What is (h\circ h^{-1})(5) ?

The solution to (h\circ h^{-1})(5) is 5\circ

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simplify (h\circ h^{-1})(5)

simplify

(h \circ h^{- 1}) (5)

5 \circ

Decimal Notation

5

Solution steps

(h \circ h^{- 1}) (5)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

h h^{- 1} = h^{1 - 1}

= \circ h^{1 - 1} \cdot 5

Subtract the numbers:

1 - 1 = 0

= \circ h^{0} \cdot 5

Apply rule

a^{0} = 1, a \neq = 0

h^{0} = 1

= 1 \cdot 5 \circ

Multiply the numbers:

1 \cdot 5 = 5

= 5 \circ

(5 x^{2} y) (2 x y^{4})

4 e^{- 0.2}

e^{- 0} 0

- 2 (0.5)^{- 1}

- 6 e^{- 6 (l n (2))}