a x ( b x c ) = ( a x b ) x c. The given statement illustrates _____ property.

closure

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commutative

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distributive

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associative

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Solution

The correct option is D associative

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Similar questions

Euclid's which axiom illustrates the statement that when x+y=15,then x+y+z=15+z ?

(a) first

(b) second

(d) four

∣ ∣ ∣ ∣ \begin{matrix} x + a & b & x c & x + b & a a & b & x + c \end{matrix} ∣ ∣ ∣ ∣ = 0

x \neq 0

\frac{1}{1 + x^{a - b}} + \frac{1}{1 + x^{b - a}} = 1

{[\{\frac{x^{a (a - b)}}{x^{a (a + b)}}\} \div \{\frac{x^{b (b - a)}}{x^{b (b + a)}}\}]}^{a + b} = 1

{(x^{\frac{1}{a - b}})}^{\frac{1}{a - c}} {(x^{\frac{1}{b - c}})}^{\frac{1}{b - a}} {(x^{\frac{1}{c - a}})}^{\frac{1}{c - b}} = 1

{(\frac{x^{a^{2} + b^{2}}}{x^{a b}})}^{a + b} {(\frac{x^{b^{2} + c^{2}}}{x^{b c}})}^{b + c} {(\frac{x^{c^{2} + a^{2}}}{x^{a c}})}^{a + c} = x^{2 (a^{3} + b^{3} + c^{3})}

{(x^{a - b})}^{a + b} {(x^{b - c})}^{b + c} {(x^{c - a})}^{c + a} = 1

{\{{(x^{a - a^{- 1}})}^{\frac{1}{a - 1}}\}}^{\frac{a}{a + 1}} = x

{(\frac{a^{x + 1}}{a^{y + 1}})}^{x + y} {(\frac{a^{y + 2}}{a^{z + 2}})}^{y + z} {(\frac{a^{z + 3}}{a^{x + 3}})}^{z + x} = 1

{(\frac{3^{a}}{3^{b}})}^{a + b} {(\frac{3^{b}}{3^{c}})}^{b + c} {(\frac{3^{c}}{3^{a}})}^{c + a} = 1

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