The equation t^2 = a^3 shows the relationship between a planet’s orbital period, t, and the planet’s mean distance from the sun, a, in astronomical units, au. if the orbital period of planet y is twice the orbital period of planet x, by what factor is the mean distance increased? a 2^1/3 b 2^1/2 c 2^2/3 d 2^3/2

Hello, Your answer is below.

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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step-by-step explanation:

once you graph it hit 2nd (the blue button) and graph

that takes you to a table of coordinates

on the left is x coordinates and on the right is y coordinates and an empty box

(you may have to scroll up/down in order to get ( x, 0)

∠lno ≅ ∠lnm

∠oln ≅ ∠mln

step-by-step explanation:

given   that triangle lno is congruent to triangle lnm.

we have to fill up the blanks in the given question.

according to above order of congruency   kept in mind we can say that

∠lno ≅ ∠lnm

∠oln ≅ ∠mln

ln ≅ ln due to common in both.

hence, by sas congruency,

the side angle side rule   states that if two sides and the included angle of one triangle are congruent to another triangle, then these two triangles are congruent.