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

step-by-step explanation:

in the absence of parentheses, i will have to make some assumptions:   i interpret your (3x+5/y^6)(4y^3/5)​ to mean:

3x + 5     4y^(3/5)       (3x + 5)(4)

* =

    y^6             1               y^(6-3/5)

if this is not what you expected, repost your question using parentheses to indicate whether you meant

4y^3

    or   4y^(3/5)

    5

answer

a the answer hopes this

step-by-step explanation: