Match the following properties.

a + b = b + a

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a+(b+c)=(a+b)+c

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a(b+c)=ab+ac

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Solution

Following are the definition for each of the property.

Commutative - a + b = b + a
Associative- a+(b+c)=(a+b)+c
Distributive - a(b+c)=ab+ac

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Similar questions

The determinant
$∣ ∣ ∣ ∣ ∣ \begin{matrix} b^{2} - a b & b - c & b c - a c a b - a^{2} & a - b & b^{2} - a b b c - a c & c - a & a b - a^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣$ equals to:

(a) abc(b-c)(c-a)(a-b)
(b) (b-c)(c-a)(a-b)
(c) (a+b+c)(b-c)(c-a)(a-b)
(d) None of these

$a^{2} + b c + a b + a c = ? (a) (a + b) (a + c) (b) (a + b) (b + c) (c) (b + c) (c + a) (d) a (a + b + c)$

In the adjoining figure, if $B C = a, A C = b, A B = c$ and $∠ C A B = 120^{\circ}$ ,
then which of the following is the correct relation?

Q. If

{tan}^{- 1} \sqrt{(\frac{a (a + b + c)}{b c})} + {tan}^{- 1} \sqrt{(\frac{b (a + b + c)}{a c})} + {tan}^{- 1} \sqrt{(\frac{c (a + b + c)}{a b})} = x π

where

a, b, c

are positive.
Fine the vale of x.

Q. The value of

{tan}^{- 1} \sqrt{\frac{a (a + b + c)}{b c}} + {tan}^{- 1} \sqrt{\frac{b (a + b + c)}{a c}} + {tan}^{- 1} \sqrt{\frac{c (a + b + c)}{a b}}

is equal to

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Match the following properties.

Following are the definition for each of the property. Commutative - a + b = b + a Associative- a+(b+c)=(a+b)+c Distributive - a(b+c)=ab+ac

Following are the definition for each of the property.

Commutative - a + b = b + a
Associative- a+(b+c)=(a+b)+c
Distributive - a(b+c)=ab+ac