SOLUTION: CAn someone please help me? I keep trying these but I also keep getting different answers:(
a) [16^(1/2) +(2/3)^0]^-2
b)(27/8)^(1/3) + 4^-1
c)(7^0 - 4^-1)^-3
Algebra ->
Exponents
-> SOLUTION: CAn someone please help me? I keep trying these but I also keep getting different answers:(
a) [16^(1/2) +(2/3)^0]^-2
b)(27/8)^(1/3) + 4^-1
c)(7^0 - 4^-1)^-3
Log On
Question 271060: CAn someone please help me? I keep trying these but I also keep getting different answers:(
a) [16^(1/2) +(2/3)^0]^-2
b)(27/8)^(1/3) + 4^-1
c)(7^0 - 4^-1)^-3 Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! Remember any number raised to the power of 0 = 1
:
a) [16^(1/2) +(2/3)^0]^-2
[16^(1/2) + 1]^-2
The exponent of 1/2 is the square root of the number
(4 + 1)^-2
5^-2
The reciprocal gets rid of the neg =
:
:
b)(27/8)^(1/3) + 4^-1
Take the cube root of both numerator and denominator + = + =
:
:
c)(7^0 - 4^-1)^-3
(1 - )^-3 = ()^-3 = ()^3
cube both the numerator and denominator
