University:
Glendale Community College
Course:
MAT142 | College Mathematics
Academic year:

2020
Views:

330

Pages:

4
Author:

Deandre Simmons

Section P.2: Properties of Negative and Zero Exponents Chapter P – Polynomials #1 - 12: Simplify the expression that involve 0 exponents. (There is a second page to this section.) 1a) 𝑥2 𝑥2 =1 Solution: 1 1b) x0 = 1 Solution: 1 3a) 34 34 81 = 81 = 1 3b) 30 = 1 5a) 2*30 Solution: 1 Solution: 1 = 2*1 = 2 Solution: 2 5b) (2*3)0 =1 Solution: 1 7a) -1*30 = -1 * 1 = -1 Solution: -1 =1 Solution: 1 7b) (-1*3)0 9a) -30 = -1 * 30 = -1*1 = -1 Solution: -1 9b) (-3)0 =1 Solution: 1 11a) -1*y0 = -1*1 = -1 Solution: -1 11b) (-1*y)0 =1 Solution: 1 13a) -y0 = -1*y0 = -1*1 = -1 Solution: -1 13b) (-y)0 = 1 Solution: 1 15a) 5*60 = 5*1 = 5 Solution: 5 15b) (5*6)0 Solution: 1 =1 17a) 2x0 = 2*1 = 2 Solution: 2 17b) (2x)0 = 1 Solution: 1 19a) 3xy0 Solution: 3x = 3x*1 = 3x 19b) (3xy)0 = 1 Solution: 1 21a) 3x0y = 3*x*y = 3y Solution: 3y 21b) (3x0y)0 Solution: 1 =1 23) x0 = 1 Solution: 1 25) 30 = 1 Solution: 1 27) -30 = -1 * 30 = -1 * 1 = -1 Solution: -1 29) (-2)0 = 1 Solution: 1 31) 2c0 = 2 * 1 = 2 Solution: 2 33) (2x)0 = 1 Solution: 1 #35 - 90: Simplify the expressions that involve negative exponents. Write the answer with positive exponents only. 35) 3−2 = 1 32 =9 1 Solution: 1 9 37) 3−5 = 1 35 = 243 Solution: 1 243 39) 𝑥 −2 = 1 𝑥2 Solution: 1 1 𝑥2 𝑥2 41) 𝑥 −5 = 1 𝑥5 Solution: 1 𝑥5 1 43) 2𝑦 −5 = 2 𝑦5 Solution: 2 𝑦5 45) 5𝑦 −2 = 5 𝑦2 Solution: 5 𝑦2 Solution: 𝟗 𝟒 Solution: 81 16 Solution: 𝑦 12 𝑥8 Solution: 𝒚𝟒 𝒙𝟖 Solution: 9 4𝑥 2 Solution: 9𝑦 2 4𝑥 4 2 −2 47) (3) 2 −4 49) (3) 3 2 32 = (2) = 22 3 4 34 = (2) = 24 𝑥2 −4 𝑥2 𝑦 −4 51) (𝑦3 ) 𝑦3 4 𝑦 3∗4 = (𝑥 2 ) = 𝑥 2∗4 53) ( ) 𝑦 4 𝑥 = ( 2) = 2𝑥 −2 3 2 𝑦4 𝑥 2∗4 32 55) ( 3 ) = (2𝑥) = 22 𝑥 2 2𝑥 2 −2 57) ( 3𝑦 ) 3𝑦 2 32 𝑦 2 = (2𝑥2 ) = 22 𝑥 2∗2 59) 𝑥2 𝑥5 1 = 𝑥 5−2 61) 𝑦 𝑦6 63) 12𝑥 14𝑥 2 = 65) 36𝑧 3 12𝑧 5 = 67) 𝑥𝑦 4 𝑥2𝑦 69) 𝑥 2𝑦5 𝑥 5𝑦2 71) 4𝑥 3 𝑦 5 12𝑥 4 𝑦 2 73) 24𝑥 4 𝑦 5 18𝑥 5 𝑦 7 75) 2𝑥 2 𝑦 5 𝑧 18𝑥𝑦 5 𝑧 2 77) 40𝑥 7 𝑦 6 𝑧 3 24𝑥 5 𝑦 7 𝑧 3 = Solutions: 1 𝑦 6−1 = = Solution: 𝟏 𝒚𝟓 6 7𝑥 2−1 Solution: 𝟔 𝟕𝒙 3 𝑧 5−3 Solution: 𝟑 𝒛𝟐 Solution: 𝒚𝟑 𝒙 Solution: 𝒚𝟑 𝒙𝟑 Solution: 𝒚𝟑 𝟑𝒙 Solution: 𝟒 𝟑𝒙𝒚𝟐 = 9𝑧2−1 Solution: 𝒙 𝟗𝒛 5𝑥 7−5 3𝑦 7−6 Solution: 𝟓𝒙𝟐 𝟑𝒚 𝑦 4−1 𝑥 2−1 𝑦 5−2 𝑥 5−2 = 𝟏 𝒙𝟑 = 1𝑦 5−2 3𝑥 4−3 = 4 3𝑥 5−4 𝑦 7−5 1𝑥 2−1