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3 years ago

Two angles are supplementary.The larger angle is 15 more than 10 times the smaller angle. Find the measure of each angle

Mathematics

1 answer:

DaniilM [7]3 years ago

4 0

<h2>The smaller angle = 15° and The larger angle = 165° </h2>

Step-by-step explanation:

Let the smaller angle = x and

The larger angle = 10x + 15°

To find, the measure of each angle = ?

We know that,

The sum of two supplementary anges = 180°

∴ x + (10x + 15°) = 180°

⇒ x + 10x + 15° = 180°

⇒ 11x = 180° - 15°

⇒ 11x = 165°

Dividing both sides by 11, we get

$\dfrac{11x}{11} =\dfrac{165}{11}$

⇒ x = 15°

∴ The smaller angle = 15° and

The larger angle = 10(15° ) + 15° = 165°

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Step-by-step explanation:

<h3>Appropriate Question :-</h3>

Find the limit

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$\large\underline{\sf{Solution-}}$

Given expression is

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On substituting directly x = 1, we get,

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Consider again,

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

can be rewritten as

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 3x + 2)}\right]$

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$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 2 - 1)(x - 2 + 1)}{x(x - 2) \: (x - 1))}\right]$

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$\rm\implies \:\boxed{ \rm{ \:\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right] = 2 \: }}$

$\rule{190pt}{2pt}$

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Read 2 more answers

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Step-by-step explanation:

Given

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3 years ago