An object is launched straight up into
1 answer:
Answer:
t ∈ {1, 3}
Step-by-step explanation:
You want to find t such that ...
h = 27
27 = -8t^2 +32t +3 . . . . . . substitute the expression for h
24 = -8t^2 +32t . . . . . . . . . subtract 3
-3 = t^2 -4t . . . . . . . . . . . . . divide by -8
1 = t^2 -4t +4 = (t -2)^2 . . . . add 4 to complete the square
±√1 = t -2 . . . . . . . . . . . . . . take the square root
t = 2 ± 1 . . . . . . . . . . . . . . . . add 2
t = 1 or 3
The object is 27 ft off the ground at t = 1 and again at t = 3.
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Answer:
Step-by-step explanation:
Number of students 10
Problem 1. $625 for the bus hire per friday, So 625*4=$2500
Problem 2. 2500/25=$100 each for the whole 4 weeks
Problem 3.
10 students tickets 220= 2200 for all tickets. The bus, 625/10 = $62.5*4= $250 dollars for the whole 4 weeks for the bus so in all each student pays $470 each
20 students, tickets 220=4400 for all tickets. The bus, 625/20=$31.25*4=$125 for the whole 4 weeks for the bus, so in all each student must pay $345 each
30 students, tickets 220 = 6600 for all tickets. The bus, 625/30 =$20.83*4=$83.32 for the whole 4 weeks for the bus, so in all each student must pay $303.32 each
41 students, tickets $160=$6560 for all tickets. The bus, because you need 2 buses at 625 each so $1250 for both buses 1250/41= 30.49*4=$121.96 for the whole 4 weeks for the bus. So in al each student must pay $281.96 each
Hope this is correct
