anonymous
  • anonymous
Need help, part and c http://prntscr.com/8cwxa2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
@Michele_Laino
Michele_Laino
  • Michele_Laino
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: \[\Large \begin{gathered} A = - \frac{{B{C^2}}}{2} + \frac{{\pi {r^2}}}{2} = - \frac{{B{C^2}}}{2} + \frac{{\pi B{C^2}}}{4} = \hfill \\ \hfill \\ = \frac{{B{C^2}}}{2}\left( {\frac{\pi }{2} - 1} \right) \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
oops... I have made a typo: \[\Large \frac{{AB \cdot BC}}{2} = \frac{{B{C^2}}}{2}\]

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More answers

anonymous
  • anonymous
14cm^2
anonymous
  • anonymous
part c also.
Michele_Laino
  • Michele_Laino
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?
anonymous
  • anonymous
@Michele_Laino
anonymous
  • anonymous
idk...
Michele_Laino
  • Michele_Laino
please try, you have to substitute the coordinates of your points into my formula above: (x1,y1)=(-4,6), and (x2, y2) = (8,-3)
anonymous
  • anonymous
x and y ???
Michele_Laino
  • Michele_Laino
x, and y are the variables, you have only to susbstitute x1, x2, y1, y2 with the coordinates of your points
anonymous
  • anonymous
1 min plz..
Michele_Laino
  • Michele_Laino
ok!
anonymous
  • anonymous
12y - 3x = 60
anonymous
  • anonymous
@Michele_Laino
Michele_Laino
  • Michele_Laino
I got this: \[\Large y = - \frac{3}{4}x + 3\]
anonymous
  • anonymous
How?
Michele_Laino
  • Michele_Laino
by substitution int my formula: \[\Large \frac{{y - 6}}{{ - 3 - 6}} = \frac{{x - \left( { - 4} \right)}}{{8 - \left( { - 4} \right)}}\]
Michele_Laino
  • Michele_Laino
into*
anonymous
  • anonymous
lemme solve plz .. 1 min..
Michele_Laino
  • Michele_Laino
ok!
anonymous
  • anonymous
4y - 3x = 12
anonymous
  • anonymous
what next?
Michele_Laino
  • Michele_Laino
please wait there is a sign error
Michele_Laino
  • Michele_Laino
the slope of your equation is 3/4 which is positive, whereas the slope of the requested line has to be negative
Michele_Laino
  • Michele_Laino
anyway: the requested distance, is: \[\Large \begin{gathered} d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} = \hfill \\ \hfill \\ = \sqrt {{{\left( {8 - \left( { - 4} \right)} \right)}^2} + {{\left( { - 3 - 6} \right)}^2}} = \hfill \\ \hfill \\ = \sqrt {{{12}^2} + {9^2}} = ...? \hfill \\ \end{gathered} \]
anonymous
  • anonymous
15 units?
Michele_Laino
  • Michele_Laino
correct!
anonymous
  • anonymous
what next?
Michele_Laino
  • Michele_Laino
the equation of the y-axis is \( \Large x=0 \), so the requested intersection point is given by the solution of this algebraic system: \[\Large \left\{ \begin{gathered} y = - \frac{3}{4}x + 3 \hfill \\ \hfill \\ x = 0 \hfill \\ \end{gathered} \right.\]
anonymous
  • anonymous
3
Michele_Laino
  • Michele_Laino
yes! it is the point (0,3)
anonymous
  • anonymous
http://prntscr.com/8cxltt how did you gte that formula *get
Michele_Laino
  • Michele_Laino
it is a standard formula
anonymous
  • anonymous
never learnt of it. Can you give me details about the same??
Michele_Laino
  • Michele_Laino
ok! the equation of the line which passes at point (x1,y1) is: \[\Large y - {y_1} = m\left( {x - {x_1}} \right)\]
Michele_Laino
  • Michele_Laino
where m is the slope of our line
anonymous
  • anonymous
okay.. what next?
Michele_Laino
  • Michele_Laino
next I require that line has to pass at point (x2,y2) too, so I can write this: \[{y_2} - {y_1} = m\left( {{x_2} - {x_1}} \right) \qquad \qquad (*)\]
anonymous
  • anonymous
now we equate them , right?
Michele_Laino
  • Michele_Laino
not exactly, I solve equation (*) for m, so I get: \[\Large m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\]
Michele_Laino
  • Michele_Laino
then I substitute such expression for m into the first equation: \[\Large y - {y_1} = \left( {\frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}} \right)\left( {x - {x_1}} \right)\]
anonymous
  • anonymous
next?
Michele_Laino
  • Michele_Laino
we have finished, since we got the standard formula
anonymous
  • anonymous
thanks.
Michele_Laino
  • Michele_Laino
here is the situation of your exercise: |dw:1441473727673:dw|
Michele_Laino
  • Michele_Laino
next we have to compute the subsequent distances: \[\Large \begin{gathered} d(A,C) = \sqrt {{{\left( { - 4 - 0} \right)}^2} + {{\left( {6 - 3} \right)}^2}} = \sqrt {{4^2} + {3^2}} = ...? \hfill \\ \hfill \\ d(B,C) = \sqrt {{{\left( {8 - 0} \right)}^2} + {{\left( { - 3 - 3} \right)}^2}} = \sqrt {{8^2} + {6^2}} = ...? \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
\[\large \begin{gathered} d(A,C) = \sqrt {{{\left( { - 4 - 0} \right)}^2} + {{\left( {6 - 3} \right)}^2}} = \sqrt {{4^2} + {3^2}} = ...? \hfill \\ \hfill \\ d(B,C) = \sqrt {{{\left( {8 - 0} \right)}^2} + {{\left( { - 3 - 3} \right)}^2}} = \sqrt {{8^2} + {6^2}} = ...? \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
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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