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Name: ___________________________________________

Essential Skills Practice for students entering

Geometry or Accelerated Geometry Use this document to review the mathematics that you have learned previously. Completion of the Essential Skills Practice and comprehension of these topics will help you to be more successful in Geometry and Accelerated Geometry. A. basic perimeter and area B. line and angle relationships C. basic equation solving D. proportions E. slope F. writing linear equations G. graphing linear equations H. parallel and perpendicular Lines I. systems of linear functions J. operations with fractions K. solving quadratics (factoring and Quadratic Formula) L. radicals M. Pythagorean Theorem Parents: We are committed to the academic success of our students and see a benefit for them to keep their skills sharp over their summer vacations. Research provides evidence to support that practice over an extended break will help students retain and apply knowledge and lessen the need for remediation at the beginning of each school year. Our goal is to maximize the growth in all of our students while we have them with us at Dakota. Our teachers are committed to creating and analyzing the work that students will complete and return at the beginning of the year. The overall intent is to help students maintain and expand their basic skills in mathematics, as appropriate to the course your child is signed up for in the fall. Summer work is not designed to be “tricky” or to introduce new material. It is also not designed to “hurt” anyone’s grade. It will instead be a way to increase foundational skills needed to be successful in courses while keeping students in the right mindset while away from the school. We see this as a way to support all students and help clarify what skills will be necessary to be successful in our classes, because of the rigor and high standards we want our students to understand and achieve. These assignments will help our teachers gauge the knowledge and skill-base of all of their students to help guide instruction that is appropriate for all learners. While these assignments are not mandatory and will not be graded, the benefit for students is research proven and will also help our teachers provide appropriate differentiated instruction for all of their students. Assignments will be collected and analyzed, because we are committed to understanding our student’s abilities and comprehension levels in order to provide the most appropriate, individualized education that we can. Thank you for your continued support towards our shared goals of maximizing every students’ educational experience at Dakota High School.

-Paul Sibley, Dakota High School Principal

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A. basic perimeter and area

The perimeter of a figure refers to the distance around a two-dimensional figure. The area of a figure refers to the amount of space inside a two-dimensional figure. Students should be able to calculate the perimeter of any polygon and be able to calculate the circumference of a circle. Incoming algebra students should be able to calculate the area of rectangles, triangles and circles proficiently. The following table includes information about how to calculate these measurements.

shape perimeter or circumference area

rectangle add all sides length times width

triangle add all sides ½ length times width

circle 2π times the radius π times the radius squared

Examples:

A.

perimeter = add all sides

P = 9 + 12 + 15

P = 36

area = ½ l • w

A = ½ • 12 • 9

A = 54

B.

perimeter = add all sides

P = 14 + 8 + 14 + 8

P = 44

area = l • w

A = 14 • 8

A = 112

C.

circumference = 2πr

C = 2 • π • 5

C = 10π in

area = πr2

A = π • 52

A = 25π in2

D.

perimeter = add all sides

P = 2x + 3 + 5x + 2x + 3 + 5 P = 14x + 6

area = l • w

A = 5x(2x+ 3) A = 10x2 + 15x

5x

2x + 3

12 15

9

8

14

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***Practice for basic perimeter and area*** Find the perimeter or circumference and the area for numbers. You may leave your answers in terms of pi.

1.

Perimeter = Area=

2. Circumference = Area =

3. Perimeter = Area=

4. Perimeter = Area =

7y

-3y + 12

16.1 ft 16.1 ft

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B. line and angle relationships

Vocabulary you should remember from elementary and middle school:

ray

line

line segment

adjacent

vertical angles

complementary angles

supplementary angles

linear pair

triangle sum theorem

***Practice for line and angle relationships****

Fill in the blanks with ray, line or line segment.

1. OA is a ______________

2. ED is a ______________

3. JH is a ______________

4. IB is a ______________

5. OJ is a ______________

6. Find MN.

7. Find SU.

Identify each pair of angles as adjacent, vertical, complementary, supplementary or a linear pair.

8.

9. 10.

11.

12. 13.

Find the missing angle.

14. 15. 16.

L M N 11

21

S T U 5 16

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C. solving basic equations When we solve an equation we are looking for the value of the variable that will make the equation true. We

use opposite operations to isolate the variable. Sometimes, we will need to distribute, combine like terms, and

move variables from one side to another in order to solve.

Example: Solve 3x – 12 = 72

+12 + 12 add 12 to each side of the equation

3𝑥

3=

84

3 divide each side of the equation by 3

x = 28

To check our answer we can plug 28 into our original equation for x and show that it makes the equation true.

3(28) – 12 = 72

84 – 12 = 72

72 = 72 The x value of 28 makes the equation true, thus our solution is correct.

Example: 1

3(6𝑥 − 12) − 10𝑥 = 20

2x – 4 – 10x = 20 distribute the 1/3

-8x – 4 = 20 combine the like terms

+4 = + 4 add 4 to each side

−8𝑥

−8=

24

−8 divide both sides by -8

x = -3

Check the answer: 1

3(6(−3) − 12) − 10(−3) = 20

1

3(−18 − 12) + 30 = 20

1

3(−30) + 30 = 20

-10 + 30 = 20

20 = 20

Example: 5x + 34 = -2 (1 – 7x)

5x + 34 = -2 + 14x distribute the -2

-5x -5x subtract 5x from each side to move the variables onto the same side

34 = -2 + 9x combine the like terms

+2 = +2 add 2 to each side of the equation

36

9=

9𝑥

9 combine like terms and then divide both sides by 9

4 = x

The x value of -3 makes the equation true,

thus our solution is correct.

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***Practice solving basic equations****

Solve each equation.

1. -10 = -14x – 6x

2. x – 1 = 5x + 3x - 8

3. -8 = -(x + 4) 4. -(7 – 5x) = 9

5. 2(4x – 3) – 8 = 4 + 2x

6. -3(4x + 3) + 4(6x + 1) = 43

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D. proportions Students will need to be able to write and solve proportions. A proportion is used when two ratios (fractions) are equal. There are numerous ways to solve a proportion, but most students will remember that when a situation is proportional, the cross products are equal. These cross products can be used to solve for missing pieces of the proportion. Examples:

A. 35

21=

20

𝑥 Cross multiply so 35x = 20 • 21

simplify 35x = 420 solve x = 12

B. 9

𝑥+1=

5

𝑥 cross multiply so 9x = 5(x+1)

simplify 9x = 5x + 5 solve 4x = 5 x = 1.25

C. Jill decides to bring cookies from Vito’s Bakery for all of the students in her Geometry class. She knows that last time she bought cookies her bill was $3.95 and she bought 5 cookies. There are 32 people in her Geometry class and everyone wants a cookie. Write and solve a proportion to calculate what Jill’s bill will be. This is a proportion because the cost per cookie will be equal. So, set up a proportion making sure to match the units.

$3.95

5 𝑐𝑜𝑜𝑘𝑖𝑒𝑠=

𝑥

32 𝑐𝑜𝑜𝑘𝑖𝑒𝑠

Now, cross multiply and solve! 5x = 126.4 x = 25.28 So, Jill’s bill will be $25.28 for 32 cookies

***Practice for proportions*** Solve each proportion.

1. 5

4=

𝑘

8

2. 5

9=

2

𝑎

3. 8

10=

2

𝑝+3

4. 𝑥−1

6=

6

7

Write and solve a proportion to solve each problem. 5. Jack is going to Florida! The first 325 miles took 5 hours. Jack knows the entire trip is about 1170 miles. If

they can travel at the same rate, how many hours should the trip take? 6. Mr. Jones wants to buy his geometry students protractors to start the year. He found a pack of 12

protractors for $9.66. The company said they will also sell him the protractors individually for the same unit price. If Mr. Jones has $75 to spend, how many protractors can he buy?

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(x1, y1) (x2, y2)

E. slope

The slope of a line refers to how steep the line is. We represent this by showing 𝑟𝑖𝑠𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑖𝑛𝑒

𝑟𝑢𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑖𝑛𝑒.

We can count slope by counting the rise and run on the graph, or we can calculate it using the x

and y values of two points on the line. We use the formula 𝑦2−𝑦1

𝑥2−𝑥1 . By subtracting the y values

we find the rise of the line and by subtracting the x values we find the run of the line. You will

see the letter m used to represent slope.

Example using a graph:

The slope of the line is −3

2.

We can determine the slope by counting the

rise (down 3, so we use -3 as the rise) and

counting the run (right 2 so we use positive

2 as the run.)

Example using the formula:

Find the slope of the line containing the points (5, -2) and (-10, 8).

Substitute the points into our formula m = 𝑦2−𝑦1

𝑥2−𝑥1 so m =

8−(−2)

−10−5 =

𝟏𝟎

−𝟏𝟓 = −

𝟐

𝟑

What about horizontal and vertical lines?

A horizontal line will always have a slope of 0, since the rise of the

line is 0.

m = 0

𝑟𝑢𝑛 = 0

A vertical line will always have an undefined slope since the run of the line is 0 and division by

zero is undefined.

m = 𝑟𝑖𝑠𝑒

0 = undefined

Make sure you simplify

your fractions.

x

y

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***Practice finding slope***

Find the slope for each of the following graphs.

1.

2.

3.

4.

Find the slope of each line using the points given.

5. (0, 0) and (-2, 8)

6. (-1, 2) and (-9, 6)

7. (1, -2) and (6, -2) 8. (1, -2) and (-6, 3)

(10, 120)

(5, 60)

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F. writing equations of lines

You can write an equation of a line by using the slope and at least one point on the line.

The slope intercept form of a linear equation is y = mx + b where m is the slope and b is the y-intercept.

Remember, horizontal lines always have a slope of 0 and are represented by equations of the form y = a

number. Vertical lines always have an undefined slope and are represented by equations of the form x = a

number.

Example A:

The table shows the coordinates of two points on the graph of a line.

Write an equation for the line.

Step 1: Find the slope of the line.

m = −4−(−1)

6−(−3) = −

3

9 = −

1

3

Step 2: Substitute the slope and one of the points into y = mx + b. Solve for the y-intercept.

(You can choose either point).

−4 = −1

3 (6) + b

−4 = −2 + b

b = −2

Step 3: Write the equation. Substitute the slope and y-intercept into the slope-intecept form of the equation.

y = −1

3 x − 2

Example B:

Write the equation of a line through (10, -6) and (10, 2).

Step 1: Find the slope of the line.

m = −6−(2)

10−(10) =

−8

0 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑

Step 2: Since the slope is undefined, this line is vertical. Vertical lines cannot be written in slope-intercept

form. Vertical lines are represented by equations of the form x = a number.

x = 10 Use 10 as the value of x because both points have an x-coordinate of 10.

x y

-3 -1

6 -4

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***Practice Writing Equations***

Write an equation of each line. 1. through (3, 7), slope of -3

2. through (5, -2), slope of zero

3. through (5, 1) and (8, -2)

4. x-intercept at (6, 0) and y-intercept at (0, 4)

Write an equation in slope-intercept form to represent the lines in the following graphs. 5. 6. 7. Mikayla and her friends rented a canoe for 3 hours and paid a total of $45. Jeremy and his friends paid $ 85

to rent a canoe for 7 hours. The canoe rental service charges a daily rental fee and an hourly rate to rent a

canoe. Write an equation to model the relationship between the total cost to rent a canoe and the number of

hours. Let x = number of hours and let y = total cost.

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x - y = 5 is in standard form. Convert the

equation into slope-intercept form by solving

for y.

x – y =5 -x -x

-y = -x + 5

y = x – 5

Now graph as slope intercept form.

G. graphing linear equations Example 1: Sketch the graph given the information provided. Example 2: Sketch the graph given the information provided.

Example 3: Graph the line x = -3.

Since this equation is not in y = mx + b form, consider what the

equation tells you about the line. All points on the line have an

x-value of -3. The line is vertical.

When an equation is given in slope-intercept

form (y = mx + b) we will graph using the y-

intercept (the b value) and the slope (the m

value.)

We start by plotting the y-intercept, which is

(0,1).

Use the slope’s numerator to move up or

down and denominator to move left or right.

Since the slope for this line is −3

2 we moved 3

down and 2 right from the y-intercept.

-3 +2

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*** Practice Graphing Equations***

Sketch the graph of each line. 1. through (2, 5) with a slope of -3

2. through (-1, 7) with an undefined slope

3. y = 8x + 4

4. y = −5

4x + 1

5. x – 4y = 16 6. y = 5

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H. Slopes of Parallel and Perpendicular Lines

Two lines are parallel if their slopes are equal. For example, y = 2x + 6 and y = 2x – 10 are parallel.

Two lines are perpendicular if their slopes are opposite (negative) reciprocals. y = 3x + 6 and y = −1

3 x are

perpendicular.

Example: Determine whether lines MN and RS are parallel, perpendicular or neither. M(-1, 3), N(0, 5), R(2, 1), S(6, -1) Calculate the slope of MN and the slope of RS. Compare the slopes. slope MN = 2

slope RS = −1

2

Since the slopes are opposite reciprocals, lines MN and RS are perpendicular.

*** Practice Slopes of Parallel and Perpendicular lines ***

Determine if the lines are parallel, perpendicular or neither.

1. y = 4

3x + 10, y =

3

4x + 10

2. y = x, y = -x

3. x + y = 10, 2x + 2y = 5

4. y = 5, x = 5

5. line MN and line RS

M (-1, 3)

N (4, 4)

R (3, 1)

S (-2, 2)

6. line AB and line CD

A (0, -3)

B (-2, -3)

C (1, -3)

D (1, -7)

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I. Solving Systems of Equations Two equations, such as y = 3x + 2 and y = -6x + 12, together are called a system of equations.

A solution of a system is an ordered pair that satisfies both equations. A system of two linear equations can

have one solution, infinitely many solutions or no solutions. Example A: Solve by graphing Graph both lines. The point where the lines intersect is the solution. y = -x + 8 and y = 4x – 7 The solution is (3, 5). If the lines are parallel, there is no solution to the system. If the lines are the same, there are infinitely many solutions to the system.

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Example B: Solve by substitution x + 3y = -6 and 0.5y – x = – 4.5 Step 1: Isolate a variable in one of the equations. (Consider which one would be easier.) x + 3y = -6 x = -3y – 6 Step 2: Substitute -3y – 6 for x in the other equation. 0.5y – (-3y – 6) = -4.5 Step 3: Solve for the remaining variable. 0.5y + 3y + 6 = -4.5 3.5y + 6 = -4.5 3.5y = -10.5 y = -3 Step 4: Solve for the other variable. Substitute the value into one of the original equations. x + 3(-3)= -6 x = 3 Step 5: Write your answer as an ordered pair. (3, -3) Example C: Solve by elimination (addition/subtraction) 2x – 4y = -26 and 3x – y = -24 Step 1: Align the variables 2x – 4y = -26 3x – y = -24 Step 2: Multiply one of the equations so that the coefficients of one of the variables are equal. 2x – 4y = -26 2x – 4y = -26 4(3x – y) = 4(-24) 12x – 4y = -96 Step 3: Add or subtract the equations in order to eliminate one of the variables. Solve for the

remaining variable. 2x – 4y = -26 -(12x – 4y = -96) -10x = 70 x = -7 Step 4: Solve for the other variable. Substitute the value into one of the original equations. 2(-7) – 4y = -26 y = 3 Step 5: Write your answer as an ordered pair. (-7, 3)

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***Practice: Solving by Graphing *** Solve each system of equations by graphing. 1. y = 2x – 3

y = 1

2x

2. y = – 1

2x + 2

y = –3x–3

3. y = −1

3x – 1

2x + y = 4

***Practice: Solving by Substitution ***

Solve by substitution.

1. y = -3x + 7

4x + 2y = 16

2. 2x + y = 5

1.5x – 1.5y = 1.5

3. 2x + 3y = -3

x = -2y + 2

***Practice: Solving by Elimination ****

Solve by elimination.

1. 2x – y = 7

3x + y = 8

2. 3x + 4y = -10

2x – 8y = 4

3. 5x – 2y = -15

3x + 3y = 12

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J. Operations with Fractions When adding and subtracting fractions, you must ALWAYS have common denominators. What do you do if the denominator is NOT the same?

4

7+

2

3 =

****Practice adding/subtracting fractions***

Students entering Algebra 1 should be able to do these problems without the use of a calculator.

1. 2

5+

1

5=

2. 2

5−

3

4=

3. 4

9−

7

9=

4. 8

9+

1

3=

The denominators are NOT the same, so you need to multiply to make them the

same. The smallest number that both 7 and 3 will go into (the LCM) is 21. So, we

will multiply the first fraction by 3

3. The second fraction will be multiplied by

7

7 .

Then we can add the fractions.

4

7+

2

3 =

4∙3

7∙3+

2∙7

3∙7 =

12

21+

14

21 =

26

21

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Multiplying and Dividing Fractions: To multiply or divide you do NOT need common denominators.

Multiplication: Multiply the numerators, multiply the denominators and simplify your answer.

Division: Flip the second fraction and multiply.

Example: Multiplication

4

7●

3

8=

12

56=

3

14

Example: Division

4

7÷

3

8=

4

7●

8

3=

32

21

or

Convert fractions to have common denominators. Divide the numerators.

4

7÷

3

8=

32

56÷

21

56=

32215656

= 32

21

Multiply the numerators, 4 ● 3 = 12

Multiply the denominators, 7 ● 8 = 56

12

56 simplifies to

3

14 by dividing 12 and 56 by dividing 12 and 56 by 4

Flip the second fraction and change to

multiplication

Multiply the numerators and denominators to get 32

21, which cannot

be reduced further.

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****Practice multiplying/dividing fractions***

Students entering Algebra 1 should be able to do these problems without the use of a calculator.

1. 8

9●

3

4 =

2. 8

9÷

3

4 =

3. 1

9●

1

18 =

4. 1

9÷

1

18 =

5. 2

7●

4

7 =

6. 2

7÷

4

7 =

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K. Quadratics

A quadratic equation is an equation with a degree (highest exponent) of 2. You can solve

quadratic equations by factoring, graphing, completing the square, or using the quadratic

formula. Example A: solve with quadratic formula Solve 24x2 – 14x = 6 Step 1: Rewrite the equation in standard form.

24x2 – 14x – 6 = 0

Step 2: Use the values of a, b and c in the quadratic formula.

a = 24, b = -14 and c = -6

x = −(−14) ±√(−14)2− 4(24)(−6)

2(24)

Step 3: Simplify to find solutions.

x = 14 ±√772

48 x =

14+27.8

48= -0.29 and x =

14−27.8

48=0.87

*** Practice: Solving Quadratics using Quadratic Formula *** 1. x2 + 10x + 12 2. 3x2 + 7x - 20

quadratic formula: x = −𝑏 ±√𝑏2− 4𝑎𝑐

2𝑎

When 0 = ax2 + bx + c

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Examples B and C: solve by factoring

Solve x2 – 9x + 20 = 0 (x – 5) (x – 4) = 0 factor zero product property x – 5 = 0 or x – 4 = 0 x = 5 x = 4

Solve 2x2 – 7x = 0 x (2x – 7) = 0 factor zero product property x = 0 or 2x – 7 = 0 x = 3.5

*** Practice: Solving by Factoring *** 1. x2 + 11x + 24 = 0

2. x2 – 3x + 2 = 0

3. x2 + 13x – 48 = 0

4. x2 – 2x – 35 = 0

5. x2 + 20x = 0

6. 3x2 + 27x = 0

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L. Simplifying Radicals

List of perfect squares: 0, 1, 4, 9, 16, 25, 36, … How to simplify a radical expression to its exact value.

Example A: Simplify √𝟏𝟗𝟐. Step 1: Decompose the number under the radical into the product of a perfect square

(the largest value evenly divisible into that number) and another number. Use the list of perfect squares above for reference.

√192 = √64 ∙ 3 Step 2: Simplify the perfect square. The number that is not a perfect square will stay “as is”

under the radical.

√64 ∙ 3 = 8√3 Examples B and C:

Simplify √𝟐𝟓𝒙𝟗.

√25𝑥9 = √(25)(𝑥8)𝑥 = 5𝑥4√𝑥

Simplify √𝟕𝟐

𝟐.

√72

2 =

√36∙2

2 =

6√2

2 = 3√2

***Practice Simplifying Radicals*** Students entering Geometry should be able to simplify each radical expression to its exact value without

the use of a calculator.

1. √75

2. √80 3. √8

√36

4. √32 5. √45

√20

6. √175

7. √144𝑥2 8. √36𝑥7 9. √18𝑥4

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How to simplify a radical with a radical in the denominator (rationalizing the denominator).

Example D: Simplify 𝟐

√𝟓.

Step 1: Multiply the numerator and denominator by the radical in the denominator.

2

√5 ∙

√5

√5

Step 2: Simplify.

2

√5 ∙

√5

√5 =

2√5

√25 =

2√5

5

***Practice rationalizing the denominator*** Students entering Geometry should be able to simplify each radical expression to its exact value without

the use of a calculator.

1. 6

√3 2.

4√5

√2 3.

20

3√5 4.

24

√6

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M. Pythagorean Theorem In a right triangle, the sum of the squares of the measures of the legs equals the square of the measure of the hypotenuse. ∆ ABC is a right triangle so a2 + b2 = c2.

Example: Find the missing side in ∆ ABC. x2 + 102 = 122 Apply the Pythagorean Theorem. x2 + 100 = 144 x2 = 44

x = √44 or 2 √11

***Practice Pythagorean Theorem*** Use the Pythagorean Theorem to solve for the exact measure of the missing side. Leave your answer as a simplified radical. You may assume each triangle is a right triangle. 1.

2.

3. 4.

x 12

10

c

a

b

A

C B

7 x

10

x

15

9

x

8

√264

x

√5 √10

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Geometry Essential Skills Practice - Answer Key basic perimeter and area

1. 34 m, 60 m2

2. 14 cm, 49 cm2 3. 48.2 ft, 112 ft2 4. 8y + 24, -21y2 + 84y basic lines and angles

1. ray

2. line

3. segment

4. ray

5. segment

6. 10

7. 21

8. vertical

9. adjacent, complementary

10. adj, linear pair, suppl

11. adjacent

12. adj, linear pair, suppl

13. vertical

14. 143

15. 58

16. 64 solving basic equations 1. 0.5 2. 1 3. 4 4. 3.2 5. 3 6. 4 proportions 1. 10 2. 3.6 3. -0.5 4. 6.14 5. 18 hrs 6. 92 protractors… use proportion with unit price

$0.81

1 𝑝𝑟𝑜𝑡𝑟𝑎𝑐𝑡𝑜𝑟=

$75.00

𝑥 𝑝𝑟𝑜𝑡𝑟𝑎𝑐𝑡𝑜𝑟𝑠

* or 93 protractors if not using unit price

slope 1. 1

2. - 1

3

3. undefined 4. 12 5. -4

6. - 1

2

7. 0

8. - 5

7

writing equations 1. y = -3x + 16 2. y = -2 3. y = -x + 6

4. y = −2

3 x + 4

5. y = 1

4 x - 4

6. y = -5 7. y = 10x + 15

graphing lines

1. 2. 3. 4. 5. 6.

parallel and perpendicular 1. neither 2. perpendicular 3. parallel 4. perpendicular 5. neither 6. perpendicular solving systems by graphing 1. (2, 1) 2. (-2, 3) 3. (3, -2) solving systems by substitution 1. (-1, 10) 2. (2, 1) 3. (-12, 7) solving systems by elimination 1. (3, -1) 2. (-2, -1) 3. (-1, 5) . adding fractions

1. 3

5

2. −7

20

3. −1

3

4. 11

9

multiplying fractions

1. 2

3 2.

32

27

3. 1

162 4. 2

5. 8

49 6.

1

2

solving quadratic equations by quadratic formula 1. x = -1.4 and - 8.6

2. x = 5

3 and - 4

solving quadratic equations by factoring 1. x = -3 and -8 2. x = 2 and 1 3. x = -16 and 3 4. x = 7 and -5 5. x = 0 and -20 6. x = 0 and -9 simplifying radicals

1. 5√3

2. 4√5

3. √2

3

4. 4√2

5. 3

2

6. 5√7 7. 12x

8. 6x3√𝑥

9. 3x2√2 rationalizing the denominator

1. 2√3

2. 2√10

3. 4√5

3

4. 4√6 Pythagorean Theorem

1. √149

2. 10√2 3. 12

4. √15

27