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2 years ago

(T 4 Marks + C 1 Marks)

Mathematics

1 answer:

bija089 [108]2 years ago

4 0

Answer:

<h2>length= 40cm</h2><h2>width= 120cm</h2><h2>Step-by-step explanation:</h2>

Area of a rectangle= lenght× width

Dimensions of 1st rectangle: L,W

Dimensions of 2nd rectangle: L+10; W+4

Hence,

LW = 480__________(1)

(L-10)(W+4) = 480_______(2)

LW = LW+4L - 10W - 40

4L = 10W +40

2L = 5W + 20

L = (5/2)W + 10_________(3)

But LW = 480 (from equation one)

So, W = 480/L

substitute as= 480/L into equation (3)

L = (5/2)(480/L)+10

L = 1200/L + 10

L^2 - 10L - 1200 = 0

factorising.

we're have,

(L-40)(L+30) = 0

therefore, since L can't be negative , L = 40

L = 40 cm (length)

Then 480/L = W = 120 cm (width)

<h2>therefore L= 40cm and W= 120cm</h2>

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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)

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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,

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
32/3 b² = 4b² + 128/3
32/3 b² = (12b² + 128)/3
32b² = 12b² + 128
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

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3 years ago

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Answer:

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Step-by-step explanation:

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