(-4)*(5/7)*(-3/4)*(7/15)

Your friend says do the multiplication in order from left to right. You see that changing the order of the numbers being multiplied will be easier to solve
It is impossible to multiply the numbers in their current order
The commutative property says changing the order in which you multiply does not change the answer
The order of operations can be ignored
The associative property says that the way factors are grouped does not change the answer

1 answer

To solve the expression \((-4) \cdot \frac{5}{7} \cdot \left(-\frac{3}{4}\right) \cdot \frac{7}{15}\), we can utilize the commutative property and the associative property of multiplication. Let's break this down step by step.

  1. Commutative Property: This property states that changing the order of the factors does not change the product. Therefore, we are free to rearrange the numbers to make the multiplication easier.

  2. Associative Property: This property states that the way in which factors are grouped does not change the product. We can group the factors in any way that may simplify our calculations.

Let's simplify the expression by grouping and rearranging the factors. We notice that multiplying certain factors can simplify our work:

\[ (-4) \cdot \left(-\frac{3}{4}\right) \cdot \frac{5}{7} \cdot \frac{7}{15} \]

Combine \(-4\) and \(-\frac{3}{4}\):

\[ (-4) \cdot \left(-\frac{3}{4}\right) = 3 \]

Now the expression becomes:

\[ 3 \cdot \frac{5}{7} \cdot \frac{7}{15} \]

Next, notice that \(\frac{5}{7} \cdot \frac{7}{15}\) can be simplified:

\[ \frac{5 \cdot 7}{7 \cdot 15} = \frac{5}{15} = \frac{1}{3} \]

Now we multiply by 3:

\[ 3 \cdot \frac{1}{3} = 1 \]

Thus, the final result of the expression \((-4) \cdot \frac{5}{7} \cdot \left(-\frac{3}{4}\right) \cdot \frac{7}{15}\) is:

\[ \boxed{1} \]

In conclusion, your friend's suggestion to multiply from left to right may not be the best approach due to the potential for complexity. The commutative property allows us to rearrange factors for easier multiplication, leading us to the correct answer efficiently.