Answer:
If u hook t up
Step-by-step explanation:
on google u find it :)
Answer:
B. 21
Step-by-step explanation:
The fractions are there to just make the problem seem complicated, but just rewrite it without the fractions.
"What is a common multiple for 7 and 3?"
In this problem it's easy, all you have to do is guess and check.
A: A doesn't work. 7 is not a factor of 18.
B: B works! Both 3 and 7 are factors of 21.
C: C doesn't work. 7 is not a factor of 3.
D: D doesn't work. Neither are a factor of 4.
They only leave us with one solution, B.
Volume a sphere: [4/3]π(r^3)
Space between the spheres = Volume of the larger sphere - Volume of the smaller sphere
= [4/3]π (R^3) - [4/3π](r^3) = [4/3]π(R^3 - r^3) = [4/3]π {(5cm)^3 - (4cm)^3} =
= 255.5 cm^3
Answer: 255.5 cm^3
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.
