Question

Looking to get help for this practice question, thank you

Answer 1

`\begin{gathered} 3m\text{ - }(2)/(3)(3m)\text{ - \lparen}(2)/(3))^x(3m)=\text{ 0} \\ 3m\text{ - }2m\text{ }^{\text{ }}\text{- \lparen}(2)/(3))^x(3m)\text{ = 0} \\ 1m=\text{\lparen}(2)/(3))^x(3m) \\ (1)/(3)=((2)/(3))^x \\ log(1)/(3)=x\text{ }log((2)/(3)) \\ (log(1)/(3))/(log(2)/(3))=x \\ x=\text{ 2.70} \end{gathered}`

So it will bounce2/3 to the 2.71 times.

So the distance is:

`\begin{gathered} 3m+2m+((2)/(3))^(2.71)(3m) \\ =5m\text{ + \lparen0.33\rparen\lparen3m\rparen} \\ =5m+0.99m \\ =5.99m \end{gathered}`