What is the answer to a question similar to this one, where the absolute value bars are only around the numerator of the fraction?
$$\frac{|2+4(2)|}{5-10}$$
Would the fraction be equal to $\frac{10}{-5}$ and would the answer just be $-2$?
Thanks
What is the answer to a question similar to this one, where the absolute value bars are only around the numerator of the fraction?
$$\frac{|2+4(2)|}{5-10}$$
Would the fraction be equal to $\frac{10}{-5}$ and would the answer just be $-2$?
Thanks
On
You've done right! But, this is from where the concept comes from: $$| x| =\begin{cases} x & x >0 \\ 0 & x = 0 \\ -x & x < 0 \end{cases} $$
So, when you have: $|-1|$ you will simply write this as: $1$ as this obeys the third condition of the above general case. [as an example]
Once you understand this general case, it will be easy for you to move on in this topic.
Here are the possible configurations: $$\frac{|1-2|}{1-2} = \frac{|-1|}{-1}=\frac{1}{-1} = -1$$
$$\frac{|1-2|}{|1-2|} = \frac{|-1|}{|-1|}=\frac{1}{1} = 1$$
$$\frac{1-2}{|1-2|} = \frac{-1}{|-1|}=\frac{-1}{1} = -1$$
$$\frac{1-2}{1-2} = \frac{-1}{-1}=\frac{-1}{-1} = 1$$