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DD. 4 Evaluate a nonlinear function HK6 Use the following function rule to find f(56). {:[f(x)=11sqrt(x-20)],[f(56)=◻]:} Submit

DD. 44 Evaluate a nonlinear function HK66\newlineUse the following function rule to find f(56) f(56) .\newlinef(x)=11x−20f(56)=□ \begin{array}{l} f(x)=11 \sqrt{x-20} \\ f(56)=\square \end{array} \newlineSubmit

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Q. DD. 44 Evaluate a nonlinear function HK66\newlineUse the following function rule to find f(56) f(56) .\newlinef(x)=11x−20f(56)=□ \begin{array}{l} f(x)=11 \sqrt{x-20} \\ f(56)=\square \end{array} \newlineSubmit
  1. Identify Function and Input: Identify the function and the input value.\newlineWe are given the function f(x)=11(x−20)f(x) = 11\sqrt{(x - 20)} and we need to find the value of f(56)f(56). This means we will substitute xx with 5656 in the function.
  2. Substitute Input into Function: Substitute the input value into the function.\newlinef(56)=11(56−20)f(56) = 11\sqrt{(56 - 20)}\newlineNow we need to calculate the value inside the square root.
  3. Calculate Value Inside Square Root: Calculate the value inside the square root. \newline56−20=3656 - 20 = 36\newlineNow we have f(56)=1136f(56) = 11\sqrt{36}.
  4. Calculate Square Root: Calculate the square root. 36=6\sqrt{36} = 6 Now we can substitute this back into the function to get f(56)=11×6f(56) = 11 \times 6.
  5. Multiply by 1111: Multiply the result by 1111.\newlinef5656 = 11×6=6611 \times 6 = 66\newlineThis is the final value of the function f(x) when x is 5656.

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