hectictar

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 #4
avatar+9041 
0

2.

 

Let  (x, y)  be the solution to the system. We know it is true that

 

y  =  (3x + 15) / 4

 

We also know it is true that:

 

x2  +  y2   =   36

                                                          Since   y  =  \(\frac{3x+15}{4}\)  we can substitute  \(\frac{3x+15}{4}\)  in for  y

x2  +  \((\frac{3x+15}{4})^2\)   =   36

 

x2  +  \((\frac{3x+15}{4})(\frac{3x+15}{4})\)   =   36

 

x2  +  \(\frac{9x^2+90x+225}{16}\)   =   36

                                                          Multiply through by  16  to eliminate the denominator

16x2  +  9x2 + 90x + 225   =   576

                                                          Combine like terms

25x2  +  90x  +  225   =   576

                                                          Subtract  576  from both sides of the equation

25x2  +  90x  -  351   =   0

                                                         We can use the quadratic formula to solve for  x

x   =  \(\dfrac{-9\pm12\sqrt3}{5}\)

 

We can use the equation  y  =  (3x + 15) / 4   to find the y-coordinates of the intersection points.

 

When   x  =  \(\frac{-9\ +\ 12\sqrt3}{5}\)   ,   y  =  \(\frac14\cdot(\ 3(\ \frac{-9\ +\ 12\sqrt3}{5}\ )+15\ )\ =\ \frac{12\ +\ 9\sqrt3}{5}\)

 

When   x  =  \(\frac{-9\ -\ 12\sqrt3}{5}\)   ,   y  =  \(\frac14\cdot(\ 3(\ \frac{-9\ -\ 12\sqrt3}{5}\ )+15\ )\ =\ \frac{12\ -\ 9\sqrt3}{5}\)

 

So the intersection points are:

 

\(\big(\frac{-9\ +\ 12\sqrt3}{5},\ \frac{12\ +\ 9\sqrt3}{5}\big)\)     and     \(\big(\frac{-9\ -\ 12\sqrt3}{5},\ \frac{12\ -\ 9\sqrt3}{5}\big)\)

 

Check:  https://www.desmos.com/calculator/jfndjchrhd

 

We can find the distance between those two points using the distance formula.

 

distance   =   \(\sqrt{\Big(\frac{-9\ +\ 12\sqrt3}{5}\ -\ \frac{-9\ -\ 12\sqrt3}{5}\Big)^2\ +\ \Big(\frac{12\ +\ 9\sqrt3}{5}\ -\ \frac{12\ -\ 9\sqrt3}{5}\Big)^2}\)

 

distance   =   \(\sqrt{\Big(\frac{24\sqrt3}{5}\Big)^2\ +\ \Big(\frac{18\sqrt3}{5}\Big)^2}\)

 

distance   =   \(\sqrt{\frac{1728}{25}\ +\ \frac{972}{25}}\)

 

distance   =   \(\sqrt{\frac{2700}{25}}\)

 

distance   =   \(6\sqrt3\)

.
May 14, 2020
 #3
avatar+9041 
0

1.

 

P  is a point on the line   y = x   and   Q  is a point on the line  y = 7x.

 

So we can say...

 

Let  P  be the point  (a, a)           where  a > 0

Let  Q  be the point  (b, 7b)        where  b > 0

And  O  is the point  (0, 0)

 

Now we can make this equation:

 

the distance between  O  and  P   =   the distance between  O  and  Q

 

√[ a2 + a2 ]   =   √[ b2 + (7b)2 ]

                                                    Square both sides of the equation

a2 + a2   =   b2 + (7b)2

 

a2 + a2   =   b2 + 49b2

                                                    Combine like terms

2a2   =   50b2

                                                    Divide both sides by  2

a2   =   25b2

                                                    Take the positive square root of both sides

a   =   5b

 

the slope of  PQ   =   the slope between the points  (a, a)  and  (b, 7b)

 

the slope of  PQ   =   \(\dfrac{a-7b}{a-b}\)

                                                         Substitute  5b  in for  a

the slope of  PQ   =   \(\dfrac{5b-7b}{5b-b}\)

                                                         Combine like terms

the slope of  PQ   =   \(\dfrac{-2b}{4b}\)

                                                         Reduce the fraction by  2b

the slope of  PQ   =   \(-\dfrac{1}{2}\)

 

Here is a graph for this problem:  https://www.desmos.com/calculator/c9sjxvalqw

May 14, 2020
 #1
avatar+9041 
+2
May 14, 2020
 #1
avatar+9041 
+2

 

m∠ACD   =   m∠BCD  -  m∠BCA

 

Since a tangent line forms a right angle with a line drawn from the center of the circle to the point of tangency:

 

m∠BCD   =   90°

 

Note that △ABC is an isosceles triangle because  BC  is the same length as  BA.

And so the base angles,  ∠BCA  and  ∠BAC ,  have the same measure.

Since the sum of the interior angles of a triangle is  180°:

 

m∠ABC  +  m∠BCA  +  m∠BAC   =   180°

                                                                    We can substitute  168°  in for  m∠ABC  since  m∠ABC  =  168°

168°  +  m∠BCA  +  m∠BAC   =   180°

                                                                    We can substitute  m∠BCA  in for  m∠BAC  since  m∠BAC  =  m∠BCA

168°  +  m∠BCA  +  m∠BCA   =   180°

                                                                    Subtract  168°  from both sides of the equation.

m∠BCA  +  m∠BCA   =   180°  -  168°

                                                                    Combine like terms.

2 * m∠BCA   =   180°  -  168°

                                                      Divide both sides of the equation by  2

m∠BCA   =   (180° - 168°) / 2

                                                      Simplify.

m∠BCA   =   6°

 

Now we can go back to the original equation and plug in the information we know:

 

m∠ACD   =   m∠BCD  -  m∠BCA

 

m∠ACD   =   90°  -  6°

 

m∠ACD   =   84°

May 14, 2020