NUMBER SYSTEMS – Exercise 1.6 – (MATHEMATICS) – 9th Class

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Exercise  1.6

1.  Find :
(i) {64^{{1 \over 2}}}

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Ans. – {64^{{1 \over 2}}} = {({8^2})^{{1 \over 2}}} = {(8)^{2 \times {1 \over 2}}} = 8

(ii) {32^{{1 \over 5}}}

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Ans. – {32^{{1 \over 5}}} = {({2^5})^{{1 \over 5}}} = {(2)^{5 \times {1 \over 2}}} = 2

(iii) {125^{{1 \over 3}}}

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Ans. – {125^{{1 \over 3}}} = {({5^3})^{{1 \over 3}}} = {(5)^{3 \times {1 \over 3}}} = 5

2. Find  :
(i) {9^{{3 \over 2}}}

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Ans. – {9^{{3 \over 2}}} = {({3^2})^{{3 \over 2}}} = {(3)^{2 \times {3 \over 2}}} = {(3)^3} = 27.

(ii) {32^{{2 \over 5}}}

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Ans. – {32^{{2 \over 5}}} = {({2^5})^{{2 \over 5}}} = {2^{5 \times {2 \over 5}}} = {(2)^2} = 4.

(iii) {16^{{3 \over 4}}}

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Ans. – {16^{{3 \over 4}}} = {({2^4})^{{3 \over 4}}} = {(2)^{4 \times {3 \over 4}}} = {(2)^3} = 4.

(iv) {125^{ - {3 \over 4}}}

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Ans. – {(125)^{ - {3 \over 4}}} = {({5^3})^{ - {1 \over 3}}} = {(5)^{3 \times \left( { - {1 \over 3}} \right)}} = {(5)^{ - 1}} = {1 \over 5}

3. Simplify :
(i) 2{2 \over 3}.2{1 \over 5}

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Ans. – {2^{{2 \over 3}}}{.2^{{1 \over 5}}} = {2^{{2 \over 3} + {1 \over 5}}} = {2^{{{13} \over {15}}}}

(ii) {\left( {{1 \over {{3^3}}}} \right)^7}

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Ans. – {\left( {{1 \over {{3^3}}}} \right)^7} = {1 \over {{{({3^3})}^7}}} = {1 \over {{3^{3 \times 7}}}} = {1 \over {{3^{21}}}} = {3^{ - 21}}

(iii) {{{{11}^{{1 \over 2}}}} \over {{{11}^{{1 \over 4}}}}}

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Ans. –  {{{{11}^{{1 \over 2}}}} \over {{{11}^{{1 \over 4}}}}} = {11^{{1 \over 2} - {1 \over 4}}} = 11{1 \over 4}

(iv) {7^{{1 \over 2}}}{.8^{{1 \over 2}}}

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Ans. – {7^{{1 \over 2}}}{.8^{{1 \over 2}}} = {(7.8)^{1/2}} = {56^{1/2}}

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