Get Just In Time Algebra (Just in Time Series) PDF

Get Just In Time Algebra (Just in Time Series) PDF

By LearningExpress Editors

This Cram sequence serves the reader who must assessment and study particular fabric in a brief time period for the aim of passing an enormous try, comparable to an go out examination, admissions try out, vocational examination, or certification examination. In ten streamlined chapters, readers examine the basics of algebra and observe time saving research talents and crucial test-taking suggestions. each one bankruptcy has precise gains designed to make learning speedy and effective, together with pointers on dealing with a number of selection questions, utilizing a calculator, and simple to recollect phrases, principles, and shortcuts.

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Extra resources for Just In Time Algebra (Just in Time Series)

Example text

Solve the equation for x: ᎏ14ᎏ(x + 4) = 10. a. 5 b. 9 c. 24 d. 36 e. 40 19. 5. a. 8 b. 4 c. 18 d. 24 e. 30 3(x + 4) ᎏ = 6. 20. Solve the equation for x: ᎏ –4 a. –12 b. 3 c. –18 d. –2 e. 2 Team-LRN N U M B E R P R O P E RTI E S AN D E Q UATI O N S O LVI N G 33 21. Solve the equation for x: –4 (x + 8) + 7x = 2x + 32. a. 0 b. 4 c. 16 d. 32 e. 64 22. Solve the equation for w in terms of A and l: A = lw. a. w = Al b. w = A – l c. w = A + l d. w = ᎏAlᎏ e. w = 2Al 23. Solve the equation for a in terms of b and c: 7ab = c.

What is the equation of the graph in the figure below? Team-LRN COOR DI NATE G EOM ETRY AN D G RAPH I NG LI N EAR EQUATIONS a. y = 3x – 1 b. y = –3x + 1 c. y = ᎏ13ᎏx + 1 d. y = –ᎏ13ᎏx + 1 e. y = ᎏ13ᎏx – 1 9. Which of the following graphs represents the line y = –2x + 2? a. b. Team-LRN 41 42 J U ST I N TI M E ALG E B R A c. d. Team-LRN COOR DI NATE G EOM ETRY AN D G RAPH I NG LI N EAR EQUATIONS 43 e. 10. What is the slope of the line that contains the points (–2,3) and (3,4)? a. –5 b. –ᎏ15ᎏ c. 0 d.

A. There are a few different ways to approach solving this problem. Method I: 3(x + 4) ᎏ = 6 • –4. Multiply both sides of the equation by –4; –4 • ᎏ –4 This simplifies to 3(x + 4) = –24. Divide each side of the equation 3(x + 4) –24 ᎏ = ᎏᎏ. This simplifies to x + 4 = –8. Subtract 4 from by 3; ᎏ 3 3 both sides of the equation; x + 4 – 4 = –8 – 4. The variable is now alone; x = –12. Method II: Another way to look at the problem is to still multiply each side by –4 in the first step to get 3(x + 4) = –24.

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