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The problem in the image is about bacteria growth in refrigerated food, modeled by a quadratic function for the number of bacteria \( N(x) \) and a linear function for the temperature change over time. Let's break it down and solve it step by step.

### Problem Setup:
- The number of bacteria \( N(x) \) is given by:
  \[
  N(x) = 10x^2 - 20x + 60
  \]
  where \( x \) is the temperature of the food in °C, and \( 2 \leq x \leq 20 \).

- The temperature of the food after being removed from the refrigerator is modeled by:
  \[
  x(t) = 3t + 2
  \]
  where \( t \) is the time in hours, for \( 0 \leq t \leq 6 \).

### (i) **Find and interpret \( N(x(t)) \):**

To find \( N(x(t)) \), we substitute \( x(t) \) into the function for \( N(x) \):
\[
N(x(t)) = 10(3t + 2)^2 - 20(3t + 2) + 60
\]

We will now expand and simplify this expression:
\[
(3t + 2)^2 = 9t^2 + 12t + 4
\]

Substituting this back into the equation:
\[
N(x(t)) = 10(9t^2 + 12t + 4) - 20(3t + 2) + 60
\]
\[
= 90t^2 + 120t + 40 - 60t - 40 + 60
\]
\[
= 90t^2 + 60t + 60
\]

Thus, the number of bacteria as a function of time \( t \) is:
\[
N(t) = 90t^2 + 60t + 60
\]

**Interpretation**: This quadratic equation models the growth of the bacteria count over time as the temperature increases according to the function \( x(t) = 3t + 2 \).

---

### (ii) **Find the bacteria count after 0 hours:**

To find the bacteria count at \( t = 0 \), substitute \( t = 0 \) into \( N(t) \):
\[
N(0) = 90(0)^2 + 60(0) + 60 = 60
\]

So, after 0 hours, the bacteria count is **60**.

---

### (iii) **Find the time when the bacteria count reaches 1500:**

We need to solve for \( t \) when \( N(t) = 1500 \):
\[
90t^2 + 60t + 60 = 1500
\]
First, subtract 1500 from both sides:
\[
90t^2 + 60t + 60 - 1500 = 0
\]
\[
90t^2 + 60t - 1440 = 0
\]
Now, divide the entire equation by 30 to simplify:
\[
3t^2 + 2t - 48 = 0
\]
This is a quadratic equation, and we can solve it using the quadratic formula:
\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 3 \), \( b = 2 \), and \( c = -48 \).

Substitute the values:
\[
t = \frac{-2 \pm \sqrt{2^2 - 4(3)(-48)}}{2(3)}
\]
\[
t = \frac{-2 \pm \sqrt{4 + 576}}{6}
\]
\[
t = \frac{-2 \pm \sqrt{580}}{6}
\]
\[
t = \frac{-2 \pm 24.08}{6}
\]
This gives two solutions:
\[
t = \frac{-2 + 24.08}{6} \approx 3.67 \quad \text{or} \quad t = \frac{-2 - 24.08}{6} \approx -4.01
\]
Since time cannot be negative, the valid solution is:
\[
t \approx 3.67 \, \text{hours}
\]

Thus, the bacteria count reaches 1500 after approximately **3.67 hours**.

---

Would you like further details on any of these steps?

Here are five related questions:
1. What happens to the bacteria count after 6 hours?
2. Can you generalize the bacteria count equation for different temperature functions?
3. What is the maximum bacteria count within the given time frame?
4. How would the bacteria growth change if the temperature increased at a slower rate?
5. How does the initial bacteria count change if the temperature function is altered?

**Tip:** When solving quadratic equations, always simplify them by dividing out common factors if possible to make calculations easier!

The number N of bacteria in refrigerated food is given by N(x) = 10x^2 - 20x + 60, where 2 ≤ x ≤ 20 is the temperature of the food in °C. When the food is removed from the refrigerator, its temperature is given by x(t) = 3t + 2, for 0 ≤ t ≤ 6, where t is time in hours. 
(i) Find and interpret N(x(t)). 
(ii) Find the bacteria count after 0 hours. 
(iii) Find the time when the bacteria count reaches 1500.

This problem explores bacteria growth in refrigerated food using a quadratic function for the number of bacteria N(x) and a linear function for temperature change over time. Learn how to find and interpret N(x(t)), calculate the bacteria count after 0 hours, and determine when the bacteria count reaches 1500.

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