Mathematics, 08.04.2020 04:45 lol4373
The area of a playground is 20 yd.². The length of the playground at five times longer than it's worth. Find the length and the width of the playground.
Answers: 1
![20=5x\times x\\20=5x^{2} \\Dividing\ both \ sides\ by\ 5\\\\ x^{2} =4\\Taking\ root\ both\ sides\\ \\ \sqrt[2]{x^{2} } =\sqrt[2]{4} \\ \\ x=\sqrt[2]{2\times2} \\ \\ x=2](/tpl/images/0588/9534/4ec98.png)
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The area of a playground is 20 yd.². The length of the playground at five times longer than it's wor...
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