Answer:
65
50
Step-by-step explanation:
yeah-ya........ right?
(cube root of 5) * sqrt(5)
--------------------------------- = ?
(cube root of 5^5)
This becomes easier if we switch to fractional exponents:
5^(1/3) * 5^(1/2) 5^(1/3 + 1/2) 5^(5/6)
------------------------ = --------------------- = ------------- = 5^[5/6 - 5/3]
[ 5^5 ]^(1/3) 5^(5/3) 5^(5/3)
Note that 5/6 - 5/3 = 5/6 - 10/6 = -5/6.
1
Thus, 5^[5/6 - 5/3] = 5^(-5/6) = --------------
5^(5/6)
That's the correct answer. But if you want to remove the fractional exponent from the denominator, do this:
1 5^(1/6) 5^(1/6)
---------- * ------------- = -------------- (ANSWER)
5^(5/6) 5^(1/6) 5
The answer would be 2010 I think! im SO sorry if its wrong! :3
Both expression have the same denominator: 9x²-1. Thus it must not be 0.
9x²-1=(3x-1)(3x+1)=0, resulting x=+-1/3.
Restrictions: x in R\{-1/3, 1/3}
Adding those expressions:
E=(-x-2)/(9x²-1 ) + (-5x+4)/(9x²-1)=
(-x-2-5x+4)/(9x²-1)=(-6x+2)/(9x²-1)=
(-2)(3x-1)/(9x²-1)=-2/(3x+1)
E=-2/(3x+1)
