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dimaraw [331]
2 years ago
7

Which angles are corresponding angles with angle 8?

Mathematics
2 answers:
Basile [38]2 years ago
7 0

Answer:

<4 and <8

Step-by-step explanation:

Corresponding angles are at the same part of the intersection of

8 and 12 are corresponding angles

8 and 4 are corresponding angles

natali 33 [55]2 years ago
4 0

Answer:

\angle 12 \cong \angle 8\\\angle 4 \cong \angle 8

Step-by-step explanation:

When two parallels lines are crossed by a transversal, certain pair of angles are called corresponding angles, specifically, those that are placed in the same side of the transversal, on inside the parallels and the other outside the parallels. The image attached shows an example of corresponding angles.

So, we observe in the given image that all corresponding angles with angle 8 are

\angle 12\\\angle 4

These angles are form corresponding angle with \angle 8, that means they are congruent, that is

\angle 12 \cong \angle 8\\\angle 4 \cong \angle 8

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Find the following: F(x, y, z) = e^(xy) sin z j + y tan^−1(x/z)k Exercise Find the curl and the divergence of the vector field.
natulia [17]

\vec F(x,y,z)=e^{xy}\sin z\,\vec\jmath+y\tan^{-1}\dfrac xz\,\vec k

Divergence is easier to compute:

\mathrm{div}\vec F=\dfrac{\partial(e^{xy}\sin z)}{\partial y}+\dfrac{\partial\left(y\tan^{-1}\frac xz\right)}{\partial z}

\mathrm{div}\vec F=xe^{xy}\sin z-\dfrac{xy}{x^2+z^2}

Curl is a bit more tedious. Denote by D_t the differential operator, namely the derivative with respect to the variable t. Then

\mathrm{curl}\vec F=\begin{vmatrix}\vec\imath&\vec\jmath&\vec k\\D_x&D_y&D_z\\0&e^{xy}\sin z&y\tan^{-1}\frac xz\end{vmatrix}

\mathrm{curl}\vec F=\left(D_y\left[y\tan^{-1}\dfrac xz\right]-D_z\left[e^{xy}\sin z\right]\right)\,\vec\imath-D_x\left[y\tan^{-1}\dfrac xz\right]\,\vec\jmath+D_x\left[e^{xy}\sin z}\right]\,\vec k

\mathrm{curl}\vec F=\left(\tan^{-1}\dfrac xz-e^{xy}\cos z\right)\,\vec\imath-\dfrac{yz}{x^2+z^2}\,\vec\jmath+ye^{xy}\sin z\,\vec k

5 0
3 years ago
Polygon D is a scaled copy of polygon C using a scale factors of 6
Vladimir79 [104]

Answer: The area of the Polygon D is 36 times larger than the area of the Polygon C.

Step-by-step explanation:

<h3> The complete exercise is: "Polygon D is a scaled copy of Polygon C using a scale factor of 6. How many times larger is the area of Polygon D than the area Polygon C"?</h3>

 In order to solve this problem it is important to analize the information provided in the exercise.

You know that the Polygon D was obtained by multiplying the lengths of the Polygon C by the scale factor of 6.

Then, you can identify that the Length scale factor used is:

Length\ scale\ factor=k=6

Now you have to find the Area scale factor.

Knowing that the Length scale factos is 6, you can say that the Area scale factor is:

Area \ scale\ factor=k^2=6^2

Finally, evaluating, you get that this is:

Area \ scale\ factor=36

Therefore, knowing the Area scale factor, you can determine that the area of the Polygon D is 36 times larger than the area of the Polygon C.

8 0
3 years ago
6th grade math please help because my grade went down.. and ion got time to be
son4ous [18]

Answer:

first you subtract 5 from both sides

your left with 3x=11

divide by 3 on both sides

x=3.666 repeating

Step-by-step explanation:

3 0
3 years ago
The length of an edge of a cube is 9 centimeters. What is the volume of the cube?
KATRIN_1 [288]

Answer:

729

Step-by-step explanation:

Volume of cube=a^3

Volume of cube=9^3=729

3 0
2 years ago
Read 2 more answers
Find the length of MN.<br> A) 7 <br> B) 25 <br> C) 32 <br> D) 39
MrRissso [65]

Step-by-step explanation:

Answer C)

Explanation:

MN + NP = MP

MP = 39

Therefore, 4(x + 5) + 2x + 1= MP

So, 4(x + 5) + 2x + 1= 39

(4x + 20) + 2x +1 = 39

So, (4x + 2x) + (20 + 1) = 39

6x + 21 = 39

Addition changes to Subtraction

So, 6x = 39 - 21

6x = 18

Multiplication changes to Division

x= 18/6

x=3

Therefore MN = 4(x + 5)

So, 4(3 + 5) = 4(8)

4(8) = 32

Therefore, MN = 32

Hope this helps!

3 0
3 years ago
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