The quarterback of a football team releases a pass at a height of 6 feet above the playing field, and the football is caught by a receiver 78 yards directly downfield at a height of 4 feet. The pass is released at an angle of 75 degrees with the horizontal. Find the time in seconds the receiver has to reach the proper position after the quarterback releases the football. Round your answer to one decimal place.

t = 25.1seconds

Step-by-step explanation:

X(t)= 16cos75°t

Y(t)= 6 + (Vosin75°t - 16t^2)

Converting 78 yards to feet :

1 yard = 3 feet

78feet =?

78×3=234 feet

Y(t)= number of feet above the ground at t seconds

VoCos75° t = 234

t = 234/(VoCos75°)

At this time,4 = y(234/VoCos75°)

4 = 6 + VoSin75°(234/VoCos75) - 16(234/VoCos75°)^2

Vo= 140.66ft/s

Time,t= 234/(140.66× Cos75°)

t = 25.087seconds

t= 25.1 seconds (1 decimal place)

d

step-by-step explanation:

by observation, we can see that the the graph is simply the graph of y = |x| that has been shifted 4 units downwards and 3 units to the left.

we can recognize in choice d :

y = | x + 3 |   - 4 (rewriting in general form)

y = | x - (- 3) |   - 4

the "-4 " term means   that the graph has been shifted by 4 units in the negative y direction

the "-3 " term means   that the graph has been shifted by 3 units in the negative x direction

this matches our observation of the graph described above.