Question about Monte Cristo Games 7 Sins

Episode in which Miss Lemon Loses her Keys is titled 'Double Sin'.

Sep 04, 2016 | Agatha Christie's Poirot (1989)

How to save 7sins in ps2 plz help me out

Mar 05, 2013 | Monte Cristo 7 Sins for PlayStation 2

The sine of 30 radians is about -0.988 while the sine of 30 degrees is 0.5 .

To change the angular setting, press MODE, down-arrow to the "Radian Degree" line, highlight Degree, press ENTER, then 2nd QUIT to exit the mode screen.

To change the angular setting, press MODE, down-arrow to the "Radian Degree" line, highlight Degree, press ENTER, then 2nd QUIT to exit the mode screen.

Jul 26, 2012 | Texas Instruments TI-84 Plus Calculator

I'm sorry but this is not a warez site. We don't give out activation keys or codes.

Jun 04, 2012 | Stardock Sins of A Solar Empire Trinity

Sorry but this is not a warez site....

May 31, 2012 | Stardock Sins of A Solar Empire Trinity

If you can find a copy of 7 Sins for the PS2, then yes. However, if you mean the disk you use with your pc, no it will not. PS2 uses a different OS than Windows to execute games- which means you cannot natively play a ps2 game in Windows, conversely you cannot play a windows game on the PS2

Dec 16, 2010 | Atari 7 Sins for Windows

Use the rule for differentiating products of functions: ()' signifies derivative

(29*sin(2X)*sin(X))'= (29)'*sin(2X)*sin(X) +29* (sin(2X))'*sin(X) +29*sin(2X)*(sin(X))'

But

(29*sin(2X)*sin(X))'= 29*2*cos(2X)*sin(X)+29*sin(2X)*cos(X)

You could also have cast your formula in the form

sin(2X)*sin(X)= 1/2[ cos(2X-X)-cos(2X+X)]=1/2[cos(X)-cos(3X)]

then calculated the derivative of

29/2*[cos(X)-cos(3X)]

which is

29/2*[-si(X) +3*sin(3X)]

The challenge for you is to prove that the two forms are equivalent

29*2*cos(2X)*sin(X)+29*sin(2X)*cos(X)=29/2*[-si(X) +3*sin(3X)]

(29*sin(2X)*sin(X))'= (29)'*sin(2X)*sin(X) +29* (sin(2X))'*sin(X) +29*sin(2X)*(sin(X))'

But

- (29)'=0 derivative of a constant is zero
- (sin(2X))'=cos(2X)*(2X)'=2*cos(2X)
- (sin(X))'=cos(X)

(29*sin(2X)*sin(X))'= 29*2*cos(2X)*sin(X)+29*sin(2X)*cos(X)

You could also have cast your formula in the form

sin(2X)*sin(X)= 1/2[ cos(2X-X)-cos(2X+X)]=1/2[cos(X)-cos(3X)]

then calculated the derivative of

29/2*[cos(X)-cos(3X)]

which is

29/2*[-si(X) +3*sin(3X)]

The challenge for you is to prove that the two forms are equivalent

29*2*cos(2X)*sin(X)+29*sin(2X)*cos(X)=29/2*[-si(X) +3*sin(3X)]

Jun 21, 2010 | Vivendi Excel@ Mathematics Study Skills...

Hii..this can be done as follow

cos^2 x = 1 - sin^2 x--------(1)

2cos^2 x=1+ sin x

2cos^2 x - sin x -1=0

Substituting formula (1)

2(1 - sin^2 x) - sin x - 1 = 0

2sin^2 x + sin x - 1 = 0

Factor this

(2 sin x - 1)(sin x +1) = 0

2 sin x - 1 = 0 or sin x +1 = 0

sin x = 1, and sin x = -1

cos^2 x = 1 - sin^2 x--------(1)

2cos^2 x=1+ sin x

2cos^2 x - sin x -1=0

Substituting formula (1)

2(1 - sin^2 x) - sin x - 1 = 0

2sin^2 x + sin x - 1 = 0

Factor this

(2 sin x - 1)(sin x +1) = 0

2 sin x - 1 = 0 or sin x +1 = 0

sin x = 1, and sin x = -1

Apr 26, 2010 | SoftMath Algebrator - Algebra Homework...

Hi there did the system requirements of the game meet your system?... if yes try to ask your local computer shop to try it on their computer then you'll see if it is defected or not...

Jan 03, 2009 | Atari 7 Sins for Windows

Change csc to 1/sin. Find a common denominator and add the two left terms.

1/sin - sin = (1 -sin^2)/sin. Rewrite formula

(1 - sin^2)/sin = cos^2/sin Divide out the /sin.

1 - sin^2 = cos^2 Rearange.

1 = cos^2 + sin^2 Yes, that's true. It's like the Pythagorean formula.

1/sin - sin = (1 -sin^2)/sin. Rewrite formula

(1 - sin^2)/sin = cos^2/sin Divide out the /sin.

1 - sin^2 = cos^2 Rearange.

1 = cos^2 + sin^2 Yes, that's true. It's like the Pythagorean formula.

May 22, 2008 | Super Tutor Trigonometry (ESDTRIG) for PC

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