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2 months ago

6

The oblique rectangular prism below has a length of 6 cm, a width of 8 cm, and a height of 7 cm.What is the area of the base of

the prism?

2 answers:

Inessa [12.5K]2 months ago

5 0

The area at the base measures 48 square centimeters.

Zina [12.3K]2 months ago

5 0

The area of the base of the prism is 48 cm². A rectangular prism has a rectangular base whose area can be calculated using the formula A = length x width. Here, A = 6 cm x 8 cm, leading to the result of 48 cm².

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On a coordinate plane, 2 polygons are shown. Polygon A B D C has points (negative 5, 2), (negative 3, negative 1), (negative 3,

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Response: A on edg 2021

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1 month ago

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The height h and collar size c , both in centimeters, measured from a sample of boys were used to create the regression line cˆ=

zzz [12365]

Response:

This is the C=intercept of the regression line

Step-by-step clarification:

The regression line's slope is logically interpreted as the average increase in collar size for each 1cm increment in height.

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1 month ago

Let D be the smaller cap cut from a solid ball of radius 8 units by a plane 4 units from the center of the sphere. Express the v

PIT_PIT [12445]

Answer:

Step-by-step explanation:

The equation representing the sphere, which has its center at the origin, can be written as $x^2+y^2+z^2 = 64$ . For z equal to 4, we find

$x^2+y^2= 64-16 = 48$ .

This results in a circle with a radius of $4\sqrt[]{3}$ in the x-y plane.

c) We will build on the analysis from earlier to set limits in both Cartesian and polar coordinates. Initially, we recognize that x spans from $-4\sqrt[]{3}$ to $4\sqrt[]{3}$ . This determination is made by fixing y = 0 and identifying the extreme x values that fall on the circle. For y, we observe that it ranges between $-\sqrt[]{48-x^2}$ and $\sqrt[]{48-x^2}$ , which holds because y must reside within the interior of the identified circle. Lastly, z will extend from 4 up to the sphere; hence, it varies from 4 to $\sqrt[]{64-x^2-y^2}$ .

The respective triple integral representing the volume of D in Cartesian coordinates is

$\int_{-4\sqrt[]{3}}^{4\sqrt[]{3}}\int_{-\sqrt[]{48-x^2}}^{\sqrt[]{48-x^2}} \int_{4}^{\sqrt[]{64-x^2-y^2}} dz dy dx$ .

b) Remember that the cylindrical coordinates are expressed as $x=r\cos \theta, y = r\sin \theta,z = z$ , where r denotes the radial distance from the origin projected onto the x-y plane. Also note that $x^2+y^2 = r^2$ . We will derive new limits for each of the transformed coordinates. Recall that due to the prior circular constraint, $\theta[\tex] is the angle between the projection to the x-y plane and the x axis, in order for us to cover the whole circle, we need that [tex]\theta$ varies between 0 and $2\pi$ . Furthermore, r starts from the origin and extends to the edge of the circle, with r reaching a maximum of 4\sqrt[]{3}. Lastly, Z increases from the plane z=4 up to the sphere, where it is constrained by \sqrt[]{64-r^2}. Thus, the integral that computes the desired volume is as follows:

$\int_{0}^{2\pi}\int_{0}^{4\sqrt[]{3}} \int_{4}^{\sqrt[]{64-r^2}} rdz dr d\theta$ . It’s important to note that the r factor arises from the Jacobian associated with the transition from Cartesian to polar coordinates, ensuring the integral maintains its value. (Explaining how to calculate the Jacobian exceeds the scope of this response).

a) When dealing with spherical coordinates, keep in mind that $z = \rho \cos \phi, y = \rho \sin \phi \sin \theta, x = \rho \sin \phi \cos \theta$ , where $\phi$ denotes the angle formed between the vector and the z axis, varying from 0 to pi. It is crucial to recognize that at z=4, this angle remains constant along the circle we previously identified. Let’s determine the angle by selecting a point on the circle and employing the angle formula between two vectors. Setting z=4 and x=0 gives us y=4 $\sqrt[]{3}$ by taking the positive square root of 48. We will now compute the angle between the vector $a=(0,4\sqrt[]{3},4)$ and vector b =(0,0,1), which represents the unit vector along the z axis. We apply the following formula

$\cos \phi = \frac{a\cdot b}{||a||||b||} = \frac{(0,4\sqrt[]{3},4)\cdot (0,0,1)}{8}= \frac{1}{2}$

Consequently, across the circle, $\phi = \frac{\pi}{3}$ . Observe that rho transitions from the plane z=4 to the sphere, with rho reaching up to 8. Given $z = \rho \cos \phi$ , we have that $\rho = \frac{4}{\cos \phi}$ at the plane. Thus, the corresponding integral is

$\int_{0}^{2\pi}\int_{0}^{\frac{\pi}{3}}\int_{\frac{4}{\cos \phi}}^{8}\rho^2 \sin \phi d\rho d\phi d\theta$ , where the new factor incorporates the Jacobian for the spherical coordinate system.

d) Let’s work with the integral in cylindrical coordinates

$\int_{0}^{2\pi}\int_{0}^{4\sqrt[]{3}} \int_{4}^{\sqrt[]{64-r^2}} rdz dr d\theta=\int_{0}^{2\pi}\int_{0}^{4\sqrt[]{3}} r (\sqrt[]{64-r^2}-4) dr d\theta=\int_{0}^{2\pi} d \theta \cdot \int_{0}^{4\sqrt[]{3}}r (\sqrt[]{64-r^2}-4)dr= 2\pi \cdot (-2\left.r^{2}\right|_0^{4\sqrt[]{3}})\int_{0}^{4\sqrt[]{3}}r \sqrt[]{64-r^2} dr$ .

It’s important to observe that the integral can be separated since the inner part remains independent of theta. By implementing the substitution $u = 64-r^2$ , we achieve $\frac{-du}{2} = r dr$ , leading to

$=-2\pi \cdot \left.(\frac{1}{3}(64-r^2)^{\frac{3}{2}}+2r^{2})\right|_0^{4\sqrt[]{3}}=\frac{320\pi}{3}$

3 0

2 months ago

G Find non-zero vectors x and y that are both orthogonaland orthogonal to each other.

babunello [11817]

Answer:

x = <1,5>

y = <-5,1>

Step-by-step explanation:

Let x, y represent two vectors in 2D

defined as:

x = <a, b>

y = <c, d>

for x, y to be orthogonal, we understand that their dot or scalar product should equal zero, therefore

x. y = 0

<a, b>. <c, d> = 0

a*c + b*d = 0 ------ (A)

Now, select any four values for a, b, c, and d that fulfill equation (A)

for example, I choose (this is subject to personal choice):

a=1, b=5, c=-5, d=1

Verifying Equation (A):

(1)*(5) + (-5)*(1) =0

5 - 5 = 0

0 = 0 (True)

Thus,

The identified vectors x, y are:

x = <1, 5>

y = <-5, 1>

4 0

2 months ago