Document Sample

Five-Minute Check (over Chapter 7) Then/Now New Vocabulary Example 1: Monomial in Factored Form Example 2: GCF of a Set of Monomials Example 3: Real-World Example: Find a GCF Over Chapter 7 A. A. A B. 3abc B. B C. 4a2b C. C 0% 0% D. 0% D 0% D. 5a3b A B C D Over Chapter 7 Simplify (–2x4y)(–3x2)2. A. –18x8y B. –6x8y A. A C. 6x6y B. B C. C D. 18x8y 0% 0% D. 0% D 0% A B C D Over Chapter 7 Simplify (3x – 2)2. A. 9x – 4 B. 9x2 – 4 A. A C. 9x2 – 12x + 4 B. B C. C D. 9x2 + 6 – 4 0% 0% D. 0% D 0% A B C D Over Chapter 7 Simplify (2x – 6)(2x + 6). A. 4x B. 4x2 – 36 A. A C. 4x2 – x + 6 B. B C. C D. 4x2 – x 0% 0% D. 0% D 0% A B C D Over Chapter 7 The perimeter of a plot of land is represented by 36x2 – 54x + 18. If the measures of three sides are given, what is the measure of the fourth side? A. 20x2 – 40x + 10 A. A B. 18x2 – 9x + 9 B. B C. C C. 10x2 – 15x + 5 0% 0% D. 0% D 0% A B C D D. 5x2 – 12x + 13 Over Chapter 7 Which value of n makes (c6 – 3)2 = cn – 6c6 + 9 true? A. 12 B. 8 A. A C. 6 B. B C. C D. 4 0% 0% D. 0% D 0% A B C D You multiplied monomials and divided a polynomial by a monomial. (Lessons 7–1 and 7–2) • Factor monomials. • Find the greatest common factors of monomials. • factored form • greatest common factor (GCF) Monomial in Factored Form Factor 18x2y3 completely. 18x2y3 = 2 ● 9 ● x ● x ● y ● y ● y 18 = 2 ● 9, x2 = x ● x, and y3 = y ● y ● y =2●3●3●x●x●y●y●y 9=3●3 Answer: 18x2y3 in factored form is 2 ● 3 ● 3 ● x ● x ● y ● y ● y. Factor 15a3b2 completely. A. 15 ● a ● a ● a ● b ● b B. 3 ● 5 ● a ● a ● a ● b ● b A. A C. 3 ● 5 ● a3b2 B. B D. 3 ● 5 ● a ● a ● a ● b2 C. C 0% 0% D. 0% D 0% A B C D GCF of a Set of Monomials Find the GCF of 27a2b and 15ab2c. 27a2b = 3 ● 3 ● 3 ● a ● a ● b Factor each number. 15ab2c = 3 ● 5 ● a ● b ● b ● c Circle the common prime factors. Answer: The GCF of 27a2b and 15ab2c is 3 ● a ● b or 3ab. Find the GCF of 39x2y3 and 26xy4. A. 2xy B. 13xy A. A C. 39x2y3 B. B D. 13xy3 C. C 0% 0% D. 0% D 0% A B C D Find a GCF GEOMETRY The lengths of the sides of a triangle are 12wz2, 8wz, and 16w2z. Find the GCF of the three lengths. 12wz2 = 3 ● 22 ● w ● z2 8wz = 23 ● w ● z 16w2z = 24 ● w2 ● z The common prime factors are 22 ● w ● z. Answer: So, the GCF is 4wz. Mary is making bracelets with large and small beads. She has 20 large beads and 96 small beads. What is the greatest number of bracelets she can make with having any beads left over? A. 3 bracelets B. 4 bracelets A. A B. B C. 5 bracelets C. C D. 8 bracelets 0% 0% D. 0% D 0% A B C D

DOCUMENT INFO

Shared By:

Categories:

Tags:

Stats:

views: | 4 |

posted: | 6/13/2012 |

language: | |

pages: | 17 |

OTHER DOCS BY jennyyingdi

How are you planning on using Docstoc?
BUSINESS
PERSONAL

By registering with docstoc.com you agree to our
privacy policy and
terms of service, and to receive content and offer notifications.

Docstoc is the premier online destination to start and grow small businesses. It hosts the best quality and widest selection of professional documents (over 20 million) and resources including expert videos, articles and productivity tools to make every small business better.

Search or Browse for any specific document or resource you need for your business. Or explore our curated resources for Starting a Business, Growing a Business or for Professional Development.

Feel free to Contact Us with any questions you might have.