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anonymous

  • one year ago

Need help, part and c http://prntscr.com/8cwxa2

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  1. anonymous
    • one year ago
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    @Michele_Laino

  2. Michele_Laino
    • one year ago
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    question b) the area of the right triangle, is: \[\Large \frac{{AC \cdot BC}}{2} = \frac{{B{C^2}}}{2}\] Using the theorem of Pitagora, we have this: \[\Large r = \frac{{BC}}{{\sqrt 2 }}\] where \( r \) is the radius of the half-circumference. So the requested area is: \[\Large \begin{gathered} A = - \frac{{B{C^2}}}{2} + \frac{{\pi {r^2}}}{2} = - \frac{{B{C^2}}}{2} + \frac{{\pi B{C^2}}}{4} = \hfill \\ \hfill \\ = \frac{{B{C^2}}}{2}\left( {\frac{\pi }{2} - 1} \right) \hfill \\ \end{gathered} \]

  3. Michele_Laino
    • one year ago
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    oops... I have made a typo: \[\Large \frac{{AB \cdot BC}}{2} = \frac{{B{C^2}}}{2}\]

  4. anonymous
    • one year ago
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    14cm^2

  5. anonymous
    • one year ago
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    part c also.

  6. Michele_Laino
    • one year ago
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    we have to write the equation of your line, in order to do that, we can apply this equation: \[\Large \frac{{y - {y_1}}}{{{y_2} - {y_1}}} = \frac{{x - {x_1}}}{{{x_2} - {x_1}}}\] what do you get?

  7. anonymous
    • one year ago
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    @Michele_Laino

  8. anonymous
    • one year ago
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    idk...

  9. Michele_Laino
    • one year ago
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    please try, you have to substitute the coordinates of your points into my formula above: (x1,y1)=(-4,6), and (x2, y2) = (8,-3)

  10. anonymous
    • one year ago
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    x and y ???

  11. Michele_Laino
    • one year ago
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    x, and y are the variables, you have only to susbstitute x1, x2, y1, y2 with the coordinates of your points

  12. anonymous
    • one year ago
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    1 min plz..

  13. Michele_Laino
    • one year ago
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    ok!

  14. anonymous
    • one year ago
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    12y - 3x = 60

  15. anonymous
    • one year ago
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    @Michele_Laino

  16. Michele_Laino
    • one year ago
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    I got this: \[\Large y = - \frac{3}{4}x + 3\]

  17. anonymous
    • one year ago
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    How?

  18. Michele_Laino
    • one year ago
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    by substitution int my formula: \[\Large \frac{{y - 6}}{{ - 3 - 6}} = \frac{{x - \left( { - 4} \right)}}{{8 - \left( { - 4} \right)}}\]

  19. Michele_Laino
    • one year ago
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    into*

  20. anonymous
    • one year ago
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    lemme solve plz .. 1 min..

  21. Michele_Laino
    • one year ago
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    ok!

  22. anonymous
    • one year ago
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    4y - 3x = 12

  23. anonymous
    • one year ago
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    what next?

  24. Michele_Laino
    • one year ago
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    please wait there is a sign error

  25. Michele_Laino
    • one year ago
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    the slope of your equation is 3/4 which is positive, whereas the slope of the requested line has to be negative

  26. Michele_Laino
    • one year ago
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    anyway: the requested distance, is: \[\Large \begin{gathered} d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} = \hfill \\ \hfill \\ = \sqrt {{{\left( {8 - \left( { - 4} \right)} \right)}^2} + {{\left( { - 3 - 6} \right)}^2}} = \hfill \\ \hfill \\ = \sqrt {{{12}^2} + {9^2}} = ...? \hfill \\ \end{gathered} \]

  27. anonymous
    • one year ago
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    15 units?

  28. Michele_Laino
    • one year ago
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    correct!

  29. anonymous
    • one year ago
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    what next?

  30. Michele_Laino
    • one year ago
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    the equation of the y-axis is \( \Large x=0 \), so the requested intersection point is given by the solution of this algebraic system: \[\Large \left\{ \begin{gathered} y = - \frac{3}{4}x + 3 \hfill \\ \hfill \\ x = 0 \hfill \\ \end{gathered} \right.\]

  31. anonymous
    • one year ago
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    3

  32. Michele_Laino
    • one year ago
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    yes! it is the point (0,3)

  33. anonymous
    • one year ago
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    http://prntscr.com/8cxltt how did you gte that formula *get

  34. Michele_Laino
    • one year ago
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    it is a standard formula

  35. anonymous
    • one year ago
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    never learnt of it. Can you give me details about the same??

  36. Michele_Laino
    • one year ago
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    ok! the equation of the line which passes at point (x1,y1) is: \[\Large y - {y_1} = m\left( {x - {x_1}} \right)\]

  37. Michele_Laino
    • one year ago
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    where m is the slope of our line

  38. anonymous
    • one year ago
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    okay.. what next?

  39. Michele_Laino
    • one year ago
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    next I require that line has to pass at point (x2,y2) too, so I can write this: \[{y_2} - {y_1} = m\left( {{x_2} - {x_1}} \right) \qquad \qquad (*)\]

  40. anonymous
    • one year ago
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    now we equate them , right?

  41. Michele_Laino
    • one year ago
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    not exactly, I solve equation (*) for m, so I get: \[\Large m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\]

  42. Michele_Laino
    • one year ago
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    then I substitute such expression for m into the first equation: \[\Large y - {y_1} = \left( {\frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}} \right)\left( {x - {x_1}} \right)\]

  43. anonymous
    • one year ago
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    next?

  44. Michele_Laino
    • one year ago
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    we have finished, since we got the standard formula

  45. anonymous
    • one year ago
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    thanks.

  46. Michele_Laino
    • one year ago
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    here is the situation of your exercise: |dw:1441473727673:dw|

  47. Michele_Laino
    • one year ago
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    next we have to compute the subsequent distances: \[\Large \begin{gathered} d(A,C) = \sqrt {{{\left( { - 4 - 0} \right)}^2} + {{\left( {6 - 3} \right)}^2}} = \sqrt {{4^2} + {3^2}} = ...? \hfill \\ \hfill \\ d(B,C) = \sqrt {{{\left( {8 - 0} \right)}^2} + {{\left( { - 3 - 3} \right)}^2}} = \sqrt {{8^2} + {6^2}} = ...? \hfill \\ \end{gathered} \]

  48. Michele_Laino
    • one year ago
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    \[\large \begin{gathered} d(A,C) = \sqrt {{{\left( { - 4 - 0} \right)}^2} + {{\left( {6 - 3} \right)}^2}} = \sqrt {{4^2} + {3^2}} = ...? \hfill \\ \hfill \\ d(B,C) = \sqrt {{{\left( {8 - 0} \right)}^2} + {{\left( { - 3 - 3} \right)}^2}} = \sqrt {{8^2} + {6^2}} = ...? \hfill \\ \end{gathered} \]

  49. Michele_Laino
    • one year ago
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    the requested ratio, part (i), is given by the subsequent expression: \[\Large r = \frac{{d(A,C)}}{{d(B,C)}}\] or by the subsequent ratio: \[\Large {r_1} = \frac{{d(B,C)}}{{d(A,C)}} = \frac{1}{r}\]

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