A right triangle has to have a right angle, which is equal to \(\displaystyle 90^{\circ}\). The image below is an example:

This triangle is a scalene triangle because all of the side lengths are different lengths.

This is not an equilateral triangle because an equilateral triangle has to have all sides equal in length. 

This is not an isosceles triangle because an isosceles triangle has two equal sides. 

This is not a right triangle because a right triangle has to have a right angle, an angle that measures \(\displaystyle 90^{\circ}\).

We can use base ten blocks to help us solve this problem. First, we want to use base ten blocks to represent \(\displaystyle .8\)

Because we are dividing\(\displaystyle .8\) by \(\displaystyle .4\), we need to split up our \(\displaystyle .8\)  into groups of \(\displaystyle 4\)

As you can see, we have \(\displaystyle 2\) groups. Thus the answer is \(\displaystyle 2\). 

We can use base ten blocks to help us solve this problem. First, we want to use base ten blocks to represent \(\displaystyle .4\)

Because we are dividing \(\displaystyle .4\) by \(\displaystyle .2\) , we need to split up our \(\displaystyle .4\) into groups of \(\displaystyle 2\)

We can see that we have 2 groups of 2, thus the answer is 2. 

We can use base ten blocks to help us solve this problem. First, we want to use base ten blocks to represent .2

Because we are dividing .2 by .2, we need to split up our .2 into groups of .2:

We can see that we have 1 group, thus our answer is 1. 

We can use base ten blocks to help us solve this problem. First, we want to use base ten blocks to represent \(\displaystyle .9\textup:\)

Because we are dividing \(\displaystyle .9\) by \(\displaystyle .3\), we need to split up our \(\displaystyle .9\) into groups of \(\displaystyle .3\):

As you can see, we have \(\displaystyle 3\) groups; thus, \(\displaystyle .9\div.3=3\)

We can use base ten blocks to help us solve this problem. First, we want to use base ten blocks to represent \(\displaystyle .8\textup:\)

Because we are dividing \(\displaystyle .8\) by \(\displaystyle .2\), we need to split up our \(\displaystyle .8\) into groups of \(\displaystyle .2\):

As you can see, we have \(\displaystyle 4\) groups; thus, \(\displaystyle .8\div.2=4\)

We can use base ten blocks to help us solve this problem. First, we want to use base ten blocks to represent \(\displaystyle .6\textup:\)

Because we are dividing \(\displaystyle .6\) by \(\displaystyle .3\), we need to split up our \(\displaystyle .6\) into groups of \(\displaystyle .3\)

As you can see, we have \(\displaystyle 2\) groups; thus, \(\displaystyle .6\div.3=2\) 

The problem that you are challenged to solve is \(\displaystyle 7.8 \div 2=?\). 

\(\displaystyle 7.8\) is the dividend, this is what is being broken up into groups. \(\displaystyle 2\) is our divisor which is the number of groups you are making. We need to split \(\displaystyle 7.8\) in half to see how many are in each group.

The first step is to place your decimal above your equation in the same place. It will line up with the decimal inside of your "long-division house". 

Next, we need to use or multiplication facts to determine what \(\displaystyle 2\) can be multiplied by to make \(\displaystyle 7\) or get close to it without going over. \(\displaystyle 2\times 3=6\) is the fact that works best (\(\displaystyle 2\times 4=8\) is too large). We will place the numeral \(\displaystyle 3\) directly above the \(\displaystyle 7\) in the ones place to indicate that \(\displaystyle 3\) groups of \(\displaystyle 2\) fit into the \(\displaystyle 7\). We will put the product of \(\displaystyle 2\times 3\) which was \(\displaystyle 6\) underneath the \(\displaystyle 7\) in the ones place and subtract the difference. The numbers above the "house" are our quotient or answer to the division problem.

Next, we will carry the \(\displaystyle 8\) in the tenths place down and put it next to the \(\displaystyle 1\). We will work with the numbers as if they were \(\displaystyle 18\) when thinking of multiplication facts, but it should be noted this is actually \(\displaystyle 1.8\) when you consider the decimal placement. \(\displaystyle 2\times 9=18\) so we place the \(\displaystyle 9\) above the "house" in the tenths place of our quotient and subtract the \(\displaystyle 18\). We are left with \(\displaystyle 0\) remaining so there is no remainder. 

Our final answer is \(\displaystyle 3.9\), which means that half of \(\displaystyle 7.8\) is \(\displaystyle 3.9\)