The answer shows that the 3 in front of the $\displaystyle y$ has been cancelled, but not removed from the denomitator for the $\displaystyle y$.

Apply the distributive property:

$\displaystyle -7 (3x - 11) = -7 \cdot 3x - \left (-7 \right ) \cdot 11 = -7 \cdot 3x + 7\cdot 11= -21x + 77$

Apply the distributive property:

$\displaystyle -5 (-3x + 13) = -5 (-3x) + \left (-5 \right ) \cdot 13= 5\cdot 3x - 5\cdot 13 = 15x -65$

Apply the distributive property:

$\displaystyle -6 \left ( 5x - 3y - 9 \right )= -6 \cdot 5x - \left (-6 \right ) \cdot 3y - \left (-6 \right ) \cdot 9$

$\displaystyle = -6 \cdot 5x + 6 \cdot 3y + 6 \cdot 9$

$\displaystyle = -30x + 18y + 54$ 

Apply the distributive property:

$\displaystyle 6 \left ( 5x - 3y + 7 \right ) = 6 \cdot 5x - 6 \cdot 3y + 6 \cdot 7=30x - 18y + 42$

In order to figure out who has the most money, we must organize the data we have. Since Shaun has more money than Jessica, let us use Jessica as our baseline. So, if the money that Jessica has is represented by $\displaystyle m$, the money Shaun has will be $\displaystyle 2m$ because he has twice as much money as Jessica.

Jessica = $\displaystyle m$

Shaun = $\displaystyle 2m$

Next, we know that Jessica has one third as much money as Chris. In other words, Chris has three times as much money as Jessica. This can be represented by $\displaystyle 3m$. We then learn that Shaun has half as much money as Carmen or, in other words, Carmen has two times the amount of money Shaun has. Since Shaun has $\displaystyle 2m$ dollars, that must mean Carmen has $\displaystyle 4m$ dollars because $\displaystyle 2\times 2m = 4m$. So, by comparing everyone side-by-side, we can see that Carmen has the most money as she has four times the amount of money that Jessica has.

Jessica = $\displaystyle m$

Shaun = $\displaystyle 2m$

Chris = $\displaystyle 3m$

Carmen = $\displaystyle 4m$

Carmen is the answer.

$\displaystyle xyz$ refers to the product of the three variables: $\displaystyle 3\cdot 6\cdot 3=54$.

To solve, you can use the commutative and associative properties of multiplication to group like-terms together.

$\displaystyle 4m\cdot3m=4\times m\times 3\times m=4\times3\times m\times m$

The 4 and 3 should be first multiplied, resulting in 12.

$\displaystyle 12\times m\times m$

Next $\displaystyle m$ should be multiplied by $\displaystyle m$, giving us $\displaystyle m^{2}$.

$\displaystyle 12\times m^2$

12 times $\displaystyle m^{2}$ is equal to $\displaystyle 12m^{2}$. 

Therefore, the correct answer is $\displaystyle 12m^{2}$.

When simplifying this expression, the first step is to apply the distributive property. 

$\displaystyle 3a(2a+4)-7$

$\displaystyle (3a\times2a)+(3a\times4)-7$

$\displaystyle 6a^{2}+12a-7$

Next, we assert whether the expression can be reduced further. It cannot, as there are no like-terms to combine.

Therefore, the correct answer is $\displaystyle 6a^{2}+12a-7$.

By the order of operations, the operation inside grouping symbols, which here is addition, takes precendence, followed by, in order, squaring and multiplication.