So using BIMDAS(bracket,indexes,multiplied,divided,aadition,subtraction) first multiply the brackets to get 12.
Next do the minus two squared to get 4.
Next add 5+1 to get 6.
Next add 6 to 12 to get 18.
Lastly add the 18 to the 4
Sorry, I am not sure about this answer
You can report it
Since the question is cut out, I am assuming that it is asking for the GCF or LCM.
The GCF is: 3a^2
The LCM is: 27a^4b^3
To solve that question we can think by the following way:
- The first letter can be any of the letters (A, C, T or G). Therefore there are 4 possibilites.
- To choose the second letter of the sequence, any letter also can be chose. So, one more time, there are 4 possibilites.
Using the fundamental principle of counting, we will find out that can be formed
differents sequences (for example, AA, AG, TT, TG, CT...).
Know we know that, we can generalize. If I have
elements and I must choose one of them
times to do a sequence, the total number of sequence that I can form will be
.
In our question, we want to know how many DNA sequence are exactly 29 letters long. As indicated above, the answer for that is
(29 times)
.
Therefore, there are
DNA sequences with exactly 29 letters long.
I hope I've helped you. =D
Enjoy your studies. \o/
Step-by-step explanation:
![x + 12 = 0](https://tex.z-dn.net/?f=x%20%20%2B%2012%20%3D%200)
![x = - 12](https://tex.z-dn.net/?f=x%20%3D%20%20-%2012)
hope it help you.