Pls help me pls it's my homework

What is the value of x?<br> pls help im so confused

Makovka662 [10]

In the proportional theory,

28 units is the total measurement that includes 8 units.

So, 10 + 3x - 5 is the total measurement that includes 10 units proportionally.

This means,

28 = 10 + 3x - 5

Now solve for x,

28 = 10 + 3x - 5

28 - 10 = 3x - 5

18 + 5 = 3x

23 = 3x

23 / 3 = x

So x = 23/3 or 7.6666....

5 0

2 years ago

(-4,r) (-8,3) m=5 Find r please!

viva [34]

Answer:

r = 23

Step-by-step explanation:

m = (y2 - y1)/(x2 - x1)

5 = (3 - r)/(-8 - (-4))

5 = (3 - r)/(-8 + 4)

3 - r = 5 × -4

3 - r = -20

-r = -23

r = 23

4 0

2 years ago

Please help. I don't understand how to do this. I need this asap. please

Schach [20]

9514 1404 393

Answer:

r = 1/9

Step-by-step explanation:

First of all, solve the equation for r:

y = rx

y/x = r . . . . . . . divide by x

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Since r is a constant, it will be the same for any corresponding pairs of x and y. It is convenient to choose both x and y as integers, as in the third table entry.

r = y/x = 5/45

r = 1/9 . . . . . . . . . reduced fraction

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<em>Additional comment</em>

It is not a bad idea to check to see that this works with other values of x and y. For the first line of the table, we have x = 11:

y = rx = (1/9)(11) = 11/9 = 1 2/9 . . . . matches the table value

6 0

2 years ago

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.