SVS2652

Correct, Good Job!

Look at this similar problem for help with this one https://www.quora.com/If-1-a-1-b-1-c-1-then-what-is-the-value-of-a-b-and-c

https://math.stackexchange.com/questions/829831/find-the-number-of-2-element-subsets-a-b-of-1-cdots-1000-such-tha 

Use this link as reference

Consider that the three are anchored as a "block' .....  they can be seated in 3! = 6 ways

And the other 5 people can be arranged  in 5! =120 ways

So

6 x 120  = 720 different arrangements

Write z = (a + ib), then z2 = a2 - b2 +2abi

For this to equal 2i we must have

a2 - b2 = 0  and ab = 1

There are four solution sets for these: a = 1, b = 1; a = -1, b= -1; a = i, b=-i; a = -i, b = i.

 These give rise to just two different solutions:

1 + i  and -1 - i

1) Count the total number of games that each team plays; then, since that counts each game twice, divide by 2:

(number of teams) times (number of opposing teams) times (number of times per season)
6*5*9 = 270

then divide by 2
270/2 = 135

OR, to be a bit more sophisticated mathematically....

(2) Determine the number of pairings and multiply by the number of times per season that each pairing occurs:

(6 choose 2) times (number of times per season)
6C2 * 9 = 15*9 = 135

.5(31024)−(3512+1)(3256+1)(3128+1)(364+1)(332+1)(316+1)(38+1)(34+1)(32+1)(3+1)

=(0.5)(Infinity)−(3512+1)(3256+1)(3128+1)(364+1)(332+1)(316+1)(38+1)(34+1)(32+1)(3+1)

=Infinity−(3512+1)(3256+1)(3128+1)(364+1)(332+1)(316+1)(38+1)(34+1)(32+1)(3+1)

=Infinity−(1.9323349832288915e+244+1)(3256+1)(3128+1)(364+1)(332+1)(316+1)(38+1)(34+1)(32+1)(3+1)

=Infinity−1.9323349832288915e+244(3256+1)(3128+1)(364+1)(332+1)(316+1)(38+1)(34+1)(32+1)(3+1)

=Infinity−1.9323349832288915e+244(1.3900845237714473e+122+1)(3128+1)(364+1)(332+1)(316+1)(38+1)(34+1)(32+1)(3+1)

=Infinity−(1.9323349832288915e+244)(1.3900845237714473e+122)(3128+1)(364+1)(332+1)(316+1)(38+1)(34+1)(32+1)(3+1)

=Infinity−Infinity(3128+1)(364+1)(332+1)(316+1)(38+1)(34+1)(32+1)(3+1)

=Infinity−Infinity(1.1790184577738583e+61+1)(364+1)(332+1)(316+1)(38+1)(34+1)(32+1)(3+1)

=Infinity−(Infinity)(1.1790184577738583e+61)(364+1)(332+1)(316+1)(38+1)(34+1)(32+1)(3+1)

=Infinity−Infinity(364+1)(332+1)(316+1)(38+1)(34+1)(32+1)(3+1)

=Infinity−Infinity(3.4336838202925124e+30+1)(332+1)(316+1)(38+1)(34+1)(32+1)(3+1)

=Infinity−(Infinity)(3.4336838202925124e+30)(332+1)(316+1)(38+1)(34+1)(32+1)(3+1)

=Infinity−Infinity(332+1)(316+1)(38+1)(34+1)(32+1)(3+1)

=Infinity−Infinity(1853020188851841+1)(316+1)(38+1)(34+1)(32+1)(3+1)

=Infinity−(Infinity)(1853020188851842)(316+1)(38+1)(34+1)(32+1)(3+1)

=Infinity−Infinity(316+1)(38+1)(34+1)(32+1)(3+1)

=Infinity−Infinity(43046721+1)(38+1)(34+1)(32+1)(3+1)

=Infinity−(Infinity)(43046722)(38+1)(34+1)(32+1)(3+1).

or the solution could be 

3^1024*.5-8.21504....E21

Gavin is correct. This problem is about converting 9% per year to how many percents per quarter. The rate is still the same. It only differed in the time period. So, through dimensional analysis,

9% per year × (1 year / 4 quarters) = 2.25% per quarter

If he was asked to convert 9% per year to how many percents per quarter compounded quarterly, then you will not simply divide by four. This is because compounding is not linear.

135 times in all.

 is the answer.