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boyakko [2]
2 years ago
12

PLEASE HELP

Mathematics
2 answers:
Kipish [7]2 years ago
8 0

Answer:

m=0.7404, b=3.9117m=0.7404,b=3.9117

y=0.7404x+3.9117y=0.7404x+3.9117

x=18x=18

y=0.7404(18)+3.9117=17.2389y=0.7404(18)+3.9117=17.2389

Residual

16-17.2389= -1.238916−17.2389=−1.2389

-

lord [1]2 years ago
3 0
The residual value for x = 18 is -1.2389
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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

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<h2>To Find:</h2>

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<h2>Solution:</h2>

The equation of an ellipse is given by,

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

Given that the ellipse passes through the point (-3, 1)

Hence,

\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,

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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,

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Subtracting equations 2 and 1,

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  • 3a² = 32
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Substituting in 2,

  • 32/3 × b² = 4b² + 4 × 32/3
  • 32/3 b² = 4b² + 128/3
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  • 32b² = 12b² + 128
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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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Answer:

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Step-by-step explanation:

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When you add a and b then divide by a, it will be the same as a divided by b. This will hold true only for specific lengths of a and b. This means you must divide the line segment in such a way that a and b meet this requirement.

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