We determine whether the two variable quantities can be represented by the equation for the inverse variation.

If the product of the two variables is always equal to the constant of variation, then the table of values represents an inverse variation.

Note that the two quantities have inverse variation if they can be represented by the equation y=\dfrac{k}{x} where k is the constant of variation and the variable quantities are x and y.

Observe that y=\dfrac{k}{x} implies k=xy. So we multiply the values of x and y to find k.

If the product of the two variables is always equal to the constant of variation k, then the table of values represents an inverse variation.

Now,

\qquadIf x=1 and y=120, then xy=1 \cdot 120=120.

\qquadIf x=2 and y=60, then xy=2 \cdot 60=120.

\qquadIf x=3 and y=40, then xy=3 \cdot 40=120.

\qquadIf x=4 and y=30, then xy=4 \cdot 30=120.

\qquadIf x=5 and y=24, then xy=5 \cdot 24=120.

This means that k=120 since xy=120 for all (x,y).

Therefore, the table of values represents an inverse variation between x and y.

We simply substitute the obtained value of the constant variation in part (a) to the equation for the inverse of variation.

From part (a), we obtained the constant variation k=120.

Substituting k=120 to the equation y=\dfrac{k}{x}, we get y=\dfrac{120}{x}

Therefore, the table of values is represented by the equation y=\dfrac{120}{x}.

We substitute each value of R to the inverse variation equation C = \dfrac {200}{R} to find the value of C.

Now,

\qquadIf R=5 , then C= \dfrac {200}{5}=40.

\qquadIf R=10 , then C= \dfrac {200}{10}=20.

\qquadIf R=20 , then C= \dfrac {200}{20}=10.

\qquadIf R=25 , then C= \dfrac {200}{25}=8.

\qquadIf R=40 , then C= \dfrac {200}{40}=5.

Therefore, the complete table of values is:

We assign the horizontal axis for variable R and the vertical axis for variable C.

We plot first the coordinates (R,C) on the coordinate plane. The points are: (5,40),(10,20),(20,10),(25,8) \text{ and }(40,5)

Then we graph the function C = \dfrac {200}{R}

The graph is given below.

The relationship between the current C and the resistance R can be modeled by C=\dfrac{200}{R}

We substitute C=50 to the equation to find R. So we get 50=\dfrac{200}{R}

Multiplying both sides by \dfrac{R}{50}, we have R=\dfrac{200}{50}=4

Therefore, the resistance in an electrical circuit when the current is 50 amperes is 4 ohms.

First, we relate the two variables by finding equation for their relationship. To find the constant of variation, we simply substitute the given values for the number of pizzas and the price of each.

We express n in terms of p using the constant of variation k.

The number of pizzas, n, is in inverse variation to the price of each pizza, p. So we have n=\dfrac{k}{p}

Since n=15 when p=8, we substitute these values to the equation to find the constant of variation k and we get 15=\dfrac{k}{8}

Multiplying both sides by 8, we obtain k=120

We use the equation which relates the two variables. To express n in terms of p, we simply substitute the value of the constant of variation obtained in part (a).

The number of pizzas, n, is in inverse variation to the price of each pizza, p. So we have n=\dfrac{k}{p}

From part (a), we obtain the constant of variation k=120.

Substituting k=120 to the equation, the expression for n in terms of p is: n=\dfrac{120}{p}

We simply use the equation obtained in part (b). To find the number of pizzas, we substitute the price of each pizza to the equation.

From part (b), we have expressed the number of pizzas, p, in terms of the price of each pizza, p.

That is, n=\dfrac{120}{p}

We substitute p=12 to the equation to find n. So we get n=\dfrac{120}{12}=10

Therefore, Charlotte can buy 10 pizzas that costs \$12.00 each.