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3 years ago

At which points are the equations y = x- + 3x + 2 and y = 2x + 3 approximately equal?

Mathematics

2 answers:

mariarad [96]3 years ago

7 0

Answer:

The equations are approximately equals at the points (0.618,4.236)(0.618,4.236) and (-1.618,-0.236)(−1.618,−0.236)

jarptica [38.1K]3 years ago

4 0

Select all the correct locations on the graph. At which points are the equations y = x2 + 3x + 2 and y = 2x + 3 approximately equal? 2.

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3 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

Find the length of a line segment with endpoints of 19 and –39. A. –58 B. –20 C. 20 D. 58

Lisa [10]

19 - (-39) = 19 + 39 = 58

<span>length of a line segment is 58 units.</span>

5 0

3 years ago

Read 2 more answers