distance between two points is [(x2-x1)^2 +(y2-y1)^2]^1/2
I.e,[{5-1)^2+(0-3)^2]^1/2
[16+9]^1/2
[25]^1/2
5
if a = b and b = c, then a = c.
In this case if y = - 7 and -7 = z then y = z .................>(y = a, -7 = b and z = c)
Answer: B)
transitive property of equality
Make use of the known limit,
We have
since 
, and the limit of a product is the same as the product of limits.
is continuous at 
, and 
. The remaining limit is also 1, since
so the overall limit is 1.
![\bf ~~~~~~\textit{parabola vertex form} \\\\ \begin{array}{llll} \stackrel{\textit{we'll use this one}}{y=a(x- h)^2+ k}\\\\ x=a(y- k)^2+ h \end{array} \qquad\qquad vertex~~(\stackrel{2}{ h},\stackrel{-1}{ k}) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} h=2\\ k=-1 \end{cases}\implies y=a(x-2)^2-1 \\\\\\ \textit{we also know that } \begin{cases} y=0\\ x=5 \end{cases}\implies 0=a(5-2)^2-1\implies 1=9a \\\\\\ \cfrac{1}{9}=a\qquad therefore\qquad \boxed{y=\cfrac{1}{9}(x-2)^2-1}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~%5Ctextit%7Bparabola%20vertex%20form%7D%20%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7Bllll%7D%20%5Cstackrel%7B%5Ctextit%7Bwe%27ll%20use%20this%20one%7D%7D%7By%3Da%28x-%20h%29%5E2%2B%20k%7D%5C%5C%5C%5C%20x%3Da%28y-%20k%29%5E2%2B%20h%20%5Cend%7Barray%7D%20%5Cqquad%5Cqquad%20vertex~~%28%5Cstackrel%7B2%7D%7B%20h%7D%2C%5Cstackrel%7B-1%7D%7B%20k%7D%29%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20%5Cbegin%7Bcases%7D%20h%3D2%5C%5C%20k%3D-1%20%5Cend%7Bcases%7D%5Cimplies%20y%3Da%28x-2%29%5E2-1%20%5C%5C%5C%5C%5C%5C%20%5Ctextit%7Bwe%20also%20know%20that%20%7D%20%5Cbegin%7Bcases%7D%20y%3D0%5C%5C%20x%3D5%20%5Cend%7Bcases%7D%5Cimplies%200%3Da%285-2%29%5E2-1%5Cimplies%201%3D9a%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B1%7D%7B9%7D%3Da%5Cqquad%20therefore%5Cqquad%20%5Cboxed%7By%3D%5Ccfrac%7B1%7D%7B9%7D%28x-2%29%5E2-1%7D)
now, let's expand the squared term to get the standard form of the quadratic.
