at t=30, the temperature of an oven is 120 degrees and increasing by 2 degrees/minute. What is the temperature at t=40?

I suppose the unit of t is min.

Temperature at (t = 40 min)
= 120° + (2°/min) × [(40 - 30) min]
= 120° + (2°/min) × (10 min)
= 120° + 20°
= 140°

(40-30min) * (2deg/min) = 20 degree change
120 + 20 = 140°
You could model this as y =  2(t-30) + 120

120+20=140                       .

Mistake: t=30 min. or hr. or sec.?
Assume t is in minutes, then
at t=40 min. the temperature
T=120+2(40-30)=140*

Assume oven has a start temperature of x° before being turned on.;
Then temperature t minutes after starting = T(t) = x° + 2t°.;
T(30) = (x+2t)° = (x+2*30)° = (x+60)° = 120° and x = 60.;
T(40) = (x+2*40)° = (60+80)° = 140°.

Increase of 2 degrees per minute for 10 mins is a 20 degree increase on120
Result is 140 degrees, 
but if you want an equation,
T(t) = 2t + 60
T(40) = 140

Since the rate of increase is constant, you can use a linear equation:

T(t) = mt + b

You are given the time (t = 30), the temp (T() = 120) and the slope (m = 2).

Solve for the remaining unknown:

120 = 2(30) + b
120 = 60 + b
60 = b

So now your base equation is:

T(t) = 2t + 60

What's the temp at t = 40?

T(40) = 2(40) + 60
T(40) = 80 + 60
T(40) = 140

If at t = 30, the temperature of an oven is 120° 
and increasing by 2°/minute, the temperature at t = 40 is 140°.