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3 years ago

Solve for d. | - 2d|<10

Mathematics

1 answer:

Romashka-Z-Leto [24]3 years ago

7 0

Your question: Find D in | - 2d|<10

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SOLVING....

Solve the absolute value

|−2d| < 10

−2d < 10

−2d > −10

Solve −2d < 10 for D

Divide each side by -2

-2d ÷ -2 < 10 ÷ -2

d > -5

Now Solve −2d > −10

Divide each side by -2

-2d ÷ -2 > -10 ÷ -2

d < 5

So D is d < 5 and d > -5

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A square has a side of 6.25. What is the area in square feet

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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)

The equation of the ellipse

<h2>Solution:</h2>

The equation of an ellipse is given by,

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

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

Hence,

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

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a²b² = 9b² + a²

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,

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Subtracting equations 2 and 1,

3a²b² = 32b²
3a² = 32
a² = 32/3----(3)

Substituting in 2,

32/3 × b² = 4b² + 4 × 32/3
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20b² = 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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