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ALGEBRA Semester 1 Final Exam Review

Algebra

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Semester 1 Final Exam Review. Algebra. Agenda. Vission/Mision Classroom Norms Where should you be? Linear Combination Applications Questions. Vision and Mission. - PowerPoint PPT Presentation

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Page 1: Algebra

ALGEBRASemester 1 Final Exam Review

Page 2: Algebra

Agenda

Vission/Mision Classroom Norms Where should you be? Linear Combination Applications Questions

Page 3: Algebra

Vision and Mission Vision: Through a combination of research-based, individualized, and specialized curriculum and instruction in the home; online conferencing with a certified teacher; and access to a community of experts in science and technology, our students will emerge as confident leaders of the digital age.

Mission: The mission of the Michigan Virtual Charter Academy is to provide an innovative, intensive academic preparation that inspires and educates students to achieve the highest levels of academic knowledge and skills. Michigan Virtual Charter Academy embraces a collaborative partnership between teachers and parents in order to empower students to reach extraordinary heights. Extraordinary results require extraordinary efforts! Through commitment, hard work, consistency, and responsibility, every student will meet the challenge of mastering high expectations.

Page 4: Algebra

Classroom Norms

*Have something ready to take notes with*Stay appropriate and on topic in chat*Complete the Setup Wizard before every class*Participate!!!*If you come in late, just roll with it and figure out where we are.

Page 5: Algebra

Semester Final Exam

Must take the exam on the assigned day.

2 parts to the test (just like your unit tests)

15 multiple-choice questions 3 multi-part questions on teacher-

scored test If you know how to do everything on

this review you will do great on the final!

Page 6: Algebra

What algebraic property does this statement show?3+(-7)=(-7)+3

Associative Commutative Closure symmetric

Page 7: Algebra

What algebraic property does this statement show?3+(-7)=(-7)+3

Associative Commutative Closure symmetric

Page 8: Algebra

Solve: 3x-5=7x-21

2 3 4 5

Page 9: Algebra

Solve: 3x-5=7x-21

2 3 4 5

3x-5=7x-21 add 5 to both sides3x=7x-16 subtract 7x from both sides-4x=-16 divide both sides by -4x=4

Page 10: Algebra

Solve: |x|+4=7

-3, +3 3 11 -11, +11

Page 11: Algebra

Solve: |x|+4=7

-3, +3 3 11 -11, +11

|x|+4=7 subtract 4 from both sides|x|=3 take out absolute value signx=-3, 3

Page 12: Algebra

Combine like terms to simplify: 7x+3y-2+6x-1+2y

x+1+5y

13x+2y -3

X+2y-3 13x+5

y-3

Page 13: Algebra

Combine like terms to simplify: 7x+3y-2+6x-1+2y

x+1+5y

13x+2y -3

X+2y-3 13x+5

y-3

7x+3y-2+6x-1+2y13x+5y-3

Page 14: Algebra

Which equation can be used to solve the following word problem: Emilio has 3 more nickels than dimes, and the total value of his coins is $1.15. How many nickels does he have? Let n represent the number of nickels.

.10n+.05n+3=$1.15 .10n+.05(n+3)=$1.1

5 .10(n+3)+.05n=$1.1

5 .10(n-3)+.05n=$1.15

Page 15: Algebra

Which equation can be used to solve the following word problem: Emilio has 3 more nickels than dimes, and the total value of his coins is $1.15. How many nickels does he have? Let n represent the number of nickels.

.10n+.05n+3=$1.15 .10n+.05(n+3)=$1.1

5 .10(n+3)+.05n=$1.1

5 .10(n-3)+.05n=$1.15

n= nickelsn-3=dimes.10(n-3)+.05n=$1.15

Page 16: Algebra

Evaluate the expression when a=5 and b=-2: |a|-b 3 7 -7 -3

Page 17: Algebra

Evaluate the expression when a=5 and b=-2: |a|-b 3 7 -7 -3

|a|-b|5|-(-2)5-(-2)5+27

Page 18: Algebra

Select the pair of equations whose graphs are perpendicular.

5x-8y=9 and 12x-5y=7 x+6y=8 and y=2x-8 y=2x-7 and x+2y=3 2y=-3x+5 and 2x+3y=4

Page 19: Algebra

Select the pair of equations whose graphs are perpendicular.

5x-8y=9 and 12x-5y=7

x+6y=8 and y=2x-8

y=2x-7 and x+2y=3

2y=-3x+5 and 2x+3y=4See next slide for solution

Page 20: Algebra

5x-8y=9 and 12x-5y=7 -8y=-5x+9 -5y=-12x+7 y=5/8x-9/8 y=12/5x-7/5

x+6y=8 and y=2x-8 6y=-x+8 y=-1/6x+8/6

y=2x-7 and x+2y=3 2y=-x+3 y=-1/2x+3

2y=-3x+5 and 2x+3y=4

Page 21: Algebra

Write an equation in slope-intercept form for the line passing through (-2,8) and (3,4).

Page 22: Algebra

Write in equation in slope-intercept form for the line passing through (-2,8) and (3,4).

m = (8-4)/(-2-3) = 4/-5y-(4)=-4/5(x-3)y-4=-4/5x+12/5y=-4/5x+12/5+4y=-4/5x+32/5

Page 23: Algebra

Given the linear equation 2x+y=6, find the slope of its graph.

Page 24: Algebra

Given the linear equation 2x+y=6, find the slope of its graph.

2x+y=6 y=-2x+6m=-2

Page 25: Algebra

Given the linear equation 3y=5x-4, find the y-intercept.

Page 26: Algebra

Given the linear equation 3y=5x-4, find the y-intercept.

3y=5x-43y/3=5x/3-4/3y=5/3x-4/3

y-intercept is -4/3

Page 27: Algebra

Solve the system of equations:3x+3y=-6 and x+3y=0

Page 28: Algebra

Solve the system of equations:3x+3y=-6 and x+3y=0

3x+3y=-6-x-3y=02x =-6x=-3(-3)+3y=03y=3y=1

Page 29: Algebra

Solve for x: 6 = 5 x-4 2

Page 30: Algebra

Solve for x: 6 = 5 x-4 2

(6)(2)=(5)(x-4)12=5x-2032=5x32/5=x6 2/5=x

Page 31: Algebra

The length of a rectangle is 5 mm longer than its width. Its perimeter is more than 30 mm. Let w equal the width of the rectangle.

Write an expression for the length in terms of the width.

Use expressions for the length and width to write an inequality for the perimeter, on the basis of the given information.

Solve the inequality, clearly indicating the width of the rectangle.

Page 32: Algebra

a) Let w = width of the rectangle. Let (w + 5) = length of the rectangle.

b) Substitute these expressions into the formula for the perimeter of a rectangle. Perimeter = 2w + 2(w + 5) The perimeter is more than 30 mm, so the inequality is 2w + 2(w + 5) > 30.

c) Solve for w.2w+2(w+5)>302w+2w+10>304w+10>304w>20w>5The width of the rectangle is greater than 5 mm.

Page 33: Algebra

Suppose you pay $45.00 for an electronic game that has been discounted 15%. What is the original price of the electronic game to the nearest cent?

Page 34: Algebra

If the game is discounted 15%, then the price you pay for the game is 85% of the original price. Use this information to set up a percent proportion. Reduce fractions to lowest terms before solving for x.

45/x=85/10045/x=17/20(45)(20)=(17)x900=17xx=52.94

The original price is $52.94.

Page 35: Algebra

Seth and George both worked hard over the summer. Together, they earned a total of $425. George earned $25 more than Seth. How much did each of them earn?

Page 36: Algebra

Let s be the money Seth earned and g be the money George earned.

g + s = 425g = 25 + s Substitute for g in the first equation. g + s = 425 (25 + s) + s = 425

25 + 2s = 4252s = 400s = 200

 Substitute 200 for s in the equation to solve for g. g = 25 + (200) = 225

Seth earned $200 and George earned $225.

Page 37: Algebra

Semester Final Exam

Must take the exam on the assigned day.

2 parts to the test (just like your unit tests)

15 multiple-choice questions 3 multi-part questions on teacher-

scored test If you know how to do everything on

this review you will do great on the final!