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  1. Chapter 1 Class 9 Number Systems
  2. Concept wise

Transcript

Find (2/3)^2 ×(2/5)^(−3) × (3/5)^2 (2/3)^2 ×(2/5)^(−3) × (3/5)^2 = (2/3)^2 ×(5/2)^3 × (3/5)^2 = 2^2/3^2 × 5^3/2^3 × 3^2/5^2 = 𝟓/𝟐 Evaluate (i) ∛((343)^(−2) ) ∛((343)^(−2) ) = [(343)^(−2) ]^(1/3) = (343)^(−2 × 1/3) = (7^3 )^( (−2)/3) = 7^[3 × ((−2)/3)] = 7^(−2) = 1/7^2 = 1/49 (ii) √(5&(32)^(−3) ) √(5&(32)^(−3) ) = [(32)^(−3) ]^(1/5) = (32)^(−3 × 1/5) = (2^5 )^( (−3)/5) = 2^[5 × ((−3)/5)] = 2^(−3) = 1/2^3 = 1/8 Which of the following is not equal to [(5/6)^(1/5) ]^((−1)/6) (a) 1/[(5/6)^(1/5) ]^(1/6) (b) (6/5)^(1/30) (c) (5/6)^((−1)/30) (d) (5/6)^(1/5 − 1/6) Value of (256)0.16 × (256)0.09 is (A) 4 (B) 16 (C) 64 (D) 256.25 (256)0.16 × (256)0.09 = (256)0.16 + 0.09 = (256)0.25 = (256)^(1/4) = (2^8 )^(1/4) Which of the following is equal to x (a) 𝑥^(12/7)−𝑥^(5/7) (b) √(12&(𝑥^4 )^(1/3) ) (c) (√(𝑥^3 ))^(2/3) (d) 𝑥^(12/7) × 𝑥^(5/7) Find value of x, if 5x−3 × 32x−8 = 225 Given 5x − 3 × 32x − 8 = 225 5x − 3 × 32x − 8 = 25 × 9 5x − 3 × 32x − 8 = 52 × 32 Comparing powers of 5 and 3 If 〖25〗^(𝑥 − 1)=5^(2𝑥 − 1)−100, find the value of x Given 〖25〗^(𝑥 − 1)=5^(2𝑥 − 1)−100 〖(5^2)〗^(𝑥 − 1)=5^(2𝑥 − 1)−100 5^(2𝑥 − 2)=5^(2𝑥 − 1)−100 5^(2𝑥 − 2)− 5^(2𝑥 − 1)=−100 5^2𝑥 × 5^(−2)− 5^2𝑥 × 5^(−1)=−100 5^2𝑥/5^2 −5^2𝑥/5 =−100 5^2𝑥 (1/5^2 −1/5) =−100 5^2𝑥 ((1 − 5)/5^2 ) =−100 5^2𝑥 ((−4)/5^2 ) =−100 5^2𝑥 =100 ×5^2/4 5^2𝑥 =25 × 25 5^2𝑥 =5^2 × 5^2 Prove that (2^30 + 2^29+ 2^28)/(2^31 + 2^30 − 2^29 ) = 7/10 We have (2^30 + 2^29+ 2^28)/(2^31 + 2^30 − 2^29 ) = (2^28 (2^2 + 2 + 1))/(2^29 (2^2 + 2 − 1)) = (2^28 (4 + 2 + 1))/(2^29 (4 + 2 − 1)) = 2^((28 − 29))× 7/5 Simplify: [5 (8^(1/3)+〖27〗^(1/3) )^3 ]^(1/4) [5 (8^(1/3)+〖27〗^(1/3) )^3 ]^(1/4) = [5 (〖(2^3)〗^(1/3)+〖(3^3)〗^(1/3) )^3 ]^(1/4) = [5 (2+3)^3 ]^(1/4) = [5 (5)^3 ]^(1/4) = [5^4 ]^(1/4) Simplify (81/16)^(−3/4)×[(25/9)^((−3)/2)÷(5/2)^(−3) ] (81/16)^(−3/4)×[(25/9)^((−3)/2)÷(5/2)^(−3) ] = [(3/2)^4 ]^(−3/4)×[[(5/3)^2 ]^((−3)/2)÷(5/2)^(−3) ] = (3/2)^(−3)×[(5/3)^(−3)÷(5/2)^(−3) ] =(2/3)^3×[(3/5)^3÷(2/5)^3 ] = 2^3/3^3 ×[3^3/5^3 ÷2^3/5^3 ] = 2^3/3^3 ×[3^3/5^3 ×5^3/2^3 ] = 2^3/3^3 ×3^3/2^3 = 1

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.