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2 months ago

Which statements are true regarding the diagram? Check all that apply.

Mathematics

2 answers:

PIT_PIT [12.4K]2 months ago

6 0

Answer:

2, 4, 6

B, D, F

Step-by-step explanation:

Ray C E is located on line m.

Ray AD is equivalent to ray AC.

∠ECB is formed by Ray C E and Ray C B.

Zina [12.3K]2 months ago

5 0

Answer:

2.CE lies on line m.

4.Ray AD is identical to ray AC.

6.Angle ECB is formed by CE and CB.

Step-by-step explanation:

A diagram has been provided.

Our task is to identify the valid statements regarding the diagram.

1.CB is found on line n.

CB does not align with line n since points C and B do not lie on that line.

Thus, this statement is false.

2.CE is situated on line m.

CE is indeed positioned on line m because points C and E are both on this line.

This statement is accurate.

3.Ray BC is equivalent to ray CB.

A ray begins at a single point and continues indefinitely without a terminal point.

Ray CB starts at point C and stretches out from point B.

However, ray BC does not exist.

This statement is therefore false.

4.Ray AD is equivalent to ray AC.

Both AD and AC are rays.

They are equivalent since both originate from point A and extend in the same direction.

5.Angle EAD is formed from AE and DA.

This is false because angle EAD is created by ray AE and ray AD, not by ray DA.

6.Angle ECB is generated by CE and CB.

This is true as angle ECB is made by ray EC and ray BC.

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The domain of f(x) is the set of all real values except 7, and the domain of g(x) is the set of all real values except –3. Which

Svet_ta [12734]

Answer: All real numbers except for x ≠ 7 and those x for which f(x)≠-3

Step-by-step explanation:

The domain of function f is defined as R - {7}

In other words, if x belongs to the domain of f,

then, x cannot equal 7

(gof)(x) = g(f(x))

Considering that g has a domain of R - {-3}

Therefore, if f(x) fits within the domain of g,

it follows that f(x)≠ -3

Consequently, the fourth option is the right choice.

8 0

2 months 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

A toy airplane climbs 30 vertical feet while traveling 50 feet horizontally. A toy helicopter climbs 20 feet while traveling 40

Inessa [12570]

Answer:

Step-by-step explanation:

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