Question

h(x)=-x^2/90+10 ajudaaaa

Answer 1

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.