Vera is using her phone. Its battery life is down to \dfrac25 5 2 start fraction, 2, divided by, 5, end fraction , and it drains another \dfrac19 9 1 start fraction, 1, divided by, 9, end fraction every hour.

6 hrs

Step-by-step explanation:

Given:-

- The current battery, xi = 2/5

- The drain rate, r = 1/9 per hour

Find:-

how many hours will the battery last

Solution:-

- The battery after every (nth) hour would be (an) with initial value of (xi) at a rate of (r). A geometric sequence can be developed:

                               an = xi*r^(n-1)

- Hence,

                               an = (2/5)*(1/9)^( n - 1 )

- When Vera is out of battery, an = 0:

                               0 = (2/5)*(1/9)^( n - 1 )

- We will use an approximation for 0% battery upto 5 decimal places:

                               n = 6

                              an = 0.00000677404

yes

step-by-step explanation:

-2

7 - 9 = -2

reversed:

-2 + 9 = 7