Compound interest of $1000 is invested at 10% compounded continuously, the future value s at any time t in years is given by s=1
000 e^0.1t how log before the investment doubles?
1 answer:
Answer: 6.93 years
Step-by-step explanation:
Given
Rate of interest is 
Future value is given by 
For the investment to double itself, i.e. 

It takes around 6.93 years to double the investment.
You might be interested in
Answer:
(-7,0) and (-5,0)
Step-by-step explanation
1. 0= (x + 7)
-7 -7
-7=x
(-7,0)
2. 0= (x + 5)
-5 -5
-5=x
(-5,0)
(-5/3)•(-6/1)=(30/3)
(30/3) simplified is (10/1)
Is there more to this question?
Perimeter of rectangle = 2(l+w)
So, 54cm = 2(l+9cm)
=> 54cm = 2l + 18cm
=> 54cm - 18cm = 2l
=> 36cm = 2l
=> 36cm/2 = l
=> 18cm = l