geno3141

The domain of  sqrt(25 - x²)  is  [-5, 5].

The domain of  sqrt(x + 25)  is  [-25, infinity)

Since the domain of  sqrt(25 - x²)  is completely contained in the domain of  sqrt(x + 25), the domain of the sum

is the intersection of these two domains:  [-5, 5].

Call the length of each side of the square 'x'.

Draw a line segment from the center of the circle to either of the two points where a square intersects the circle.

There is a right triangle formed by this line segment, one side of a square, and the side formed by two sides of the square.

Since the area of the circle is  60·pi in²  and the formula for the area of a circle is  pi·r²:

     pi·r²  =  60·pi in²     --->     r  =  sqrt(60) in      

The right triangle (described above) has sides  x,  2x,  and sqrt(60).

Using the Pythagorean Theorem:  x² + (2x)²  =  [ sqrt(60) ]²     --->     x² + 4x²  =  60

                                                                                                    --->            5x²  =  60

                                                                                                    --->              x²  =  12

                                                                                                    --->                x  =  sqrt(12)  inches

Because you have a reglular polygon, it can be inscribed in a circle.

The angle A₁A₄A₈ cuts off the arc A₈A₉A₁₀A1.

Thre are three sections to this arc:  A₈A₉, A₉A₁₀, and A₁₀A₁.

Since each portion of this arc is each to every other portion and there are 10 portions in the polygon, the 360^o circumference

of the circle is divided into 10 each parts, each 36^o.

The three sections have a total of 3 x 36^o  =  108^o.

The measure of the angle that you want is one-half of this intercepted arc or 54^o.

[There are other ways to solve this problem.]

1)  Find the equation of the bounday line:

     a)  Two points on this line are  (4,0)  and  (0,1)

     b)  Slope of this line:  m  =  (1 - 0) / (0 - 4)  =  -1/4

     c)  y-intercept:  1

     d)  equation of the line:  y  =  -1/4x + 1

2)  Find the inequality:

     Since the shaded area is below the line:  y <  -1/4x + 1

The circumcenter of a right triangle occurs at the midpoint of the hypotenuse.

The radius of the circumcircle will be the distance from the midpoint to either end of the hypotenuse;

in this case, the radius is 5.

If you find the vertex, you will have the highest point.

One way to find the vertex is to complete the square:

         y  =  -12x² + 48x

         y  =  -12(x² - 4x)         

         y  =  -12(x² - 4x + 4) + 48

         y  =  -12(x - 2)² + 48

When  x = 2  the height will be  y = 48.

3ax² + 3bx + 3c

Factor out the 3:           3[ ax² + bx + c ]

Factor out the a:           3[ a( x² + (b/a)x ) + c ]

Complete the square:  3[ a( x² + (b/a)x + b²/(4a)² ) + c - b²/(4a) ]

Factor:                         3[ a( (x + b/(2a) )² ) + ( c - b²/(4a) )]

Multiply back the 3:         3a( (x + b/(2a) )² ) + ( 3c - 3b²/(4a) )

n = 3a         h  =  -b/(2a)     k  =  3c - 3b²/(4a)

y  =  -6t² + 51t

y = 108

Combining:                        -6t² + 51t  =  108

                                 -6t² + 51t - 108  =  0

Multiplying by -1:      6t² -  51t + 108  =  0

Dividing by 3:              2t² - 17t + 36  =  0

Factoring:                      (2t - 9)(t - 4)  =  0

Answer:  Either  t = 4.5  or  t = 4

For the first one:

1)  Extend QRS to point T on the circle.

2)  By the "power of a point":  QP²  =  QR·QT   --->   6²  =  3·QT   --->   QT = 12

      Since QR = 3, RT = 9.

      Since RS = 3, ST = 6.

3)  Draw OT and OR, creating triangle(TSO) and triangle(RSO).

      OT and OR are radii; Let the radius = r.

4)  Use the Law of Cosines on these two triangles.

      OT²  =  r²  =  6² + 2² - 2·6·2·cos(TSO)

      OR²  =  r²  =  3² + 2² - 2·3·2·cos(RSO)

      Since these are equal:  6² + 2² - 2·6·2·cos(TSO)  =  3² + 2² - 2·3·2·cos(RSO)

                                                        40 - 24·cos(TSO)  =  13 - 12·cos(RSO)

      Since angle(RSO) is supplementary to angle(TSO),  cos(RSO) = - cos(TSO)

           --->                                       40 - 24·cos(TSO)  =  13 + 12·cos(TSO)

                                                                                27  =  36·cos(TSO)

                                                                              0.75  =  cos(TSO)

5)  Since  OT²  =  r²  =  6² + 2² - 2·6·2·cos(TSO)

           --->            r²  =  40 - 24·0.75  

                            r²  =  22

                             r  =  sqrt(22)

a + b + c  =  19(97)

a + n  =  b - n     --->     2n  =  b - a     --->     a  =  b - 2n

b - n  =  c/n     --->     n(b - n)  =  c     

Substituting:     a + b + c  =  19(97)     --->     (b - 2n) + (b) + ( n(b - n ) )  =  19(97)

       --->                                                                          2b - 2n + bn - n²  =  19(97)

                                                                                      2b + bn - 2n - n²  =  19(97)

                                                                                    b(2 + n) - n(2 + n)  =  19(97)

                                                                                            (b - n)(2 + n)  =  19(97)

Using the hint  19(97), let's try:       2 + n  =  19     and     b - n  =  97

--->                                                        n  =  17      and     b - 17  =  97     --->     b  =  114

a  =  b - 2n     --->     a  =  114 - 2(17)  =  80

c  =  n(b - n)  =  17(114 - 17)  =  1649

a + b + c  =  80 + 114 + 1649  =  1843     and     19(97)  =  1843