(-1.84-1.56)/5=8

I got the -3.40/5 but i can not figure out how it equals 8.

Math is not my strongest point, but i would appreciate any kind of help.

Guest Apr 30, 2017

#1**+2 **

(-1.84-1.56)/5 \(\stackrel{?}{=}\) 8

(-3.4)/5 \(\stackrel{?}{=}\) 8

-0.68 ≠ 8

You are right, it doesn't equal 8!

Maybe there is more to the question? What do the instructions say?

hectictar
May 1, 2017

#2**+1 **

Thanks hectictar

I've never seen your LaTex command before

\stackrel{?}{=}

I shall try and remember and I will add it to our LaTex thread :)

I was trying to work out how to put an arc over 2 letters, like for arc length... do you know how to do that?

I could not work it out. :/

Melody
May 1, 2017

#4**+2 **

Here are a few ways to indicate an angle.

**The first two seem the most clear, to me.**

The “\degree” command does not work in Tex, but the explicit **°** sign renders properly.

\(\begin{array}{|rcl|} \hline \angle {ABC} &=& 27° \\ \measuredangle{ABC} &=& 27° \\ \stackrel {\; \frown} {ABC} &=& 27° \\ \stackrel { \hspace{.1em} \wedge} {AC} &=& 27° \\ \stackrel {\, \hspace{.1em} \frown} {AC} &=& 27° \\ \hline \end{array}\\\ \hspace{1.5em} \overset{\; \frown}{AC} = 27° \\ \)

The third and the last option look identical but use different commands. The “\overset” command does not work inside an array for some reason.

Here are the Tex commands in text form.

\begin{array}{|rcll|}

\hline

\angle {ABC} &=& 27° \\

\measuredangle{ABC} &=& 27° \\

\stackrel {\; \frown} {ABC} &=& 27° \\

\stackrel { \hspace{.1em} \wedge} {AC} &=& 27° \\

\stackrel {\, \hspace{.1em} \frown} {AC} &=& 27° \\

\hline \end{array}\\\

\overset{\; \frown}{AC} = 27° \\

GingerAle
May 2, 2017