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skad [1K]
3 years ago
14

What is the midpoint of AB?

Mathematics
1 answer:
Lana71 [14]3 years ago
6 0

Answer:

the answer is 0

Step-by-step explanation:

just count from -6 all the way to 8 then you will find out that the there are 15 numbers from point a to point b then count from both sides the center

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Help me with this,i really forgot how to do these type of questions
Montano1993 [528]
Work=speed times time (my equation to help me figure it out)

ok, 10 men take 84 days to complete a job
therefor, each man completes 1/10 of the job in 84 days
each man does 1/840 of the job per 1 day

so, if we had 5 more, or 15 men

15 times 1/840 times xdays=1 job complete
15/840 times xdays=1job
times both sides by 840/15
xdays=840/15
xdays=56
56 days it would take
5 0
3 years ago
hi guys the problem is i don't know how to use this so i will post this for fun lol i hope this help your day get better because
ValentinkaMS [17]

Answer: thats fun i guess keep Vibin doe

Step-by-step explanation:

4 0
3 years ago
Prime factorization of 52​
lara [203]

Answer:

The first prime factor we test is 2:

52 / 2 = 26

26 / 2 = 13

13 is a prime number so the prime factorization is

2 * 2 * 13 = 52

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
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

7 0
3 years ago
Solve the following equations for 0 ≤ theta ≤ 360º cosec(2 — 45°) = 2​
lions [1.4K]

Step-by-step explanation:

here ,

cosec \alpha  =  \frac{1}{ \sin( \alpha ) }

now,

cosec(2-45°)=2

or,

1/sin(2-45°) =2

or,

1/sin2cos45°-cos2sin45=2

or,

\frac{1}{ \sin(2)   \frac{1}{2}   -  \cos(2) \frac{1}{2}  }  = 2

or,

sorry I have that much qualifications to work on this question and hoping this much will make bit easy for you to solve it further

4 0
2 years ago
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