Which comparison is correct for the values of f(x) and g(x) when x = -3
function a : f(x) = -x^2 +4
function b

a. f(-3) < g(-3)
b. f(-3)> g (-3)
c. f(-3) = g(-3)

A) g(-3) < g(-3)

Step-by-step explanation:

Given: f(x) = -x^2 + 4

When x = -3, f(-3) = - (-3)^2 + 4

Now we have to simplify it.

f(-3) = -9 + 4

f(-3) = -5

Now we have to find g(-3) from the given table.

When x = -3, g(-3) = 1 from the given table values.

We found f(-3) = -5 and g(-3) = 1

Here -5 < 1

Therefore, f(-3) < g(-3)

Answer is A) g(-3) < g(-3)

Hope you will understand the concept.

Thank you.

Choice A is the answer.

Step-by-step explanation:

We have given the values of function and a equation of function.

f(x) = -x² +4

We have to find the relation between both function when x = -3.

From table, we observe that

g(-3) = 1

Now, Putting x = -3 in f(x), we have

f(-3) = -(-3)²+4

f(-3) = -9+4

f(-3) =  -5

Hence, -5 < 1

f(-3) < g(-3) which is the answer.

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

the correct option is b

step-by-step explanation: