Step 1; To solve the question we need two variables. P which represents the number of years a planet takes to complete a revolution around the Sun. This is given as 13.2 years in the question so P = 13.2 years. The other variable is the distance between the planet and the sun in astronomical units. We need to determine the value of this using the given equation.

Step 2; So we have to calculate the value of 'a' in Kepler's equation. But the exponential power $\frac{3}{2}$ is on the variable we need to find so we multiply both the sides by an exponential power of $\frac{2}{3}$ to be able to calculate 'a'.

P = $a^{\frac{3}{2} }$ , $P^{\frac{2}{3} }$ = $a^{\frac{3}{2}*\frac{2}{3} }$ , $P^{\frac{2}{3} }$ = a, $13.2^{\frac{2}{3} }$ = a = 5.58533 astronomical units.

Rounding it over to nearest hundredth we get 5.59 astronomical units.

6 0

3 years ago

Lamar is writing a coordinate proof to show that a segment from the midpoint of the hypotenuse of a right triangle to the opposi

Zanzabum

1. N is a midpoint of the segment KL, then N has coordinates

$\left(\dfrac{x_K+x_L}{2},\dfrac{y_K+y_L}{2} \right) =\left(\dfrac{0+2a}{2},\dfrac{2b+0}{2} \right) =(a,b).$

2. To find the area of △KNM, the length of the base MK is 2b, and the length of the height is a. So an expression for the area of △KNM is

$A_{KMN}=\dfrac{1}{2}\cdot \text{base}\cdot \text{height}=\dfrac{1}{2}\cdot 2b\cdot a=ab.$

3. To find the area of △MNL, the length of the base ML is 2a and the length of the height is b. So an expression for the area of △MNL is

$A_{MNL}=\dfrac{1}{2}\cdot \text{base}\cdot \text{height}=\dfrac{1}{2}\cdot 2a\cdot b=ab.$

4. Comparing the expressions for the areas you have that the area $A_{KMN}$ is equal to the area $A_{MNL}$ . This means that the segment from the midpoint of the hypotenuse of a right triangle to the opposite vertex forms two triangles with equal areas.

6 0

2 years ago

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