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Aleks [24]
3 years ago
9

How much taller was Mt. Everest determined to be in 1999, than in 1852? 11 meters 12 meters 22 meters 147 meters

Mathematics
1 answer:
katen-ka-za [31]3 years ago
7 0

Answer:

11 meters

Step-by-step explanation:

In 1852 its height was 29.002ft or 8.839m, while the measuring in 1999 said that it was 29.035ft or 8.850m. That means that it is taller by some 11 meters or 33 feet. That is because it keeps increasing its size since its positioned on such plates that hit each other.

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Q.6. The equation of the ellipse whose centre is at the origin and the x-axis, the major axis, which passes
azamat

<h3>Answer:</h3>

Equation of the ellipse = 3x² + 5y² = 32

<h3>Step-by-step explanation:</h3>

<h2>Given:</h2>

  • The centre of the ellipse is at the origin and the X axis is the major axis

  • It passes through the points (-3, 1) and (2, -2)

<h2>To Find:</h2>

  • The equation of the ellipse

<h2>Solution:</h2>

The equation of an ellipse is given by,

\sf \dfrac{x^2}{a^2} +\dfrac{y^2}{b^2} =1

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\sf \dfrac{(-3)^2}{a^2} +\dfrac{1^2}{b^2} =1

Cross multiplying we get,

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Multiply by 4 on both sides,

  • 4a²b² = 36b² + 4a²------(1)

Also by given the ellipse passes through the point (2, -2)

Substituting this,

\sf \dfrac{2^2}{a^2} +\dfrac{(-2)^2}{b^2} =1

Cross multiply,

  • 4b² + 4a² = 1 × a²b²
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Subtracting equations 2 and 1,

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Substituting in 2,

  • 32/3 × b² = 4b² + 4 × 32/3
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Substituting the values in the equation for ellipse,

\sf \dfrac{x^2}{32/3} +\dfrac{y^2}{32/5} =1

\sf \dfrac{3x^2}{32} +\dfrac{5y^2}{32} =1

Multiplying whole equation by 32 we get,

3x² + 5y² = 32

<h3>Hence equation of the ellipse is 3x² + 5y² = 32</h3>
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Step-by-step explanation:

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