Answer:
Original position =6,800 feet.
800 (ft per hour) x 6(hours) =4800 ft
11,600-4,800=6,800 feet
Step-by-step explanation:
Answer:
50.265448
Step-by-step explanation:
Answer: A picture of the graph would help a ton but I'm pretty sure its "<"
Answer:
Receiver path:
x-5 = t(0)
y-50 = -10t
defender path:
x-10 = -0.9t
y-54 =-10.72t
Yes, he will reach the goal line (y = 0) before the defensive player catches him.
Step-by-step explanation:
Vector equation: (x-x1, y-y1) = t(a1, a2)
Horizontal component : x-x1= ta1
Vertical component: y-y1= ta2
(x-5, y-50) = t(0, -10)
x-5 = t(0)
y-50 =t(-10)
Parametric equation for the receiver path:
x-5 = t(0)
y-50 = -10t
(x - 10, y - 54) = t(-0.9, -10.72)
x - 10 = t(-0.9)
y - 54 = t(-10.72)
Parametric equation for the defender path:
x-10 = -0.9t
y-54 =-10.72t
At the 50yard line, using the receiver parametric equation for vertical component:
y-50 = -10t
At y= 50
50-50 = -10t
0= -10t
t= 0/-10 = 0
At y= 0
0-50 = -10t
t = -50/-10 = 5
Defender at y = 50
y-54 =-10.72t
50-54 =-10.72t
t = -4/-10.72 = 0.37
at y = 0
0-54 =-10.72t
t = -54/-10.72 = 5.04
t of receiver < t of defender
Since time of receiver at y=0 is less than time of defender, he will reach the goal line (y = 0) before the defensive player catches him.
Answer:
2990.54
Step-by-step explanation:
The answer would be 2990.54.