CPhill

Let  the  time  for  one pipe to fill the tank be  = T  (in minutes)

Let  the  time for the other pipe  to fill the T + 40    (in minutes)

So....in one  minute  the  first pipe fills  1/T  of the tank  and the second fills   1 / (T + 40)  of the tank

And both pipes fill1/48  of the tank in one minute

So....we  have  this equation

1 /T  +  1/ (T + 40)   = 1/48      simplify

(T + 40 + T ) / [ (T) (T + 40) ] =  1/48        cross-multiply

48 (2T + 40)  =  T(T +  40)

96T + 1920   = T^2  + 40T       rearrange as

T^2 - 56T  - 1920  =  0         factor  as

(T - 80)  (T + 24)  = 0

The first factor  set to  0 and solved   produces  a positive solution  for T  = 80  (min)

The  time  for the smaller pipe to fill the tank  =  T + 40  =  120  min

(2x + 3) ( x + 6)  =  95

2x^2 + 3x  + 12x + 18  =  95    simpify

2x^2  + 15x  - 77  =  0      factor as

(2x  - 7) ( x + 11)  =   0

Only   the first  factor set to  0  produces  what we want

2x  - 7  =   0

2x   =  7

x  = 7/2 

One  dimension =  (2(7/2) + 3)  =  ( 7 + 3)  =  10 ft

The  other  is  95  /10   = 9.5 ft

4 choices  for  the first answer  and 3 choices each for the other 4 answers   =

4 * 3^4  = 

4 * 81  = 

324  ways

sec (3x)  =  -2√3/3  =  -2/√3

Notice  that  this  is  the same as

cos (3x)  = -√3/2

Note  that  cos (x)  = -√3/2   at     (5/6)pi , (7/6)pi , (17/6)pi, (19/6)pi, (29/6)pi  and (31/6)pi

Dividing  each  of these angles  by 3  gives  us  the solutions we  want  =

(5/18)pi , (7/18)pi, (17/18)pi, (19/18) pi, (29/18)pi, (31/18)pi

2sin^2 x  + 5sin x  - 3    =  0      factor as

(2sin x   - 1) ( sin x   + 3)  =  0

The only  solution(s)  come(s) from setting the  first factor to 0 and solving

So   we  have

2sin x  - 1   = 0

2sin x  =  1

sin x  = 1/2

This happens  at   

pi/6  + n *2pi 

And

(5pi)/6  + n* 2 pi                where n is an integer

2, 3, 4, 5 , 6 , 7

If  6  is chosen first , we have  3 possibilities for the  last digit (3, 5 or 7)

And we  have 4!  = 24  ways to permute the remaining digits

So  1 * 4! * 3  = 1* 24 * 3  =  72

If either  5 or 7 is chosen first, we  have 2 ways to choose  the last digit  and 4!  ways to permute  the remaining digits  =    2 * 4! * 2   =  4 * 24  =  96

[For example....if  we  choose 5 first, we  have two choices for the last digit, either 3 or 7.....and 4!  ways = 24 ways  to permute  the middle four digits  '

So....the total possibiities =   72  + 96   =  168

Note that   3 1/2  =  7/2        1  1/3   =  4/3       2  1/4  =   9/4

The  area  =   2 ( the sum of the products  of all three dimensions taken two at a  time )  =

( 7/2 * 4/3   +  4/3 * 9/4  +  7/2* 9/4 )  =

( 28/6  + 36/12  +  63/8)   =

(14/3  + 3   + 63/8)   =

 ( 8 * 14  + 3 * 63)  / 24   + 3  =

301/24  +  3   =

12 + 13/24  +  3  =

15 + 13/24  ft^2

The  object is  to get  x by itself  and  then "switch" x  and  y......we can write

y = 2x^2  - 3      add  3 to both sides

y + 3  =   2x^2      divide both sides  by  2

[ y + 3 ] / 2    =  x^2         take both roots

±√ [ (y + 3) / 2 ]  =  x   "switch" x and y

±√ [ ( x + 3)  / 2  ]  =  y   =  the inverse

11:30 AM  +  25 minutes  =   11:55AM

11:55 AM  +  40 minutes  =  12:35 PM  ⇒  the time she met her friend

Let  the  number  of 2 point shots  made   =  N

Then the number of 3 points  shots  made  =  35 - N

And we  have  this

2N + 3(35 - N)  =  79      simplify

2N + 105 - 3N  =  79       subtract 105 from each side

-1N  =    -26        divide throug by -1

N =  26  =  the  number of 2 point shots

35 - N =   35 - 26  =  9  =  number of 3 point shots