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Talja [164]
3 years ago
11

The vertices of a quadrilateral in the coordinate plans are known. How can the perimeter of the figure be found?

Mathematics
1 answer:
Serggg [28]3 years ago
3 0

Answer:

The perimeter can be found by calculating lengths of sides using distance formula and then adding up the lengths

Step-by-step explanation:

If the vertices of a quadrilateral are known in the coordinate plane, the vertices can be used to determine the lengths of sides of quadrilateral. The distance formula is used for calculating the distance between two vertices which is the length of the side

d=\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}

after calculating all the lengths of four sides using their vertices, they can be summed up to find the perimeter ..

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How do you create and solve equations in one variable given a real–world situation?
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Step-by-step explanation:

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2 years ago
Use the given transformation to evaluate the given integral, where r is the triangular region with vertices (0, 0), (8, 1), and
Jlenok [28]
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For the line with points (0, 0) and (1, 8), the equation is given by:

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For the line with points (8, 1) and (1, 8), the equation is given by:

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\left|\begin{array}{cc} \frac{\partial x}{\partial u} &\frac{\partial x}{\partial v}\\\frac{\partial y}{\partial u}&\frac{\partial y}{\partial v}\end{array}\right| = \left|\begin{array}{cc} 8 &1\\1&8\end{array}\right| \\  \\ =64-1=63

The integrand x - 3y is transformed as 8u + v - 3(u + 8v) = 8u + v - 3u - 24v = 5u - 23v

Therefore, the integration is given by:

63 \int\limits^1_0 \int\limits^{1}_0 {(5u-23v)} \, dudv =63 \int\limits^1_0\left[\frac{5}{2}u^2-23uv\right]^{1}_0 \\  \\ =63\int\limits^1_0(\frac{5}{2}-23v)dv=63\left[\frac{5}{2}v-\frac{23}{2}v^2\right]^1_0=63\left(\frac{5}{2}-\frac{23}{2}\right) \\  \\ =63(-9)=|-576|=576
6 0
3 years ago
Which shows the graph of the solution set of y &lt; 1/3 x – 2?
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Answer:

The graph in the attached figure

Step-by-step explanation:

we have

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The slope of the line is positive m= \frac{1}{3}

The y-intercept of the line is the point (0,-2)

The x-intercept of the line is the point (6,0)

so

The graph in the attached figure

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2 years ago
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