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Sindrei [870]
3 years ago
11

What is the solution to this equation? 6+7(b−1)=5+b

Mathematics
2 answers:
Ratling [72]3 years ago
6 0
6 + 7b - 7 = 5 + b
7b - 1 = 5 + b
7b = 5 + b + 1
7b = b + 6
7b - b = 6
6b = 6
b = 1

Answer: B)
Mariana [72]3 years ago
5 0
I need points and the answer is B
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Answer:

a) True

<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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Answer:

d

Step-by-step explanation:

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