Question 260292
We start with our coordinates in disguise:
(1920, 45.8) and (1959,44.9), however, time is years "since" 1920, so we readjust to get
(0, 45.8), (39, 44.9).
step 1 - we can find the slope of these as
{{{m = (44.9-45.8)/(39-0)}}}
{{{m = -.9/39}}}
{{{m = -.0230769}}}
step 2 - using this slope and a coordinate, we can find the linear equation as
{{{y = mx + b}}}
{{{45.8 = -.0230769*0 + b}}}
{{{45.8 = b}}}
step p 3 - write the linear equation as
{{{R(t) = -.0230769t + 45.8}}}
--
find time for 2003? t = 83, so we get
{{{R(t) = -.0230769*(83) + 45.8}}}
{{{R(t) = 43.88}}}
--
find time for 2006? t = 86, so we get
{{{R(t) = -.0230769*(86) + 45.8}}}
{{{R(t) = 43.82}}}
--
In what year will t = 43.10?
{{{43.10 = -.0230769t + 45.8}}}
subtract 45.8 to get
{{{-2.7 = -.0230769t}}}
divide by -0.0230769 to get
t = 117
or 2037.