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kolbaska11 [484]
2 years ago
8

Arrange the reasons for the proof in the correct order. Prove: If the diameter of a circle is 6 meters and the formula for diame

ter is d = 2r, then the radius of the circle is 3 meters.
Mathematics
1 answer:
Misha Larkins [42]2 years ago
3 0
Given the question: Arrange the reasons for the proof in the correct order. Prove: If the diameter of a circle is 6 meters and the formula for diameter is d = 2r, then the radius of the circle is 3 meters.
A. If r ≠ 3 m, then d ≠ 6 m. Since the contrapositive is true, the original statement must also be true. Therefore, if the diameter of a circle is 6 meters, then the radius is 3 meters.
B. Multiplication of real numbers shows that d = 2(2 m) = 4 m.
C. Substitute r = 2 m into d = 2r.
D. Assume that r ≠ 3 m. For example, the radius equals another length, such as r = 2 m.

To prove that i<span>f the diameter of a circle is 6 meters and the formula for diameter is d = 2r, then the radius of the circle is 3 meters by contradiction, we assume that the radius in not equal to 3 meters, for </span>example, the radius equals another length, such as r = 2 m.
Next, we substitute the value: r = 2m nto the original equation that says that d = 2r, i.e. d = 2(2m) = 4m which is not true and contradicts the original statement that the diameter of the circle is 6m.

Therefore, the arrangement of the proof is as follows:
 D. Assume that r ≠ 3 m. For example, the radius equals another length, such as r = 2 m.
C. Substitute r = 2 m into d = 2r.
B. Multiplication of real numbers shows that d = 2(2 m) = 4 m.
A. If r ≠ 3 m, then d ≠ 6 m. Since the contrapositive is true, the original statement must also be true. Therefore, if the diameter of a circle is 6 meters, then the radius is 3 meters.

D C B A.
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Find the mass and center of mass of the lamina that occupies the region D and has the given density function rho. D = {(x, y) |
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Answer:

M=168k

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Step-by-step explanation:

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Now, we know the limits of the integrals of x and y, and also know that ρ = ky², so we will have:

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Let's solve this integral:

M=k\int^{9}_{1}\frac{y^{3}}{3}|^{4}_{1}dx

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So the mass will be:

M=21k*8=168k

Now we need to find the x-coordinate of the center of mass.

\bar{x}=\frac{1}{M}\int\int x*\rho dydx

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\bar{x}=\frac{k}{168k}\int^{9}_{1}\int^{4}_{1}x*y^{2} dydx

\bar{x}=\frac{1}{168}\int^{9}_{1}x*\frac{y^{3}}{3}|^{4}_{1}dx

\bar{x}=\frac{1}{168}\int^{9}_{1}x*21 dx

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Now we need to find the y-coordinate of the center of mass.

\bar{y}=\frac{1}{M}\int\int y*\rho dydx

\bar{y}=\frac{1}{M}\int^{9}_{1}\int^{4}_{1}y*ky^{2} dydx

\bar{y}=\frac{k}{168k}\int^{9}_{1}\int^{4}_{1}y^{3} dydx

\bar{y}=\frac{1}{168}\int^{9}_{1}\frac{y^{4}}{4}|^{4}_{1}dx

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\bar{y}=\frac{85}{28}

Therefore the center of mass is:

(\bar{x},\bar{y})=(5,\frac{85}{28})

I hope it helps you!

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