Um I think if you multiply the amount of hersheys by the ribbon you would get youre answer
Hoped i helped i couldn't see the number of ribbons
v₀ = initial velocity of the freight train while it approach a road crossing = 16 km/h = 16 (5/18) m/s = 4.44 m/s
v = final velocity of the freight train after it crosses a road crossing = 65 km/h = 65 x 5/18 m/s = 18.06 m/s
t = time to do so = 10 min = 10 x 60 sec = 600 sec
acceleration is given as
a = (v - v₀ )/t
a = (18.06 - 4.44)/(600)
a = 0.023 m/s²
a = 294
Answer:
I recommend using a graph that can help. For slope afind two solid points and make it a fraction. Y intercept is where it passes through the y axis.
Step-by-step explanation:
Answer:
Maximum area possible
f(max) = 3906,25 ft²
Dimensions:
a = 62,5 ft
w = 62,5 ft
Step-by-step explanation:
Perimeter of the rectangular fencing P = 250 feet
And sides of the rectangle a and w (width of rectangle)
Then
A = a*w
2a + 2w = 250 ⇒ a = (250 -2w)/ 2 ⇒ a = 125 - w
f(w) = (125 - w ) *w f(w) = 125w - w²
Taking derivatives both sides of the equation
f´(w) = 125 - 2w f´(w) = 0 125 - 2w = 0
w = 125/2
w = 62,5 ft ⇒ a = 125 - 62,5
a = 62,5 ft
f(max) = ( 62,5)²
f(max) = 3906,25 ft²
6 1/6, 5.5, 4 3/8, 4.3, 15/4
