CPhill

The probability is given  by :

C (10, 5) * (1/2)^5 * (1/2)^5  = 63 / 256 =  about    24.6 %

For the first one, we have

a =  log ( 1 + 1/3)   →  log (4/3)    →   log (4) - log (3)   →  log(2^2) - log (3)  =  2log 2 - log (3)

Add  log(3) to both sides

a + log(3) =  2log(2)    

Divide both sides by 2

[ a + log (3) ] / 2    = log (2)

For the second we have

b = log (1 + 1/9)   →  log(10/9)  →  log (10) - log (9)  →    1 - log(3^2)   →   1 - 2log(3)

Add 2log(3)  to both sides  and subtract b from both sides

2log(3) =  1 - b      divide both sides by 2

log (3)   = [ 1 - b ] / 2

What's there to answer ???.................

  ​

1/4(a + 10) =2a -4         multiply both sides by 4 to clear the fraction on the left

a + 10  = 8a - 16          subtract a from both sides and add 16 to both sides

26 = 7a      divide both sides by 7

26 / 7   =   a

The product of positive integers x, y and z equals 2004. What is the minimum possible value of the sum x+y+z?

The  divisors of 2004 are  1 | 2 | 3 | 4 | 6 | 12 | 167 | 334 | 501 | 668 | 1002 | 2004

The minimum sum will come from the factors of the middle two divisors, 12 and 167

167 is prime....it will be included in the sum.......the divisors of 12   are 1 | 2 | 3 | 4 | 6 | 12.......

The remaining two integers involved in minimizing the sum will come from the middle two divisors of 12, i.e, 3 and 4

So....the minumum sum is given when  the three integers x, y and z  are   3 , 4 and 167   ....the sum  =  174

cos (theta)  =  [ u (dot) v] /  [ ll u ll * ll v ll ]

u (dot) v  = -1*3 + 4 * -4   =  -19

ll u ll = sqrt [ (-1)^2 + 4^2 ] = sqrt (17)

ll v ll  = sqrt [ (3)^2 + (-4)^2  ]   = 5

So

cos (theta)  =  -19 / [ 5 sqrt (17)]      so using the cosine inverse to find theta, we have

arccos (  -19 / [ 5 sqrt (17)] )  = theta = about 157.166° ......   "e"   is correct

1,000,000,000,000,000     =    "one quadrillion "

We can find the angle thusly :

cos θ   = [ u (dot) v ] / [ ll u ll * ll v ll  ]

u (dot) v  = 1*-1  + 4 * 2 =   7

ll u ll =  √[  1*2 + 4^2 ] = √17

ll v ll  = √ [(-1)^2 + 2^2]  = √5

So   ll u ll *  ll v ll   = √[17 * 5 ]  = √85

So

cos θ   =   7 /  √85    and we can find θ  using the cosine inverse

arccos [  7 /  √85 ] =  θ   =   40.6°

The sum, S, is given by

S = (1/2) [ 1 - r^n ] [ 1 - r ]      where (1/2)  is the first term   [  1/2 =  2^1 ]      and r is the common ratio between terms = 1/2                                                 and n is the number of terms = 10

So we have

S = (1/2) [ 1 - r^10 ] / [1/2]   =

 [ 1 - (1/2)^10 ]  =

[ 2^10 - 1 ] / 2^10  = 

1023 / 1024

We can solve these two equations to find a₁ [the first term]  and d [ the common difference between terms]

9 = a₁ + d(5 -1)  →           9 =  a₁ + 4d      (1)

-84 = a₁ + d(32 -1)   →   -84 = a₁ + 31d    (2)

Multiply (1)  by -1 and add it to (2)....this gives

-93  = 27d      divide both sides by  27

-93/27   = d  = -31/9

Using (1), we can find a₁ thusly

9 = a₁ + 4( -31/9)

81/9 = a₁ - 124/9       [ add 124/9 to both sides ]

205/9 = a₁ 

So........the 23rd term, a₂₃,  is given by

a₂₃  = 205/9 + (-31`/9)(23 -1)

a₂₃ = 205/9  + (-31/9)(22)   = -53