((16.39-10.19)^2+(5.15-10.19)^2 〖+(5.22-10.19)〗^2+(14.0-10.19)^2)/(4-1)

asked by guest
on Nov 17, 2024 at 6:20 am



You asked:

Evaluate the expression: (16.39−10.19)2+(5.15−10.19)2(5.22−10.19)2+(14.0−10.19)24−1\frac{{\left( 16.39 - 10.19 \right)}^{2} + {\left( 5.15 - 10.19 \right)}^{2} {\left( 5.22 - 10.19 \right)}^{2} + {\left( 14.0 - 10.19 \right)}^{2}}{4 - 1}

MathBot Answer:

(16.39−10.19)2+(5.15−10.19)2(5.22−10.19)2+(14.0−10.19)24−1=425249050918750000=226.799493813‾\frac{{\left( 16.39 - 10.19 \right)}^{2} + {\left( 5.15 - 10.19 \right)}^{2} {\left( 5.22 - 10.19 \right)}^{2} + {\left( 14.0 - 10.19 \right)}^{2}}{4 - 1} = \frac{4252490509}{18750000} = 226.79949381\overline{3}

(16.39−10.19)2+(5.15−10.19)2(5.22−10.19)2+(14.0−10.19)24−1=(1639100−10.19)2+(5.15−10.19)2(5.22−10.19)2+(14.0−10.19)24−1=(1639100−1019100)2+(5.15−10.19)2(5.22−10.19)2+(14.0−10.19)24−1=(315)2+(5.15−10.19)2(5.22−10.19)2+(14.0−10.19)24−1=96125+(5.15−10.19)2(5.22−10.19)2+(14.0−10.19)24−1=96125+(10320−10.19)2(5.22−10.19)2+(14.0−10.19)24−1=96125+(10320−1019100)2(5.22−10.19)2+(14.0−10.19)24−1=96125+(−12625)2(5.22−10.19)2+(14.0−10.19)24−1=96125+15876625(5.22−10.19)2+(14.0−10.19)24−1=96125+15876625(26150−10.19)2+(14.0−10.19)24−1=96125+15876625(26150−1019100)2+(14.0−10.19)24−1=96125+15876625(−497100)2+(14.0−10.19)24−1=96125+15876625⋅24700910000+(14.0−10.19)24−1=96125+9803787211562500+(14.0−10.19)24−1=10404412211562500+(14.0−10.19)24−1=10404412211562500+(14−10.19)24−1=10404412211562500+(14−1019100)24−1=10404412211562500+(381100)24−1=10404412211562500+145161100004−1=425249050962500004−1=425249050962500003=425249050918750000\begin{aligned}\frac{{\left( 16.39 - 10.19 \right)}^{2} + {\left( 5.15 - 10.19 \right)}^{2} {\left( 5.22 - 10.19 \right)}^{2} + {\left( 14.0 - 10.19 \right)}^{2}}{4 - 1}&=\frac{{\left( \frac{1639}{100} - 10.19 \right)}^{2} + {\left( 5.15 - 10.19 \right)}^{2} {\left( 5.22 - 10.19 \right)}^{2} + {\left( 14.0 - 10.19 \right)}^{2}}{4 - 1}\\&=\frac{{\left( \frac{1639}{100} - \frac{1019}{100} \right)}^{2} + {\left( 5.15 - 10.19 \right)}^{2} {\left( 5.22 - 10.19 \right)}^{2} + {\left( 14.0 - 10.19 \right)}^{2}}{4 - 1}\\&=\frac{{\left( \frac{31}{5} \right)}^{2} + {\left( 5.15 - 10.19 \right)}^{2} {\left( 5.22 - 10.19 \right)}^{2} + {\left( 14.0 - 10.19 \right)}^{2}}{4 - 1}\\&=\frac{\frac{961}{25} + {\left( 5.15 - 10.19 \right)}^{2} {\left( 5.22 - 10.19 \right)}^{2} + {\left( 14.0 - 10.19 \right)}^{2}}{4 - 1}\\&=\frac{\frac{961}{25} + {\left( \frac{103}{20} - 10.19 \right)}^{2} {\left( 5.22 - 10.19 \right)}^{2} + {\left( 14.0 - 10.19 \right)}^{2}}{4 - 1}\\&=\frac{\frac{961}{25} + {\left( \frac{103}{20} - \frac{1019}{100} \right)}^{2} {\left( 5.22 - 10.19 \right)}^{2} + {\left( 14.0 - 10.19 \right)}^{2}}{4 - 1}\\&=\frac{\frac{961}{25} + {\left( \frac{-126}{25} \right)}^{2} {\left( 5.22 - 10.19 \right)}^{2} + {\left( 14.0 - 10.19 \right)}^{2}}{4 - 1}\\&=\frac{\frac{961}{25} + \frac{15876}{625} {\left( 5.22 - 10.19 \right)}^{2} + {\left( 14.0 - 10.19 \right)}^{2}}{4 - 1}\\&=\frac{\frac{961}{25} + \frac{15876}{625} {\left( \frac{261}{50} - 10.19 \right)}^{2} + {\left( 14.0 - 10.19 \right)}^{2}}{4 - 1}\\&=\frac{\frac{961}{25} + \frac{15876}{625} {\left( \frac{261}{50} - \frac{1019}{100} \right)}^{2} + {\left( 14.0 - 10.19 \right)}^{2}}{4 - 1}\\&=\frac{\frac{961}{25} + \frac{15876}{625} {\left( \frac{-497}{100} \right)}^{2} + {\left( 14.0 - 10.19 \right)}^{2}}{4 - 1}\\&=\frac{\frac{961}{25} + \frac{15876}{625} \cdot \frac{247009}{10000} + {\left( 14.0 - 10.19 \right)}^{2}}{4 - 1}\\&=\frac{\frac{961}{25} + \frac{980378721}{1562500} + {\left( 14.0 - 10.19 \right)}^{2}}{4 - 1}\\&=\frac{\frac{1040441221}{1562500} + {\left( 14.0 - 10.19 \right)}^{2}}{4 - 1}\\&=\frac{\frac{1040441221}{1562500} + {\left( 14 - 10.19 \right)}^{2}}{4 - 1}\\&=\frac{\frac{1040441221}{1562500} + {\left( 14 - \frac{1019}{100} \right)}^{2}}{4 - 1}\\&=\frac{\frac{1040441221}{1562500} + {\left( \frac{381}{100} \right)}^{2}}{4 - 1}\\&=\frac{\frac{1040441221}{1562500} + \frac{145161}{10000}}{4 - 1}\\&=\frac{\frac{4252490509}{6250000}}{4 - 1}\\&=\frac{\frac{4252490509}{6250000}}{3}\\&=\frac{4252490509}{18750000}\end{aligned}

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