Question

Determinar la resistencia​

Answer 1

Respuesta:

Explicación paso a paso:

$16x^{2} +25y^{2} +64x+50y-311=0$

$16x^{2} +26y^{2} +64x + 50y-311=0$

$16x^{2} +64x+25y^{2} +50y-311=0$

$x=\\\frac{-64+\sqrt{64{2} -4 x 16 (25y^{2} + 50y -311 } }{2x16}$

$x= \\\frac{-64 + \sqrt{4096 - 4x 16 ( 25y^{2} + 50y - 311) } }{2x16}$

$x= \\\frac{-64 + \sqrt{4096 - 64(25y^{2} + 50y - 311)} }{2x16}$

$x=\\\frac{-64+ \sqrt{4096 + 19904 - 3200y - 1600y^{2} } }{2x16}$

$x= \\\frac{-64 + \sqrt{1600 (3-y)(y+5)} }{2x16}$

$x= \\\frac{-64+40 \sqrt{(3-y)(y+5)} }{32}$

$x= \frac{40\sqrt{(3-y)(y+5) - 64} }{32}$

$x= \frac{5\sqrt{(3-y)(y+5) } -2}{4}$

$x=\\\frac{-40 \sqrt{(3-y)(y+5)} -64}{32}$$x= \frac{5\sqrt{(3-y)(y+5)}-2 }{4}$

$x= \\\frac{-40\sqrt{(3-y)(y+5)} -64}{32}$

$x=- \frac{(3-y)(y+5)}-2{4}$

$x= \frac{5\sqrt{(3-y)(y+5)} -2}{4} \\x=-\frac{5\sqrt{(3-y)(y+5)}-2 }{4}$