What is (8^{-1}+8^0)-(2^{-1}+2^0) ?

The solution to (8^{-1}+8^0)-(2^{-1}+2^0) is -3/8

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simplify (8^{-1}+8^0)-(2^{-1}+2^0)

simplify

(8^{- 1} + 8^{0}) - (2^{- 1} + 2^{0})

- \frac{3}{8}

Decimal Notation

- 0.375

Solution steps

(8^{- 1} + 8^{0}) - (2^{- 1} + 2^{0})

Apply rule:

(a) = a

(8^{- 1} + 8^{0}) = 8^{- 1} + 8^{0}

= 8^{- 1} + 8^{0} - (2^{- 1} + 2^{0})

8^{- 1} = \frac{1}{8}

8^{0} = 1

- (2^{- 1} + 2^{0}) = - \frac{3}{2}

= \frac{1}{8} + 1 - \frac{3}{2}

\frac{1}{8} + 1 - \frac{3}{2} = - \frac{3}{8}

= - \frac{3}{8}

2 x^{3} + 3 x^{2} - 26 x - 39

(x + 4)^{2} + 3

4 0^{2} + 5 0^{2}

\frac{1}{9} + \frac{2}{8}

44 1^{\frac{1}{2}}

What is (8^{-1}+8^0)-(2^{-1}+2^0) ?
The solution to (8^{-1}+8^0)-(2^{-1}+2^0) is -3/8