Math Bacteria Question ?

Question

I tried, but I just CANT get this one. Beleive me

Refrigeration slows down the growth of bacteria in food. The number of bacteria in a certain food is approximated by B(T)=15T^2-70T+600, where T represents the temperature in degree Celsius and 3 </= T </= 12. Once the food is removed from the refrigeration, the temperature, T(t) is given by T(t)=3.5T+3, where t is the time in hours and 0 </= t </= 3.

a) Write the expression for the number of bacteria in the food, t hours after it is removed from refrigeration

b) How many bacteria are in the food at 1.5 hours?

c) When does the bacteria count reach approximately 1200?

Answer 1

&lt;&lt; Write the expression for the number of bacteria in the food, t hours after it is removed from refrigeration &gt;&gt;

Since

T = 3.5t + 3 then the number of bacteria in the food after it is taken out of the refrigerator is given by the equation

B(t) = 15(3.5t + 3)^2 - 70(3.5t + 3) + 600

&lt;&lt;  How many bacteria are in the food at 1.5 hours? &gt;&gt;

Substitute t = 1.5 in the above equation

B(1.5) = 15(3.5*1.5 + 3)^2 - 70(3.5*1.5 +3) + 600

B(1.5) = 1043.438



&lt;&lt; When does the bacteria count reach approximately 1200? &gt;&gt;

For the bacteria count to be around 1200,

1200 = 15(3.5t + 3)^2 - 70(3.5t + 3) + 600

Rearranging the above,

15(3.5t + 3)^2 - 70(3.5t + 3) + 600 - 1200 = 0

15(3.5t + 3)^2 - 70(3.5t + 3) - 600 = 0

Use the quadratic formula to solve for (3.5t + 3), thus

(3.5t + 3) = 9.07

3.5t = 9.07 - 3 = 6.07

Solving for &quot;t&quot;,

t = 1.73 hours

Hope this helps.

Answer 2

a)    B(3.5t + 3)

b)  plug in t = 1.5 into answer from a.

c) B = 1200 and solve for t.