geno3141

There are 9 "sections" from exit 41 to exit 50.

If 8 of these exits are 6 km each, this would account for 8 x 6  =  48 km.

Since the total distance from exit 41 to exit 50 is 100 km, the maximum distance from exit 47 to exit 48

is  100 km - 48 km  =  52 km.

Let  A  =  adult ticket price

       C  =  child ticket price

First day:         6A +   7C  =  162.00   (multiply by -2)   --->   -12A - 14C  =  -324.00

Second day:  11A + 14C  =  311.00                                        11A + 14C  =   311.00

                          add down the columns:                                   -A             =   -13.00

                                                                                                               A  =  13.00

Since  A = 13.00  --->  6A + 7C  =  162.00   --->   6(13.00) + 7C  =  162.00

                                                                                   78.00 + 7C  =  162.00

                                                                                                7C  =  84.00

                                                                                                 C  =  12.00

There are 7 choices for the first digit (2, 3, 4, 5, 6, 7, 8).

There are 4 choices for the second digit (4, 5, 6, 7)

There are 9 choices for the third digit (0, 1, 2, 3, 4, 6, 7, 8, 9).

Since these choices are independent, we need to multiply the three numbers:  7 x 4 x 9  to get the correct answer.

Solve:  f² - 4f  =  8

            f² - 4f - 8  =  0

Using the quadratic formula:  a = 1  b = -4  c = -8

f  =  [ 4 + sqrt(16 +32) ] / 2   --->   [ 4 + 4sqrt(3) ] / 2  =  2 + 2sqrt(3)

f  =  [ 4 - sqrt(16 +32) ] / 2   --->   [ 4 - 4sqrt(3) ] / 2  =  2 - 2sqrt(3)

An equation for a parabola is  y  =  a(x - r)(x -s)  where r and s are x-intercepts.

Placing the two intercepts into this formula, we get:  y  =  a( x - (5 + sqrt(3) )( x - (5 - sqrt(3) )

   --->   y  =  a( x - 5 - sqrt(3) )( x - 5 + sqrt(3) )

           y  =  a(x² - 10x + 22)

Since the point (0, 4) is on this parabola:

           4  =  a(0² -10·0 + 22)

           4  =  22a

           a  =  2/11

y  =  (2/11)(x² - 10x + 22)

Number of apple buckets = a.

Amount of money earned by selling apple buckets = 4a.

Number of peach buckets = p.

Amount of money earned by selling peach buckets = 6p.

Number of strawberry buckets = s.

Amount of money earned by selling strawberry buckets = 9s.

4a + 6p + 9s  =  258.00

Half as many strawberry buckets as peach buckets:  s = ½p

8 fewer apple buckets than peach buckets:  a = p - 8

Substitution:  4(p - 8) + 6p + 9(½p)  =  258.00

Solve for p, and then find s and a.

The positive multiples of 3 are 3, 6, 9, 12, ...

Which of these numbers, when cubed has the largest value that is less than 1000?

What you have are these numbers:  19, 19, 20, 20, 21, 21, ... 45, 48    [Wrtie down all of these.]

To find the mean, x, add all these numbers and divide by 27.

To find the median, y, find the number that is the middle number.

Fo find the mode, z, find which number is written the most often.

Place these numbers in the formula:  x(y - z).

3 of the 8 sectors do not allow the player to move the token. This area is 35.9 in².

(3/8)·pi·r²  =  35.9 in²

             r  =  sqrt( 35.9 / pi · 8/3 )

2x + 7 <= 16

       2x <=  9                                  (subtract 7)

         x  <= 4.5                               (divide by 2)

The greatest integer value for x is 4; there is no least value.