Technology can be used in a wide variety of constructions. When it comes to math, for example, you can use a calculator. Hope this helps
Answer:
0.128rad/sec
Step-by-step explanation:
Let x represent the between the man and the point on the path
θ = the angle
dx/dt = 4 ft/s
dθ/dt = rate is the searchlight rotating when the man is 15 ft from the point on the path closest to the searchlight
tan θ = x/20 ft
Cross Multiply
20tan θ = x
dx/dt = 20sec² θ dθ/dt
dθ/dt = 1/20 × cos² θ dx/dt
dθ/dt= 1/20 × cos² θ × 4
dθ/dt = 1/5 × cos² θ
Note : cos θ = 4/5
dθ/dt = 1/5 × (4/5)²
dθ/dt = 16/125
dθ/dt = 0.128rad/sec
4/8 simplified to 1/2 because out of eight there are 4 even and 4 odd numbers 0.50
What we know:
Football field is 100 yards in length
End zones are 10 yards each in length
Perimeter between pylons is 306 2/3 yards
What we need to find:
a. Perimeter and area of one end zone
b. Perimeter and area of without end zones
c. Perimeter and area of playing field with end zones
First we need to find the measurements of the field between pylons using the perimeter of 306 2/3 yards. We already know the length is 100 yards so we need to find width (w).
P=2l + 2w
306 2/3=2 (100) + 2w
306 2/3= 200 + 2w
300 2/3-200=200-200+2w
106 2/3=2w
106 2/3/2=2/2w
53 1/3=w
a. Perimeter=2 (10)+2 (53 1/3)=126 2/3 yards
Area=10×53 1/3=533 1/3 yd²
b. Perimeter=2 (100)+2 (53 1/3)=306 2/3 yards
Area=100×53 1/3=5333 1/3 yd²
c. Perimeter=2 (120) + 2 (53 1/3)=346 2/3 yards
Area=120×53 1/3=6400 yd²
