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.

Thank You.

~

Your friend Papaguy.

step-by-step explanation:

the area of a triangle is determined by the formula: base x height * 1/2

you'll need to assign a base to the triangle. based on the vertices given, the simplest base would be the line segment from (-2,1) to (2,1). the distance between any two numbers is found by just subtracting them. here, you should subtract the x values to find the distance of this side.

2 - (-2) = 4.

so the distance (aka the length of the base) is 4. the height of the triangle is the distance between the base and the highest point on the triangle which is (3,4). we should subtract the y values this time to find the height/distance.

4 - 1 = 3

so the height is 3.

base = 4

height = 3

hope this if needed hmu