Janice will be 20 years old in 12 years, so she is $\displaystyle 20-12= 8$ years old now. Marcia is four times as old, or $\displaystyle 4 \times 8 = 32$ years old.

Let $\displaystyle N$ be the number of years in which Marcia's age will be three times Janice's age. Janice will be $\displaystyle N + 8$ years old, and Marcia will be $\displaystyle N + 32$ years old. Since Marcia will be three times as old as Janice, we can set up and solve the equation:

$\displaystyle 3\left (N + 8 \right ) = N + 32$

$\displaystyle 3N + 24 = N + 32$

$\displaystyle 3N + 24-N - 24 = N + 32 -N - 24$

$\displaystyle 2N = 8$

$\displaystyle N = 4$

The total cost of beans is the price per pound times the number of pounds. The total cost of the mixture is equal to that of the Strawberry Dream beans and that of the Vanilla Madness beans. Each can be found by multiplying the cost per pound by the number of pounds:

Strawberry Dream: $\displaystyle \$ 9 \cdot S$

Vanilla Madness: $\displaystyle \$14 \cdot (80-S)$, since the total number of pounds is eighty.

Total mixture: $\displaystyle \$12 \cdot 80 = \$960$

Add to get the equation:

$\displaystyle 9S + 14 (80-S)= 960$

There are $\displaystyle f - g - 2$ pigeons remaining on the wire. We start with $\displaystyle f$ pigeons, then subtract $\displaystyle (g + 2)$ pigeons. $\displaystyle f - (g + 2) = f - g - 2$.

Assume there are x pennies.  Hence the number of nickels is $\displaystyle 70-x$. 

Now you have to set up an equation for the dollar value of the pennies and nickels which will be $\displaystyle x+5(70-x)=150$. 

Now solving for $\displaystyle x$ results in $\displaystyle x=50$ (number of pennies).  Hence the number of nickels will be $\displaystyle 70-50=20$.

A box is a rectangular prism that has a length, width, and height dimension.

The volume of a box is: $\displaystyle V=LWH$

Substitute the values into the formula.

$\displaystyle V=(12 \textup{ units})(3\textup{ units})(5 \textup{ units}) = 180 \textup{ units}^3$

There are four meats from which to choose two; the number of ways to do this is

$\displaystyle C(4,2)$

$\displaystyle = \frac{4!}{(4-2)! 2!}$

$\displaystyle = \frac{4!}{2! 2!}$

$\displaystyle = \frac{1 \cdot 2 \cdot 3 \cdot 4 }{1 \cdot 2 \cdot 1 \cdot 2 }$

$\displaystyle = \frac{ 3 \cdot 4 }{ 1 \cdot 2 }$

$\displaystyle = \frac{12}{2}$

$\displaystyle = 6$

There are five vegetables from which Kathy can choose two (six minus the one to which Kathy is allergic); the number of ways to do this is

$\displaystyle C(5,2)$

$\displaystyle = \frac{5!}{(5-2)! 2!}$

$\displaystyle = \frac{5!}{3! 2!}$

$\displaystyle = \frac{1 \cdot 2 \cdot 3 \cdot 4\cdot 5 }{1 \cdot 2 \cdot 3 \cdot 1 \cdot 2 }$

$\displaystyle = \frac{ 4 \cdot 5 }{ 1 \cdot 2 }$

$\displaystyle = \frac{20}{2}$

$\displaystyle = 10$

The number of ways to make a Special Pizza that leaves out green peppers is 

$\displaystyle C(4,2) \cdot C(5,2) = 6 \cdot 10 = 60$

Clara has three meats from which to choose (the four available, minus the one she will not eat); she has five vegetables from which to choose (the six available minus the one to which she is allergic). She can do one of the following:

Case 1: She can choose one meat and three vegetables.

She can choose one of three meats. 

She can choose three of five vegetables; the number of ways to do so is

$\displaystyle C(5,3)$

$\displaystyle = \frac{5!}{(5-3)! 3!}$

$\displaystyle = \frac{5!}{2! 3!}$

$\displaystyle = \frac{1 \cdot 2 \cdot 3 \cdot 4\cdot 5 }{1 \cdot 2 \cdot 1 \cdot 2\cdot 3 }$

$\displaystyle = \frac{ 4 \cdot 5 }{ 1 \cdot 2 }$

$\displaystyle = \frac{20}{2}$

$\displaystyle = 10$

The number of possible pizzas with one meat and three vegetables is 

$\displaystyle 3 \cdot C(5,3) = 3 \cdot 10 = 30$

Case 2: She can choose two meats and two vegetables.

She can choose two of three meats. There are three ways to do this, since choosing two meats means leaving one out; there are three ways to choose which meat to leave out.

She can choose two of five vegetables; the number of ways to do so is

$\displaystyle C(5,2)$

$\displaystyle = \frac{5!}{(5-2)! 2!}$

$\displaystyle = \frac{5!}{3! 2!}$

$\displaystyle = \frac{1 \cdot 2 \cdot 3 \cdot 4\cdot 5 }{1 \cdot 2 \cdot 3 \cdot 1 \cdot 2 }$

$\displaystyle = \frac{ 4 \cdot 5 }{ 1 \cdot 2 }$

$\displaystyle = \frac{20}{2}$

$\displaystyle = 10$

The number of possible pizzas with two meats and two vegetables is 

$\displaystyle 3 \cdot C(5,2) = 3 \cdot 10 = 30$

The total number of combinations of toppings from which Clara can choose is 

$\displaystyle 30+30 = 60$

Melissa paid $575 rent for each month from March to December - ten months - for a total of 

$\displaystyle \$ 575 \times 10 = \$5, 750$.

Add to this her prorated February rent and her security deposit (she was not late with any rent payments, hence no penalties); Melissa paid

$\displaystyle \$( 5, 750 + 250 +300)= \$6, 300$

Henry's rent is $645 per month; multiply this by 12 to get yearly rent before the penalty.

$\displaystyle \$ 645 \times 12 =\$ 7,740$

Henry paid his November rent three days late, so multiply the daily penalty of $20 by 3 to get the total penalty.

$\displaystyle \$20 \times 3 = \$ 60$

Add the regular yearly rent and the November penalty:

$\displaystyle \$ 7,740 + \$60 = \$7, 800$

The increment that is added to each successive term to obtain the next term is doubled each time.

$\displaystyle 7 + 1 = 8$

$\displaystyle 8 + 2 = 10$

$\displaystyle 10 + 4 = 14$

$\displaystyle 14+ 8 = 22$

$\displaystyle 22+ 16 = 38$

$\displaystyle 38 + 32 = 70$, the number in the circle

$\displaystyle 70 + 64 = 134$, the number in the square, which is the correct choice.