At heart we're being asked for a line through two points,
[tex](40^\circ \textrm{ C}, 355 \textrm { m/s}) \quad \textrm{and} \quad (49^\circ \textrm{ C}, 360 \textrm { m/s}) [/tex]
In general the line through (a,b) and (c,d) is
[tex](y-b)(c-a)=(x-a)(d-b)[/tex]
Check that you understand why both (a,b) and (c,d) are on this line.
Here our indepedent variable, instead of x, is T, temperature. Our dependent variable is v, velocity. Substituting,
[tex](v - 355)(49 - 40) = (T - 40)(360 - 355)[/tex]
[tex]9(v - 355) = 5(T - 40)[/tex]
[tex]v-355 = \frac 5 9 T - \frac{200}{9}[/tex]
[tex]v= \frac 5 9 T - \frac{200}{9} + 355[/tex]
[tex]v= \frac 5 9 T + \frac{2995}{9}[/tex]
That's our answer; let's check it.
When T=40, v = (5/9)40 + (2995/9) = 355 good
When T=49, v= (5/9)49 + (2995/9) = 360 good
