The temperature of a chemical solution is originally 21^\circ\text{C}21 ? C21, degree, C. A chemist heats the solution at a cons
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Answer:
Step-by-step explanation:
The given system of linear equations is:
We have to solve the equations for the values of .
Let us use elimination method in which we eliminate one of the variables from the two variables.
For this, let us multiply the first equation by and second equation by
Now, the equations become:
Now, let us add (1) and (2):
Using the equation:
Putting value of :
So, the answer to the system of linear equations is:
Let r = (t,t^2,t^3)
Then r' = (1, 2t, 3t^2)
General Line integral is:
The limits are 0 to 1
f(r) = 2x + 9z = 2t +9t^3
|r'| is magnitude of derivative vector
Fortunately, this simplifies nicely with a 'u' substitution.
Let u = 1+4t^2 +9t^4
du = 8t + 36t^3 dt
After integrating using power rule, replace 'u' with function for 't' and evaluate limits:
Then r' = (1, 2t, 3t^2)
General Line integral is:
The limits are 0 to 1
f(r) = 2x + 9z = 2t +9t^3
|r'| is magnitude of derivative vector
Fortunately, this simplifies nicely with a 'u' substitution.
Let u = 1+4t^2 +9t^4
du = 8t + 36t^3 dt
After integrating using power rule, replace 'u' with function for 't' and evaluate limits: