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skelet666 [1.2K]

3 years ago

10

A, B & C form a triangle where ∠ BAC = 90°. AB = 4.3 mm and CA = 3.1 mm. Find the length of BC give your ANSWR rounded to 1

DP

1 answer:

Anika [276]3 years ago

6 0

Answer:

4.3^2+3.1^2= square root of 28.1= 5.3

Step-by-step explanation:

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Which statement is true regarding the graphed functions?

RoseWind [281]

Answer:c. F(2)=g(-2)

Step-by-step explanation:

4 0

3 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

3 years ago

Please help fast <br> Question is below<br> v v v v v v

zvonat [6]

They would cost $8.75. If you take out 30% from the original price, that is how much they would be.

4 0

3 years ago

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marin [14]

Answer:

X = $\frac{L}{6\pi r }$ - $\frac{5r}{6}$

Step-by-step explanation:

4 0

3 years ago

Based only on the information given in the diagram, it is guaranteed that JKL ~ WXY.

Alika [10]

Answer:

Option A True

Step-by-step explanation:

we know that

The sum of the internal angles of the triangle is equal to $180\°$

<u>In the triangle JKL</u>

m∠L= $180\°-90\°-27\°=63\°$

so

The measurements of the angles of triangle JKL are $27\°-90\°-63\°$

<u>In the triangle WXY</u>

m∠W= $180\°-90\°-63\°=27\°$

so

The measurements of the angles of triangle WXY are $27\°-90\°-63\°$

therefore

Triangle JKL and triangle WXY are similar by AAA ( The AAA postulate states that if you can prove that all three angles of two triangles are congruent, you can prove the two triangles are similar)

6 0

3 years ago

Read 2 more answers