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Dennis_Churaev [7]
2 years ago
12

What is the midpoint of CD? • (1/2,-5/2) •(1,-5/2) •(1/2,-3/2) •(1,-3/2)

Mathematics
2 answers:
kodGreya [7K]2 years ago
8 0

Answer:

\displaystyle \left(\frac{1}{2},-\frac{5}{2}\right)

Step-by-step explanation:

Given a segment defined by the points (x1,y1) and (x2,y2), the midpoint (xm,ym) is calculated as follows:

\displaystyle x_m=\frac{x_1+x_2}{2}

\displaystyle y_m=\frac{y_1+y_2}{2}

The segment CD is defined by the points in graph C(-3,-4) and D(4,-1). The midpoint is:

\displaystyle x_m=\frac{-3+4}{2}=\frac{1}{2}

\displaystyle x_m=\frac{1}{2}

\displaystyle y_m=\frac{-4-1}{2}=\frac{-5}{2}

\displaystyle y_m=-\frac{5}{2}

Thus, the midpoint is:

\mathbf{\displaystyle \left(\frac{1}{2},-\frac{5}{2}\right)}

polet [3.4K]2 years ago
7 0

Answer:

(1/2,-5/2)

Step-by-step explanation:

\frac{ - 3 + 4 }{2}  = \frac{1}{2}  \\   \frac{ - 4 +  - 1}{2}  =   \frac{ - 5}{2}

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AlladinOne [14]

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Maria by 1 and 2/5 pounds of lunch meat for $3.36 how much would 2 3/4 lb of lunch meat cost
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Step-by-step explanation:

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Can someone help me please
masya89 [10]

Answer:

A) (3,-8); Reflection

Step-by-step explanation:

The question kind of gave you a hint. Look at how the coordinate point is arranged...

(-y,-x)

That means all positive become negatives and all negatives become positives. This gives you the coordinate of (3,-8).  

As for reflection or rotation, the image's coordinates FLIPPED, like how images do in a mirror's reflection.

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If 30.0 moles of oxygen are used, how many grams of water will form?
liberstina [14]

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Step-by-step explanation:

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3 years ago
Find the curl of ~V<br> ~V<br> = sin(x) cos(y) tan(z) i + x^2y^2z^2 j + x^4y^4z^4 k
ch4aika [34]

Given

\vec v =  f(x,y,z)\,\vec\imath+g(x,y,z)\,\vec\jmath+h(x,y,z)\,\vec k \\\\ \vec v = \sin(x)\cos(y)\tan(z)\,\vec\imath + x^2y^2z^2\,\vec\jmath+x^4y^4z^4\,\vec k

the curl of \vec v is

\displaystyle \nabla\times\vec v = \left(\frac{\partial h}{\partial y}-\frac{\partial g}{\partial z}\right)\,\vec\imath - \left(\frac{\partial h}{\partial x}-\frac{\partial f}{\partial z}\right)\,\vec\jmath + \left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right)\,\vec k

\nabla\times\vec v = \left(4x^4y^3z^4-2x^2y^2z\right)\,\vec\imath \\\\ - \left(4x^3y^4z^4-\sin(x)\cos(y)\sec^2(z)\right)\,\vec\jmath \\\\ + \left(2xy^2z^2+\sin(x)\sin(y)\tan(z)\right)\,\vec k

\nabla\times\vec v = \left(4x^4y^3z^4-2x^2y^2z\right)\,\vec\imath \\\\ + \left(\sin(x)\cos(y)\sec^2(z)-4x^3y^4z^4\right)\,\vec\jmath \\\\ + \left(2xy^2z^2+\sin(x)\sin(y)\tan(z)\right)\,\vec k

7 0
3 years ago
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