Brandon has a fish tank that can hold 38 gallons of water.
- on monday (1st day of the month) there were 38 gallons of water in the tank.
- the fish tank loses 0.75 gallons of water on every odd numbered day.
- brandon adds 0.5 gallons of water to the fish tank on days that are multiples of 3
- how much total water will brandon need to add on the last day of the month (31 days in this month) to fill the tank to 38 gallons
- show 2 different ways to solve this problem

7 gallons

Step-by-step explanation:

The first method

You will  start with 38 gallons of water on 1st day then as you count the days, subtract 0.75 gallons of water in every odd number day, and add 0.5 gallons of water for everyday that is a multiple of 3 up to the 31st day.Then find the difference of the value you get with the initial amount of water that the tank could hold.

The second method

Given 0.75 gallons of water are lost in every odd numbered day and 0.5 gallons of water are added in every day that is a multiple  of 3 then;

⇒ identify all days that are odd in this month.

They are 16 days

⇒Find the amount of water that will be lost in 16  odd number days of that month

16×0.75=12 gallons

⇒Find the number of days that are multiple of 3

They are 10 days

⇒Find amount of water that can be added in 10 days

10×0.5=5 gallons

⇒The water that will be in tank on 31st will be

38-12+5

(38+5)-12

43-12=31 gallons

⇒Amount of water to add

38-31=7gallons

x^6

step-by-step explanation:

this is correct because:

x-6=0

0*x=0

so

x^6

it’s the first one your welcome