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3 years ago

I need help with this

Mathematics

1 answer:

Kobotan [32]3 years ago

5 0

The answers are in the picture. Good luck

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Geometric objects can be grouped by their __________.

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A rectangular beam will be cut from a cylindrical log of radius 10 inches.

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Answer:

Step-by-step explanation:

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2 years ago

Please answer, I’ve been stuck on this the whole day, so basically I only need the four digit code. You can read the instruction

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2 years ago

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

Please help! I have not seen a correct answer yet :(

Naily [24]

Answer:

60 possible sequences.

Step-by-step explanation:

That would be 6! / 3! 2!

6*5*4*3*2*1

= ------------------

3*2*1*2*1

= 5*4*3

= 60.

8 0

2 years ago