15 + (20)(0.10) = 17.

The difference is 17 – 3 = 14. Now we try the given formulas. A: (–0.05)(20) – 15 = –16 B: (–0.05)(20) + 15 = 14 C: (15)(20) + 0.05 = 300.05 D: (15)(20) – 0.05 = 299.95 Only choice B predicted the difference in the real situation, so it must be the correct choice. If, when we made up a situation, two equations worked, we would have to create a new real situation and try out the two remaining equations to see which one still functioned correctly.

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INTERPRETING FUNCTIONS

ISP B costs (0.15)(20) = 3.

Chapter Three: Functions InterpretIng FunCtIOnS:

Don'T overThink iT

Don’t Overthink It Have you ever spent extra time deliberating over a question and changed your answer, only to find out you erased the correct answer and went with a wrong one instead? Many high-performing students cannot help themselves and waste time overthinking. It’s important to learn how to go with your gut.

5.00

990.00

megawatt hours cost $

. 4950.00

This question doesn’t require any calculations at all. Simply choose the answer choice that rephrases how the function is described in the question. The question says the function uses the number of megawatt hours to determine the cost. Because 5 and 990.00

INTERPRETING FUNCTIONS

are available as answer choices, all you have to do is fill in the meaning of the equation that you are already given: S(5) = 990 . The correct answers are 5 and 990.00 . For this problem, if you spend time overanalyzing the answer choices or attempting to perform calculations, you are wasting your time and might end up with a wrong answer! Go with your gut and don’t second-guess yourself.

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Chapter Three: Functions InterpretIng FunCtIOnS:

uSe The AnSwer choiceS

Use the Answer Choices When you get stuck, you can try plugging numbers into the given equation to see what happens. This process is far more straightforward when you don’t have to pick those numbers yourself. When the answer choices give numbers, use them!

Determine all zeros for the function g ( x) = ( x 2 + 4 x − 12)( x − 1) . Circle all zeroes of the function.

–12

–6

–4

–2

–1

4 6

If you aren’t sure how to factor the first polynomial, just plug in each of the answers provided and see which ones give you a zero. It’s a lot of arithmetic, but it’s worth it if you get to the right answer: –12: (144 – 48 –12)(–12 – 1) = (84)(–13). Not zero.

INTERPRETING FUNCTIONS

–6: (36 – 24 – 12)(–6 – 1) = (0)(–7). Zero! –4: (16 – 16 – 12)(–4 – 1) = (–12)(–5). Not zero. –2: (4 – 8 – 12)(–2 – 1) = (–16)(–3). Not zero. –1: (1 – 4 – 12)(–1 – 1) = (–15)(–2). Not zero. 0: (0 – 0 – 12)(0 – 1) = (–12)(–1). Not zero. 1: (1 + 4 – 12)(1 – 1) = (–7)(0). Zero! 2: (4 + 8 – 12)(2 – 1) = (0)(1). Zero! 4: (16 + 16 – 12)(4 – 1) = (20)(3). Not zero. 6: (36 + 24 – 12)(6 – 1) = (48)(5). Not zero. 12: (144 + 48 – 12)(12 – 1) = (180)(11). Not zero. If you are willing to do the work, there are many questions on the test where you can earn points, even if you aren’t sure of the “right” way to solve them. Plug in the answer choices provided to brute force your way to a higher score.

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Chapter Three: Functions InterpretIng FunCtIOnS:

Mini-TeST exPlAnATionS

Mini-Test Explanations MINI-TEST ONE 1. The correct answer is B. Create expressions for the costs for ISP A and ISP B, and then subtract. ISP A: 0.10x + 15 ISP B: 0.15x f(x) = ISP A – ISP B f(x) = 0.10x + 15 – 0.15x f(x) = –0.05x + 15 2. The correct answer is A. The y-intercept of f(x) is 7. Find the linear equation for g(x) by first finding the slope of the function. m=

3 4 −1 = 2 −3 − ( −5 )

3 . Now, insert the slope and one point from the table into the slope-intercept form of the 2 function and solve for the y-intercept. 3 (1) + b 2 3 20 3 17 – b = 10 – = = 2 2 2 2

10 =

Finally, find the difference between the y-intercept of g(x) and the y-intercept of f(x). 17 17 14 3 –7= – = = 1.5 2 2 2 2 3. Part A: The correct answer is D. At the x value –12, the y value is 0. A distance of 0 from the ground means that the projectile was on the ground. It can therefore be deduced that the projectile launched from the ground. Choice A is incorrect because the line goes above a y value of 20. Choice B is incorrect because the point at an x value of –6 is not the highest point on the line. Choice C is incorrect because over the interval between –8 and –2, the y value increases and then decreases, and the y value at the start of the interval is the same as the y value at the end of the interval. Part B: The correct answers are A, D, and F. A zero occurs when the graph of the function intersects the x-axis, so the function has a zero at (–12, 0). The value of the function goes down from 24 to 12 on the interval –4 < x < 0, so the function is decreasing on this interval. The greatest value y of the function occurs between x = –6 and x = –4, so the function has a maximum between these values. 4. The correct answers are 5 and 990.00, respectively. The function S(5) gives the cost to generate 5 megawatt hours of electricity. Since S(5) = 990, we know that 5 megawatt hours cost $990.00.

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INTERPRETING FUNCTIONS

The slope of g(x) is

Chapter Four: Statistics & Probability

Chapter 4 Solving Graphically and Rate of Change

Chapter Four: Statistics & Probability SOLvIng graphICaLLy and rate OF Change:

overview

Solving Graphically and Rate of Change Overview The Solving Graphically and Rate of Change conceptual category tests your proficiency over a broad range of algebra skills. The top skills that will be tested on your exam include the following:

Graphing Solutions •

Graphing solutions to inequalities and systems of inequalities

•

Representing constraints

•

Understanding lines as solution sets

Rate of Change and Comparing Functions •

Determine the slope or average rate of change of a line or function

•

Identify changes and transformations to a line’s slope and intercepts

In this chapter, we will also review how to summarize data from frequency tables.

SOLVING GRAPHICALLY

NOTES:

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Chapter Four: Statistics & Probability

Solving Graphically and Rate of Change - Mini-Test One 1.

The table below shows the orders at an ice cream shop during the month of June. Ice Cream Shop Sales - June Chocolate Syrup

Sprinkles

No Topping

Total

Mint Chocolate

101

146

Café Mocha

176

Cookie Dough

203

Total

250

104

171

525

Which statements about the ice cream shop sales for the month of June are true? Select all that apply.

The percentage of mint chocolate sales was less than the percentage of café mocha sales.

The percentage of sales of cookie dough with chocolate syrup was greater than 25% of all sales.

The percentage of sales of a café mocha with no topping was less than 1% of all sales.

The number of sales of cookie dough ice cream with sprinkles was 44.

The number of ice cream sales with no topping was less than the number of mint chocolate ice cream sales.

What is the average rate of change for the function f(x) = 4x 2 – 3 on the interval –2 ≤ x ≤ 1? Write the average rate of change in the space below.

Solving Graphically - Mini-Test One

GO ON TO THE NEXT PAGE.

Chapter Four: Statistics & Probability

The table below shows the weight of an algae bloom after several days of growth. Time (days)

Weight (lb)

0.13

0.17

0.23

0.32

0.51

What is the average rate of change in weight of the algae bloom from day 2 to day 6?

0.032 pounds per day

0.095 pounds per day

0.190 pounds per day

0.380 pounds per day

Graph the solution set of 8x – 6y ≥ 12.

y 12 10 8 6 4 2 –12 –10 –8 –6 –4 –2 –2

8 10 12

–4 –6 –8 –10 –12

Solving Graphically - Mini-Test One

GO ON TO THE NEXT PAGE.

Chapter Four: Statistics & Probability

Which graph best represents the solution to the following system of inequalities? 2x – 7y ≥ 21 2x – 5y ≥ 15

Solving Graphically - Mini-Test One

STOP! END OF TEST. YOU MAY GO BACK AND CHECK YOUR WORK.

Chapter Four: Statistics & Probability

SOLvIng graphICaLLy and rate OF Change:

ProceSS oF eliMinATion

Process of Elimination When you get stuck on a question and need to guess, always try to avoid making a random guess when you can. Instead, narrow down the choices to increase your chance of guessing correctly.

The table below shows the weight of an algae bloom after several days of growth. Time (days)

Weight (lb)

0.13

0.17

0.23

0.32

0.51

What is the average rate of change in weight of the algae bloom from day 2 to day 6? A.

0.032 pounds per day

0.095 pounds per day

0.190 pounds per day

0.380 pounds per day

Looking at the table, you can see that the rate of growth increases slightly each day. The lowest growth rate is between the first two days, days 2 and 3, when the plant grows 0.17 – 0.13 = 0.040 lb. The highest growth rate is between the last two days, days 5 and 6, when the plant grows 0.51 – 0.32 = 0.190 lb. Since the question asks for the average rate of change and an average is a way of measuring the “middle” of a

SOLVING GRAPHICALLY

set of data, you can eliminate answer choices that do not reflect a number between 0.040 and 0.190 lb per day. Choice A can be eliminated because 0.032 is less than 0.040. Choices C and D can be eliminated because 0.190 and 0.380 are equal to or greater than 0.190. Since choice B, 0.095 lb per week, is the only option that falls between the highest and lowest rates of change, it is the best guess and also the correct answer.

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Chapter Four: Statistics & Probability SOLvIng graphICaLLy and rate OF Change:

Mini-TeST exPlAnATionS

Mini-Test Explanations MINI-TEST ONE 1. The correct answers are A and D. Choice A is correct because the percentage of mint chocolate sales is 146 176 · 100 ≈ 28%, which is less than the percentage of café mocha sales, · 100 ≈ 34%. Choice D is 525 525 correct because the value in the row labeled Cookie Dough under the column Sprinkles is 44. Choices B, C, and E do not accurately describe the data. The percentage of cookie dough with chocolate syrup is 60 27 · 100 ≈ 11%, which is not over 25%. The percentage of café mocha with no topping is · 100 ≈ 5%, 525 525 which is not less than 1% of all sales. The number of ice cream sales with no topping is 171, which is not less than the number of mint chocolate ice cream sales, 146. f ( b ) − f ( a )

, where b is the b − a greatest value on the interval and a is the least value on the interval. Evaluate the expression for the given

2. The correct answer is –4. The average rate of change is given by the expression

function.

( 4 (1) − 3) − ( 4 ( −2 ) − 3) = ( 4 − 3) − (16 − 3) = 1 − 13 = −12 = –4 2

1 − ( −2 )

The average rate of change of the function on the interval –2 ≤ x ≤ 1 is –4.

3. The correct answer is B. Use the expression for average rate of change

f ( b) − f ( a) : b−a

0.51 − 0.13 0.38 = = 0.095 pounds per week 6 − 2 4

SOLVING GRAPHICALLY

4. The correct response should resemble the graph below. First put the inequality into slope-intercept form: y

8x – 6y > 12

10 8

–6y > –8x + 12

y < 8x – 2 6 y<

4 x–2 3

The graph needs a slope of 4 , a y-intercept of –2, 3 and shading below a solid line because the inequality used is less than or equal to.

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4 2 –10 –8 –6 –4 –2

2 –2 –4 –6 –8

–10

Chapter Four: Statistics & Probability SOLvIng graphICaLLy and rate OF Change:

Mini-TeST exPlAnATionS

5. The correct answer is A. Put both inequalities into slope-intercept form to determine the properties of their lines. 2x – 7y > 21 –7y > –2x + 21 y<

2x – 5y > 15 and

2 x–3 7

–5y > –2x + 15 y<

2 x–3 5

Both inequalities have positive slope, so eliminate choices C and D. Both inequalities use the less than or equal to sign, which means that the shaded region should be below both lines, so eliminate choice B. Only choice A matches the information given by the inequalities in slope-intercept form.

MINI-TEST TWO 1. Part A: The correct answer is C. First, create the two inequalities. x and y cannot total greater than 40, and the number of hours Darrel spends cutting grass, times 15, plus the number of hours he spends at the library, times 9, must add together to be greater than or equal to 400. x + y < 40 15x + 9y > 400 The first inequality can be rewritten as y < x. The second inequality can also be converted into slope-intercept form: y> −

4 15 x + 44 9 9

The line for the first inequality appears on all four answer options, so focus on the second inequality. The slope is somewhat steep and downward, so we can eliminate choices A and B for having lines that are too shallow. Choice D can be eliminated because this inequality is greater than or equal to, which means that the shaded portion should be above the line, not below. This leaves choice C, which also correctly shows the shading beneath the line for the first inequality. Part B: The correct answers are A and D. Use the second part of the system of inequalities and determine if the amount of money earned for each input is greater than or equal to 400.

B. (18, 14): 15(18) + 9(14) = 396. No C. (15, 19): 15(15) + 9(19) = 396. No D. (12, 28): 15(12) + 9(28) = 432. Yes E. (5, 35): 15(5) + 9(35) = 390. No Part C: The correct answer is 15. Substitute the known value for y, 20, into the second inequality and solve for x. 15x + 9y > 400

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SOLVING GRAPHICALLY

A. (25, 12): 15(25) + 9(12) = 483. Yes

LEAP 2025 Algebra I: Wrap-Up

Chapter 5 Wrap-Up

LEAP 2025 Algebra I: Wrap-Up

WORKKEYS OVERVIEW

Wrap-Up

Wrap-Up Remember These Key Test-Taking Techniques •

Process of Elimination

•

Plug It In

•

Word Problem Translation

•

Negative Paranoia

•

Create a Visual

•

Don’t Overthink It

•

Read the Question

NOTES:

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LEAP 2025 Algebra I: Further Practice

aCLgebra I eOC hapter 6

Further Practice

LEAP 2025 Algebra I: Further Practice Further praCtICe:

PrAcTice SeT one

Practice Set One 1.

Which of the following expressions is equivalent to 5x 2 – (x + 3) 2 + 8x – 12? A.

4x 2 + 2x – 21

4x 2 + 2x – 3

4x 2 + 14x – 21

4x 2 + 14x – 3

The graph represents the change in internal temperature of a pie as it bakes in an oven for one hour. y Pie Internal Temperature

Temperature (°F)

200 150 100 50 0

30 45 Time (min)

Which unit would be appropriate for the rate of change in the graph? A.

degrees hour

degrees minute

C. D.

hours degree minutes degree

Which of the following is a zero of the polynomial expression 4x + 32? A.

–32

–8

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LEAP 2025 Algebra I: Further Practice Further praCtICe: 4.

PrAcTice SeT one

Which equation could be used to find the zeros of the function 15x – 3x – 12? 2

(x – 4)(x + 3) = 0

(x + 5)(x + 3) = 0

(3x + 5)(4x – 3) = 0

(5x + 4)(3x – 3) = 0

Three times Sasha’s age plus two times Paulo’s age equals 52. Paulo’s age is also five times Sasha’s age. How old is Paulo?

Force is related to the mass of two objects by the formula F = •

Gm1 m2 . d2

G is the gravitational constant.

•

m1 and m2 are the mass of two objects.

•

d is the distance between the objects.

Which equation finds d, given F, G, m1, and m2 ?

F Gm1 m2

Gm1 m2 F

F Gm1 m2

Gm1 m2 F

The table below shows the cost of buying protein bars from a health food store. Protein Bars

120

150

Cost (in dollars)

100

125

What is the meaning of the slope of the linear model for the data? A.

The cost of 5 bars is 1 dollar.

The cost of 5 bars is 6 dollars.

The cost of 6 bars is 1 dollar.

The cost of 6 bars is 5 dollars.

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LEAP 2025 Algebra I: Further Practice Further praCtICe: 8.

PrAcTice SeT one

Use the equation to answer the question. 2x2 + 12x + 5 = 7 Valentina is completing the square to rewrite the equation. Which equation could be her result?

(x + 3)2 = 5

(x + 3)2 = 7

(x + 3)2 = 10

(x + 3)2 = 13

An acrobat’s height as she jumps from a platform above a trampoline, in feet, is modeled by the function h(x) = –x 2 + 5x + 36, where x represents the distance of the acrobat from the platform. How far from the platform will the acrobat be when she reaches the trampoline? A.

0 feet

4 feet

5 feet

9 feet

10. Jessica wants to earn at least $145 dollars from her two jobs next week. She can work 18 hours at most. Her first job pays $8 per hour, and her second job pays $9 per hour. Let b represent the number of hours worked at the first job and s represent the number of hours worked at the second job. Which system of linear inequalities models Jessica’s situation? A.

b + s < 18 8b + 9s > 145

b + s ≤ 18 8b + 9s ≥ 145

b + s ≥ 18 8b + 9s ≤ 145

b + s > 18 8b + 9s < 145

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LEAP 2025 Algebra I: Further Practice Further praCtICe:

PrAcTice SeT exPlAnATionS

Practice Set Explanations PRACTICE SET ONE 1. The correct answer is A. Simplify the expression by using FOIL on (x + 3)2, distributing the negative sign, and combining like terms. 5x2 – (x + 3)2 + 8x – 12 5x2 – (x + 3)(x + 3) + 8x – 12 5x2 – (x2 + 6x + 9) + 8x – 12 5x2 – x2 – 6x – 9 + 8x – 12 4x2 + 2x – 21 2. The correct answer is B. In the figure, the y-axis represents temperature (°F) and the x-axis represents time (minutes). The rate of change, or slope, of the linear equation is

change in y degrees or . change in x minute

3. The correct answer is B. Find the zero of the polynomial by setting the expression equal to zero and solving for x. 4x + 32 = 0 4(x + 8) = 0 x+8=0 x = –8 4. The correct answer is D. Factor the function and use the process of elimination to quickly rule out incorrect answer options. Because 15x2 appears in the function, the x terms in the answer choices must multiply to equal this. The only answer choice where this occurs is choice D, where (3x)(5x) = 15x2. 5. The correct answer is 20. Let Sasha’s age be s and let Paulo’s age be p. Use the first sentence to create an equation. Then, use the second sentence to create another equation. 3s + 2p = 52 p = 5s Substitute 5s for p in the first equation and solve for s.

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LEAP 2025 Algebra I: Further Practice Further praCtICe:

PrAcTice SeT exPlAnATionS

3s + 2(5s) = 52 3s + 10s = 52 13s = 52 s=4 Sasha is 4 years old. So Paulo is p = 5(4) = 20 years old. 6. The correct answer is D. Use the formula to solve for d. F=

Gm1m2 d2

Fd = Gm1m2 2

d2 =

Gm1m2 F

7. The correct answer is D. The slope of the linear model is in the form

bars . Take two points from the table, dollars

(25, 30) and (50, 60), and use the slope formula. m=

60 − 30 30 6 6 bars = = = 50 − 25 25 5 5 dollars

The cost of 6 bars is 5 dollars. 8. The correct answer is C. Isolate the x terms to the left side of the equation, simplify, and complete the square 2

 b  by adding   to both sides of the equation.  2a 

2x2 + 12x + 5 = 7 2x2 + 12x = 2 x2 + 6x = 1 2

 6  x2 + 6x +   =  2 (1)  x2 + 6x + 9 = 9 + 1

 6   2 (1)  + 1  

(x + 3)2 = 10 9. The correct answer is D. Set the quadratic function equal to zero. Then, factor and solve for x. –x2 + 5x + 36 = 0 x2 – 5x – 36 = 0 (x – 9)(x + 4) = 0 x = –4, 9

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Leap 2025 geometry Better Scores in ONE Day

Our Louisiana LEAP 2025 Geometry Boot Camp is designed specifically to increase the number of students scoring Basic and Mastery on the LEAP 2025 Geometry assessment.

In just one day, students will learn: • Core skills for success in geometry • Pacing and time management • Test-taking and guessing strategies that really work • How to overcome test anxiety and put their best foot forward on test day

Why schedule a LEAP Geometry Boot Camp? • Authentic, up-to-date practice questions • Students review exactly what they need in the “final hours” before the test. • Improves student confidence • Easy to schedule, during the school day or on the weekend • Makes test prep fun and less overwhelming for students

Implementation Models • Full-day workshop during school hours • After-school or Saturday programming • Virtual and in-person programs available

Table of Contents

Table of Contents Chapter 1: LEAP 2025 GEomEtry ovErviEw ..................................................7 Chapter 2: ConGruEnCE trAnsformAtions And simiLArity................13 CongruenCe TransformaTions and similariTy overview............................. 14 whaT are BooT Camp mini-TesTs?...................................................................... 15 mini-TesT one................................................................................................................ 16 dilaTions........................................................................................................................... 18 refleCTions and roTaTions........................................................................................ 19 mini-TesT one explanaTions....................................................................................21

Chapter 3: simiLArity in triGonomEtry And modELinG & APPLyinG.......23 similariTy in TrigonomeTry and modeling & applying overview............. 24 mini-TesT Two...............................................................................................................25 sohCahToa.............................................................................................................28 assume values.............................................................................................................. 30 mini-TesT Two explanaTions...................................................................................32 Chapter 4: AdditionAL & suPPortinG ContEnt...............................................33 addiTional & supporTing ConTenT overview..................................................34 mini-TesT Three............................................................................................................ 35 arCs and angles......................................................................................................... 38 geomeTry definiTions.................................................................................................40 mini-TesT Three explanaTions.................................................................................41

Table of Contents

Chapter 5: ExPrEssinG mAthEmAtiCAL rEAsoninG..........................................43 expressing maThemaTiCal reasoning overview.............................................. 44 mini-TesT four..............................................................................................................45 mulTi-sTep paniC......................................................................................................... 46 ComponenTs and parTs: how The TesT is sCored......................................... 48 mini-TesT four explanaTions..................................................................................49 Chapter 6: modELinG & APPLiCAtion.........................................................................51 modeling & appliCaTion overview........................................................................52 mini-TesT five................................................................................................................ 53 word proBlem TranslaTion.....................................................................................55 hidden similar Triangles...........................................................................................56 mini-TesT five explanaTions....................................................................................58 Chapter 7: mixEd PrACtiCE.............................................................................................59 mini-TesT six..................................................................................................................60 angle rules....................................................................................................................63 mini-TesT six explanaTions.......................................................................................64 Chapter 8: Wrap-Up...........................................................................................................65

Chapter One: LEAP 2025 Geometry Overview

Chapter 1 LEAP 2025 Geometry Overview

LEAP 2025 GEOMETRY OVERVIEW

Chapter One: LEAP 2025 Geometry Overview LEAP 2025 GEomEtry ovErviEw

What Is End-of-Course Testing? End-of-Course (EOC) testing measures your aptitude in a given subject after you have finished a course. Consider it a subject understanding checkup. Teachers use it to identify both your strengths and areas for improvement. This helps ensure you are on track in developing the knowledge and skills needed for the next grade and, eventually, college and a career. In this Boot Camp, we’ll focus on the major concepts tested on the LEAP 2025 Geometry test.

Why Should You Care? •

Many schools require the EOC as part of your final grade in the course.

•

If your school uses the EOC as a final exam for the course, then doing well on this test can boost your GPA.

•

A good EOC score is a positive indicator that you are on track for college.

•

Mastering the foundational skills taught in this Boot Camp will help you succeed in more difficult math courses in the future.

•

Put in the effort now and save yourself from repeating a course or taking summer school.

NOTES:

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leap 2025 geomeTry BooT Camp

LEAP 2025 GEomEtry ovErviEw

Orientation The five conceptual categories tested on the LEAP 2025 Geometry test are Congruence Transformations, Similarity in Trigonometry, Additional & Supporting Content, Expressing Mathematical Reasoning, and Modeling & Application. The test is timed. All sessions except the first one allow you to use a calculator. To do well on this test, it is important to be comfortable performing mathematical operations with and without a calculator. Here is a breakdown of possible ways the LEAP 2025 Geometry assessment will test you in each of its main content categories. Each test session features questions from each category. Congruence Transformations and Similarity ask you to use similarity and congruence criteria to determine the nature of transformed figures, to prove theorems and geometric relationships, and to solve problems related to congruence. Similarity in Trigonometry and Modeling & Applying measure your ability to understand and use trigonometric functions and the Pythagorean Theorem. You are also asked to apply geometric concepts in modeling situations, on the coordinate plane, and with equations to solve problems. Additional & Supporting Content includes problems related to understanding geometric definitions, transforming figures, and the geometry of circles and solids. Expressing Mathematical Reasoning asks you to construct mathematical justifications, arguments, and critiques. Modeling & Application assesses how well you solve real-world problems by practicing modeling.

NOTES:

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LEAP 2025 GEOMETRY OVERVIEW

Chapter One: LEAP 2025 Geometry Overview

LEAP 2025 GEOMETRY OVERVIEW

Chapter One: LEAP 2025 Geometry Overview LEAP 2025 GEomEtry ovErviEw

Orientation In Louisiana, the LEAP 2025 test score counts as a percentage of your final grade for your Geometry course. The percentage is always between 15 and 30 percent of your grade, depending on your school district. The table below gives the approximate contribution of each conceptual category to your score. This is just an estimate and can vary greatly from one test to another. If you know what areas you struggle in, compare them to the most important skills needed for the test. Category

Total Points

Percentage of Points

Congruence Transformation and Similarity

20%

Similarity in Trigonometry

20%

Additional & Supporting Content

30%

Expressing Mathematical Reasoning

10%

Modeling & Application

20%

Total

100%

The test is administered in four sessions. The sessions are timed, so it is important to manage your minutes and avoid spending too much time on any one question. Usually, each single-part question is worth one point, and each multi-part question is worth one point per part. Questions that require you to show work or justify your answers are typically worth 2, 3, or even 4 points each. Each test session may include multiple choice, multiple select, constructed response, fill-in-the-blank, and a variety of technology-assisted answering methods. For example, you might have to actually draw a graph on your computer. We provide practice for all of these different question types during the boot camp.

10 |

Test Session

Calculator?

Number of Points

Time Limit

Session 1a

25 minutes

Session 1b

Yes

55 minutes

Session 2

Yes

80 minutes

Session 3

Yes

80 minutes

Total

–

240 minutes

leap 2025 geomeTry BooT Camp

Chapter Two: Congruence Transformations and Similarity

Chapter 2 Congruence Transformations and Similarity

Chapter Two: Congruence Transformations and Similarity CongrUenCe transformations and similarity:

overview

Congruence Transformations and Similarity Overview CONGRUENCE TRANSFORMATIONS

The Congruence Transformations and Similarity conceptual category tests your proficiency over a broad range of geometry skills. The skills that will be tested on your exam include but are not limited to the following:

Dilations •

Dilate a figure given a center and a scale factor

Congruence and Similarity •

Determine if two figures are similar

•

Use congruence and similarity criteria to solve problems and prove relationships

Transformations •

Transform figures using geometric descriptions of rigid motions

Geometric Theorems •

Prove geometric theorems

NOTES:

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Chapter Two: Congruence Transformations and Similarity CongrUenCe transformations and similarity: whaT are mini-TesTs?

What Are Boot Camp Mini-Tests? During this Boot Camp you will take several mini-tests, which are small segments of a Geometry test. While taking these mini-tests, it’s important to imagine that you are in an actual testing environment. The time limits assigned as you complete the mini-tests. In the mini-tests, we are focusing on only one category of question at a time, but on the real assessment each test session will include questions from several conceptual categories. For these mini-tests, you have 10 to 15 minutes to answer several questions. Your instructor will signal when you are out of time. Try to get through all the questions within the time limit. Unless your instructor has provided you with an answer sheet, circle your answers directly in this book. The real test does not allow the use of cell phones, watches, or computers, so you shouldn’t use them on the mini-tests, either.

NOTES:

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CONGRUENCE TRANSFORMATIONS

match the pace that you should try to keep during the actual test. Practice all of the skills that you have learned

Chapter Two: Congruence Transformations and Similarity

Mini-Test One 1.

In the coordinate plane, line q has slope 6 and y-intercept (0, 8). Line p is the result of dilating line q by a factor of 4 with center (0, 4). What is the y-intercept and slope of line p? A.

Line p has slope 2 and y-intercept (0, 4).

Line p has slope 6 and y-intercept (0, 8).

Line p has slope 6 and y-intercept (0, 20).

Line p has slope 10 and y-intercept (0, 12).

The figure gives rhombus ABCD on the coordinate plane. Diagonals AC and BD intersect at point E.

PART A Find the coordinates of point A in terms of r, s, and t. Write your answer in the box provided below. Write only your answer in the box.

Mini-Test One

GO ON TO THE NEXT PAGE.

Chapter Two: Congruence Transformations and Similarity

PART B Since ABCD is a rhombus, AC and BD bisect one another. Use the coordinates of the rhombus to prove that AC and BD bisect one another. Enter your justification in the space provided below.

Vertices X(2, 3), Y(5, 7), and Z(5, 3) form triangle XYZ on the coordinate plane. The triangle is reflected across the y-axis and then rotated 90 degrees counterclockwise about the origin to form triangle X'Y'Z'. What is the sum of the x values of the vertices of triangle X'Y'Z'? A.

–13

–12

–11

STOP! END OF TEST. Mini-Test One

YOU MAY GO BACK AND CHECK YOUR WORK.

Chapter Two: Congruence Transformations and Similarity CongrUenCe transformations and similarity:

dilaTions

Dilations In the Congruence Transformations and Similarity category, dilations are the most important topic. A dilation is a transformation that produces an image that is the same shape, but a different size.

CONGRUENCE TRANSFORMATIONS

In the coordinate plane, line q has slope 6 and y-intercept (0, 8). Line p is the result of dilating line q by a factor of 4 with center (0, 4). What is the y-intercept and slope of line p? A.

Line p has slope 2 and y-intercept (0, 4).

Line p has slope 6 and y-intercept (0, 8).

Line p has slope 6 and y-intercept (0, 20).

Line p has slope 10 and y-intercept (0, 12).

There are a few rules that need to be memorized about dilations. When a line or line segment is dilated, no matter where the center of dilation is, the slope remains the same. In the above question, since line q has a slope of 6, line p must also have a slope of 6. We can eliminate choices A and D. Every point on the line, line segment, or shape being dilated has a distance from the center of the dilation. The dilation factor gives you what to multiply that distance by to determine the distance of the new points on the dilated image. Each new point is always along the line formed by the center and the original point. For example, if point Z is part of a dilation and it has a distance of 2 from center, then a dilation factor of 3 would mean that Z’ should have a distance of 6 from center. If the dilation factor is 0.5, Z’ would be 1 from center. A dilation factor of 1 means that you end up with the same image. In the above question, the y-intercept of q has a distance of 4 from the center. Since the dilation factor is 4, the new corresponding point should have a distance of 16 from center. Since the center is on the y-axis, the new point has coordinates of (0, 20), which gives us line p’s y-intercept. Thus, the correct answer is C. A line that that passes through the center of a dilation remains unchanged. A line segment is shortened or lengthened by the dilation factor, regardless of where the center is. For example, a line segment with a length of 6 that gets dilated by a factor of 2 has a new length of 12. If the dilation factor is 0.5, the new length is 3. To dilate a shape, use the dilation factor and the location of the center to determine the location of each vertex in the new image.

18 |

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Chapter Two: Congruence Transformations and Similarity CongrUenCe transformations and similarity: refleCTions and roTaTions

Reflections and Rotations

Some questions will ask you to reflect or rotate figures in the coordinate plane. With items of this type, it's important to take one step at a time, to make sure you are following instructions about clockwise or counterclockwise

–13

–12

–11

In this question, we are first asked to make a reflection across the y-axis. Here are the rules for reflections: • When reflecting across the x-axis, multiply all y-coordinates by –1. • When reflecting across the y-axis, multiply all x-coordinates by –1. With a reflection, you multiply by –1 the coordinates that don't belong to the reflection axis (x-axis means y, y-axis means x). After the reflection, our new triangle is X(–2, 3), Y(–5, 7), Z(–5, 3). Next, we must rotate the triangle 90° counterclockwise about the origin. • The rule for rotating 90° counterclockwise is to multiply the y-coordinate by –1, then swap the x- and y-coordinates. You can memorize this table if you want to be very fast with rotations around the origin: Counterclockwise Rotation

Clockwise Rotation

New Coordinates

90°

270°

(–y, x)

180°

(–x, –y)

270°

90°

(y, –x)

However, if all you remember is the 90° counterclockwise rotation rule, you can slowly figure out everything on the table. If you are given a clockwise rotation, convert it to counterclockwise. (90° CW = 270° CCW, 270° CW = 90° CCW). Then do one 90° rotation at a time until you've reached the correct number of degrees.

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CONGRUENCE TRANSFORMATIONS

direction, and to keep some essential rules in mind.

Chapter Two: Congruence Transformations and Similarity CongrUenCe transformations and similarity: mini-TesT explanaTions

Mini-Test Explanations 1. The correct answer is C. When a line is dilated with center not on the line, the resulting line is parallel (same slope). Line p must therefore have a slope of 6. Since the center of the dilation is on the y-axis and the distance from the y-intercept of line q to the center is 8 – 4 = 4, then the distance from the center to the y-intercept of line p must be 4 · 4 = 16. Add 16 to the y value of the center to find the y-intercept of line p. 16 + 4 = 20. The slope of line p is 6 and its y-intercept is (0, 20).

Since the shape is a rhombus, and the base of the shape runs along the x-axis, AB must also be parallel to the x-axis. Therefore, the y value for A must be the same as the y value for B: 2t. A rhombus has four side lengths that are the same. Since the distance from C to D is 2r, then the distance from A to B must also be 2r. The x-coordinate of B is 4s, so the x-coordinate of A must be 4s + 2r. The coordinates of point A are (4s + 2r, 2t). Part B: Sample correct response: (One point for stating that the midpoint of AC must be the same as the midpoint of BD. One point for providing reasoning to verify that AC and BD bisect one another.) AC bisects BD because AC has the same midpoint as BD. I know this because I can use the midpoint formula on the given coordinates.  4 s + 2 r + 0 2t + 0  , The midpoint of AC =   = (2s + r, t) 2 2    4 s + 2 r 2t + 0  , The midpoint of BD =   = (2s + r, t) 2 2   Since both line segments have the same midpoint, they bisect one another. 3. The correct answer is A. When a point is reflected across the y-axis, the y-coordinate stays the same and the x-coordinate is multiplied by –1. The reflected points are X(–2, 3), Y(–5, 7), and Z(–5, 3). To rotate counterclockwise 90 degrees, multiply the y-coordinate by –1 and then swap the x- and y-coordinates. The transformed vertices are thus X'(–3, –2), Y'(–7, –5), and Z'(–3, –5). Add the x-coordinates together. –3 + – 7 + – 3 = –13.

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CONGRUENCE TRANSFORMATIONS

2. Part A: The correct answer is (4s + 2r, 2t).

Chapter Three: Similarity in Trigonometry and Modeling & Applying

Chapter 3 Similarity in Trigonometry and Modeling & Applying

Chapter Three: Similarity in Trigonometry and Modeling & Applying similarity in trigonometry:

overview

Similarity in Trigonometry and Modeling & Applying Overview SIMILARITY IN TRIGONOMETRY

The Similarity in Trigonometry conceptual category tests your proficiency over a broad range of trigonometry and geometry skills. The skills that will be tested on your exam include but are not limited to the following:

Solving Right Triangles •

Use trigonometric ratios and the Pythagorean Theorem to solve right triangles

Line Segment Ratios •

Find a point on a line segment that splits the segment in a given ratio

Similarity and Trigonometric Ratios •

Use similarity, side ratios, and the properties of special right triangles to solve problems

•

Understand and apply the definitions of trigonometric ratios for acute angles

NOTES:

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Chapter Three: Similarity in Trigonometry and Modeling & Applying

Mini-Test Two 1.

Point A, with coordinate 1.6, point B, with coordinate 3.4, and point C are on a number line. Point B partitions AC into two parts such that the ratio of the length of AB to BC is 6:5. What is the coordinate of point C? Write your answer in the box provided below.

Right triangles LMN and QRS are similar to one another.

Which ratios are equal to cos S? Select all that apply. A.

QR SQ

SR SQ

QR SR

LM NL

NM NL

NM LM

Mini-Test Two

GO ON TO THE NEXT PAGE.

Chapter Three: Similarity in Trigonometry and Modeling & Applying

Use the triangle depicted below to answer the question.

In right triangle XYZ, m∠X ≠ m∠Y . Let cos X = 2a and sin X = 3b. What is cos Y – sin Y? A.

3b – 2a

2a – 3b

2a + 3b

3b 2a

Mini-Test Two

GO ON TO THE NEXT PAGE.

Chapter Three: Similarity in Trigonometry and Modeling & Applying

A projector screen is mounted from the ceiling at point A. The bottom of the screen is connected to a motorized hook at point B, which can move up and down a pole running between points C and D. The angle θ is the angle created as the hook moves between points C and D.

PART A When the screen is extended to a distance of 11.4 meters, what is the value of θ to the nearest tenth of a degree? A.

72.1°

69.5°

60.0°

20.5°

PART B When θ = 47°, what is the distance from point B to point A to the nearest tenth of a meter? Write your answer in the box provided below. meters

STOP! END OF TEST. Mini-Test Two

YOU MAY GO BACK AND CHECK YOUR WORK.

Chapter Three: Similarity in Trigonometry and Modeling & Applying similarity in trigonometry:

mini-TesT explanaTions

Mini-Test Explanations 1. The correct answer is 4.9. The length of AB is 3.4 – 1.6 = 1.8. Use the ratio 6:5 to set up a proportion to find the length of BC. x 5 = 1.8 6

SIMILARITY IN TRIGONOMETRY

6x = 9 x = 1.5 Add the length to the coordinate of B to find the coordinate of C. 3.4 + 1.5 = 4.9. 2. The correct answers are B and E. The cosine of an angle is defined as the length ratio between the adjacent side and the hypotenuse. The adjacent side is SR, and the hypotenuse is SQ. cos S =

SR SQ

Therefore, B is correct. NM Since LMN is similar to QRS, m∠N = m∠S. It follows that cos N = cos S. Thus, is also equal to cos S. NL Choice E is correct. 3. The correct answer is A. sin X = cos Y, and cos X = sin Y. Since cos X = 2a, sin Y = 2a. Likewise, sin X = 3b, so cos Y = 3b. Therefore, cos Y – sin Y = 3b – 2a. 4. Part A: The correct answer is B. Since AB is 11.4 and AC is 4, the adjacent side and the hypotenuse have known lengths. Set up an equation using cosine and solve for θ. cos θ =

4 11.4

θ = cos–1

θ ≈ 69.5°

4 11.4

Part B: The correct answer is 5.9. AC is 4 and the measure of θ is 47°. Set up an equation using cosine and solve for the length of AB. cos 47° = x=

4 x

4 cos 47

x ≈ 5.9

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Chapter Four: Additional & Supporting Content

Chapter 4 Additional & Supporting Content

Chapter Four: Additional & Supporting Content additional & sUpporting Content:

overview

Additional & Supporting Content Overview The Additional & Supporting Content item category tests your proficiency over a broad range of geometry skills.

ADDITIONAL & SUPPORTING CONTENT

The skills that will be tested on your exam include but are not limited to the following:

Definitions •

Know precise definitions of angle, circle, perpendicular line, parallel line, and line segment

Rotations and Transformations •

Given a shape, such as a parallelogram, trapezoid, or regular polygon, describe the rotations and reflections that map it unto itself

•

Perform rotations, reflections, and translations on geometric figures

Circles, Angles, Radii, and Chords •

Find the center and radius of a circle using the equation of a circle and derive the equation of a circle using the Pythagorean Theorem

•

Solve problems involving inscribed angles, radii, and chords

Shapes and Solids •

Solve for the volume of cylinders, pyramids, cones, and spheres

•

Identify cross-sections of three-dimensional objects and determine 3-D objects created from rotations of 2-D objects

NOTES:

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Chapter Four: Additional & Supporting Content

Mini-Test Three 1.

 is congruent to ST  is equal to  . Five times the measure of QR Points Q, R, S, and T lie on circle A. QR . four times the measure of QT R Q A S

Part A Select from the choices near the box to complete the statement: four-ninths

five-ninths

equal to

The measure of angle ∠RST is

the measure of ∠QRS . four times

five times

nine times

Part B Select from the choices near the box to complete the statement. four-ninths

five-ninths

equal to the measure of ∠QRS .

The measure of ∠QSR is four times

Mini-Test Three

five times

nine times

GO ON TO THE NEXT PAGE.

Chapter Four: Additional & Supporting Content

Select from the choices near the boxes to complete each sentence. A shape formed by two rays diverging from a common endpoint is called a(n) triangle

tangent .

angle

perpendicular line

The common endpoint is called the center

vertex .

axis

radius

Regular hexagon ABCDEF is given below with center at point G. D

G E

Which tranformations will map hexagon ABCDEF onto itself? Select each correct transformation. A.

reflecting over CF

reflecting over BF

reflecting over AD

rotating 120° clockwise around point G

rotating 90° clockwise around point G

rotating 240° counterclockwise around point G

Mini-Test Three

GO ON TO THE NEXT PAGE.

Chapter Five: Expressing Mathematical Reasoning

Chapter 5 Expressing Mathematical Reasoning

Chapter Five: Expressing Mathematical Reasoning expressing mathematiCal reasoning:

overview

Expressing Mathematical Reasoning Overview The Expressing Mathematical Reasoning task category tests your proficiency over a broad range of geometry

EXPRESSING MATH REASONING

skills. The skills that will be tested on your exam include but are not limited to the following:

Coordinate Reasoning •

Apply geometric reasoning in the coordinate plane and use coordinates to draw geometric conclusions

Geometric Propositions and Conjectures •

Construct chains of reasoning to justify or refute geometric propositions and conjectures

Multi-Step Problems •

Solve multi-step geometry problems, or identify and correct solutions for multi-step problems, and explain your reasoning

Algebra and Geometry •

Use a combination of algebraic and geometric reasoning to justify or refute statements about geometric figures

NOTES:

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Chapter Six: Modeling & Application

Chapter 6 Modeling & Application

Chapter Six: Modeling & Application modeling & appliCation:

word proBlem TranslaTion

Word Problem Translation The secret to solving a word problem is translating it into math. When translating word problems to algebraic equations, it is important to know which words translate to which

MODELING & APPLICATION

operations. is, equal to, is the same as → =

times, product, each, per, of → ·

minus, without, less, difference, change → –

plus, together, and, combined, both, more → +

divided into, split between or among, divvied up → ÷

Let’s take a look at how word problem translation can help you solve a problem on your exam. 1.

A company offers blankets in two sizes: twin and double. The twin blanket is 24 inches longer than it is wide. The double blanket has the same length as the twin blanket, but it is 14 inches wider. The area of the twin blanket is 1,260 square inches less than the area of the double. PART A Write and equation that can be used to determine the area of the double blanket. Define all variables used. Write your equation in the space provided below.

The area of the blanket is length times width. To find the equation of the blanket's area, you must first find its dimensions. You will need to translate the sentence the double blanket has the same length as the twin blanket, but it is 14 inches wider into an algebraic equation. Let x represent the width of the twin blanket and A represent the area of the double blanket. The length of the twin blanket, and thus the length of the double blanket, is x + 24, since the twin is 24 inches longer than it is wide. The width of the double blanket is x + 14 because it is 14 inches wider than the twin blanket. We use the plus operand in both cases because we are dealing with the concept of more: more width and more length. Area is determined by multiplying width by height, so we can create this final equation to answer Part A: A = (x + 14)(x + 24) Be sure to define the variables you use when the question asks for it. Otherwise you will not receive points!

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Chapter Seven: Mixed Practice

Chapter 7 Mixed Practice

Chapter Seven: Mixed Practice

Mini-Test Six 1.



In the figure shown below, GH intersects AB and CF at points E and D, respectively. G

A E B C D

F H

PART A Given: ∠CDH ≅ ∠AED Prove: ∠EDF ≅ ∠AED Fill in the blanks using the following options: • • • • • •

Transitive Property of Congruence Symmetric Property of Congruence Reflexive Property of Congruence Definition of Congruent Angles Vertical Angles are Congruent Given

Statement: ∠CDH ≅ ∠AED . Reason:

Statement: ∠EDF ≅ ∠CDH . Reason:

Statement: ∠EDF ≅ ∠AED . Reason:

Mini-Test Six

GO ON TO THE NEXT PAGE.

LEAP 2025 Geometry: Wrap-Up

Chapter 8 Wrap-Up

LEAP 2025 Geometry: Wrap-Up

WORKKEYS OVERVIEW

Wrap-Up

Wrap-Up Remember These Key Tips •

Dilations

•

Reflections and Rotations

•

SOHCAHTOA

•

Assume Values

•

Arcs and Angles

•

Geometry Definitions

•

How the Test is Scored

•

Multi-Step Panic

•

Word Problem Translation

•

Hidden Similar Triangles

NOTES:

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LEAP 2025 Geometry: Wrap-Up

WORKKEYS OVERVIEW

Wrap-Up

Study Plan •

7 Days Before Test: Review Chapter 2, Congruence Transformations and Similarity.

•

6 Days Before Test: Review Chapter 3, Similarity in Trigonometry and Modeling & Applying.

•

5 Days Before Test: Review Chapter 4, Additional & Supporting Content.

•

4 Days Before Test: Review Chapter 5, Expressing Mathematical Reasoning.

•

3 Days Before Test: Review Chapter 6, Modeling & Application.

•

2 Days Before Test: Quickly review the entire boot camp book, paying special attention to the strategies you learned.

•

1 Day Before Test: Be sure to get a good night’s rest and line up a nutritious breakfast in the morning.

NOTES:

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LEAP 2025 Geometry: Wrap-Up

WORKKEYS OVERVIEW

Wrap-Up

Before Your Test Date •

Get enough sleep the entire week before the test.

•

Eat well, especially on the day of the test.

•

Bring a snack with you on test day. Fruit and/or protein bars work wonders. Avoid sugar and junk food. A bottle of water is a good suggestion.

•

If you can’t do without caffeine, allow about one month before test time to minimize your intake. Being jittery during the test can affect your test performance.

•

It is important to feel as healthy as possible both mentally and physically on the day of the exam.

•

Reduce distractions! Stay away from social media for 12–24 hours before the test starts.

NOTES:

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