CPhill

18

Note that the  first term  is 2(1) + 1 =  3   and the last term  is 2(12) + 1  = 25

So...the sum is

[ first term + last term ]  *  [ number of terms ] / 2  =

[ 3 + 25 ] *  [ 12]  / 2  =

28 *  6  =

168

17

5th term  = 140

7th term = 35

We have that  

(140)r^2  = (35)       where r  is the common ratio

Divide both sides by  140

r^2  = 35/140  =  1/4        take both roots

r = +/-  sqrt  (1/4)   =  +/- 1/2

So....note that the 6th term could be  (140)(1/2)  = 70

Or.....the 6th term could be (140) (-1/2)  =  -70

Thus.....  ± 70   is correct

1: Joe has exactly enough paint to paint the surface of a cube whose side length is 2. It turns out that this is also exactly enough paint to paint the surface of a sphere. If the volume of this sphere is {Ksqrt6}/{sqrt(pi)}, then what is K?

The total surface area  of the cube  is   6 * 2^2  = 24 units^2

So......for the sphere, we can find the radius as

24   = 4 pi ^ r^2     

6/pi  = r^2

√[6/pi]  = r

So.....the volume of the sphere  is  

K√[6/pi]  =  (4/3) pi * [ √[6/pi] ] ^3

K = (4/3) pi  * √[6/pi]^2

K = (4/3) pi *[  6 / pi ]  

K = (4/3) * 6

K  =  8

2: A pipe with inside diameter 10'' is to carry water from a reservoir to a small town in an arid land. Neglecting the friction and turbulence of the water against the inside of the pipes, what is the minimum number of 2''-inside-diameter pipes of the same length needed to carry the same volume of water to the arid town?

Note that the  volume of water carried by  one 10 in  diameter pipe of  some length, L, is :

pi (10/2)^2  * L  =  25 L  pi    units ^3

Note that the  volume of water carried  by one 2 in diameter pipe of the same length, L, is :

pi (2/2)^2 * L  =  L pi  units^3

So.....we would need 25 times the number of 2 in diameter pipes of some length L to carry the same amount of water as one 10 in diameter pipe of the same length

a)   f(2)  = [ 2 + 3 ] /2  =  5/2

b)  for f(x), write y

y  = [ x + 3 ] / 2     multiply both sides by 2

2y  = x +  3     subtract 3 from both sides

2y  - 3  = x        "swap" x and y

2x - 3  =  y   =  f^-1(x)

c)  f^-1 (12) =    2(12)  - 3   =  24  - 3     =      21

We can use the Law of Cosines, here

AB^2  = 18^2 + 12^2  - 2 (18)(12)cos(79)

AB^2   = 324 + 144 - 432*cos(79)

AB^2  = 468 - 432*cos(79)

AB  = √ [ 468 - 432*cos(79) ]  ≈   19.6

We have this equation

Total on day 6  =  initial value *  [ 1  -  common ratio^number of days  ] /  [ 1 - common ratio]

The initial value is 6

The common ratio  is  2  [ the ratio "doubles" the number each day ]

The number of days  = 6

So....

Total on day 6  =  6  [1 - 2^6] /[1 - 2 ]   = 

6 [ 1 - 64] / [-1]  =

6 [ 64 -  1]  =

6 [63]   =

378   total students on at the end of day 6

Let one of the integers   = a

Let the other integer  = b

And 1 less than a multiple of 5 can be written as  5n  - 1   where n is an integer ≥ 1

So....we have this equation

  S          +   P   =  5n - 1

(a + b )   +  (ab)  = 5n - 1      (1)

Rearranging (1), we have

a + ab +  b  = 5n -1

a + ab + b + 1  = 5n

a (b + 1)  + 1 ( b + 1)   = 5n

(a + 1)(b + 1)  = 5n

Note that if  "a" ends in a "4"  or a "9' then  (a + 1)  is a multiple of 5, and no matter the integer value of b, the left side is always a multiple of 5. And the right side is definitely a multiple of 5

So...the  "a's"  that  end in either 4 or  9 from 1-50 inclusive are :

4, 9 ,14, 19, 24, 29,34, 39, 44, 49

Notice that we can pair 4 with any of the other 49 integers and (1) will be true

Likewise, we can can pair 9 with any of 48 integers  [we've already paired it with 4 ], and (1)  will be true

And 19 can be paired with any of 47 other integers  [ we've already paired it with 4 and 9 ] and (1)  will be true

Continuing this reasoning with each successive number, we finally arrive at the fact that 49 can  be paired with any of 40 other integers and (1) will be true

So....the number of pairs  is just   49 + 48  + 47 +  .....+ 41  + 40

And totalling these, we have that the number of pairs is just :

( 49 + 40)  *  10  / 2   =

89 * 5   =

445  pairs a,b   that make (1)  true

And the number of possible pairs of a and b is  C(50,2)   = 1225

So, the probability that P + S  is one less than a multiple of 5, is just

445  / 1225  =

89  / 245    ≈  36.3 %

The graph  of   y  =x^2   has a vertex  of  (0, 0)

The graph  of   y  = (x + h) - k      shifts this vertex to  the  left  by  h units and down by  k units

So

y  =  ( x + 1)^2  - 2     shifts   (0,0)  to the left 1 unit   and down 2 units.....so.....the "new" vertex  is  (-1,-2)

The graph with this vertex  is the first one

1. B is the correct answer.....the first term, a₁, is  7

And a_n  =  -4*a _n-1

2.   7pi / 6   converted to  degrees is   

7pi / 6  *  180 / pi  =

[ 7pi ]  / [ 6 pi ] * 180  =

(7/6) * 180  =

7 ( 180 / 6 )  =

7 * 30  =

210°.....and this is a 3rd quadrant angle

3.  We can use the tangent here

tan 62  =  x  / 112   multiply both sides by  112

112 * tan 62  = x  ≈  210.6 ft  =  211 ft

4.  None of the answers are correct

cos C  = adjacent side  / hypotenuse  =  15  / 17