Questions: Question n(t) = t^2 / 2 - 20t + k There was a 100-day period when the number of bees in a certain hive could be modeled by the function n above. In the function, k is a constant and n(t) represents the number of bees on day number t for 0 ≤ t ≤ 99. On what number day was the number of bees in the hive the same as it was on day number 10 ? 20 30 40 50

Question

n(t) = t^2 / 2 - 20t + k

There was a 100-day period when the number of bees in a certain hive could be modeled by the function n above. In the function, k is a constant and n(t) represents the number of bees on day number t for 0 ≤ t ≤ 99. On what number day was the number of bees in the hive the same as it was on day number 10 ?

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50

Transcript text: Question \[ n(t)=\frac{t^{2}}{2}-20 t+k \] There was a 100 -day period when the number of bees in a certain hive could be modeled by the function $n$ above. In the function, $k$ is a constant and $n(t)$ represents the number of bees on day number $t$ for $0 \leq t \leq 99$. On what number day was the number of bees in the hive the same as it was on day number 10 ? 20 30 40 50

Solution

Solution Steps

Step 1: Define the Function

The number of bees in the hive is modeled by the function \[ n(t) = \frac{t^{2}}{2} - 20t + k \]

Step 2: Calculate $ n(10) $

To find the number of bees on day 10, we substitute $ t = 10 $ into the function: \[ n(10) = \frac{10^{2}}{2} - 20 \cdot 10 + k = \frac{100}{2} - 200 + k = 50 - 200 + k = k - 150 \]

Step 3: Set Up the Equation

We need to find the day $ t $ when the number of bees is the same as on day 10. Thus, we set up the equation: \[ n(t) = n(10) \implies \frac{t^{2}}{2} - 20t + k = k - 150 \]

Step 4: Simplify the Equation

By eliminating $ k $ from both sides, we simplify the equation to: \[ \frac{t^{2}}{2} - 20t = -150 \] Multiplying through by 2 to eliminate the fraction gives: \[ t^{2} - 40t + 300 = 0 \]

Step 5: Solve the Quadratic Equation

Using the quadratic formula $ t = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} $ where $ a = 1, b = -40, c = 300 $: \[ t = \frac{40 \pm \sqrt{(-40)^{2} - 4 \cdot 1 \cdot 300}}{2 \cdot 1} = \frac{40 \pm \sqrt{1600 - 1200}}{2} = \frac{40 \pm \sqrt{400}}{2} = \frac{40 \pm 20}{2} \] This results in two potential solutions: \[ t = \frac{60}{2} = 30 \quad \text{and} \quad t = \frac{20}{2} = 10 \]

Step 6: Identify Valid Solutions

Since we are looking for a day $ t $ in the range $ 0 \leq t \leq 99 $ that is not day 10, the valid solution is: \[ t = 30 \]