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I've assumed this is meant to be:   (x^(1/4) + 16)^2 = 144 + x^(3/2)

The real number solution is obtained as follows:

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BRILLIANT heureka!!. Congrats.

Here's some info :    http://www.purplemath.com/modules/binomial.htm

Well what do you think your caclulator is for :)

To continue where Melody left off :

8x^3  - x^2  - 96x  - 1792   =  0 

There is only one real solution that occurs at about  x  =6.7755

Here are the real and non-real solutions : http://www.wolframalpha.com/input/?i=8x^3++-x^2++-+96x++-+1792++%3D+0

We have 10 standard 6-sided dice, all different colors.

In how many ways can we roll them to get a sum of 20?

1.

The number of ways of partitioning 20 into exactly 10 distinct parts with  the 6 numbers {1,2,3,4,5,6} we have:

30 partitions of 20 into ten distinct parts:

The 30 distinct partitions with the sum 20 are:

   1.) {2,2,2,2,2,2,2,2,2,2}
   2.) {1,2,2,2,2,2,2,2,2,3}
   3.) {1,1,2,2,2,2,2,2,3,3}
   4.) {1,1,2,2,2,2,2,2,2,4}
   5.) {1,1,1,2,2,2,2,3,3,3} 
   6.) {1,1,1,2,2,2,2,2,3,4}
   7.) {1,1,1,2,2,2,2,2,2,5}
   8.) {1,1,1,1,2,2,3,3,3,3}
   9.) {1,1,1,1,2,2,2,3,3,4}
  10.) {1,1,1,1,2,2,2,2,4,4}
  11.) {1,1,1,1,2,2,2,2,3,5}
  12.) {1,1,1,1,2,2,2,2,2,6}
  13.) {1,1,1,1,1,3,3,3,3,3}
  14.) {1,1,1,1,1,2,3,3,3,4}
  15.) {1,1,1,1,1,2,2,3,4,4}
  16.) {1,1,1,1,1,2,2,3,3,5}
  17.) {1,1,1,1,1,2,2,2,4,5}
  18.) {1,1,1,1,1,2,2,2,3,6}
  19.) {1,1,1,1,1,1,3,3,4,4}
  20.) {1,1,1,1,1,1,3,3,3,5}
  21.) {1,1,1,1,1,1,2,4,4,4}
  22.) {1,1,1,1,1,1,2,3,4,5}
  23.) {1,1,1,1,1,1,2,3,3,6}
  24.) {1,1,1,1,1,1,2,2,5,5}
  25.) {1,1,1,1,1,1,2,2,4,6}
  26.) {1,1,1,1,1,1,1,4,4,5}
  27.) {1,1,1,1,1,1,1,3,5,5}
  28.) {1,1,1,1,1,1,1,3,4,6}
  29.) {1,1,1,1,1,1,1,2,5,6}
  30.) {1,1,1,1,1,1,1,1,6,6}

2.

In how many ways can we roll them to get a sum of 20?
We need the permutations of each partitions and the sum is the answer.

\(\begin{array}{rcll} 1.)& \{2,2,2,2,2,2,2,2,2,2\} & \frac{10!}{10!} &=& 1 \\ 2.)& \{1,2,2,2,2,2,2,2,2,3\} & \frac{10!}{1!8!1!} &=& 90 \\ 3.)& \{1,1,2,2,2,2,2,2,3,3\} & \frac{10!}{2!6!2!} &=& 1260 \\ 4.)& \{1,1,2,2,2,2,2,2,2,4\} & \frac{10!}{2!6!1!} &=& 360 \\ 5.)& \{1,1,1,2,2,2,2,3,3,3\} & \frac{10!}{3!4!3!} &=& 4200 \\ 6.)& \{1,1,1,2,2,2,2,2,3,4\} & \frac{10!}{3!5!1!1!} &=& 5040 \\ 7.)& \{1,1,1,2,2,2,2,2,2,5\} & \frac{10!}{3!6!1!} &=& 840 \\ 8.)& \{1,1,1,1,2,2,3,3,3,3\} & \frac{10!}{4!2!4!} &=& 3150 \\ 9.)& \{1,1,1,1,2,2,2,3,3,4\} & \frac{10!}{4!3!2!1!} &=& 12600 \\ 10.)& \{1,1,1,1,2,2,2,2,4,4\} & \frac{10!}{4!4!2!} &=& 3150 \\ 11.)& \{1,1,1,1,2,2,2,2,3,5\} & \frac{10!}{4!4!1!1!} &=& 6300 \\ 12.)& \{1,1,1,1,2,2,2,2,2,6\} & \frac{10!}{4!5!1!} &=& 1260 \\ 13.)& \{1,1,1,1,1,3,3,3,3,3\} & \frac{10!}{5!5!} &=& 252 \\ 14.)& \{1,1,1,1,1,2,3,3,3,4\} & \frac{10!}{5!1!3!1!} &=& 5040 \\ 15.)& \{1,1,1,1,1,2,2,3,4,4\} & \frac{10!}{5!2!1!2!} &=& 7560 \\ 16.)& \{1,1,1,1,1,2,2,3,3,5\} & \frac{10!}{5!2!2!1!} &=& 7560 \\ 17.)& \{1,1,1,1,1,2,2,2,4,5\} & \frac{10!}{5!3!1!1!} &=& 5040 \\ 18.)& \{1,1,1,1,1,2,2,2,3,6\} & \frac{10!}{5!3!1!1!} &=& 5040 \\ 19.)& \{1,1,1,1,1,1,3,3,4,4\} & \frac{10!}{6!2!2!} &=& 1260 \\ 20.)& \{1,1,1,1,1,1,3,3,3,5\} & \frac{10!}{6!3!1!} &=& 840 \\ 21.)& \{1,1,1,1,1,1,2,4,4,4\} & \frac{10!}{6!1!3!} &=& 840 \\ 22.)& \{1,1,1,1,1,1,2,3,4,5\} & \frac{10!}{6!1!1!1!1!} &=& 5040 \\ 23.)& \{1,1,1,1,1,1,2,3,3,6\} & \frac{10!}{6!1!2!1!} &=& 2520 \\ 24.)& \{1,1,1,1,1,1,2,2,5,5\} & \frac{10!}{6!2!2!} &=& 1260 \\ 25.)& \{1,1,1,1,1,1,2,2,4,6\} & \frac{10!}{6!2!1!1!} &=& 2520 \\ 26.)& \{1,1,1,1,1,1,1,4,4,5\} & \frac{10!}{7!2!1!} &=& 360 \\ 27.)& \{1,1,1,1,1,1,1,3,5,5\} & \frac{10!}{7!1!2!} &=& 360 \\ 28.)& \{1,1,1,1,1,1,1,3,4,6\} & \frac{10!}{7!1!1!1!} &=& 720 \\ 29.)& \{1,1,1,1,1,1,1,2,5,6\} & \frac{10!}{7!1!1!1!} &=& 720 \\ 30.)& \{1,1,1,1,1,1,1,1,6,6\} & \frac{10!}{8!2!} &=& 45 \\ \hline && \text{sum} &=& 85228 \end{array} \)

3.

The probability is:

\(\begin{array}{rcll} \frac{ 85228 } { 6^{10} } &=& \frac{ 85228 } { 60466176 } \\ &=& 0.00140951530 \end{array}\)

11 is just a strange number... 

1 x 11 = 11

2 x 11 = 22 etc. 

12 x 11 = 132

13 x 11 = 143

14 x 11 = 154 etc. 

11 x 11 = 121

111 x 111 = 12321 etc. 

111,111,111 x 111,111,111=12345678987654321

111,111,111,111 x 111,111,111,111 = 12345679012320987654321 which ends with 987654321

it's just wonderful

26° 20'   = [26 + 1/3]°

The wire forms the the hypotenuse of a right triangle, with the 18.6 ft being the leg opposite the given angle.......so.....we can use the sine function to find the length of the wire

sin [26 + 1/3]   =  18.6  / L     where L is the wire length   .....rearrange to solve for L

L = 18.6 / sin[26 + 1/3]  = about  41.93 ft

26 degrees 20 minutes  is 26.333 degrees

   sin 26.333 = 18.6/hyp

hyp = 18.6/sin26.33 =41.92 ft

log25+log2xlog50+(log2)^2=?

log 25   + [ log 2 ] [log 50] + [log 2 ] [log 2]   =

log [100/4]  + [log2] [ log 100/2] + [log2] [log2]   =

log100  - log 4  + [log2] [ log 100 - log 2] + [log2] [log 2]   =

log100  - log 4  + [log100] [ log 2] - [ log 2] [ log 2] + [log2] [log 2]

2  - log 2^2  + 2log2 -   [ log 2] [log2]  + [log2] [ log 2]  =

2 - log2^2 + log2^2  =

2

Good question. (I really don't know!)

well 45*1004=45180 

We can use the arc length formula,  S  = r θ       where S is the arc length, r is the radius and  θ   is in radians.......so we have.......

40.3 =  96.4 θ       divide both sides by 96.4

40.3 / 96.4   = θ  = about 0.418 rads  = about 23.95°

The total  possible orders of finish  = 8!  = 40320

The order we're looking for is for the three individuals to finish in the top three  [ in any order]  = 3!   = 6 and the other 5 to finish in any order = 5!   = 120

So....the probability of this occurence =     [6 * 120] / 40320   =  720/ 40320   =  1/56

(x^1/4 + 16)^2 = 144 + x^3/2

\((\frac{x^1}{4} + 16)^2 = 144 + \frac{x^3}{2}\\ \frac{1}{16}*(x+ 64)^2 = 144 + \frac{x^3}{2}\\ (x+ 64)^2 = 144*16 + 8x^3\\ x^2+128x+4096=2304+8x^3\\ x^2+96x+1792=8x^3\\ 8x^3-x^2-96x-1792=0\\ \)

Have fun solving that :)

what is the units digit of 178^83

8, 4, 2, 6, 8, ......

so

178^1 ends in 8                                             

178^2 ends in the last digit of 8*8 = 4

178^3 ends in the last digit of 4*8 = 2

178^4 ends in the last digit of 2*8 = 6

178^5 ends in the last digit of 6*8 = 8        

so now the pattern will repeat

178^(1+4n) ends in 8                                             

178^(2+4n) ends in the last digit of 8*8 = 4

178^(3+4n) ends in the last digit of 4*8 = 2

178^(4+4n) ends in the last digit of 2*8 = 6

where n is an positive integer  greater or equal to 0.

83/4=20 reminder 3

178^(3+4*20) last digit = 2

Divide both sides by 3:
x^2-(22 x)/3+140/3 = 0

Subtract 140/3 from both sides:
x^2-(22 x)/3 = -140/3

Add 121/9 to both sides:
x^2-(22 x)/3+121/9 = -299/9

Write the left hand side as a square:
(x-11/3)^2 = -299/9

Take the square root of both sides:
x-11/3 = (i sqrt(299))/3 or x-11/3 = 1/3 (-i) sqrt(299)

Add 11/3 to both sides:
x = 11/3+(i sqrt(299))/3 or x-11/3 = 1/3 (-i) sqrt(299)

Add 11/3 to both sides:
Answer: |  x = 11/3+(i sqrt(299))/3     or      x = 11/3-(i sqrt(299))/3

6.0934070630681014546073838247379e+186

1/24 is the correct answer.

Be aware though, that if your teacher is marking the test, you need to be able to calculate the wrong answer, (else otherwise your correct answer will be marked wrong).

A person would need to know something about golf to answer this question.

write different: 

(4 1/2 + 2 1/2) + 2 2/3 + 3 1/4 

= (7 + 2 2/3) + 3 1/4

= 9 2/3 + 3 1/4 -> can be written as 9 8/12 + 3 3/12 

= 12 11/12

77-16 = 61

Still 5, but now 3 are on the ground and 2 in the tree, instead of 0 on the ground and 5 in the tree.