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Question

Find the value of.

(i) (30 + 4−1) × 22 (ii) (2−1 × 4−1) ÷2−2

(iii) (iv) (3−1 + 4−1 + 5−1)0

(v)

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Solution

(i)

(ii) (2−1 × 4−1) ÷ 2− 2 = [2−1 × {(2)2}− 1] ÷ 2− 2

= (2− 1 × 2− 2) ÷ 2− 2

= 2−1+ (−2) ÷ 2−2 (am × an = am + n)

= 2−3 ÷ 2−2

= 2−3 − (−2) (am ÷ an = amn)

= 2−3 + 2 = 2 −1

(iii)

(iv) (3−1 + 4−1 + 5−1)0

= 1 (a0 = 1)

(v)


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