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2 years ago

X f(x) = x2 + 4x . x2_4

Mathematics

1 answer:

PtichkaEL [24]2 years ago

6 0

The answer is 1/(x+4)
Explanation:
You would factor out the denominator
So,
(X-4)(x+4)=x^2-16
So, x-4/(x+4)(x-4)
Then x-4 cancels each other out from the numerator and denominator
Leaving 1/x+4

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2 years ago

The histogram shows the number of miles driven by a sample of automobiles in New York City. What is the maximum possible number

hjlf

Answer:

The maximum possible number of miles traveled by an automobile included in the histogram is 40,000 miles.

Step-by-step explanation:

A histogram is a bar graph reprinting the frequency of various categories.

In a histogram the y-axis represents the frequency and the x-axis represents the categories.

Consider the histogram provided for the number of miles driven by a sample of automobiles in New York City.

The y-axis represents the frequency and the x-axis represents the number of miles driven.

The category 10,000 miles has the highest frequency at 30.

And the maximum possible number of miles traveled by an automobile is 40,000 miles.

Thus, the maximum possible number of miles traveled by an automobile included in the histogram is 40,000 miles.

6 0

2 years ago

Read 2 more answers

Surface integrals using an explicit description. Evaluate the surface integral \iint_{S}^{}f(x,y,z)dS using an explicit represen

Jobisdone [24]

Parameterize $S$ by the vector function

$\vec r(x,y)=x\,\vec\imath+y\,\vec\jmath+f(x,y)\,\vec k$

so that the normal vector to $S$ is given by

$\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=\left(\vec\imath+\dfrac{\partial f}{\partial x}\,\vec k\right)\times\left(\vec\jmath+\dfrac{\partial f}{\partial y}\,\vec k\right)=-\dfrac{\partial f}{\partial x}\vec\imath-\dfrac{\partial f}{\partial y}\vec\jmath+\vec k$

with magnitude

$\left\|\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}\right\|=\sqrt{\left(\dfrac{\partial f}{\partial x}\right)^2+\left(\dfrac{\partial f}{\partial y}\right)^2+1}$

In this case, the normal vector is

$\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=-\dfrac{\partial(8-x-2y)}{\partial x}\,\vec\imath-\dfrac{\partial(8-x-2y)}{\partial y}\,\vec\jmath+\vec k=\vec\imath+2\,\vec\jmath+\vec k$

with magnitude $\sqrt{1^2+2^2+1^2}=\sqrt6$ . The integral of $f(x,y,z)=e^z$ over $S$ is then

$\displaystyle\iint_Se^z\,\mathrm d\Sigma=\sqrt6\iint_Te^{8-x-2y}\,\mathrm dy\,\mathrm dx$

where $T$ is the region in the $x,y$ plane over which $S$ is defined. In this case, it's the triangle in the plane $z=0$ which we can capture with $0\le x\le8$ and $0\le y\le\frac{8-x}2$ , so that we have

$\displaystyle\sqrt6\iint_Te^{8-x-2y}\,\mathrm dx\,\mathrm dy=\sqrt6\int_0^8\int_0^{(8-x)/2}e^{8-x-2y}\,\mathrm dy\,\mathrm dx=\boxed{\sqrt{\frac32}(e^8-9)}$