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

1 + (6 / x - 1)

step-by-step explanation:

divide using polynomial division.

x =-.20752

step-by-step explanation:

4^(x+2)=12

4^(x+2)=12

take the log base 4 on each side

log 4 (4^(x+2) = log 4(12)

we know log b(a^y) = y log b (a)

(x+2) log 4(4) = log 4 (12)

x+2 = log 4 (12)

we know   logb c = loga c/loga b

we want to convert to base 10   so a = 10

by default, we do not write the 10   c = 4   and b = 10

log4 10 = log 12/log 4

log4 10 = log 12/ log 4

x+2 = log 12/ log 4

subtract 2 from each side

x +2 -2 = log 12/ log 4 -2

x = 1.79248125 -2

x =-.20752