This is a linear differential equation of first order. Solve this by integrating the coefficient of the y term and then raising e to the integrated coefficient to find the integrating factor, i.e. the integrating factor for this problem is e^(6x).
<span>Multiplying both sides of the equation by the integrating factor: </span>
<span>(y')e^(6x) + 6ye^(6x) = e^(12x) </span>
<span>The left side is the derivative of ye^(6x), hence </span>
<span>d/dx[ye^(6x)] = e^(12x) </span>
<span>Integrating </span>
<span>ye^(6x) = (1/12)e^(12x) + c where c is a constant </span>
<span>y = (1/12)e^(6x) + ce^(-6x) </span>
<span>Use the initial condition y(0)=-8 to find c: </span>
<span>-8 = (1/12) + c </span>
<span>c=-97/12 </span>
<span>Hence </span>
<span>y = (1/12)e^(6x) - (97/12)e^(-6x)</span>
(i just changed my answer)
sqrt is the square root of
v=sqrt(2(ke)/m)
Answer:
4032 different tickets are possible.
Step-by-step explanation:
Given : At a race track you have the opportunity to buy a ticket that requires you to pick the first and second place horses in the first two races. If the first race runs 9 horses and the second runs 8.
To find : How many different tickets are possible ?
Solution :
In the first race there are 9 ways to pick the winner for first and second place.
Number of ways for first place - 
Number of ways for second place - 
In the second race there are 8 ways to pick the winner for first and second place.
Number of ways for first place -
Number of ways for second place - 
Total number of different tickets are possible is
Therefore, 4032 different tickets are possible.
C = Total Daily Cost
Total Daily Cost = (Salary per employee per day)×(Total no. of employees on that day) + Fixed Expenses.
C = 35x + 100
