Question

Multiple Answer question
All the expression given below gives '$k^8$' as a result of simplification except

• A. $(k^0{)}^8$
• B. $\frac{k^{16}\times k^{-8}\times k^7}{k^9\times k^{-2}}$
• C. $k÷k^{-7}$
• D. $\frac{(k^{-1}{)}^{18}}{k^2}$

Answer 1

Answer: (B), (C)

The correct options are
B $\frac{k^{16}\times k^{-8}\times k^7}{k^9\times k^{-2}}$
C $k÷k^{-7}$

$(k^0{)}^8=1^8=1$

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

$k÷k^{-7}=k^{1-(-7)}=k^8$

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