Answer:
The attached file contains the solution to the question
Answer:
x1Rules=true;
for (i=0; i<2; i++)
for (j=0; j<3; j++)
if (x1[i][j] <= x2[i][j])
{
x1Rules=false;
break;
}
Step-by-step explanation:
The code is written above .
The answer is C!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Answer:
Therefore the concentration of salt in the incoming brine is 1.73 g/L.
Step-by-step explanation:
Here the amount of incoming and outgoing of water are equal. Then the amount of water in the tank remain same = 10 liters.
Let the concentration of salt be a gram/L
Let the amount salt in the tank at any time t be Q(t).

Incoming rate = (a g/L)×(1 L/min)
=a g/min
The concentration of salt in the tank at any time t is =
g/L
Outgoing rate =



Integrating both sides

[ where c arbitrary constant]
Initial condition when t= 20 , Q(t)= 15 gram


Therefore ,
.......(1)
In the starting time t=0 and Q(t)=0
Putting t=0 and Q(t)=0 in equation (1) we get









Therefore the concentration of salt in the incoming brine is 1.73 g/L
Answer:
x = 45
Step-by-step explanation:
Well here's what we know:
2x + 45 + x = 180
Solve from there:
3x = 135
x = 45
You can plug it back in to check:
2(45) + 45 + 45 = 180
90 + 80 = 180, so it checks out :)