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Aleksandr [31]
3 years ago
5

$17.25 for 5 yards of frabic

Mathematics
1 answer:
Neko [114]3 years ago
8 0
D onser is 1.0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000...   KFC

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find an equation of the tangent plane to the given parametric surface at the specified point. x = u v, y = 2u2, z = u − v; (2, 2
Alexxandr [17]

The surface is parameterized by

\vec s(u,v) = x(u, v) \, \vec\imath + y(u, v) \, \vec\jmath + z(u, v)) \, \vec k

and the normal to the surface is given by the cross product of the partial derivatives of \vec s :

\vec n = \dfrac{\partial \vec s}{\partial u} \times \dfrac{\partial \vec s}{\partial v}

It looks like you're given

\begin{cases}x(u, v) = u + v\\y(u, v) = 2u^2\\z(u, v) = u - v\end{cases}

Then the normal vector is

\vec n = \left(\vec\imath + 4u \, \vec\jmath + \vec k\right) \times \left(\vec \imath - \vec k\right) = -4u\,\vec\imath + 2 \,\vec\jmath - 4u\,\vec k

Now, the point (2, 2, 0) corresponds to u and v such that

\begin{cases}u + v = 2\\2u^2 = 2\\u - v = 0\end{cases}

and solving gives u = v = 1, so the normal vector at the point we care about is

\vec n = -4\,\vec\imath+2\,\vec\jmath-4\,\vec k

Then the equation of the tangent plane is

\left(-4\,\vec\imath + 2\,\vec\jmath - 4\,\vec k\right) \cdot \left((x-2)\,\vec\imath + (y-2)\,\vec\jmath +  (z-0)\,\vec k\right) = 0

-4(x-2) + 2(y-2) - 4z = 0

\boxed{2x - y + 2z = 2}

6 0
2 years ago
Similar figures always have .....?
rewona [7]

Answer: Similar figures are always the same shape, but not the same size. They have equal angles but not equal side lengths. Check out a big square and a small square. They're both squares because they have four sides and four equal angles, but the sides aren't the same length.

Step-by-step explanation:

5 0
4 years ago
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Vlad1618 [11]

Answer:

can u please tell what is 158° here

8 0
2 years ago
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What part of the circumference is an arc whose measure is 30°?
Radda [10]
Circumference of a circle:
C = 2 r π;
Length of an arc:
L = r π α / 180°
L = r π · 30° / 180° = r π /6
r π /6   :   2 r π = 1/6 : 2 = 1/12
Answer: A ) 1/12 
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3 years ago
The population of a city
julsineya [31]

Answer:

Step-by-step explanation:

what is the name of the city ?

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