Posted on 21 March 2017 - 07:21 PM
Jinxful wrote
Hm, I'm not sure on the details, but I'll give you a good guess.
Long ago, there were small chemicals compounds scattered across the planet. As time went by, these compounds combined to form more complex systems. The presence of water on earth was a big help in assisting in this, since many biochemical reactions depend on water.
I'm thinking an incredibly important compound that eventually formed was RNA and DNA. RNA is nearly the same as DNA in its composition (DNA lacks a 2' hydroxyl -OH group). However, RNA can serve as both a carrier of genetic information and as a catalyst for biochemical reactions. A particular type of RNA called ribosomal RNA makes up a great deal of ribosomes. If I recall correctly, the entire process of translation can occur without the protein parts of the ribosome acting.
If you recall from biology, to create proteins, you need to go through a process called translation, which uses a ribosome as a facilitator. With RNA in play, the manufacturing of proteins could begin, and more complex systems form. In time, these structures formed small organelles like the mitochondria and chloroplasts (see endosymbiotic theory). It's been theorized that mitochondria and chloroplasts were once precursors to modern cells. After all, they're studded with their own ribosomes, and as mentioned previously, ribosomes are pretty important.
Then more and more complex cells and structures formed until life began. I'm in no way an expert in this, but this is the best explanation you'll get from me about that.
Posted on 24 March 2017 - 06:50 PM
ok so I'm supposed to find the roots of this equationsx^2/x^2-4 = 5x-6/x^2-4
Posted on 24 March 2017 - 07:02 PM
whitpillow wrote
x^2/x^2-4 = 5x-6/x^2-4
Let's factor.
x^2/(x^2-4) = x^2/(x+4)(x-4). (RHS)
(5x-6)/(x+4)(x-4) (LHS).
We can cancel out the denominators on both sides.
Then, we have
x^2 = 5x - 6.
x^2 - 5x + 6 = 0. Factor now.
(x-3)(x-2) = 0.
Then, x= 3 or x=2.
Last edited on 24 March 2017 - 07:32 PM by whitpillow
AlbertEinstein wrote
whitpillow wrote...
Let's factor.
x^2/(x^2-4) = x^2/(x+4)(x-4). (RHS)
(5x-6)/(x+4)(x-4) (LHS).
We can cancel out the denominators on both sides.
Then, we have
x^2 = 5x - 6.
x^2 - 5x + 6 = 0. Factor now.
(x-3)(x-2) = 0.
Then, x= 3 or x=2.
Oh alright for those problems I factor out but for problems like these, what do I do before I do the quadratic formula
5y+1/y+1 = y+2/y
10/2x-3 = x-1
Posted on 24 March 2017 - 07:35 PM
whitpillow wrote
AlbertEinstein wrote...
Oh alright for those problems I factor out but for problems like these, what do I do before I do the quadratic formula
5y+1/y+1 = y+2/y
10/2x-3 = x-1
Try to cross multiply and get everything without fractions (try the LCD).
(1) Multiply both sides by y(y+1):
y(5y+1) = (y+2)(y+1)
5y^2 + y = y^2 + 3y + 2, etc.
(2) Same idea… Multiply both sides by 2x-3:
10(1) = (2x-3)(x-1) = 2x^2 -3x -2x +3 = 2x^2 -5x +3
Posted on 24 March 2017 - 07:44 PM
AlbertEinstein wrote
whitpillow wrote...
Try to cross multiply and get everything without fractions (try the LCD).
(1) Multiply both sides by y(y+1):
y(5y+1) = (y+2)(y+1)
5y^2 + y = y^2 + 3y + 2, etc.
(2) Same idea… Multiply both sides by 2x-3:
10(1) = (2x-3)(x-1) = 2x^2 -3x -2x +3 = 2x^2 -5x +3
The second problem would be no solution right because if you put everything on the left it would be 0=0
Posted on 24 March 2017 - 07:58 PM
whitpillow wrote
AlbertEinstein wrote...
The second problem would be no solution right because if you put everything on the left it would be 0=0
Remember that the ten is still there. 10 = 2x^2 -5x +3
Last edited on 26 March 2017 - 11:03 PM by sdfsdgsdgsfgafdg
Does anyone know how to program a TI-84 calculator? It's a math project that I need to do. Any help would be greatly appreciated.Also, "AlbertEinstein", you seem very knowledgeable in multiple sectors of academia. Are you in college? Just wondering.
Posted on 26 March 2017 - 11:15 PM
PresumptionV2 wrote
Also, "AlbertEinstein", you seem very knowledgeable in multiple sectors of academia. Are you in college? Just wondering.
Yes.
Posted on 26 March 2017 - 11:16 PM
AlbertEinstein wrote
PresumptionV2 wrote...
Yes.
Are you competent in progamming TI-84 calculators though?
Posted on 26 March 2017 - 11:35 PM
PresumptionV2 wrote
AlbertEinstein wrote...
Are you competent in progamming TI-84 calculators though?
No, I'm not! Lots of people on here enjoy programming though. So maybe they could chime in.
Last edited on 27 March 2017 - 05:37 AM by nandrolone
Employ a matrix multiplication method with the following enciphering matrix:A = [5 6]
- - - [1 3]
to encode the message "DO YOUR BEST"
Plain: - - - -A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
Cipher: - - D E F G H I J K L M N O P Q R S T U V W X Y Z A B C
(The dashes are used in places of space, because Badlion excludes multiple spaces)
Posted on 27 March 2017 - 07:08 AM
What's 2+2?Posted on 27 March 2017 - 10:49 AM
McFlops wrote
Very original post. Nobody's asked a question on here like that before.
Posted on 28 March 2017 - 09:57 PM
So, anyone know how to program a TI-84 calculator?Posted on 31 March 2017 - 01:07 AM
how would you find all possible values of the equation 1/y if5<y<8 and 0.125<y<0.25
answer should be something like blank < 1/y < blank
Posted on 31 March 2017 - 01:31 AM
SporkHandles wrote
5<y<8 and 0.125<y<0.25
answer should be something like blank < 1/y < blank
Let's take the reciprocal of each of these inequalities. Note that when you do this, you must flip the signs of the inequalities, similar to what you do when you multiply by negatives:
.2 > 1/y > .125 and 8 > 1/y > 4.
Now, if you mean AND in a strict sense of how 1/y must satisfy both those pairs of inequalities, then there is no such 1/y.
But individually, those are the answers. Just take the reciprocal of each and flip the inequalities.
Posted on 31 March 2017 - 01:35 AM
AlbertEinstein wrote
SporkHandles wrote...
Let's take the reciprocal of each of these inequalities. Note that when you do this, you must flip the signs of the inequalities, similar to what you do when you multiply by negatives:
.2 > 1/y > .125 and 8 > 1/y > 4.
Now, if you mean AND in a strict sense of how 1/y must satisfy both those pairs of inequalities, then there is no such 1/y.
But individually, those are the answers. Just take the reciprocal of each and flip the inequalities.
what would you do if the answer has to be in the form of blank < 1/y < blank
Posted on 31 March 2017 - 01:44 AM
SporkHandles wrote
AlbertEinstein wrote...
what would you do if the answer has to be in the form of blank < 1/y < blank
Just turn it around for each one:
.125 < 1/y < .2 and 4 < 1/y < 8.
Posted on 01 April 2017 - 08:04 AM
@AlbertEinstein, I request a response to my previous post. Please and thank you!