कोलकाता मे कौन सा उत्सव धूमधाम से मनाया जाता है?​

Answer:

ANSWER IS IN PICTURE

Step-by-step explanation:

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Answer:

C = 90° and D = 125°

Step-by-step explanation:

Made a extended imaginary straight line from D

Now,

AB || CD

angle BAD = Angle ADE = 55° (alternate angles)

angle ADE + angle ADC = 180° (linear pair)

55 + ADC = 180

Angle ADC or D = 180-55=125°

Now

<A+<B+<C+<D = 360° (quadrilateral)

55+90+<C+125 = 360

<C = 360-270 = 90°

Answer:

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Answer:

Are ; starting

Explanation:

Please find the answer given above

Answer:

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Answer:

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Answer:

[tex]\qquad\qquad\boxed{ \sf{ \: \bf \: {a}^{2} + {b}^{2} = \:98 \: \: }} \\ \\ [/tex]

Step-by-step explanation:

Given that,

[tex]\sf \:a = 5 + 2 \sqrt{6} - - - (1) \\ \\ [/tex]

and

[tex]\sf \:b = \dfrac{1}{a} \\ \\ [/tex]

[tex]\sf \:b = \dfrac{1}{5 + 2 \sqrt{6} } \\ \\ [/tex]

On rationalizing the denominator, we get

[tex]\sf \:b = \dfrac{1}{5 + 2 \sqrt{6} } \times \dfrac{5 - 2 \sqrt{6} }{5 - 2 \sqrt{6} } \\ \\ [/tex]

[tex]\sf \:b = \dfrac{5 - 2 \sqrt{6} }{(5)^{2} - (2 \sqrt{6})^{2} } \\ \\ [/tex]

[tex]\boxed{ \sf{ \: \because \: (x + y)(x - y) = {x}^{2} - {y}^{2} \: }} \\ \\ [/tex]

[tex]\sf \:b = \dfrac{5 - 2 \sqrt{6} }{25 - 24} \\ \\ [/tex]

[tex]\sf \:b = \dfrac{5 - 2 \sqrt{6} }{1} \\ \\ [/tex]

[tex]\sf\implies \sf \:b = 5 - 2 \sqrt{6} - - - (2) \\ \\ [/tex]

Now, Consider

[tex]\bf \: {a}^{2} + {b}^{2} \\ \\ [/tex]

[tex]\sf \: = \: {(5 + 2 \sqrt{6})}^{2} + {(5 - 2 \sqrt{6}) }^{2} \\ \\ [/tex]

We know,

[tex]\qquad\boxed{ \sf{ \: {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2}) \: }} \\ \\ [/tex]

So, on substituting the values, we get

[tex]\sf \: = \: 2[ {(5)}^{2} + {(2 \sqrt{6}) }^{2}] \\ \\ [/tex]

[tex]\sf \: = \:2(25 + 24) \\ \\ [/tex]

[tex]\sf \: = \:2 \times 49 \\ \\ [/tex]

[tex]\sf \: = \:98 \\ \\ [/tex]

Hence,

[tex]\sf\implies \bf \: {a}^{2} + {b}^{2} = \:98 \\ \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Additional Information

[tex]\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: identities}}}} \\ \\ \bigstar \: \bf{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {x}^{2} - {y}^{2} = (x + y)(x - y)}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}\:\\ \\ \bigstar \: \bf{ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y) }\:\\ \\ \bigstar \: \bf{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}[/tex]

Answer:

Answer in photo

Step-by-step explanation:

Hope this helps you