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Ymorist [56]
3 years ago
14

PLEASE HELP! ILL MARK BRAINLIEST Find f(x) and g(x) so that the function can be described as y = f(g(x)). y = nine divided by x

squared + 2
Mathematics
1 answer:
Usimov [2.4K]3 years ago
4 0

y = (9 / (x^2))  + 2

Rewrite in terms of f(x):

y = (9/(g(x))) + 2

To get

g(x) = x^2

f(x) = 9/x + 2

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Let S be the surface defined by x 2 + 2y 3 + 3z 4 = 6. Let T be the surface defined parametrically by r(u, v) = (1+ln u, 2e v+u−
aleksandrvk [35]

The tangent to C through (1, 1, 1) must be perpendicular to the normal vectors to the surfaces S and T at that point.

Let f(x,y,z)=x^2+2y^3+3z^4. Then S is the level curve f(x,y,z)=6. Recall that the gradient vector is perpendicular to level curves; we have

\nabla f(x,y,z)=(2x,6y,12z^2)

so that the gradient of f at (1, 1, 1) is

\nabla f(1,1,1)=(2,6,12)

For the surface T, we have

\begin{cases}1+\ln u=1\\2e^v+u-2=1\\uv+1=1\end{cases}\implies u=1,v=0

so that \vec r(1,0)=(1,1,1). We can obtain a vector normal to T by taking the cross product of the partial derivatives of \vec r(u,v), and evaluating that product for u=1,v=0:

\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}=\left(u-2ve^v,-1,\dfrac{2e^v}u\right)

\left(\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}\right)(1,0)=(1,-1,2)

Now take the cross product of the two normal vectors to S and T:

(2,6,12)\times(1,-1,2)=(24,8,-8)

The direction of vector (24, 8, -8) is the direction of the tangent line to C at (1, 1, 1). We can capture all points on the line containing this vector by scaling it by t\in\Bbb R. Then adding (1, 1, 1) shifts this line to the point of tangency on C. So the tangent line has equation

\vec\ell(t)=(1,1,1)+t(24,8,-8)=(1+24t,1+8t,1-8t)

7 0
3 years ago
What is the range for the function y=x
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3 years ago
What is the area, in square meters, of the figure below?
Veseljchak [2.6K]

Answer:

0.55 square meters

Step-by-step explanation:

The area of a triangle can be calculated by multiplying its height and base, then dividing the result by 2. As an algebraic expression, that would be \frac{bh}{2}, where b is the base and h is the height. In this case, b=2.2 and h=0.5. Therefore, the area of the triangle is:

\frac{bh}{2}\\= \frac{2.2*0.5}{2}\\=\frac{1.1}{2} \\= 0.55

Hope this helps!

5 0
3 years ago
Read 2 more answers
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