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makkiz [27]
3 years ago
14

What's the value for x makes this equation true 3/5 X-0.5=19 ​

Mathematics
1 answer:
masya89 [10]3 years ago
7 0

Answer:

what's the value for x makes this equation true 3/5 X-0.5=19 ​  HERE YOU GO!!

Step-by-step explanation:

-5x - (-7 - 4x) = -2(3x - 4)

 

-5x + 7+ 4x = -6x + 8

 

-x + 7= -6x + 8

 

5x = 8 - 7

 

5x = 1

 

x = 1/5

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Solve the equation. Check the solution.<br> 6/7x= - 18
dexar [7]

Answer:

x= -21

Step-by-step explanation:

6/7x=-18

divide both sides by 6/7

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x= -21

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Mrrafil [7]

\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}

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so, major angle P = 60°

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A bank offers a savings account that currently pays 2% interest per year compounded monthly. The amount of money in the account
LekaFEV [45]
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6 0
3 years ago
WHO CAN solve it Please !
Mariulka [41]

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>

5 0
2 years ago
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liubo4ka [24]

Answer:

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5 0
2 years ago
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