geno3141

If it is a linear function, it must have a constant slope:  m  =  [ y₂ - y₁ ] / [ x₂ - x₁ ]

For this problem:  if  x₂ = 6  and  x₁ = 2,  then  y₂ = f(6)  and  y₁ = f(2)

     --->     m  =  [ y₂ - y₁ ] / [ x₂ - x₁ ]     --->     m  =  12 / [ 6 - 2 ]

                                                             --->     m  =  12 / 4

                                                             --->     m  =  3

--->     m  =  [ y₂ - y₁ ] / [ x₂ - x₁ ]     --->     3    =  [ f(12) - f(2) ] / [ 12 - 2 ]

                                                        --->     3    =  [ f(12) - f(2) ] / [ 10

                                                        --->     30  =   f(12) - f(2)

1)  Use the Law of Cosines on triangle(ABC) to find the size of angle(A).

2)  Using this answer, find the size of angle(ABD).

3)  Now, use the Law of Cosines on triangle(ABD) to find the size of AD.

Since this is arithmetic, the middle term will be the average of the outside two:

       12  =  [ (y + 6) + (y) ] / 2

Solve ...

Let V be the midpoint of arc(AB) and Z the midpoint of arc(CD).

Draw VWXYZ, where W is the point of intersection with AB (W will be the midpoint of AB);

X is the point of intersection with AC;

Y is the point of intersection with CD (Y will be the midpoint of CD).

VWXYZ will be perpendicular to both AB and CD.

Since AB = 2, AW = 1 and WX = 1.

Since CD = 6, CY = 3 and XY = 3.

Call XW = a and call YZ = b.

Since WXYZ is a diameter and AB is a chord of the circle:  AW · WB  =  VW · WZ.

     --->     1 · 1  =  a · (1 + 3 + b)     --->     1  =  a · (4 + b)     --->     4a + ab  =  1     --->     ab  =  1 - 4a

Since WXYZ is a diameter and CD is a chord of the circle:  CY · YD  =  YZ · VY.

     --->     3 · 3  =  b · (1 + 3 + a)     --->     9  =  b · (4 + a)     --->     4b + ab  =  9     --->     ab  =  9 - 4b

Combining these two results:  1 - 4a  =  9 - 4b    --->     4b - 4a  =  8     --->     b - a  =  2

     --->     b  =  2 + a

Since     a · (4 + b)  =  1     --->     a · (4 + (2 + a) )  =  1     --->     a · (6 + a)   =  1 

                  --->     6a + a²  =  1     --->     a² + 6a - 1  =  0

                  --->     a  =  -3 + sqrt(10)      [quadratic formula]

Since      b · (4 + a)  =  9     --->     b · (4 + (b - 2) )  =  9     --->     b · (2 + b)   =  9

                 --->     2b + b²  =  9     --->     b² + 2b - 9  =  0

                 --->     b  =  -1 + sqrt(10)     [quadratic formula]

The diameter will be:     (a) + (1) + (3) + (b)  =  ( -3 + sqrt(10) ) + (1) + (3) + ( -1 + sqrt(10) )

                                                                      =  2sqrt(10)

So, the radius is:     sqrt(10)

A circle has diameter XY = 28.

Chord AB is perpendicular to XY, has length 12, and intersects XY at point P.

By theorem:  AP · PB  =  XP · PY

                           6 · 6  =  x · (28 - x)

                              36  =  28X - x²

             x² - 28x + 36  =  0

Now, use the quadratic formula:  x  =  [ - -28  +/-  sqrt( 28² - 4·1·36 ) ] / (2·1)

           x  =  ( 28 + sqrt(640) ) / 2     or     x  =  ( 28 - sqrt(640) ) / 2 

           x  =  14 + 4sqrt(10)                       x  =  14 - 4sqrt(10)

Distance between the chord and the center of the circle:  14 - ( 14 - 4sqrt(10) )  =  4sqrt(10)

Geometric sequence:  x - 2,  x + 2,  x + 4

Since this sequence is geometric, there is a value  r  such that:

     r(x - 2)  =  x + 2                        and     r(x + 2)  =  x + 4

               r  =  (x + 2) / (x - 2)                             r  =  (x + 4) / (x + 2)

Setting these two equal to each other:     (x + 2) / (x - 2)  =  (x + 4) / (x + 2)

                                                                   (x + 2)(x + 2)  =  (x - 2)(x + 4)

                                                                      x² + 4x + 4  =  x² + 2x - 8

                                                                                    2x  =  -12

                                                                                      x  =  -6

1 + i  is  not real, so  f(1 + i)  =  (1 + i)²  =  1² + 2i + i²  =  1 + 2i - 1  =  2i

2i  is  not real, so  f( f(1 + i) )  =  f(2i)  =  4i²  =  -4

-4  is  real, so  f( f( f(1 + i) ) )  =  -(-4)²  =  -16

-16  is  real, so  f( f( f( f(1 + i) ) ) )  =  -(-16)²  =  -256  

To make the calculations easier, let angle(B) = 2b and angle(A) = 2a.

To make the description easier, let point X be on BC and point Y be on AC.

In triangle(AXB), write angle(AXB) in terms of a and b. Then, write angle(CXI) in terms of a and b.

In triangle(AYB), write angle(AYB) in terms of a and b. Then, write angle(CYI) in terms of a and b.

Looking at triangle(AIB), how many degrees are in a + b?

How many degrees are in angle(XIY)?

Finally, look at quadrilateral(CXIY) --- you can get the answer!

The formula to be used:  z-score  =  ( score - mean ) / (standard deviation)

Exam A:  you know the score, the mean, and the standard deviation; use these values to find

                       the z-score.

Exam B:  you know the mean, the standard deviation, and the z-score (this is what you calculated for

                       Exam A. Use these numbers to find the score that is needed.

z-score  =  ( score - mean ) / (standard deviation )

Exam A:  z-score  =  ( 565 - 650 ) / 50  =  -1.7

To do as well on Exam B, she'll need the same z-score:

Exam B:  -1.7  =  ( score - 50 ) / 10

--->           -17  =  score - 50

--->            33  =  score