Website Statistics Bonjour jai léquation 5tcarré20t16 Je dois calculer ht0 Merci de votre aide

Répondre :

Bonsoir
h(t) = -5t²+20t +1.6 
h(t) = 0  
-5t²+20t+1.6 = 0
delta = 432    Vdelta = V432
x ' = (-20-V432)/ -10 = 4.078  arrondi
x " = (-20+V432)/-10 = 0.078 arrondi
[tex]-5t^2+20t+1,6 = 0 \\ -5(t^2-4t-0,32) =0 \\ t^2-4t-0,32 =0 \\ (t-2)^2 = t^2 -4t + 4 \\ (t-2)^2 - \frac{432}{100} =0 \\ (t-2)^2 - (\frac{12\sqrt{3}}{10})^2 =0 \\ (t-2- \frac{12\sqrt{3}}{10})(t-2+ \frac{12\sqrt{3}}{10}) =0 \\ (t- \frac{10+6\sqrt{3}}{5})(t- \frac{10-6\sqrt{3}}{5}) =0\\ donc\,2\,solutions: t=\frac{10+6\sqrt{3}}{5}\,ou\,\frac{10-6\sqrt{3}}{5}[/tex]

En espérant t'avoir aidé.



D'autres questions