19-03-2020
Mathematics

contestada

what is the maximum value of h(t)=−16t^2+32t+64

Respuesta :

surjithayer10 surjithayer10

19-03-2020

Answer:

Step-by-step explanation:

h(t)=-16t²+32t+64

=-16(t²-2t)+64

=-16(t²-2t+1-1)+64

=-16(t-1)²+16+64

=-16(t-1)²+80

maximum value=80 and is obtained at t=1

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