Match the following.

b+a

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(a+b)+c

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(a×b)+(a×c)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

Commutative Property:
a + b = b + a
Example: 3 + 4 = 7 = 4 + 3

Associative Property:
a + (b + c) = (a + b) + c
Example: 5 + (10 + 3) = 18 = (5 + 10) +3

Distributive Property:
a $\times$ (b + c) = (a $\times$ b) + (a $\times$ c)
Example: 10 $\times$ (5 + 4) = 90 = 10 $\times$ 5 + 10 $\times$ 4

Suggest Corrections

Similar questions

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

|\begin{matrix} a + b + c & - c & - b \\ - c & a + b + c & - a \\ - b & - a & a + b + c \end{matrix}| = 2 (a + b) (b + c) (c + a)

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

a, b, c > 0

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

a, b, c

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

tan θ

Join BYJU'S Learning Program