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Powers and Exponents

The value of ...

Question

The value of $[2^{3} - 3^{2} - 1^{3}]^{0}$ is 0.

True

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False

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Solution

The correct option is B
False

$G i v e n, [2^{3} - 3^{2} - 1^{3}]^{0}$

Now, $(2^{3} - 3^{2} - 1^{3})^{0} = (8 - 9 - 1)^{0}$

$= (- 2)^{0}$

$= 1$ $(∵ a^{0} = 1)$

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\frac{3 \times 1^{3}}{1^{2}} + \frac{5 \times (1^{3} + 2^{3})}{1^{2} + 2^{2}} + \frac{7 \times (1^{3} + 2^{3} + 3^{3})}{1^{2} + 2^{2} + 3^{2}} + \dots upto 10 terms

x^{2} + k (4 x + k - 1) + 2 = 0

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\frac{3}{2}, \frac{1}{3}

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A = ∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 & - 1 3 & 4 & 5 0 & 6 & 7 \end{matrix} ∣ ∣ ∣ ∣

A^{- 1} = ∣ ∣ ∣ ∣ \begin{matrix} a_{11} & a_{12} & a_{13} a_{21} & a_{22} & a_{23} a_{31} & a_{32} & a_{33} \end{matrix} ∣ ∣ ∣ ∣

a_{23}

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A^{- 1} = ⎡ ⎢ ⎣ \begin{matrix} a_{11} & a_{12} & a_{13} a_{21} & a_{22} & a_{23} a_{31} & a_{32} & a_{33} \end{matrix} ⎤ ⎥ ⎦

a_{23} =

Find the value of $θ$ in each of the following

(1) 2 sin 2 $θ$ = $\sqrt{3}$

(2) 2 cos 3 $θ$ = 1

(3) $\sqrt{3}$ tan 2 $θ$ - 3 = 0

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