Step-by-step explanation:
Volume of cubical tin = [tex]10m \times 10m \times 10m[/tex] = [tex]1000 {m}^{3}[/tex]
therefore, Volume of oil that can be poured into cubical tin whose edge measures [tex]10m[/tex] is [tex]1000 {m}^{3}[/tex]
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Step-by-step explanation:
Volume of cubical tin = [tex]10m \times 10m \times 10m[/tex] = [tex]1000 {m}^{3}[/tex]
therefore, Volume of oil that can be poured into cubical tin whose edge measures [tex]10m[/tex] is [tex]1000 {m}^{3}[/tex]