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Mathematics 24 Online
OpenStudy (anonymous):

simplify

OpenStudy (anonymous):

\[2x \sqrt{20x^20}\]

OpenStudy (anonymous):

4x^2\[\sqrt{5x^5}\] ?

OpenStudy (phi):

if that is x^20 then its square root is x^10 x*x*x.... twenty times. take out pairs (10 of them) the easy way is divide the exponent by 2. if it is odd (for example) sqrt(x^21) write it as sqrt(x^20 * x) then divide the 20 by 2 to get x^10 * sqrt(x)

OpenStudy (phi):

so \[ 2x \sqrt{20x^{20} } = 2x\sqrt{2*2*5*x^{10}*x^{10} }= 2x*2*x^{10}\sqrt{5}\] or \[ 4x^{11}\sqrt{5} \]

OpenStudy (phi):

remember \[ x^{10}\cdot x^{10}= x^{20} \] you add exponents. the easy way to remember is that x*x*...*x (times itself 10 times) * x*x*..*x (x times itself 10 times) give us x*x*...*x (x times itself 20 times)

OpenStudy (anonymous):

k thankyou

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