## College Algebra (10th Edition)

$\mathrm{a}$. $A(\displaystyle \frac{1}{3})\approx 1.26\ \mathrm{f}\mathrm{t}^{2}$ $\mathrm{b}$. $A(\displaystyle \frac{1}{2})\approx 1.73\ \mathrm{f}\mathrm{t}^{2}$ $\mathrm{c}$. $A(\displaystyle \frac{2}{3})\approx 1.99\ \mathrm{f}\mathrm{t}^{2}$
Substitute x=(given value) into the expression for A. $\mathrm{a}$. $A(\displaystyle \frac{1}{3})=4\cdot\frac{1}{3}\sqrt{1-(\frac{1}{3})^{2}}$ $=\displaystyle \frac{4}{3}\sqrt{\frac{8}{9}}$ $=\displaystyle \frac{4}{3}\cdot\frac{2\sqrt{2}}{3}$ $=\displaystyle \frac{8\sqrt{2}}{9}\approx 1.26\ \mathrm{f}\mathrm{t}^{2}$ $\mathrm{b}$. $A(\displaystyle \frac{1}{2})=4\cdot\frac{1}{2}\sqrt{1-(\frac{1}{2})^{2}}$ $=2\sqrt{\dfrac{3}{4}}$ $=2\displaystyle \cdot\frac{\sqrt{3}}{2}$ $=\sqrt{3}\approx 1.73\ \mathrm{f}\mathrm{t}^{2}$ $\mathrm{c}$. $A(\displaystyle \frac{2}{3})=4\cdot\frac{2}{3}\sqrt{1-(\frac{2}{3})^{2}}$ $=\displaystyle \frac{8}{3}\sqrt{\frac{5}{9}}$ $=\displaystyle \frac{8}{3}\cdot\frac{\sqrt{5}}{3}$ $=\displaystyle \frac{8\sqrt{5}}{9}\approx 1.99\ \mathrm{f}\mathrm{t}^{2}$