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A:

The amount received from the currency option hedge (after considering the premium paid) is $325,500. 

This is calculated by taking the 500,000 Australian dollars, multiplying it by the spot rate of $0.67, and then subtracting the cost of the options ($0.02 * 500,000 = 10,000). 

500,000 * 0.67 - 10,000 = 325,500.

What is the 2009th term in the arithmetic sequence whose first four terms are 2, 10, 18, 26?      

Let's see if we can get a little pattern going here 

We can see that the increment is 8, so, tabulating    

Term

Number (n)         Value    

         1                 2 + (8)(1 – 1)    

         2                 2 + (8)(2 – 1)    

         3                 2 + (8)(3 – 1)    

         4                 2 + (8)(4 – 1)    

         n                 2 + (8)(n – 1)    

   2009                 2 + (8)(2009 – 1)    

The 2009^th term therefore is 16,066    

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Let $a_1, a_2, a_3,\dots$ be an increasing arithmetic sequence of integers. If $a_4a_5 a_6 = 6$, what is $a_3a_7$?      

Do you mean a₄ + a₅ + a₆ = 6 ?    

If so, I think that would be 1 + 2 + 3 = 6    

which would make a₃ = 0 and a₇ = 4     

if I understand the problem correctly.   

_.

How many integers belong to the arithmetic sequence 13, 20, 27, 34, $\dots$, 97?    

13, 20, 27, 34, 41, 48, 55, 62, 69, 76, 83, 90, 97    I count 13 of them.   

_.

The sequence 24, $a$, $b$, $c$, $d$ 60 is arithmetic. What is the value of $a+b+c+d$?    

31.2 + 38.4 + 45.6 + 52.8  =  168.0    

The increment in the arithmetic sequence is add 7.2 each time.    

_.

I suspect mingus means \(\frac{-47+69i}{7+6i}\) though only he/she will know for certain!

A

         12

                   X

                          4

B                               C

BC^2 - XC^2 = BX^2

AB^2 - AX^2 =  BX^2

BC^2  - 4^2  = AB^2 - 12^2

BC^2 = AB^2 - 144 + 16

BC^2  = AB^2 - 128

BC^2 + AB^2  = AC^2

AB^2 -128 + AB^2  = 16^2

2AB^2 - 128 = 256

2AB^2  = 384

AB^2 = 192

AB^2 = AX^2 + BX^2

192 = 144 + BX^2

48 = BX^2

sqrt (48) = BX  =  4 sqrt (3)

Area =  (1/2) AB * BC * sin (ABC)  =

(1/2) (6)(8) * sqrt (2) / 2 =

12 sqrt (2)

This is a right triangle

Area = (1/2) (15) (20)  =  150

The shortest altitude is  drawn  to  the  longest side

Area =  (1/2) (25) (altitude)

150 = (1/2) (25) (altitude)

300 = 25 * altitude

300 / 25 =   12  = altitude

Triangle PUQ is similar to  triangle TUS

ST = 3 / (2 + 3) * SR  =  (  3/5) * (20) =  12

So ST / PQ = 12 / 20  =  3/ 5

Then  the altitude of PUQ =  QV

And the altitude of TUS = RV

So  RV  /  QV =  3 / 5

And QV / QR =  QV / ( QV + RV)  =   5 / ( 5 + 3) =  5 / 8

So QV = (5/8)QR

And triangle  QUV is similar to  triangle QSR

So

UV / QV =  SR / QR 

UV /  (5/8)QR = 20 / QR

UV  / (5/8) = 20

UV  = (5/8) (20)    =  12.5

Using similar triangles (twice)....let AD = x  and  side of the square=  s

AB / AE =  BC / DE

sqrt (10)  / sqrt (s^2 + x^2)  =  3 / s

10 / (s^2 + x^2)  = 9 / s^2

10s^2 = 9s^2 + 9x^2

s^2 = 9x^2

s = 3x    → x = s/3

And

AC / FG  = AB / BG

1 / s  =  sqrt (10)  / sqrt [ (sqrt 10 - 4x)^2 + s^2 ]

1/s^2 =  10 / [ (sqrt 10 - (4/3)s)^2 + s^2 ]

(sqrt 10 - (4/3)s )^2  + s^2  =  10s^2

9s^2 =  (sqrt 10 - (4/3)s)^2

9s^2  = 10 - (8sqrt10)s/3 + 16s^2/9

(65/9)s^2 - (8/3)sqrt (10)s - 10 = 0

Solving this for  s produces   

s = (3/13)sqrt (10)  =   sqrt (90) / 13

Area of the  square =  90 / 169

You have just posted 6 questions within 20 minutes. I understand that you may feel stuck and it is totally ok to ask here, but it really looks like you are just spamming your homework on here for us to solve.