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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Hello!
The problem has asked that we simplify the given expression using the Distributive Property. The Distributive Property states the following:
a(b + c) = ab + ac
a(b – c) = ab – ac
In this case, we’ll use the second of the two formulas listed above. Let’s begin by inserting any known values given in the original problem:
a(b – c) = ab – ac
5(2x – 5) = 5(2x) – 5(5)
Now simplify the right side of the equation:
5(2x) – 5(5) = 10x – 25
We have now proven that 5(2x – 5) is equal to 10x – 25. Therefore, the correct answer is C.
I hope this helps!
Answer:
1.16 in
2.24 in
3.12.8 in
Step-by-step explanation:
We are given that
Width of photo=5 in
Length of photo=8 in
We have to find the length of new photo in each case.
1. Let x represent the width and y represents the length of photo.
When width increase then length of photo is also increases then it is direct proportion.
Substitute the values then we get
Hence, the length of photo=16 in
2.Width=x=15 in
Again using the formula
Hence, the length of new photo=24 in
3. Width =x=8 in
Substitute the values in the given formula
Hence, the length of new photo=12.8 in
<h2>y = -0.25x + 2</h2><h3></h3><h3><em>Please let me know if I am wrong.</em></h3>
