+130561 There are infinite possibilites for this....
The equation will be in the form
x^2 / 121 - y^2 / b^2 = 1 where "a" = 11 and a^2 = 121
Here's a graph when b = 5.........https://www.desmos.com/calculator/ptqorqve0j
Here's a graph when b = 13.........https://www.desmos.com/calculator/tmv8vqrcl6
As "a" stays constant and "b" increases, the branches of the hyperbola are less "curved," and the focal points move further from the center.......(as expected)
Finally....here's a graph when b = 100........https://www.desmos.com/calculator/26r6fevyc3
Notice that the hyperbola appears to be almost "upright" !!!
+130561 I'm going to use the following function to model this
T = abx where T is the temperature after some "x" minutes have elapsed
So we have, when x = 2
323 = ab2 → 323/b2 = a
And when x = 5 we have
288 = ab5 = (323/b2)b5 = 323 b3 divide both sides by 323
288/323 = b3 take the 3rd root of each side
(288/323)^(1/3) = b = about .9624
So, a = 323/(.9624)2 = about 348.66
So, our function is
T = 348.66(.9624)x
So, to find out when they will be at 120 degrees, we have
120 = 348.66(.9624)x divide both sides by 348.66
120/348.66 = (.9624)x take the log of each side
log ( 120/348.66 ) = log (.9624)x and by a log property, we have
log ( 120/348.66 ) = x log (.9624) divide both sides by log(.9624)
log ( 120/348.66 ) / log (.9624)= x = about 27.83 minutes
Here's a graph of the function.......https://www.desmos.com/calculator/logb5i0pwa
