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neonofarm [45]
3 years ago
8

Determine which postulate can be used

Mathematics
1 answer:
Kruka [31]3 years ago
3 0

Answer:

SAS postulate

Step-by-step explanation:

The triangles have two congruent sides and one congruent angle. The congruent angle is the included angle. This meets SAS criteria.

<u>Hope this helps :-)</u>

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Find the following: F(x, y, z) = e^(xy) sin z j + y tan^−1(x/z)k Exercise Find the curl and the divergence of the vector field.
natulia [17]

\vec F(x,y,z)=e^{xy}\sin z\,\vec\jmath+y\tan^{-1}\dfrac xz\,\vec k

Divergence is easier to compute:

\mathrm{div}\vec F=\dfrac{\partial(e^{xy}\sin z)}{\partial y}+\dfrac{\partial\left(y\tan^{-1}\frac xz\right)}{\partial z}

\mathrm{div}\vec F=xe^{xy}\sin z-\dfrac{xy}{x^2+z^2}

Curl is a bit more tedious. Denote by D_t the differential operator, namely the derivative with respect to the variable t. Then

\mathrm{curl}\vec F=\begin{vmatrix}\vec\imath&\vec\jmath&\vec k\\D_x&D_y&D_z\\0&e^{xy}\sin z&y\tan^{-1}\frac xz\end{vmatrix}

\mathrm{curl}\vec F=\left(D_y\left[y\tan^{-1}\dfrac xz\right]-D_z\left[e^{xy}\sin z\right]\right)\,\vec\imath-D_x\left[y\tan^{-1}\dfrac xz\right]\,\vec\jmath+D_x\left[e^{xy}\sin z}\right]\,\vec k

\mathrm{curl}\vec F=\left(\tan^{-1}\dfrac xz-e^{xy}\cos z\right)\,\vec\imath-\dfrac{yz}{x^2+z^2}\,\vec\jmath+ye^{xy}\sin z\,\vec k

5 0
3 years ago
How would you find the area of the shaded region?
Whitepunk [10]

Answer:

The Area of the shaded region = (Area of the largest circle) – (Area of the circle with radius 3) – (Area of the circle with radius 2). Whatever is left over is the shaded region. The diameter of the largest circle is 10, so its radius is 5 and thus its area is 25π. SO do this in that problem and you will get the answer.

Step-by-step explanation:

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3 years ago
A garden in the shape of an equilateral triangle
Strike441 [17]

Answer:

area is 25\sqrt{3}

Step-by-step explanation:

the hight is 5\sqrt{3}

5\sqrt{3}*10/2=25\sqrt{3}

8 0
2 years ago
The figure below shows a rope that circles the Earth. The radius of the Earth is 637163716371 kilometers. Assuming the rope make
lutik1710 [3]
The length of the rope for this case is given by the length of the circumference.
 We have then:
 C = 2 * pi * R
 Where,
 R: Earth radio.
 We have then:
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 C = 12742 * pi
 Answer:
 the length of this rope is:
 C = 12742 * pi
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The mean price of a textbook is $82, and the range of prices are $16. What are the possible prices for the textbook?
Akimi4 [234]
82$ and 74$ since there is a variety of the same numbers
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