Answer: so i think for part a , the first term is a subscrpit 0 which is 10. Then the formula for any term in the sequence is Ao x R ^n. So R =2 so
1st term = 10 ( 2^0)= 1
2nd term = 10 (2^1) =20
3rd term = 10 (2^2)= 40
4th = 80
5th = 160
Part B
The limit as n approaches infinity for a sub n when R is 0, 1 and 2. THE LIMIT FOR R of 0, 1 and 2 is infinity.
Part C...i will have to research...sorry
Step-by-step explanation:
Answer:
hello
Step-by-step explanation:
it must be 0,125
hope it helps
If you mean you get them by diving them by 100 or something over 100(as in a fraction) that's the answer, I'm incredibly sorry if it is not what you meant!
Answer:
12
Step-by-step explanation:
Answer:
![3m\sqrt[5]{2m^4p^4}](https://tex.z-dn.net/?f=3m%5Csqrt%5B5%5D%7B2m%5E4p%5E4%7D)
Step-by-step explanation:
We want to find the fifth root of 
. In order to do so, we need to factorise
Let's factorise 486 first:
486 = 2 * 243 = 2 * 
Ah, we see that can be taken out and becomes 3 outside of the 5th root since the 5th root of
Now look at the variables. We see that since we have p^"4", whose exponent is less than 5, it's impossible for us to write it as a power of 5, so we leave this in the root.
We also have m^"9", which can be written as m^"5" * m^"4". Again, we see that the m^"4" term will have to remain inside the root, but we can take out the m^"5", which becomes m.
Our final answer is thus: .
<em>~ an aesthetics lover</em>
