Let after n week both tanks have same quantity of water .
for first tank
=============
tank hold 540 gallon of water ,
leaking of water 4 gallon /week
so, in n week water leak from the tank =4× n = 4n
now, after n week quantity of water in the tank = 540 -4n
for second tank
==============
quantity of water in the tank =720 gallon
leaking of water at the rate = 7gallon/week
so, after n week water leak from the tank = 7× n =7n
after n week quantity of water rest in the tank = 720 -7n
now ,
according to question ,
water in first tank = water in second tank
540 -4n = 720 -7n
7n -4n = 720 -540
3n = 180
n = 60
hence , after 60 week both tank have same quantity of water.
Yes this is the right answer
The answer to this is log14 4
This is an exponential equation. We will solve in the following way. I do not have special symbols, functions and factors, so I work in this way
2 on (2x) - 5 2 on x + 4=0 =>. (2 on x)2 - 5 2 on x + 4=0 We will replace expression ( 2 on x) with variable t => 2 on x=t =. t2-5t+4=0 => This is quadratic equation and I solve this in the folowing way => t2-4t-t+4=0 => t(t-4) - (t-4)=0 => (t-4) (t-1)=0 => we conclude t-4=0 or t-1=0 => t'=4 and t"=1 now we will return t' => 2 on x' = 4 => 2 on x' = 2 on 2 => x'=2 we do the same with t" => 2 on x" = 1 => 2 on x' = 2 on 0 => x" = 0 ( we know that every number on 0 gives 1). Check 1: 2 on (2*2)-5*2 on 2 +4=0 => 2 on 4 - 5 * 4+4=0 => 16-20+4=0 =. 0=0 Identity proving solution.
Check 2: 2 on (2*0) - 5* 2 on 0 + 4=0 => 2 on 0 - 5 * 1 + 4=0 =>
1-5+4=0 => 0=0 Identity provin solution.
Answer:
60
Step-by-step explanation:
