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3 years ago

3x +y = –15 -3x - 4y = 30 Which variable will eliminate in the system?

Mathematics

1 answer:

Anestetic [448]3 years ago

5 0

Answer:

x should be eliminated

hope it helps

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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

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

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

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

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2 years ago

What is 3/4 +3/5 as a mixed number

Dmitry_Shevchenko [17]

Answer:

27 /20 =1 and 7 over 20= 1.35

Step-by-step explanation:

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Irina-Kira [14]

Based on your investigation, what is the value of b for the point (0,b)?
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julia-pushkina [17]

Answer:

(a - 6)²

(3c + 1)²

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Step-by-step explanation:

a^2-12a+36

a² - 6a - 6a + 36

a(a - 6) - 6(a - 6)

(a - 6)(a - 6)

(a - 6)²

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9c² + 3c + 3c + 1

3c(3c + 1) + 1(3c + 1)

(3c + 1)(3c + 1)

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30e(e² - 2e + 1)

30e(e² - e - e + 1)

30e[e(e - 1) - (e - 1)]

30e(e - 1)(e - 1)

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jenyasd209 [6]

Answer:

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