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Murrr4er [49]
3 years ago
10

Please help with this​

Mathematics
1 answer:
SSSSS [86.1K]3 years ago
4 0

Answer:

G

Step-by-step explanation:

It is g as it is directly in the middle of the points

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\sqrt\frac{387}{20}

Step-by-step explanation:

Arc Length =\int\limits^a_b {\sqrt{1+(\frac{dy}{dx})^2 } } \, dx

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1+(\frac{dy}{dx})^2 }=1+(x^{\frac{2}{3}}-\dfrac{1}{4}x^{-\frac{2}{3}})^2\\=1+(x^{\frac{4}{3}}-\dfrac{1}{2}+ \dfrac{1}{16}x^{-\frac{4}{3}})

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For the Interval 1\leq x\leq 8

Length of the Curve =\int\limits^8_1 {\sqrt{\dfrac{1}{2}+x^{\frac{4}{3}}+ \dfrac{1}{16}x^{-\frac{4}{3}} } } \, dx\\

Using T1-Calculator

=\sqrt\frac{387}{20}

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Is there anybody was good at math? If so please go look at my most recent question​
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