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Which game?

NINJA SQUAD 24/7

THE MAN OF BATS

what up with bateman

BATEMAN!!!

Shades.......we have to be careful about multiplying across these identities like the guest did......we might end up with something that isn't true  = not an identity.....here's another (better) approach

(2 + cos2y - 5 sin y)/(3 + sin y) = 1 - 2 sin y        [ remember cos2y = 1 - 2sin^2y]

(2 + 1 - 2sin^2y - 5siny) / (3 + siny)      simplify the numerator

(-2sin^2y - 5siny + 3) / ( 3 + siny)     factor the numerator

(siny + 3)(-2siny + 1) / (3 + siny)     rearrange the factors slightly

( 3 + siny) (1 -2siny) / (3 + siny)            3 + siny  "cancels" on top/bottom

1 - 2siny    =   1 - 2siny

I do not really like batman. I like superman more.

a.k.a, 0.5. That's what my friend here was trying to say.

??? Angry?

Ok, got it. 

1500

batman !

Ohh? Confused with my school's teacher's explanations I guess... I can't even get a single question correct...

There's a calculator, you know? 

60*25 = 1500

No Jack, it is 0.5

y not batman like really

BATMAN!!

I go for superman!!!!!!!!!!1

Why would you care? You are the "Superwoman"!.

1.5

In the Batman vs Superman comic, Alfred beats the s**t out of Superman, so I'm gonna go with Batman since his BUTLER can beat the s**t out of Superman.

Shades....most of this is straight out of Euclid's Elements......Book VI....Proposition 2

Here it is: http://aleph0.clarku.edu/~djoyce/java/elements/bookVI/propVI2.html

I won't repeat it verbatim, because I can't do a better job than Eulclid....LOL!!!!

You need to use the second part  under the long dashed line that begins with...

"Next, let the sides AB and AC of the triangle ABC be cut proportionally, so that BD is to AD as CE is to AE. Join DE."

At the end of the proof....it is shown that the lines are parallel.....now....all we need to add is this :

m<ADE  = m<ABC       the corresponding angles formed by a transversal  [in this case, the transversal AB]  cutting two parallel lines are congruent

Thus, by AA congruency.....triangle ADE ≈ triangle ABC

And.....similar triangles are similar in all respects......thus AD  is half of AB....therefore....DE  is half of BC  [ thus....the line segment joining the sides  = 1/2 of the base]

Let me know if you get stuck......I'll see if I can help you out....!!!!

Verify the following identity:

(2+cos(2 y)-5 sin(y))/(3+sin(y)) = 1-2 sin(y)

Multiply both sides by 3+sin(y):

2+cos(2 y)-5 sin(y) = ^?(1-2 sin(y)) (3+sin(y))

cos(2 y) = 1-2 sin(y)^2:

2+1-2 sin(y)^2-5 sin(y) = ^?(1-2 sin(y)) (3+sin(y))

2+1-2 sin(y)^2-5 sin(y) = 3-5 sin(y)-2 sin(y)^2:

3-5 sin(y)-2 sin(y)^2 = ^?(1-2 sin(y)) (3+sin(y))

(1-2 sin(y)) (3+sin(y)) = 3-5 sin(y)-2 sin(y)^2:

3-5 sin(y)-2 sin(y)^2 = ^?3-5 sin(y)-2 sin(y)^2

The left hand side and right hand side are identical:

Answer: |(identity has been verified)

1 4/6

Stop trolling!!!!!!!!!!!!!!!!!!!