Question

multiply.
\\(\frac{-2}{5} cdot 7 cdot \frac{-7}{-6}\\)
write your answer in simplest form.

Explanation:

Step1: Multiply the numerators and denominators

First, multiply the numerators: \(-2 \times 7 \times -7\) and the denominators: \(5 \times 1 \times -6\) (note that 7 can be written as \(\frac{7}{1}\) for multiplication of fractions).
So we have \(\frac{(-2) \times 7 \times (-7)}{5 \times 1 \times (-6)}\)
Calculating the numerator: \((-2)\times7 = -14\), \(-14\times(-7)=98\)
Calculating the denominator: \(5\times1 = 5\), \(5\times(-6)= -30\)
Now the fraction is \(\frac{98}{-30}\)

Step2: Simplify the fraction

We can simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 2.
\(\frac{98\div2}{-30\div2}=\frac{49}{-15}=-\frac{49}{15}\) Wait, no, let's check the signs again. Wait, in the original problem, we have three terms: \(\frac{-2}{5}\cdot7\cdot\frac{-7}{-6}\)
Let's re - evaluate the signs:
The product of two negative numbers is positive, and the product of a positive and a negative is negative.
We have \(\frac{-2}{5}\times7\times\frac{-7}{-6}=\frac{(-2)\times7\times(-7)}{5\times(-6)}\)
\((-2)\times(-7) = 14\), \(14\times7 = 98\)
Denominator: \(5\times(-6)=-30\)
So \(\frac{98}{-30}\), but also, we can cancel before multiplying. Let's try that approach for accuracy.
\(\frac{-2}{5}\times7\times\frac{-7}{-6}=\frac{-2}{5}\times\frac{7}{1}\times\frac{-7}{-6}\)
Notice that \(-2\) and \(-6\) can be simplified. \(\frac{-2}{-6}=\frac{1}{3}\) (dividing numerator and denominator by - 2)
So now we have \(\frac{1}{5}\times7\times\frac{-7}{3}\) (Wait, no, let's do the sign correctly. The original fractions: \(\frac{-2}{5}\), \(7=\frac{7}{1}\), \(\frac{-7}{-6}=\frac{7}{6}\) (because \(\frac{-7}{-6}=\frac{7}{6}\))
Wait, I made a mistake in the sign of \(\frac{-7}{-6}\). \(\frac{-7}{-6}=\frac{7}{6}\) (negative divided by negative is positive)
So now the expression is \(\frac{-2}{5}\times\frac{7}{1}\times\frac{7}{6}\)
Now, multiply the numerators: \(-2\times7\times7=-98\)
Multiply the denominators: \(5\times1\times6 = 30\)
So the fraction is \(\frac{-98}{30}\)
Now, simplify by dividing numerator and denominator by 2: \(\frac{-98\div2}{30\div2}=\frac{-49}{15}\)

Wait, let's start over with correct sign handling:

\(\frac{-2}{5}\times7\times\frac{-7}{-6}\)

First, handle the signs:
The number of negative factors: \(\frac{-2}{5}\) has one negative, \(\frac{-7}{-6}\) has two negatives (which is positive), and 7 is positive.
Total number of negative factors: 1 (from \(\frac{-2}{5}\)) + 0 (from 7) + 0 (from \(\frac{-7}{-6}\) as two negatives make positive) = 1. So the result will be negative.

Now, multiply the numerators: \(2\times7\times7 = 98\) (we take the absolute values for now)
Multiply the denominators: \(5\times1\times6=30\) (since \(\frac{-7}{-6}=\frac{7}{6}\), so denominator is \(5\times6 = 30\) and numerator is \(2\times7\times7 = 98\))

Since there is 1 negative factor, the result is \(-\frac{98}{30}\)

Simplify \(\frac{98}{30}\) by dividing numerator and denominator by 2: \(\frac{49}{15}\)

So the final result is \(-\frac{49}{15}\)? Wait, no, wait:

Wait \(\frac{-2}{5}\times7\times\frac{-7}{-6}=\frac{(-2)\times7\times(-7)}{5\times(-6)}\)

\((-2)\times(-7)=14\), \(14\times7 = 98\)

Denominator: \(5\times(-6)=-30\)

So \(\frac{98}{-30}=-\frac{98}{30}=-\frac{49}{15}\)

But let's check with another approach:

\(\frac{-2}{5}\times7\times\frac{-7}{-6}=\frac{-2}{5}\times\frac{7}{1}\times\frac{-7}{-6}\)

Multiply the first and the third fractions first: \(\frac{-2}{5}\times\frac{-7}{-6}=\frac{(-2)\times(-7)}{5\times(-6)}=\frac{14}{-30}=-\frac{7}{15}\)

Then multiply by 7: \(-\frac{7}{15…

Answer:

\(-\frac{49}{15}\)