Evaluate- - 4 14- 2 12- 5 15 -0

The correct option is D 1 = [ ( 4 )^14× ( 2 )^12× ( 5 )^12 ]^0 = 4^0× 2^0× 5^0 = 1× 1× 1 = 1 So, option C is correct. ...

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Algebraic Identities

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Evaluate: ${[{(4)}^{\frac{1}{4}} \times {(2)}^{\frac{1}{2}} \times {(5)}^{\frac{1}{5}}]}^{0}$

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x = {(2 + \sqrt{5})}^{1 / 2} + {(\sqrt{5} - 2)}^{1 / 2}

y = {(2 + \sqrt{5})}^{1 / 2} - {(\sqrt{5} - 2)}^{1 / 2}

x^{2} + y^{2}

[x 1] [\begin{matrix} 1 & 0 \\ - 2 & - 3 \end{matrix}] [\begin{matrix} x \\ 5 \end{matrix}] = 0

[1 2 1] [\begin{matrix} 1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2 \end{matrix}] [\begin{matrix} 0 \\ 2 \\ x \end{matrix}] = 0

[x - 5 - 1] [\begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix}] [\begin{matrix} x \\ 4 \\ 1 \end{matrix}] = 0

Evaluate the determinants..

$∣ ∣ ∣ ∣ \begin{matrix} 3 & - 1 & - 2 0 & 0 & - 1 3 & - 5 & 0 \end{matrix} ∣ ∣ ∣ ∣$

$∣ ∣ ∣ ∣ \begin{matrix} 3 & - 4 & 5 1 & 1 & - 2 2 & 3 & 1 \end{matrix} ∣ ∣ ∣ ∣$

$∣ ∣ ∣ ∣ \begin{matrix} 0 & 1 & 2 - 1 & 0 & - 3 - 2 & 3 & 0 \end{matrix} ∣ ∣ ∣ ∣$

$∣ ∣ ∣ ∣ \begin{matrix} 2 & - 1 & - 2 0 & 2 & - 1 3 & - 5 & 0 \end{matrix} ∣ ∣ ∣ ∣$

Solve: $5 \frac{1}{3} - [2 \frac{1}{3} \div {\frac{3}{4} - \frac{1}{2} \times (\frac{7}{10} - \frac{3}{5})}]$

Distance(km)	$1 - 11$	$11 - 21$	$21 - 31$	$31 - 41$	$41 - 51$
Number of persons	$4$	$6$	$10$	$12$	$18$

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