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Leviafan [203]
4 years ago
15

If f(x)=2x^2+5√(x-2), complete the following statement: f(3) = _____

Mathematics
1 answer:
dsp734 years ago
5 0
Given:
f(x) = 2x^{2} +5 \sqrt{x-2}

Asked:
f(3)

Solution:
Subtitute x with 3 in the function
f(x) = 2x^{2} +5 \sqrt{x-2}
f(3) = 2(3)^{2} +5 \sqrt{3-2}

Then simplify to find the value of f(3)
f(x) = 2(9) +5 \sqrt{1}
f(x) = 18 +5
f(x) = 23

The final answer is 23
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Answer:

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Step-by-step explanation:

Given:

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\begin{aligned}\textsf{Area of a rectangle}&=\sf width \times length\\& = \sf 75 \times 105\\& = \sf 7875\:\: yd^2\end{aligned}

To maintain the <u>same perimeter</u>, but <u>change the area</u>, either:

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In geometry, length pertains to the <u>longest side</u> of the rectangle while width is the <u>shorter side</u>.  Therefore, we should choose:

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<u>Define the variables</u>:

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Therefore:

\implies \sf width \times length < 7875\:\:yd^2

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Solve the inequality:

\begin{aligned}(75-x)(105+x) & < 7875\\7875-30x-x^2 & < 7875\\-x^2-30x & < 0\\-x(x+30) & < 0\\x(x+30) & > 0\\\implies x & > 0 \:\: \textsf{ or }\:\:x < - 30\end{aligned}

Therefore, as distance is positive only and the maximum width is 75 yd (since we are subtracting from the original width):

\begin{cases}\textsf{width} = 75 - x\\\textsf{length} = 105 + x\end{cases}

\textsf{where } 0 < x < 75

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⇒ Length = 105 + 3 = 108 yd

⇒ Perimeter = 2(72 + 108) = 360 yd

⇒ Area = 72 × 108 = 7776 yd²

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⇒ Width = 75 - 74 = 1 yd

⇒ Length = 105 + 74 = 179 yd

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