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anonymous
 one year ago
square root of 20.591
anonymous
 one year ago
square root of 20.591

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Jadeishere
 one year ago
Best ResponseYou've already chosen the best response.0use a calculator >.< type in 20.951 then the square root number

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0Lets first do a linear approximation of it. This is \(\large\color{black}{ \displaystyle f(x)=\sqrt{x} }\), and value we are going to use is \(a=25\). In other words, we are drawing a tangent line (at x=25) to this function, and use this line to estimate the value of \(\sqrt{20.951}\). (This process is also known as \(\normalsize \color{blue}{\rm linearization.}\)) \(\large\color{black}{ \displaystyle L(x)=f(a)+f'(a)(xa) }\) (this is just the tangent line, to a curve f(x), at x=a, in general)  \(\large\color{black}{ \displaystyle \frac{d}{dx}\sqrt{x} ~=\frac{d}{dx}x^{\LARGE ½}~=\frac{1}{2}x^{{\LARGE ½}1}~=\frac{1}{2}x^{\LARGE ½} ~=\frac{1}{2\sqrt{x}} }\) \(\large\color{black}{ \displaystyle f(a)~=\sqrt{a}~=\sqrt{25}=5 }\) \(\large\color{black}{ \displaystyle f'(a)~=\frac{1}{2\sqrt{a}} ~=\frac{1}{2\sqrt{25}}~=\frac{1}{10} }\)  \(\large\color{black}{ \displaystyle L(x)=5+\frac{1}{10}(x25) }\) Now, you can use this equation to approximate your square root (for values near x=25) \(\large\color{black}{ \displaystyle L(20.951)=5+\frac{1}{10}(\color{red}{20.951} 25) }\) \(\large\color{black}{ \displaystyle L(20.951)=5+\frac{1}{10}(4.049) }\) \(\large\color{black}{ \displaystyle L(20.951)=50.4049 }\) \(\large\color{black}{ \displaystyle L(20.951)=4.5951 }\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0You can get a beter approximation, by drawing a bigger polynomial .... \(\color{black}{ \displaystyle f(a)+f'(a)(xa) +\frac{f''(a)}{2!}(xa)^2+\frac{f'''(a)}{3!}(xa)^3 ....}\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0Or, this can be written entirely as: \(\color{black}{ \large \displaystyle \sum_{{\rm n=0}}^{\infty}\frac{\left(\displaystyle \frac{d^{\rm n}~f(x)}{dx^{\rm n}}~{\Huge }_{x=a}\right) }{{\rm n}!}\times (xa)^{\rm n}}\) More traditionally written as: \(\color{black}{ \large \displaystyle \sum_{{\rm n=0}}^{\infty}\frac{f^{\rm n}(a) }{{\rm n}!}\times (xa)^{\rm n}}\)
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