If an object is propelled upward from a height of s feet at an initial velocity of v feet per second, then its height h after t seconds is given by the equation h=-16t^2 +vt+ s , where h is in feet. if the object is propelled from a height of 8 feet with an initial velocity of 64 feet per seconds , it's height h is given by the equation h=-16t^2+64t+8 after how many seconds is the height 68 feet? the time is .

At 1.5 seconds and 2.5 seconds

Step-by-step explanation:

Because this is parabolic motion, the object will reach the height of 68 feet when it's going up AND when it's coming down.  In order to find out those times, you set the h on the left side of the equation equal to 68, because h stands for height.

Set that equal to 0 and factor to solve for t:

Using the quadratic formula, we get t = 1.5 and t = 2.5

a) 0

step-by-step explanation:

i^413 x i^0 = 0 as any number multiplied by a zero is

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asnwer is (g o h o f)(x) = 2x - 6/x + 7

step-by-step explanation:

given,

f(x) = x - 3/x

g(x) = x + 3

h(x) = 2x + 1

we need to find : (g o h o f)(x)

we can write (g o h o f)(x) = g[ h{ f(x) } ]

so first we find h{f(x)}

h{f(x)} = f(x) + 3 = (x - 3/x) + 3

h{f(x)} = x - 3/x + 3

now we find g[ h{ f(x) } ].

g[ h{ f(x) } ] = g(x - 3/x + 3)       {plug h{f(x)} = x - 3/x + 3 in g[h{f(x)}] }

g(x - 3/x + 3) = 2 (x - 3/x + 3) + 1

                    = 2x - 6/x + 6 + 1

                    = 2x - 6/x + 7

hence we get (g o h o f)(x) = 2x - 6/x + 7