An object hangs from a spring at a stationary rest position 9 metres above the ground. The object is pulled 3 m downward and released. The object
begins to oscillate (move up and down), taking 4 seconds to first return
back to the stationary rest position before continuing upwards in its
movement. a) Graph the motion of the object over 2 full periods. (Make
sure any key points are labelled) (2 marks) b) Write an equation for the
motion of this object where h(t) represents the height of the object and t
represents the time in seconds. (2 marks) c) How far is the object from it's
rest position 1.5 seconds after release. (1 mark) d) When is the NEXT time
that the object will be at this point again? (2 marks). Show all work where
possible for full marks. *

x =-.20752

step-by-step explanation:

4^(x+2)=12

4^(x+2)=12

take the log base 4 on each side

log 4 (4^(x+2) = log 4(12)

we know log b(a^y) = y log b (a)

(x+2) log 4(4) = log 4 (12)

x+2 = log 4 (12)

we know   logb c = loga c/loga b

we want to convert to base 10   so a = 10

by default, we do not write the 10   c = 4   and b = 10

log4 10 = log 12/log 4

log4 10 = log 12/ log 4

x+2 = log 12/ log 4

subtract 2 from each side

x +2 -2 = log 12/ log 4 -2

x = 1.79248125 -2

x =-.20752

answer:   35 % is the probability of a late departure or an early arrival.

step-by-step explanation:

let l represents the late departure and e represents the early departure.

then, according to the question,

p(l) = 12 %,   p(e) = 27 % and = 4 %

and, we have to find out,

but, we know that,

which is the required equation that will use to find the probability of a late departure or an early departure.

therefore,   the probability of a late departure or an early arrival =   35 %