+26381 How do i find the height?
\(\begin{array}{|rcll|} \hline h &=& 62\cdot \sqrt[3]{t} + 76 \quad & | \quad -76 \\ h - 76 &=& 62\cdot \sqrt[3]{t} \quad & | \quad :62 \\ \frac{h - 76}{62} &=& \sqrt[3]{t} \\ \frac{h - 76}{62} &=& t^{\frac13} \quad & | \quad \text{cube both sides} \\ \left({\frac{h - 76}{62}} \right)^3 &=& t^{\frac33} \\ \left({\frac{h - 76}{62}} \right)^3 &=& t^1 \\ \left({\frac{h - 76}{62}} \right)^3 &=& t \\ \mathbf{t} & \mathbf{=} & \mathbf{ \left({\frac{h - 76}{62}} \right)^3 } \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline t &=& \left({\frac{h - 76}{62}} \right)^3\quad & | \quad h = 231\ cm \qquad t \text{ in years}\\ t &=& \left({\frac{231 - 76}{62}}\right)^3 \\ t &=& \left({\frac{155}{62}}\right)^3 \\ t &=& {2.5}^3 \\ \mathbf{t} & \mathbf{=} & \mathbf{15.625 \text{ years}} \\ \hline \end{array}\)
+26381 How do i find the height?
\(\begin{array}{|rcll|} \hline h &=& 62\cdot \sqrt[3]{t} + 76 \quad & | \quad -76 \\ h - 76 &=& 62\cdot \sqrt[3]{t} \quad & | \quad :62 \\ \frac{h - 76}{62} &=& \sqrt[3]{t} \\ \frac{h - 76}{62} &=& t^{\frac13} \quad & | \quad \text{cube both sides} \\ \left({\frac{h - 76}{62}} \right)^3 &=& t^{\frac33} \\ \left({\frac{h - 76}{62}} \right)^3 &=& t^1 \\ \left({\frac{h - 76}{62}} \right)^3 &=& t \\ \mathbf{t} & \mathbf{=} & \mathbf{ \left({\frac{h - 76}{62}} \right)^3 } \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline t &=& \left({\frac{h - 76}{62}} \right)^3\quad & | \quad h = 231\ cm \qquad t \text{ in years}\\ t &=& \left({\frac{231 - 76}{62}}\right)^3 \\ t &=& \left({\frac{155}{62}}\right)^3 \\ t &=& {2.5}^3 \\ \mathbf{t} & \mathbf{=} & \mathbf{15.625 \text{ years}} \\ \hline \end{array}\)
+36923 Just substitue in the values you are given...
231 cm = 62 \(\sqrt[3]{t}\)+76
(231-76)/(62) = \(\sqrt[3]{t }\)
15.63 years = t
