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3 years ago

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If m (x) = StartFraction x + 5 Over x minus 1 EndFraction and n(x) = x – 3, which function has the same domain as (m circle n) (

x)?
h (x) = StartFraction x + 5 Over 11 EndFraction
h (x) = StartFraction 11 Over x minus 1 EndFraction
h (x) = StartFraction 11 Over x minus 4 EndFraction
h (x) = StartFraction 11 Over x minus 3 EndFraction

2 answers:

Andrew [12]3 years ago

4 0

Answer:

The function h (x) = StartFraction 11 Over x minus 4 EndFraction

has the domain of h (x) = ( - ∞, 4) U (4, ∞ ) which is same as the

domain of (m ∘ n)(x) = ( - ∞, 4) U (4, ∞ ).

Step-by-step explanation:

m (x) = StartFraction x + 5 Over x minus 1 and n(x) = x – 3

As m (x) could be written as:

$m(x) = \frac{x+5}{x-1}$

n(x) = x – 3

(m ∘ n)(x) = m(n(x)) = m(x – 3)

= $\frac{(x-3) + 5}{(x-3) - 1}$

= $\frac{x+2}{x-4}$

In order to find the domain of (m ∘ n)(x) = $\frac{x+2}{x-4}$ , we need to

make sure that denominator can not be zero,

So,

x - 4 = 0

x = 4

So, our domain can be anything except for 4.

Hence, Domain = D = ( - ∞, 4) U (4, ∞ )

Now, compare the domain of (m ∘ n)(x) i.e D = ( - ∞, 4) U (4, ∞ ) with all the options:

Option A: h (x) = x + 5 / 11

Option A has the Domain of h (x) = Set of all real numbers

So, Option A is false.

Option B: h (x) = 11 / x - 1

x - 1 = 0

x = 1

Option B has the domain of h (x) = ( - ∞, 1) U (1, ∞ )

So, Option B is false.

Option C: h (x) = 11 / x - 4

x - 4 = 0

x = 4

Domain C has the domain of h (x) = ( - ∞, 4) U (4, ∞ )

So, Option C is true.

Option D: h (x) = 11 / x - 3

x - 3 = 0

x = 3

Option D has the domain of h (x) = ( - ∞, 3) U (3, ∞ )

So, option D is also false.

Hence, only option C is true as the option C i.e. h (x) = StartFraction 11 Over x minus 4 EndFraction has the domain of h (x) = ( - ∞, 4) U (4, ∞ ) which is same as the domain of (m ∘ n)(x) = ( - ∞, 4) U (4, ∞ ).

keywords: Domain, composite function

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zhuklara [117]3 years ago

4 0

Answer:

c

Step-by-step explanation:

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Mrrafil [7]

We are asked to express the diagonal of a rectangle in terms of the width in which the area of the polygon is equal to 13 square feet. The area of teh rectangle is equal to length times width. That is 13 = lw; diagonal using pythagorean theorem is sqrt ( l2 + w2). Thus d = sqrt ( (13/w)2 + w2) = sqrt ( (169/w2 + w2) = sqrt ( (169 + w4)/w2) = 1/w sqrt (169 + w4)

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kotegsom [21]

So,

$\frac{24}{48}$

Write in prime factored form.
$\frac{2*2*2*3}{2*2*2*2*3}$

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$\frac{1}{2}$

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3 years ago

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Name/ Uid:1. In this problem, try to write the equations of the given surface in the specified coordinates.(a) Write an equation

Gemiola [76]

To find:

(a) Equation for the sphere of radius 5 centered at the origin in cylindrical coordinates

(b) Equation for a cylinder of radius 1 centered at the origin and running parallel to the z-axis in spherical coordinates

Solution:

(a) The equation of a sphere with center at (a, b, c) & having a radius 'p' is given in cartesian coordinates as:

$(x-a)^{2}+(y-b)^{2}+(z-c)^{2}=p^{2}$

Here, it is given that the center of the sphere is at origin, i.e., at (0,0,0) & radius of the sphere is 5. That is, here we have,

$a=b=c=0,p=5$

That is, the equation of the sphere in cartesian coordinates is,

$(x-0)^{2}+(y-0)^{2}+(z-0)^{2}=5^{2}$

$\Rightarrow x^{2}+y^{2}+z^{2}=25$

Now, the cylindrical coordinate system is represented by $(r, \theta,z)$

The relation between cartesian and cylindrical coordinates is given by,

$x=rcos\theta,y=rsin\theta,z=z$

$r^{2}=x^{2}+y^{2},tan\theta=\frac{y}{x},z=z$

Thus, the obtained equation of the sphere in cartesian coordinates can be rewritten in cylindrical coordinates as,

$r^{2}+z^{2}=25$

This is the required equation of the given sphere in cylindrical coordinates.

(b) A cylinder is defined by the circle that gives the top and bottom faces or alternatively, the cross section, & it's axis. A cylinder running parallel to the z-axis has an axis that is parallel to the z-axis. The equation of such a cylinder is given by the equation of the circle of cross-section with the assumption that a point in 3 dimension lying on the cylinder has 'x' & 'y' values satisfying the equation of the circle & that 'z' can be any value.

That is, in cartesian coordinates, the equation of a cylinder running parallel to the z-axis having radius 'p' with center at (a, b) is given by,

$(x-a)^{2}+(y-b)^{2}=p^{2}$

Here, it is given that the center is at origin & radius is 1. That is, here, we have, $a=b=0,p=1$ . Then the equation of the cylinder in cartesian coordinates is,

$x^{2}+y^{2}=1$

Now, the spherical coordinate system is represented by $(\rho,\theta,\phi)$

The relation between cartesian and spherical coordinates is given by,

$x=\rho sin\phi cos\theta,y=\rho sin\phi sin\theta, z= \rho cos\phi$

Thus, the equation of the cylinder can be rewritten in spherical coordinates as,

$(\rho sin\phi cos\theta)^{2}+(\rho sin\phi sin\theta)^{2}=1$

$\Rightarrow \rho^{2} sin^{2}\phi cos^{2}\theta+\rho^{2} sin^{2}\phi sin^{2}\theta=1$

$\Rightarrow \rho^{2} sin^{2}\phi (cos^{2}\theta+sin^{2}\theta)=1$

$\Rightarrow \rho^{2} sin^{2}\phi=1$ (As $sin^{2}\theta+cos^{2}\theta=1$ )

Note that $\rho$ represents the distance of a point from the origin, which is always positive. $\phi$ represents the angle made by the line segment joining the point with z-axis. The range of $\phi$ is given as $0\leq \phi\leq \pi$ . We know that in this range the sine function is positive. Thus, we can say that $sin\phi$ is always positive.

Thus, we can square root both sides and only consider the positive root as,

$\Rightarrow \rho sin\phi=1$

This is the required equation of the cylinder in spherical coordinates.

Final answer:

(a) The equation of the given sphere in cylindrical coordinates is $r^{2}+z^{2}=25$

(b) The equation of the given cylinder in spherical coordinates is $\rho sin\phi=1$

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3 years ago

Peter is trying to set a new record for pizza deliveries. His previous record is 20 pizzas in one hour. He has already delivered

Arlecino [84]

We know that
<span>Peter has already delivered </span><span> 2 pizzas in 5 minutes
so
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therefore
in the next 55 minutes

if the

<span>average per minute = (18/55) pizzas /minute
</span>Pete beat his previous record
<span>
</span>average per minute > (18/55) pizzas /minute
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average per minute < (18/55) pizzas /minute
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stealth61 [152]

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2 years ago