Find the value of:
Question:

Find the value of:

(i) (20 + 3−1) × 32

(ii) (2−1 × 3−1) ÷ 2−3

(iii) $\left(\frac{1}{2}\right)^{-2}+\left(\frac{1}{3}\right)^{-2}+\left(\frac{1}{4}\right)^{-2}$

Solution:

(i) $\left(2^{0}+3^{-1}\right) \times 3^{2}=\left(1+\frac{1}{3}\right) \times 3^{2} \quad$ (because $2^{0}=1$ and $\left.3^{-1}=\frac{1}{3}\right)$

$=\left(\frac{1 \times 3}{1 \times 3}+\frac{1 \times 1}{3 \times 1}\right) \times 3^{2}=\left(\frac{3}{3}+\frac{1}{3}\right) \times 3^{2}=\left(\frac{4}{3}\right) \times 3^{2}=4 \times 3^{(2-1)}=4 \times 3=12$

$\left(2^{-1} \times 3^{-1}\right) \div 2^{-3}=\left(\frac{1}{2} \times \frac{1}{3}\right) \div\left(\frac{1}{2}\right)^{3}$

$\left(\frac{1}{6}\right) \div \frac{1^{3}}{2^{3}}=\left(\frac{1}{6}\right) \div\left(\frac{1}{8}\right)=\frac{1}{6} \times 8=\frac{8}{6}=\frac{4}{3}$

(iii) $\left(\frac{1}{2}\right)^{-2}+\left(\frac{1}{3}\right)^{-2}+\left(\frac{1}{4}\right)^{-2}=\left(\frac{2}{1}\right)^{2}+\left(\frac{3}{1}\right)^{2}+\left(\frac{4}{1}\right)^{2}=2^{2}+3^{2}+4^{2}=4+9+16=29$

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