Answer:
c,d,e
Step-by-step explanation:
Since

The square root of 11 is somewhere between 3 and 4. In order to round it to the nearest tenth, we have to try all numbers between 3 and 4 with one decimal digit, and see which is closest to 11 when squared. We have
![\begin{array}{c|c}n&n^2\\3&9\\3.1&9.61\\3.2&10.24\\3.3&10.89\\3.4&11.56\end{array}\right]](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bc%7Cc%7Dn%26n%5E2%5C%5C3%269%5C%5C3.1%269.61%5C%5C3.2%2610.24%5C%5C3.3%2610.89%5C%5C3.4%2611.56%5Cend%7Barray%7D%5Cright%5D)
So, the square root of 11 is somewhere between 3.3 and 3.4.
The mode is 50 because it occurs the most frequently (three times) out of any of the values.
The combined distance between CD and DE is CE, so...
2x+4 + 6 = x+14 Simplify like terms..
2x + 10 = x + 14 Subtract x from both sides..
x + 10 = 14 Subtract 10 from both sides...
<h2><u><em>
x = 4</em></u></h2>
Proof: 2 (4) + 4 + 6 = 4 + 14
8 + 10 = 18
<em> Check</em>