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Anna35 [415]
3 years ago
6

Find a solution of inequality 2x+y<10

Mathematics
1 answer:
Vitek1552 [10]3 years ago
7 0

Option C:

The solution of inequality 2x + y < 10 is (5, –3).

Solution:

Given inequality is 2x + y < 10.

Lets substitute the given points in the inequality and find the solution.

Option A: (6, –1)

2(6) + (–1) < 10

12 – 1 < 10

11 < 10

This is false because 11 > 10.

Therefore (6, –1) is not the solution of the inequality.

Option B: (1, 10)

2(1) + (10) < 10

2 + 10 < 10

12 < 10

This is false because 12 > 10.

Therefore (1, 10) is not the solution of the inequality.

Option C: (5, –3)

2(5) + (–3) < 10

10 – 3 < 10

7 < 10

This is true.

Therefore (5, –3) is the solution of the inequality.

Option D: (5, 5)

2(5) + (5) < 10

10 + 5 < 10

15 < 10

This is false because 15 > 10.

Therefore (5, 5) is not the solution of the inequality.

Hence Option C is the correct answer.

The solution of inequality 2x + y < 10 is (5, –3).

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Now that we know the vertices of the surface \mathcal S, we can parameterize it by

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\mathrm dS=\|\mathbf s_u\times\mathbf s_v\|\,\mathrm du\,\mathrm dv=2\sqrt{406}(1-v)\,\mathrm du\,\mathrm dv

With respect to our parameterization, we have x(u,v)=(1-u)(1-v), so the surface integral is

\displaystyle\iint_{\mathcal S}x\,\mathrm dS=2\sqrt{406}\int_{u=0}^{u=1}\int_{v=0}^{v=1}(1-u)(1-v)^2\,\mathrm dv\,\mathrm du=\frac{\sqrt{406}}3
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Divide both sides by t^2 to solve for <em>y</em> :

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or

y(t)=\dfrac{4\sin t-4t\cos t+9\pi^2-4}{4t^2}

6 0
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