How do you plot #sqrt46# on the decimal number line as accurately as possible?

1 Answer
Aug 11, 2018

Answer:

Here's how you could do it geometrically...

Explanation:

Draw a circle with centre #(0, 45/2)# with radius #47/2#.

This will pass though the points #(0, -1)# and #(0, 46)# on the #y# axis and the points #(-sqrt(46), 0)# and #(sqrt(46), 0)# on the #x# axis.

graph{(x^2 + (y-45/2)^2 - 47^2/4)((x-sqrt(46))^2+y^2-0.01)((x+sqrt(46))^2+y^2-0.01) = 0 [-9.83, 10.17, -3.28, 6.72]}