Answer:
a=2
Step-by-step explanation:
-15a-19-35=-84
-15a-19=-49
-15a=-30
a=2
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
You re write the equation with parentheses as (8x5)+(4x2)+(3x2).
Answer:

Step-by-step explanation:
A complex number is a number which has some real part and some imaginary part.
Standard form of a complex number is represented as

Where
is the real part,
and
is the imaginary part.
And 
Given complex number:

Hence, the standard form is
.
Well this is pretty simple. So the first thought is that the peanut butter would be 10$ and the jam would be 0.20$, however, the peanut butter would not be 10$ more. Instead, subtract the 10$ from the total, which gives you 0.20$, and then divide that by two. Now you have 0.10$ for each, along with another 10$ for the peanut butter. The peanut butter would be $10.10, and the jam would be 0.10$ (that's pretty cheap!).