\frac{\left(121-100\right)^2}{100}+\frac{\left(288-150\right)^2}{150}+\frac{\left(91-250\right)^2}{250}

asked by guest
on Jan 15, 2025 at 5:43 pm



You asked:

Evaluate the expression: (121−100)2100+(288−150)2150+(91−250)2250\frac{{\left( 121 - 100 \right)}^{2}}{100} + \frac{{\left( 288 - 150 \right)}^{2}}{150} + \frac{{\left( 91 - 250 \right)}^{2}}{250}

MathBot Answer:

(121−100)2100+(288−150)2150+(91−250)2250=116247500=232.494\frac{{\left( 121 - 100 \right)}^{2}}{100} + \frac{{\left( 288 - 150 \right)}^{2}}{150} + \frac{{\left( 91 - 250 \right)}^{2}}{250} = \frac{116247}{500} = 232.494

(121−100)2100+(288−150)2150+(91−250)2250=212100+(288−150)2150+(91−250)2250=441100+(288−150)2150+(91−250)2250=441100+1382150+(91−250)2250=441100+19044150+(91−250)2250=441100+317425+(91−250)2250=13137100+(91−250)2250=13137100+−1592250=13137100+25281250=116247500\begin{aligned}\frac{{\left( 121 - 100 \right)}^{2}}{100} + \frac{{\left( 288 - 150 \right)}^{2}}{150} + \frac{{\left( 91 - 250 \right)}^{2}}{250}&=\frac{{21}^{2}}{100} + \frac{{\left( 288 - 150 \right)}^{2}}{150} + \frac{{\left( 91 - 250 \right)}^{2}}{250}\\&=\frac{441}{100} + \frac{{\left( 288 - 150 \right)}^{2}}{150} + \frac{{\left( 91 - 250 \right)}^{2}}{250}\\&=\frac{441}{100} + \frac{{138}^{2}}{150} + \frac{{\left( 91 - 250 \right)}^{2}}{250}\\&=\frac{441}{100} + \frac{19044}{150} + \frac{{\left( 91 - 250 \right)}^{2}}{250}\\&=\frac{441}{100} + \frac{3174}{25} + \frac{{\left( 91 - 250 \right)}^{2}}{250}\\&=\frac{13137}{100} + \frac{{\left( 91 - 250 \right)}^{2}}{250}\\&=\frac{13137}{100} + \frac{{-159}^{2}}{250}\\&=\frac{13137}{100} + \frac{25281}{250}\\&=\frac{116247}{500}\end{aligned}

+ Submit Question