An aquifer holds 146,000 cubic feet of water. water exits through a well at a steady rate of 500 cubic feet per day, and the aquifer is replenished at a constant rate of x cubic feet per day. if the water in the aquifer is used up in a year (365 days), what is the value of x? ignore evaporation and other losses.

The value of x = 100  ft^3/ day

Step-by-step explanation:

We need to look at what amount of water comes in and what amount of water leaves

What comes in:  x ft^3/ day

What leaves : 500x ft^3/ day

We start with 146 000 ft^3

Water in tank = what we start with + what comes in - what leaves

water in tank = 146000 + x*d - 500*d  where d is the number of days

After 365 days , the tank has 0 ft^3 of water

0  = 146000 + x*365 - 500*365

Multiply

0 = 146000 + 365x - 182500

Combine like terms

0 = -36500+365x

Add 36500 to each side

36500 = 36500-36500+365x

36500 = 365x

Divide by 365

36500/365 = 365x/365

100=x

The value of x = 100

y = 2x

step-by-step explanation:

if you divide each number in the sequence, you'll see that they all have a multiple factor of 2. each number in the series, and any other numbers which may continue, will be multiplied by 2. therefore, y is the next number in the sequence, and x is the current one, so, our formula will be y = 2x.