Respostas

2014-08-03T11:38:09-03:00
A )

Dada a função pede-se a altura máxima usando Δ e Yv

h(x)=- \frac{x^2}{90}+10

\Delta=b^2-4*a*c\\
\\\Delta=0^2-4*(- \frac{1}{90} )*10\\
\\\Delta= \frac{4*1\not0}{9\not0} \\
\\\Delta= \frac{4}{9}

Agora aplica no Yv

Yv= \frac{-\Delta}{4a} \\
\\Yv= \frac{ -\frac{4}{9} }{4*(- \frac{1}{90} )} \\
\\Yv= \frac{ -\frac{4}{9} }{- \frac{4}{90} } \\
\\
\\Yv=(- \frac{4}{9} )* (-\frac{90}{4} )\\
\\Yv= \frac{90}{10} \\
\\\boxed{Yv=10~m}

B )

Como já temos o valor de Δ

x= \frac{\pm \sqrt{\Delta} }{2a} \\
\\x= \frac{\pm \sqrt{ \frac{4}{9} } }{2*(- \frac{1}{90} )} \\
\\x= \frac{\pm \frac{2}{3} }{- \frac{1}{45} }

x'= \frac{ \frac{2}{3} }{- \frac{1}{45} }\\
\\x'= \frac{2}{3}*(- \frac{45}{1} ) \\
\\x'= -\frac{90}{3} \\
\\\boxed{x'=-30}

x''= \frac{- \frac{2}{3} }{- \frac{1}{45} }\\
\\x''=(- \frac{2}{3})*(- \frac{45}{1} ) \\
\\x''= \frac{90}{3} \\
\\\boxed{x''=30}

\Delta_x=|x'|+|x''|\\
\\\Delta_x=|-30|+|30|\\
\\\Delta_x=30+30\\
\\\boxed{\Delta_x=60~m}

PS- Considerei as unidades em metros.
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