<span>$140.25. Formula, I=Prt. Solution, I=$5000 x 8.5% x 120/360. I=5000x8.5%x0.33. I=$140.25 </span>
Answer:
1/3
Step-by-step explanation:
pic bleow
also can u help or someone
draw three cells in parallel, two light bulbs (loads) in series, with one switch.
6/7 in fraction
0.85714285 in decimal form
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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Your friend Papaguy.