F(x)= 2/x-2 find (fof) (i)

Mathematics

1 answer:

brilliants [131]2 years ago

5 0

Answer:

Step-by-step explanation:

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Please show all work

Mekhanik [1.2K]

Answer:

4.8 minutes

Step-by-step explanation:

2.5/2.5=1

12/2.5=4.8

6 0

3 years ago

Read 2 more answers

Use double integrals to calculate the area inside the ellipse whose semiaxes have length a and b

Stells [14]

The general equation of an ellipse centered at the origin with its semiaxes coinciding with the coordinate axes is given by

$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$

Substituting $x=a\cos\theta$ and $y=b\sin\theta$ into the above equation gives the Pythagorean identity, so we can use polar coordinates quite nicely to our advantage.

If $\mathcal E$ is the ellipse with the equation above, the area is given by the double integral

$\displaystyle\iint_{\mathcal E}\mathrm dA=\iint_{(x/a)^2+(y/b)^2\le1}\mathrm dx\,\mathrm dy$

Let $x(r,\theta)=ar\cos\theta$ and $y(r,\theta)=br\sin\theta$ , so that the Jacobian matrix is

$\mathbf J=\begin{bmatrix}x_r&x_\theta\\y_r&y_\theta\end{bmatrix}=\begin{bmatrix}a\cos t&-ar\sin t\\b\sin t&br\cos t\end{bmatrix}$

and the magnitude of its determinant is $|\det\mathbf J|=|abr|=abr$

since in polar coordinates we use the convention that $r\ge0$ , and $a,b>0$ because they are lengths.

Now, the area is given by

$\displaystyle\iint_{\mathcal E}\mathrm dA=\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}abr\,\mathrm dr\,\mathrm d\theta$
$=\displaystyle2\pi a b\int_{r=0}^{r=1}r\,\mathrm dr$
$=\pi a b$

4 0

4 years ago

Regina’s work to find the surface area of a square pyramid from its net is shown below.

Y_Kistochka [10]

She put 12.8 in place of 11.9 in the formula of the triangle. Then the correct option is A.

<h3>What is Geometry?</h3>

It deals with the size of geometry, region, and density of the different forms both 2D and 3D.

Regina’s work to find the surface area of a square pyramid from its net is shown below.

A square pyramid. The square base has side lengths of 9.4 inches. The triangular sides have a height of 11.9 inches.

Surface area of the pyramid = 4 [one-half (9.4) (12.8)] + 9.4² = 329 square inches

Then the error done by Regina will be

She should have used One-half (9.4) (11.9) to represent the area of a single triangle.