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I HAVE A QUESTION!
TYPE OF QUESTION:Sciences-Chemistry
YOUR QUESTION:
What is the formula for Alkanes? is it going by ?

thanks!

[Galifey]

ooo cool idea! Bop for later

[~ moth ~]

I HAVE A QUESTION!
TYPE OF QUESTION: fst
YOUR QUESTION: we're working with the qudratic formula and graphing, so we need to find the zeroes. i can't figure out how to find the zeroes

the formula is 3x² - 6x + 2 = 0

i know how to start

6 ± √(-6)² - 4 (3) (2)
------------------------------
2 (3)

I can get to

6 ± 2 √3
-------------
6

But I can’t figure out how to find the zeroes, which are (1.6, 0) and (.42, 0)

[captain_stardust]

I HAVE A QUESTION!
TYPE OF QUESTION:Sciences-Chemistry
YOUR QUESTION:
What is the formula for Alkanes? is it going by ?

thanks!

You got it right! That's the alkane formula.
Alkene is CnH2n, and alkine is CnH2n-2

[CᴀɴᴅʏNᴜᴛᴍᴇɢ]

I HAVE A QUESTION!
TYPE OF QUESTION: fst
YOUR QUESTION: we're working with the qudratic formula and graphing, so we need to find the zeroes. i can't figure out how to find the zeroes

the formula is 3x² - 6x + 2 = 0

i know how to start

6 ± √(-6)² - 4 (3) (2)
------------------------------
2 (3)

I can get to

6 ± 2 √3
-------------
6

But I can’t figure out how to find the zeroes, which are (1.6, 0) and (.42, 0)

The +/- sign indicates that there are two numbers in one expression.
First calculate (6 + 2 √3)/6, then (6 - 2 √3)/6, and 1.6 and .42 should be the two values you get.

[--------------]

I HAVE A QUESTION!
TYPE OF QUESTION:Sciences-Chemistry
YOUR QUESTION:
What is the formula for Alkanes? is it going by ?

thanks!

You got it right! That's the alkane formula.
Alkene is CnH2n, and alkine is CnH2n-2

thanks! i was curious about

[Blumenkranz]

I HAVE A QUESTION!
TYPE OF QUESTION: Trigonometry
YOUR QUESTION:If tan(A + B) = 3, and tanB = 1/2, find tanA
I know I'm supposed to use the sum formula for tan but I'm having some trouble figuring out exactly how to get tanA. Thank you for the help!

[kh.aotic]

I HAVE A QUESTION!
TYPE OF QUESTION: Trigonometry
YOUR QUESTION:If tan(A + B) = 3, and tanB = 1/2, find tanA
I know I'm supposed to use the sum formula for tan but I'm having some trouble figuring out exactly how to get tanA. Thank you for the help!

The sum formula for tan(A + B) is as follows:
(tanA + tanB)/(1 - tanA*tanB)

Therefore:
tan(A + B) = (tanA + tanB)/(1 - tanA*tanB) = 3

Plug in 1/2 for tanB:
(tanA + (1/2))/(1 - tanA*(1/2)) = 3

Multiply both sides by (1 - tanA*(1/2):
tanA + (1/2) = 3*(1 - tanA*(1/2))

Distribute:
tanA + (1/2) = 3 - (3/2)tanA

Move like terms together. Subtract 1/2 from both sides and add (3/2)tanA to both sides:
tanA + (3/2)tanA = 3 - (1/2)

Simplify:
(5/2)tanA = 5/2

Multiply both sides by 2/5:
tanA = 1

[Blumenkranz]

@kh.aotic, okay that makes sense. Thank you so much for the help!

[bellamare]

I HAVE A QUESTION!
TYPE OF QUESTION: ap biology
YOUR QUESTION:how is the maintenance of a highly disordered living organisms possible, given the tendency of the universe to tend toward increasing disorder? provide and example for the order of an organism, and an explanation for the order of successive generations of organisms.