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Sagot :
well, if the water is coming out at 0.44 ft³ per minute, and you take a shower for 8 mins, that simply means you used 8*0.44 ft³, or namely 3.52 ft³.
what if you instead fill the bathtub with 0.45 yd³? is that more or less than 3.52ft³?
[tex]\begin{array}{ccll} yard&(yard)^2&(yard)^3\\ \cline{1-3} (3~ft)&(3~ft)^2&(3~ft)^3\\ 3~ft&3^2~ft^2&3^3~ft^3\\ &9~ft^2&27~ft^3 \end{array}[/tex]
so then, there are that many ft³ in a yd³, so how many will there be in 0.45yd³ anyway?
[tex]\begin{array}{ccll} yd^3&ft^3\\ \cline{1-2} 1&27\\ 0.45&x \end{array}\implies \cfrac{1}{0.45}=\cfrac{27}{x}\implies x=27(0.45)\implies x=12.15~ft^3[/tex]
well, then you'd use more water with the 0.45yd³ and some soap btw, by how much more? 12.15 - 3.52 = 8.63 ft³.
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