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Tju [1.3M]
3 years ago
7

Use the diagram to find the measure of arc. AB and CD and the diameters of the circle F

Mathematics
1 answer:
dybincka [34]3 years ago
7 0

Answer:

11). m(\widehat{DE}) = 90°

12). m(\widehat{AEC}) = 212°  

Step-by-step explanation:

A circle F with AB and CD are the diameters has been given in the figure attached.

11). Since, m(\widehat{AB}) = 180°

    and m(\widehat{AB})=m(\widehat{AD})+m(\widehat{DE})+m(\widehat{BE})

    Therefore, m(\widehat{AD})+m(\widehat{DE})+m(\widehat{BE}) = 180°

    32° + m(\widehat{DE}) + 58° = 180°

    m(\widehat{DE}) = 180° - 90°

                = 90°

12. Since, m(\widehat{AD})=m(\widehat{BC}) = 32°

                 m(\widehat{AEC}) = m(\widehat{AB})+m(\widehat{BC})

                                = 180° + 32°

                                = 212°

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Alona [7]

Answer:

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Step-by-step explanation:

Let \rho(x,y) be a continuous density function of a lamina in the plane region D,then the mass of the lamina is given by:

m=\int\limits \int\limits_D \rho(x,y) \, dA.

From the question, the given density function is \rho (x,y)=xye^{x+y}.

Again, the lamina occupies a rectangular region: D={(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1}.

The mass of the lamina can be found by evaluating the double integral:

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Since D is a rectangular region, we can apply Fubini's Theorem to get:

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Let u=xy, the partial derivative of u wrt y is

\implies du=xdy

and

dv=\int\limits e^{x+y} dy, integrating wrt y, we obtain

v=\int\limits e^{x+y}

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\int\limits xye^{x+y}dy=xye^{x+y}-\int\limits e^{x+y}\cdot xdy

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I_0=xe^x

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I=\int\limits^1_0(xe^x)dx.

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I=\int\limits^1_0xe^xdx.

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I=\int\limits^1_0xe^xdx=e^1(1-1)-e^0(0-1).

I=\int\limits^1_0xe^xdx=0-1(0-1).

I=\int\limits^1_0xe^xdx=0-1(-1)=1.

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