Simplify $\frac{16}{- 21} \times \frac{14}{5}$

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Solution

Simplification:
$\begin{array}{rcl} \frac{16}{- 21} \times \frac{14}{5} & = & \frac{16 \times 14}{- 21 \times 5} [∵ \frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}] \\ = & \frac{- 224}{105} \end{array}$
Hence, the value of the given expression is $\frac{- 224}{105}$

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Similar questions

Solve for x:
$(i) \frac{x - 1}{x - 2} + \frac{x - 3}{x - 4} = 3 \frac{1}{3}; x \neq 2, 4 (i i) \frac{1}{x} + \frac{2}{2 x - 3} = \frac{1}{x - 2}, x \neq 0, \frac{3}{2}, 2 (i i i) x + \frac{1}{x} = 3, x \neq 0 (i v) \frac{16}{x} - 1 = \frac{15}{x + 1}, x \neq 0, - 1 (v) \frac{1}{x - 3} - \frac{1}{x + 5} = \frac{1}{6}, x \neq 3, - 5,$

Find the missing numeral $:$

$\frac{2}{3} =$ $\frac{14}{□}$

Q. The least integer greater than

{log}_{2} 15 \cdot {log}_{\frac{1}{6}} 2 \cdot {log}_{3} \frac{1}{6}

is equal to

Q. The smallest integer satisfying the equation

∣ ∣ ∣ {log}_{\frac{1}{6}} x - 1 ∣ ∣ ∣ + 2 = ∣ ∣ ∣ {log}_{\frac{1}{6}} x - 3 ∣ ∣ ∣

Match the elements of List I with List II

List I	List II
A) lf $λ$ be the number of terms which are integers, in the expansion of $(5^{\frac{1}{6}} + 7^{\frac{1}{9}})^{1824}$ , then $λ$ is divisible by	P) 2
B) lf $λ$ be the number of terms which are rational in the expansion of $(5^{\frac{1}{6}} + 2^{\frac{1}{8}})^{100}$ , then $λ$ is divisible by	Q) 3
C) lf $λ$ be the number of terms which are irrational in the expansion of $(3^{\frac{1}{4}} + 4^{\frac{1}{3}})^{99}$ , then $λ$ is divisible by	R) 7
	S) 13
	T) 17

The correct option which matches all the elements correctly, is :

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Simplify 16-21×145

Simplification:16-21×145=16×14-21×5∵ab×cd=a×cb×d=-224105Hence, the value of the given expression is -224105

Simplify $\frac{16}{- 21} \times \frac{14}{5}$

Simplification:
$\begin{array}{rcl} \frac{16}{- 21} \times \frac{14}{5} & = & \frac{16 \times 14}{- 21 \times 5} [∵ \frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}] \\ = & \frac{- 224}{105} \end{array}$
Hence, the value of the given expression is $\frac{- 224}{105}$