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3.7006

Yes I agree. Max has already answered at length.  You are being very rude.

If you want to ask questions about what Max has done then ask on the original thread.

You can post a new thread but only to direct people back to the original thread.

It must have the address of the original question in it!

Perhaps the real problem is that you cannot find your original question and answer.

Easy fix.     BECOME A MEMBER,  then your questions will be stored in your watchlist and you can get email alerts for new answers!

Now I have written all this you probably have already 'lost' this thread.     

14.1421356237

Totally agree with CPhill :D

(d) is correct

Every time x changes by a positive 1,  y  changes by  positive 2  = rate of change of f(x) is 2

So......we have the points  (3, −10)  (4, − 8) (5, −6) (6, −4)

Well, I should have specified no REAL solution.....

Yikes...you DO like those long calcs don't you ???

Thanx !!

( 1 + 2 + 4 + 16)^2  =

(7 + 16)^2 

23^2  =

529

\(10sqrt{2}\)

The answer in red is the water answer, silly!   10.19 m

e = 2.7182818284590452

or  Euler's number

\(\sum_{1}^{infin}\) (1+1/n)^n

Subtract 'A' from both sides and you are left with:

0 <= B     SO B    must  'B' .... greater than or equal to zero.

Ninja's answer is correct in absolute dollars. As a percentage, you have:

3,292,822 / 2,500,000 =1.3171288 - 1 x 100 =31.71% - percentage that you exceeded your goal by.

P(3 girls)   = C(5,3)(1/2)^5  = 5/16

P(4 girls)  = C(5,4)(1/2)^5  = 5/32

P(5 girls)= C(5,5)(1/2)^5  = 1/32

So  P (3 or more girls)  =  5/16 + 5/32  + 1/32  =   10/32 + 5/32 + 1/32  = 16/32  = 1/2

Just as  NinjaA found.....!!!!!

3292822-2500000 = 792822<<<------Answer

In all cases, at least half of John's kids will be boys or at least half will be girls. Furthermore, since John has an odd number of children, these conditions are mutually exclusive--that is, they are never true at the same time. Since a boy is equally likely to be born as is a girl, our answer is therefore 1/2.

but the answer has to be with water not mercury

Incorrect

Well, obiously you can't have 2 1/5 children. So the probability that John has 3 girls and 2 boys is:

((3 + 2)!)/(3! 2! 2^(3 + 2)) = 5/16 ≈ 0.3125 ≈ 1/3.2
(assuming children are independent and male and female are equally likely)

10(.80) in  by 7(.80)in  =   8 in   by  5.6 in

Let x be the number of purple ones to be added.......and we have....

40 / [60 +  x]   = 5/12    cross multiply

40 [ 12]  = [60 + x] 5     simplify

480  = 300 + 5x         

180  =  5x

36   = x

Proof :

40 / [60+36]   =  40/ 96  = 5/12

Solve for x:
cos(x) + sec(x) = 1

Multiply both sides of cos(x) + sec(x) = 1 by cos(x):
1 + cos^2(x) = cos(x)

Subtract cos(x) from both sides:
1 - cos(x) + cos^2(x) = 0

Subtract 1 from both sides:
cos^2(x) - cos(x) = -1

Add 1/4 to both sides:
1/4 - cos(x) + cos^2(x) = -3/4

Write the left hand side as a square:
(cos(x) - 1/2)^2 = -3/4

Take the square root of both sides:
cos(x) - 1/2 = (i sqrt(3))/2 or cos(x) - 1/2 = 1/2 (-i) sqrt(3)

Add 1/2 to both sides:
cos(x) = 1/2 + (i sqrt(3))/2 or cos(x) - 1/2 = 1/2 (-i) sqrt(3)

Take the inverse cosine of both sides:
x = cos^(-1)(1/2 + (i sqrt(3))/2) + 2 π n_1 for n_1 element Z or x = 2 π n_2 - cos^(-1)(1/2 + (i sqrt(3))/2) for n_2 element Z
 or cos(x) - 1/2 = 1/2 (-i) sqrt(3)

Add 1/2 to both sides:
x = cos^(-1)(1/2 + (i sqrt(3))/2) + 2 π n_1 for n_1 element Z
 or x = 2 π n_2 - cos^(-1)(1/2 + (i sqrt(3))/2) for n_2 element Z
 or cos(x) = 1/2 - (i sqrt(3))/2

Take the inverse cosine of both sides:
Answer: |
 | x = cos^(-1)(1/2 + (i sqrt(3))/2) + 2 π n_1 for n_1 element Z
 or x = 2 π n_2 - cos^(-1)(1/2 + (i sqrt(3))/2) for n_2 element Z
 or x = cos^(-1)(1/2 - (i sqrt(3))/2) + 2 π n_3 for n_3 element Z or x = 2 π n_4 - cos^(-1)(1/2 - (i sqrt(3))/2) for n_4 element Z

....another thought...the secant and cosine of x are always of the same sign  (sec = 1/cos )   so for this equation to be answered they both need to be POSITIVE.    Now Secant is ALWAYS >= 1     so this question has no solution!

Rauhan: Look at here and see how it is done correctly:

1) -

FV = PV x [1 + R]^N, FV=Future value, PV=Present value, R=Interest rate, N=Number of periods.

FV =661 x [1 + 0.031/12]^(15*12]

FV =661 x [ 1.0025833]^180

FV =661 x        1.59106

FV =$1,051.69 The balance in Amy's account in 15 years.

2) -

PV = FV / [1 + R]^N

PV = 419.21 / [ 1 + 0.055/4]^(7*4)

PV = 419.21 / [ 1.01375]^28

PV = 419.21 /      1.465765

PV =$286.00 The amount Ricardo deposited 7 years ago.