a) `-16t^2+78t+6=0` we find t value by applying the quadratic formula `t=(-78+-sqrt(78^2-4(-16)(6)))/((2)(-16))=(-78+-sqrt(6084+384))/(-32)=`

`(-78+-sqrt(6468))/(-32)=(-78+-80.42)/(-32)=2.42/-32, -158.42/-32=-.08, 4.95` Common sense tells us that t cannot have a negative value, so t=4.95 sec.

b)` ``-16t^2+78t+6>20` We solve this by subtracting 20 from both sides, and then re-applying the quadratic formula to the inequality `-16t^2+78t-14>0` The interval is between the two values of t for `t=(-78+-sqrt(78^2-4(-16)(-14)))/((2)(-16))=(-78+-sqrt(6084-896))/(-32)=`

`(-78+-sqrt(5188))/-32=(-78+-72.03)(-32)=(-5.97)/-32, (-150.03)/-32=`

`0.19, 4.69`So the poodle's height exceeds 20 ft from 0.19sec, to 4.69 sec.

c) The formula for the Max/min of a quadratic equation is `t=-(78/(-32))=2.44` The poodle will reach max height at 2.44sec, and plugging that value of t into our equation finds a maximum height. `f(2.44)=-16(2.44)^2+78(2.44)+6=`

`-95.26+190.32+6=` 101.06 ft.

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