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.
Answer:
27.92 percent
Step-by-step explanation:
Answer:
dooky
Step-by-step explanation:
boiiiii dwivwejrfiuejbjfd
Let's take length of HK = n
By using trigonometry,
Hence, value of HK = 44 cm
We first get the gradient of the equation of a line by taking the
change in y over change in x
Hence our gradient will be
(6-9.5)/(8-1)=-3.5/7=0.5
Taking an arbitrary point(x,y) and using our gradient we get
and one of the points above we get
(y-6)/(x-8)=0.5
y-6=0.5*(x-8)
y-6=0.5x-4
y=0.5x+2 or y=1/2x+2
which is the line of our equation the answer we are
asked for.
<span> </span>
