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zhannawk [14.2K]
2 years ago
13

Without using a calculator, fill in the blanks with two consecutive integers to complete the following inequality.

Mathematics
1 answer:
Mariana [72]2 years ago
7 0

Answer:

11≤√134≤12

Step-by-step explanation:

11^2 is 121

and 12^2 is 144

so √134 would have to fall between these numbers

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kumpel [21]

Answer: 37%

Step-by-step explanation:

4 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.

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
2 years ago
12.5z + 19.4 = -2.7z + 93.9
Nikitich [7]

Answer:

z =  4.90131578

Step-by-step explanation:

3 0
3 years ago
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A sports ball has a diameter of 27 cm. Find the volume of the ball. cm.) Round your answer to the nearest tenth if necessary.​
Anestetic [448]
To calculate the volume of a sphere you must use the formula
v=4/3pir^3

The radius (r) is half of the diameter
r=13.5

v=4/3pi13.5^3
v=10305.99

Therefore the volume of the sports ball is 10306cm^3 (to the nearest tenth)
8 0
2 years ago
Please help I need this question ASAP please help I am begging
MaRussiya [10]

bill nye is in fact the science guy

3 0
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