Respuesta:
46creo que es así
[tex]area=4a^{4}+2a^{3}+6a^{2} \\\\base=2a^{2}\\\\area=\frac{base*altura}{2} \\\\4a^{4}+2a^{3}+6a^{2}=\frac{2a^{2} *altura}{2} \\\\altura*2a^{2}=2( 4a^{4}+2a^{3}+6a^{2})\\\\altura=\frac{8a^{4}+4a^{3}+12a^{2}}{2a^{2} } \\\\altura=\frac{2a^{2}( 4a^{2}+2a+6)}{2a^{2} }\\\\altura=4a^{2}+2a+6[/tex]
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Respuesta:
46creo que es así
Verified answer
Respuesta:
[tex]area=4a^{4}+2a^{3}+6a^{2} \\\\base=2a^{2}\\\\area=\frac{base*altura}{2} \\\\4a^{4}+2a^{3}+6a^{2}=\frac{2a^{2} *altura}{2} \\\\altura*2a^{2}=2( 4a^{4}+2a^{3}+6a^{2})\\\\altura=\frac{8a^{4}+4a^{3}+12a^{2}}{2a^{2} } \\\\altura=\frac{2a^{2}( 4a^{2}+2a+6)}{2a^{2} }\\\\altura=4a^{2}+2a+6[/tex]