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

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Answer 1 · 2018-06-26T02:40:48

The answer is $5,508$ .

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

For

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

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

$=4900 + 560 + 48$

$= 5,508$

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