formula is 
where A=final amount
P=principal
r=interest rate in decimal
n=number of times per year it is compounded
t=time in years
we want to find where
A=2P
and P=5745
and r=6.5%=0.065
n=monthly=12
remember that 
also that 
, solving for t
divide both sides by 5745 to simplify things a bit
I'd rather not simplify this because it give us a decimals and those aren't exact, if we combine, we get 12.065/12 for inside parenthases

take ln of both sides


divide both sides by 

using our calculator, t≈10.6927
so rounded, we get 10.7 years
Answer:
A) 1/12
Step-by-step explanation:
The sample space of rolling a die is { 1,2,3,4,5,6}
P (getting 3)= The number of "3"'s in the sample space over the number of items in the sample space
P (getting a 3} = 1/6
P (getting even)= The number of evens in the sample space over the number of items in the sample space
P (getting an even} = 3/6= 1/2
Since these are independent events, we multiply the probabilities
P(3, even) = 1/6 * 1/2 = 1/12
And its 6 600 and 202 and add 100 which means it will be 12 THATS your answer

(a)
![f'(x) = \frac{d}{dx}[\frac{lnx}{x}]](https://tex.z-dn.net/?f=f%27%28x%29%20%3D%20%5Cfrac%7Bd%7D%7Bdx%7D%5B%5Cfrac%7Blnx%7D%7Bx%7D%5D)
Using the quotient rule:


For maximum, f'(x) = 0;


(b) <em>Deduce:
</em>

<em>
Soln:</em> Since x = e is the greatest value, then f(e) ≥ f(x) > f(0)


, since ln(e) is simply equal to 1
Now, since x > 0, then we don't have to worry about flipping the signs when multiplying by x.



Taking the exponential to both sides will cancel with the natural logarithmic function in the right hand side to produce:


, as required.
\[\sum_{n=1}^{7} 2(-2)^{n-1}\]