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3 years ago

Find the circumference of a circle with each 18.2cm diameter

Mathematics

1 answer:

e-lub [12.9K]3 years ago

6 0

Answer:

22/7×18.2

=400.4/7

=57.2cm

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Step-by-step explanation:

The value of f(2) is -5.

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3 years ago

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A town is planning a playground. It wants to fence in a rectangular space using an existing wall. What is the greatest area it c

777dan777 [17]

Answer:

1250 $ft^{2}$

Step-by-step explanation:

We are given that the playground is fenced on three sides of the playground and the four side has an existing wall.

Let the length of the rectangle be 'X' feet and width be 'Y' feet.

As the fencing is done using 100 feet of fence. We get the relation between the sides and the fence as, X + 2Y = 100.

As, X + 2Y = 100 → X = 100 - 2Y

Now, the area of the rectangle = XY = X × ( 100 - 2Y ).

i.e Area of the rectangle = $-2Y^{2} +100Y$ .

The general form of a quadratic equation is $y=ax^{2}+bx+c$ .

The maximum value of a quadratic equation is given by $x=\frac{-b}{2a}$ .

Therefore, the greatest value of $-2Y^{2} +100Y$ is at Y = $\frac{-100}{2 \times -2}$ = $\frac{-100}{-4}$ = 25.

Thus, Y = 25 and X = 100 - 2Y → X = 100 - 2 × 25 → X = 50.

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3 years ago

Please answer the question attached, marking brainliest (serious answers only)

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Step-by-step explanation:

By Exterior Angle Theorem,

81° + b = 117°.

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Rationalise the denominator of:<br>1/(√3 + √5 - √2)

Paul [167]

Step-by-step explanation:

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

Given expression is

$\rm :\longmapsto\:\dfrac{1}{ \sqrt{3} + \sqrt{5} - \sqrt{2} }$

can be re-arranged as

$\rm :\longmapsto\:\dfrac{1}{ \sqrt{3} - \sqrt{2} + \sqrt{5} }$

$\rm \: = \: \dfrac{1}{( \sqrt{3} - \sqrt{2} ) + \sqrt{5} }$

On rationalizing the denominator, we get

$\rm \: = \: \dfrac{1}{( \sqrt{3} - \sqrt{2} ) + \sqrt{5} } \times \dfrac{( \sqrt{3} - \sqrt{2} ) - \sqrt{5} }{( \sqrt{3} - \sqrt{2} ) - \sqrt{5} }$

We know,

$\rm :\longmapsto\:\boxed{\tt{ (x + y)(x - y) = {x}^{2} - {y}^{2} \: }}$

So, using this, we get

$\rm \: = \: \dfrac{ \sqrt{3} - \sqrt{2} - \sqrt{5} }{ {( \sqrt{3} - \sqrt{2} )}^{2} - {( \sqrt{5}) }^{2} }$

$\rm \: = \: \dfrac{ \sqrt{3} - \sqrt{2} - \sqrt{5} }{3 + 2 - 2 \sqrt{6} - 5}$

$\rm \: = \: \dfrac{ \sqrt{3} - \sqrt{2} - \sqrt{5} }{5 - 2 \sqrt{6} - 5}$

$\rm \: = \: \dfrac{ \sqrt{3} - \sqrt{2} - \sqrt{5} }{ - 2 \sqrt{6}}$

$\rm \: = \: \dfrac{ - ( - \sqrt{3} + \sqrt{2} + \sqrt{5}) }{ - 2 \sqrt{6}}$

$\rm \: = \: \dfrac{- \sqrt{3} + \sqrt{2} + \sqrt{5}}{2 \sqrt{6}}$

On rationalizing the denominator, we get

$\rm \: = \: \dfrac{- \sqrt{3} + \sqrt{2} + \sqrt{5}}{2 \sqrt{6}} \times \dfrac{ \sqrt{6} }{ \sqrt{6} }$

$\rm \: = \: \dfrac{- \sqrt{18} + \sqrt{12} + \sqrt{30}}{2 \times 6}$

$\rm \: = \: \dfrac{- \sqrt{3 \times 3 \times 2} + \sqrt{2 \times 2 \times 3} + \sqrt{30}}{12}$

$\rm \: = \: \dfrac{- 3\sqrt{2} + 2 \sqrt{3} + \sqrt{30}}{12}$

Hence,

$\boxed{\tt{ \rm \dfrac{1}{ \sqrt{3} + \sqrt{5} - \sqrt{2} } =\dfrac{- \sqrt{3 \times 3 \times 2} + \sqrt{2 \times 2 \times 3} + \sqrt{30}}{12}}}$

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<h3><u>More Identities to </u><u>know:</u></h3>

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$\purple{\boxed{\tt{ {(x + y)}^{3} = {x}^{3} + 3xy(x + y) + {y}^{3}}}}$

$\purple{\boxed{\tt{ {(x - y)}^{3} = {x}^{3} - 3xy(x - y) - {y}^{3}}}}$

$\pink{\boxed{\tt{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}}}$

$\pink{\boxed{\tt{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}}}$

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3 years ago

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valentinak56 [21]

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Find the circumference of a circle with each 18.2cm diameter​

Find the circumference of a circle with each 18.2cm diameter