Username*
E-Mail*
Password*
Confirm Password*
Username or email*
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
You must login to ask question.
(i) 64^(1/2) = (8^2)^1/2 = 8 (ii) 32^1/2 = (2^5)^1/5 = 2^(5×1/5) = 2 (iii) 125^1/3 = (5^3)^1/3 = 5^(3×1/3) = 5
4x² – 3x + 7 Polynomials in one variable as it contains only one variable x.
(i) 2^(2/3).2^(1/5) = 2^((2/3)+(1/5)) = 2^((10+3)/15) = 2^(13/15) (ii) (1/3^3)^7 = (3^-3)^7 = 3^-21 (iii) (11^1/2)/(11^1/4) = 11^1/2 × 11^-1/4 = 11^((1/2)-(1/4)) = 11^(2-1 / 4) = 11^1/4 (iv) 7^1/2 . 8^1/2 = (7 × 8)^1/2 = 56^1/2
(i) 9^(3/2) = (3^2)^(3/2) = 3^(2×3/2) = 3^2 = 9 (ii) 32^(2/5) = (2^5)^(2/5) = 2^(5×2/5) = 2^2 = 4 (iii) 16^(3/4) = (2^4)^(3/2) = 2^(4×3/4) = 2^3 = 8 (iv) 125^(-1/3) = (5^3^(-1/3) = 5^(3×-1/3) = 5^-1 = 1/5 = 5
It is an important Question. See here 👀👇
Find: (i) 64^(1/2) (ii) 32^(1/5) (iii) 125^(1/3)
(i) 64^(1/2) = (8^2)^1/2 = 8 (ii) 32^1/2 = (2^5)^1/5 = 2^(5×1/5) = 2 (iii) 125^1/3 = (5^3)^1/3 = 5^(3×1/3) = 5
(i) 64^(1/2) = (8^2)^1/2 = 8
(ii) 32^1/2 = (2^5)^1/5 = 2^(5×1/5) = 2
(iii) 125^1/3 = (5^3)^1/3 = 5^(3×1/3) = 5
Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer. 4x² – 3x + 7
4x² – 3x + 7 Polynomials in one variable as it contains only one variable x.
4x² – 3x + 7 Polynomials in one variable as it contains only one variable x.
Simplify: (i) 2^(2/3). 2^(1/5) (ii) (1/3^3)^7 (iii) (11^1/2)/(11^1/4) (iv) 7^(1/2) . 8^(1/2)
(i) 2^(2/3).2^(1/5) = 2^((2/3)+(1/5)) = 2^((10+3)/15) = 2^(13/15) (ii) (1/3^3)^7 = (3^-3)^7 = 3^-21 (iii) (11^1/2)/(11^1/4) = 11^1/2 × 11^-1/4 = 11^((1/2)-(1/4)) = 11^(2-1 / 4) = 11^1/4 (iv) 7^1/2 . 8^1/2 = (7 × 8)^1/2 = 56^1/2
(i) 2^(2/3).2^(1/5) = 2^((2/3)+(1/5)) = 2^((10+3)/15) = 2^(13/15)
(ii) (1/3^3)^7 = (3^-3)^7 = 3^-21
(iii) (11^1/2)/(11^1/4) = 11^1/2 × 11^-1/4 = 11^((1/2)-(1/4)) = 11^(2-1 / 4) = 11^1/4
(iv) 7^1/2 . 8^1/2 = (7 × 8)^1/2 = 56^1/2
Find : (i) 9^(3/2) (ii)32^(2/5) (iii) 16^(3/4) (iv) 125^(-1/3)
(i) 9^(3/2) = (3^2)^(3/2) = 3^(2×3/2) = 3^2 = 9 (ii) 32^(2/5) = (2^5)^(2/5) = 2^(5×2/5) = 2^2 = 4 (iii) 16^(3/4) = (2^4)^(3/2) = 2^(4×3/4) = 2^3 = 8 (iv) 125^(-1/3) = (5^3^(-1/3) = 5^(3×-1/3) = 5^-1 = 1/5 = 5
(i) 9^(3/2) = (3^2)^(3/2) = 3^(2×3/2) = 3^2 = 9
(ii) 32^(2/5) = (2^5)^(2/5) = 2^(5×2/5) = 2^2 = 4
(iii) 16^(3/4) = (2^4)^(3/2) = 2^(4×3/4) = 2^3 = 8
(iv) 125^(-1/3) = (5^3^(-1/3) = 5^(3×-1/3) = 5^-1 = 1/5 = 5
Obtain all other zeroes of 3x⁴ + 6x³ – 2x² – 10x – 5, if two of its zeroes are √(5/3) and – √(5/3).
It is an important Question. See here 👀👇
It is an important Question.
See here 👀👇