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Alex17521 [72]
2 years ago
8

5. The two figures are scaled copies of each other.

Mathematics
1 answer:
geniusboy [140]2 years ago
4 0

The quadrilaterals are similar and are related by a scale factor, taken from

a specific location, that maps the points on one figure to the other.

a. The scale factor that takes Figure 1 to Figure 2, is 3

b. The scale factor that takes Figure 2 to Figure 1 is \dfrac{1}{3}

Reasons:

From the question, we have that Figure 1 is a scaled copy of Figure 2, therefore;

Let ABCD represent Figure 1, we have;

ABCD ~ PQRS

Length of \overline{AD} = √(2² + 3²) = √(13)

Length of \overline{PS} = √(6² + 9²) = √(117) = √(9 × 13) = 3·√(13)

Therefore;

a. The scale factor that takes Figure 1 to Figure 2, SF₁₂, is therefore;

SF_{12} =  \mathbf{\dfrac{\overline{PS}}{\overline{AD}}}

Which gives;

SF_{12} = \dfrac{3 \cdot \sqrt{13} }{\sqrt{13} } = 3

The scale factor that takes Figure 1 to Figure 2, SF₁₂ = <u>3</u>

(<em>3 times the lengths of Figure 1 gives the lengths on Figure 2</em>)

b. The scale factor that takes Figure 2 to Figure 1, SF₂₁, is given as follows;

SF_{21} = \mathbf{\dfrac{\overline{AD}}{\overline{PS}}}

Which give;

SF_{21} = \dfrac{\sqrt{13} }{3 \cdot \sqrt{13} } =  \mathbf{\dfrac{1}{3}}

The scale factor that takes Figure 2 to Figure 1, SF₂₁ is \underline{\dfrac{1}{3}}

Learn more here:

brainly.com/question/17207164

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insens350 [35]
The given function is 
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The derivative of f is
f'(x) = 1 - 1/x
The second derivative is 
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A local maximum or minimum occurs when f'(x) = 0.
That is,
1 - 1/x = 0  => 1/x = 1  => x =1.
When x = 1, f'' = 1 (positive).
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The minimum value is
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The maximum value of f is
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A graph of f(x) confirms the results.

Answer: 
Minimum value  = 1 - ln(8)
Maximum value = 2 - ln(16)


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Alexxx [7]

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I don’t think I need to explain, it’s just common knowledge.

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