CPhill

1

  C

                4

                       X

                                 12

B                                           A

triangle  XBA similar to triangle BCA

triangle  XCB similar to triangle BCA

So

triangle XCB is similar to triangle XBA

So

CX / XB  = XB / AX

4 / XB   XB / 12

XB^2  =  48

XB  = sqrt (48)  =  4 sqrt (3)

2

[ABC]  =  (1/2) (AB) ( BC) sin (ABC)   =

(1/2) ( 6) (8)  (sqrt 2) / 2  =

12 sqrt 2

Question 1

Six children are each offered a single scoop of any of 3 flavors of ice cream from the Combinatorial Creamery. In how many ways can each child choose a flavor for their scoop of ice cream so that some flavor of ice cream is selected by exactly three children?

We can select 3 children from any  of the  6  to have the same  flavor  =  C (6,3) =  20  different sets  of  children

And each of these sets  has  three ways to  choose the same  flavor  = C (3, 1) = 3

The remaining 3 children are by default.....one  of then  can choose any  of  the  2  remaining flavors and  this  child can be any one  of the remaining  3...the other 2 children are stuck with the  remaining  flavor not yet selected...so the  total sets here  =  C(3,1) * C(2,1)

So....the total  number of ways = 

C(6,3) * C (3,1) * C(3,1) * C(2,1)  =

20  * 3  *  3 * 2  = 

360  ways

11111111 = 1  identifiable   integer

1111112 =    C(7,1)  = 7    identifiable integers

111122  =    C(6,2)   =   15  identifiable integers

11222 =  C (5,3)   =  10     identifiable integers

2222  = 1   identifiable integer

34  integers

\(a_4a_5 a_6 = 6\)

What is    \(a_3a_7\) 

Let  a4  = 1       a5 = 2   and   a6 = 3

a3  =  0        a7    =    4

a3a7  = (0) (4)   =   0

Question 2

Here's what I get

Choose any 3 of the 4 "suites"  =  C (4,3)

In any one of  of  these choose  any 2 cards of 7  in  one  "suite"  = C(7,2)

And we  can choose any one  of the  three "suites" for  choosing 2  cards of the same  color =  C(3,1)

In the two remaining  suites choose  1 of  these  =  C(2,1)

And in this  suite  choose any 1  of the  7 cards

Total hands =  C(4,3) * C (7,2) * C (3,1) * C( 2,1) * C (7,1)   = 3528

Question 3

Anchor   any adult in seat 1   and  any child in seat 2

We have 3 ways to do the first and  3  ways to do the second

In seat 3 choose any 2 adults

In seat 4 choose any 2 children

In seats 5 and 6 we have one way to choose an adult and one way to choose a child

Total arrangemets  =  3 * 3 * 2 * 2 * 1 * 1 =   36

                            A

                F                   G

B             D                    E                   C

Let AB = 3   and AC   =  1

The hypotenuse  BC  =  sqrt ( AB^2 + AC^2)  = sqrt (3^2 + 1^2)  =  sqrt (10)

Let the side of the square  = s

Let  BD =  x

Using similar triangles

triangle ABC  similar to triangle  DBF

BD / DF =   AB / AC

x / s =  3 / 1

x = 3s

Let EC  = y

Using similar triangles again

triangle ABC similar to triangle EGC

CE / EG = AC / AB

y / s = 1/3

y = (1/3)s

BD  +  DE +  EC =   sqrt 10

x  +   s   +   y  =  sqrt (10)

3s  +  s  + (1/3)s  = sqrt (10)

(13/3)s =sqrt (10)

s =    3sqrt (10)  / 13

Angle  DAB   = 30°

Angle ABC = Angle ABD =  45°

Angle ADB = 180 - 30 - 45 = 105°

Using the Law of Sines

AB / sin (ADB)  = AD / sin ( ABD )

AB / sin (105°) = 24 / sin (45°)

AB =  24 sin (105°) / sin (45°)  = 12 (1 + sqrt 3)

Angle ADC  = 180 - 105 =  75°

Angle CAD = 30°

Angle ACD  = 180 - 75 - 30 =  75°

Since  angle ADC = angle ACD  then

AC  = AD =  24              

[ ABC ]  =  (1/2) ( AB) (AC) sin (BAC)  = 

(1/2) ( 12 ( 1 + sqrt 3) (24) sin (60°)  =

12 * 12 * ( 1 + sqrt 3) * sqrt (3)  / 2  = 

72 ( sqrt (3) + 3)  ≈  340.7

Let p   = the number of  people

So

10 chocolates for each person but the last + 2 for the  last person =  8 chocolates for each person + 20 left over

10 (p - 1)  + 2   =  8p + 20 

10p  - 10 + 2  =   8p+ 20

10p - 8  = 8p + 20

2p  = 28

p = 14

See Tiggsy's answer here : https://web2.0calc.com/questions/help_21682

\( $\displaystyle \frac{1}{2 - \sqrt[3]{2}}$ \)

\($\displaystyle \frac{\sqrt[3]{A} + \sqrt[3]{B} + \sqrt[3]{C}}{D}$ \)

Note that

(2^3 - 2) =  (2 - 2^(1/3)) ( 4 + 2^(4/3) + 2^(2/3) )

So....multiply  numerator / denominator   by   ( 4 + 2^(4/3) + 2^(2/3) )  and we  have

4 + 2^(4/3) + 2^(2/3)             ∛64  + ∛16 + ∛4                

________________    =    ____________________    

     2^3  - 2                                          6

A  + B + C + D =   64  + 16 + 4  + 6  =      90 

This is a  right triangle with legs of 15 and  20

Area = (1/2) (product of the  legs)   =(1/2) (15)(20) =  150

Also....the shortest altitude is  drawn to  the  longest side

So

150  = (1/2) (25) (altitude)

300  =25 (altitude)

300 / 25  =  12    =   the shortest altitude

Let the normal cost of  a  ticket =  C

At a 20% discount the cost of a  ticket = .80C

So....80 people bought tickest at the  normal price and  60  bought them at the  discounted price

80C + 60 * .80C   = 1920

80C + 48C  =1920

128C  = 1920

C = 1920 / 128 =    15 (dollars)  =  the normal price

6y + 2x  = -1

6y = -2x -1

y = (-1/3)x   - 1/6

The slope of this  line =  -1/3

y= (1/2)(3x^2 - k)

y= (3/2)x^2 - k/2        the derivative of  this is

y' = 3x

We are looking for the  point on this  function where  the the slope =  3

So

3x = 3

x =1

When x = 1,    y =(-1/3)x - 1/6 =   (-1/3)(1) - 1/6  =  -1/3 - 1/6 =  -1/2

So....to find k....

-1/2 = (1/2)(3(1)^2  - k)

-1/2 = 3/2 - k/2

-4/2 = -k/2

4  =  k

Second part

The equation of the tangent line at  A = (1, -1/2)   is

y = 3(x -1) -1/2

y = 3x - 3 -1/2

y= 3x - 7/2

This line will intercept the x axis  when y  = 0

0 =  3x -7/2

7/2 = 3x

x = 7/6

B= (7/6 , 0)