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Artyom0805 [142]
3 years ago
10

Find m∠PQR. A. 81 B. 90 C. 77 D. 72

Mathematics
1 answer:
AnnZ [28]3 years ago
5 0
<h3>Answer: D. 72</h3>

Work Shown:

Angle PQR = (far arc - near arc)/2

Angle PQR = ( (major arc PR) - (minor arc PR) )/2

Angle PQR = ( (2x+252) - (2x+108) )/2

Angle PQR = 144/2

Angle PQR = 72

Notice how we didn't need to find the value of x at all.

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Square root of 512m^3. show work
ElenaW [278]

The solution would be like this for this specific problem:

√512m³ = √256m² × √2m

= 16 × √2m

I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

6 0
3 years ago
Set up a double integral for calculating the flux of the vector field ????⃗ (????⃗ )=????⃗ , where ????⃗ =⟨x,y,z⟩, through the p
Tema [17]

Answer:

-937.5π

Step-by-step explanation:

F (r) = r = (x, y, z) the surface equation z = 3(x^2 + y^2) z_x = 6x, z_y = 6y the normal vector n = (- z_x, - z_y, 1) = (- 6x, - 6y, 1)

Thus, flux ∫∫s F · dS is given as;

∫∫ <x, y, z> · <-z_x, -z_y, 1> dA

=∫∫ <x, y, 3x² + 3y²> · <-6x, -6y, 1>dA , since z = 3x² + 3y²

Thus, flux is;

= ∫∫ -3(x² + y²) dA.

Since the region of integration is bounded by x² + y² = 25, let's convert to polar coordinates as follows:

∫(θ = 0 to 2π) ∫(r = 0 to 5) -3r² (r·dr·dθ)

= 2π ∫(r = 0 to 5) -3r³ dr

= -(6/4)πr^4 {for r = 0 to 5}

= -(6/4)5⁴π - (6/4)0⁴π

= -937.5π

4 0
3 years ago
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Mademuasel [1]

Answer: 3/4 : 8/4

Step-by-step explanation:

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make the denominators the same and you are good

5 0
2 years ago
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Roman55 [17]

Answer:

I believe its -800, but i'm not sure.

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insens350 [35]

Answer:

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280 divided by 20

3 0
3 years ago
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