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4 years ago

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

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:

Width = 75 yards
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Area of the field with the given values:

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

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decrease the width and increase the length by the same amount, or
increase the width and decrease the length by the same amount.

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

decrease the width and increase the length by the same amount.

Define the variables:

Let x = the amount by which to decrease/increase the width and length.

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

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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):

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

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Therefore, to find the width and length of another rectangular field that has the same perimeter but a smaller area than the first field, simply substitute a value of x from the restricted interval into the found expressions for width and length:

Example 1:

Let x = 3

⇒ Width = 75 - 3 = 72 yd

⇒ Length = 105 + 3 = 108 yd

⇒ Perimeter = 2(72 + 108) = 360 yd

⇒ Area = 72 × 108 = 7776 yd²

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Let x = 74

⇒ Width = 75 - 74 = 1 yd

⇒ Length = 105 + 74 = 179 yd

⇒ Perimeter = 2(1 + 179) = 360 yd

⇒ Area = 1 × 179 = 179 yd²

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