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ludmilkaskok [199]
3 years ago
11

Answer the equation 1/4(2(x-1)+10=xI will mark first answer brainliest!

Mathematics
1 answer:
ElenaW [278]3 years ago
3 0

Answer:

Step-by-step explanation:1/4(2(x-2)+10=x

1/4(2x-2)+10=x

1/2x-1/2+10=x

1/2x-x=-10+1/2

-1/2x=-8

X=16

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

Step-by-step explanation:

<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>

<em>            cos²∝ = 1-sin²∝</em>

<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>

<em>                                = sin π - sin 0</em>

<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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2 years ago
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zhuklara [117]

Answer:

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Step-by-step explanation:

hope this helps :)

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