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Lorico [155]
3 years ago
14

This graph shows the solution to which inequality?

Mathematics
1 answer:
Ksenya-84 [330]3 years ago
7 0

Answer:

The graph shows the solution of the inequality y > \frac{4}{3} x - 2 ⇒ D

Step-by-step explanation:

In the inequality,

  • If the sign of inequality is ≤ or ≥, then the line that represents it must be a solid line
  • If the sign of inequality is < or >, then the line that represents it must be a dashed line
  • If the sign of inequality is > or ≥, then the shaded area must be over the line
  • If the sign of inequality is < or ≤, then the shaded area must be under the line

From the given graph

∵ The slope of the line = \frac{2--6}{3--3} = \frac{2+6}{3+3} = \frac{8}{6} = \frac{4}{3}

∵ The y-intercept is (0, -2)

∵ The line is dashed and the shaded area is over the line

→ By using the 2nd and 3rd notes above, the line is dashed and

   the sign of inequality is >

∴ The inequality is y > \frac{4}{3} x - 2

∴ The graph shows the solution of the inequality y > \frac{4}{3} x - 2

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Answer:

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Step-by-step explanation:

Given that,

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L=\int_{a}^{b}{\sqrt{1+(\dfrac{dx}{dy})^2}dy}

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L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4}-y^{-3})^2}dy}

L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-2\times\dfrac{y^3}{4}\times y^{-3}}dy}

L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-\dfrac{1}{2}}dy}

L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4})^2+(y^{-3})^2+\dfrac{1}{2}}dy}

L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4}+y^{-3})^2}dy}

L= \int_{1}^{2}{(\dfrac{y^3}{4}+y^{-3})dy}

L=(\dfrac{y^{3+1}}{4\times4}+\dfrac{y^{-3+1}}{-3+1})_{1}^{2}

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Put the limits

L=(\dfrac{2^4}{16}+\dfrac{2^{-2}}{-2}-\dfrac{1^4}{16}-\dfrac{(1)^{-2}}{-2})

L=\dfrac{21}{16}

Hence, The arc length is \dfrac{21}{16}

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