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:
1
Step-by-step explanation:
Given triangle ABC with vertices at points A(2,-4), B(4,-4) and C(4,-2).
1. Rotate this triangle anticlockwise by 90° angle over the origin. This rotation has the rule
then
2. Reflect triangle A'B'C' across the y-axis. This reflection has the rule
then
These are vertices of triangle 1
Answer:
The fourth option, 5x + 11y
Given:
(7x + 8y) - (2x - 3y)
Distribute the negative:
7x + 8y - 2x + 3y
Reorder terms (please note you cannot always do this, but we can here):
7x - 2x + 8y + 3y
Combine like terms:
5x + 11y
Answer:
5x + 11y
Step-by-step explanation:
the third of 7 consecutive numbers is 12.
so, this looks like
..., ..., 12, ..., ..., ..., ...
that means the numbers have to be
10, 11, 12, 13, 14, 15, 16
the sum is then
16+10 + 15+11 + 14+12 + 13 = 3×26 + 13 = 91