One business set up security cameras to discourage burglars. The cameras are set beside a path, and can record anything that is up to 30m away. The cameras are set to track movement and have a pan of 90 degrees. If a typical burglar walks at about 2m per second, how much footage will each camera capture as they walk along the path?.

answer: it is 1: 47 pm

f(x)=2/3x-8, which give g(x) the greatest value.

step-by-step explanation:

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