Quiz #2 with Answer Key - Basic Algebra III | MATH 10024, Quizzes of Mathematics

Material Type: Quiz; Class: BASIC ALGEBRA IV; Subject: Mathematics; University: Kent State University; Term: Spring 2011;

Typology: Quizzes

2010/2011

Uploaded on 05/22/2011

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MATH 10024 Circle one: 9:55 12:05 Name: Core Mathematics IV KEY 7 Spring 2011/H2 Ms, Kracht Quiz Score Quiz 2: Version A Show your reasoning. Use si 1. (4 pts) Solve the equation = 5(£) '3 [5 13 eitt (Zxr2 (<1 5 & 3 = [x12] & : Bxso= 7B oe Ext2= ea 5(Zxv2)= C$) | 2) Bxtio= “6 ben © 2x2 —b-!D Bx = 6-1 Bea THk 3x = it _ lb < ~it x= 3 OR x= 3 2. (5 pts) Solve the inequality, Express the solution set (we want bxtF to be in interva! notation. 6x +7) 210 far frm 0.5 bxt¢S-l0 oe = bxX4I FY éx 4-0-4 Gx Ze (o- bx € -14+ ox 2 B 2 ~it Xe xs 6 ow “¢ . *F5 dem ——__\-_ mY, “1 o tandard notation correctly. Simplify your answers. NO CALCULATORS. 3. (B pts} Write a formula for an exponential function satisfying the following conditions. (a) initial value: 300 and growth factor: 2.5 AUD= B00 (2.5)" (2) initial value: AUD = Boo( Ato. osreye = 300 CL og4e) * (c) initial value: 300 and decay factor: 0.63 A= 306 (0.63% 300 and growth rate: 5.75% (d) initial vaiue: 300 and decay rate: 10% Aid= Boo U~ On 10) = 200 (040) % 4. (3 pts) Suppose the population of black squirrels on campus is currently 1200 and is increasing at an an- nual rate of 5%. Then after one year, the population will be PA) = 1200 1 0.05 x 1200 = 1200(1 4 0.053 = 1200(2.03). DERIVE and simplify an exoression for P(2), the population after 2 years, | PlAD= PLA) + 0.05 PEAY =fire0lt03)|e 0.05 [120001.08)] “fiasolos)]( 4 + 0.05) * 4900(4.05) 4.05) = 4200 (14.05)*