How many unique ways are there to arrange the letter in the word PRIOR?
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1. The distance between the perihelion and the aphelion is 116 million miles
2. The distance from the center of Mercury’s elliptical orbit and the Sun is 12 million miles
3. The equation of the elliptical orbit of Mercury is
4. The eccentricity of the ellipse is 0.207 to the nearest thousandth
5. The value of the eccentricity tell you that the shape of the ellipse is near to the shape of the circle
Step-by-step explanation:
Let us revise the equation of the ellipse is
, where the major axis is parallel to the x-axis
- The length of the major axis is 2a
- The coordinates of the vertices are (± a , 0)
- The coordinates of the foci are (± c , 0) , where c² = a² - b²
∵ The Sun is located at a focus of the ellipse
∴ The sun located ate c
∵ The perihelion is the point in a planet’s orbit that is closest to the
Sun ( it is the endpoint of the major axis that is closest to the Sun )
∴ The perihelion is located at the vertex (a , 0)
∵ The closest Mercury comes to the Sun is about 46 million miles
∴ The distance between a and c is 46 million miles
∵ The aphelion is the point in the planet’s orbit that is furthest from
the Sun ( it is the endpoint of the major axis that is furthest from
the Sun )
∴ The aphelion is located at the vertex (-a , 0)
∵ The farthest Mercury travels from the Sun is about 70 million miles
∴ The distance from -a to c is 70 million miles
∴ The distance between the perihelion and the aphelion =
70 + 46 = 116 million miles
1. The distance between the perihelion and the aphelion is 116 million miles
∵ The distance between the perihelion and the aphelion is the
length of the major axis of the ellipse
∵ The length of the major axis is 2 a
∴ 2 a = 116
- Divide both sides by 2
∴ a = 58
∴ The distance from the center of Mercury’s elliptical orbit to the
closest end point to the sun is 58 million miles
∵ The distance between the sun and the closest endpoint is
46 million miles
∴ The distance from the center of Mercury’s elliptical orbit and
the Sun = 58 - 46 = 12 million miles
2. The distance from the center of Mercury’s elliptical orbit and the Sun is 12 million miles
∵ The major axis runs horizontally
∴ The equation is
∵ a = 58
∵ c is the distance from the center to the focus of the ellipse
∴ c = 12
∵ c² = a² - b²
∴ (12)² = (58)² - b²
- Add b² to both sides
∴ (12)² + b² = (58)²
- Subtract (12)² from both sides
∴ b² = (58)² - (12)² = 3220
- Substitute these values in the equation
∴
3. The equation of the elliptical orbit of Mercury is
The eccentricity (e) of an ellipse is the ratio of the distance from the
center to the foci (c) and the distance from the center to the
vertices (a) ⇒
∵ c = 12
∵ a = 58
∴ = 0.207
4. The eccentricity of the ellipse is 0.207 to the nearest thousandth
If the eccentricity is zero, it is not squashed at all and so remains a circle.
If it is 1, it is completely squashed and looks like a line
∵ The eccentricity of the ellipse is 0.207
∵ This number is closed to zero than 1
∴ The shape of the ellipse is near to the shape of the circle
5. The value of the eccentricity tell you that the shape of the ellipse is near to the shape of the circle
Learn more:
You can learn more about conics section in brainly.com/question/4054269
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Answer:
49.5 square inches
Step-by-step explanation:
We can split the figure into two easier shapes to find the area of; a triangle and a square.
The formula for a triangle is (b*h)1/2. Once we solve we get 24.5 or 24 1/2.
The formula for a square is l*w. Once we solve we get 25.
To find the area of the shaded region, we would add both area and get 49.5 or 49 1/2 square inches.
