Before watching: (a/b)^c = (a^c)/(b^c). Further, (a^m)^n = a^(mn). Since 9 = 3^2 and 4 = 2^2... (9/4)^(9/4) = ((3^2)/(2^2))^(9/4) = ((3/2)^2)^(9/4) = (3/2)^(2(9/4)) = (3/2)^(9/2) = (3/2)^4.5 = (3/2)^4 * (3/2)^(1/2) = (81/16)√(3/2) = (81√3)/(16√2) If your teacher demands that there no be radicals in the denominator, multiply by √2/√2 to get (81√6)/32.
(√1.5)(81/16)
Before watching:
(a/b)^c = (a^c)/(b^c). Further, (a^m)^n = a^(mn). Since 9 = 3^2 and 4 = 2^2...
(9/4)^(9/4) = ((3^2)/(2^2))^(9/4)
= ((3/2)^2)^(9/4)
= (3/2)^(2(9/4))
= (3/2)^(9/2)
= (3/2)^4.5
= (3/2)^4 * (3/2)^(1/2)
= (81/16)√(3/2)
= (81√3)/(16√2)
If your teacher demands that there no be radicals in the denominator, multiply by √2/√2 to get (81√6)/32.
= (9/4)^(9/4)
= (3²/2²)^(9/4)
= [(3/2)^(2)]^(9/4) → recall: [x^(a)]^(b) = x^(ab)
= (3/2)^[2 * (9/4)]
= (3/2)^(9/2)
= (3/2)^[(8 + 1)/2]
= (3/2)^[(8/2) + (1/2)]
= (3/2)^[4 + (1/2)] → recall: x^(a + b) = x^(a) * x^(b)
= (3/2)^(4) * (3/2)^(1/2)
= (3⁴/2⁴) * (3/2)^(1/2)
= (81/16) * (3/2)^(1/2) → recall: x^(1/2) = √x
= (81/16) * √(3/2)
= (81/16) * (√3)/(√2) → you multiply by (√2) the bottom, so the top too (to eliminate √ at denominator)
= (81/16) * [(√3).(√2)/(√2).(√2)]
= (81/16) * (√6)/2
= (81 * √6)/(16 * 2)
= (81 * √6)/32
= (81/32).√6
{20.1+9}=21.0 10^20^10 10^2^10^10 2^5^2^2^5^2^5 1^1^1^1^1^2^1 2^1 (x ➖ 2x+1).