Answer:
length = 200 m
width = 400 m
Step-by-step explanation:
Let the length of the plaing area is L and the width of the playing area is W.
Length of fencing around three sides = 2 L + W = 800
W = 800 - 2L ..... (1)
Let A is the area of playing area
A = L x W
A = L (800 - 2L)
A = 800 L - 2L²
Differentiate with respect to L.
dA/dL = 800 - 4 L
It is equal to zero for maxima and minima
800 - 4 L = 0
L = 200 m
W = 800 - 2 x 200 = 400 m
So, the area is maximum if the length is 200 m and the width is 400 m.
Answer:
isosceles :)
Step-by-step explanation:
V^2/(1-v^2/c^2)=R
v^2=R(1-v^2/c^2)
v^2=R-Rv^2/c^2
v^2-Rv^2/c^2=R
v^2(1-R/c^2)=R
v=sqrt(R/(1-R/c^2))
where R was original right side, dont forget plus minus
Answer:
17.07m/s
Step-by-step explanation:
v=d/t
d=956km
t=56sec
v=956/56
v=17.07m/s
I thought it was the one above it but I think you are right tho
