geno3141

If they consist of single digit integers, they must be in the set:  { -9, -8, ..., -1, 0, 1, ..., 8, 9 }

If there is only one mode and that mode is 2, there must be no other digit used more than the 2 is used.

To get the largest possible median, try:  2, 2, 6, 7, 8, 9  which has a median of 6.5.

3sin(x) + 5cos(x) = 5

Using the identity:  sin²(x) + cos²(x)  =  1:

                    3sin(x) + 5cos(x) = 5

  3sin(x) + 5·sqrt( 1 - sin²(x) )  =  5

                 5·sqrt( 1 - sin²(x) )  =  5 - 3sin(x)

           [ 5·sqrt( 1 - sin²(x) ) ]²  =  [ 5 - 3sin(x) ]² 

                    25( 1 - sin²(x) ) )  =  25 - 30sin(x) + 9sin²(x)

                         25 - 25sin²(x)  =  25 - 30sin(x) + 9sin²(x)

                              - 25sin²(x)  =  - 30sin(x) + 9sin²(x)

                                            0  =  -30sin(x) + 34sin²(x)

               34sin²(x) - 30sin(x)  =  0

             sin(x)[34sin(x) - 30 ]  =  0

Either  six(x)  =  0     or     34sin(x) - 30  =  0

                  x  =  0                    34sin(x)  =  30

                                                    sin(x)  =  30/34

                                                           x  =  sin^-1(x)

For all the solutions, you will need to add  2·n·pi  to each of the above solutions.

Draw square(ABCD).

On the diagonal AC, find point E such that AE = 1.

Find point X on BC such that EX is perpendicular to CB.

Since AC is a diagonal of a square whose sides are each 1, the length of AC = sqrt(2).

Since AE = 1, EC = sqrt(2) - 1..

Triangle(ACB) is similar to triangle(ECX)      [by AA].

This makes                       CE/CA = EX/AB

--->              (sqrt(2) - 1) / sqrt(2)  =  EX / 1

--->              (sqrt(2) - 1) / sqrt(2)  =  EX                  

--->                                         EX  =  0.293  (approximately)

[It would be the same istance to side CD.]  

a)  There are no zeros:  the vertex of the parablola is (0,24) and this parabola rises from there.

Another way:  setting the equation equal to zero:  0.005x² + 24  =  0

                                                      --->                            0.005x²  =  - 24

                                                      --->                                      x²  =  - 24/0.005                      

                                                      --->                                      x²  =  - 4800  (there is no real solution)

b)  24 meters because this is the minimum value of the quadratic.

Bonus:  Setting the equation equal to 30:  0.005x² + 24  =  30

                                               --->                      0.005x²  = 6

                                               --->                               x²  =  6/0.005

                                              --->                                 x  =  sqrt(1200)  =  34.64

Total distance:  34.64 ·2  =  69.3 meters.

With 4 black socks and 6 red socks, he can end without two red socks in these ways:

1) grabbing two black socks:                             (4/10)·(3/9)  =  12/90

2) grabbing a red sock and then a black sock:  (6/10)·(4/9)  =   24/90

3) grabbing a black sock and then a red sock:  (4/10)·(6/9)  =  24/90

Since these are independent:  12/90 + 24/90 + 24/90  =  60/90  =  1/3

The probability that he will not have a pair of red socks:  1/3

Because this is a right triangle (10² + 24²  =  26²), the center of the circumcircle is the midpoint of the

hypotenuse. This means that the radius of the circumcircle = 26/2  =  13.

Because this is a right triangle, the radius of the incircle can be found using this formula:

   r  =  ( a + b - c ) / 2           where c is the hypotenuse and a and b are the other two sides.

From these radii, you can calculate the areas.

If you can't do this by graphing, it's a painful process ...

If there are any rational solutions, the numerator is a divisor of 45:

     1, -1, 3, -3, 5. -5. 9. -9. 15. -15. 45. -45

and the denominator is a divisor of 6:

     1, -1, 2, -2, 3, -3, 6, -6

If you try enough of these, you'll find that both  -5/3  and  -3/2  are solutions.

This means that  3x + 5  and  2x + 3  are factors.

If you divide  6x4 + 49x3  128x2 + 132x + 45  by  (3x + 5)(2x + 4) 

you'll get the quotient  x2 + 5x + 3.

Then, apply the quadratic formula to  x2 + 5x + 3  to get the final two solutions.

To get a double root, you need the discriminant of the quadratic formula to be zero.

Discriminant:  b² - 4·a·c

Quadratic:  2x² + 3x + k     --->     a  =  2      b  =  3     c  =  k

--->     b² - 4·a·c  =  0     --->     3² - 4·2·k  =  0     --->    9 - 8k  =  0

You'll need to solve this equation for k.

Since you want  f( f(x) )  =  x, you'll want the first part to be the inverse of the second part.

Since the second part is  8 - 3x   --->     y  =  8 - 3x

(interchange both x and y)                     x  =  8 - 3y

(solve for y)                                        x - 8  =  -3y

                                                      (8 - x)/3  =  y

So:  ax + b  =  (8 - x)/3  =  8/3  - x/3  =  -(1/3)x + (8/3)

You need to eliminate all the possibilities for a denominator to be zero.

Since there is a  tan(x)  term, (and since  tan(x) = sin(x)/cos(x) ), you must eliminate all the values that

make  cos(x) = 0.

Since there is a  cot(x)  term, (and since  cot(x) = cos(x)/sin(x) ), you must eliminate all the values that

make  sin(x) = 0.

Since there is a denominator of  1 + cot(x),  you must eliminate all the values that make  cot(x) = -1.

Since there is a denominator of  cot(x) - 1,  you must eliminate all the values that make  cot(x) = 1.

[And, remember that both  tan(x)  and  cot(x)  have a period of  pi.]