Question

cuál es el área de una caja cuyas medidas son alto:x+3, largo:x+6 y ancho:x-3 por favor urge!​

Answer 1

Respuesta:

 $A = 6x^{2} +24x -18$

Explicación paso a paso:

$Largo: L = x+6$

$Alto: h = x+3$

$Ancho: a = x -3$

$Area: A$

$A =[ 2 ( x+6 ) ( x-3 ) ] + [ 2 ( x+3 ) ( x+6 ) ] + [ 2 ( x + 3 ) (x -3 ) ]$

$A = [ 2 ( x^{2} -3x + 6x - 18 ) ] + [ 2 ( x^{2} + 3x + 6x + 18 ) ] + [ 2 ( x^{2} -3x + 3x -9 ) ]$

$A = 2( x^{2} +3x-18 ) + 2 ( x^{2} +9x+18 ) + 2 ( x^{2} -9 )$

$A = 2x^{2} +6x-36+2x^{2} +18x+36+2x^{2} -18$

$A = 6x^{2} +24x -18$

Answer 2

Explicación paso a paso:

Largo:L=x+6

Alto: h = x+3Alto:h=x+3

Ancho: a = x -3Ancho:a=x−3

Area: AArea:A

A =[ 2 ( x+6 ) ( x-3 ) ] + [ 2 ( x+3 ) ( x+6 ) ] + [ 2 ( x + 3 ) (x -3 ) ]A=[2(x+6)(x−3)]+[2(x+3)(x+6)]+[2(x+3)(x−3)]

\begin{gathered}A = [ 2 ( x^{2} -3x + 6x - 18 ) ] + [ 2 ( x^{2} + 3x + 6x + 18 ) ] + [ 2 ( x^{2} -3x + 3x -9 ) ]\\\end{gathered}

A=[2(x

2

−3x+6x−18)]+[2(x

2

+3x+6x+18)]+[2(x

2

−3x+3x−9)]

A = 2( x^{2} +3x-18 ) + 2 ( x^{2} +9x+18 ) + 2 ( x^{2} -9 )A=2(x

2

+3x−18)+2(x

2

+9x+18)+2(x

2

−9)

A = 2x^{2} +6x-36+2x^{2} +18x+36+2x^{2} -18A=2x

2

+6x−36+2x

2

+18x+36+2x

2

−18

A = 6x^{2} +24x -18A=6x

2

+24x−18