3}} \right)}^{z}}\] \[={{2}^{x+2y+3z}}\] \[\THEREFORE x+2y+3z=\FRAC{1}{9}+\frac{8}{9}+\frac{18}{9}\] \[=\frac{27}{9}=3\] \[\therefore A\times B\times C={{2}^{3}}=8.\]

"> 3}} \right)}^{z}}\] \[={{2}^{x+2y+3z}}\] \[\THEREFORE x+2y+3z=\FRAC{1}{9}+\frac{8}{9}+\frac{18}{9}\] \[=\frac{27}{9}=3\] \[\therefore A\times B\times C={{2}^{3}}=8.\]

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If A=\[{{\mathbf{2}}^{x}}\], B= \[{{\mathbf{4}}^{y}}\]c=\[{{\mathbf{8}}^{z}}\] where x=\[\mathbf{0}.\overline{\mathbf{1}}\], \[\mathbf{y}=\mathbf{0}.\overline{\mathbf{4}},\mathbf{z}=\mathbf{0}.\overline{\mathbf{6}},\]then \[\mathbf{A}\times \mathbf{B}\times \mathbf{C}\] is

7th Class Mathematics in 7th Class 11 months ago

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(a): \[A\times B\times C={{2}^{x}}\times {{\left( {{2}^{2}} \right)}^{y}}\times {{\left( {{2}^{3}} \right)}^{z}}\] \[={{2}^{x+2y+3z}}\] \[\THEREFORE x+2y+3z=\FRAC{1}{9}+\frac{8}{9}+\frac{18}{9}\] \[=\frac{27}{9}=3\] \[\therefore A\times B\times C={{2}^{3}}=8.\]

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Posted on 23 Aug 2024, this text provides information on 7th Class related to Mathematics in 7th Class. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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