Question

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f(t) = \

$$\begin{cases} t^2 - 5t , & t \\leq -10 \\\\ t + 19 , & -10 < t < -2 \\\\ \\dfrac{t^3}{t + 9} , & t \\geq -2 \\end{cases}$$

f(-10) = \boxed{}
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Explanation:

Step1: Determine the applicable function

Since \( t = -10 \) and the first piece of the piece - wise function is defined for \( t\leq - 10 \), we use the function \( f(t)=t^{2}-5t \) when \( t = - 10 \).

Step2: Substitute \( t=-10 \) into the function

Substitute \( t=-10 \) into \( f(t)=t^{2}-5t \). We get \( f(-10)=(-10)^{2}-5\times(-10) \).

First, calculate \( (-10)^{2}=100 \) and \( 5\times(-10)=- 50 \), so \( - 5\times(-10) = 50 \).

Then, \( f(-10)=100 + 50=150 \).

Answer:

\( 150 \)