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ExtremeBDS [4]
3 years ago
11

Squareroot of

bsmiddle" class="latex-formula">
​
Mathematics
2 answers:
olganol [36]3 years ago
5 0

The squareroot of 12 \frac14 is 3 \frac{1}{2}.

\large\mathfrak{{\pmb{\underline{\red{Step-by-step\:explanation}}{\orange{:}}}}}

\sqrt{12 \frac{1}{4} }  \\ \\  =  \sqrt{ \frac{49}{4} }  \\ \\ =  \sqrt{ \frac{7 \times 7}{2 \times 2} }  \\ \\ =  \sqrt{ \frac{( {7})^{2} }{( {2})^{2} } }  \\  \\=  \frac{7}{2}  \\ \\ = 3 \frac{1}{2}

\large\mathfrak{{\pmb{\underline{\orange{Happy\:learning }}{\orange{!}}}}}

icang [17]3 years ago
3 0

Answer:

7/2

Step-by-step explanation:

12 1/4 is equivalent to the improper fraction 49/4.

The square root of 49/4 is 7/2.

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<u><em>Step(i):-</em></u>

<em>Given that the definite integration</em>

<em>            </em>\int\limits^\pi _ {0} \,(\sqrt{1-sin^{2}\alpha  }   )d\alpha<em></em>

<em>we know that the trigonometric formula</em>

<em> sin²∝+cos²∝ = 1</em>

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<u><em>step(ii):-</em></u>

<em>Now the  integration</em>

<em>         </em>\int\limits^\pi _ {0} \,(\sqrt{1-sin^{2}\alpha  }   )d\alpha = \int\limits^\pi _0 {(\sqrt{cos^{2} \alpha } } \, )d\alpha<em></em>

<em>                                      = </em>\int\limits^\pi _0 {cos\alpha } \, dx<em></em>

<em>Now, Integrating </em>

<em>                                  </em>= ( sin\alpha )_{0} ^{\pi }<em></em>

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<em>                               = 0-0</em>

<em>                              = 0</em>

<u><em>Final answer:-</em></u>

<em>      </em>\int\limits^\pi _ {0} \,(\sqrt{1-sin^{2}\alpha  }   )d\alpha =0<em></em>

<em></em>

<em></em>

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