The given equation for the relationship between a planet's orbital period, T and the planet's mean distance from the sun, A is T^2 = A^3.
Let the orbital period of planet X be T(X) and that of planet Y = T(Y) and let the mean distance of planet X from the sun be A(X) and that of planet Y = A(Y), then
A(Y) = 2A(X)
[T(Y)]^2 = [A(Y)]^3 = [2A(X)]^3
But [T(X)]^2 = [A(X)]^3
Thus [T(Y)]^2 = 2^3[T(X)]^2
[T(Y)]^2 / [T(X)]^2 = 2^3
T(Y) / T(X) = 2^3/2
Therefore, the orbital period increased by a factor of 2^3/2
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Answer:
radius=2, height=3
Step-by-step explanation:
diameter= 2x the radius therefore 4/2=2=radius
cylinder volume formula=πr^2h
volume=12π
height= v/πr^2=12π/π*2^2=12π/π*2^2=12π/4π=3
hope this helped!
Answer:
oop hi
Step-by-step explanation:
Answer:
B I guess is the answer
Step-by-step explanation:
I'm not really sure
Answer:
y = x + 6
x = 1
y = ¼(x - 5) + 3
Step-by-step explanation:
Vetices are;
A(1,5), B(5,3) and C(-3, -2)
Thus;
Median of AB is; D = (1 + 5)/2, (5 + 3)/2
D = (3, 4)
Median of BC is; E = (5 + (-3))/2, (3 + (-2))/2
E = (1, 0.5)
Median of AC is; F ; (-3 + 1)/2, (-2 + 5)/2
F = (-1, 1.5)
Thus, the median lines will be;
CD, AE & BF.
Thus;
Equation of CD is;
(y - (-3))/(x - (-2)) = (-2 - 4))/(-3 - 3)
(y + 4)/(x + 2) = -6/-6
y - 4 = 1(x + 2)
y = 4 + x + 2
y = x + 6
Equation of AE;
(y - 5)/(x - 1) = (0.5 - 5)/(1 - 1)
(y - 5)/(x - 1) = -4.5/0
Cross multiply to get;
0(y - 5) = -4.5(x - 1)
-4.5x = -4.5
x = 1
Equation of BF;
(y - 3)/(x - 5) = (1.5 - 3)/(-1 - 5)
(y - 3)/(x - 5) = -1.5/-6
(y - 3)/(x - 5) = 1/4
y - 3 = ¼(x - 5)
y = ¼(x - 5) + 3
