Question

expand -5/6 (9b-12)

Answer 1

expand -5/6 (9b-12) and make sure its right

Answer 2

Find the GCF of 20 x - 4

Answer 3

To expand the expression \(-\frac{5}{6} (9b - 12)\), you need to distribute the term \(-\frac{5}{6}\) to both terms inside the parentheses. Here's how you do it:

First, multiply \(-\frac{5}{6}\) by \(9b\):
\(-\frac{5}{6} \times 9b = -\frac{45}{6}b\)
Since \(45\) and \(6\) have a common factor of \(3\), you can simplify the fraction:
\(-\frac{45}{6}b = -\frac{45 \div 3}{6 \div 3}b = -\frac{15}{2}b\)
So, \(-\frac{5}{6} \times 9b = -\frac{15}{2}b\)

Next, multiply \(-\frac{5}{6}\) by \(-12\):
\(-\frac{5}{6} \times -12 = \frac{5}{6} \times 12 = \frac{60}{6}\)
Since \(60\) and \(6\) are both divisible by \(6\), you can simplify the fraction:
\(\frac{60}{6} = \frac{60 \div 6}{6 \div 6} = \frac{10}{1} = 10\)
So, \(-\frac{5}{6} \times -12 = 10\)

Now, put the two results together to get the expanded form of the original expression:
\(-\frac{15}{2}b + 10\)

Therefore, \(-\frac{5}{6} (9b - 12)\) expands to \(-\frac{15}{2}b + 10\).

Answer 4

Certainly! Let's expand the expression \(-\frac{5}{6} (9b - 12)\) by distributing \(-\frac{5}{6}\) to both terms within the parentheses.

First, distribute \(-\frac{5}{6}\) to \(9b\):
\[
-\frac{5}{6} \cdot 9b = \left(-\frac{5 \cdot 9}{6}\right) b = -\frac{45}{6} b
\]
This fraction can be simplified by dividing both the numerator and the denominator by their greatest common divisor, which is 3:
\[
-\frac{45}{6} b = -\frac{45 \div 3}{6 \div 3} b = -\frac{15}{2} b
\]

Next, distribute \(-\frac{5}{6}\) to \(-12\):
\[
-\frac{5}{6} \cdot -12 = \frac{5 \cdot 12}{6} = \frac{60}{6}
\]
This simplifies to:
\[
\frac{60}{6} = 10
\]

Now, combine the two results to get the expanded form:
\[
-\frac{15}{2} b + 10
\]

However, on re-evaluation, it seems I made an error in the initial response, as the incorrect multiplication of fractions was performed. I apologize for the confusion, and the correct expansion is indeed:

\[
-\frac{15}{2} b + 10
\]

Therefore, \(-\frac{5}{6} (9b - 12)\) expands correctly to \(-\frac{15}{2} b + 10\).