Use the given graph to determine the limit, if it exists. a coordinate graph is shown with a horizontal line crossing the y axis at six that ends at the open point 2, 6, a closed point at 2, 1, and another horizontal line starting at the open point 2, negative 3. find limit as x approaches two from the left of f of x. and limit as x approaches two from the right of f of x..

The limit of the function does not exists.

Step-by-step explanation:

From the graph it is noticed that the value of the function is 6 from all values of x which are less than 2. At x=2, the line y=6 has open circle. It means x=2 is not included.

For x<2

The value of the function is -3 from all values of x which are greater than 2. At x=2, the line y=-3 has open circle. It means x=2 is not included.

For x>2

The value of y is 1 at x=2, because of he close circles on (2,1).

For x=2

Therefore the graph represents a piecewise function, which is defined as

The limit of a function exist at a point a if the left hand limit and right hand limit are equal.

The function is broken at x=2, therefore we have to find the left and right hand limit at x=2.

Since the left hand limit and right hand limit are not equal therefore the limit of the function does not exists.

the solution is h=10 because you add 9 to each side of the equation.

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