The height of a falling object is given by h(t) = 400 − 39t − 157e−t/4 with h in meters and t in seconds. assuming that the ground is at height h = 0, how fast is the object moving at the instant it hits the ground? your answer must be accurate to one decimal place.

This is the concept of algebra; given that the height of the house at time t is given by:
h(t)= 400 − 39t − 157e−t/4
the velocity of the house is given by:
dh/dt
velocity of our object is:
dh/dt=-39+1/4*157e^-(t/4)
=-39+157/4 e^(-t/4)
thus the velocity at h=0 will be as follows:
at h=0, the value of t will be:
0=400-39t-157e^(-t/4)
solving for t we get:
t=-5.4 or t=10.1
since we don't have negative time, t=10.1
thus the velocity at this point will be:
dh/dt=-38+1/4*157e^(-10.1/4)
=-38.86 sec
this means that the object dropped by a velocity of 38.86 sec

2,4,5,8,10 are divisible by 40

You need to include the original coordinates of the triangle to answer this. and how much does it translate (move left right up down)