All the expression given below gives $k^{s}$ (s is a positive number) as a result of simplification except

(k^{0})^{8}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{k^{16} \times k^{- 8} \times k^{7}}{k^{9} \times k^{- 2}}

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

k \div k^{- 7}

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

\frac{(k^{- 1})^{18}}{k^{2}}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct options are
A $(k^{0})^{8}$
D $\frac{(k^{- 1})^{18}}{k^{2}}$
$\Rightarrow$ $(k^{o})^{8} = k^{0}$

$\Rightarrow$ $\frac{(k^{- 1})^{18}}{k^{2}} = \frac{k^{- 18}}{k^{2}} = k^{- 18 - 2} = k^{- 20}$

When we compare both the expressions with $k^{s}$ , we see that $s$ is not a positive number.
$∴$ The correct answers are option A and D.

Suggest Corrections

Similar questions

k^{8}

A = ⎡ ⎢ ⎢ ⎣ \begin{matrix} 2 k - 1 & 2 \sqrt{k} & 2 \sqrt{k} 2 \sqrt{k} & 1 & - 2 k - 2 \sqrt{k} & 2 k & - 1 \end{matrix} ⎤ ⎥ ⎥ ⎦

B = ⎡ ⎢ ⎢ ⎣ \begin{matrix} 0 & 2 k - 1 & \sqrt{k} 1 - 2 k & 0 & 2 \sqrt{k} - \sqrt{k} & - 2 \sqrt{k} & 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

(A d j (A)) + d e t (A d j (B)) = 10^{o}

[k]

n \sum k = 0 (\frac{n}{k}) (\frac{n + k}{k}) = n \sum k = 0 2^{k} {(\frac{n}{k})}^{2} .

k

5, k, k^{2} - 8

n

k = 5.1 \times 10^{n}

k

6, 000 < k < 500, 000

k^{2} = 2.601 \times 10^{9}

Join BYJU'S Learning Program