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Q1.Simplify:
(I) {1³+2³}×(1/3) ²
(II) {5^-1×4^-1}²
(III) {(1/3) ^-1×(1/2)^-2}÷(1/4) ^-3

Give correct answer I will mark as brianlist. It wrong will deleated.

Q1Simplify I 1213 II 5141 III 13 112214 3 Give correct answer I will mark as brianlist It wrong will deleated class=

Respuesta :

loneguy

[tex]\textbf{(i)}\\\\\{1^3 +2^3 \} \times \left(\dfrac 13 \right)^2\\\\=(1+8) \left(\dfrac 19 \right)\\\\=9\left(\dfrac 19 \right)\\\\=1\\\\[/tex]

[tex]\textbf{(ii)}\\\\\{5^{-1}\times 4^{-1}\}^2\\\\=\left(5^{-1} \right)^2 \times \left(4^{-1} \right)^2~~~~~~~~~~~;[(ab)^m = a^mb^m]\\\\=5^{-2}\times 4^{-2}~~~~~~~~~~~~~~~~~~~~;[(a^m)^n = a^{mn}]\\\\=\dfrac 1{5^2} \times \dfrac 1{4^2}~~~~~~~~~~~~~~~~~~~~~~;\left[a^{-m} = \dfrac 1{a^m},~ a\neq 0 \right]\\\\=\dfrac{1}{400}\\\\=0.0025\\\\[/tex]

[tex]\textbf{(iii)}\\\\\left\{ \left(\dfrac 13 \right)^{-2} \times \left( \dfrac 12 \right)^{-2}\right\}\div \left(\dfrac 14 \right)^{-3}\\\\\\=\left( \dfrac 13 \times \dfrac 12 \right)^{-2} \div\left(4^{-1}\right)^{-3}\\\\\\=\left(\dfrac 16 \right)^{-2} \div 4^3\\\\\\=\left(6^{-1} \right)^{-2}\div 4^3\\\\\\=6^2 \div 4^3\\\\\\=36\div 64\\\\\\=\dfrac{9}{16}\\\\\\=0.5625[/tex]

TwinklingStar

[tex]\purple{\maltese}\large\underline{\underline{\red{\sf\:\: Question :}}} \\ \\ [/tex]

Q1.Simplify:

[tex] \\ \sf \: 1) \: \: \: ( \: 1 ^{3} +2 ^{3} \: ) \: × {\bigg( \frac{1}{3} \bigg) }^{2} \\ [/tex]

[tex] \sf \: 2) \: \: \: {{( \: \: 5 ^{ - 1} ×4^{ - 1} } } \: ) \: ^{2} \\ [/tex]

[tex] \sf \: 3) \: \: \: \bigg(\bigg( \frac{1}{3}\bigg) ^{-1}×\bigg( \frac{1}{2} \bigg)^{-2} \bigg)÷\bigg( \frac{1}{4} \bigg)^{-3} \\\\ [/tex]

[tex]\\\purple{\maltese}\large\underline{\underline{\red{\sf\:\: Solution :}}} \\ \\ [/tex]

[tex] \\ \bf \: 1) \: \: \: ( \: 1 ^{3} +2 ^{3} \: ) \: × {\bigg( \frac{1}{3} \bigg) }^{2} \\ [/tex]

[tex]\\ \sf \implies\:\: \: \: ( \: 1 ^{3} +2 ^{3} \: ) \: × {\bigg( \frac{1}{3} \bigg) }^{2} [/tex]

[tex]\\ \sf \implies\:\: \: \: ( \: 1 + 8\: ) \: × {\bigg( \frac{1}{3} \bigg) }^{2} [/tex]

[tex]\\ \sf \implies\:\: \: \: ( \: 1 + 8\: ) \: × {\bigg( \frac{1}{9} \bigg) } [/tex]

[tex]\\ \sf \implies\:\: \: \: ( \: 9\: ) \: × {\bigg( \frac{1}{9} \bigg) } [/tex]

[tex]\\ \sf \implies\:\: \: \: ( \: \cancel{9}\: ) \: × {\bigg( \frac{1}{\cancel{9}} \bigg) } [/tex]

[tex]\\ \sf \implies\:\: \: \: 1[/tex]

[tex]\implies\sf \underline{ \boxed{\sf \: \: \: \: 1 \: \: \: }} \: \: \bigstar[/tex]

[tex]\\\\\qquad\rule{150pt}{2pt}\\\\[/tex]

[tex] \bf \: 2) \: \: \: {{( \: \: 5 ^{ - 1} ×4^{ - 1} } } \: ) \: ^{2} \\ [/tex]

[tex]\\ \sf \implies\:\: \: \: {{( \: \: 5 ^{ - 1} ×4^{ - 1} } } \: ) \: ^{2} \\ [/tex]

[tex]\\ \sf \implies\:\: \: \: {{ {\bigg( \: \: \frac{1}{5} × \frac{1}{4} } } \: \bigg) } ^{2} \\ [/tex]

[tex]\\ \sf \implies\:\: \: \: {{ {\bigg( \: \: \frac{1}{20} } } \: \bigg) } ^{2} \\ [/tex]

[tex]\\ \sf \implies\:\: \: \: {{ {\bigg( \: \: \frac{1}{400} } } \: \bigg) } \\ [/tex]

[tex]\implies\sf \underline{ \boxed{\sf \: \: \: \: \frac{1}{400} \: \: \: }} \: \: \bigstar \\ \\ [/tex]

[tex]\\\\\qquad\rule{150pt}{2pt}\\\\[/tex]

[tex] \\ \bf \: 3) \: \: \: \bigg(\bigg( \frac{1}{3}\bigg) ^{-1}×\bigg( \frac{1}{2} \bigg)^{-2} \bigg)÷\bigg( \frac{1}{4} \bigg)^{-3} \\ [/tex]

[tex]\\ \sf \implies\:\: \: \: \bigg(\bigg( \frac{1}{3}\bigg) ^{-1}×\bigg( \frac{1}{2} \bigg)^{-2} \bigg)÷\bigg( \frac{1}{4} \bigg)^{-3}\\ [/tex]

[tex]\\ \sf \implies\:\: \: \:(\:\: 3^{2}- 2^{3} \:\:)÷4^{3}\\ [/tex]

[tex]\\ \sf \implies\:\: \: \:(\:\: 27- 8\:\:)÷64\\ [/tex]

[tex]\\ \sf \implies\:\: \: \:19÷64\\ [/tex]

[tex]\\ \sf \implies\:\: \: \:\frac{19}{64}\\ \\[/tex]

[tex]\implies\sf \underline{ \boxed{\sf \: \: \: \: \frac{19}{64} \: \: \: }} \: \: \bigstar \\ \\ [/tex]

[tex]\\\\\qquad\rule{150pt}{2pt}\\\\[/tex]

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