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A basketball is thrown upwards. The height f(t), in feet, of the basketball at time t, in seconds, is given by the following function: f(t) = -16t^2 + 44t + 12 Which of the following is a reasonable domain of the graph of the function when the basketball falls from its maximum height to the ground? -0.25 < t < 2 0 < t < 3 1 < t < 2.4 1.4 < t < 3
It reaches a maximum height at F(t) = 0 [dont ask me why] so make the function equal to zero and solve for t
I got 3 and -1/4
my mistake @gabymarie313 , @DanJS had probably worked that out, its at dy/dt=0 that it reaches a maximum height
so how would I solve it?
first we will find it at the bottom i.e. let t=0 f(0)= -16(0)^2 +44 + 12
f(0) = 44 + 12 f(0) = 56
so is the the maximum height?
maximum height occurs where balls stops, ball stops at dy/dt = 0
dy/dt = 2 (-16t) + 44 = -18t + 44
-18t + 44 = 0 -18t = -44 18 t = 44 t=44/18 t = 2.44
I get it now. Thank you for your time.
do you think you can help me with a couple more @BPDlkeme234
go for it!
Matt sells burgers and sandwiches. The daily cost of making burgers is $520 more than the difference between the square of the number of burgers sold and 30 times the number of burgers sold. The daily cost of making sandwiches is modeled by the following equation: C(x) = 2x^2 - 40x + 300 C(x) is the cost in dollars of selling x sandwiches. Which statement best compares the minimum daily cost of making burgers and sandwiches? It is greater for sandwiches than burgers because the approximate minimum cost is $250 for burgers and $292 for sandwiches. It is greater for sandwiches than burgers because the approximate minimum cost is $100 for burgers and $295 for sandwiches. It is greater for burgers than sandwiches because the approximate minimum cost is $295 for burgers and $100 for sandwiches. It is greater for burgers than sandwiches because the approximate minimum cost is $292 for burgers and $250 for sandwiches.
so how would I solve it ?
When I modelled the burgers I got C(burger) = [(burgers sold)^2 - 30(burgers sold)] + $520
this is very similar to the equation for the sandwiches
now, you can put these under each other and solve: C(x) = 2x^2 -40x +30 C(x) = x^2 - 30x +520
now these are two quafratic equations , and will intersect each other
do I solve for x?
or do I use the quadratic formula?
are you still there @BPDlkeme234 ?
do I solve for x or do I use the quadratic formula to solve?