What is the solution to the equation

a. m = 3
b. m = 6
c. all real numbers
d. no solution

The answer is "D" NO SOLUTION.

Step-by-step explanation:

I must correct your given equation because it must read like this:

(3/(m+3)) - (m/(3-m)) = (m∧2+9)/(m∧2-9)  

Now we can solve it:

First we will transfer the fraction from the right to the left side of the equation.

(3/(m+3)) - (m/(3-m)) - (m∧2+9)/(m∧2-9) = 0

We will change binomial sign (3-m) = - (m-3) and that change sign of the second fraction from - to + , also we will change (m∧2-9) which is difference squared to (m-3)(m+3) and we get:

(3/(m+3)) + (m/(m-3)) - (m∧2+9)/(m-3)(m+3) = 0

The smallest common denominator for this equation is (m+3)(m-3)

when we multiply entire equation with (m+3)(m-3) we get:

3·(m-3) + m·(m+3)-(m∧2+9) = 0 => 3m-9+m∧2+3m-m∧2-9=0 =>

6m-18=0 => 6m=18 => m=18/6 = 3 => m=3

But we have conditions (m+3)(m-3) ≠ 0 => m+3≠0 and m-3≠0 => m≠-3 and m≠3

We conclude that this equation does not have solution.

Good luck!!!

nothjing

step-by-step explanation:

step-by-step explanation:

we can use this expression to calculate the energy gain as a particle moves with a field or the energy loss as it moves against it. (note that it is also used to define the electron-volt, a unit of energy equal to that gained by an electron as it falls through a pd of 1 v, i.e. w = 1 ev = 1.6 10-19 c 1 v = 1.6 10-19j).in fact this expression is really nothing more than the statement that potential is the potential energy per unit charge. in this case, v= w/qis simply stating that the change in potential energy(or work done) per unit charge is equal to the change in potential (or potential difference), v.student questions: uniform electric fields2electron beamcathode+5 kvanode0 v