How do you solve $x^{2} + 2 x y + 3 y^{2}$ $x^2 + 2xy + 3y^2$ for x = 70 & y = 4?

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Answer 1 · 2018-06-26T02:40:48

The answer is $5, 508$ $5,508$ .

$x^{2} + 2 x y + 3 y^{2}$ $x^2 + 2xy + 3y^2$

For

Substitute the values given for $x$ $x$ and $y$ $y$ in the equation

${(70)}^{2} + 2 (70) (4) + 3 {(4)}^{2}$ $( 70)^2 + 2( 70)(4)+ 3( 4 )^2$

$= 4900 + 560 + 48$ $=4900 + 560 + 48$

$= 5, 508$ $= 5,508$

How do you solve x^2 + 2xy + 3y^2x2+2xy+3y2x^2 + 2xy + 3y^2 for x = 70 & y = 4?